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4700 | https://www.youtube.com/watch?v=0qgJnkb5aoQ | The Auxiliary Angle Method part 1
Hoang Maths
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Posted: 18 Jul 2019
The Auxiliary Angle Method
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Transcript:
let's say a trick equation given is a sin theta plus b cos thet is equal a constant to solve this equation it can be very difficult so in order to solve this kind of trick equation so what we can do is we can convert a sin Theta + B cos Theta is equal to R sin of theta + Alpha now Alpha is an acute angle and it's called the auxiliary angle so if this is a minus okay this is a minus let's say and then this Theta instead of plus Alpha we put Theta minus Alpha if it is I cos theta plus b s of theta is equal to R so this is C the first term so we use C just remember that if it's s the first term we use S so that is equal to R COS of theta now remember for cost if you expand positive become negative so therefore positive we change into negative Alpha and then if it is negative here okay this become positive instead of negative so please take note on that so let's focus on um a sin theta plus b COS of theta let's convert into r s of theta + Alpha for instance so let me expand the right hand side first so remember s of a + b is equal to S of a COS of B plus COS of i s of B that's a trick identity so to expand this out that become s of theta COS of alpha plus COS of theta I just clear this first and S of Alpha and then I expand further this will give me R s of theta COS of alpha plus r COS of theta s of alpha so this is equal to I just try this out so a s of theta plus b COS of theta is equal to so I just write R sin Theta cos Alpha in a different order so they become R COS of alpha s of theta plus similarly r s of alpha COS of theta so let's compare the coefficient of sin Theta so this on the left hand side here I sin Theta and R cos Alpha and then sin Theta now remember Alpha is a constant so that mean cos Alpha is a constant so as I so I just compare the um two terms first so that is I side Theta is the same as R COS of alpha sin Theta s Theta the coefficient is I and the coefficient of right hand side is R cos Alpha so therefore I is equal to R COS of alpha let name this equ number one so now I can compare the um coefficient of cos Theta and the coefficient of cos Theta of course on this side is equal so that is B cos Theta is equal to R sin of alpha COS of theta so that means the coefficient of this is B and the coefficient of the right hand side for cos Theta is r s of alpha let's name this equation number two so now we have these two equation here if I use equation number two divided by equation number one that is equation 2 over equation 1 so that is on the left hand side b/ a is equal to r s of alpha over R COS of alpha so equation number two divide by equation number one so now this give me b/ I is equal to the r cancel out so sin Alpha over cos Alpha is tan of alpha so that is Alpha equal tan inverse of b/ a so please take note on this so now on equation number one now if I Square both side okay if I Square both side that mean I squ is equal to r² COS of square Alpha and I Square the equation number two this mean that b² is equal to r² and S of square Alpha so now if I add them up so that means i² + b² is = to r² Square COS of alpha square + r² sin squ Alpha so now i² + b² is equal to r² I take the common factor out so that's leave with cos Square Alpha plus s of square Alpha now please note that cos Square Alpha plus sin Square Alpha is = 1 this mean that I 2 + b² is = to r² so therefore square root of both side so R = A squ + B squ and square root of that and of course every time you square root a number you get a plus or minus but in this case we just find the magnitude of R so this is we ignore the negative sign so therefore we can convert i s of theta + b COS of theta is equal to r s of theta + alha whereas R is a square + b squ square root of that and Alpha is equal to tan inverse of B / I please stop the video and try this one yourself so let's expand right hand side first so this is = to R now remember COS of I + B is equal to COS of I COS of B minus s of i s of B this is the the trick identity so I expand this out so pleas give me cos of theta COS of alpha minus s of theta s of alpha so now this is equal to R COS of theta COS of alpha minus r s of theta s of alpha so this mean that I COS of Theta minus B sin of theta is equal to R COS of alpha COS of theta minus r s of alpha s of theta so now to compare the coefficient of cos Theta so just write this is a COS Theta is equal to R cos Alpha because R cos Alpha and a are constant so that is I is equal to R COS of alpha so I name this is equation number one and then the coefficients of s Theta that means B is equal to r s of alpha equation number two so I divide by equation two with equation number one so that is b/ I is equal to R sin Alpha over R COS of alpha so that is B over I because the negative both side negative I can just ignore and the r cancel out tan of alpha so therefore Alpha is tan inverse of B over a so now if I Square the um equation number one so that become i² is = to r² cos Square Alpha so Square on equation number two so that b² = to negative square is positive anyway so I can ignore that so that R square s square of alpha let's name this I equation I this is equation B so just write equation I underneath here so i² = r² cos Square Alpha so now if I add I and B together that means I square + b² is = r² cos² Alpha + r² sin squ Alpha so that means I S + b² is equal to r² I take common factor out so which is left with cos Square Alpha plus s Square Alpha so this one here it give me one because cos Square Theta + sin Square Theta = 1 so cos Square Alpha + sin Square alpha = 1 as well so now a² + b² = r² that that is R is equal to square root of a² + b² therefore a COS Theta minus B sin Theta is equal to R COS of theta + Alpha where Alpha is tan inverse of b/ i and r is equal to a² + b² square root of that thank you so much for watching so if you King you can try these two question yourself thank you again |
4701 | https://pubmed.ncbi.nlm.nih.gov/21568864/ | IgG-binding proteins of bacteria - PubMed
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. 2011 Mar;76(3):295-308.
doi: 10.1134/s0006297911030023.
IgG-binding proteins of bacteria
E V Sidorin1,T F Solov'eva
Affiliations Expand
Affiliation
1 Pacific Institute of Bioorganic Chemistry, Far-Eastern Division of the Russian Academy of Sciences, Vladivostok, Russia. sev1972@mail.ru
PMID: 21568864
DOI: 10.1134/s0006297911030023
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IgG-binding proteins of bacteria
E V Sidorin et al. Biochemistry (Mosc).2011 Mar.
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. 2011 Mar;76(3):295-308.
doi: 10.1134/s0006297911030023.
Authors
E V Sidorin1,T F Solov'eva
Affiliation
1 Pacific Institute of Bioorganic Chemistry, Far-Eastern Division of the Russian Academy of Sciences, Vladivostok, Russia. sev1972@mail.ru
PMID: 21568864
DOI: 10.1134/s0006297911030023
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Abstract
Proteins capable of non-immune binding of immunoglobulins G (IgG) of various mammalian species, i.e. without the involvement of the antigen-binding sites of the immunoglobulins, are widespread in bacteria. These proteins are located on the surface of bacterial cells and help them to evade the host's immune response due to protection against the action of complement and to decrease in phagocytosis. This review summarizes data on the structure of immunoglobulin-binding proteins (IBP) and their complexes with IgG. Common and distinctive structural features of IBPs of gram-positive bacteria (staphylococci, streptococci, peptostreptococci) are discussed. Conditions for IBP expression by bacteria and their functional heterogeneity are considered. Data on IBPs of gram-negative bacteria are presented.
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4702 | https://www.youtube.com/watch?v=pjmfWmD8gyU | What is invariance (in math competitions)?
Shefs of Problem Solving
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Description
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Posted: 24 Oct 2021
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Transcript:
Intro hello fellow problem solvers so today we're going to be talking about invariance and i've made a video on and i've sold a couple of problems with invariants but now i want to focus just on the general concept overall what is it how do you do it and the first really step is what is invariance it's really a higher order problem solving technique it's saying that when you're working on a problem and you see there's something that's changing in the problem say there's some transformation that's happening again and again it might be happening differently at different times but there might be some thing that's the same as it's happening you might see this in games or whatever and or any sort of like algorithmic changes that are happening you ask yourself while this change is happening what is something that remains constant or something i can really control and when i say control what i really mean is something that you can model or evaluate algebraically evaluate like a value of something End states and say a good example of invariance actually before we get into the example like here's a couple of questions that you can ask yourself while you're seeing these changes happening in problems one of them is what are the all the possible end states so a state of like there's a state that the problem is in before a transformation then after a transformation is in another state and then another state and then you might come to a point where you can't do any more transformations and you ask yourself well what does that point look like what are all the possibilities for that end state you might also look at periods do these transformations create any periods of any sort in the problem for example if you're looking at say the fibonacci sequence you have periods so this is not invariance or monovariance in a strict sense but you have periods modulo any natural number right and you can actually prove this but an example with invariance we're actually looking at end states is say a problem where say n is a positive integer and you have numbers one through four n plus two written on the blackboard so you have one two three four all the way till four times then plus two Example and an example let's say you have numbers one two three four five six written on the board and at every turn your friend takes two of these numbers and writes their absolute difference on the board so takes two numbers deletes them and writes their absolute difference on the board and the problem says can a zero be written on the board at the end right and the answer is well there's an invariance here happening so what is something that remains constant while you're writing a minus b and the thing that does remain constant here is say if you write if you take two if you have one and two you'll have one if you have numbers one two three four five six you take six and four you have a two two one three and five you take a two and two you have a one three and five you take a five three ever one two but something while you're doing this we can have a one at the end you can have e3 at the end and as you try out different transformations you may find that say if you take five four you get one two three you get one one one you get one so you have one sixth you can get one free and five on the board and once you try different stuff out you see that the only numbers you can get on the board are one three and five if you try say n equals two and you have numbers 1 through 10 written on the board you'll see that what you have you can get numbers 1 3 5 7 and 9. i believe you can get all these numbers on the board at the end and now the question is why can't you get on your defense now it's sort of hinting out you're not going to be able to get zero here at all and the question is why can't you get any of the evens and what remains the same Common invariance is and this is a common invariance that parity remains the same if you delete the numbers a and b and you write a minus being absolute values what you're left with say if the sum of the numbers before was of all the numbers aside from a and b if it was s then your total sum before you made the move was s plus a plus b after you make the move it's s plus a minus b in absolute value terms now a minus b and a plus b have the same parity okay you can test it out you can prove this they have the same parity so s plus a plus b and s plus a minus b in absolute value terms have the same parity now given at the beginning that your whole that this whole sum is equal to n plus 2 over 4 n plus 2 over so it's two n plus one times or n plus three if i'm not mistaken because that's the sum at the beginning and it's odds that means at the end because the parity remains the same at the end you'll also have an odd sum which means an odd number so that's why you can't have a zero at the end this is an example of you looking at okay what's my end state how can what are all my possible end states how can i reach them and then you're looking for different invariants different things that don't change as the situation the numbers written on the board change so that's invariance in a nutshell and how you practice this is with every technique you have this little nugget i gave you and now go out try to solve maybe try to if you haven't solved the free problems i've covered on invariants please do i might cover some more in the future and try to solve them and try to see if you come up with problems if you see problems which use invariants try to see if you come across problems which when you try to solve them they have something that's changing and with that change try to figure out is there something that remains the same something i can control a good example of this is also say a game where you have 20 coins on a board on a table and two players take alternate turns and at each point you can remove one two or three coins from the table who can win is basically a question and this is a bit of a copying maybe strategy so the situation as well but the potential invariant that one of the players can take advantage of is they can always have the sum decrease by four after their turn so that's another cool invariant and you come across these a lot but the thing is solve problems and then you'll get better at this and most importantly here that's the basic principle and and this is the basic principle now it gets applied it can be applied to simple problems like the one i described here or it can also be applied to very very difficult problems i can spoil a problem for you now and the problem is from the 2021 imo a free difficult problem was just an invariance it was one of variance that if you figure it out you'll solve the problem i covered it on my channel or you have also an invariance even there was a problem that was a third or sixth that the imo which used a bit of a complicated monovariance it looked at something that changes in a controllable manner to come up with something and solve the problem but this principle can be applied in many situations and now you're aware of it and as always thanks for problem solving |
4703 | https://proofwiki.org/wiki/Summation_over_k_of_Floor_of_k_over_2 | Summation over k of Floor of k over 2 - ProofWiki
Summation over k of Floor of k over 2
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Theorem
∑k=1 n⌊k 2⌋=⌊n 2 4⌋∑k=1 n⌊k 2⌋=⌊n 2 4⌋
Proof
By Permutation of Indices of Summation:
∑k=1 n⌊k 2⌋=∑k=1 n⌊n+1−k 2⌋∑k=1 n⌊k 2⌋=∑k=1 n⌊n+1−k 2⌋
and so:
∑k=1 n⌊k 2⌋=1 2∑k=1 n(⌊k 2⌋+⌊n+1−k 2⌋)∑k=1 n⌊k 2⌋=1 2∑k=1 n(⌊k 2⌋+⌊n+1−k 2⌋)
First take the case where n n is even.
For k kodd:
⌊k 2⌋=k 2−1 2⌊k 2⌋=k 2−1 2
and:
⌊n+1−k 2⌋=n+1−k 2⌊n+1−k 2⌋=n+1−k 2
Hence:
⌊k 2⌋+⌊n+1−k 2⌋⌊k 2⌋+⌊n+1−k 2⌋==k 2−1 2+n+1−k 2 k 2−1 2+n+1−k 2
==k−1+n+1−k 2 k−1+n+1−k 2
==n 2 n 2
For k keven:
⌊k 2⌋=k 2⌊k 2⌋=k 2
and:
⌊n+1−k 2⌋=n+1−k 2−1 2=n−k 2⌊n+1−k 2⌋=n+1−k 2−1 2=n−k 2
Hence:
⌊k 2⌋+⌊n+1−k 2⌋⌊k 2⌋+⌊n+1−k 2⌋==k 2+n−k 2 k 2+n−k 2
==k+n−k 2 k+n−k 2
==n 2 n 2
So:
∑k=1 n⌊k 2⌋∑k=1 n⌊k 2⌋==1 2∑k=1 n(⌊k 2⌋+⌊n+1−k 2⌋)1 2∑k=1 n(⌊k 2⌋+⌊n+1−k 2⌋)
==1 2∑k=1 n(n 2)1 2∑k=1 n(n 2)
==1 2 n n 2 1 2 n n 2
==n 2 4 n 2 4
==⌊n 2 4⌋⌊n 2 4⌋as n 2 4 n 2 4 is an integer
□◻
Next take the case where n n is odd.
For k kodd:
⌊k 2⌋=k 2−1 2⌊k 2⌋=k 2−1 2
and:
⌊n+1−k 2⌋=n+1−k 2−1 2⌊n+1−k 2⌋=n+1−k 2−1 2
Hence:
⌊k 2⌋+⌊n+1−k 2⌋⌊k 2⌋+⌊n+1−k 2⌋==k 2−1 2+n+1−k 2−1 2 k 2−1 2+n+1−k 2−1 2
==k−1+n+1−k−1 2 k−1+n+1−k−1 2
==n−1 2 n−1 2
For k keven:
⌊k 2⌋=k 2⌊k 2⌋=k 2
and:
⌊n+1−k 2⌋=n+1−k 2⌊n+1−k 2⌋=n+1−k 2
Hence:
⌊k 2⌋+⌊n+1−k 2⌋⌊k 2⌋+⌊n+1−k 2⌋==k 2+n−k+1 2 k 2+n−k+1 2
==k+n−k+1 2 k+n−k+1 2
==n+1 2 n+1 2
Let n=2 t+1 n=2 t+1.
Then:
∑k=1 n⌊k 2⌋∑k=1 n⌊k 2⌋==1 2∑k=1 n(⌊k 2⌋+⌊n+1−k 2⌋)1 2∑k=1 n(⌊k 2⌋+⌊n+1−k 2⌋)
==1 2∑k=1 2 t+1(⌊k 2⌋+⌊2 t+2−k 2⌋)1 2∑k=1 2 t+1(⌊k 2⌋+⌊2 t+2−k 2⌋)
==t 2(2 t+1)+1 2+t+1 2(2 t+1)−1 2 t 2(2 t+1)+1 2+t+1 2(2 t+1)−1 2 there are t teven terms and t+1 t+1odd terms
==2 t 2+2 t 4+2 t 2+2 t 4 2 t 2+2 t 4+2 t 2+2 t 4 multiplying out
==4 t 2+4 t 4+1 4−1 4 4 t 2+4 t 4+1 4−1 4
==(2 t+1)2 4−1 4(2 t+1)2 4−1 4
==n 2 4−1 4 n 2 4−1 4
==⌊n 2 4⌋⌊n 2 4⌋
■◼
Sources
1997:Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms(3rd ed.)... (previous)... (next): §1.2.4§1.2.4: Integer Functions and Elementary Number Theory: Exercise 36 36
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12: Output Profit Maximization
Intermediate Microeconomics with Excel (Barreto)
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Thu, 22 Jun 2023 06:38:05 GMT
12.1: Initial Solution
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Contents
1. Home
2. Bookshelves
3. Economics
4. Microeconomics
5. Intermediate Microeconomics with Excel (Barreto)
6. 12: Output Profit Maximization
7. 12.1: Initial Solution
Expand/collapse global location
Intermediate Microeconomics with Excel (Barreto)
Front Matter
1: Budget Constraint
2: Satisfaction
3: Optimal Choice
4: Compartive Statics
5: Endowment Models
6: Bads
7: Search Theory
8: Behavioral Economics
9: Rational Addiction
10: Production Function
11: Input Cost Minimization
12: Output Profit Maximization
13: Input Profit Maximization
14: Consistency
15: Monopoly
16: Game Theory
17: Partial Equilibrium
18: General Equilibrium
19: Conclusion
Back Matter
12.1: Initial Solution
Last updated Jun 22, 2023
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12: Output Profit Maximization
12.2: Deriving the Supply Curve
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Page ID 58498
Humberto Barreto
DePauw University
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Table of contents
1. Perfectly Competitive Market Structure
2. Setting Up the Problem
3. Finding the Initial Solution
4. Representing the Optimal Solution with Graphs
5. The Shutdown Rule
1. What’s Normal about Zero Profits?
2. Shutdown Rule and Corner Solution
3. Finding and Displaying the Initial Solution
Exercises
References
With a total cost function, TC(q), and its associated average and marginal cost curves, we are ready to solve the the firm’s output profit maximization problem. The firm chooses the amount of output that maximizes profit, defined as total revenue minus total cost. This is the second of three optimization problems that make up the Theory of the Firm.
All firms face this profit maximization problem, but this chapter works with a perfectly competitive (PC) firm in the short run (SR). There are, of course, many other market structures and types of firms, but perfect competition is the first step from which more sophisticated scenarios arise.
The firm’s market structure tells us the environment in which it operates. Its market structure determines the firm’s revenue function. A PC firm is the simplest case because it takes price as given. Thus, revenues are simply price times quantity and the revenue function is linear.
Remember that we are not trying to describe the actual operation of a business. In fact, a truly perfectly competitive firm does not exist in the real world. The concept is an abstraction that enables derivation of the supply curve. This is our goal.
Remember also that the short run is defined by the fact that at least one input (usually K) is fixed. In the long run, the firm is free to choose how much to use of every factor. K is fixed not because it is immovable (like a pizza oven or a building), but because the firm has contracted to rent a certain amount. It cannot increase or decrease the amount of K in the short run.
Profit maximization and its graphs may be familiar from introductory economics. This experience will help you, but do not be complacent. Keep your eye on how the economic way of thinking is being applied in this case and make connections with other optimization problems we have explored.
Perfectly Competitive Market Structure
A perfectly competitive firm sells a product provided by countless other firms selling that homogeneous (which means identical) product to perfectly informed consumers. Because the product is homogeneous, there are no quality differences or other reasons for consumers to care about who they buy from. Because consumers are perfectly informed, they know the price of every seller.
Thus, the PC firm’s market structure is one of intense price competition. Every firm sells the product at the exact same price because if anyone tried to sell at even a tiny bit higher than the market price, no one would buy from them.
The shorthand term for this environment is price taking. The PC firm must take the price and cannot choose its priceprice is exogenous to the firm.
In addition to price taking, the market structure of the PC firm is characterized by an assumption about the movement of other firms into and out of the industry: free entry and exit. Firms can enter or leave the market, selling the same good as everyone else, at any time.
These two ideas, price taking and free entry, distinguish the PC firm from its polar opposite, monopoly. A monopolist chooses price and has a barrier to entry. Between these two extremes are many other market structures in which real-world firms actually exist.
The PC firm’s market structure means that an individual PC firm does not worry about what other firms are doing. Each firm simply chooses its own output to maximize profit and does not watch the other firms to gain a strategic advantage. In this sense, there is no rivalry in perfect competition.
Setting Up the Problem
As usual, we organize the optimization problem into three parts:
Goal: maximize profits (π, Greek letter pi), which equal total revenues (TR) minus total costs (TC).
Endogenous variable: output (q).
Exogenous variables: price of the product (P), input prices (the wage rate (w) and the rental rate of capital (r)), and technology (parameters in the production function).
Unlike the consumer’s utility maximization and the firm’s input cost minimization problems, this profit maximization problem is unconstrained. The firm does not have a restriction, like a budget constraint or isoquant, that limits its choice of output to a particular range. It can choose any non-negative level of output.
This greatly simplifies the optimization problem. For the analytical method, it means we do not need the Lagrangean method. All we need to do is take a single derivative and set it equal to zero.
Finding the Initial Solution
Suppose the cost function is: (12.1.1)TC(q)=aq 3+bq 2+cq+d Then we can form the PC firm’s profit function and optimization problem like this: (12.1.2)max qπ=TR−TC(12.1.3)max qπ=Pq−(aq 3+bq 2+cq+d) As usual, we have two ways to solve this optimization problem: numerically and analytically.
STEP Open the Excel workbook OutputProfitMaxPCSR.xls and look over the Intro sheet.
The Intro sheet is not meant to be immediately understood. It offers highlights of material that will be explained and prints as one landscaped page. It provides a compact summary of the optimal solution of the output profit maximization problem for a perfectly competitive firm in the short run.
STEP Proceed to the OptimalChoice sheet to find the initial solution.
The sheet is organized into the components of an optimization problem, with goal, endogenous, and exogenous variable cells.
Initially, the firm is producing nine units of output and making $11.74 of profit. Is this the highest profit it can possibly make?
No. The sheet reveals the information needed to give this answer. By comparing marginal revenue (MR) and marginal cost (MC), we immediately know that the firm would make a mistake (we would say it is inefficient) if it produced just nine units.
The MC of the ninth unit is $3.52 as shown in cell B22, but what about MR? Perhaps you remember from introductory economics that P=MR for perfectly competitive firms? We can see that the additional revenue produced by the last unit, 7(theprice),isgreaterthantheadditionalcost,3.52 (cell B22). Thus, the firm should produce more. How much exactly should the firm produce?
STEP Run Solver to find out.
Look carefully at B22. At the optimal solution, q≈13.09, MC = $7 per unit. P=MC, a special case of MR=MC for a PC firm, is the equimarginal condition in this problem, analogous to MRS=p 1 p 2 and TRS=w r. When the equimarginal condition is met, the firm is guaranteed to be maximizing profits.
To find the optimal solution via the analytical method, we take the derivative of the profit function with respect to q, set it equal to zero, and solve for q. Our cubic cost function introduces the complication that the solution has two roots so we have to use the quadratic formula.
STEP Click the button to see how to solve this problem with calculus.
Cell AC17’s formula has the root that maximizes profits (the other root minimizes profitsmore on this in the next section). As usual, Solver and calculus agree (not exactly, but they give effectively the same answer).
Representing the Optimal Solution with Graphs
Since this is an unconstrained optimization problem (unlike utility maximization and input cost minimization), the graphical display of the optimal solution is different.
The firm’s output profit maximization problem is usually represented by a graph that depicts the family of cost curves along with marginal and average revenue. Figure 12.1 and the Intro sheet shows this canonical graph for a perfectly competitive firm (signaled by the fact that firm demand is horizontal, so marginal revenue equals demand).
Figure 12.1: The canonical output profit maximization graph.
Source: _OutputProfitMaxPCSR.xls!Intro_.
Figure 12.1 is the usual display of the optimal solution, but it is actually part of a much larger graphical display.
STEP Proceed to the Graphs sheet to see how Figure 12.1 fits into the bigger picture, also shown in Figure 12.2. Zoom out to see all four graphs.
Figure 12.2: Four graphs of output profit maximization.
Source: _OutputProfitMaxPCSR.xls!Graphs_.
Each of the four graphs in Figure 12.2 and on your screen can be used to show the firm’s optimization problem and its solution. We will walk through each one.
The top left graph plots total revenue and total cost. TR is linear because the firm’s market structure is perfect competition, hence, it is a price taker. The cubic total function produces the shape of TC. The firm wants to choose q to maximize the difference between revenues and costs.
The top right graph shows the profit function, which is TR−TC. The firm wants to choose q so that it is at the highest point on the profit hill.
The bottom right graph displays marginal profit, which can be expressed as the derivative of the profit function with respect to q. The firm can find the maximum profit by choosing q so that marginal profit is zero. This is the first-order condition from the analytical solution.
Finally, the bottom left graph is the usual display. The firm chooses q where MR (which equals P given that the firm is a price taker) equals MC. Profits can be calculated as the area of the rectangle (AR−ATC)q.
To be clear, all four graphs in Figure 12.2 show the same optimal q and maximum profits, but the graph that is most often used is the bottom left. It highlights the comparison of MR and MC and the family of cost curves provides information about the firm’s cost structure. We can also find profits as the area of the rectangle (with blue top and dashed line bottom).
STEP Move the output with the slider control (in the middle of the four charts) to the left and right of q to see how the profit rectangle changes.
Only when q is such that MR=MC do you get the maximum area of the profit rectangle. Moving left from optimal q, you can make the rectangle taller, but you must make it shorter to do this and you end up with less area. You can make the rectangle longer by moving right from optimal q, but ATC rises and the rectangle gets thinner, so once again the area falls.
The intersection of MR and MC immediately reveals the optimal q. Profits at any q are also easily seen as the area of a rectangle, length times width, with units in dollars. Because the y axis is a rate, $/unit, and the x axis is in units of the product, multiplying the two leaves dollars. In other words, say the product is milk in gallons. Then price, average total, and average variable cost are all in /gallon.Supposethatatapriceof 2/gallon, MR = MC at an output of 7,000 gallons and ATC = 1.50/gallonatthisoutput.Clearly,profitsare(2/gallon - 1.50/gallon)x7,000gallons,whichequals 3,500.
We can compute profits from the profit rectangle at any level of output. The height of the rectangle is always average revenue (which equals price) minus average total cost. This vertical distance is average profit. When multiplied by the level of output, we get profits, in dollars, at that level of output.
The bottom left graph has another advantage over the other graphs. It can be used to explain a curious and puzzling feature of a firm’s short run profit maximization problem. The story revolves around a firm with negative profits and what it should do in this situation.
The Shutdown Rule
The firm has an option when maximum profits are negative: it can simply shut down, close its doors, hire no workers, and produce nothing. The Shutdown Rule says the the firm will maximize profits by producing nothing (q=0) when P<AVC.
The key to whether the firm shuts down or continues production in the face of negative profits lies in its fixed costs. If the firm can do better by shutting down and paying its fixed costs instead of producing and choosing the level of output where MR=MC, then it should produce nothing.
Continuing production in the face of negative profits versus shutting down are actually the last two of four possible profit positions for the firm.
Excess Profits: π>0 and P>ATC
Normal Profits: π=0 and P=ATC
Negative Profits, Continuing Production: π<0 and P≥AVC
Shutdown: π<0 and P<AVC
Case 1, excess profits, occurs whenever maximum profits are positive. The example we have been working on is this case. With P = 7, we know that q=13.09 and π=$20.23.
STEP In the Graphs sheet, click on the pull down menu (over cell R5) and select the Zero Profits option.
Your screen now looks like Figure 12.3.
Figure 12.3: Case 2: Normal (zero) profits.
Source: OutputProfitMaxPCSR.xls!Graphs.
Notice that the price ($5.373) in the bottom left chart just touches the minimum of the average total cost curve. The profit rectangle has zero area because it has zero height. The best the firm can do is zero profitsall other choices of q lead to lower (negative) profits.
In the top left graph, you can see that TR just touches TC. In the top right graph, the top of the profit hill just touches the x axis. These charts confirm what the bottom left chart tells uswith P = $5.373, q yields π=0.
The third and fourth profit cases are the flip side of the first two in the sense that price is so low that profits are now negative. This means firms will leave in the long run, but another question arises: should the firm shut down immediately or continue production?
STEP Click on the pull down menu (over cell R5) and select the Neg Profits, Cont Prod option.
With the Neg Profits, Cont Prod option selected, P = 5.10. The firm produces q = 11.43 and suffers negative maximum profits of −$3.16. Notice that price is below ATC in the bottom left graph, so that the profit rectangle, (AR - ATC)q, will be a negative number. (The area is not negative, but it is interpreted as a negative amount since revenues are below costs.) In the top left graph, the TR line is below the TC curve. In the top right graph, the profit function is below the x axis. There is a maximum, or top of the hill, but it is negative, like a mountain under water.
Keep your eye on the top right graph, reproduced as Figure 12.4. Notice that the top of the profit function is higher than the intercept (where q = 0). It is better for the firm to continue production, even though it is earning negative profits of −$3.16 at the optimal output level, because it would make an even lower negative profit of −$5 (the fixed cost) if it shut down.
Figure 12.4: Case 3: Negative profits, continuing production.
Source: OutputProfitMaxPCSR.xls!Graphs.
The canonical graph of profit maximization can be used to determine whether the firm should produce or shut down by comparing price to average variable cost. The Shutdown Rule is easy: hire no labor and produce nothing if P<AVC.
STEP Look at the bottom left graph on your screen. It confirms that the Shutdown Rule works. Profits are negative because price is below average total cost, but the firm will continue production because P>AVC. When the relationship between P and AVC is such that price is greater than average variable cost, it means that the top of the profit function is higher than the y intercept, as in Figure 12.4.
STEP Click on the pull down menu (over cell R5) and select the Neg Profits, Shutdown option. Figure 12.5 displays the top right graph.
Figure 12.5: Case 4: Negative profits, shut down.
Source: OutputProfitMaxPCSR.xls!Graphs.
In this case, the top of the profit function is below the y intercept. In other words, the maximum profit if the firm produces, −$9.81, is worse than the negative profit incurred if the firm shuts down, −$5. The firm optimizes by choosing q = 0, that is, shutting down.
STEP Look at the bottom left graph on your screen. Once again, we have confirmation of the Shutdown Rule. With P = 4.5, P<AVC and the firm should shut down.
STEP Carefully watch the canonical (bottom left) and profit function (top right) graphs as you change the price (with the pull down menu over cell R5).
As long as P>AVC, the top of the profit hill is above the y intercept. If P=AVC, the two are exactly equal and the firm is indifferent between producing and shutting down.
P<AVC is the magic cutoff point. When this happens, the top of the hill is below the y intercept (which is the negative profit suffered if the firm produces nothing). Thus, the firm’s best choice is to produce nothing.
Here is why the rule works. Multiply the Shutdown Rule by q to get: \begin{gathered} %star suppresses line # (P<AVC)q \ Pq<AVCq\ TR<TVC\end{gathered}TR<TVC is a restatement of the Shutdown Ruleproduce nothing if total revenue cannot cover total variable costs. This makes sense. Why produce if you can’t even pay for the variable expenses? You are better off not producing at all.
If total revenue is less than average total cost, then profits are negative. However, the firm can be in a situation where TRTVC. If so, then production makes sense because you will be able to reduce some of the fixed costs you have to pay no matter what you do. Profits are negative, but it is better to produce than not produce because variable costs are covered and fixed costs are at least partially reduced.
STEP For a summary of the four cases and what the Shutdown Rule is doing, click the button (over cell AC5).
What’s Normal about Zero Profits?
In economics, zero profits are called normal profits. This is confusing. Zero sounds bad, not normal. There is a logical explanation, but it requires a clear separation of accounting versus economic profits. They differ because economists include opportunity costs when calculating economic profits.
Accounting profits = revenues - explicit costs
Economic profits = revenues - explicit costs - opportunity costs
In economics, without an adjective, "profits" means economic profits. So, when profits are zero that means economic profits are zero. Economic profits have had an extra item subtracted, the opportunity costs of using firm resources to make this particular product.
An accountant would subtract explicit (out-of-pocket) costs (wages, rent, etc.) from revenues and if this number is positive, announce that the firm is making money. The economist would then subtract the cost of the profits that could be made by the next best alternative industry that the firm could be in. If economic profits are zero, it means the opportunity costs are exactly equal to the accounting profit and the firm cannot do better by switching to its next best alternative.
Although this may seem needlessly contorted at first, there is a nice interpretation of economic profits: If positive, the firm will stay in the industry and new firms will enter in the long run; if negative, the firm will exit in the long run; and if zero, there will be neither exit nor entry in the long run. It is in this sense of equilibrium that we say zero profits are normal. With π=0, there is stability and no tendency to change in the movement of firms.
The distinction between economic and accounting profits also explains why positive profits are excess profits. It is not meant as a pejorative term, but to indicate that the firm is earning greater profits than are needed to keep producing that product in the long run. Excess profits also mean that others are attracted and will enter that industry.
Economists are not concerned with how much money the firm made, but with profits as a signal to entry and exit. Defining economic profits as accounting profits minus opportunity costs gives us a profit measure that tells us whether the firm will stay or leave in the long run.
Shutdown Rule and Corner Solution
The Shutdown Rule is usually covered in introductory economics. Memorization is often all that is achieved. We can do better by properly situating the Shutdown Rule in the landscape of mathematical and economic conceptsit is a corner solution.
Recall that, in the Theory of Consumer Behavior, there are situations in which the MRS does not equal the price ratio, yet the solution is optimal. This is a corner solution.
Food stamps are an example. The fact that food stamps can only be used to buy food creates a horizontal segment on the budget constraint so that a consumer might not be able to make MRS = p 1 p 2. At the kink in the constraint, the consumer is optimizing even though the equimarginal condition is not met.
Corner solutions are a general phenomenon. They can be seen whenever a restriction or border blocks further improvement in the objective function. Consider Figure 12.6 which sketches a maximization problem to highlight the difference between an interior and a corner solution. In panel B, the agent cannot choose negative values of the x variable and, therefore, the function is cut off by the y axis.
Figure 12.6: Understanding the corner solution.
In panel B, although the marginal condition is not met, we have an optimal solution, defined as doing the best we can without violating any constraints.
Shutting down is another example of a corner solution because, once again, the equimarginal condition is not met at q=0, yet producing nothing is the optimal solution. Shutting down is an unusual example of a corner solution because there is a place where the marginal condition is met (there is an output where MR=MC), but it is not optimal. The profit function twists in such a way (see Figure 12.5) that profit is decreasing as output rises from zero. This means that profits would go up if we were able to produce negative output. Since we are not allowed to choose q<0, we have a corner solution.
How can we know if we should choose q at MR=MC, the interior solution, or shut down, the corner solution? The only way is to compare the profit positions at the two quantities. The good news is that no checking is required for cases 1 and 2. As long as profits are non-negative, there is no way that a profit of minus total fixed cost can be better than the interior solution of q where MR=MC. But, whenever, MR=MC yields negative maximum profits, comparing those negative profits to TFC is necessary. Or, you could just use the Shutdown Rule and see if P < AVC, which will give the same, correct answer.
The complexity of the firm’s profit maximization problem in the short run, with its shutdown possibility, should increase your sensitivity to lurking problems with analytical and numerical methods. We know neither is perfect so there may be glitches in applying these methods to the firm’s profit maximization problem. The Q&A sheet provides an example. Be sure to look carefully at questions 2 and 3.
Finding and Displaying the Initial Solution
The output profit maximization problem for a PC firm in the short run is a single-variable (q) unconstrained problem. It can be solved with numerical and analytical methods. The equimarginal rule applied is that MR=MC and since price taking behavior means that P=MR for a PC firm, the equimarginal rule is often shown as P=MC.
The firm’s profit maximization problem contains a complication in the short run. If maximum profits are negative, it is possible that the firm is better off not producing anything. A shortcut to determine whether or not to produce when π<0 is the Shutdown Rule, P<AVC.
The initial optimal solution is displayed by a canonical graph that superimposes the firm’s revenue side (average and marginal revenue) over its cost structure (average and marginal costs). Optimal output is easily found where MR intersects MC (as long as P>AVC) and maximum profit is displayed as the area of the appropriate rectangle. The ability to instantly show the optimal solution, maximum profits, and whether or not to shut down explains the popularity of this graph.
You can think of the firm as walking through a series of three steps when solving its profit maximization problem:
Choose q where MR=MC in the canonical graph.
Compute profits at q via (AR−ATC)q (the profit rectangle).
If profits are negative, shut down if P<AVC.
The PC firm’s profit maximization is simpler in the long run. If π<0, firms exit the industry; π>0 (also known as excess profits) lead to entry. Thus, in long run equilibrium (a state never actually attained), P=ATC and π=0 for all firms. This is why zero economic profits are called normal profits.
Exercises
Use Excel’s Solver to find the optimal output and profit for a firm with cost function TC=2q 2+10q+50 and P=40. Take a screen shot of your optimal solution (including output and profits) and paste it in a Word document.
Use analytical methods to solve the problem in the previous question.
For what price range will the firm in question 1 shut down? Explain.
If fixed costs are higher, will this influence the firm’s shutdown decision? Explain.
References
The epigraph is from the foreword (p. vi) of Joan Robinson, The Economics of Imperfect Competition (first edition, 1933, followed by many reprints). In a male-dominated profession, Joan Robinson established herself as a well-known, important economist. She helped create the Theory of the Firm, including the canonical graph with average and marginal revenue and cost that is used to this day.
Ironically, however, much of her work was critical of mainstream economics. Her famous Richard T. Ely lecture at the 1971 American Economics Association conference pulled no punches:
For once the president of the AEA was a dissident. This was the veteran institutionalist and Keynesian John Kenneth Galbraith, a longtime friend of Robinson’s and celebrated critic of US capitalism and its apologists in academic economics. Galbraith now offered her the most important platform she had ever occupied. Robinson took full advantage of it, delivering an abrasive, challenging, deliberately provocative indictment of neoclassical economics that was designed to polarize her audience between the old and conservative and the young and progressive. (John Edward King, A History of Post Keynesian Economics Since 1936 (2002), p. 123.)
This page titled 12.1: Initial Solution is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Humberto Barreto.
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Key takeaways
• Decimal numbers get smaller the farther away from the decimal point they are. Decimal places closer to the decimal point are bigger.
• Adding zeros after a decimal number does not change the number’s value.
• To round decimal places, look at the number that comes after the decimal place value you are rounding to. Numbers 5 or more are rounded up. Numbers 4 or less can be kept the same.
When you’re buying something from the store or measuring something with a ruler, you’ve probably noticed that you don’t always get whole numbers. For example, a candy bar might cost $1.98 or a shelf in your room might be 15.25 inches long.
But how do you work with numbers like this?
In this article, we’ll go over decimal place values, decimal comparison, rounding decimals, and the decimal place value chart.
What is Decimal Place Value?
Decimals are numbers made up of a whole number followed by a fractional number. These two numbers are separated by a dot called a decimal point.
For example, 13.5 is a decimal number because of the decimal point in the middle. The 13 is the whole number and the 5 is a fractional part.
Numbers that come after decimal points should be thought of as fractions. The position a number falls in relation to the decimal point is called its decimals place value.
Let’s take a look at a decimal places chart to better understand each decimal place.
Table of contents
What is it?
Decimal places
Comparing values
How to round
Practice problems
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What are the Decimal Places?
Decimal place values have different names and values according to how close or far away they are from the decimal point. The decimal placesare all powers of 10. This is visualized on the decimal chart.
Place value chart with decimals:
Take a look at the decimals chartabove.
Decimal numbers get smaller as they move farther away from the decimal point, and larger as they move closer. Whole numbers are the opposite.
Let’s compare the tens place and the tenths place. When a whole number reaches the tens place, it means there are tens of something. When a decimal number reaches thetenths place, it means there is a tenth of something or a fraction of a whole.
Tenths Place
The tenths place is one decimal placeafter the decimal point and is the largest value decimal place. Numbers in this place are called tenths because when turned into fractions, they are put over a 10.
Ex. 13.5 = 13 5/10 or 13 ½ when simplified
Hundredths Place
The hundredths placecomes second after the decimal point. When decimal numbers reach this place, it means they will be written over 100 in fraction form.
Ex. 12.25 = 12 25/100 or 12 ¼
Thousandths Place
The thousandths place comes third after the decimal point. When turned into a fraction, these numbers will be written over 1,000.
Ex. 3.825 = 3 825/1,000 or 3 33/44
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Comparing Decimal Values
We already mentioned that negative numbers appear on thermometers to show us that the temperature is “below zero”, but that’s not the only real-life example of negative numbers.
Negative numbers are all around us! When we make purchases, the money or credit due is a negative number. Negative numbers also appear on elevation maps to indicate places that are “below sea level”.
Negative numbers can be used as penalties in games, quizzes, and tests. When you get an answer wrong, you may get -5 points, meaning 5 points are being taken away from your total score.
In sports, too, negative numbers can be used to denote the scores of teams. One great example is golf. The golfer with the lowest score, which represents the least amount of strokes, wins!
As you can see, there are many real-world applications of negative numbers. They may not be as common as positive numbers, but they’re just as important.
Now, let’s go over the rules for working with negative numbers.
How to Round Decimals Using Decimal Place Values
Pi is a rare number that has 31.4 trillion decimal place valuesand counting!
When doing math with pi, mathematicians definitely don’t use 31.4 trillion decimal places. They round pi down to a more manageable number. At NASA, pi is rounded down to 15 digits: 3.14159265358979.
Let’s try rounding to the one decimal place.
What is 12.57 rounded to the nearest tenths place?
We’re rounding to the tenths place, so look at the hundredths place. In this case, that’s 7.
Next, determine whether this number is 5 or more, or 4 or less. Numbers 5 or more are rounded up. Numbers 4 or less are kept the same. In this case, round up since 7 falls under the category of 5 or more.
So, 12.57 rounded to the nearest tenth is 12.6.
Now let’s try to round to three decimal places. What is 2.4983 rounded to the nearest thousandths place?
The number that comes after the thousandths place is 3. Since 3 falls under the category of 4 or less, we will be keeping the number in the thousandths placethe same.
So, 2.4983 rounded to the nearest thousandth is 2.498.
For more help with reading, writing, and rounding decimals, there are great online math apps and websites you can use that will provide you with detailed instruction, a decimals place value chart, practice questions, and explanations to strengthen your understanding of decimals.
Practice Problems
Click on the boxes below to see the answers!
Question 1: How would you write 5.34 as a mixed number based on what you know about decimal place values?
Answer: 5 34/100.
Since the number 5.34 ends in the hundredths decimal place, the number after the decimal point can be written over 100 in fraction form.
Question 2: Look at the number 8.723. Round to 2 decimal places.
Answer: 8.72
To round to the second decimal place, look at the number that comes after it in the thousandths place, 3. Since 3 is smaller than 5, you keep the hundredths place the same. So, 8.723 rounded to the second decimal place is 8.72.
Question 3: Which is the larger number? 1.54 or 1.6?
Answer: 1.6
To compare the size of decimal numbers, start by comparing the whole numbers and work your way to the right. The number 1.54 has a 5 in the tenths place while 1.6 has a 6 in the tenths place. Six is larger than 5, so 1.6 is the larger number.
FAQs about Decimal Place Value
What is decimal place value in math?
Decimal place value is the value of numbers according to how far away they are from a decimal point. Place values after a decimal point represent fractions of a whole number.
How do you write place value in decimals?
Place value in decimals starts with the tenths place and grows by powers of 10 as you move to the right. The second place value is the hundredths placeand the third is the thousandths place.
Are negative numbers whole numbers?
No, negative numbers are not whole numbers. Whole numbers are only positive round numbers, including 0.
Where is the hundredths place in a decimal?
The hundredths placein a decimal is the second decimal place valueafter the decimal point.
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4706 | https://totalmetrics.com/help/Managing_your_Function_Point_Counts/Performing_a_Function_Point_Count/Setting_up_the_Count/Basic_Steps_of_an_FPA_Count_.htm | Basic Steps of an FPA Count
Managing your Function Point Counts>Performing a Function Point Count>Setting up the Count> Basic Steps of an FPA Count
Basic Steps of an FPA Count
Set Up New Count-Baseline Count
Set Up New Count-Enhancement Count
First Function Point Count
When you first open a Count Session the Function Hierarchy Tree is displayed. This is where your software is modelled as a hierarchical structure. Your first step is to determine the main functional areas within the software application to be measured and enter them into SCOPE as Functions (use Alt INS Key) in the Function Tree Hierarchy. Continue breaking down each Function until you get to the lowest elementary Process to be performed within the software application. Enter each Process (use Ins Key) into SCOPE.
Working from the application’s functional specification, in the Detail screen classify each elementary Process as either an:
Input i.e. something that enables the user to input data into the software to be stored
Output i.e. something that enables the user to extract derived information from the software
Inquiry i.e. something that enables the user to query stored data
Select to display the Data Tree Hierarchy by selecting the Data Tab at the top of the screen on the RHS window. Determine the categories of logical files in your software and create some File Folders. Under each Data GroupFolder identify the logical groups of data in your software and enter each Data Group (ILF or EIF) into SCOPE under the appropriate Folder the Data Tree Hierarchy. Data groups may be further decomposed into sub-groupings of Record Element Types (RETs). You can also enter the fields (Data Element Types-DETs) under each RET sub-grouping.
Working from the application’s functional specification, in the Detail Screen classify each Data Group as either an:
Internal File (ILF) i.e. something that stores data input from the user’s transactions, i.e.Processes on the Function Tree update ILFs
External File (EIF) i.e. something that stores data accessed by the user’s transactions. i.e.Processes on the Function Tree only read EIFs
Use the detail screen to classify Processes and data groups as either low, average or high complexity.
SCOPE will assign a weighting in function point units to the software’s functional and data components using a set of prescribed formula’s, based on the type and complexity you have selected.
NOTE: SCOPE defaults Processes to be Inputs of average complexity and data groups to be of a type undefined and low complexity. You can change these defaults by selecting VIEW, Function Point Count Default Values, from the Main Menu.
After the assessment of each Process and Data Group is complete, the Functional Size in Unadjusted Function Points can be reported for the whole application, or any selected part of it or just for those functions and data impacted by a change request. The size is reported in the status bar at the bottom of the screen or can be seen in detail by selecting the Reporting Selection option under the Main Menu.
SCOPE:
Can be used to measure the size of a functional branch of the hierarchy by selecting a node at any level
Can use Flags to select sets of function and or Process nodes, for selective measurement
Can be used to selectively record and report functions impacted by a project using an impact Count Session
SCOPE allows you to perform the optional step of evaluating an application’s General Systems Characteristics to calculate the Value Adjustment Factor for the application for that Release. The Functional Size can then be combined with a Value Adjustment Factor to take into account quality and technical characteristics, which will then give a product size in Adjusted Function Points. |
4707 | https://artofproblemsolving.com/wiki/index.php/Rational_root_theorem?srsltid=AfmBOop2Ed5wfOqYnsM5C4g9Mt37J6mdAg-3zLqxuMsBOrhTOcthW3hU | Art of Problem Solving
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Rational root theorem
In algebra, the rational root theorem states that given an integer polynomial with leading coefficient and constant term , if has a rational root in lowest terms, then and .
This theorem is most often used to guess the roots of polynomials. It sees widespread usage in introductory and intermediate mathematics competitions.
Contents
[hide]
1 Proof
2 Examples
2.1 Example 1
2.2 Example 2
2.3 Example 3
3 See also
Proof
Let be a rational root of , where every is an integer; we wish to show that and . Since is a root of , Multiplying by yields Using modular arithmetic modulo , we have , which implies that . Because we've defined and to be relatively prime, , which implies by Euclid's lemma. Via similar logic in modulo , , as required.
Intro to Rational Roots theorem:
Examples
Here are some problems with solutions that utilize the rational root theorem.
Example 1
Find all rational roots of the polynomial .
Solution: The polynomial has leading coefficient and constant term , so the rational root theorem guarantees that the only possible rational roots are , , , , , , , and . After testing every number, we find that none of these are roots of the polynomial; thus, the polynomial has no rational roots.
Example 2
Factor the polynomial .
Solution: After testing the divisors of 8, we find that it has roots , , and . Then because it has leading coefficient , the factor theorem tells us that it has the factorization .
Example 3
Using the rational root theorem, prove that is irrational.
Solution: The polynomial has roots . The rational root theorem guarantees that the only possible rational roots of this polynomial are , and . Testing these, we find that none are roots of the polynomial, and so it has no rational roots. Then because is a root of the polynomial, it cannot be a rational number.
See also
Root
Polynomial
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4709 | https://www.basic-mathematics.com/weight-word-problems.html | Home
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Weight word problems
Learn how to calculate the weight when the mass is given with some weight word problems. Also learn how to get the mass when the weight is given.
Question #1:
If a scale says 80 kg after you step on it, is that a measure of your weight?
A common misconception is that a scale always gives you your weight.
When the scale says 80 kg, it is telling you about your mass.
To get your weight, you either need to convert the mass to pounds or Newtons.
Question #2:
If 2 kg of feather weigh 19.6 Newtons, does 2 kg of nails weigh 19.6 Newtons?
Yes, at the earth's surface 2 kg of anything will have the same weight.Because feathers are much lighter than nails, some may have a tendency to think that it will weigh less. However, once you pack 2 kg of each material, 2 kg is 2 kg and it will weigh the same for both materials.
Problem #1:
What is the weight of a baby with a mass of 3 kg?
Solution
To find the weight, use the formula below:
w = m × g w = 3 kg × 9.8 w = 29.4 Newtons
Problem #2:
You step on the scale and it says 100 kg. You brother steps on the scale and it says 220 pounds. Who is heavier you or your brother?
Solution:
To get your weight, multiply 100 kg by 2.2. ( 1 kg weighs 2.2 pounds)
2.2 multiplied by 100 equals 220.
You and your brother have the same weight.
Weight word problems with a little more math
The weight word problems below are just a little more challenging. They show how to get the mass from the weight.
Problem #3:
An orange weighs about 1 N. what is the mass of the orange in grams?
(Use g = 10 Newtons instead of 9.8 to simplify the math.)
Solution:
mass =
weight / g
mass =
1 / 10
m = 0.1 kg 1 kg = 1000 grams
Problem #4:
If your weight is 240 pounds, what is your mass?
What is your mass on the moon?
What is your weight in Newtons ?
How much would you weight on the moon?
Solution:
The mass is expressed in kilograms ( kg ) and 1 kg weighs 2.2 pounds.
Let's us use chain-link conversion and the concept of conversion factor to solve this problem to find your weight.
Solution:
1)
The conversion factor is
1 kg / 2.2 pounds
Multiply the conversion factor by 240 pounds
1 kg / 2.2 pounds
× 240 pounds
This simplify to
240 kg / 2.2
Your mass is equal to 109.09 kg.
2) Your mass is the same on the moon!
3) w = m × g w = 109.09 × 9.8 w = 1069.082 Newtons
4) On the moon, gravity is about one-sixth that of the earth.
Therefore, just get one-sixth of 1069.082 to get your weight on the moon.
One-sixth of 1069.082 is the same as 1069.082 divided by 6.
1069.082 divided by 6 equals 178.18.
Weight of an object
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4710 | https://www.mathway.com/popular-problems/Algebra/674361 | Simplify square root of 3 | Mathway
Enter a problem...
[x]
Algebra Examples
Popular Problems
Algebra
Simplify square root of 3
√3 3
Step 1
The result can be shown in multiple forms.
Exact Form:
√3 3
Decimal Form:
1.73205080…1.73205080…
2√3 3 2
⋅√3⋅3⋅√2⋅2+√3+3
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⎡⎢⎣x 2 1 2√π∫x d x⎤⎥⎦[x 2 1 2 π∫x d x]
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4711 | https://me304.cankaya.edu.tr/uploads/files/ZME304%206%20Dimensional%20Analysis(1).pdf | DIMENSIONAL ANALYSIS AND SIMILITUDE Many real fluid flow problems can be solved, at best, only approximately by using analytical or numerical methods. Therefore experiments play a crucial role in verifying these approximate solutions.
Solutions of real problems usually involve a combination of analysis and experimental work. First, the real physical flow situation is approximated with a mathematical model that is simple enough to yield a solution. Then experimental measurements are made to check the analytical results. Based on the measurements, refinements are made in the mathematical model and analysis.
The experimental results are an essential link in this iterative process.
The experimental work in the laboratory is both time consuming and expensive.
Hence, the goal is to obtain the maximum information from the fewest experiments.
The dimensional analysis is an important tool that often helps in achieving this goal. Dimensional analysis is a packaging or compacting technique used to reduce the complexity of experimental programs and at the same time increase the generality of experimental information.
Consider the drag force on a stationary smooth sphere immersed in a uniform stream. What experiments must be conducted to determine the drag force on the sphere?
F V µ ρ D ME304 6 1 We would expect the drag force, F, depends on diameter of the sphere, D, the fluid velocity, V, fluid viscosity, µ and the fluid density ρ. That is, Let us imagine a series of experiments to determine the dependence of F on the variables D, V, ρ and µ. To obtain a curve of F versus V for fixed values of ρ, µ and D, we might need tests at 10 values of V. To explore the diameter effect, each test would be repeat for spheres of ten different diameters. If the procedure were repeated for 10 values of ρ and µ in turn, simple arithmetic shows that 104 separate tests would be needed. Also we would have to find 100 different fluids.
Because we need 10 different ρ’s and 10 different µ’s. Assuming each test takes ½ hour and we work 8 hours per day, the testing will require 2.5 years to complete.
Dimensional analysis comes to rescue. If we apply dimensional analysis, it reduces to the following equivalent form: The form of function still must be determined experimentally. However, rather than needing to conduct 104 experiments, we would establish the nature of function as accurately with only 10 tests.
ME304 6 2 BUCKINGHAM PI THEOREM The dimensional analysis is based on the Buckingham Pi theorem.
Suppose that in a physical problem, the dependent variable q1 is a function of n‐1 independent variables q2, q3, ….., qn. Then the relationship among these variables may be expressed in the functional form as Mathematically, we can express the functional relationship in the equivalent form.
Where g is an unspecified function, and it is different from the function f. For the drag on sphere we wrote the symbolic equation We could just as well have written The Buckingham Pi theorem states that, the n parameters may be grouped into n‐m independent dimensionless ratios, or π parameters, expressible in functional form by or The number m is usually, but not always, equal to the minimum number of independent dimensions required to specify the dimensions of all the parameters, q1, q2, ….., qn.
ME304 6 3 DETERMINING THE π GROUPS To determine the π parameters, the steps listed below should be followed.
Step 1 Select all the parameters that affect a given flow phenomenon and write the functional relationship in the form or If all the pertinent parameters are not included, a relation may be obtained, but it will not give the complete story. If parameters that actually have no effect on the physical phenomenon are included, either the process of dimensional analysis will show that these do not enter the relation sought, or experiments will indicate that one or more non‐dimensional groups are irrelevant.
Step 2 List the dimensions of all parameters in terms of the primary dimensions which are the mass, M, the length, L, and the time, t (MLt), or the force, F, the length, L, and the time, t (FLt). Let “r” be the number of primary dimensions.
Step 3 Select a number of repeating parameters, equal to the number of primary dimensions, r, and including all the primary dimensions. As long as, the repeating parameter may appear in all of the non‐dimensional groups that are obtained, then do not include the dependent parameter among those selected in this step.
Step 4 Set up dimensional equations, combining the parameters selected in step 3 with each of the remaining parameters in turn, to form dimensionless groups. (There will be n‐m equations). Solve the dimensional equation to obtain the (n‐m) dimensionless groups.
Step 5 Check to see that each group obtained is dimensionless.
ME304 6 4 Example: The drag force, F, on a smooth sphere, which is mowing comparatively slowly through a viscous fluid, depends on the relative velocity, V, the sphere diameter, D, the fluid density, ρ, and the fluid viscosity, µ. Obtain a set of dimensionless groups that can be used to correlate experimental data.
ME304 6 5 To be completed in class Example: When a small tube is dipped into a pool liquid, surface tension causes a meniscus to form at the free surface, which is elevated or depressed depending on the contact angle at the liquid‐solid‐gas interface. Experiments indicate that the magnitude of the capillary effect, Δh, is a function of the tube diameter, D, liquid specific weight, γ, and surface tension, σ. Determine the number of independent π parameters that can be formed and obtain a set.
ME304 6 6 DIMENSIONLESS GROUPS OF SIGNIFICANCE IN FLUID MECHANICS There are several hundred dimensionless groups in engineering. Following tradition, each such group has been given the name of a prominent scientist or engineer, usually the one who pioneered its use.
Forces encountered in the flowing fluids include those due to inertia, viscosity, pressure, gravity, surface tension, and compressibility. The ratio of any two forces will be dimensionless. We can estimate typical magnitudes of some of these forces in a flow as follow: / force ility Compressib force tension Surface force Gravity force Pressure force Viscous force Inertia 2 3 2 2 2 2 3 d dp E L E A E L L g mg pL A p VL L L V A dy du A V L L V V L ...
x u u m Dt V D m a m v v v ME304 6 7 Inertia force is important in most fluid mechanics problems. The ratio of the inertia force to each of other forces listed above leads to five fundamental groups encountered in fluid mechanics.
The Reynolds number is the ratio of inertia forces to the viscous forces, and it is named after Osbourne Reynolds (1842 ‐ 1912).
force force Ma number Mach force force We number Weber force force Fr number Froude force force Eu number Euler force Viscous force Inertia Re number Reynolds 2 2 f f f f VL VL VL L V ME304 6 8 FLOW SIMILARITY AND MODEL STUDIES When an object, which is in original sizes, is tested in laboratory it is called prototype. A model is a scaled version of the prototype. A model which is typically smaller than its prototype is economical, since it costs little compared to its prototype. The use of the models is also practical, since environmental and flow conditions can be rigorously controlled. However, models are not always smaller than the prototype. As an example, the flow in a carburetor might be studied in a very large model.
There are three basic laws of similarity of model and prototype flows. All of them must be satisfied for obtaining complete similarity between fluid flow phenomena in a prototype and in a model. These are a) The geometric similarity, b) the kinematic similarity, and c) the dynamic similarity.
Geometric Similarity: The geometric similarity requires that the model and prototype be identical in shape but differ in size. Therefore, ratios of the corresponding linear dimensions in the prototype and in the model are the same.
Kinematic Similarity: The kinematic similarity implies that the flow fields in the prototype and in the model must have geometrically similar sets of streamlines.
The velocities at corresponding points are in the same direction and are related in magnitude by a constant scale factor.
Dynamic Similarity: When two flows have force distributions such that identical types of forces are parallel and are related in magnitude by a constant scale factor at all corresponding points, the flows are dynamically similar.
ME304 6 9 By using Buckingham theorem, we can find which dimensionless groups are important for a given flow phenomenon. To achieve dynamic similarity between geometrically similar flows, we must duplicate all of these dimensionless groups.
For example, in considering the drag force on sphere we found that Thus, in considering a model flow and prototype flow about a sphere, the flows will be dynamically similar if The results determined from the model study can be used to predict the drag on the full scale prototype.
Re 1 1 2 2 f VD f D V F then Re Re is that prototype 2 2 model 2 2 prototype model prototype model D V F D V F VD VD ME304 6 10 Example: A one‐tenth‐scale model of a derby car, shown in the figure, is tested in a wind tunnel. The air speed in the wind tunnel is 70 m/s, the air drag on the model car is 240 N, and the air temperature and pressure are identical those expected when the prototype car is racing. Find the corresponding racing speed in still air and the drag on the car.
ME304 6 11 ME304 6 12 Example: A jet plane travelling at a velocity of 900 m/s at 6 km altitude, where the temperature and the pressure are ‐24 oC and 47.22 kPa, respectively. A one‐tenth scale model of the jet is tested in a wind tunnel in which carbon dioxide is flowing. The gas constant for air and carbon dioxide are 287 J/kg K and 18.8 J/kgK, respectively. The specific heat ratios for air and carbon dioxide are 1.4 and 1.28, respectively. Also the absolute viscosities of the air at ‐24 oC and carbon dioxide at 20 oC are 1.6x10‐5 Pa s and 1.47x10‐5 Pa s, respectively. Determine a) The required velocity in the model, and b) The pressure required in the wind tunnel.
ME304 6 13 ME304 6 14 INCOMPLETE SIMILARITY To achieve complete dynamic similarity between geometrically similar flows all of the dimensionless numbers in prototype and in the model (that is Re, Eu, Fr, We, …) should be equal. Fortunately, in most engineering problems, the equality of all of dimensionless groups is not necessary. Since some of forces i.
may not act ii.
may be negligible magnitude or iii.
may oppose other forces in such a way that the effect of both is reduced.
In some cases complete dynamic similarity may not be attainable. Determining the drag force of surface ship is an example of such a situation. The viscous shear stress and surface wave resistance cause the drag. So that for complete dynamic similarity, both Reynolds and Froude numbers must be equal between model and prototype.
This requires that to ensure dynamically similar surface wave patterns.
From the Reynolds number requirement ME304 6 15 If we use the velocity ratio obtained from matching Froude numbers, equality of Reynolds number leads to a kinematic viscosity ratio of If Lm/Lp equals 1/100 (a typical length scale for ship model tests), then νm/ νp must be 1/1000.
Mercury, which is the only liquid, its kinematic viscosity is less than water. Thus, we cannot simultaneously match Reynolds number and Froude number in the scale‐model test. Then one is forced to choose either the Froude number similarity, or the Reynolds number similarity. For this reason, the experiments with the model are performed so that Frp=Frm which results Rep>>Rem. The test results are then corrected by using the experimental data which is dependent on the Reynolds number.
ME304 6 16 Example: The drag force on a submarine, which is moving on the surface, is to be determined by a test on a model which is scaled down to one‐twentieth of the prototype. The test is to be carried in a towing tank, where the model submarine is moved along a channel of liquid. The density and the kinematic viscosity of the seawater are 1010 kg/m3 and 1.3x10‐6 m2/s, respectively. The speed of the prototype is 2.6 m/s.
a) Determine the speed at which the model should be moved in the towing tank.
b) Determine the kinematic viscosity of the liquid that should be used in the towing tank.
c) If such a liquid is not available, then the test may be carried out with seawater by neglecting the viscous effects. In this case, determine the ratio of the drag force due to the surface waves in the prototype to the drag force in the model.
Solution: a) Because of low speed of the submarine, the compressibility has no effect on the dynamic similarity, and the Mach number plays no role. The Froude numbers for the prototype and the model may be equated to yield.
ME304 6 17 b) To determine the kinetic viscosity of the liquid that should be used in the towing tank, one may equate the Reynolds number in the model and prototype.
Rearranging one may obtain c) However, one should note that a liquid with a given kinematic viscosity cannot be practically formed. Then the test in towing tank may be carried out with sea water by neglecting the viscous effects. In this case, only the equality of the Froude number is sufficient for the dynamic similarity and the drag force is only due to the surface waves.
By using Buckingham π theorem one may obtain.
But in this case only the equality of the Froude number is sufficient, than Frmodel=Frprototype ME304 6 18 This result must be corrected for viscous effects.
ME304 6 19 |
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Market Segment Size and Growth Analysis
Goal of the analysis:
The goal of market segment size and growth analysis is to evaluate the current size and historical growth trends of specific market segments. This analysis helps organizations identify lucrative segments, assess growth potential, and align business strategies to capitalize on expanding opportunities.
Data required:
Market Data:
Total market size in terms of revenue, units sold, or customer base.
Historical market size data for the segment over multiple periods.
Segment-Specific Data:
Revenue, units sold, or customer base for the specific segment.
Key demographic, geographic, or psychographic attributes of the segment.
Growth Data:
Annual growth rates for the segment and overall market.
Competitor presence and performance within the segment.
Industry Benchmarks:
Average segment growth rates in the industry.
Historical performance data for the organization within the segment.
Detailed step-by-step instruction on how to conduct the analysis:
Define the Market Segment:
Identify the specific market segment to analyze (e.g., geographic region, customer demographic, product type).
Determine whether the analysis will focus on revenue, units sold, or customer count.
Collect Market Data:
Gather total market size and segment-specific data from reliable sources (e.g., industry reports, government data, market intelligence platforms).
Include historical data for at least 3-5 years to observe trends.
Calculate Segment Size:
Determine the segment’s contribution to the overall market using the formula: Segment Share=(Segment Size÷Total Market Size)×100Segment
Analyze Historical Growth:
Calculate the annual growth rate for the segment using the formula: Growth Rate=[(Segment Size in Current Year−Segment Size in Previous Year)÷Segment Size in Previous Year]×100
Repeat this calculation for multiple years to identify growth trends.
Compare with Market Growth:
Assess the segment’s growth rate relative to the overall market’s growth.
Identify whether the segment is growing faster, slower, or in line with the total market.
Benchmark Against Competitors:
Compare the organization’s performance within the segment to key competitors.
Identify any competitive advantages or gaps.
Summarize Insights:
Highlight the segment’s size, historical growth trends, and growth potential.
Identify emerging opportunities or risks within the segment.
Format of the output of analysis:
Summary Report:
Total size of the market segment.
Segment’s contribution to the overall market.
Annual growth rates and historical trends.
Charts and Graphs:
Line charts showing historical growth trends for the segment and overall market.
Bar charts comparing segment size to other segments.
Pie charts illustrating the segment’s contribution to the total market.
Benchmarks Comparison:
Table comparing segment growth rates with industry averages and competitor performance.
How to interpret results:
Large Segment Size with High Growth Rate:
Indicates a lucrative and expanding opportunity.
Prioritize investment and strategic focus in this segment.
Small Segment Size with High Growth Rate:
Signals a niche opportunity with potential for future expansion.
Consider strategies for market entry or increased presence.
Stagnant or Declining Segment:
Suggests limited opportunities or potential risks in this segment.
Evaluate the causes of stagnation and consider reallocating resources.
Benchmark Comparison:
Growth rates above benchmarks indicate competitive strength and alignment with market trends.
Below-average growth rates may highlight gaps in strategy or execution.
Steps a company can take to improve on this measure:
Target High-Growth Segments:
Focus on segments with above-average growth potential.
Develop tailored products or services to meet segment-specific needs.
Enhance Market Research:
Conduct regular market research to identify emerging trends and opportunities.
Use advanced analytics to predict segment growth trajectories.
Strengthen Competitive Positioning:
Develop unique value propositions to differentiate the organization within the segment.
Address competitive weaknesses by improving product quality, pricing, or customer service.
Optimize Marketing Strategies:
Use targeted campaigns to reach high-potential customers within the segment.
Leverage digital marketing tools to track and engage specific audience groups.
Invest in Innovation:
Develop innovative solutions to meet evolving demands within the segment.
Monitor technology and consumer trends to stay ahead of competitors.
Expand Distribution Channels:
Explore new sales channels to improve access to the segment.
Partner with distributors or retailers who specialize in serving the target segment.
Benchmark Continuously:
Compare segment size and growth trends with industry benchmarks to measure progress.
Adjust strategies based on performance relative to competitors and market conditions.
Monitor Emerging Segments:
Identify adjacent or emerging segments with high growth potential.
Explore opportunities to diversify offerings and capture additional market share.
Benchmark Comparisons:
General Market Segment Size and Growth Metrics
Overall Market Size: The total market size can vary significantly by industry but is often measured in terms of revenue or unit sales. A common approach is to express market size in billions of dollars or millions of units sold.
Growth Rate: The average annual growth rate (AAGR) for various markets typically ranges from 3% to 10%, with emerging markets often experiencing higher growth rates.
Industry-Specific Market Segment Size and Growth Benchmarks
Technology Sector:
In the technology sector, particularly software, the market size can reach $500 billion with an average growth rate of 8% to 12% annually. For example, the cloud computing segment has shown rapid growth, expanding from $100 billion in 2020 to an expected $200 billion by 2025, reflecting a compound annual growth rate (CAGR) of about 15%.
E-commerce:
The e-commerce market is estimated to be around $4 trillion, with specific segments like fashion or electronics growing at rates of 10% to 15% annually. For instance, online fashion retail has been projected to grow from $600 billion in 2021 to over $1 trillion by 2025, indicating robust demand and changing consumer behaviors.
Healthcare:
In healthcare, the overall market size can be approximately $8 trillion, with segments like telehealth experiencing significant growth rates of 20% to 30% annually.For example, the telemedicine market was valued at about $40 billion in 2020 and is expected to reach $130 billion by 2025, driven by increased acceptance and technological advancements.
Automotive Industry:
The global automotive market size is around $3 trillion, with electric vehicles (EVs) growing at an impressive rate of 15% to 25% per year. The EV segment alone is projected to grow from approximately $150 billion in 2020 to over $800 billion by 2025 as consumer preferences shift towards sustainable options.
Food and Beverage:
In the food and beverage sector, the overall market size is estimated at about $2 trillion, with organic food segments growing at rates of 5% to 10% annually. For instance, the organic food market was valued at around $50 billion in 2020 and is projected to reach approximately $70 billion by 2025 due to rising health consciousness among consumers.
Key Metrics for Measuring Market Segment Size and Growth
Total Addressable Market (TAM): The total revenue opportunity available if a product or service achieves 100% market share.
Serviceable Available Market (SAM): The segment of the TAM targeted by your products or services that is within your reach.
Market Growth Rate Calculation: Calculated using the formula:
Growth Rate=[(Current Year Size−Previous Year Size)÷Previous Year Size]×100
4. Segmented Analysis: Breaking down overall market data into specific segments based on demographics, geography, or behavior helps identify niche opportunities.
5. Trends Over Time: Monitoring changes in segment sizes and growth rates over time can provide insights into emerging trends and shifts in consumer preferences.
Contact us at [email protected] for a full list of sources.
Find a marketing consultant
Menu of the Marketing Analyses:
INDEX FOR THIS RESOURCE
Digital Marketing
Email Open Rate
Email Click-Through Rate
Email Conversion Rate
Email Bounce Rate
Email Unsubscribe Rate
Email Return on Investment
Audience Demographics
Audience Growth Rate
Organic Search Traffic
Keyword Rankings
Click-Through Rate (CTR) from Search
Bounce Rate on Search Landing Pages
Domain Authority
Backlinks Analysis
Conversion Rate from Organic Traffic
Social Media Engagement
Social Media Follower Reach
Social Media Follower Growth Rate
Social Media Impressions
Social Media Referral Traffic
Conversion Rate from Social Media
Content Marketing
Average Time on Website Page
Audience Engagement Rate
Website Content Return Rate
Video View Count
Video Watch Time
Video Engagement Rate
Video Completion Rate
Conversions from Video Content
Blog Traffic Analysis
Blog Average Time on Page
Conversions from Blog Content
Brand Marketing
Brand Awareness Analysis
Brand Recall Analysis
Brand Reach Analysis
Brand Search Volume
Brand Messaging Analysis
Brand Loyalty Score Analysis
Brand Equity Analysis
Advertising
Campaign Impressions Analysis
Campaign Reach Analysis
Campaign Conversion Rate Analysis
Ad Click-through Rate Analysis
Ad Cost Per Impression Analysis
Ad Cost Per Click Analysis
Ad Cost Per Conversion Analysis
Ad Cost Per Acquisition Analysis
Return on Ad Spend Analysis
Demand Generation & Customer Marketing
Cost per lead (CPL) analysis
Lead-to-Customer Ratio Analysis
Lead Quality Score Analysis
Lead Conversion Rate Analysis
Cost Per Acquisition Analysis
Customer Persona Analysis
Loyalty Enrollment Rate Analysis
Repeat Purchase Frequency Analysis
Average Order Value from Referrals Analysis
Redemption Rate Analysis
Customer Lifetime Value from Loyalty Program Analysis
Cost Per Referral Analysis
Referral Engagement Rate Analysis
Referral Conversion Rate Analysis
Average Order Value from Referrals Analysis
Account Conversion Rate Analysis
Account Engagement Score Analysis
ROI Per Account Analysis
Marketing Communications & Events
Media Reach and Impressions Analysis
Sentiment Analysis
Share of Voice Analysis
PR-driven Site Traffic Analysis
PR-generated Leads Analysis
Press Release Reach and Impressions Analysis
Influencer Reach Analysis
Influencer Engagement Rate Per Post Analysis
Influencer Cost Per Engagement Analysis
Conversions from Influencer Content Analysis
Influencer ROI Analysis
Event Attendance Rate Analysis
Event Engagement Analysis
Lead Conversion Rate from the Event Analysis
Post-event ROI Analysis
Market Research & Analytics
Market Share Analysis
Market Segment Size and Growth Analysis
Revenue Contribution by Segment Analysis
Customer Satisfaction by Segment Analysis
Market Penetration Rate by Segment Analysis
Competitive Reach and Engagement Analysis
Share of Voice Analysis
Competitor Pricing Comparison
Competitive Product Feature Benchmarking
Market Demand Forecasting Analysis |
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Grade 7
5.01 Simplify expressions with integer coefficients
LessonPractice
Introduction
In the previous year, we have looked at how to represent algebraic expressions , and how to generate equivalent expressions by applying properties of operations. We are now ready to simplify algebraic expressions.
Ideas
Simplify algebraic expressions
Simplify algebraic expressions
The image represents one box containing p apples, and then we get another box containing p apples:
We can write p apples plus p more apples as: Number of apples=p+p
Remember that adding the same number multiple times is the same as multiplying it. So two boxes of p apples can be written as: Number of apples=p+p=2p
This is a very simple case of what is known as combining like terms. If we wanted to then add another 3 boxes of p apples, that is we want to add 3p to 2p, we can see that we would have a total of 5p apples.
| | | |
---
| 2p+3p | = | {p+p}+{p+p+p} |
| | = | p+p+p+p+p |
| | = | 5p |
But what if we wanted to now add 4 boxes, each containing n bananas to our existing boxes of apples?
| | | |
---
| 2p+3p+4n | = | {p+p}+{p+p+p}+{n+n+n+n} |
| | = | p+p+p+p+p+n+n+n+n |
| | = | 5p+4n |
Can we simplify this addition any further?
We cannot add 5 boxes of apples and 4 boxes of bananas into one combined term, because we wouldn't have 9 boxes of apples, nor would we have 9 boxes of bananas. What would we have? 9 boxes of Bapples? Bapples don't exist.
We cannot simplify this expression any further, because p and n are not like terms. If you replace p and n with any other variables the same logic applies.
Two algebraic terms are called like terms if they have exactly the same combination of variables.
This includes exponents: x and x2 are not like terms, in the same way that 4 and 42 are not equal.
Let's look at the expression: 9x+4y−5x+2y What does this mean, and how can we simplify it?
Remember that we leave out multiplication signs between numbers and variables. So we can read the expression as follows:
| 9x | +4y | −5x | +2y |
--- --- |
| 9 groups of x | plus 4 groups of y | minus 5 groups of x | plus 2 groups of y |
Thinking about it this way, we can see that 9x and −5x are like terms (they both represent groups of the same unknown value x). We can now rearrange the equation, ensuring the sign attached the left of any term remains with it.
| 9x | −5x | +4y | +2y |
--- --- |
| 9 groups of x | minus 5 groups of x | plus 4 groups of y | plus 2 groups of y |
If we have "9 groups of x" and subtract "5 groups of x", then we will be left with "4 groups of x". That is 9x−5x=4x.
Similarly, 4y and 2y are like terms, so we can add them: 4y+2y=6y.
Putting this together, we have: 9x+4y−5x+2y=4x+6y
Notice that we can't simplify 4x+6y any further. The variables x and y represent different unknown values, and they are not like terms.
To combine like terms means to simplify an expression by combining all like terms together through addition and/or subtraction.
Examples
Example 1
Are the following like terms: 9y and 10y?
Worked Solution
Apply the idea
9y and 10y are like terms, because they have the same variables and the same exponents.
Example 2
Are the following like terms: 10x2y and 9y2x?
Worked Solution
Apply the idea
In 10x2y,x is raised to the power of 2.
In 9y2x,y is raised to the power of 2.
So 10x2y and 9y2x are not like terms, because they do not have the same variables raised to the same exponent.
Example 3
Simplify the expression: 4m+8m+9m
Worked Solution
Create a strategy
We can add the terms all together by finding the sum of their coefficients.
Apply the idea
| | | | |
--- --- |
| 4m+8m+9m | = | (4+8+9)m | Find the sum of the coefficients |
| | = | 21m | Evaluate the addition |
Example 4
Simplify the following expression: 8x+6y−2y−4x
Worked Solution
Create a strategy
To simplify an expression we combine all the like terms.
Apply the idea
Let's rearrange the expression and group the like terms together so we can clearly see which terms we need to sum.
| | | | |
--- --- |
| 8x+6y−2y−4x | = | 8x−4x+6y−2y | Rearrange the expression |
| | = | (8x−4x)+(6y−2y) | Group the like terms |
| | = | 4x+4y | Simplify |
Reflect and check
We identified like terms and then combined them until no like terms remained. We can add or subtract like terms regardless of the ordering of the expression.
Idea summary
Two algebraic terms are called like terms if they have exactly the same combination of variables.
This includes the exponents: x and x2 are not the same variables, in the same way that 4 and 42 are not equal.
To combine like terms means to simplify an expression by combining all like terms together through addition and/or subtraction.
Outcomes
7.EE.A.1
Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
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Change of base formula
The change of base formula is a formula for expressing a logarithm in one base in terms of logarithms in other bases.
For any positivereal numbers such that neither nor are , we have
This allows us to rewrite a logarithm in base in terms of logarithms in any base . This formula can also be written
Proof
Let .
Then .
And, taking the of both sides, we get
By the properties of logarithms,
Substituting for y,
Use for computations
The change of base formula is useful for simplifying certain computations involving logarithms. For example, we have by the change of base formula that
The formula can also be useful when calculating logarithms on a calculator. Many calculators have only functions for calculating base-10 and base-e logarithms. But you can still calculate logs in other bases, you just need to use the change of base formula to put in in base 10. For example, if you wanted to calculate , you would first convert it to the form . Then you would evaluate it using the base-10 log function on the calculator.
Special cases and consequences
Many other logarithm rules can be written in terms of the change of base formula. For example, we have that . Using the second form of the change of base formula gives .
One consequence of the change of base formula is that for positive constants , the functions and differ by a constant factor, for all .
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4716 | https://math.libretexts.org/Bookshelves/Analysis/Complex_Variables_with_Applications_(Orloff)/01%3A_Complex_Algebra_and_the_Complex_Plane/1.06%3A_Euler's_Formula | eiθ
1.6.1
i
1.6.2
1.6.3
N
1.6.4
1.6.5
1.6.6
1.6.7
1.6.8
N
1.6.1
deitdt=ddt(cos(t)+isin(t))=−sin(t)+icos(t)=i(cos(t)+isin(t))=ieit.
1.6.1
ei⋅0=cos(0)+isin(0)=1
1.6.1
eia⋅eib=(cos(a)+isin(a))⋅(cos(b)+isin(b))=cos(a)cos(b)−sin(a)sin(b)+i(cos(a)sin(b)+sin(a)cos(b))=cos(a+b)+isin(a+b)=ei(a+b).
ex
cos(x)
sin(x)
excos(x)sin(x)=1+x+x22!+x33!+x44!+...=1−x22!+x44!−x66!+…=x−x33!+x55!+...
eiθ
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1.6: Euler's Formula
Last updated
: May 2, 2023
Save as PDF
1.5: Polar Coordinates
1.7: The Exponential Function
Page ID
: 6471
Jeremy Orloff
Massachusetts Institute of Technology via MIT OpenCourseWare
( \newcommand{\kernel}{\mathrm{null}\,})
Euler’s (pronounced ‘oilers’) formula connects complex exponentials, polar coordinates, and sines and cosines. It turns messy trig identities into tidy rules for exponentials. We will use it a lot. The formula is the following:
eiθ=cos(θ)+isin(θ).
eiθ=cos(θ)+isin(θ).(1.6.1)
There are many ways to approach Euler’s formula. Our approach is to simply take Equation 1.6.11.6.1 as the definition of complex exponentials. This is legal, but does not show that it’s a good definition. To do that we need to show the eiθeiθ obeys all the rules we expect of an exponential. To do that we go systematically through the properties of exponentials and check that they hold for complex exponentials.
eiθeiθ behaves like a true exponential
P1
eiteit differentiates as expected:
deitdt=ieit.
deitdt=ieit.
Proof
: This follows directly from the definition in Equation 1.6.1:
deitdt=ddt(cos(t)+isin(t))=−sin(t)+icos(t)=i(cos(t)+isin(t))=ieit.
P2
ei⋅0=1.
ei⋅0=1.
Proof
: This follows directly from the definition in Equation 1.6.1:
ei⋅0=cos(0)+isin(0)=1.
P3
The usual rules of exponents hold:
eiaeib=ei(a+b).
eiaeib=ei(a+b).
Proof
: This relies on the cosine and sine addition formulas and the definition in Equation 1.6.1:
eia⋅eib=(cos(a)+isin(a))⋅(cos(b)+isin(b))=cos(a)cos(b)−sin(a)sin(b)+i(cos(a)sin(b)+sin(a)cos(b))=cos(a+b)+isin(a+b)=ei(a+b).
P4
The definition of eiθeiθ is consistent with the power series for exex.
Proof
: To see this we have to recall the power series for ex, cos(x) and sin(x). They are
ex=1+x+x22!+x33!+x44!+...cos(x)=1−x22!+x44!−x66!+…sin(x)=x−x33!+x55!+...
Now we can write the power series for eiθ and then split it into the power series for sine and cosine:
eiθ=∞∑0(iθ)nn!=∞∑0(−1)kθ2k(2k)!+i∞∑0(−1)kθ2k+1(2k+1)!=cos(θ)+isin(θ).
So the Euler formula definition is consistent with the usual power series for ex.
Properties P1-P4 should convince you that eiθ behaves like an exponential.
Complex Exponentials and Polar Form
Now let’s turn to the relation between polar coordinates and complex exponentials.
Suppose z=x+iy has polar coordinates r and θ. That is, we have x=rcos(θ) and y=rsin(θ). Thus, we get the important relationship
z=x+iy=rcos(θ)+irsin(θ)=r(cos(θ)+isin(θ))=reiθ.
This is so important you shouldn’t proceed without understanding. We also record it without the intermediate equation.
z=x+iy=reiθ.
Because r and θ are the polar coordinates of (x,y) we call z=reiθ the polar form of z.
Let’s now verify that magnitude, argument, conjugate, multiplication and division are easy in polar form.
Magnitude
|eiθ|=1.
Proof
: |eiθ|=|cos(θ)+isin(θ)|=√cos2(θ)+sin2(θ)=1.
In words, this says that eiθ is always on the unit circle - this is useful to remember!
Likewise, if z=reiθ then |z|=r. You can calculate this, but it should be clear from the definitions: |z| is the distance from z to the origin, which is exactly the same definition as for r.
Argument
If z=reiθ then arg(z)=θ.
Proof
: This is again the definition: the argument is the polar angle θ.
Conjugate
¯(z=reiθ)=re−iθ.
Proof
: ¯(z=reiθ)=¯r(cos(θ)+isin(θ))=r(cos(θ)−isin(θ))=r(cos(−θ)+isin(−θ))=re−iθ.
In words: complex conjugation changes the sign of the argument.
Multiplication
If z1=r1eiθ1 and z2=r2eiθ2 then
z1z2=r1r2ei(θ1+θ2).
This is what mathematicians call trivial to see, just write the multiplication down. In words, the formula says the for z1z2 the magnitudes multiply and the arguments add.
Division
Again it's trivial that
r1eiθ1r2eiθ2=r1r2ei(θ1−θ2).
Example 1.6.1: Multiplication by 2i
Here’s a simple but important example. By looking at the graph we see that the number 2i has magnitude 2 and argument π/2. So in polar coordinates it equals 2eiπ/2. This means that multiplication by 2i multiplies lengths by 2 and add π/2 to arguments, i.e. rotates by 90∘. The effect is shown in the figures below
Example 1.6.2: Rasing to a power
Let's compute (1+i)6 and (1+i√32)3
Solution
1+i has magnitude = √2 and arg=π/4, so 1+i=√2eiπ/4. Rasing to a power is now easy:
(1+i)6=(√2eiπ/4)6=8e6iπ/4=8e3iπ/2=−8i.
Similarly, 1+i√32=eiπ/3, so (1+i√32)3=(1⋅eiπ/3)3=eiπ=−1
Complexification or Complex Replacement
In the next example we will illustrate the technique of complexification or complex replacement. This can be used to simplify a trigonometric integral. It will come in handy when we need to compute certain integrals.
Example 1.6.3
Use complex replacement to compute
I=∫excos(2x) dx.
Solution
We have Euler's formula
e2ix=cos(2x)+isin(2x),
so cos(2x)=Re(e2ix). The complex replacement trick is to replace cos(2x) by e2ix. We get (justification below)
Ic=∫excos2x+iexsin2x dx
with
I=Re(Ic)
Computing Ic is straightforward:
Ic=∫exei2x dx=∫ex(1+2i) dx=ex(1+2i)1+2i.
Here we will do the computation first in rectangular coordinates. In applications, for example throughout 18.03, polar form is often preferred because it is easier and gives the answer in a more useable form.
Ic=ex(1+2i)1+2i⋅1−2i1−2i=ex(cos(2x)+isin(2x))(1−2i)5=15ex(cos(2x)+2sin(2x)+i(−2cos(2x)+sin(2x)))
So,
I=Re(Ic)=15ex(cos(2x)+2sin(2x)).
Justification of complex replacement. The trick comes by cleverly adding a new integral to I as follows, Let J=∫exsin(2x) dx. Then we let
Ic=I+iJ=∫ex(cos(2x)+isin(2x)) dx=∫ex22ix dx.
Clearly, by construction, Re(Ic)=I as claimed above.
Alternative using polar coordinates to simplify the expression for Ic:
In polar form, we have 1+2i=reiϕ, where r=√5 and ϕ=arg(1+2i)=tan−1(2) in the first quadrant. Then:
Ic=ex(1+2i)√5eiϕ=ex√5ei(2x−ϕ)=ex√5(cos(2x−ϕ)+isin(2x−ϕ)).
Thus,
I=Re(Ic)=ex√5cos(2x−ϕ).
Nth roots
We are going to need to be able to find the nth roots of complex numbers, i.e., solve equations of the form
zN=c,
where c is a given complex number. This can be done most conveniently by expressing c and z in polar form, c=Reiϕ and z=reiθ. Then, upon substituting, we have to solve
rNeiNθ=Reiϕ
For the complex numbers on the left and right to be equal, their magnitudes must be the same and their arguments can only differ by an integer multiple of 2π. This gives
r=R1/N(Nθ=ϕ+2πn),wheren=0,±1,±2,...
Solving for θ, we have
θ=ϕN+2πnN.
Example 1.6.4
Find all 5 fifth roots of 2.
Solution
For c=2, we have R=2 and ϕ=0, so the fifth roots of 2 are
zn=21/5e2nπi/5, where n=0,±1,±2,...
Looking at the right hand side we see that for n=5 we have 21/5e2πi which is exactly the same as the root when n=0, i.e. 21/5e0i. Likewise n=6 gives exactly the same root as n=1, and so on. This means, we have 5 different roots corresponding to n=0,1,2,3,4.
zn=21/5,e1/5e2πi/5,e1/5e4πi/5,e1/5e6πi/5,e1/5e8πi/5
Similarly we can say that in general c=Reiϕ has N distinct N th roots:
zn=r1/Neiϕ/N+i2π(n/N) for n=0,1,2,...,N−1.
Example 1.6.5
Find the 4 forth roots of 1.
Solution
We need to solve z4=1, so ϕ=0. So the 4 distinct fourth roots are in polar form
zn=1,eiπ/2,eiπ,ei3π/2
and in Cartesian representation
zn=1,i,−1,−i.
Example 1.6.6
Find the 3 cube roots of -1.
Solution
z2=−1=eiπ+i2πn. So, zn=eiπ+i2π(n/3) and the 3 cube roots are eiπ/3, eiπ, ei5π/3. Since π/3 radians is 60∘ we can simplify:
eiπ/3=cos(π/3)+isin(π/3)=12+i√32⇒zn=−1,12±i√32
Example 1.6.7
Find the 5 fifth roots of 1+i.
Solution
z5=1+i=√2ei(π/4+2nπ)
for n=0,1,2,.... So, the 5 fifth roots are
21/10eiπ/20, 21/10ei9π/20, 21/10ei17π/20, 21/10ei25π/20, 21/10ei33π/20.
Using a calculator we could write these numerically as a+bi, but there is no easy simplification.
Example 1.6.8
We should check that our technique works as expected for a simple problem. Find the 2 square roots of 4.
Solution
z2=4ei2πn. So, zn=2eiπn, with n=0,1. So the two roots are 2e0=2 and 2eiπ=−2 as expected!
geometry of Nth roots
Looking at the examples above we see that roots are always spaced evenly around a circle centered at the origin. For example, the fifth roots of 1+i are spaced at increments of 2π/5 radians around the circle of radius 21/5.
Note also that the roots of real numbers always come in conjugate pairs.
1.5: Polar Coordinates |
4717 | https://www.triangle-calculator.com/?a=5&b=12&c=12 | Triangle calculation using three given side lengths: a=5 b=12 c=12
5 12 12 triangle
Acute isosceles triangle.
The lengths of the sides of the triangle:
a = 5
b = 12
c = 12
Area:T = 29.34 2 17364858
Perimeter:p = 29
Semiperimeter: s = 14.5
Angle∠ A = α = 24.04 9 93983611° = 24°2'58″ = 0.4 2 197411845 rad
Angle∠ B = β = 77.97 5 53008194° = 77°58'31″ = 1.36 1 09257345 rad
Angle∠ C = γ = 77.97 5 53008194° = 77°58'31″ = 1.36 1 09257345 rad
Altitude (height) to the side a:h a = 11.73 7 66945943
Altitude (height) to the side b:h b = 4.8 9 902894143
Altitude (height) to the side c:h c = 4.8 9 902894143
Median:m a = 11.73 7 66945943
Median:m b = 6.96 4 41941386
Median:m c = 6.96 4 41941386
Inradius:r = 2.02 4 35680335
Circumradius:R = 6.13 5 46062489
Vertex coordinates: A[12; 0] B[0; 0] C[1.04 2 16666667; 4.8 9 902894143]
Centroid:CG[4.34 7 72222222; 1.6 3 300964714]
Coordinates of the circumscribed circle:U[6; 1.27 8 80429685]
Coordinates of the inscribed circle:I[2.5; 2.02 4 35680335]
Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 155.95 1 06016389° = 155°57'2″ = 0.4 2 197411845 rad
∠ B' = β' = 102.02 5 46991806° = 102°1'29″ = 1.36 1 09257345 rad
∠ C' = γ' = 102.02 5 46991806° = 102°1'29″ = 1.36 1 09257345 rad
Calculate another triangleShow more digits
How did we calculate this triangle?
We know the lengths of all three sides of the triangle, so the triangle is uniquely specified.
a=5 b=1 2 c=1 2
1. The triangle perimeter is the sum of the lengths of its three sides
p=a+b+c=5+1 2+1 2=2 9
2. The semiperimeter of the triangle
The semiperimeter of the triangle is half its perimeter. The semiperimeter frequently appears in formulas for triangles to be given a separate name. By the triangle inequality, the longest side length of a triangle is less than the semiperimeter.
s=2 p=2 2 9=1 4.5
3. The triangle area using Heron's formula
Heron's formula gives the area of a triangle when the length of all three sides is known. There is no need to calculate angles or other distances in the triangle first. Heron's formula works equally well in all cases and types of triangles.
T=s(s−a)(s−b)(s−c)T=14.5(14.5−5)(14.5−12)(14.5−12)T=14.5⋅9.5⋅2.5⋅2.5 T=860.94=29.342 T = \sqrt{ s(s-a)(s-b)(s-c) } \ \ T = \sqrt{ 14.5(14.5-5)(14.5-12)(14.5-12) } \ \ T = \sqrt{ 14.5 \cdot \ 9.5 \cdot \ 2.5 \cdot \ 2.5 } \ \ T = \sqrt{ 860.94 } = 29.342 T=s(s−a)(s−b)(s−c)T=1 4.5(1 4.5−5)(1 4.5−1 2)(1 4.5−1 2)T=1 4.5⋅9.5⋅2.5⋅2.5T=8 6 0.9 4=2 9.3 4 2
4. Calculate the heights of the triangle from its area.
There are many ways to find the height of the triangle. The easiest way is from the area and base length. The triangle area is half of the product of the base's length and height. Every side of the triangle can be a base; there are three bases and three heights (altitudes). Triangle height is the perpendicular line segment from a vertex to a line containing the base.
T=a h a 2 h a=2 T a=2⋅29.342 5=11.737 h b=2 T b=2⋅29.342 12=4.89 h c=2 T c=2⋅29.342 12=4.89 T = \dfrac{ a h _a }{ 2 } \ \ \ \ h _a = \dfrac{ 2 \ T }{ a } = \dfrac{ 2 \cdot \ 29.342 }{ 5 } = 11.737 \ \ h _b = \dfrac{ 2 \ T }{ b } = \dfrac{ 2 \cdot \ 29.342 }{ 12 } = 4.89 \ \ h _c = \dfrac{ 2 \ T }{ c } = \dfrac{ 2 \cdot \ 29.342 }{ 12 } = 4.89 T=2 a h ah a=a 2 T=5 2⋅2 9.3 4 2=1 1.7 3 7 h b=b 2 T=1 2 2⋅2 9.3 4 2=4.8 9 h c=c 2 T=1 2 2⋅2 9.3 4 2=4.8 9
5. Calculation of the inner angles of the triangle using a Law of Cosines
The Law of Cosines is useful for finding a triangle's angles when we know all three sides. The cosine rule, also known as the Law of Cosines, relates all three sides of a triangle with an angle of a triangle. The Law of Cosines extrapolates the Pythagorean theorem for any triangle. Pythagorean theorem works only in a right triangle. Pythagorean theorem is a special case of the Law of Cosines and can be derived from it because the cosine of 90° is 0. It is best to find the angle opposite the longest side first. With the Law of Cosines, there is also no problem with obtuse angles as with the Law of Sines because the cosine function is negative for obtuse angles, zero for right, and positive for acute angles. We also use an inverse cosine called arccosine to determine the angle from the cosine value.
a 2=b 2+c 2−2 b c cos α α=arccos(2 b c b 2+c 2−a 2)=arccos(2⋅1 2⋅1 2 1 2 2+1 2 2−5 2)=2 4°2′5 8"b 2=a 2+c 2−2 a c cos β β=arccos(2 a c a 2+c 2−b 2)=arccos(2⋅5⋅1 2 5 2+1 2 2−1 2 2)=7 7°5 8′3 1"γ=1 8 0°−α−β=1 8 0°−2 4°2′5 8"−7 7°5 8′3 1"=7 7°5 8′3 1"
6. Inradius
An incircle of a triangle is a tangent circle to each side. An incircle center is called an incenter and has a radius named inradius. All triangles have an incenter, and it always lies inside the triangle. The incenter is the intersection of the three-angle bisectors. The product of a triangle's inradius and semiperimeter (half the perimeter) is its area.
T=r s r=T s=29.342 14.5=2.024 T = rs \ \ r = \dfrac{ T }{ s } = \dfrac{ 29.342 }{ 14.5 } = 2.024 T=r s r=s T=1 4.5 2 9.3 4 2=2.0 2 4
7. Circumradius
The circumcircle of a triangle is a circle that passes through all of the triangle's vertices, and the circumradius of a triangle is the radius of the triangle's circumcircle. The circumcenter (center of the circumcircle) is the point where the perpendicular bisectors of a triangle intersect.
R=a b c 4 r s=5⋅12⋅12 4⋅2.024⋅14.5=6.135 R = \dfrac{ a b c }{ 4 \ r s } = \dfrac{ 5 \cdot \ 12 \cdot \ 12 }{ 4 \cdot \ 2.024 \cdot \ 14.5 } = 6.135 R=4 r s a b c=4⋅2.0 2 4⋅1 4.5 5⋅1 2⋅1 2=6.1 3 5
8. Calculation of medians
A median of a triangle is a line segment joining a vertex to the opposite side's midpoint. Every triangle has three medians, and they all intersect each other at the triangle's centroid. The centroid divides each median into parts in the ratio of 2:1, with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex. We use Apollonius's theorem to calculate a median's length from its side's lengths.
m a=2 2 b 2+2 c 2−a 2=2 2⋅1 2 2+2⋅1 2 2−5 2=1 1.7 3 7 m b=2 2 c 2+2 a 2−b 2=2 2⋅1 2 2+2⋅5 2−1 2 2=6.9 6 4 m c=2 2 a 2+2 b 2−c 2=2 2⋅5 2+2⋅1 2 2−1 2 2=6.9 6 4
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4718 | https://www.concurrences.com/IMG/pdf/oecd_-_glossary_of_industrial_organisation_economics_and_competition_law.pdf?39924/61543ab059ef02f25a5b58d7b8b4636a8fe2232efa57c3b86700b24cdb1da9ca | Published Time: 1999-08-11T11:41:19.000Z
GLOSSARY OF INDUSTRIAL ORGANISATION ECONOMICS AND COMPETITION LAW
-ORGANISATION FOR ECONOMIC CO-OPERATION AND DEVELOPMENT
2
FOREWORD
The Centre for Co-operation with European Economies in Transition, created in March 1990, is the focal point for co-operation between the OECD and central and eastern European countries and the former Soviet republics. Its major responsibility is to design and manage a programme of policy advice, technical assistance and training which puts the expertise of the Secretariat and Member countries at the disposal of countries engaged in economic reform. In December 1990, the Council adopted a programme "Partners in Transition" for the purpose of providing more focused assistance to those countries that are more advanced in introducing market-oriented reforms and desire to become members of OECD. Additional activities which the Centre co-ordinates under this programme include reviews of the country’s economic situation and prospects; reviews of specific policy areas and the participation of the Partner countries in a number of OECD committees. In all these activities, the Centre maintains close relations with other multilateral bodies with the mutual objective of ensuring the complementarity of respective efforts to support economic reforms in Central and Eastern Europe and the former Soviet Union. This Glossary of Industrial Organisation Economics and Competition Law has been commissioned by the Directorate for Financial, Fiscal and Enterprise Affairs in the framework of the Centre’s work programme, to assist officials, academics and policy makers in the reforming central and eastern European economies in their understanding of the basic concepts of modern micro-economics. 3The Glossary has been compiled by R. S. Khemani, Adjunct Professor at the Faculty of Commerce and Business Administration, University of British Columbia, B. C., Canada and D. M. Shapiro, Principal, School of Community and Public Affairs, Concordia University, Montreal P.Q. The Glossary is published on the responsibility of the Secretary-General of the OECD. Salvatore Zecchini Director of the Centre for Co-operation with the European Economies in Transition 4
Index of Terms
Terms in bold type are defined and discussed in this glossary. Those in
italics are cross-referenced or incorporated in the discussion of other related terms and concepts. (The column of figures below right indicates the corresponding term in the French version of the Glossary). 1. Abuse of Dominant Position ............................................................... 22. Acquisition ............................................................................................ 73. Administered Prices ............................................................................. 152 4. Advertising ............................................................................................ 168 5. Aggregate Concentration (See Concentration )................................... 27 6. Agreement ............................................................................................. 47. Allocative Efficiency (See Pareto Efficiency )...................................... 79 8. Alternative Costs (See Opportunity Costs )......................................... 48 9. Amalgamation (See Merger ) ................................................................ 110. Anticompetitive Practices ................................................................... 147 11. Anti-Monopoly Policy (See Antitrust ) ................................................. 143 12. Antitrust ................................................................................................ 813. Average Costs (See Costs ) .................................................................... 54 14. Barriers to Entry .................................................................................. 11 15. Basing Point Pricing ............................................................................ 153 16. Bertrand (Nash) Equilibrium ............................................................. 90 17. Bid Rigging ............................................................................................ 194 18. Bilateral Monopoly/Oligopoly ............................................................ 129 19. Brand Competition (Inter- and Intra-)................................................. 38 20. Bundling ................................................................................................ 196 21. Buyer Concentration (See Concentration ) .......................................... 30 22. Buyout .................................................................................................... 173 23. Cartel ..................................................................................................... 15, 83 24. Cartelization (See Cartel, Collusion, Monopolization )..................... 19 25. Collusion ................................................................................................ 23 26. Collusive bidding (tendering) (See Bid Rigging )................................. 202 27. Combination ......................................................................................... 20 28. Common Control (See Control of Enterprises, Holding 5
Company ).............................................................................................. 44 29. Competition .......................................................................................... 33 30. Compulsory Licensing (See Licensing )................................................ 119 31. Concentration ....................................................................................... 26 32. Concentration Indexes ......................................................................... 114 33. Concentration Measures (See Concentration Indexes ) ..................... 126 34. Concentration Ratio (See Concentration Indexes )............................. 126 35. Concerted Action or Practice (See Cartel, Collusion )........................ 148 36. Conglomerate ........................................................................................ 42 37. Conglomerate Merger (See Merger ).................................................... 102 38. Conscious Parallelism .......................................................................... 139 39. Consolidation ........................................................................................ 104 40. Conspiracy ............................................................................................ 941. Constant Returns to Scale (See Economies of Scale ).......................... 182 42. Consumers’ Surplus ............................................................................. 198 43. Consumer Welfare ............................................................................... 13 44. Contestability ........................................................................................ 43 45. Contestable Markets (See Contestability ) ........................................... 121 46. Control of Enterprises ......................................................................... 45 47. Costs ....................................................................................................... 47 48. Countervailing Power (See Bilateral Monopoly,
Buyer Concentration, Monopsony ) ...................................................... 145 49. Cournot (Nash) Equilibrium .............................................................. 91 50. Crisis Cartel (See Cartel ) ..................................................................... 16 51. Cross Price Elasticity of Demand ....................................................... 81 52. Cut-Throat Competition ..................................................................... 34 53. Deadweight Welfare Loss .................................................................... 142 54. Deconcentration ................................................................................... 58 55. Deep Pockets ......................................................................................... 201 56. Delivered Pricing (See Basing Point Pricing ) .................................... 161 57. Demonopolization (See Anti-Monopoly Policy, Antitrust, Deconcentration ) .................................................................................. 62 58. Depression Cartel (See Cartel ) ............................................................ 16 59. Deregulation (See Regulation )............................................................. 63 60. Destructive Competition (See Cut-Throat Competition )................... 34 61. Differentiated Products (See Product Differentiation )...................... 163 62. Discrimination (See Price Discrimination )......................................... 67 63. Diseconomies of Scale (See Economies of Scale ) ............................... 64 64. Distributor’s Mark (See Trade Mark ) ................................................. 124 65. Diversification ....................................................................................... 70 66. Divestiture ............................................................................................. 61 67. Dominant Firm ..................................................................................... 89 668. Dominant Market Position (See Dominant Firm ) .............................. 144 69. Dominant Price Leadership (See Dominant Firm, Price Leadership ) ........................................................................................... 100 70. Dumping ................................................................................................ 72 71. Duopoly .................................................................................................. 73 72. Economies of Scale ............................................................................... 74 73. Economies of Scope .............................................................................. 75 74. Efficiency ............................................................................................... 77 75. Elasticity of Demand (Price) ............................................................... 81 76. Enterprise .............................................................................................. 86 77. Entropy (See Concentration Indexes )................................................. 21 78. Excess Capacity .................................................................................... 197 79. Excessive Competition (See Cut-Throat Competition ) ..................... 35 80. Excess Prices ......................................................................................... 151 81. Exclusive Dealing (See Vertical Restraints )....................................... 69 82. Export Cartel ........................................................................................ 84 83. External Economies/Diseconomies (See Externalities ) ...................... 76 84. Externalities .......................................................................................... 94 85. Extraterritoriality ................................................................................ 97 86. Failing Firm .......................................................................................... 88 87. Fighting Brand ..................................................................................... 195 88. Fixed Costs (See Costs )......................................................................... 51 89. Foreclosure of Competition (See Anticompetitive Practices ) ........... 98 90. Franchising ........................................................................................... 101 91. Free Rider or Riding ............................................................................ 140 92. Full Cost Pricing ................................................................................... 127 93. Full-Line Forcing (See Tied Selling ) ................................................... 204 94. Gentlemen’s Agreement (See Collusion ) .............................................. 107 95. Gini Coefficient (See Concentration Indexes ).................................... 22 96. Herfindahl-Hirschman Index (See Concentration Indexes ) .............. 111 97. Heterogenous Products (See Homogenous Products, Product Differentiation ) ..................................................................................... 164 98. Holding Company ................................................................................ 192 99. Homogenous Products ......................................................................... 165 100. Horizontal Integration (See Merger ) ................................................... 117 101. Horizontal Mergers (See Merger )........................................................ 105 102. Income Elasticity of Demand .............................................................. 82 103. Increasing Returns to Scale (See Economies of Scale ) ....................... 183 104. Industry Concentration (See Concentration ........................................ 32 105. Integration (See Vertical Integration ) ................................................ 116 106. Intellectual Property Rights ................................................................ 71 7107. Inter- and Intra-Brand Competition (See Brand Competition ) ......... 36 108. Interlocking Directorate ...................................................................... 57 109. International Cartel (See Cartel ).......................................................... 17 110. Inverse Index (See Concentration Indexes )........................................ 113 111. Joint Monopoly Profits (See Joint Profit Maximization ) .................. 167 112. Joint Profit Maximization ................................................................... 125 113. Joint Venture ........................................................................................ 87 114. Lerner Index ......................................................................................... 112 115. Leveraged Buyout (See Buyout ) ........................................................... 172 116. Licensing ................................................................................................ 136 117. Limit Pricing ......................................................................................... 160 118. Lorenz Curve (See Concentration Indexes )........................................ 46 119. Loss-Leader Selling .............................................................................. 203 120. Management Buyout (See Buyout )....................................................... 173 121. Marginal Cost (See Costs ) .................................................................... 53 122. Marginal Revenue (See Revenue ) ........................................................ 175 123. Market ................................................................................................... 120 124. Market Concentration (See Concentration )........................................ 29 125. Market Definition ................................................................................. 60 126. Market Failure ..................................................................................... 59 127. Market for Corporate Control ........................................................... 122 128. Market Power ....................................................................................... 169 129. Market Share ........................................................................................ 141 130. Merger ................................................................................................... 102 131. Mobility Barriers .................................................................................. 12 132. Monopolistic Competition ................................................................... 37 133. Monopolization ..................................................................................... 133 134. Monopoly ............................................................................................... 128 135. Monopoly Power (See Market Power ) ................................................ 146 136. Monopoly Rents (See Rent )................................................................... 186 137. Monopsony ............................................................................................ 134 138. Nash Equilibrium ................................................................................. 92 139. Natural Monopoly ................................................................................ 130 140. Negative Externality (See Externalities ).............................................. 95 141. Non-Price Predation ............................................................................ 93 142. Oligopoly ............................................................................................... 137 143. Oligopsony (See Monopsony ) .............................................................. 138 144. Opportunity Costs (or Alternative Costs ) ........................................... 49 145. Ownership Concentration (See Concentration ).................................. 28 146. Package Tie-in (See Bundling )............................................................. 205 147. Parent ..................................................................................................... 193 148. Pareto Efficiency .................................................................................. 78 8149. Patents ................................................................................................... 14 150. Perfect Competition ............................................................................. 39 151. Per Se Illegal (See Rule of Reason )..................................................... 110 152. Positive Externality (See Externalities ) ............................................... 96 153. Predatory Pricing ................................................................................. 156 154. Preemption of Facilities (See Barriers to Entry, Anticompetitive Practices ) .................................................................. 3155. Price Cartel (See Cartel ) ...................................................................... 85 156. Price Discrimination ............................................................................ 68 157. Price-Fixing Agreement ...................................................................... 85 158. Price Leadership .................................................................................. 157 159. Price Regulation ................................................................................... 181 160. Producers’ Surplus (See Deadweight Welfare Loss ).......................... 199 161. Privatization .......................................................................................... 150 162. Product Differentiation ....................................................................... 65 163. Profit ...................................................................................................... 66 164. Profitability ........................................................................................... 184 165. Quasi-rents (See Rent ).......................................................................... 170 166. Rationalization Agreement ................................................................. 5167. Reciprocity ............................................................................................ 177 168. Recommended or Suggested Price ..................................................... 154 169. Refusal to Deal/Sell .............................................................................. 178 170. Regulation ............................................................................................. 180 171. Rent ........................................................................................................ 185 172. Rent Seeking ......................................................................................... 176 173. Resale Price Maintenance (RPM) ...................................................... 159 174. Restriction of Entry to the Market (See Barriers to Entry, Limit Pricing ) ........................................................................... 187 175. Restriction of Technology (See Licensing )........................................... 190 176. Restriction on Exportation .................................................................. 188 177. Restriction on Importation ................................................................. 189 178. Revenues ................................................................................................ 174 179. Ruinous Competition (See Cut-Throat Competition )........................ 35 180. Rule of Reason ...................................................................................... 179 181. Second Best, Theory of ........................................................................ 200 182. Self-Regulation (See Regulation ) ......................................................... 10 183. Seller Concentration (See Concentration ) .......................................... 31 184. Selling Below Cost ................................................................................ 203 185. Shared or Joint Monopoly .................................................................. 131 186. Shipping Conferences .......................................................................... 41 187. Specialization Agreements .................................................................. 69188. Standards .............................................................................................. 135 189. Strategic Behaviour ............................................................................. 25 190. Subsidiary ............................................................................................. 99 191. Substantial Lessening of Competition (See Market Power ) ............... 66 192. Sunk Costs ............................................................................................. 52 193. Sustainable Monopoly (See Contestability ) ........................................ 132 194. Tacit Collusion (See Collusion, Conscious Parallelism ) ................... 24 195. Takeover ................................................................................................ 149 196. Tied Selling ............................................................................................ 205 197. Total costs (See Costs ) .......................................................................... 55 198. Trade Mark ........................................................................................... 123 199. Transaction Costs ................................................................................. 50 200. Uniform Delivered Pricing (See Basing Point Pricing )...................... 162 201. Variable Costs (See Costs ).................................................................... 56 202. Vertical Integration ............................................................................. 118 203. Vertical Merger (See Merger ).............................................................. 106 204. Vertical Restraints (or Restrictions) .................................................. 191 205. Workable Competition ........................................................................ 40 206. X-Efficiency (See Efficiency, X-Inefficiency ) ..................................... 206 207. X-Inefficiency ........................................................................................ 207
Abuse of Dominant Position
Anticompetitive business practices in which a dominant firm may engage in order to maintain or increase its position in the market. These business practices by the firm, not without controversy, may be considered as "abusive or improper exploitation" of monopolistic control of a market aimed at restricting competition. The term abuse of dominant position has been explicitly incorporated in competition legislation of various countries such as Canada, EEC and Germany. In the United States, the counterpart provisions would be those dealing with monopoly and attempts to monopolize or monopolization of a market. Which of the different types of business practices are considered as being abusive will vary on a case by case basis and across countries. Some business practices may be treated differently in different jurisdictions as well. However, the business practices which have been contested in actual cases in different countries, not always with legal success, have included the following: charging unreasonable or excess prices , price discrimination , predatory pricing , price squeezing by integrated firms, refusal to deal/sell , tied selling or product bundling and pre-emption of facilities . See Anticompetitive practices . See also I. Schmidt, "Different Approaches and Problems in Dealing With Control of Market Power: A Comparison of German, European and U.S. Policy Towards Market-Dominating Enterprises", Antitrust Bulletin , Vol. 28, 1983, pp. 417-460. And, F.M. Scherer 10 and D. Ross, Industrial Market Structure and Economic Performance, Houghton Mifflin Co., Boston, 1990, Ch. 12, especially pp. 483-488.
Acquisition
Refers to obtaining ownership and control by one firm, in whole or in part, of another firm or business entity. As distinct from a merger , an acquisition does not necessarily entail amalgamation or consolidation of the firms. An acquisition, even when there is complete change in control, may lead the firms involved to continue to operate as separate entities. Nevertheless, joint control implies joint profit maximization and is a potential source of concern to antitrust authorities. See also Takeover .
Administered Prices
Administered prices are prices set by firms that do not vary in response to short-run fluctuations in demand and supply conditions. This price rigidity has been viewed by some economists as arising from the exercise of market power .Various research studies have been conducted attempting to link administered prices to concentration and inflation. What emerges from the findings is that there are differences across industries (and across countries) in the degree of price flexibility which simple models of market clearing cannot fully explain. However, researchers have been confronted with serious measurement difficulties, notably the fact that official price indices often do not reflect price discounts. For further details, see D.W. Carlton, "The Theory and the Facts of How Markets Clear", in R. Schmalensee and R. Willig (eds.), The Handbook of Industrial Organization ,North Holland, Amsterdam, 1989.
Advertising
Advertising helps manufacturers differentiate their products and provides information about products to consumers. As information, advertising provides many benefits to consumers. Price advertising, for example, lowers market prices. Advertising that tells consumers about the existence of new products facilitates entry. On the other hand, by contributing to product differentiation , advertising may create market power by raising barriers to entry . Much empirical work has been carried out about the competitive effects of advertising, with no definitive results. 11
Aggregate Concentration
See Concentration .
Agreement (to lessen or restrict competition) Agreement refers to an explicit or implicit arrangement between firms normally in competition with each other to their mutual benefit. Agreements to restrict competition may cover such matters as prices, production, markets and customers. These types of agreements are often equated with the formation of
cartels or collusion and in most jurisdictions are treated as violations of competition legislation because of their effect of increasing prices, restricting output and other adverse economic consequences. Agreements may be arrived at in an extensive formal manner, and their terms and conditions are explicitly written down by the parties involved; or they may be implicit, and their boundaries are nevertheless understood and observed by convention among the different members. An explicit agreement may not necessarily be an "overt" agreement, that is one which can be openly observed by those not party to the agreement. Indeed, most agreements which give rise to
anticompetitive practices tend to be covert arrangements that are not easily detected by competition authorities. Not all agreements between firms are necessarily harmful of competition or proscribed by competition laws. In several countries, competition legislation provides exemptions for certain cooperative arrangements between firms which may facilitate efficiency and dynamic change in the marketplace. For example, agreements between firms may be permitted to develop uniform product standards in order to promote economies of scale , increased use of the product and diffusion of technology. Similarly, firms may be allowed to engage in cooperative research and development (R&D), exchange statistics or form joint ventures to share risks and pool capital in large industrial projects. These exemptions, however, are generally granted with the proviso that the agreement or arrangement does not form the basis for price fixing or other practices restrictive of competition.
Allocative Efficiency
See Pareto Efficiency. 8. Alternative Costs 12 See Opportunity Costs. 9. Amalgamation
See Merger. 10. Anticompetitive Practices
Refers to a wide range of business practices in which a firm or group of firms may engage in order to restrict inter-firm competition to maintain or increase their relative market position and profits without necessarily providing goods and services at a lower cost or of higher quality. The essence of competition entails attempts by firm(s) to gain advantage over rivals. However, the boundary of acceptable business practices may be crossed if firms contrive to artificially limit competition by not building so much on their advantages but on exploiting their market position to the disadvantage or detriment of competitors, customers and suppliers such that higher prices, reduced output, less consumer choice, loss of economic efficiency and misallocation of resources (or combinations thereof) are likely to result. Which types of business practices are likely to be construed as being anticompetitive and, if that, as violating competition law, will vary by jurisdiction and on a case by case basis. Certain practices may be viewed as per se illegal
while others may be subject to rule of reason . Resale price maintenance , for example, is viewed in most jurisdictions as being per se illegal whereas exclusive dealing may be subject to rule of reason. The standards for determining whether or not a business practice is illegal may also differ. In the United States, price fixing agreements are per se illegal whereas in Canada the agreement must cover a substantial part of the market. With these caveats in mind, competition laws in a large number of countries examine and generally seek to prevent a wide range of business practices which restrict competition. These practices are broadly classified into two groups: horizontal and vertical restraints on competition. The first group includes specific practices such as cartels, collusion, conspiracy, mergers, predatory pricing, price discrimination and price fixing agreements .The second group includes practices such as exclusive dealing , geographic market restrictions, refusal to deal/sell, resale price maintenance and tied selling .Generally speaking, horizontal restraints on competition primarily entail other competitors in the market whereas vertical restraints entail supplier-distributor relationships. However, it should be noted that the distinction between horizontal 13 and vertical restraints on competition is not always clear cut and practices of one type may impact on the other. For example, firms may adopt strategic behaviour
to foreclose competition . They may attempt to do so by pre-empting facilities
through acquisition of important sources of raw material supply or distribution channels, enter into long term contracts to purchase available inputs or capacity and engage in exclusive dealing and other practices. These practices may raise
barriers to entry and entrench the market position of existing firms and/or facilitate anticompetitive arrangements.
Anti-Monopoly Policy
See Antitrust. 12. Antitrust
Antitrust refers to a field of economic policy and laws dealing with
monopoly and monopolistic practices. Antitrust law or antitrust policy are terms primarily used in the United States, while in many other countries the terms competition law or policy are used. Some countries have utilized the phrases Fair Trading or Antimonopoly law. The intellectual basis for antitrust economics or policy is the sub-field of industrial organization economics which addresses issues arising from the behaviour of firms operating under different market structure conditions and the effect this has on economic performance. Most antitrust or competition laws have provisions dealing with structure such as mergers ,
monopoly, dominant market position and concentration, as well as behaviour, such as collusion , price fixing , and predatory pricing .
Average Costs
See Costs. 14. Barriers to Entry
Barriers to entry are factors which prevent or deter the entry of new firms into an industry even when incumbent firms are earning excess profits. There are two broad classes of barriers: structural (or innocent) and strategic. These two classes are also often referred to as economic and behavioural barriers to entry. Structural barriers to entry arise from basic industry characteristics such as 14 technology, costs and demand. There is some debate over what factors constitute relevant structural barriers. The widest definition, that of Joe Bain, suggests that barriers to entry arise from product differentiation , absolute cost advantages of incumbents, and economies of scale . Product differentiation creates advantages for incumbents because entrants must overcome the accumulated brand loyalty of existing products. Absolute cost advantages imply that the entrant will enter with higher unit costs at every rate of output, perhaps because of inferior technology. Scale economies restrict the number of firms which can operate at minimum costs in a market of given size. A narrower definition of structural barriers is given by George Stigler, who suggests that barriers to entry arise only when an entrant must incur costs which incumbents do not bear. This definition excludes scale economies as a barrier. There is some debate as to whether Stigler’s definition includes costs not currently being incurred by incumbents or costs which have never been incurred by incumbents. Other economists would emphasize the importance of sunk costs as a barrier to entry. Since such costs must be incurred by entrants, but have already been borne by incumbents, a barrier to entry is created. In addition, sunk costs reduce the ability to exit and thus impose extra risks on potential entrants. Strategic barriers to entry arise from the behaviour of incumbents. In particular, incumbents may act so as to heighten structural barriers or threaten to retaliate against entrants if they do enter. Such threats must, however, be credible in the sense that incumbents must have an incentive to carry them out if entry does occur. See Strategic Behaviour .Strategic entry deterrence often involves some kind of pre-emptive behaviour by incumbents. One example is the pre-emption of facilities by which an incumbent over-invests in capacity in order to threaten a price war if entry occurs. Another would be the artificial creation of new brands and products in order to limit the possibility of imitation. This possibility remains subject to considerable debate. It should also be noted that governments can be a source of entry barriers through licensing and other regulations. A concise discussion, with examples, is found in P. Geroski and A. Jacquemin, "Industrial Change, Barriers to Mobility and European Industrial Policy", Economic Policy , Nov. 1985, section 3. For more detail see R. Gilbert. "Mobility Barriers and the Value of Incumbency", in R. Schmalensee and R. Willig (eds), The Handbook of Industrial Organization, North Holland, 15 Amsterdam, 1989.
Basing Point Pricing
Basing point pricing (also known as delivered pricing ) refers to a system in which a buyer must pay a price for a product inclusive of freight costs that does not depend on the location of the seller. The freight costs may be calculated from a specific location or "basing point" from standard published freight rate schedules. Under this system, customers located near or far from the basing point pay the same price. Thus nearby customers are discriminated against or are charged "phantom" freight that they would not incur if they had a choice of paying separately for the product and for the freight charges. Conversely, the freight costs of distant customers are absorbed by the sellers. This practice has been extensively used in industries such as steel and cement and has been viewed as a method to facilitate collusion among firms. In competition prices are expected to reflect costs. Economists therefore expect FOB (free on board) plus actual freight costs to emerge in competition. However, firms even in competition may adopt a system of delivered pricing because it is simple and saves administrative costs. This is particularly the case when firms establish price zones within which transportation distances and costs do not vary very much. Moreover FOB pricing plus actual freight costs may be a better means of collusion because it facilitates allocation of customers geographically. In addition the practice may be adopted in order to deter locational entry by otherwise competing firms.
Bertrand (Nash) Equilibrium
In a Bertrand model of oligopoly, firms independently choose prices (not quantities) in order to maximize profits. This is accomplished by assuming that rivals’ prices are taken as given. The resulting equilibrium is a Nash equilibrium
in prices, referred to as a Bertrand (Nash) equilibrium. When the industry is symmetric, i.e., comprising firms of equal size and identical costs, and the costs are constant and the product homogenous, the Bertrand equilibrium is such that each firm sets price equal to marginal cost, and the outcome is Pareto efficient . This result holds regardless of the number of firms and stands in contrast to the Cournot equilibrium where the deviation from Pareto efficiency increases as the number of firms decreases. However, when products are differentiated even the Bertrand model results in prices which exceed marginal cost, and the difference increases as products become more differentiated. 16 For details, and a comparison of Bertrand and Cournot models, see C. Shapiro, "Theories of Oligopoly Behavior", in R. Schmalensee and R. Willig (eds),
The Handbook of Industrial Organization , North Holland, Amsterdam, 1989.
Bid Rigging
Bid rigging is a particular form of collusive price-fixing behaviour by which firms coordinate their bids on procurement or project contracts. There are two common forms of bid rigging. In the first, firms agree to submit common bids, thus eliminating price competition. In the second, firms agree on which firm will be the lowest bidder and rotate in such a way that each firm wins an agreed upon number or value of contracts. Since most (but not all) contracts open to bidding involve governments, it is they who are most often the target of bid rigging. Bid rigging is one of the most widely prosecuted forms of collusion .17
Bilateral Monopoly/Oligopoly
A situation where there is a single (or few) buyer(s) and seller(s) of a given product in a market . The level of concentration in the sale of purchase of the product results in a mutual inter-dependence between the seller(s) and buyer(s). Under certain circumstances the buyer(s) can exercise countervailing power to constrain the market power of a single or few large sellers in the market and result in greater output and lower prices than would prevail under monopoly or
oligopoly . This would particularly be the case when: the "upstream" supply of the product is elastic, i.e. fairly responsive to price changes and not subject to production bottlenecks; the buyers can substantially influence downwards the prices of monopolistic sellers because of the size of their purchases; and the buyers themselves are faced with price competition in the "downstream" markets (see
vertical integration for discussion of terms upstream-downstream). Such a situation is particularly likely in the case of purchase of an intermediate product. However, if the supply of the product upstream is restricted and there is no effective competition downstream, the bilateral monopoly/oligopoly may result in joint profit maximization between sellers-buyers to the detriment of consumers.
Brand Competition (Inter- and Intra-) Firms marketing differentiated products frequently develop and compete on the basis of brands or labels. Coca Cola vs. Pepsi-Cola, Levi vs. GWG jeans, Kellogg’s Corn Flakes vs. Nabisco’s Bran Flakes are a few examples of inter-brand
competition. Each of these brands may be preferred by different buyers willing to pay a higher price or make more frequent purchases of one branded product over another.
Intra-brand competition is competition among retailers or distributors of the same brand. Intra-brand competition may be on price or non-price terms. As an example, a pair of Levi jeans may be sold at a lower price in a discount or specialty store as compared to a department store but without the amenities in services that a department store provides. The amenities in services constitute intra-brand non-price competition. Some manufacturers seek to maintain uniform retail prices for their products and prevent intra-brand price competition through business practices such as resale price maintenance (RPM) , in order to stimulate intra-brand non-price competition if it will increase sales of their product. 18
Bundling
This term is also referred to as package tie-in and tends to occur when one product is sold in proportion to another as a requirement for the sale. It is related to the concept of tied selling . For example, a computer manufacturer may require customers to purchase along with the computer all or a specified amount of ancillary products such as floppy disks and printing paper. Alternatively, the sale may be made as a complete package such as an automobile equipped with all options including automatic transmission, cassette-radio and air conditioning. Bundling of products may be a source of economies or efficiencies for the manufacturer, part of which may be reflected in a lower composite price for the buyer than if all the different products were supplied or bought separately. However, bundling may also make it difficult for firms to enter different product segments of the market. The competition implications of bundling, including that of tied selling generally, are complex and need to be evaluated on a case by case basis adopting a rule of reason approach. See also Tied Selling .
Buyer Concentration
See Concentration. 22. Buyout
Refers to a situation where the existing owners of a firm are "bought out" by another group, usually management and/or workers of that firm. A buyout may be for the whole firm or a division or a plant as the case applies. The financing of the buyout can be structured in various ways such as bank loans or through the issuance of bonds. In a leveraged buyout for example, a fairly large proportion of debt in relation to the asset value of the firm is incurred. Because buyouts lead to replacing publicly traded equity with debt (in the form of bonds backed by assets and other guarantees) the firms are often viewed as "going private" since its shares may no longer be listed on the stock exchange. Buyouts are viewed as an integral part of the market for corporate control and the re-deployment of assets from lower to higher valued uses.
Cartel
A cartel is a formal agreement among firms in an oligopolistic industry. Cartel members may agree on such matters as prices, total industry output, market shares , allocation of customers, allocation of territories, bid-rigging ,19 establishment of common sales agencies, and the division of profits or combination of these. Cartel in this broad sense is synonymous with "explicit" forms of collusion . Cartels are formed for the mutual benefit of member firms. The theory of "cooperative" oligopoly provides the basis for analyzing the formation and the economic effects of cartels. Generally speaking, cartels or cartel behaviour attempts to emulate that of monopoly by restricting industry output, raising or fixing prices in order to earn higher profits. A distinction needs to be drawn between public and private cartels. In the case of public cartels, the government may establish and enforce the rules relating to prices, output and other such matters. Export cartels and shipping conferences are examples of public cartels. In many countries depression cartels
have been permitted in industries deemed to be requiring price and production stability and/or to permit rationalization of industry structure and excess capacity . In Japan for example, such arrangements have been permitted in the steel, aluminum smelting, ship building and various chemical industries. Public cartels were also permitted in the United States during the depression in the 1930s and continued to exist for some time after World War II in industries such as coal mining and oil production. Cartels have also played an extensive role in the German economy during the inter-war period. International commodity agreements covering products such as coffee, sugar, tin and more recently oil (OPEC: Organization of Petroleum Exporting Countries) are examples of
international cartels which have publicly entailed agreements between different national governments. Crisis cartels have also been organized by governments for various industries or products in different countries in order to fix prices and ration production and distribution in periods of acute shortages. In contrast, private cartels entail an agreement on terms and conditions from which the members derive mutual advantage but which are not known or likely to be detected by outside parties. Private cartels in most jurisdictions are viewed as being illegal and in violation of antitrust laws. Successful cartels, be they public or private, require "concurrence", "coordination" and "compliance" among members. This means that cartel members need to be able to detect when violations of an agreement take place and be able to enforce the agreement with sanctions against the violators. These conditions are not easily met and this often explains why cartels tend to break down over time. See Agreement , Collusion . Refer also to G.J. Stigler, "A Theory of Oligopoly," Journal of Political Economy , Vol. 72(1), February, 1964, pp. 44-61; D.K. Osborn, "Cartel Problems," American Economic Review , Vol. 66, September, 1976, pp. 835-844; and F.M. Scherer and D. Ross, Industrial Market Structure and 20
Economic Performance , Houghton Mifflin Co., Boston, 1990, Chs. 7 and 8.
Cartelization
See Cartel , Collusion , Monopolization. 25. Collusion
Adam Smith observed in his book An Inquiry Into the Nature and Causes of the Wealth of Nations published in 1776 that: ...people of the same trade seldom meet together, even for merriment and diversion, but the conversation ends in a conspiracy against the public, or in some contrivance to raise prices. Collusion refers to combinations , conspiracies or agreements among sellers to raise or fix prices and to reduce output in order to increase profits. As distinct from the term cartel , collusion does not necessarily require a formal agreement, whether public or private, between members. However, it should be noted that the economic effects of collusion and a cartel are the same and often the terms are used somewhat interchangeably. Collusion between firms to raise or fix prices and reduce output are viewed by most authorities as the single most serious violation of competition laws. Collusive arrangements are known to have been arrived at and enforced in ways which are as varied as the human imagination itself. Cases drawn from across different countries reveal that collusion may be reached through informal
gentlemen’s agreements where mutual regard, social convention and personal contacts and connections provide sufficient basis for ensuring adherence to agreed prices and related business practices by members. While collusion is generally easier when sellers are few and produce homogenous products , price fixing conspiracies have also arisen in the sale of complex products. An example is the electrical equipment industry in the United States which involved 29 different companies selling diverse technical products such as turbine generators, transformers, switch gears, insulators, controls and condensers. Similarly, through agreement on product specification details and standards , American steel producers were able to collude successfully for some time. In one bid-rigging
conspiracy firms used the "phases of the moon" to take turns and determine which amongst them would submit the "low" bid to win the contracts. In yet other types of cases, collusion entailed market sharing agreements. 21 Collusion does not necessarily have to involve an explicit agreement or communication between firms. In oligopolistic industries, firms tend to be interdependent in their pricing and output decisions so that the actions of each firm impact on and result in a counter response by the other firm(s). In such circumstances, oligopolistic firms may take their rivals’ actions into account and coordinate their actions as if they were a cartel without an explicit or overt agreement. Such coordinated behaviour is often referred to as tacit collusion or
conscious parallelism .Various factors may facilitate the formation of price-fixing conspiracies. These include: a) Ability to raise and maintain industry price. If barriers to entry are low or there exist substitute products, collusion will not be successful and firms will not have an incentive to remain in or join the price conspiracy. b) Firms do not expect collusion to be easily detected or severely punished. If such is the case, the profits from collusion may be significantly higher than the costs of fines and of the firms’ loss of reputation. c) Organizational costs are low. If the negotiations between firms are protracted and enforcement and monitoring costs of the conspiracy are high, it may be difficult to form a combination. d) Homogenous or very similar products are produced. Uniform price agreements are not easily reached if the products differ in attributes such as quality and durability. It becomes difficult for firms in such circumstances to detect whether variations in sales are due to changing buyer preferences or cheating by firms in the form of secret price cuts. e) Industry is highly concentrated or a few large firms provide the bulk of the product. When the number of firms is few, the costs of organizing collusion will tend to be low. Also, the probability of detecting firms which do not respect the fixed prices will be correspondingly higher. f) The existence of an industry or trade association. Associations tend to provide a basis for coordinating economic activities and exchange of information which may facilitate collusion. They may also reduce the organizational and monitoring costs of the combination. Collusion does not necessarily arise in the presence of all or some of the above mentioned factors in a given market. In addition, there are a number of factors which may limit collusion. Such factors include product heterogeneity, 22 inter-firm cost differences, cyclical business conditions, existence of sophisticatedcustomers, technological change, infrequent product purchases,differing expectations of firms, and incentives to secretly cut prices and increase market share. The last of these is a particularly important reason why collusion among firms tends to break down over time. For further details see D.W. Carlton and J.M. Perloff, Modern Industrial Organization , Scott, Foresman/Little Brown. Glenview, Il., 1990, Ch. 9; F.M. Scherer and D. Ross, Industrial Market Structure and Economic Performance ,Houghton Mifflin Co., Boston, 1990, Chs. 7 and 8; G.J. Stigler, "A Theory of Oligopoly," Journal of Political Economy , 1964, Vol. 72(1), pp. 44-61. See also
agreement , cartel .
Collusive bidding (tendering)
See Bid Rigging. 27. Combination
In the parlance of competition law and policy, the term combination refers to firms organized together to form a monopoly , cartel or agreement to raise or fix prices and restrict output in order to earn higher profits. This term has been interchangeably used with conspiracy and collusion as well. For further details see discussion under these headings.
Common Control
See Control of Enterprises , Holding Company .
Competition
A situation in a market in which firms or sellers independently strive for the patronage of buyers in order to achieve a particular business objective, e.g., profits, sales and/or market share. Competition in this context is often equated with rivalry. Competitive rivalry between firms can occur when there are two firms or many firms. This rivalry may take place in terms of price, quality, service or combinations of these and other factors which customers may value. 23 Competition is viewed as an important process by which firms are forced to become efficient and offer greater choice of products and services at lower prices. It gives rise to increased consumer welfare and allocative efficiency . It includes the concept of "dynamic efficiency" by which firms engage in innovation and foster technological change and progress. See also Cut-Throat Competition ,
Contestability , Perfect Competition , Efficiency , Pareto Efficiency , Workable Competition .
Compulsory Licensing
See Licensing. 31. Concentration
Concentration refers to the extent to which a small number of firms or enterprises account for a large proportion of economic activity such as total sales, assets or employment. There are at least four distinct concepts embodied within the term concentration:
Aggregate Concentration which measures the relative position of large enterprises in the economy. This measure has interested economists, sociologists and political scientists mainly in the context of theories relating to actual (and potential) economic-political power which big business may be able to exercise because of their economic importance in a country/industrial sector/geographic region.
Industry or Market Concentration (also often referred to as seller concentration) which measures the relative position of large enterprises in the provision of specific goods or services such as automobiles or mortgage loans. The rationale underlying the measurement of industry or market concentration is the industrial organization economic theory which suggests that, other things being equal, high levels of market concentration are more conducive to firms engaging in monopolistic practices which leads to misallocation of resources and poor economic performance. Market concentration in this context is used as one possible indicator of market power .
Buyer Concentration which measures the extent to which a large percentage of a given product is purchased by relatively few buyers. At the extreme, a single purchaser of all the production of a good or service would give rise to a situation of
monopsony . Buyer concentration may result in countervailing power that offsets the market power that may otherwise arise from high levels of market or seller 24
concentration . See also discussion under bilateral monopoly/oligopoly .
Ownership Concentration which measures the extent to which shares of stock exchange listed companies are widely or narrowly (closely) held. This last concept is often extended to describe the wealth or control of corporate assets among individual families or business entities. See Concentration Indexes .
Concentration Indexes
Various concentration indexes or measures have been suggested in the field of industrial organization economics. These measures are used to describe market structure and/or as a prima facie indicator of market power or competition
among firms. Essentially, concentration indexes attempt to measure the number and relative size inequality of firms. The most frequently used measures are:
Concentration Ratio : The percentage of total industry output (or other such measure of economic activity, e.g., sales revenue, employment) which a given number of large firms account for. The four-firm concentration ratio (CR 4)measures the relative share of total industry output accounted for by the four largest firms. Similarly, CR 3, CR 5, CR 8, etc. measures may be computed. The number of large firms are ranked and grouped in order to avoid disclosure of confidential economic information pertaining to individual firms. A disadvantage of the concentration ratio is that it does not indicate the total number of firms that may be operating and competing in an industry. For example, two industries with the same high CR 4 levels of 75 percent may differ nonetheless because one industry may have few firms while the other may have many firms.
Herfindahl-Hirschman Index (HHI): This measure is based on the total number and size distribution of firms in the industry. It is computed as the sum of the squares of the relative size of all firms in the industry. Algebraically it is: n nHHI = Σ ( s i ) 2where Σ s i = 1 i=1 i=1 si is the relative output (or other measures of economic activity such as sales or capacity) of the i th firm, and n is the total number of firms in the industry. In an industry with one firm (monopoly), the HHI measure will be equal to 1. In a duopoly with two equal sized firms, the HHI measure will be: (0.5) 2+ (0.5) 2= 0.50 25 The HHI may be computed on a base of 1 (as in the above examples) or 1 000 or 10 000. The index is used, for example, in the United States Antitrust Division Merger Guidelines as an administrative criterion to screen mergers that may warrant further examination for their effects on competition . The HHI has several mathematical and economic theoretic properties which make it a desirable concentration measure. There are other measures of concentration, e.g., the Lorenz Curve , Gini Coefficient , Inverse Index and Entropy . These measures, while of different theoretic significance, are not as frequently employed in industrial organization and competition policy analysis as the Concentration Ratio and the Herfindahl-Hirschman Index. For further information see, for example, G. Rosenbluth, "Measures of Concentration" in National Bureau of Economic Research, Business Concentration and Price Policy , Princeton University Press, Princeton, 1955; E.M. Singer, "The Structure of Industrial Concentration Indexes," Antitrust Bulletin , Vol. X, January-April, 1965, pp. 75-104; and D. Encaoua and A. Jacquemin, "Degree of Monopoly, Indices of Concentration and Threat of Entry,"
International Economic Review , Vol. 21, 1980, pp. 87-105.
Concentration Measures
See Concentration Indexes. 34. Concentration Ratio
See Concentration Indexes. 35. Concerted Action or Practice
See Cartel , Collusion. 36. Conglomerate
A firm or business enterprise having different economic activities in different unrelated industries. Conglomerate firms may emerge through mergers
and acquisitions and/or investments across a diverse range of industries for a variety of reasons such as minimization of risk, increased access to financial and 26 management resources, and more efficient allocation of resources. Competition policy concerns have been raised, although without universal agreement among economists, that conglomerates facilitate anticompetitive practices through cross-subsidization of less profitable activities aimed at driving out competition and reciprocal arrangements with other conglomerates in the purchase and sale of inputs-outputs. There is increasing evidence that conglomerates are not necessarily more profitable and many conglomerate firms have in recent times been divesting different activities and focusing their operations on fewer lines of business. See also Diversification , Mergers .
Conglomerate Merger
See Merger. 38. Conscious Parallelism
Under conditions of oligopoly , the pricing and output actions of one firm have a significant impact upon that of its rivals. Firms may after some period of repeated actions become conscious or aware of this fact and without an explicit
agreement coordinate their behaviour as if they were engaged in collusive behaviour or a cartel to fix prices and restrict output. The fear that departure from such behaviour may lead to costly price cutting, lower profits and market share instability may further create incentives for firms to maintain such an implicit arrangement amongst themselves. This form of conscious parallel behaviour or
tacit collusion generally has the same economic effect as a combination ,
conspiracy or price fixing agreement. However, whether or not conscious parallel behaviour constitutes an illegal action which is restrictive of competition is a subject of controversy in both competition law and economics. Price uniformity may be a normal outcome of rational economic behaviour in markets with few sellers and homogenous products. Arguments have been advanced that the burden of proof must be higher than circumstantial evidence of concerted or parallel behaviour and uniform pricing and output policies. In other words, conscious parallelism in and of itself should not necessarily be construed as evidence of collusion. The problem arises more from the nature of the market or industry structure in which firms operate than from their respective behaviour. See also discussion under agreements , cartel , collusion .27
Consolidation
Generally refers to combination or amalgamation of two or more firms into one new firm through the transfer of net assets. The new firm may be specially organized to distinguish it from a merger.
Conspiracy
Normally a covert or secret arrangement between competing firms in order to earn higher profits by entering into an agreement to fix prices and restrict output. The terms combination , conspiracy , agreement and collusion are often used interchangeably. For further details see discussion under these headings.
Constant Returns to Scale
See Economies of Scale. 42. Consumers’ Surplus 28 Consumers’ surplus is a measure of consumer welfare and is defined as the excess of social valuation of product over the price actually paid. It is measured by the area of a triangle below a demand curve and above the observed price. In the diagram below, the market demand curve for good X is drawn as AC. At price = P 0, Q 0 units of X are purchased by all consumers. However, given the demand curve, there are some consumers who would be prepared to pay a higher price for X. These consumers receive a benefit from the fact that they actually pay only P 0. The dollar value of the benefit to all such consumers is given by the area of the triangle P 0AB which is the dollar measure of consumers’ surplus. Consumers’ surplus is a widely used measure of consumer welfare because it only requires information on the demand curve (prices and quantities). However, there is considerable debate over the degree to which it corresponds to more theoretically appealing measures of consumer welfare. In general, consumers’ surplus is more useful the lower is the income elasticity of demand. (For a useful textbook discussion and references, see R. Just, D. Hueth and A. 29 Schmitz, Applied Welfare Economics and Public Policy , Prentice Hall, Englewood Cliffs, N.J., 1982, Chs. 5 and 6.)
Consumer Welfare
Consumer welfare refers to the individual benefits derived from the consumption of goods and services. In theory, individual welfare is defined by an individual’s own assessment of his/her satisfaction, given prices and income. Exact measurement of consumer welfare therefore requires information about individual preferences. In practice, applied welfare economics uses the notion of consumer surplus to measure consumer welfare. When measured over all consumers, consumers’ surplus is a measure of aggregate consumer welfare. In anti-trust applications, some argue that the goal is to maximize consumers’ surplus, while others argue that producer benefits should also be counted. See Consumers’ Surplus ,
Deadweight Welfare Loss .
Contestability
A contestable market is one in which the following conditions are satisfied: a) there are no barriers to entry or exit; b) all firms, both incumbent and potential entrants, have access to the same production technology; c) there is perfect information on prices, available to all consumers and firms; d) entrants can enter and exit before incumbents can adjust prices. In contrast to perfect competition , a contestable market may have any number of firms (including only one or a few) and these firms need not be price-takers. The analysis of contestable markets is designed for cases in which the existence of scale economies precludes a large number of competitors. The theory of contestable markets suggests that an industry consisting of one or a few firms may be efficient. The basic idea is that incumbent firms will 30 maintain prices close to the competitive level because of the threat posed by potential entrants. If incumbents raise price, entry will occur (no barriers to entry ), and the entrants will be able to produce as efficiently as incumbents (access to technology). Moreover, if price declines as a result of the entry, the entrant will be able to exit the industry quickly and costlessly (no barriers to exit). This is known as "hit and run" entry. It is the fear of "hit and run" entry which motivates even a monopolist to maintain prices close to average cost. When incumbent firms set prices such that they make profits without providing an incentive for entry, prices are said to be sustainable. A sustainable monopoly price is one which clears the market, allows the monopolist to break even, and leaves no opportunity for profitable entry. Sustainability therefore defines the equilibrium in a contestable market. A natural monopoly may be a contestable market if there are no significant
sunk costs . This means that a natural monopoly which is contestable and sustainable need not be regulated or subject to competition policies because it is disciplined by the threat of entry. However, a natural monopoly market may be contestable, but not sustainable, in which case entry regulation may be required. It is of considerable importance to establish whether a market is contestable. Deregulation in airlines and trucking has been predicated on the belief that these industries are contestable. This view arises from the position that entry and exit are relatively quick and easy. Easy exit is associated with the absence of sunk costs, which in the above industries is indicated by the existence of second-hand markets for trucks and airplanes. The basic reference is W.J. Baumol, J.C. Panzar and R.D. Willig,
Contestable Markets and the Theory of Industry Structure , Harcourt, Brace, Jovanovich, New York, 1982, revised 1988. See also D.F. Spulber, Regulation and Markets, MIT Press, Cambridge, 1989, Ch. 4 for discussion and references.
Contestable Markets
See Contestability. 31
Control of Enterprises
Control over enterprises is generally viewed to be exercised when an individual or group of investors hold more than 50 per cent of the common voting stock of the enterprise or firm. However, "effective control" may be exercised when the investor(s) holds a large block of voting stock even when it is less than 50 per cent but the remaining shares are widely held by many smaller investors. Control of enterprises may also be exercised through interlocking directorates
and inter-corporate ownership links between firms as in the case of
conglomerates .
Costs
Costs refer to the value in alternative uses of the factors of production used by a firm (labour costs, materials costs, capital costs. See also Opportunity Costs ). Costs may be fixed or variable. Fixed costs are costs that do not vary with the amount produced. Examples are interest on debt, property taxes and rent. Economists also add to fixed cost an appropriate return on capital which is sufficient to maintain that capital in its present use. This reflects the idea that all economic costs are opportunity costs, the cost of foregone alternatives. Thus, the return to capital if employed elsewhere constitutes its opportunity cost. Variable costs are costs that vary with the amount produced. Examples are materials, fuel, production labour and maintenance. As the relevant time period is extended, more costs become variable.
Total costs refer to the sum of fixed and variable costs. Average costs refer to total costs divided by output. Marginal cost is the increment to total cost that results from producing an additional unit of output. Marginal cost is a function of variable costs alone, since fixed costs do not vary with increases in output. Marginal cost has a particular importance in economic theory. The profit maximizing firms will always produce an output such that marginal cost equals marginal revenue. See revenues ).
Countervailing Power
See Bilateral Monopoly , Buyer Concentration , Monopsony. 32
Cournot (Nash) Equilibrium
The Cournot model of oligopoly assumes that rival firms produce ahomogenous product, and each attempts to maximize profits by choosing how much to produce. All firms choose output (quantity) simultaneously. The basic Cournot assumption is that each firm chooses its quantity, taking as given the quantity of its rivals. The resulting equilibrium is a Nash equilibrium in quantities, called a Cournot (Nash) equilibrium. The Cournot model provides results which are of some importance to industrial economics. First of all, it can be shown that price will not in most cases equal marginal costs (see costs ) and Pareto efficiency is not achieved. Moreover, the degree to which each firm’s price exceeds marginal cost is directly proportional to the firm’s market share and inversely proportional to the market elasticity of demand .If the oligopoly is symmetric, that is all firms have identical products and cost conditions, then the degree to which price exceeds marginal cost is inversely related to the number of firms. Thus, as the number of firms increases, the equilibrium approaches what it would be under perfect competition .More generally, it can be shown that for the industry the degree to which price exceeds marginal cost is directly proportional to the Herfindahl-Hirschman Index of concentration. As concentration rises, industry performance deviates more from the norm of perfect competition. See Bertrand (Nash) equilibrium .For more detail, see C. Shapiro, "Theories of Oligopoly Behavior", in R. Schmalensee and R. Willig (eds), The Handbook of Industrial Organization,
North Holland, Amsterdam, 1989.
Crisis Cartel
See Cartel. 51. Cross Price Elasticity of Demand
Refers to the percentage change in the quantity demanded of a given product due to the percentage change in the price of another "related" product. If all prices are allowed to vary, the quantity demanded of product X is dependent not only on 33 its own price (see elasticity of demand ) but upon the prices of other products as well. The concept of cross price elasticity of demand is used to classify whether or not products are "substitutes" or "complements". It is also used in market definition to group products that are likely to compete with one another. If an increase in the price of product Y results in an increase in the quantity demanded of X (while the price of X is held constant), then products X and Y are viewed as being substitutes. For example, such may be the case of electricity vs. natural gas used in home heating or consumption of pork vs. beef. The cross price elasticity measure is a positive number varying from zero (no substitutes) to any positive number. Generally speaking, a number exceeding two would indicate the relevant products being "close" substitutes. If the increase in price of Y results in a decrease in the quantity demanded of product X (while the price of X is held constant), then the products X and Y are considered complements. Such may be the case with shoes and shoe laces.
Cut-Throat Competition
Also known as destructive or ruinous competition ; refers to situations when competition results in prices that do not chronically or for extended periods of time cover costs of production, particularly fixed costs. This may arise in secularly
declining or "sick" industries with high levels of excess capacity or çwhere frequent cyclical or random demand downturns are experienced. The destructive competition argument is often advanced to advocate government intervention in the form of price regulation or stabilization and structural rationalization.
Deadweight Welfare Loss
The deadweight welfare loss is a measure of the dollar value of consumers' surplus lost (but not transferred to producers) as a consequence of a price increase. Consider the following diagram: 34 It is assumed that the industry is originally in a state of perfect competition ,such that price equals marginal cost (Pc = MC), where the latter is assumed to be constant (constant returns to scale). Industry output is therefore Qc and consumers’ surplus is triangle PcAC. Now compare this with the same industry which has some degree of monopoly power such that price (Pm) exceeds marginal costs. Industry output is now reduced to Qm and consumers’ surplus is PmAB, a reduction of PcPmBC. However, a portion of the lost consumer surplus, PcPmBD, is transferred to producers in the form of excess profits, referred to as producers’ surplus (PcPmBE). The remainder, the triangle BCE, is referred to as adeadweight welfare loss and is a measure of lost allocative efficiency .35 In anti-trust economics, there is some debate over the appropriate welfare measure to be applied. Some argue that lost consumer surplus (i.e. including both deadweight loss and producers’ surplus) should be considered on the grounds that a transfer from consumers to firms does not improve social welfare. Others argue that this represents a value judgment and all decisions should be based only on the deadweight welfare loss (allocative efficiency ), with judgments regarding transfers of income left to the political process. Still others argue that producers’ surplus should be considered because much of it is dissipated in the quest for monopoly profits. See Rent Seeking .Useful textbook discussions, with applications and reference are: S. Martin,
Industrial Economics, Macmillan, New York, 1988, pp. 30-41; 274-276; F.M. Scherer and D. Ross, Industrial Market Structure and Economic Performance ,Houghton Mifflin, Boston, 1990, Ch. 18.
Deconcentration
A policy of breaking up and divesting operations of large firms in order to reduce the degree of concentration in an industry. This policy has been advocated from time to time in different countries particularly in periods of high levels of merger activity. Lower industry concentration levels and increase in the number of firms are viewed as being conducive to stimulating competition. There are however inherent risks in adopting this policy as a general approach to resolving competition problems that may be associated with high industry concentration levels. A structural deconcentration policy may result in significant loss in economic efficiency. Large firms may be large because of economies of scale, superior technology and innovation which may not be divisible without high costs. This is more likely to be the case where firms have attained their respective size in response to market conditions and opportunities. However, in several countries, particularly in Eastern European economies, growth of industrial concentration and large firm size have been encouraged by deliberate government policy. Deconcentration policies in such an environment may be appropriate in order to promote market oriented firm behaviour and efficiency.
Deep Pockets
An expression used to describe the idea that extensive financial and other resources of large firms or conglomerates can be used to sell below cost for extended periods of time. In this view, deep pockets are thought to give such firms 36 an unfair advantage over competitors particularly if the practice of selling at prices below costs imposes losses and drives out competing firms. Others argue that firms using "deep pockets" to finance anticompetitive actions impose a cost on themselves because those funds could be more profitably employed elsewhere. Moreover, if capital markets work reasonably well, target firms should have no trouble obtaining financing to sustain themselves through the anticompetitive action. See Predatory Pricing .
Delivered Pricing
See Basing Point Pricing. 57. Demonopolization
See Anti-Monopoly, Antitrust, Deconcentration. 58. Depression Cartel
See Cartel. 59. Deregulation
See Regulation. 60. Destructive Competition
See Cut-Throat Competition. 61. Differentiated Products
See Product Differentiation. 37
Discrimination
See Price Discrimination. 63. Diseconomies of Scale
See Economies of Scale. 64. Distributor’s Mark
See Trade Mark .
Diversification
The term refers to the expansion of an existing firm into another product line or market. Diversification may be related or unrelated. Related diversification occurs when the firm expands into similar product lines. For example, an automobile manufacturer may engage in production of passenger vehicles and
light trucks. Unrelated diversification takes place when the products are very different from each other, for example a food processing firm manufacturing leather footwear as well. Diversification may arise for a variety of reasons: to take advantage of complementarities in production and existing technology; to exploit
economies of scope ; to reduce exposure to risk; to stabilize earnings and overcome cyclical business conditions; etc. There is mounting evidence that related diversification may be more profitable than unrelated diversification.
Divestiture
Refers to firms selling part of their current operations, divisions or subsidiaries. Divestiture may take place as a result of firms restructuring their business in order to concentrate on certain products or markets. It may also be imposed upon them by competition authorities as a result of a merger or acquisition which is likely to reduce competition substantially. Divestiture under these latter circumstances is aimed at maintaining existing competition in the market. Divestiture may also form a part of a policy to deconcentrate an industry. 38
Dominant Firm
A dominant firm is one which accounts for a significant share of a given market and has a significantly larger market share than its next largest rival. Dominant firms are typically considered to have market shares of 40 per cent or more. Dominant firms can raise competition concerns when they have the power to set prices independently. An industry with a dominant firm is therefore often an oligopoly in that there are a small number of firms. However, it is an asymmetric oligopoly because the firms are not of equal size. Normally, the dominant firm faces a number of small competitors, referred to as a competitive fringe. The competitive fringe sometimes includes potential entrants. Thus the dominant firm may be amonopolist facing potential entrants. Like a monopolist, the dominant firm faces a downward sloping demand curve. However, unlike the monopolist, the dominant firm must take into account the competitive fringe firms in making its price/output decisions. It is normally assumed that the dominant firm has some competitive advantage (such as lower costs) as compared to the fringe. The term competitive fringe arises from the basic theory of dominant firm pricing. It is generally assumed that the dominant firm sets its price after ascribing a part of the market to the competitive fringe which then accepts this price as given. Dominant firms may be the target of competition policy when they achieve or maintain their dominant position as a result of anti-competitive practices. See
Abuse of Dominant Position .For further discussion and references, see S. Martin, Industrial Economics,
Macmillan, New York, 1988, Ch. 4.
Dominant Market Position
See Dominant Firm. 39
Dominant Price Leadership
See Dominant Firm , Price Leadership. 70. Dumping
The practice by firms of selling products abroad at below costs or significantly below prices in the home market. The former implies predatory pricing ; the latter, price discrimination. Dumping of both types is viewed by many governments as a form of international predation, the effect of which may be to disrupt the domestic market of foreign competitors. Economists argue, however, that price discriminatory dumping, where goods are not sold below their incremental costs of production, benefits consumers of the importing countries and harms only less efficient producers. Under the General Agreement on Tariffs and Trade (GATT) rules, dumping is discouraged and firms may apply to their respective government to impose tariffs and other measures to obtain competitive relief. As in the case of
predatory pricing or selling below costs (see discussion under these headings), arguments have been advanced questioning the economic feasibility of dumping at prices below costs over extended periods of time.
Duopoly
A duopoly is an industry consisting of two sellers. It is therefore a special case of oligopoly . In industrial organization economic theory, duopoly is often analysed as a simplified example of oligopoly behaviour. See oligopoly .
Economies of Scale
Refers to the phenomenon where the average costs per unit of output decrease with the increase in the scale or magnitude of the output being produced by a firm. Similarly, the opposite phenomenon, diseconomies of scale , occurs when the average unit costs of production increase beyond a certain level of output. At the point where the average costs are at a minimum, the minimum efficient scale (MES) of output of a firm or plant is reached. The maximum efficient scale of output is reached at the point just before diseconomies set in, that is unit costs of production start to increase. Between the range of minimum and maximum efficient scale of output, there may also exist constant returns to scale 40 where the average unit costs of production remain unchanged as output increases. The minimum and maximum scales of output, in relation to the total demand or market size have an important bearing on the number and size distribution of firms in an industry and on concentration .A distinction is often made between different types of economies of scale:
Product Specific Economies are associated with the volume of output of any single product made and sold. In a multi-product firm or plant, product specific economies are often realized by specializing in the manufacture of one or a few products over a larger scale of output. Such economies generally arise by avoiding the costs of interrupting production and re-tooling that is required in order to produce different products with the same machinery and equipment. Product specific economies are often the basis for specialization agreements .
Plant Specific Economies are associated with the total output (frequently encompassing many products) of an entire plant or plant complex. Economies of scope may be embodied as part of plant economies as the costs of common overheads, e.g., head office administration and accounting costs, are spread across multiple products.
Economies of Multi-plant Operations are associated with operating more than one plant and may arise for such reasons as minimizing transportation costs of raw materials and/or finished products, to better serve different geographic markets, economies of scope , specialization, etc. For more details, see F.M. Scherer and D. Ross, Industrial Market Structure and Economic Performance, Houghton Mifflin Co., Boston, 1990, Ch. 4.
Economies of Scope
Economies of scope exist when it is cheaper to produce two products together (joint production) than to produce them separately. For example, it may be less costly to provide air service from point A to points B and C with one aircraft than have two separate air flights, one to point B and another to point C. Similarly, a steer produces beef and hide and it may be inefficient to breed steers separately for beef and for hide. While many factors such as technology may explain economies of scope, of particular importance is the presence of common 41 input(s) and/or complementarities in production. Firms may often endeavor to exploit economies of scope in order to produce and offer multiple products at lower costs.
Efficiency
The term has a wide number of usages. In the context of industrial organization economics and competition law and policy, it relates to the most effective manner of utilizing scarce resources. Two types of efficiency are generally distinguished: technological (or technical) and economic (or allocative). A firm may be more technologically efficient than another if it produces the same level of output with one or fewer physical number of inputs. Because of different production processes, not all firms may be technologically efficient or comparable. Economic efficiency arises when inputs are utilized in a manner such that a given scale of output is produced at the lowest possible cost. An increase in efficiency occurs when an existing or higher scale of output is produced at lower cost. Unlike technological efficiency, economic efficiency enables diverse production processes to be compared. Competition is generally viewed by economists to stimulate individual firm(s) or economic agents in the pursuit of efficiency. Efficiency increases the probability of business survival and success and the probability that scarce economic resources are being put to their highest possible uses. At the firm level, efficiency arises primarily through economies of scale and scope and, over a longer period, through technological change and innovation. The term "efficiency" in distribution or consumption is used to describe the situation when a particular set of goods and services are divided amongst the consumers in such a way that no one individual can be made better off without making another worse off. See also Pareto Efficiency .
Elasticity of Demand (Price)
The price elasticity of demand measures the responsiveness of demand to variations in price. It is defined as the percentage change in quantity demanded divided by the percentage change in price. Since the demand curve is normally downward sloping, the price elasticity of demand is usually a negative number. However, the negative sign is often omitted. In principle, the price elasticity may vary from (minus) infinity to zero. The closer to infinity, the more elastic is demand; and the closer to zero, the more 42 inelastic is demand. In practice, elasticities tend to cluster in the range of minus 10 to zero. Minus one is usually taken as a critical cut-off point with lower values (that is less than one) being inelastic and higher values (that is greater than one) being elastic. If demand is inelastic a price increase will increase total revenues while if demand is elastic, a price increase will decrease revenues. Demand curves are defined for both the industry and the firm. At the industry level, the demand curve is almost always downward sloping. However, at the firm level the demand curve may be downward sloping or horizontal. The latter is the case of the firm in a perfectly competitive industry whose demand is infinitely elastic. When the firm’s demand curve is downward sloping, the firm has some control over its price. The price elasticity of demand is determined by a number of factors, including the degree to which substitute products exist see cross price elasticity of demand ). When there are few substitutes, demand tend. See to be inelastic. Thus firms have some power over price. When there are many substitutes, demand tends to be elastic and firms have limited control over price.
Enterprise
A term in the commercial world used to describe a project or venture undertaken for gain. It is often used with the word "business" as in "business enterprise". Usually, by extension, it refers to the business entity carrying out the enterprise and is thus synonymous with "undertaking", "company" or "firm". See also holding company. 77. Entropy
See Concentration Indexes. 78. Excess Capacity
A situation where a firm is producing at a lower scale of output than it has been designed for. It exists when marginal cost is less than average cost and it is still possible to decrease average (unit) cost by producing more goods and services. Excess capacity may be measured as the increase in the current level of output that is required to reduce unit costs of production to a minimum. Excess capacity is a 43 characteristic of natural monopoly or monopolistic competition . It may arise because as demand increases, firms have to invest and expand capacity in lumpy or indivisible portions. Firms may also choose to maintain excess capacity as a part of a deliberate strategy to deter or prevent entry of new firms.
Excessive Competition
See Cut-Throat Competition. 80. Excess Prices
Refers to prices set significantly above competitive levels as a result of
monopoly or market power . However, in practice, in absence of a conspiracy or
price fixing agreement or evidence of market power stemming from high concentration, it is very difficult to establish a threshold beyond which a price may be considered excessive or unreasonable. Because the basic method of organizing production in a market economy is through the price system, price flexibility is critical. Prices fluctuate in order to bring supply and demand into equilibrium. Temporary shortages in supply or increases in demand will cause prices to rise and provide incentives for increased production and entry of new suppliers. Moreover, it should be noted that price and/or profit comparisons between different firms, markets, or countries are fraught with legal and economic problems. Attempts by government to control or force a roll back of prices that are not a result of restrictions on competition are inconsistent with the philosophy underlying competition policy.
Exclusive Dealing
See Vertical Restraints. 82. Export Cartel
An agreement or arrangement between firms to charge a specified export price and/or to divide export markets. Many competition law statutes exempt such agreements from the conspiracy provisions provided that the cartel does not lead to injurious effects on competition in the domestic market, e.g., give rise to price fixing agreements or result in reduction in exports. The rationale for permitting 44 export cartels is that it may facilitate cooperative penetration of foreign markets, transfer income from foreign consumers to domestic producers and result in a favourable balance of trade. See also cartel .
External Economies/Diseconomies
See Externalities. 84. Externalities
Externalities refers to situations when the effect of production or consumption of goods and services imposes costs or benefits on others which are not reflected in the prices charged for the goods and services being provided. Pollution is an obvious example of a negative externality , also termed an external diseconomy . Chemicals dumped by an industrial plant into a lake may kill fish and plant life and affect the livelihood of fishermen and farmers nearby. In contrast, a
positive externality or external economy may arise from the construction of a road which opens a new area for housing, commercial development, tourism, etc. The invention of the transistor generated numerous positive externalities in the manufacture of modern telecommunication, stereo and computer equipment. Externalities arise when property rights cannot be clearly assigned. See also
Market Failure .
Extraterritoriality
Refers to the application of one country’s laws within the jurisdiction of another country. In the context of competition policy, the issue of extraterritoriality would arise if the business practices of firm(s) in one country had an anticompetitive effect in another country which the latter considered to be in violation of its laws. For example, an export cartel formed by companies which may be exempt from competition laws of country A may nevertheless be viewed as a price-fixing agreement to limit competition in markets of country B and in violation of the latter country’s antitrust laws. Another situation that could arise is a merger between two competing firms in one country resulting in substantial lessening of competition in the markets of another country. (This can arise if the merging companies are primarily export-oriented and account for the bulk of the market in the importing country.) 45 Whether or not companies can be successfully prosecuted for violations of competition laws of another country is importantly dependent, among other factors, on the nature of the sovereign relationship between the countries involved, where the alleged violation has taken place, the legal status of the business practice or action in the originating country and the existence of subsidiary operations and significant assets in the affected country against which legal actions can be brought forward.
Failing Firm
A firm that has been consistently earning negative profits and losing market share to such an extent that it is likely to go out of business. The concept becomes an issue in merger analysis when the acquiring firm argues that the acquisition of such a firm does not result in substantial lessening of competition since it is likely to exit the market anyway. If this is true, the "current" market share of the failing firm may have no "future" competitive significance and should be weighted accordingly.
Fighting Brand
Refers to a new brand of an existing or similar product which is priced very low or below costs and is made available for a limited time period in specific market areas in order to combat competition from other (usually smaller) firms. Firms introduce "fighting brands" to avoid lowering the prices charged for their established brands as this may prove to be costly for reasons such as these established brands being priced uniformly across a wide number of markets. Fighting brands’ are often viewed as a form of predation or anticompetitive practice intended to drive out competitors from a given market. As with other forms of predation, however, their chances of success are limited. See Predatory Pricing .
Fixed Costs
See Costs. 46
Foreclosure of Competition
See Anticompetitive Practices. 90. Franchising
A special type of vertical relationship between two firms usually referred to as the "franchisor" and "franchisee". The two firms generally establish acontractual relationship where the franchisor sells a proven product, trademark or business method and ancillary services to the individual franchisee in return for a stream of royalties and other payments. The contractual relationship may cover such matters as product prices, advertising, location, type of distribution outlets, geographic area, etc. Franchise agreements generally fall under the purview of competition laws, particularly those provisions dealing with vertical restraints .Franchise agreements may facilitate entry of new firms and/or products and have efficiency enhancing benefits. However, franchising agreements in certain situations can restrict competition as well. See Competition Policy and Vertical Restraints: Franchising , OECD, Paris, forthcoming in 1993.
Free Rider or Riding
Free riding occurs when one firm (or individual) benefits from the actions and efforts of another without paying or sharing the costs. For example, a retail store may initially choose to incur costs of training its staff to demonstrate to potential customers how a particular kitchen appliance works. It may do so in order to expand its sales. However, the customers may later choose to buy the product from another retailer selling at a lower price because its business strategy is not to incur these training and demonstration costs. This second retailer is viewed as "free riding" on the efforts and the costs incurred by the first retailer. If such a situation persists, the first retailer will not have the incentive to continue demonstrating the product.
Full Cost Pricing
This is a practice where the price of a product is calculated by a firm on the basis of its direct costs per unit of output plus a markup to cover overhead costs and profits. The overhead costs are generally calculated assuming less than full capacity operation of a plant in order to allow for fluctuating levels of production 47 and costs. Full cost pricing is often used by firms as it is very difficult to calculate the precise demand for a product and establish a market price. Empirical studies indicate that full cost pricing methods are widely employed by business firms. See, for example, R.B. Heflebower, "Full Costs, Cost Changes and Prices" in the National Bureau of Economic Research, Business Concentration and Pricing Policy, Princeton University Press, Princeton, 1955 pp. 361-396; and A. Silberston, "Surveys of Applied Economics: Price Behaviour of Firms," Economic Journal ,Vol. 80, September, 1970, pp. 511-582.
Full Line Forcing
See Tied Selling .
Gentlemen’s Agreement
See Collusion. 95. Gini Coefficient
See Concentration Indexes. 96. Herfindahl-Hirschman Index
See Concentration Indexes. 97. Heterogenous Products
See Homogenous Products, Product Differentiation. 98. Holding Company
A holding company is a purely financial concern which uses its capital solely to acquire interests (normally controlling interests) in a number of operating companies. Although the purpose of a holding company is mainly to gain control and not to operate, it will typically have representation on the boards of directors of 48 the operating firms. Holding companies provide a means by which corporate control can become highly concentrated through pyramiding. A holding company may gain control over an operating company which itself has several subsidiaries .
Homogenous Products
Products are considered to be homogenous when they are perfect substitutes and buyers perceive no actual or real differences between the products offered by different firms. Price is the single most important dimension along which firms producing homogenous products compete. However, empirical experience demonstrates that when the number of such firms is few, the existence of homogenous products may facilitate collusion . In various jurisdictions, collusive arrangements have been found to exist in homogenous products such as cement, flour, steel and sugar. In contrast, heterogenous products differ significantly from each other and are not easily substitutable. See also product differentiation .
Horizontal Integration
See Merger .
Horizontal Merger
See Merger .
Income Elasticity of Demand
The quantity demanded of a particular product depends not only on its own price (see elasticity of demand ) and on the price of other related products (see
cross price elasticity of demand ), but also on other factors such as income. The purchases of certain commodities may be particularly sensitive to changes in nominal and real income. The concept of income elasticity of demand therefore measures the percentage change in quantity demanded of a given product due to a percentage change in income. The measures of income elasticity of demand may be either positive or negative numbers and these have been used to classify products into "normal" or "inferior goods" or into "necessities" or "luxuries". If as a result of an increase in income the quantity demanded of a particular product decreases, it would be classified as an "inferior" good. The opposite would be the 49 case of a "normal" good. Margarine has in past studies been found to have a negative income elasticity of demand indicating that as family income increases, its consumption decreases possibly due to substitution of butter. This finding may, however, be less applicable today given health concerns regarding heart disease and cholesterol levels and new information on beneficial attributes of margarine. This illustrates the inherent risks likely to be associated with generalizations or classification of products based on income elasticity measures.
Increasing Returns to Scale
See Economies of Scale. 104. Industry Concentration
See Concentration. 105. Integration
See Vertical Integration. 106. Intellectual Property Rights
The general term for the assignment of property rights through patents, copyrights and trademarks. These property rights allow the holder to exercise a monopoly on the use of the item for a specified period. By restricting imitation and duplication, monopoly power is conferred, but the social costs of monopoly power may be offset by the social benefits of higher levels of creative activity encouraged by the monopoly earnings.
Inter- and Intra-Brand Competition
See Brand Competition. 50
Interlocking Directorate
An interlocking directorate occurs when the same person sits on the board of directors of two or more companies. There is a danger that an interlock between competing firms (direct interlocks) may be used to co-ordinate behaviour and reduce inter-firm rivalry. Direct interlocks are illegal in the U.S. under the Clayton Act, but other countries are more lenient. Empirical evidence suggests, however, that the majority of interlocking directorates are between financial and non-financial companies. Thus, representatives of banks commonly sit on the boards of competing firms. These indirect interlocks are typically not a factor in competition policy.
International Cartel
See Cartel. 110. Inverse Index
See Concentration Indexes. 111. Joint Monopoly Profits
See Joint Profit Maximization. 112. Joint Profit Maximization
A situation where members of a cartel , duopoly , oligopoly or similar market condition engage in pricing-output decisions designed to maximize the groups’ profits as a whole. In essence, the member firms seek to act as a
monopoly . Note should be made that joint profit maximization does not necessarily entail collusion or an agreement among firms. The firms may independently adopt price-output strategies which take into account rival firms’ reactions and thereby produce joint profit maximization. 51
Joint Venture
A joint venture is an association of firms or individuals formed to undertake a specific business project. It is similar to a partnership, but limited to a specific project (such as producing a specific product or doing research in a specific area). Joint ventures can become an issue for competition policy when they are established by competing firms. Joint ventures are usually justified on the grounds that the specific project is risky and requires large amounts of capital. Thus, joint ventures are common in resource extraction industries where capital costs are high and where the possibility of failure is also high. Joint ventures are now becoming more prevalent in the development of new technologies. In terms of competition policy, the problem is to weigh the potential reduction in competition against the potential benefits of pooling risks, sharing capital costs and diffusing knowledge. At present there is considerable debate in many countries over the degree to which research joint ventures should be subject to competition law. See Competition Policy and Joint Ventures , OECD, Paris, 1990.
Lerner Index
A measure proposed by economist A.P. Lerner to measure monopoly or
market power . The Lerner Index (LI) is: LI = Price - Marginal Cost = -1 Price Ewhere E is the price elasticity of demand In perfect competition, LI is equal to zero. The index defines monopoly power in terms of the slope of the demand curve. In the case of a profit maximizing firm in equilibrium, marginal revenue equals marginal cost and the LI is equal to the inverse of the elasticity of demand .The LI is a static measure and does not indicate whether the deviation between price and marginal cost is a worthwhile cost to pay for possible 52 innovation or new plant construction, or whether the disparity between marginal cost and price may reflect superior efficiency rather than the ability of a firm to charge high prices.
Leveraged Buyout
See Buyout. 116. Licensing
Refers to granting legal permission to do something, such as produce a product. The license confers a right which the person or firm did not previously possess. Some licenses are granted free of charge, but most require payment. Licenses are legal agreements which may contain restrictions as to how the license is employed. There are two broad cases of licensing which are relevant to competition policy. The first is licenses granted by governments to entrants in specific industries. Licensing systems exist in many communication industries (radio and T.V. broadcasting), professions (doctors) and services (banking, liquor outlets). The terms of licenses vary, but they are often accompanied by various restrictions on the firm. Those restrictions (or regulations) may apply to price, quality or amount of service. Government licensing represents an important barrier to entry in these industries. The second use of licensing is in patent, copyright and trademark cases whereby authority (in the form of a license) is granted by the owner to another party to make, reproduce, buy or sell the item. Copyright, trademark and patent holders may license others to use or produce the good, usually in return for a fixed payment and a royalty rate. In most cases, a patent holder has no preference between licensing and producing his invention himself because he can maximize his return through payment of the licensing fees. However, patent holders are not required to either use or license their technology. Thus, there may be a restriction of technology diffusion which also acts as a barrier to entry . In many countries there is a provision for revoking patents or imposing compulsory licensing when it can be proved that the patent has been abused through non-use or anticompetitive restrictions on licensing. In practice, compulsory licensing is seldom used. 53
Limit Pricing
Limit pricing refers to the pricing by incumbent firm(s) to deter or inhibit entry or the expansion of fringe firms. The limit price is below the short-run profit-maximizing price but above the competitive level. There are a number of models of limit pricing and a considerable debate over the issue of whether it is in fact profitable for firms to engage in such behaviour. Limit pricing implies that firms sacrifice current profits in order to deter entry and earn future profits. It is not clear whether this strategy is always superior to one where current prices (and profits) are higher, but decline over time as entry occurs. In the early literature on limit pricing, the ability of incumbents to establish such prices was linked to the existence of structural barriers to entry . However, this required rather stringent assumptions about the behaviour of incumbents, notably that incumbents would maintain output in the face of entry, and that this threat was believed by potential entrants. The more recent literature has focused on strategic barriers to entry , notably the actions which incumbents can take to persuade entrants that they will not accommodate entry. See R. Gilbert, "Mobility Barriers and the Value of Incumbency", in R. Schmalensee and R. Willig (eds), The Handbook of Industrial Organization, North Holland, Amsterdam, 1989.
Lorenz Curve
See Concentration Indexes. 119. Loss-Leader Selling
A marketing practice of selling a product or service at a loss in order to attract customers to buy other products at regular prices. Although this practice is illegal in some jurisdictions, in others it is viewed benevolently as a promotional device that has the procompetitive effect of increasing total sales. 54
Management Buyout
See Buyout. 121. Marginal Cost
See Costs. 122. Marginal Revenue
See Revenue. 123. Market
A market is where buyers and sellers transact business for the exchange of particular goods and services and where the prices for these goods and services tend towards equality. In order for a market to "clear" or function properly, the quantity of goods and services demanded and supplied must be equal at some given price. At any particular point in time, markets can be in "equilibrium" or "disequilibrium" depending on whether or not aggregate supply equals aggregate demand at the prevailing price. Markets may be local, regional, national or international in scope and do not necessarily require buyers and sellers to meet or communicate directly with each other. Business may be transacted through the use of intermediaries as well. See also market definition .
Market Concentration
See Concentration. 125. Market Definition
The starting point in any type of competition analysis is the definition of the "relevant" market. There are two fundamental dimensions of market definition: (i) the product market, that is, which products to group together and (ii) the geographic market, that is, which geographic areas to group together. Market definition takes into account both the demand and supply considerations. On the 55 demand side, products must be substitutable from the buyer’s point of view. On the supply side, sellers must be included who produce or could easily switch production to the relevant product or close substitutes. Market definition generally includes actual and potential sellers, that is, firms that can rapidly alter their production processes to supply substitute products if the price so warrants. The rationale for this is that these firms will tend to dampen or curb the ability of existing firms in the market to raise price above the competitive level. The location of buyers and sellers will determine whether the geographic market is local, regional, national or international. If markets are defined too narrowly in either product or geographic terms, meaningful competition may be excluded from the analysis. On the other hand, if the product and geographic markets are too broadly defined, the degree of competition may be overstated. Too broad or too narrow market definitions lead to understating or overstating market share and concentration measures. The U.S. Department of Justice and the Canadian Bureau of Competition Policy Merger Guidelines, for example, provide a paradigm for defining the relevant product and geographic markets that is based on the likely demand response of consumers to an anticompetitive price increase. A market is defined as a product or group of products and a geographic area in which it is sold such that a hypothetical, profit-maximizing firm that was the only seller of those products in that area could raise prices by a small but significant and non-transitory amount above prevailing levels. The result of applying this paradigm is to identify a group of products and a geographic area with respect to which sellers could exercise market power if they were able to coordinate their actions perfectly so as to act like a monopolist. See G. Werden, "Market Delineation and the Justice Department’s Merger Guidelines," Duke Law Review , June, 1983, pp. 514-579.
Market Failure
A general term describing situations in which market outcomes are not
Pareto efficient . Market failures provide a rationale for government intervention. There are a number of sources of market failure. For the purposes of competition policy, the most relevant of these is the existence of market power , or the absence of perfect competition . However, there are other types of market failure which may justify regulation or public ownership. When individuals or firms impose costs or benefits on others for which the market assigns no price, then an externality exists. Negative externalities arise when an individual or firm does not bear the costs of the harm it imposes 56 (pollution, for example). Positive externalities arise when an individual or firm provides benefits for which it is not compensated. Finally, there are cases in which goods or services are not supplied by markets (or are supplied in insufficient quantities). This may arise because of the nature of the product, such as goods which have zero or low marginal costs and which it is difficult to exclude people from using (called public goods; for example, a lighthouse or national defense). It may also arise because of the nature of some markets, where risk is present (called incomplete markets; for example, certain types of medical insurance).
Market for Corporate Control
In an economic system where the voting stock (shares) of companies are publicly bought and sold through the mechanism of a stock exchange, the term "market for corporate control" refers to the process by which ownership and control of companies is transferred from one group of investors and managers to another. The share prices of companies publicly listed on the stock exchange are often viewed as a "barometer" indicating the extent to which management is efficiently operating the corporation and maximizing shareholder wealth. Generally speaking, investors or shareholders delegate substantial authority to professional managers who are hired to make the company’s day-to-day pricing, production, investment, marketing and other business decisions. However, shareholders may not always be in a position to monitor or oversee these decisions, particularly if there are a large number of such shareholders. Under these circumstances, the company managers may not necessarily take decisions that maximize shareholder wealth. They may choose to shirk their duties by pursuing their own personal goals such as avoiding risk, maximizing their pay and fringe benefits, and spending money on prestige projects. Depending on the available information, the share prices of the company will be valued low and this would create incentives for takeover by a more efficient group of managers and shareholders. By taking control and subsequently changing management or management practices and reallocating resources, the assets of the acquired company may be put to more highly valued uses. The "market for corporate control" along with competition in the markets for products and services play an important role in reinforcing each other in promoting efficiency . See M.C. Jensen and R.S. Ruback, "The Market for Corporate Control: The Scientific Evidence," Journal of Financial Economics ,Vol. 11, 1983, pp. 5-50. 57
Market Power
The ability of a firm (or group of firms) to raise and maintain price above the level that would prevail under competition is referred to as market or monopoly power . The exercise of market power leads to reduced output and loss of economic welfare. Although a precise economic definition of market power can be put forward, the actual measurement of market power is not straightforward. One approach that has been suggested is the Lerner Index , i.e., the extent to which price exceeds
marginal cost . However, since marginal cost is not easy to measure empirically, an alternative is to substitute average variable cost . Another approach is to measure the price elasticity of demand facing an individual firm since it is related to the firm’s price-cost (profit) margin and its ability to increase price. However, this measure is also difficult to compute. The actual or potential exercise of market power is used to determine whether or not substantial lessening of competition exists or is likely to occur. An approach adopted in the administration of merger policy in the United States and Canada seeks to predict whether, post-merger, the parties can institute a non-transitory price increase above a certain threshold level (say 5 or 10 per cent) which will vary depending on the case without attracting entry of new firms or production of substitute products. Their ability to maintain or exceed this price threshold is assessed by detailed examination of quantitative and qualitative market structure and firm behaviour factors.
Market Share
Measure of the relative size of a firm in an industry or market in terms of the proportion of total output or sales or capacity it accounts for. In addition to profits, one of the frequently cited business objectives of firms is to increase market share. Market share, profits and economies of scale are often positively correlated in market economies. High levels of market share may bestow market power on firms. See also Concentration , Concentration Indexes .58
Merger
An amalgamation or joining of two or more firms into an existing firm or to form a new firm. A merger is a method by which firms can increase their size and expand into existing or new economic activities and markets. A variety of motives may exist for mergers: to increase economic efficiency , to acquire market power , to diversify , to expand into different geographic markets, to pursue financial and R&D synergies, etc. Mergers are classified into three types:
Horizontal Merger : Merger between firms that produce and sell the same products, i.e., between competing firms. Horizontal mergers, if significant in size, can reduce competition in a market and are often reviewed by competition authorities. Horizontal mergers can be viewed as horizontal integration of firms in a market or across markets.
Vertical Merger : Merger between firms operating at different stages of production, e.g., from raw materials to finished products to distribution. An example would be a steel manufacturer merging with an iron ore producer. Vertical mergers usually increase economic efficiency, although they may sometimes have an anticompetitive effect. See also Vertical Integration .
Conglomerate Merger : Merger between firms in unrelated business, e.g., between an automobile manufacturer and a food processing firm.
Mobility Barriers
Mobility barriers are factors which impede the ability of firms to enter or exit an industry, or to move from one segment of an industry to another. "Mobility barriers" is therefore a general term which includes barriers to entry , barriers to exit, and barriers to intra-industry changes in market position. More specifically, mobility barriers may refer to barriers to movement from one strategic group of firms within an industry to another group. See R. Gilbert, "Mobility Barriers and the Value of Incumbency," in R. Schmalensee and R. Willig, (eds), The Handbook of Industrial Organization , North Holland, Amsterdam, 1989.
Monopolistic Competition
Monopolistic competition describes an industry structure combining elements of both monopoly and perfect competition . As in perfect competition, 59 there are many sellers and entry and exit is relatively easy. However, unlike the situation in perfect competition, products are somewhat differentiated. As a consequence, each firm faces a downward sloping demand curve which gives it some power over price. In this sense the firm is like a monopolist, although the demand curve is more elastic than that of the monopolist (see elasticity of demand ). In essence, although the product is differentiated, it does have substitutes so that the demand curve facing the firm will depend on the prices charged by rivals producing similar products. Monopolistic competition is probably the most prevalent market structure, particularly in service industries. Although it can be shown that monopolistic competition is Pareto inefficient because equilibrium price exceeds marginal cost, this inefficiency is the result of producing a variety of products. Because there are many firms and free entry/exit, monopolistic competition is not usually considered a problem for competition policy. In equilibrium, monopolistic competitors earn zero or low economic profits.
Monopolization
Attempts by a dominant firm or group of relatively large firms to maintain or increase market control through various anticompetitive practices such as
predatory pricing , pre-emption of facilities , and foreclosure of competition . See also discussion under abuse of dominant position .
Monopoly
Monopoly is a situation where there is a single seller in the market. In conventional economic analysis, the monopoly case is taken as the polar opposite of perfect competition . By definition, the demand curve facing the monopolist is the industry demand curve which is downward sloping. Thus, the monopolist has significant power over the price it charges, i.e. is a price setter rather than a price taker. Comparison of monopoly and perfectly competitive outcomes reveals that the monopolist will set a higher price, produce a lower output and earn above normal profits (sometimes referred to as monopoly rents). This suggests that consumers will face a higher price, leading to a deadweight welfare loss . In addition, income will be transferred from consumers to the monopoly firm. 60 The preceding arguments are purely static in nature and constitute only part of the possible harm resulting from monopoly. It is sometimes argued that monopolists, being largely immune from competitive pressures, will not have the appropriate incentives to minimize costs or undertake technological change. Moreover, resources may be wasted in attempts to achieve a monopoly position However, a counter argument advanced is that a degree of monopoly power is necessary to earn higher profits in order to create incentives for innovation. Monopoly should be distinguished from market power . The latter is a term which refers to all situations in which firms face downward sloping demand curves and can profitably raise price above the competitive level. Market power may arise not only when there is a monopoly, but also when there is oligopoly ,
monopolistic competition , or a dominant firm .Monopolies can only continue to exist if there are barriers to entry .Barriers which sustain monopolies are often associated with legal protection created through patents and monopoly franchises. However, some monopolies are created and sustained through strategic behaviour or economies of scale. The latter are natural monopolies which are often characterized by steeply declining long-run average and marginal costs and the size of the market is such that there is room for only one firm to exploit available economies of scale. For purposes of competition law and policy, monopoly may sometimes be defined as a firm with less than 100 per cent market share. Different jurisdictions approach "monopoly" in different ways depending upon market share criteria. See also Dominant Firm .
Monopoly Power
See Market Power .
Monopoly Rents See Rent. 61
Monopsony
A monopsony consists of a market with a single buyer. When there are only a few buyers, the market is defined as an oligopsony . In general, when buyers have some influence over the price of their inputs they are said to have monopsony power. Monopsony (or oligopsony) in and of itself is not often of concern in competition policy, although it does imply a lack of competition. It becomes more relevant when combined with monopoly or oligopoly, that is with monopoly power. One common use of the notion of monopsony power arises in the context of defining market structure. For example, in cases where monopoly power is the issue, it may be useful to examine the extent to which such power is offset by powerful buyers. This is sometimes referred to as countervailing power . The ability of a firm to raise prices, even when it is a monopolist, can be reduced or eliminated by monopsony or oligopsony buyers. To the extent that input prices can be controlled in this way, consumers may be better off. A second important use of the concept of monopsony power arises in cases of vertical integration and merger. It is generally agreed that where monopsony power exists, there will be an incentive for vertical integration. Moreover, under some circumstances it can be shown that vertical integration, even when it occurs between a monopolist and a monopsonist (bilateral monopoly), can increase economic efficiency. For further discussion and references, see F.M. Scherer and D. Ross, Industrial Market Structure and Economic Performance, Houghton Mifflin, Boston, 1990, Ch. 14.
Nash Equilibrium
In non-cooperative oligopoly theory it is necessary to model the manner in which firms choose strategies, given the fact that their decisions will affect their rivals. The most common assumption is that each firm chooses its strategy so as to maximize profits, given the profit-maximizing decisions of other firms. The result is a Nash Equilibrium, developed by the game theorist John Nash. A Nash equilibrium is a strategy selection such that no firm can gain by altering its strategy, given the existing strategies of its rivals. Thus, a Nash 62 equilibrium represents a best response by any firm to the given strategies of the others. Consider a duopoly , with each of two firms choosing a strategy. The strategy pair chosen is a Nash equilibrium if firm A’s choice maximizes its profits, given firm B’s choice and firm 2 maximizes its profits given firm 1’s choice. Strategies refer to the decisions firms make. Strategies may involve quantities, prices, or any other relevant decisions (such as R&D, investment, or location). The choice will depend on the nature of the problem. When the strategy analysed involves quantities, the resulting equilibrium is termed a Cournot (Nash) equilibrium . When the strategy involves prices, it is called a Bertrand (Nash) equilibrium .See D. Fudenberg and J. Tirole, "Noncooperative Game Theory for Industrial Organization", and C. Shapiro, "Theories of Oligopoly", both in R. Schmalensee and R. Willig (eds), The Handbook of Industrial Organization ,North Holland, Amsterdam, 1989.
Natural Monopoly
A natural monopoly exists in a particular market if a single firm can serve that market at lower cost than any combination of two or more firms. Natural monopoly arises out of the properties of productive technology, often in association with market demand, and not from the activities of governments or rivals (see monopoly ). Generally speaking, natural monopolies are characterized by steeply declining long-run average and marginal-cost curves such that there is room for only one firm to fully exploit available economies of scale and supply the market. In essence natural monopolies exist because of economies of scale and
economies of scope which are significant relative to market demand. Natural monopolies are thought to exist in some portions of industries such as electricity, railroads, natural gas, and telecommunications. Because productive efficiency requires that only one firm exist, natural monopolies are typically subject to government regulation. Regulations may include price, quality, and/or entry conditions. For more detail, with references, see W.W. Sharkey, The Theory of Natural Monopoly , Cambridge University Press, Cambridge, UK, 1982 and M. Waterson, "Recent Developments in the Theory of Natural Monopoly," Journal of Economic Surveys , Vol. 1, No. 1, 1987. 63
Negative Externality
See Externalities. 141. Non-Price Predation
Non-price predation is a form of strategic behaviour that involves raising rivals’ costs. It is potentially less costly and hence more profitable than predatory pricing . Typical methods include using government or legal processes to disadvantage a competitor. A firm may be able to force competitors to incur significant litigation or administrative costs, at little cost to itself. See D W. Carlton and J. M. Perloff, Modern Industrial Organization , Scott, Foresman Little Brown, Glenview, Il., 1990, Ch. 13.
Oligopoly
An oligopoly is a market characterized by a small number of firms who realize they are interdependent in their pricing and output policies. The number of firms is small enough to give each firm some market power .Oligopoly is distinguished from perfect competition because each firm in an oligopoly has to take into account their interdependence; from monopolistic competition because firms have some control over price; and from monopoly
because a monopolist has no rivals. In general, the analysis of oligopoly is concerned with the effects of mutual interdependence among firms in pricing and output decisions. There are several types of oligopoly. When all firms are of (roughly) equal size, the oligopoly is said to be symmetric. When this is not the case, the oligopoly is asymmetric. One typical asymmetric oligopoly is the dominant firm . An oligopoly industry may produce goods which are homogeneous/ undifferentiated or it may produce goods which are heterogeneous/ differentiated. The analysis of oligopoly behaviour normally assumes a symmetric oligopoly, often a duopoly . Whether the oligopoly is differentiated or undifferentiated, the critical problem is to determine the way in which the firms act in the face of their realized interdependence. 64 In general, there are two broad approaches to this problem. The first is to assume that firms behave cooperatively. That is, they collude in order to maximize
joint monopoly profits . The second is to assume that firms behave independently or non-cooperatively. The analysis of oligopoly behaviour under the non-cooperative assumption forms the basis of oligopoly theory. Within non-cooperative oligopoly theory a distinction is made between models in which firms choose quantities and those in which they choose prices. Quantity-setting models are often referred to as Cournet models and price-setting models as Bertrand models. For a useful survey of non-cooperative oligopoly models, see C. Shapiro, "Theories of Oligopoly Behavior" in R. Schmalensee and R. Willig (eds.), The Handbook of Industrial Organization , North Holland, Amsterdam, 1989.
Oligopsony
See Monopsony. 144. Opportunity Costs (or Alternative Costs )An essential concept in economics whereby the cost of using a resource in one activity is measured in terms of its best alternative use. The opportunity or alternative cost of producing one unit of commodity Y is what must be sacrificed by employing resources to produce it rather than something else. If several opportunities are given up in this manner, the relevant cost is the value assigned to the best (or highest) alternative. Opportunity costs are often referred to as "implicit costs" and while the concept is central to economics, they are not easy to measure. Cash outlays are "explicit costs" and are measured in terms of conventional accounting principles. See also Costs .
Ownership Concentration
See Concentration. 65
Package Tie-in
See Bundling. 147. Parent
A company which owns or operates a number of other companies, known as
subsidiaries . A parent firm can be a holding company but it loses that status if it actively operates its subsidiaries .
Pareto Efficiency
Pareto efficiency, also referred to as allocative efficiency , occurs when resources are so allocated that it is not possible to make anyone better off without making someone else worse off. When referring to a situation as Pareto efficient, it is usually assumed that products are being produced in the most efficient (least-cost) way. Pareto optimality is sometimes used interchangeably with Pareto efficiency. Sometimes Pareto optimality is reserved for cases when both production and allocative efficiency are obtained.
Deadweight welfare loss is a measure of allocative inefficiency. In the case considered above under that heading, the total loss of consumer surplus involved in moving from competition to monopoly was P cPmBC of which BCE was deadweight loss and P cPmBE was producers’ profit. Now consider the movement from monopoly to competition. The gain in consumers’ surplus is P cPmBC, while producers lose P cPmBE. However, it is potentially possible for consumers to compensate producers by this amount and still retain BCE. Thus consumers are potentially better off, producers are no worse off and so the movement to competition represents a Pareto improvement and competition is said to be Pareto efficient. This result has been termed "the first theorem of welfare economics" and it states that an economy characterized by perfect competition in all markets will always be Pareto efficient, if there are no market failures .66
Patents
Patents give inventors property rights to the exclusive use of their invention for a specified period of time. The profits stemming from a patent are socially useful because they encourage inventive activity. In the absence of patents, competitive industries may produce too few inventions. Investments in inventive activity are sunk costs , and without the protection of patents to allow inventors to recoup those investments, inventive activity would probably decline. See discussion under Intellectual Property Rights and Licensing . See also F.M. Scherer and D. Ross, Industrial Market Structure and Economic Performance , 3rd edition, Houghton Mifflin, Boston, 1990, Ch. 17.
Perfect Competition
Perfect competition is defined by four conditions (in a well-defined market): a) There is such a large number of buyers and sellers that none can individually affect the market price. This means that the demand curve facing an individual firm is perfectly elastic. See Elasticity of Demand .b) In the long run, resources must be freely mobile, meaning that there are no barriers to entry and exit. c) All market participants (buyers and sellers) must have full access to the knowledge relevant to their production and consumption decisions. d) The product should be homogenous. When these conditions are fulfilled in any well-defined market, the market is perfectly competitive; when they are fulfilled in all markets, the economy is perfectly competitive. It is this definition of perfect competition which underlies the conclusion that a perfectly competitive economy is Pareto efficient . Under these conditions, the price of the goods produced = marginal cost and all goods will be produced in the least costly way. It is evident that this notion of competition can be highly restrictive in terms of policy-making. Some economists have therefore argued that the goal of 67 competition policy should not be perfect competition, but a more realistic target such as workable competition .Another drawback to the use of perfect competition as a policy goal is that it is not clear that perfect competition is desirable unless it can be achieved in all markets. See Second Best, Theory of .
Per Se Illegal
See Rule of Reason. 152. Positive Externality See Externalities. 153. Predatory Pricing
A deliberate strategy, usually by a dominant firm, of driving competitors out of the market by setting very low prices or selling below the firm’s incremental costs of producing the output (often equated for practical purposes with average variable costs). Once the predator has successfully driven out existing competitors and deterred entry of new firms, it can raise prices and earn higher profits. The economic literature on the rationality and effectiveness of predatory pricing is in a state of flux. Many economists have questioned the rationality of predatory pricing on grounds that: it can be at least as costly to the predator as to the victim; targets of predation are not easily driven out, assuming relatively efficient capital markets; and entry or re-entry of firms in the absence of barriers reduces the predator’s chances of recouping losses incurred during the period of predation. However, other economists have suggested that price predation might be feasible if it is undertaken to "soften" up rivals for future acquisition, or if potential targets of predation or their sources of capital have less information about costs and market demand than the predator. See Predatory Pricing , OECD, Paris, 1989, for a summary of issues, concepts, and operational approaches toward this practice. 68
Preemption of Facilities
See Barriers to Entry , Anticompetitive Practices. 155. Price Cartel
See Cartel. 156. Price Discrimination
Price discrimination occurs when customers in different market segments are charged different prices for the same good or service, for reasons unrelated to costs. Price discrimination is effective only if customers cannot profitably re-sell the goods or services to other customers. Price discrimination can take many forms, including setting different prices for different age groups, different geographical locations, and different types of users (such as residential vs. commercial users of electricity). Where sub-markets can be identified and segmented then it can be shown that firms will find it profitable to set higher prices in markets where demand is less elastic (see Elasticity of Demand ). This can result in higher total output, a pro-competitive effect. Price discrimination can also have anti-competitive consequences. For example, dominant firms may lower prices in particular markets in order to eliminate vigorous local competitors. However, there is considerable debate as to whether price discrimination is really a means of restricting competition. Price discrimination is also relevant in regulated industries where it is common to charge different prices at different time periods (peak load pricing) or to charge lower prices for high volume users (block pricing). For more details, see L. Philips, The Economics of Price Discrimination ,Cambridge University Press, Cambridge, UK, 1983; and F.M. Scherer and D. Ross, Industrial Market Structure and Economic Performance , Houghton Mifflin, Boston, 1990, Ch. 13. 69
Price Fixing Agreement
An agreement between sellers to raise or fix prices in order to restrict inter-firm competition and earn higher profits. Price fixing agreements are formed by firms in an attempt to collectively behave as a monopoly . For further details see discussion under agreement , cartel , collusion and other headings indicated therein.
Price Leadership
Prices and price changes established by a dominant firm , or a firm accepted by others as the leader, and which other firms in the industry adopt and follow. When price leadership is adopted to facilitate collusion, the price leader will generally tend to set a price high enough that the least efficient firm in the market may earn some return above the competitive level.
Price Regulation
The policy of setting prices by a government agency, legal statute or regulatory authority. Under this policy, minimum and/or maximum prices may be set. Price regulation also encompasses "guidelines" which specify the magnitude by which prices can increase as in the case of rent controls. The bases on which regulated prices are set vary. These may be on costs, return on investment, markups, etc.
Producers’ Surplus
See Deadweight Welfare Loss. 161. Privatization
Refers to transfer of ownership and control of government or state assets, firms and operations to private investors. This transfer takes the form of issue and sale or outright distribution of shares to the general public. Broadly used, the term privatization includes other policies such as "contracting out" that is, the process by which activities, while publicly organized and financed, are carried out by private sector companies, e.g., street cleaning, garbage collection, housing, 70 education. The policy of privatization has been extensively implemented in the United Kingdom and since adopted in several countries around the world. See collection of papers in J.A. Kay, C.P. Mayer and D.J. Thompson, Privatization and Regulation - the U.K. Experience , Oxford University Press, Oxford, 1986; and J. Vickers and G. Yarrow, Privatization - an Economic Analysis , MIT Press, Cambridge, 1988. See also Regulatory Reform, Privatisation and Competition Policy , OECD, Paris, 1992.
Product Differentiation
Products are considered to be differentiated when there are physical differences or attributes which may be real or perceived by buyers so that the product is preferred over that of a rival firm. Products are differentiated by firms in order to obtain higher prices and/or increased sales. Differentiation may occur in terms of physical appearance, quality, durability, ancillary services (e.g., warranties, post-sales services and information), image and geographic location. Firms will frequently engage in advertising and sales promotion activities to differentiate their products. Product differentiation can give rise to barriers to entry but then it may also facilitate entry into and penetration of markets by firms with products which buyers may prefer over existing ones. It should be noted that differentiated products are not to be confused with heterogenous products. The latter generally refers to products which are different and not easily substitutable whereas among differentiated products there is some degree of substitutability. See also Homogenous Products .
Profit
In economic theory, profit is the surplus earned above the normal return on capital. Profits emerge as the excess of total revenue over the opportunity cost of producing the good. Thus, a firm earning zero economic profits is still earning a normal or competitive return. Positive economic profits therefore indicate that a firm is earning more than the competitive norm. Economic profits are not the same as accounting profits. In accounting, profits are simply the excess of revenues over the explicit costs of obtaining the revenues. Costs are not calculated as opportunity costs and do not include a normal return on capital. Moreover, accountants calculate different categories of profits which may differ from country to country. 71 For purposes of competition policy, the problem is that positive economic profits may (but not necessarily) indicate the existence of monopoly power .However, economic profits are not observable and use must be made of accounting profits. Positive accounting profits may reflect nothing other than a normal or competitive return. See Profitability .
Profitability
Measures of profitability figure prominently in both the empirical literature in industrial organization and in the resolution of anti-trust cases. At issue is the extent to which observed (accounting) measures of profitability can indicate the presence of monopoly power. A variety of measures of profitability have been employed. Rates of return on equity or assets are defined as accounting profits divided by either equity or assets. Profits may be calculated before- or after-tax and may or may not include interest payments. Normally, interest payments are excluded when calculating the rate of return on equity, but are included when calculating the rate of return on assets. The rate of return on assets reflects operating results and, if interest rates are included, should not reflect financing decisions. Many empirical studies have employed the price-cost margin, defined as revenues less variable costs divided by revenues. This measure typically excludes various capital costs, but is defended on the grounds that it is related to the Lerner Index .Finally, some use has been made of Tobin’s "q", defined as the market value of a firm divided by the replacement costs of its tangible assets. The market value of a firm is determined in stock markets. To the extent that stock markets capture the long-run profitability of a firm, then higher "q" values reflect greater profitability. The question of whether any of these measures can be employed to measure economic profits (see profit ) has been widely debated. Moreover, even if it can be determined that they do, there is considerable controversy as to whether higher levels of profitability reflect the exercise of market power or the returns to superior efficiency and skills. See R. Schmalensee, "Inter-Industry Studies of Structure and Performance," in R. Schmalensee and R. Willig (eds), The Handbook of Industrial Organization , North Holland, Amsterdam, 1989. 72
Quasi-rents
See Rent. 166. Rationalization Agreement
An agreement (generally approved or authorized by government) between firms in an industry to close down inefficient plants, reduce excess capacity and realign production in order to increase overall industry efficiency and performance.
Reciprocity
A form of bilateral (or multilateral) arrangement between firms to bestow favourable terms on, or buy and sell from, each other to the exclusion of others. This may have the effect of limiting competition and/or preventing the entry of firms into certain markets. Concern about reciprocal arrangements has been particularly raised in the context of conglomerates. It is argued that subsidiary firms are likely to encounter each other frequently as buyers or sellers in different markets. Reciprocity may benefit firms by ensuring contract fulfillment or by facilitating secret price-cutting.
Recommended or Suggested Price
In several industries, suppliers may recommend or suggest the price at which a product may be resold. In certain cases the supplier may indicate the "maximum" price for the product in order to discourage retailers from raising prices to increase their own margins and thus reduce total sales. Such practices may be adopted in order to avoid violating laws against resale price maintenance .The specification and attempted enforcement of "minimum" prices for products is illegal in many countries.
Refusal to Deal/Sell
The practice of refusing or denying supply of a product to a purchaser, usually a retailer or wholesaler. The practice may be adopted in order to force a retailer to engage in resale price maintenance (RPM) , i.e., not to discount the 73 product in question, or to support an exclusive dealing arrangement with other purchasers or to sell the product only to a specific class of customers or geographic region. Refusal to deal/sell may also arise if the purchaser is a bad credit risk, does not carry sufficient inventory or provide adequate sales service, product advertising and display, etc. The competitive effects of refusal to deal/sell generally have to be weighed on a case-by-case basis.
Regulation
Broadly defined as imposition of rules by government, backed by the use of penalties that are intended specifically to modify the economic behaviour of individuals and firms in the private sector. Various regulatory instruments or targets exist. Prices, output, rate of return (in the form of profits, margins or commissions), disclosure of information, standards and ownership ceilings are among those frequently used. Different rationales for economic regulation have been put forward. One is to curb potential market power and increase efficiency or avoid duplication of facilities in cases of natural monopoly . Another is to protect consumers and maintain quality and other standards including ethical standards in the case of professional services provided by doctors, lawyers, etc. Regulations may also be enacted to prevent excessive competition and protect suppliers from unstable output and low price conditions, to promote employment and more equitable distribution of income. Excessive competition, sometimes also called
ruinous competition, is a controversial term without precise meaning in economics. It usually refers to a condition of excess capacity and/or declining demand in an industry, which causes prices to fall to the level of average variable costs, discouraging new investment and causing some incumbents to leave the industry until capacity is reduced to the point where supply once again intersects with demand at a price sufficient to cover all costs. When regulatory authorities interfere with this process by setting minimum price levels, excess capacity and its attendant resource misallocation will tend to persist in the industry. Many economists use this as an example of the use of regulation to promote the private interests of producers at the expense of the public interest. Not all forms of regulation have to be mandated or imposed by government. Many professions adopt self-regulation , i.e., develop and self-enforce rules commonly arrived at for the mutual benefit of members. Self-regulation may be adopted in order to maintain professional reputation, education and ethical standards. They may also act as a vehicle to set prices, restrict entry and ban certain practices (e.g., advertising in order to restrict competition). 74
Deregulation refers to the relaxation or removal of regulatory constraints on firms or individuals. Deregulation has become increasingly equated with promoting competition and market-oriented approaches toward pricing, output, entry and other related economic decisions. See P.L. Joskow and N. Rose, "The Effects of Economic Regulation," in R. Schmalensee and R. Willig (eds), Handbook of Industrial Organization , North Holland, Amsterdam, 1989. Competition Policy and the Professions OECD, Paris, 1985; Regulatory Reform, Privatisation and Competition Policy, OECD, Paris, 1992.
Rent
In modern economics, rent refers to the earnings of factors of production (land, labour, capital) which are fixed in supply. Thus, raising the price of such factors will not cause an increase in availability but will increase the return to the factor. This differs from the more common usage of the term, whereby rent refers to payments for the use of a resource. Economists use the term economic rent to denote the payment to factors which are permanently in fixed supply and quasi-rent to denote payments for factors which are temporarily in fixed supply. The presence of economic rents implies that the factor can neither be destroyed nor augmented. Quasi-rents exist when factors can be augmented over time, or when their supply can be reduced over time through depreciation. Factors which earn economic or quasi-rents typically are paid an amount in excess of their opportunity costs .In the case of economic rents the supplier receives a payment in excess of the amount required to induce the supplier to supply the factor. Quasi-rents, on the other hand, are returns in excess of that required to keep the factor active, but may not be sufficient to have induced the supplier to enter in the first place. When the availability of a good is artificially restricted (for example by laws limiting entry), then the increased earnings of the remaining suppliers are termed monopoly rents. The potential existence of monopoly rents provides an incentive for firms to pay for the right to earn these rents. See Rent-Seeking .For further details, see A.A. Alchian, "Rent," in J. Eatwell, M. Milgate and P. Newman, The New Palgrave: A Dictionary of Economics, Macmillan, London, 1987. 75
Rent Seeking
The opportunity to capture monopoly rents (see Rents ) provides firms with an incentive to use scarce resources to secure the right to become a monopolist. Such activity is referred to as rent-seeking. Rent-seeking is normally associated with expenditures designed to persuade governments to impose regulations which create monopolies. Examples are entry restrictions and import controls. However, rent-seeking may also refer to expenditures to create private monopolies. The notion of rent-seeking may be traced to economists Tullock and Krueger, who both assumed that the potential for monopoly rents would induce firms to compete for the right to earn such rents. The outcome would then be that resources equal to the monopoly rents would be expended on securing the monopoly. This, they suggested, was a social loss equivalent to the amount of monopoly rents (or profits), since the resources could have been better employed in other activities. Hence it is argued that the social costs of monopoly power should include at least some portion of monopoly profits. See Deadweight Loss .See D.W. Carlton and J.M. Perloff, Modern Industrial Organization, Scott, Foresman/Little Brown, Glenview, Il., 1990, Ch. 13.
Resale Price Maintenance (RPM)
A supplier specifying the minimum (or maximum) price at which the product must be re-sold to customers. From a competition policy viewpoint, specifying the minimum price is of concern. It has been argued that through price maintenance, a supplier can exercise some control over the product market. This form of vertical price fixing may prevent the margin from retail and wholesale prices from being reduced by competition. However, an alternative argument is that the supplier may wish to protect the reputation or image of the product and prevent it from being used by retailers as a loss leader to attract customers. Also, by maintaining profit margins through RPM, the retailer may be provided with incentives to spend greater outlays on service, invest in inventories, advertise and engage in other efforts to expand product demand to the mutual benefit of both the supplier and the retailer. RPM may also be used to prevent free riding by retailers on the efforts of other competing retailers who instead of offering lower prices expend time, money and effort promoting and explaining the technical complexities or attributes of the product. For example, one retailer may not reduce 76 price but explain and demonstrate to customers the use of a complex product such as a computer. The customer may after acquiring this information choose to buy the computer from a retailer that sells it at a lower price and does not explain or demonstrate its uses. In many countries, RPM is per se illegal with few exceptions or exempt products. Many economists now advocate adopting a less stringent approach in competition law towards RPM and other vertical restraints . See
Competition Policy and Vertical Restraints: Franchising , OECD, Paris, forthcoming in 1993, for a brief synthesis of some of the economic debates in this area.
Restriction of Entry to the Market
See Barriers to Entry , Limit Pricing. 175. Restriction of Technology
See Licensing. 176. Restriction on Exportation
Restrictions placed on the ability of firms to export. Such restrictions may come from governments, normally to protect or conserve non-renewable resources or cultural treasures. They may also come from agreements among firms to limit exports as part of a cartel arrangement. These restrictions may also arise from agreements negotiated by the importing country such as the case of "voluntary" export restrictions (VER’s) of Japanese automobiles to the United States. Finally export restrictions may be part of licensing arrangements whereby the firm granted the license is not allowed to export the good in competition with other licensees, or the firm selling the license. See also Export Cartel .
Restriction on Importation
Measures, normally adopted by governments, which restrict the ability of firms to enter foreign markets via imports. The most common restrictions are tariffs, quotas and voluntary export restraints. Tariffs serve to tax imports, thus making them expensive relative to domestic goods. Quotas affect imports directly by restricting the number of units which can come from abroad. Voluntary export 77 restraints, largely confined to the U.S., are similar to quotas in that they restrict quantities. They differ from quotas in that they are not imposed unilaterally by the importing country. Rather, they are agreed to by the exporting country or countries, usually to forestall the imposition of tariffs and/or quotas.
Revenues
Revenues (or total revenue) refer to the value of output sold, that is the number of units times the price per unit. Average revenue is revenue per unit, that is total revenue divided by the amount of output sold. Average revenue is therefore equal to price per unit. Marginal revenue is the increment in total revenue resulting from the sale of an additional unit. Marginal revenue may or may not equal average revenue. A firm operating under perfect competition has no control over price. It must sell all units at the same price. Hence marginal revenue equals price (which is average revenue) and these are constant. A firm with market power , on the other hand, faces a downward sloping demand curve. In order to sell more, it must reduce price. Thus average revenue (price) is declining. Moreover, it can be shown that marginal revenue is not only declining, but is less than average revenue. The profit maximizing firm sets marginal cost (see costs ) equal to marginal revenue. For the perfectly competitive firm, the result is that price equals marginal cost, the condition for Pareto efficiency . For a monopolist, price exceeds marginal cost and is therefore higher than that of the perfectly competitive firm. See monopoly , perfect competition .
Ruinous Competition
See Cut-Throat Competition. 180. Rule of Reason
A legal approach by competition authorities or the courts where an attempt is made to evaluate the pro-competitive features of a restrictive business practice against its anticompetitive effects in order to decide whether or not the practice 78 should be prohibited. Some market restrictions which prima facie give rise to competition issues may on further examination be found to have valid efficiency-enhancing benefits. For example, a manufacturer may restrict supply of a product in different geographic markets only to existing retailers so that they earn higher profits and have an incentive to advertise the product and provide better service to customers. This may have the effect of expanding the demand for the manufacturer’s product more than the increase in quantity demanded at a lower price. The opposite of the rule of reason approach is to declare certain business practices per se illegal, that is, always illegal. Price fixing agreements and resale price maintenance in many jurisdictions are per se illegal.
Second Best, Theory of
The theory of the second best suggests that when two or more markets are not perfectly competitive, then efforts to correct only one of the distortions may in fact drive the economy further away from Pareto efficiency . Thus, for example, if there is one industry which can never satisfy all the conditions for perfect competition , it is no longer clear that the optimal policy is to move the remaining industries towards perfect competition. Moreover, the conditions under which
Pareto efficiency can be achieved under these circumstances are complex and not likely to be implementable. Thus, the defense of competition policy often requires giving weight to more than Pareto efficiency . For example, competition policy may be defended on the grounds of equity, democracy and incentives. However, achievement towards Pareto efficiency is generally given more weight in the application of competition policy. A useful discussion, with references, is found in F.M. Scherer and D. Ross,
Industrial Market Structure and Economic Performance , Houghton Mifflin, Boston, 1990, pp. 33-38.
Self-Regulation
See Regulation. 79
Seller Concentration
See Concentration. 184. Selling Below Cost
A practice whereby a firm sells products at less than costs of manufacture or purchase in order to drive out competitors and/or to increase market share. This practice may arise partly because of deep pockets or cross-subsidization using profits derived from sale of other products. A number of measurement issues arise as to what constitutes costs but generally the practice would arise if price is below
marginal cost or average variable cost . A question also arises as to whether selling a product below costs is economically feasible over a long period of time since the firm may incur high costs in the form of loss of potential profits. See also
loss leadering , predatory pricing .
Shared or Joint Monopoly
Anticompetitive behaviour by firms, normally an oligopoly , in order to secure monopoly profits for the firms as a group. Essentially, shared monopoly requires some form of collusion but stops short of being a formal cartel . It is therefore similar to tacit collusion . In a shared monopoly firms may not compete for the same customers and have instead local monopolies. Since in theory industry profits under a non-coordinated oligopoly will be less than those under monopoly , there is some incentive for firms in an oligopoly to attempt to coordinate their actions so as to achieve profits nearer the monopoly solution.
Shipping Conferences
Refers to shipping companies that have formed an association to agree on and set freight rates and passenger fares over different shipping routes. There are different shipping conferences for different regions of the world. Shipping conferences, aside from setting rates, adopt a wide number of policies such as allocation of customers, loyalty contracts, open pricing contracts, etc. Historically, eastern bloc country shipping lines have not joined these conferences. In many 80 jurisdictions, shipping conferences are exempt from the application of competition laws but this position is being increasingly changed in order to promote greater competition and choice for shippers (exporters).
Specialization Agreements
A provision permitting firms to form an agreement to specialize in the production of a narrow or specific range of product lines in order to realize "product specific economies" (see Economies of Scale ). In several industries, firms may be manufacturing at sub-optimal scales, multiple and duplicate products. Specialization agreements are aimed towards facilitating re-alignment of production in order to achieve longer production runs of specific products and realize efficiencies. A formal provision for these agreements and exemption from the application of competition laws may be necessary in order to assure firms they will not be viewed as forming an illegal combination . Specialization agreements are particularly relevant in the context of small economies where the market may not be large enough for firms to exploit potential product specific economies of scale.
Standards
Refers to defining and establishing uniform specifications and characteristics for products and/or services. In the case of manufactured products, the standard may relate to physical measurements and dimensions, materials and performance attributes. For example, a standard bath tub may be established as measuring 162 cm by 74 cm, being made of either enamelled metal or fiberglass and having performance capability of carrying human body and water weight of minimum 180 kg. A distinction is usually made between technical and performance standards. The specifications that the bath tub should be of certain dimensions and made of enamelled metal or fiberglass are considered as technical standards. The weight and capacity capabilities are the performance standards. A distinction is also made between "voluntary" and "mandatory" standards; the former are generally developed by industry associations and member firms may voluntarily adopt them. Mandatory standards are usually those developed by government agencies or regulation and are compulsory. Standardization of products often promotes economies of scale in production, interchangeability between products of different manufacture, higher quality, complementarity between different products, and diffusion of technology. 81 Standards may also reduce product heterogeneity and facilitate collusion and/or act as a non-tariff barrier to trade. Standards may also be used by incumbent firms in favour of their own products and processes and raise barriers to entry .In the case of services, many professions and trades ranging from medical doctors to carpenters set minimum standards for granting licenses to practice. While these licenses are likely to raise the average quality of the service, they also have the effect of restricting supply and increasing prices. If standards are to enhance economic welfare, the standard setting and certification procedures must be transparent and subject to checks and balances such that the influence of any one particular interest group does not dominate. Also, standards should be reviewed periodically and updated.
Strategic Behaviour
Strategic behaviour is the general term for actions taken by firms which are intended to influence the market environment in which they compete. Strategic behaviour includes actions to influence rivals to act cooperatively so as to raise joint profits, as well as noncooperative actions to raise the firm’s profits at the expense of rivals. Various types of collusion are examples of cooperative strategic behaviour. Examples of noncooperative strategic behaviour include pre-emption of facilities , price and non-price predation and creation of artificial barriers to entry . Strategic behaviour is more likely to occur in industries with small numbers of buyers and sellers. See D.W. Carlton and J.M. Perloff, Modern Industrial Organization, Scott, Foresman/Little Brown, Glenview, Il., 1990, Ch. 13.
Subsidiary
A company controlled by another company. Control occurs when the controlling company owns more than 50 per cent of the common shares. When the parent owns 100 per cent of the common shares, the subsidiary is said to be wholly-owned. When the subsidiary operates in a different country, it is called a foreign subsidiary. The controlling company is called a holding company or
parent . A subsidiary is a corporation with its own charter and is not a division of the controlling company. 82
Substantial Lessening of Competition
See Market Power. 192. Sunk Costs
Sunk costs are costs which, once committed, cannot be recovered. Sunk costs arise because some activities require specialized assets that cannot readily be diverted to other uses. Second-hand markets for such assets are therefore limited. Sunk costs are always fixed costs (see costs ), but not all fixed costs are sunk. Examples of sunk costs are investments in equipment which can only produce a specific product, the development of products for specific customers, advertising expenditures and R&D expenditures. In general, these are firm-specific assets. The absence of sunk costs is critical for the existence of contestable markets . When sunk costs are present, firms face a barrier to exit. Free and costless exit is necessary for contestability . Sunk costs also lead to barriers to entry . Their existence increases an incumbents’ commitment to the market and may signal a willingness to respond aggressively to entry.
Sustainable Monopoly
See Contestability. 194. Tacit Collusion
See Collusion , Conscious Parallelism. 195. Takeover
The acquisition of control of one company by another or occasionally by an individual or group of investors. Takeovers are usually instituted by purchasing shares at a "premium" over existing prices and may be financed in a variety of ways including cash payment and/or with shares of the acquiring company. While the terms mergers , acquisitions and takeover are often used interchangeably, 83 there are subtle differences between them. A takeover may be complete or partial and may not necessarily involve merging the operations of the acquired and acquiring firms. The fact that joint ownership and control may arise from a takeover implies that the companies could maximize joint profits, which can be a source of concern to competition authorities. See also Market for Corporate Control .
Tied Selling
Refers to situations where the sale of one good is conditioned on the purchase of another good. One variant is full-line forcing in which a seller presses (or forces) a complete line of products on a buyer who is predominantly interested in only a specific product. Tied selling is sometimes a means of price discrimination. Competition concerns have been expressed that tying may foreclose opportunities for other firms to sell related products or may increase
barriers to entry for those that do not offer a full line of products. An opposite view is that these practices are efficiency driven i.e., used to reduce costs of producing and distributing the line of products and ensuring that like quality products are used to complement the product being sold. For example, a computer manufacturer may require purchase of disks in order to prevent damage to or poor performance of his equipment by the use of substitute lower quality disks. There is increasing recognition that depending on different market situations, tied selling arrangements may have a valid business rationale. In the administration of competition policy, an increasing number of economists suggest adopting a rule of reason approach to tied selling.
Total Costs
See Costs. 198. Trade Mark
Trade mark refers to words, symbols or other marks which are used by firms to distinguish their products or services from those offered by others. A trade mark may be registered under the Patent Act or the Trademark Act or other such intellectual property legislation as may be applicable. A trade mark may often become equated with the product itself and may be a source of competitive advantage. For example, "kleenex" as a trade mark name is used to refer to "tissue 84 "handkerchiefs"; "Xerox" in place of "photocopying"; "Coke" instead of a "cola drink". Trade marks may communicate information about the quality of a good or service to consumers. Firms which license their trade marks to retailers may thus require conditions in the licensing contract assuring uniform quality. See
Intellectual Property Rights, Licensing .
Transaction Costs
Transaction costs refer to the costs involved in market exchange. These include the costs of discovering market prices and the costs of writing and enforcing contracts. Transaction cost economics, as developed primarily by economists Coase and Williamson, suggests that economic organizations emerge from cost-minimizing behaviour (including transaction costs) in a world of limited information and opportunism. Transaction-cost analysis has been used to explain vertical integration, multinational enterprises, and franchising. See O. Williamson, "Transaction Cost Economics" in R. Schmalensee and R. Willig (eds), The Handbook of Industrial Organization , North Holland, Amsterdam, 1989.
Uniform Delivered Pricing
See Basing Point .
Variable Costs
See Costs. 202. Vertical Integration
Describes the ownership or control by a firm of different stages of the production process, e.g., petroleum refining firms owning "downstream" the terminal storage and retail gasoline distribution facilities and "upstream" the crude oil field wells and transportation pipelines. "Forward" integration refers to the production to distribution stages whereas "backward" integration refers to the 85 production to raw material stages of the operations of a firm. Vertical integration may be achieved through new investment and/or vertical mergers and acquisition
of existing firms at different stages of production. An important motive for vertical integration is efficiencies and minimization of transaction costs .
Vertical Merger
See Merger. 204. Vertical Restraints (or Restrictions)
Refers to certain types of practices by manufacturers or suppliers relating to the resale of their products. The usual practices adopted in this regard are resale price maintenance (RPM) , exclusive dealing and exclusive territory or geographic market restrictions. Under exclusive dealing and/or exclusive territory, a single distributor is the only one who obtains the rights from a manufacturer to market the product. A significant debate exists in the economic literature as to whether this confers monopoly power on the distributor. Usually, the distributor’s market power is limited by inter-brand competition. The manufacturer’s purpose is normally to provide incentives to the distributor to promote the product and provide better service to customers. See also discussion under Free Riding , RPM
and Competition Policy and Vertical Restraints: Franchising , OECD, Paris, forthcoming in 1993.
Workable Competition
Workable competition is a notion which arises from the observation that since perfect competition does not exist, theories based on it do not provide reliable guides for competition policy. The idea was first enunciated by economist J.M. Clark in 1940. He argued that the goal of policy should be to make competition "workable," not necessarily perfect. He proposed criteria for judging whether competition was workable, and this provoked a series of revisions and counter-proposals. The criteria put forward are wide ranging e.g. the number of firms should be at least large as economies of scale permit, promotional expenses should not be excessive and advertising should be informative. 86 No consensus has arisen over what might constitute workable competition but all bodies which administer competition policy in effect employ some version of it. An interesting discussion is found in G. Reid, Theories of Industrial Organization , Blackwell, Oxford, 1987, Ch. 7. See also F.M. Scherer and D. Ross,
Industrial Market Structure and Economic Performance, Houghton Mifflin, Boston, 1990, pp. 53-54.
X-Efficiency
See Efficiency, X-Inefficiency .
X-Inefficiency
In The Wealth of Nations published in 1776, Adam Smith observed that "Monopoly... is a great enemy to good management." This insight explicitly recognized that the problem of monopoly is not only one of price but also one of costs. While monopoly may provide the basis for extracting higher prices from customers, the lack of competitive stimulus may raise the costs of producing the goods and services it sells. The lack of incentives or competitive pressures may lead monopolistic firms to neglect minimizing unit costs of production, i.e., to tolerate "X-inefficiency" (phrase coined by H. Leibenstein). Included in X-inefficiency are wasteful expenditures such as maintenance of excess capacity, luxurious executive benefits, political lobbying seeking protection and favourable regulations, and litigation. See H. Leibenstein, "Allocative Efficiency vs. X-Efficiency," American Economic Review , Vol. 56, June, 1966, pp. 392-415. For a contrary viewpoint see G.J. Stigler, "The Xistence of X-Efficiency," American Economic Review , Vol. 66, March 1976, pp. 213-216. Stigler’s objection is based on his observation that there is no economic theory which predicts that monopolists will not maximize profits by producing efficiently. See also H. Leibenstein’s reply, "X-Inefficiency Xists - Reply to an Xorcist," American Economic Review , Vol. 68, March 1978, pp. 203-211. 87
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Antitrust Bulletin , Vol. X, Jan.-Apr., pp. 75-104. SPULBER, D.F. (1989), Regulation and Markets, MIT Press, Cambridge. STIGLER, G.J. (1964), "A Theory of Oligopoly", Journal of Political Economy ,Vol. 72(1), February, pp. 44-61. STIGLER, G.J. (1966), "The Xistence of X-Efficiency", American Economic Review , Vol. 66, March, pp. 213-216. VICKERS, J. and YARROW, G. (1988), Privatization - an Economic Analysis ,MIT Press, Cambridge. 90 WATERSON, M. (1987), "Recent Developments in the Theory of Natural Monopoly", Journal of Economic Surveys , Vol. 1, No. 1. WERDEN, G. (1983), "Market Delineation and the Justice Department’s Merger Guidelines", Duke Law Review , June, pp. 514-579. WILLIAMSON, O. (1989), "Transaction Cost Economics" in R. Schmalensee and R. Willig (eds.), op. cit. |
4719 | https://fiveable.me/key-terms/principles-physics-iii-thermal-physics-waves/drude-model | Drude Model - (Principles of Physics III) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Principles of Physics III
Drude Model
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key term - Drude Model
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Definition
The Drude Model is a classical theory that describes the electrical and thermal properties of metals by treating conduction electrons as a gas of free particles that can move freely through a lattice of positively charged ions. This model provides insight into the behavior of electrons in conductive materials, linking their motion to properties such as electrical conductivity and heat capacity.
5 Must Know Facts For Your Next Test
The Drude Model assumes that conduction electrons move freely and randomly between collisions with fixed ions, leading to a mean free path that describes the average distance traveled before a collision.
This model predicts that the electrical conductivity of metals is proportional to the number of free electrons available for conduction, which varies among different metals.
The Drude Model also explains thermal conductivity, stating that both electrical and thermal conductivities are related due to the movement of free electrons transferring energy through collisions.
One limitation of the Drude Model is its inability to account for quantum effects, which become significant at very small scales or high temperatures, leading to the need for more advanced models like quantum mechanics.
Despite its limitations, the Drude Model laid the foundation for modern solid-state physics and is still used as a starting point for understanding complex behaviors in materials.
Review Questions
How does the Drude Model explain the electrical conductivity of metals?
The Drude Model explains electrical conductivity by treating conduction electrons as free particles that can move through a lattice of positive ions. When an electric field is applied, these electrons gain drift velocity, causing a net current flow. The model states that conductivity depends on the density of these free electrons and their mean free path between collisions, which together determine how effectively they can conduct electricity.
Compare and contrast the Drude Model with Band Theory regarding electron behavior in solids.
The Drude Model and Band Theory both address electron behavior but from different perspectives. The Drude Model treats conduction electrons as a gas that moves freely between collisions, focusing on classical mechanics. In contrast, Band Theory utilizes quantum mechanics to describe energy bands where electrons can exist. While Drude can predict conductivity based on free electron density, Band Theory offers deeper insights into semiconductors and insulators by highlighting band gaps and allowed energy states.
Evaluate the significance of the Drude Model in modern solid-state physics and its limitations in explaining electron behavior.
The Drude Model is significant as it provides a simple yet effective framework for understanding electrical and thermal conductivity in metals. However, its limitations are notable; it fails to incorporate quantum effects and cannot adequately explain phenomena like superconductivity or semiconducting behavior. Consequently, while the model serves as a fundamental starting point in solid-state physics, advancements in quantum mechanics have necessitated more sophisticated approaches to fully understand electron dynamics in various materials.
Related terms
Free Electron Gas:A model in which conduction electrons in a metal are considered to behave like an ideal gas, moving independently in all directions and colliding with ions in the lattice.
Fermi Energy:
The highest energy level occupied by electrons at absolute zero temperature, playing a crucial role in determining the electronic properties of metals and semiconductors.
Band Theory:
A theoretical framework that explains the allowed and forbidden energy levels for electrons in solids, critical for understanding electrical conductivity in materials.
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4720 | https://m.fx361.com/news/2019/0422/5038639.html | 频数与频率：“统计学”的两个重要指标_参考网
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频数与频率：“统计学”的两个重要指标
2019-04-22 周红娟
初中生世界·八年级订阅 2019年3期 收藏
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关键词：初中生学生
同学们在小学阶段就学习过统计图，当时学过的扇形图、条形图、折线图在初中仍然有很大的用途，而且新增了一类直方图，初中阶段常常称它为频数分布直方图。直方图的主要特征是能清楚地显示各组频数分布的情况，并且方便显示各组之间频数的差别，与之相近的有两个概念：频数和频率。
频数，指在数据统计时，某个对象出现的次数。
频率，是指频数与数据总数的比。
从上面定义可以看出，频率反映了各组频数的大小在总数中所占的分量。频率×100%就是百分比。
下面我們结合例题为同学们详细讲评。
例1 有40个数据，共分成6组，第1～4组的频数分别是10、5、7、6，第5组频率为10%，则第6组频率为（）。
A.25% B.30% C.15% D.20%
【讲解】选D。前四组的频率和为（10+5+7+6）÷40=70%，所以，第6组的频率为1-70%-10%=20%。
例2 在对n个数据进行整理的频率分布表中，各小组的频数与频率之和分别等于（）。
A.n，1 B.n，n C.1，n D.1，1
【讲解】各小组的频数之和等于样本容量n，各小组的频率之和为1。故选A。
例3 某中学组织了献爱心捐款活动，该校数学兴趣小组对本校学生献爱心捐款额做了一次随机抽样调查，并绘制了不完整的频数分布表和频数分布直方图（每组含前一个边界值，不含后一个边界值）。如图表所示：
[捐款额（元） 频数 百分比 5≤x<10 5 10% 10≤x<15 a 20% 15≤x<20 15 30% 20≤x<25 14 b 25≤x<30 6 12% 总计 100% ]
（1）填空：a=__，b=____。
（2）补全频数分布直方图。
（3）该校共有1600名学生，估计这次活动中爱心捐款额不低于20元的学生有多少人？
【讲解】（1）5÷10%=50，a=50×20%=10;b=×100%=28%。
（2）如图，
（3）1600×（28%+12%）=640（人）。
答：估计这次活动中爱心捐款额不低于20元的学生有640人。
（作者单位：江苏省南通市第一初级中学）
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4721 | https://people.clas.ufl.edu/rdholt/files/295.pdf | This article was downloaded by: [University of Florida] On: 25 November 2014, At: 07:36 Publisher: Taylor & Francis Informa Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK Journal of Biological Dynamics Publication details, including instructions for authors and subscription information: Overcoming Allee effects through evolutionary, genetic, and demographic rescue Andrew R. Kanareka, Colleen T . Webbb, Michael Barfieldc & Robert D. Holtc a National Institute for Mathematical and Biological Synthesis, University of Tennessee, Knoxville, TN 37996-1527, USA b Department of Biology, Colorado State University, Fort Collins, CO 80523-1878, USA c Department of Biology, University of Florida, Gainesville, FL 32611, USA Published online: 24 Nov 2014.
To cite this article: Andrew R. Kanarek, Colleen T . Webb, Michael Barfield & Robert D. Holt (2015) Overcoming Allee effects through evolutionary, genetic, and demographic rescue, Journal of Biological Dynamics, 9:1, 15-33, DOI: 10.1080/17513758.2014.978399 To link to this article: PLEASE SCROLL DOWN FOR ARTICLE Taylor & Francis makes every effort to ensure the accuracy of all the information (the “Content”) contained in the publications on our platform. Taylor & Francis, our agents, and our licensors make no representations or warranties whatsoever as to the accuracy, completeness, or suitability for any purpose of the Content. Versions of published Taylor & Francis and Routledge Open articles and Taylor & Francis and Routledge Open Select articles posted to institutional or subject repositories or any other third-party website are without warranty from Taylor & Francis of any kind, either expressed or implied, including, but not limited to, warranties of merchantability, fitness for a particular purpose, or non-infringement. Any opinions and views expressed in this article are the opinions and views of the authors, and are not the views of or endorsed by Taylor & Francis. The accuracy of the Content should not be relied upon and should be independently verified with primary sources of information. Taylor & Francis shall not be liable for any losses, actions, claims, proceedings, demands, costs, expenses, damages, and other liabilities whatsoever or howsoever caused arising directly or indirectly in connection with, in relation to or arising out of the use of the Content. This article may be used for research, teaching, and private study purposes. Terms & Conditions of access and use can be found at It is essential that you check the license status of any given Open and Open Select article to confirm conditions of access and use.
Downloaded by [University of Florida] at 07:36 25 November 2014 Journal of Biological Dynamics, 2014 Vol. 9, No. 1, 15–33, Overcoming Allee effects through evolutionary, genetic, and demographic rescue Andrew R. Kanareka∗, Colleen T. Webbb, Michael Barfieldc and Robert D. Holtc aNational Institute for Mathematical and Biological Synthesis, University of Tennessee, Knoxville, TN 37996-1527, USA; bDepartment of Biology, Colorado State University, Fort Collins, CO 80523-1878, USA; cDepartment of Biology, University of Florida, Gainesville, FL 32611, USA (Received 2 February 2014; accepted 13 October 2014) Despite the amplified threats of extinction facing small founder populations, successful colonization sometimes occurs, bringing devastating ecological and economic consequences. One explanation may be rapid evolution, which can increase mean fitness in populations declining towards extinction, permit-ting persistence and subsequent expansion. Such evolutionary rescue may be particularly important, given Allee effects. When a population is introduced at low density, individuals often experience a reduction in one or more components of fitness due to novel selection pressures that arise from diminished intraspe-cific interactions and positive density dependence (i.e. component Allee effects). A population can avoid extinction if it can adapt and recover on its own (i.e. evolutionary rescue), or if additional immigration sustains the population (i.e. demographic rescue) or boosts its genetic variation that facilitates adaptation (i.e. genetic rescue). These various forms of rescue have often been invoked as possible mechanisms for specific invasions, but their relative importance to invasion is not generally understood. Within a spatially explicit modelling framework, we consider the relative impact of each type of rescue on the probability of successful colonization, when there is evolution of a multi-locus quantitative trait that influences the strength of component Allee effects. We demonstrate that when Allee effects are important, the effect of demographic rescue via recurrent immigration overall provides the greatest opportunity for success.
While highlighting the role of evolution in the invasion process, we underscore the importance of the ecological context influencing the persistence of small founder populations.
Keywords: Allee effects; adaptive evolution; biological invasion; individual-based model 1.
Introduction A fundamental question for the preservation of biodiversity and ecosystem health is, ‘what min-imal numbers are necessary if a species is to maintain itself in nature?’ . The issue of how population size influences extinction risk is also fundamental in community ecology, because during community assembly (especially on islands and in patchy habitats), communities build up via colonization, and colonizing propagules are typically small in number. Larger populations are less likely to become extinct [46,49], particularly when there are Allee effects (positive den-sity dependence at low densities, so fitness declines as density declines; [11,58]). Maintaining Corresponding author. Email: andrew.kanarek@gmail.com Author Emails: colleen.webb@colostate.edu; mjb01@ufl.edu; rdholt@ufl.edu © 2014 The Author(s). Published by Taylor & Francis.
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (
org/licenses/by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. The moral rights of the named author(s) have been asserted.
Downloaded by [University of Florida] at 07:36 25 November 2014 16 A.R. Kanarek et al.
sufficient population size acts as a buffer from the decrease in fitness due to Allee effects, as well as from the extinction risks of demographic and environmental stochasticity. Given genetic variation, maintaining a larger population may also enhance the opportunity for adaptive response to novel selection pressures, further facilitating persistence .
During the introduction phase of a biological invasion into a novel environment, an invad-ing population may be reduced to such a low density that it experiences a high risk of extinction.
However, if demographic constraints and stochastic effects are not too severe, the population may persist long enough to adapt and recover (i.e. experience evolutionary rescue; [24,32,35]), or be rescued by additional immigrants . Additionally, an influx of immigrants can significantly increase population viability by increasing population size, thereby reducing the demographic threats of extinction (i.e. demographic rescue), or by introducing adaptive genetic variation that facilitates selection to increase mean fitness (i.e. genetic rescue). However, introduced genetic variation can sometimes decrease mean fitness (e.g. because gene flow hampers local adapta-tion; [19,28,54,60]), reducing population viability. These various forms of rescue have often been invoked as possible mechanisms for specific invasions , but their relative importance to invasion more generally is not understood. In this paper, we explore a model that permits us to weigh the relative importance of each of these distinct mechanisms in determining the influence of population size on invasion success.
Intraspecific interactions are a primary factor influencing the relationship between these eco-logical and evolutionary processes, especially for sexually reproducing species. The ability to find a mate and reproduce in a sexual species is required to maintain and bolster population size, and moreover influences the genetic variation available for selection, and hence the evolutionary potential of the introduced population. Other examples of intraspecific mechanisms that affect population persistence and increase survival at higher densities include environmental condi-tioning, predator dilution, anti-predator behaviour, and group foraging [4,11,58 and references therein]. These mechanisms influence specific fitness components (e.g. mating success, fecun-dity, or survival), leading to a ‘component Allee effect’ . If component Allee effects have a sufficient impact on overall individual fitness, then when combined with other local interac-tions and ecological processes, a demographic Allee effect can emerge and influence population dynamics [21,33]. Hence, a particular challenge for small founder populations is the reduc-tion of these positive intraspecific interactions. This challenge can produce a systematic novel selection pressure during colonization, compared to established populations near their carrying capacity .
With genetic variability, some individuals may better cope with the negative ecological or genetic effects of low density, so natural selection can occur on component Allee effects [11,21,32]. Adaptation to low-density conditions across generations can effectively alleviate the ecological constraints of demographic Allee effects – as long as the population is not so limited by demographic stochasticity in the short term that it simply goes extinct . Such adaptation can be viewed as a form of evolutionary rescue [24,26]. Some of the clearest examples of rapid evolution come from introduced species [12,39,41,51,54], despite the fact that founder events can reduce genetic variation and thus hamper evolutionary change [2,48]. It has been suggested that, in altered environments, ‘rapid adaptation is the norm rather than the exception’ , and such adaptation can, in principle, surmount the challenges posed by Allee effects at low densities, as well as challenges posed by novel environments.
Although evolutionary rescue in a single colonizing episode can facilitate successful establish-ment, there is substantial evidence that recurrent introductions into a given invasive population significantly increase its probability of establishment . Additional migrants, if they arrive frequently enough, can bolster population size and thus help buffer against Allee effects, over-dispersal, and the impacts of demographic and environmental stochasticity . However, even if an immigration event pushes the population above its demographic Allee threshold (so that Downloaded by [University of Florida] at 07:36 25 November 2014 Journal of Biological Dynamics 17 expected births exceed expected deaths), this may not be sufficient to rescue the population as the spatial structure of a population is critical to how density is experienced by individuals and how component Allee effects scale to demographic Allee thresholds. Further, high disper-sal rates can decrease local density and thereby amplify Allee effects and so hinder invasion [15,32,33,37,45,66]. In cases where increased propagule pressure does not on its own suffice for demographic rescue, the addition of conspecifics can still potentially introduce beneficial alle-les and facilitate adaptive evolution and evolutionary rescue [25,63]. The strength of this effect depends on the degree of genetic divergence between the founder and immigrant populations, and the degree to which genetic variation is limiting evolution; the net effect of immigrants involves both demographic and genetic effects .
To better illustrate the interplay between ecological and evolutionary processes that influence persistence of a founder population subject to Allee effects, we consider a conceptual frame-work that demonstrates the relative effects and consequences of evolutionary, demographic, and genetic rescue. Since selection pressure and individual fitness are assumed to be governed by intraspecific interactions and population density, we consider the evolution of an ecologically important quantitative trait that directly influences the relationship between local population size and individual fitness at low densities (where Allee effects are expected). The trait is a phenotypic measure of the overall strength of component Allee effects, measured in terms of the minimum number of conspecifics within a given neighbourhood that an individual requires for the expected value of its lifetime offspring production to exceed 1 (i.e. it is analogous to an individual ‘Allee threshold’). This is the most direct way to model evolution with Allee effects. For example, the trait could influence the ability of an individual to detect conspecifics in its area, in which case, individuals with decreased detection ability require a higher density of conspecifics to find a mate, and therefore would have a higher individual Allee threshold if the component Allee effect is based on mate finding. As seen in Figure 1, an individual would be considered maladapted with reduced fitness if its trait value is larger than the local population size, and better adapted otherwise; hence, the trait value and local population size are measured and displayed on the same scale. Variation around the mean phenotype (Figure 1, arrow a) is important, because more heritable variation increases the probability that there will be some well-adapted individuals in the population (with trait values in region b, below the local population size), even if on average it is maladapted. Since individuals with lower trait values have higher fitness, the trait distribution should shift in the direction of a smaller threshold population size (indicated by arrow d), which could lead to evolutionary rescue (though because population size is initially decreasing, it might not). With the addition of more immigrants, rescue may be more likely and hastened; if there is a sufficient increase in population size (arrow c), so that it is beyond the mean phenotype, demo-graphic rescue is likely to occur. However, even if population size does not increase dramatically, immigrants can widen the trait distribution by infusing new variation (arrow a) or actually shift-ing the mean phenotype with the introduction of beneficial alleles (arrow d), resulting in genetic rescue. Conversely, if immigrants all have maladapted trait values, the mean phenotype could be increased, moving away from the threshold and hindering evolutionary rescue. We here consider the relative importance of these effects in contributing to rescue from extinction, given ecolog-ical constraints on population growth, and incorporating effects of mutation and recombination on the trait variation (arrow a), using a simulation modelling approach.
We developed a spatially explicit individual-based model that incorporates both demographic and genetic stochastic processes to gauge the relative importance of the different forms of res-cue in mitigating Allee effects in a small founder population. We track multi-locus genotypes of individuals to investigate how initial genetic variation, coupled with the effects of mutation and recombination, affects the rate of adaptive evolution when selection pressure on the Allee threshold varies with population size. Additionally, we assess the influence of recurrent immi-gration events on demographic and genetic processes and determine the relative contribution Downloaded by [University of Florida] at 07:36 25 November 2014 18 A.R. Kanarek et al.
Figure 1.
Conceptual representation of the basic elements driving evolutionary, genetic, and demographic rescue. A trait is assumed to determine each individual’s Allee threshold, so the population’s mean trait value (phenotype) is the average Allee threshold. The variance of the phenotype distribution determines its width (arrow a). The distance between the local population size of an individual and its phenotype is its degree of maladaptation, if the population size is lower.
The individuals in shaded region b have phenotypes below the local population size, and therefore are well adapted.
If those individuals successfully reproduce enough prior to population extinction, the population size can eventually increase (arrow c) and the mean phenotype should decrease (arrow d), beginning the trend towards evolutionary rescue.
Demographic rescue is caused by the addition of new immigrants that push the local population size (arrow c) beyond the mean phenotype. Genetic rescue occurs when the contribution of immigration increases the variance (arrow a) and primarily shifts the mean phenotype to the left (arrow d). All of these forces can operate simultaneously.
of these factors to the overall likelihood of rescue and subsequent invasion. Finally, we determine how much evolutionary rescue depends on initial genetic variance versus mutation and recombination.
2.
Model description We expanded on an individual-based framework previously used to examine the effect of endoge-nous spatial heterogeneity (the local spatial structure of individuals within a population due to their movement, births and deaths) on intraspecific interactions by explicitly incorporating quantitative genetic structure (following [9,27]). Computer simulations were performed incor-porating multiple stochastic and genetic effects (e.g. mutation, recombination, dispersal, and the demographic birth–death process) in order to better understand the relative importance of the different dimensions of rescue in invasive populations.
2.1.
Ecological assumptions In each run of the model, a small population of diploid, sexually reproducing, hermaphroditic individuals was introduced into a continuous space environment with explicit spatial loca-tions. We used Gillespie’s Direct algorithm to simulate a continuous time birth–death process [5,17,23,33,52]. For each individual (i), its current birth (bi) and death (di) rates were determined by the number of conspecifics (Ni) in its local neighbourhood (the region within a distance of 1 unit from i), its phenotypic value for strength of component Allee effects via the individual Allee threshold, ai, and the local carrying capacity, K . The birth and death rates are assumed to be given by bi = Ni ai + Ni K = Ni(ai + K) aiK , di = 1 + N2 i aiK = aiK + N2 i aiK .
(1) Downloaded by [University of Florida] at 07:36 25 November 2014 Journal of Biological Dynamics 19 These relations match a well-studied reaction–diffusion model [14,15,32,34,37,45,50] that represents the standard dynamics of strong Allee effects. Specifically, a simple model for per capita population growth rate that includes an Allee effect is r = r0(1 −N/K)(N/a −1), which is often used in deterministic models. Our intent was to craft an individual-based model that generated this particular expression for per capita growth rate, which has been used in much of the previous literature. The first part of the standard deterministic model represents logistic population growth with a carrying capacity of K. The last term models the Allee effect, making the growth rate increase with increasing N at low N; N = a separates populations that grow from populations that do not (so a is a demographic Allee threshold). For a stochastic, individual-based model, we do not need the per capita growth rate, but instead its components, the individual per capita birth rate (bi) and death rate (di), since births and deaths are to be simulated. One fairly general way to assign birth and death rates so that b −d equals the r above is given in Ackleh et al. (2007), in which d is the sum of a constant and a term that grows as N2, and b is the sum of a constant and terms that depend on N and N2. This general model has three parameters in addition to K and a that can be adjusted to model different scenarios. We chose two of these parameters so that bi is a linear function of Ni by setting the constant term to zero (certainly reasonable for a sexual species, since an individual cannot reproduce in the absence of conspecifics) and also setting to zero the N2 i term (this term can cause the birth rate to saturate and eventually fall with increasing Ni, so these possibilities are not as relevant to the initial invasion process of interest here). At low densities, a linear bi might be reasonable if births are mostly limited by difficulty in mate finding; the more mates that are in an individual’s vicinity, the faster it can find a mate and reproduce. We also set r0 to 1, which amounts to re-scaling the time variable. Future extensions to our work could include a broader span of functional forms for component density dependence, such as saturating effects of density on births. With this form of birth and death rates, both birth and death rates increase with increasing Ni. Therefore, the Allee effect is due to increasing births with increasing population size, such as when growth rates are limited by ability to find mates, and not by decreasing death rates, such as those due to cooperative behaviours.
In this model, the individual Allee threshold evolves directly with the trait while the carrying capacity remains constant. An alternative would have been to define a trait and birth or death rate functions that depend on the trait, and then calculated the individual Allee threshold and an individual carrying capacity, both of which could then change as the trait evolves. This would be an interesting extension to this work, but the current model has the advantage that the quantity we are most interested in (the individual Allee threshold) changes directly with the trait (and our results are not confounded by changes in K).
Births and deaths were assumed to be independent Poisson processes. These rates were summed over all individuals to give an overall event rate E at each time (this is also a Pois-son process, so the time until the next birth or death had an exponential distribution with mean 1/E). The event was chosen to be a birth or death based on the relative values of total birth and total death rate for the population; an individual was then chosen to reproduce or die based on the magnitude of that individual’s respective birth or death rate. If the event was a death, the chosen individual was deleted. If the event was a birth, the chosen reproducing individual, i, randomly chose a mate, j, within its local neighbourhood (the region within a distance of 1 unit from i, but see for a deeper analysis of this assumption). Each parent produced a gamete according to the genetic assumptions described below, and one offspring was produced from those gametes at the location of parent i. Each individual in the population then moved randomly, the distance in each coordinate direction having a zero-mean normal distribution with variance 2Dt (where D is the diffusion coefficient and t is the inter-event time, which was small enough to approximate continuous movement; see Table 1 for parameters; [5,33,65]). This process continued until the population either went extinct or grew sufficiently large that persistence was reasonably certain.
From preliminary results, persistence was reasonably certain when the population size exceeded Downloaded by [University of Florida] at 07:36 25 November 2014 20 A.R. Kanarek et al.
Table 1.
Parameters, definitions, and values explored and used.
Parameter Definition Range Default n Number of loci 1–10 5 nµ Mutation rate per haplotype 0,10−6–0.1 0.01 α2 Mutational variance 0.01–0.1 0.05 σ 2 g Initial genetic variance per haplotype locus 0,0.01–0.25 0.05 ¯ a Initial mean phenotype, Allee threshold 25 25 ¯ aimm Mean phenotype, Allee threshold of immigrants 20–30 20, 25, 30 It Time of immigration event 1–10 1 In Number of immigrants per event 1–25 15 K Carrying capacity 100 100 D Dispersal rate 0.001–0.1 0.01 ˆ K0 Initial modified Ripley’s K −0.2 to 0.2 0 100 individuals (which is the carrying capacity), and we used reaching this value as our cut-off for assessing establishment.
When an immigration event was incorporated, immigrants (after their introduction) repro-duced, died, and moved following the same algorithms. Since other work has investigated invasion risk with multiple introductions of varying spatial proximity to the original release point , we fixed the introduction site (to match the initial population) and manipulated the number of immigrants and arrival times to explore other dimensions of immigration effects (Section 2.3).
2.2.
Genetic assumptions We modelled the fitness-governing quantitative trait (ai) assuming multi-locus heritability because many traits of ecological importance in natural populations are polygenic . The value of ai was determined by summing over n diploid loci, with additive allelic effects within and among loci (i.e. with no dominance or epistasis). Earlier studies with similar assumptions have found little quantitative difference among simulation results with different n, as long as n ≥5 ; we thus set n equal to 5. The phenotypic value for each individual was simply the sum of the allelic values (with a negligible environmental contribution, and no plasticity).
We examined the impact of recombination by allowing loci to either be completely linked, or to freely recombine. In simulations with recombination, each parent randomly contributed one allelic value for each locus from its diploid genome to its gamete. Without recombination, one haplotype was randomly chosen from each parent. New alleles were generated through mutation.
The mutation rate per haplotype was nµ, where µ was the mutation rate per locus [9,27]. Follow-ing segregation, up to one mutation occurred (per haplotype) at a randomly chosen locus. The mutation effect size was normally distributed with mean zero and variance α2 and was added to the previous allelic value.
2.3.
Parameters and initial conditions For the results presented here, we started each simulation with 25 individuals randomly placed in the unit circle around the origin (see below for discussion on varying the initial population size and how we controlled the initial dispersion). For simulations with non-zero initial genetic variance, allelic values for each initial individual were chosen independently from a normal dis-tribution with mean ¯ a/(2n) and variance σ 2 g , which gave each individual an expected phenotypic value of ¯ a (set to 25 unless otherwise noted), and an initial genetic variance of 2nσ 2 g ; with no initial variance, all initial alleles were ¯ a/(2n). Therefore, if each individual was (initially) in Downloaded by [University of Florida] at 07:36 25 November 2014 Journal of Biological Dynamics 21 the local neighbourhood of all others, the introduced population would have been at the demo-graphic Allee threshold (the unstable equilibrium for critical size in our model; [32,33]). We kept the mutation rate and mutation effect variance constant after exploring the range of values indicated in Table 1 (and used values based on , where nµ = 0.01 and α2 = 0.05) and stan-dardized the initial genetic variance across simulations (at σ 2 g = 0.05; see Appendix 1 for the effect of varying this).
To consider the impact of immigration on genetic rescue, we present broad comparisons between the founder population alone and with a fixed immigrant population size of 15 individ-uals introduced after one time unit (see Appendix 2 for the effect of varying these parameters).
Genetic variance, mutation and recombination parameters, and initial spatial distribution (see below) for immigrants matched the founder population; however, we did vary the expected mean phenotype of the immigrant population from well adapted (i.e. ¯ aimm = 20) to the same as the founder population (i.e. ¯ aimm = 25) to maladapted (i.e. ¯ aimm = 30).
Kanarek et al. found that the interaction between dispersal rate (diffusion coefficient) and initial spatial structure can qualitatively influence population dynamics during introductions; we explored multiple combinations of these characteristics in , but use particular fixed values for the results presented here. The diffusion coefficient, D, was set at 0.01. The initial spatial structure, measured by Ripley’s K, a clustering statistic , was similar across runs and approx-imately fit complete spatial randomness for a homogeneous Poisson process. This was achieved by placing the initial individuals randomly in the unit circle, but only using initial populations for which the magnitude of the modified Ripley’s K, | ˆ K0(1)|, was less than 0.02 . These val-ues describe an ecological context in which individuals were neither initially over-dispersed, nor thereafter dispersed too quickly, giving a sufficient demographic window for evolutionary rescue (see ‘null’ model below).
The model embodies multiple sources of stochasticity. To average across such stochasticity, 1000 replicates were performed for each choice of parameter values, so as to tease apart the relative contribution of various model components and better understand the sensitivity of the outcomes to different assumptions (see appendices A and B). We then fixed particular parame-ters to draw comparisons and elucidate the primary drivers of population success (we did assess the model over the ranges of parameter values listed in Table 1, but note that we are report-ing only a subset of the simulations performed). We present results for three main model types for populations with and without initial genetic variance: (1) without mutation or recombination (what we call the ‘null’ model), (2) with mutation only, and (3) both with mutation and recom-bination (hereon referred to by just ‘recombination’). We used this same comparison structure to assess the impact of immigration. We quantified the proportion of successes (i.e. establishment and persistence through positive population growth) and assessed average time to extinction (of the populations that went extinct). We also tracked initial trait distributions, and changes in population size, mean phenotype, and phenotypic variance over time.
3.
Results and discussion We evaluated the individual fitness as a function of the trait, ai, and the local population size, Ni, to gain a better sense of the ecological and evolutionary forces influencing model results. We used the difference between the probability that the next event for an individual will be a birth [bi/(bi + di)] and the probability that it will be a death [di/(bi + di)] as a measure of individual fitness (ri). In Figure 2, the dot represents an individual with the default mean initial phenotype in the model, ai = 25, and with local population size Ni = 25 (as it would be in the centre of the introduced population). Hence, the initial probability that the individual reproduced before dying was 0.5, matching the probability that it died first, giving it a fitness of 0. The fate of an individual Downloaded by [University of Florida] at 07:36 25 November 2014 22 A.R. Kanarek et al.
Figure 2.
Individual fitness surface where fitness, the difference between birth and death probabilities, is a function of the individual phenotype (ai) and local population size (Ni). An individual, represented by the dot, has an equal probabil-ity of birth or death when the trait value equals the population size (along the dashed line). The error bar represents the range based on one standard deviation of phenotypic values an individual could be initialized with. Fluctuations in local population size will move the individual along the dotted line, indicated by the arrows. Individuals in the darker region are well adapted, while those in the lighter region are maladapted.
with this trait was highly sensitive to its local, neighbourhood population size due to the density-dependent fitness function. Because of constant fluctuations in the individual’s local population (as birth, immigration, death, and movement occur), the individual’s fitness was always changing along the dotted line (where ai remains constant). This fast-paced ecological process caused by demographic feedbacks is by definition how Allee effects are expressed (i.e. as an increase in fitness with population size, ). If the individual’s local population size exceeded its trait value, the individual had a better chance of reproducing than dying and therefore a positive fitness. The demographic processes that impact the individual described in Figure 2 scale up to impact the probability of demographic rescue at the population level (and mirrors the effect of varying the initial population size relative to the mean Allee threshold). To better understand how individual fitness affects evolutionary and genetic rescue, we must consider multiple individuals on this fitness surface. Given the default values for expected genetic variance (Table 1), at the start of a simulation with variation, the majority of individual phenotypic values would likely fall within one standard deviation of the mean shown with the error bar on the point in Figure 2 for constant Ni = 25. Because phenotypic variation in ai is generally small and does not change as rapidly as does local population size, the relative impact of phenotype (ai) and ecological processes (Ni) with respect to individual fitness foreshadows the importance of demographic rescue versus other types of rescue at the population level. Overall, an individual has an expected fitness greater than 0 if ai < Ni (the darkly shaded region of Figure 2, where bi > di).
Scaling up from individual-level fitness to population dynamics, we first highlight the rel-ative effect of each type of rescue, given various sources of genetic variation, by presenting comparisons of mean behaviour. We then illustrate major trends with representative model runs.
We used the baseline null model (no mutation or recombination) to understand the impact of the stochastic birth–death process on persistence; 10% of the introduced populations suc-ceeded with no evolution (Figure 3(A), see also ). It should be noted that the conclusions we draw on the impact of different forms of ecological and evolutionary processes on rescue, drawn from the probability of successful establishment and persistence relative to this baseline sce-nario, reflect the initial conditions of our simulations. In particular, the trade-off between initial population size and the strength of initial Allee effects primarily drives the response to selection Downloaded by [University of Florida] at 07:36 25 November 2014 Journal of Biological Dynamics 23 A B Figure 3.
Comparisons between different sources of new genetic variation without (dark grey bars) or with (light grey bars) initial genetic variance. The null model has no variance-generating processes. (A) The proportion of 1000 replicate populations that succeeded. (B) The average time to extinction of populations that went extinct (error bars are one standard deviation).
and the likelihood of success; and when we varied the initial population size above and below the mean Allee threshold, we confirmed this general behaviour – where the probability of suc-cess was, on average, inversely proportional to the degree of maladaptation. Based on the forms of the birth and death rates, the interplay between the population size, carrying capacity, and individual phenotype has a relatively straightforward effect on probability of establishment, but becomes less predictable with the added influence of spatial structure, dispersal rate, and inter-action kernel (see for the full analysis of the interaction of these parameters; and the online appendix for the results of varying parameter values for the additional genetic and immigration components). Thus, we expect our results to hold qualitatively for populations faced with similar selection pressure (e.g. functional forms for birth and death rates and demographic and genetic assumptions); however, further investigation would be needed to draw conclusions regarding the robustness of our results for a wider range of ecological and evolutionary conditions (e.g.
non-random movement, non-hermaphroditic individuals, epistasis, and dominance).
With no initial genetic variance, neither mutation nor recombination contributed strongly to evolutionary rescue as the proportion of successes did not increase meaningfully under these models (Figure 3(A), dark bars). This makes sense, given that these processes infuse genetic variance slowly, relative to the demographic processes determining persistence or extinction.
There was a small increase in mean time to extinction with recombination (Figure 3(B), dark bars), indicating that the generation of new genetic variation allowed populations that inevitably went extinct to persist slightly longer (but note that mutation alone had almost no effect).
By contrast, founder populations with substantial initial genetic variation did generate a higher proportion of successes, even with no mutation or recombination, suggesting that evolutionary Downloaded by [University of Florida] at 07:36 25 November 2014 24 A.R. Kanarek et al.
rescue does occur (i.e. null model with initial variation compared to without, Figure 3(A)). The fraction that went extinct decreased, but for those populations that did not persist, the mean time to extinction was unaffected (Figure 3(B)). Mutation added to initial variation resulted in a greater number of successes (Figure 3(A), light bars) and (possibly) a slightly increased mean time to extinction (Figure 3(B)). Not surprisingly, if there is initial genetic variation, recombina-tion facilitated success because it generates the most genetic variation over a short time scale of all the processes considered (Figure 3(A)).
In assessing the added impact of immigration of 15 individuals after one time unit, we first note that approximately 40% of the model replicates succeeded for populations without initial variation and with the addition of genetically identical immigrants (Figure 4(A)). Compared with the null model (without variation) in Figure 3(A), this means that demographic rescue accounted for a 30% point increase in the likelihood of persistence. In a similar way, we use the results from Figure 3 to interpret the additional genetic impact of immigrants and assess the occurrence of genetic rescue. Overall, similar general trends appear in the mutation and recombination mod-els with and without initial variation and with and without immigration, if immigrants have the same initial mean trait value as residents (first two bars for each model type, Figures 3 and 4).
The immigrant population labelled ‘with variation’ had allelic values drawn from the same dis-tribution as did the initial resident population and was not more or less adapted. Thus, it is not surprising that in this case, when immigrants had the same initial mean trait value and variation as residents, immigration had little evolutionary impact. We observed approximately the same Figure 4.
Comparisons between different sources of new genetic variation with or without initial genetic variation and with immigration of 15 individuals one time unit after population establishment (¯ aimm = 25 for ‘no variation’ and ‘with variation’ results, and ¯ aimm = 20 for better adapted immigrants and ¯ aimm = 30 for maladapted immigrants; there was initial variation in the latter two cases). Selection acts on the variants within the three models: (1) the null model has only standing variation, (2) mutation only, or (3) mutation and recombination contribute to variation. (A) shows the proportion of populations that have succeeded of 1000 replicates and (B) gives the average time to extinction (error bars are one standard deviation).
Downloaded by [University of Florida] at 07:36 25 November 2014 Journal of Biological Dynamics 25 30% point increases in the proportion of successes (Figure 3(A) compared with left two bars in Figure 4(A)), and we consider this a combination of demographic rescue with adaptive evolution rather than purely genetic rescue.
The effect of the genetic contribution from the immigrant population was primarily demon-strated when the immigrant initial mean trait values differed from that of the original founders. In Figure 4, the latter two bars for each model type show the results for better adapted immigrants and maladapted immigrants with the same initial variance as the founder population. In the case of better adapted immigrants, the increases in the proportion of success due to the combination of effect on mean phenotype and mean time to extinction represent genetic rescue (Figure 4(A) and (B)). The additional introduction of maladapted individuals is detrimental and constrains adap-tive evolution; the positive demographic effect is now outweighed by the negative genetic effect.
We further evaluate the impacts of phenotypic divergence in the immigrants and the effect of tim-ing of immigration in Appendix 2. Note that immigration, even in the worst case, facilitates ulti-mate persistence, compared to isolated populations (compare left sides of Figures 3(A) and 4(A)).
We specifically illustrate some of the dynamics that gave rise to the results for evolution-ary, demographic and genetic rescue in order to better understand these broad comparisons.
Figure 5 shows characteristic examples of evolutionary rescue. In each scenario, there was no initial genetic variation in order to show how variation emerges through mutation and recombi-nation. The null model population driven by demographic stochasticity only goes extinct. The A B Figure 5.
Representative trajectories for (A) population size and (B) mean phenotype (Allee threshold) with no initial genetic variance or immigration. In (B), dark lines show dynamics of the mean and light lines indicate the associated range denoting standard error.
Downloaded by [University of Florida] at 07:36 25 November 2014 26 A.R. Kanarek et al.
Figure 6.
Representative trajectories of population size over time with an immigration event of 15 individuals (with the same mean phenotype as the founders (i.e. ¯ a = 25) and no standing genetic variation) occurring at approximately time 1 as indicated by the arrow. The dashed line illustrates demographic rescue. There was no genetic variation and hence, no evolution.
A B Figure 7.
Representative trajectories of (A) population size and (B) mean phenotype and standard error (light grey lines following means) over time with immigration at time 1. These simulations included mutation and recombination and further variation was introduced by immigration. The dashed trajectories illustrate genetic rescue.
Downloaded by [University of Florida] at 07:36 25 November 2014 Journal of Biological Dynamics 27 mutation and recombination models showed extended time lags before adaptation related to the rate at which genetic variation was generated by each process. Once the population size approx-imated the carrying capacity, selection for decreased Allee threshold was negligible, allowing it in some cases to increase due to mutation and drift (Figure 5(B), recombination curve).
Demographic rescue (population rescue with no genetic variation and therefore no evolution) is not just based on the number of immigrants and time of introduction, but also hinges on the trajectory of population size and how the invasion develops (e.g. spatial distribution; Figure 6).
The impact of the ecological conditions is illustrated by the very different trajectories under potential demographic rescue of the two populations that had the same simulation parameters . Figure 7 shows an example of genetic rescue (and one of failure). In these examples, we manipulated which individuals were chosen as immigrants to clearly illustrate the consequences of the level of adaptation of the immigrants. We started with a founder population with no initial genetic variation, and, after one time unit, introduced immigrants with the same mean pheno-type as the founders (i.e. ¯ a = 25), but with standing variation. In one model run (solid lines), we chose immigrants with maladapted phenotypes (ai > 25), and in the other (dashed lines), we chose well-adapted immigrants (ai < 25). Both founder populations had similar behaviour until the immigration event (Figure 7(A)). At that time, the trait means diverged significantly with roughly the same standard error (Figure 7(B)). Maladapted immigrants (solid lines) increased the mean phenotype, which caused reduced growth rates due to an elevated necessity for intraspe-cific interactions and stronger component Allee effects, and resulted in extinction. Well-adapted immigrants (dashed lines) genetically rescued the population by introducing beneficial alleles, facilitating evolution (these immigrants also facilitated rescue through their demographic effect).
4.
Conclusions A small introduced population with Allee effects can only succeed if it is faced with favourable ecological conditions, or experiences rapid adaptive evolution, or simply has good luck. Our stochastic simulations produced all three possibilities and allowed us to quantify their relative importance for invasion success. Beyond the 10% of successes not attributable to any rescue effect (i.e. sheer luck), additional immigration had a stronger impact on overcoming density dependence than evolution alone due to the rapid and consistently favourable effect that addi-tional individuals have over changes in the mean phenotype (see individual fitness surface Figure 2). For example, the addition of 15 immigrants early on had the same effect on establishment success as a founder population with five times more additive genetic variance than the default value (compare Figure 4(A), light grey bars and Appendix 1, Figure A1 at a genetic variance of 0.25). The impact of immigration was largely through demographic rescue, as opposed to genetic rescue. Once demographic rescue occurred, additional immigration did not notably enhance local adaptation. Local adaptation ceases following demographic rescue in our model because intraspecific interactions are the source of endogenous selection pressure. Once populations are above the Allee threshold by any means, the direct selection pressure on this demographic param-eter is reduced, and it ceases entirely when numbers are at carrying capacity. Thus, there was little difference between evolutionary change with or without an immigration event, except when the immigrants were divergent enough to shift the mean phenotype and generate a genetic rescue effect. Overall, the increased effect of demographic rescue over evolutionary and genetic rescue is a general consequence of strong Allee effects as illustrated conceptually in Figures 1 and 2, and we feel that this general conclusion is broadly applicable beyond the detailed assumptions we have made in our simulation explorations.
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In our model, not only is there an abbreviated time scale during which evolution can occur, because of the high risk of extinction, there are other implicit genetic consequences of small pop-ulations and Allee effects. The most obvious is the potential reduction of diversity due to genetic drift and founder effects; however, there is much recent evidence that indicates that the typical loss of additive genetic variance in introductions is minimal [41,56,67]. Kramer and Sarnelle even suggest that Allee effects may lead to resistance to significant changes in heterozy-gosity and genetic distance by imposing limits on minimum population size. Specifically, they found that 70–75% of populations of an alpine copepod that maintained the minimal population size lost < 10% of allelic richness. Although it seems plausible that the ecological limitations that Allee effects impose on critical density can actually indirectly maintain genetic variation, we suspect that the spatial constraints influencing population growth often instead limit genetic variation through restricted mating options. This is consistent with Kramer and Sarnelle’s finding that increased habitat size of a founder population at critical density also increased the proportion of original allelic richness. Thus, in the race against time for evolution to reduce component Allee effects through heritable fitness-related traits, endogenous spatial heterogene-ity that emerges ecologically to mitigate component Allee effects may indirectly limit the amount of genetic variation available for selection, reducing the scope for the process of evolu-tionary rescue. Even though there may be sufficient genetic variance in the founder population, spatial structure may render it inaccessible to evolution, in effect leading to a tug of war between ecological and evolutionary survival mechanisms.
The mounting empirical evidence of adaptive evolution following invasions [6,7,12,22,30,42, 43,47,49,55,62] must be reconciled with these results. One reasonable reconciliation is that estab-lishment and persistence are relatively rare compared with the number of introductions that fail , yet the empirical data are necessarily biased towards the former (it is easier to record long-lasting ‘successes’, than rapid ‘failures’, in invasions). Our results then suggest that because of demographic constraints on the evolutionary dynamics, the probability of evolutionary rescue is low and the observed examples are rare events. This explanation is consistent with the paradox of evolutionary rescue, where stronger selection gives rise to faster evolution while also impos-ing a greater demographic cost and risk of extinction . We chose to exemplify this scenario by incorporating strong (as opposed to weak) Allee effects that produce an extinction threshold with negative growth . Thus, extinction is drastically hastened as soon as the population size falls below the mean phenotype (i.e. Allee threshold). Evolutionary processes would play a more dramatic role in this simulation framework if a weak Allee effect or a more substantial fitness advantage from a small phenotypic change were incorporated. Hence, an alternative reconcili-ation of our results with empirical evidence is that strong Allee effects are necessarily rare in successful introductions. Alternatively, evolution may occur following invasion, without being causally strongly responsible for the invasion success in the first place.
To further assess the relevance of evolutionary processes to colonizing success, it might be helpful to take a more empirical approach in characterizing how density dependence operates at low densities and determining the following: Which mechanisms generating Allee effects are under selection? What is the probability of adaptation given the mating system? Do these adaptations allow persistence at low density or serve to increase density [21,28]? There are a number of examples of the selective pressures that Allee effects exert on invasive species. The evolutionary response can be primarily thought of as adaptations that facilitate reproduction by altering mating systems (e.g. self-fertilization, ; reproductive timing, ; induced ovulation, ; parthenogenesis, ; masting, ; gamete morphology and performance, ; and other life history traits, ). In addition, there are adaptations that affect survival, including detection of conspecifics as well as dispersal characteristics [64,68]. In one of the most direct studies, Elam et al. found in self-incompatible invasive wild radishes that population size and genetic relatedness influence maternal reproductive success, and suggested that multi-seeded fruits are Downloaded by [University of Florida] at 07:36 25 November 2014 Journal of Biological Dynamics 29 an apparent adaptation to overcome the challenge of an Allee effect. Overall, understanding the ecological attributes of the mating system and dispersal mode can offer powerful insight into evolution, invasiveness, and establishment likelihood of small populations.
Our results are broadly applicable to a wide variety of taxa and emphasize the complex real-ity facing a small founder population. Our model captures the evolutionary phenomenon of adaptations influencing intraspecific interactions (rather than responding to the exogenous envi-ronment) in order to demonstrate the demographic challenge posed by Allee effects. When fitness is depressed at small population sizes, the ecological, evolutionary, and genetic obstacles that successful invaders need to overcome are exacerbated. Our results not only highlight potential mechanisms and conditions facilitating or hampering rapid adaptive evolution and establishment success of small founder populations but also provide phenomenological insights into how Allee effects contribute to the paradox of invasion. For species with strong Allee effects, invasion outside their basic ecological niches appears difficult.
Acknowledgements We would like to thank the Webb Lab and two anonymous reviewers for their helpful comments.
Funding ARK acknowledges funding from NSF-IGERT [grant number DGE-#0221595], administered by PRIMES at CSU; travel grants from the Global Invasions Network [grant number NSF-RCN DEB-#0541673] for facilitating this collaboration; and by the National Institute for Mathematical and Biological Synthesis, an Institute sponsored by NSF; the US Depart-ment of Homeland Security; and the US Department of Agriculture through NSF Award #EF-0832858, with additional support from The University of Tennessee, Knoxville. R.D. H and M. B thank the University of Florida Foundation, and NSF Grant DMS #1220342.
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Appendix 1. Impact of initial genetic variance In general, directional selection leads to an increase in mean fitness that is proportional to the additive genetic variance in a population . Similar to Kanarek and Webb , we found that an increase in the initial genetic variance (σ 2 g ) had a strong influence on the rate of evolution, resulting in an increased chance of survival (Figure A1). This is illustrated by a 40% point increase in the proportion of successful founder populations (out of 1000 replicates) between those that cannot evolve (null model with no genetic variance) and simulated populations that undergo recombination and mutation with σ 2 g = 0.25. Comparing the three model types provides further evidence of how mutation and recombination influ-ence genetic variation and affect the evolutionary processes in overcoming Allee effects. The light grey solid trend line indicates the null model with initial genetic variation. Because variation introduces both well-adapted and maladapted individuals around the mean phenotype, it is unsurprising that a wider spread will increase the likelihood of evolutionary rescue. The darker grey long dashed line shows that with added mutation (using defaults in Table 1), more variation was introduced and the proportion of successes was increased. The black short dashed line demonstrates that random recombination can allow more effective removal of deleterious alleles contributed by increased initial variation and accumulated with mutation, increasing mean fitness, and population growth.
Appendix 2. Time and number of immigrants Shifting the focus to the process of demographic rescue without genetic variation and evolution, Figure A2 demonstrates that the size of the immigrant population and temporal proximity to the introduction of the initial population influenced Downloaded by [University of Florida] at 07:36 25 November 2014 32 A.R. Kanarek et al.
Figure A1.
The proportion of successful populations of 1000 replicates according to the amount of initial genetic variance (varied from 0 to 0.25 incremented by 0.01).
both the likelihood of establishment and time to extinction (note, the immigrants had the same mean phenotype as the founders (i.e. ¯ a = 25) and no standing genetic variation). The proportion of successes increased with the number of immigrants as long as they were introduced before the founder population became too dispersed or depauperate, and as long as the total population size was close to or exceeded the Allee threshold just after introduction. Even if the number of immigrants did not dramatically increase the total population size, additional individuals contributed to the lag phase and extended the time to extinction.
Figure A3(A) shows the probability of success resulting from the addition of 15 individuals at different points in time for each model type. The degree of adaptation of the mean immigrant phenotypes was higher than (mean Allee threshold of 20), equal to (25), or lower than (30) that of the initial population. Figure A3(B) further indicates the implications for genetic rescue based on the mean phenotype of the immigrants and genetic processes contributing to genetic variation. In this case, even a small number of immigrants (i.e. 5), for which there is little demographic rescue effect, can provide the opportunity for adaptive evolution in population recovery, if immigrants are better adapted and introduced early (solid lines). It should be noted that because of number of simulations presented in this figure, we used the LOWESS method of smoothing over the time of immigration for clarity.
Comparison of Figure A3 with Figure Figure A2 demonstrates the additional contribution that genetic variation has on population success with evolution. For each model type in Figure A3, the original founder population and immigrant populations were initialized with expected genetic variance given in Table 1. Hence, the proportion of success increases with adaptive evolution (mean phenotype 25, long dashed lines Figure A3(A) versus dark grey short dashed line Figure A2 for 15 immigrants). A noteworthy effect is observed with the introduction of just five immigrants. Figure A3(B) shows that with added variation (in both the founder and immigrant populations), success increases compared to the low Figure A2.
The proportion of successes and the average time to extinction based on the timing and size of an immigrant population upon introduction, with no initial genetic variation or evolution.
Downloaded by [University of Florida] at 07:36 25 November 2014 Journal of Biological Dynamics 33 A B Figure A3.
The proportion of successes based on the timing and mean phenotype of the immigrant population upon introduction for each model type. The immigrant population size is 15 in (A) and 5 in (B). Trend lines were generated with LOWESS smoothing across time of immigration (with degree 0.5 with 2 iterations).
proportion of successes in Figure A2 (grey long dashed line). Even with the introduction of maladapted immigrants, success is still enhanced due to evolutionary rescue in the founder population when the five immigrants are rapidly purged from the population (short dashed black lines on Figure A3(B); the negative effect of maladapted immigrants is more pronounced with more individuals in Figure A3(A)). However, five individuals can positively affect the mean phenotype when well adapted, resulting in genetic rescue (solid curves in Figure A3(B)).
Downloaded by [University of Florida] at 07:36 25 November 2014 |
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Iterations in Geometry, an example
Let there be given a triangle ABC and a point P. First, point P moves towards A. But half way from its original position to A it makes a turn and continues towards B. However, half way from B, it turns towards C. Half way towards C, it turns towards A, and so on.
Verify and try to prove that point P eventually settles into cycling between three successive positions irrespective of its initial position. These three positions form a triangle whose area is 1/7 that of the area of triangle ABC.
(Click anywhere in the applet to start iterations.)
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Explanation
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From its initial position point P moves to (P + A)/2. From here on the next step, it moves half distance towards B. It's new position is at [(P + A)/2 + B]/2 = (P + A + 2B)/4. After that it moves half way in the direction of C to [(P + A + 2B)/4 + C]/2 = (P + A + 2B + 4C)/8. From here the process starts again: half way towards A, half way towards B, half way towards C, and so on.
We may now consider a function f of a point P in a plane defined by f(P) = (P + A + 2B + 4C)/8, or f(P) = (P + D)/8, where D = (A + 2B + 4C)/8.
This function defines an iterative process: Pn+1 = f(Pn), n = 0, 1, 2, ..., where P0 = P, the initial position of the given point. The distance between successive iterations shrinks by a factor of 8 in the following sense:
Pn+1 - Pn = f(Pn) - f(Pn-1) = (Pn + D)/8 - (Pn-1 + D)/8 = (Pn - Pn-1)/8.
In other words, the distance between Pn+1 and Pn is 1/8 the distance between Pn and Pn-1. The latter is of course 1/8 the previous distance:
Pn - Pn-1 = (Pn-1 - Pn-2)/8.
Which means that Pn+1 - Pn = (Pn-1 - Pn-2)/82. Continuing in this manner we obtain that
Pn+1 - Pn = (P1 - P0)/8n.
Remembering the definitions of P0 and P1, this is the same as
Pn+1 - Pn = (D - 7P)/8n+1.
Write a sequence of similar identities for finitely many indices n and sum them up (let m > n):
Pm - Pm-1 = (D - 7P)/8m Pm-1 - Pm-2 = (D - 7P)/8m-1 ... Pn+2 - Pn+1 = (D - 7P)/8n+2 Pn+1 - Pn = (D - 7P)/8n+1
Summing all of these up, we get
Pm - Pn = (D - 7P)(1 + 2-1 + ... + 2-(m-n))·2-n
Since the sum in parentheses is less then 2, we may claim that {Pn} is a Cauchy sequence. Since the plane along with the real line are complete metric spaces, the sequence {Pn} has a limit which we denote Q. From
Pn+1 = f(Pn) = (Pn + D)/8
we obtain Q = (Q + D)/8, or Q = D/7 = (A + 2B + 4C)/7.
Other vertices of the limiting triangle are found by cycling the symbols in that expression: (B + 2C + 4A)/7 and (C + 2A + 4B)/7.
There are several ways to prove that that triangle has the area 1/7 that of triangle ABC. The simplest is probably to recognize a construction we discussed on another occasion. One can also use the barycentric coordinates to identify the intersection A' of AD and BC. A' divides BC in a 2:1 ratio, which naturally leads to the same result.
The above problem admits a two-fold generalization.
Limits in Geometry
Two Circles and a Limit
Analytic Proof
Barbeau, Klamkin, Moser
Proof #2 by Stuart Anderson
Mariano Perez de la Cruz
Trigonometric Solution
A Geometric Limit
Iterations in Geometry, an example
Iterations in Geometry, a generalization
Iterated Function Systems
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4723 | https://www.youtube.com/watch?v=Rf05H8ogHLg | Verifying Trigonometric Identities
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This trigonometry video tutorial focuses on verifying trigonometric identities with hard examples including fractions. It contains plenty of examples and practice problems. This video tutorial contains plenty of tips and tricks that will help you on your homework or upcoming worksheet assignment. It explains when to convert secant and tangent into sine and cosine, when to multiply by the conjugate or get common denominators in addition to knowing when to factor a trigonometric trinomial equations or a difference of perfect squares expression.
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Introduction in this video we're going to talk about how to verify trigonometric identities which is not a very easy topic but here's what you need to do typically you'll be given a problem with an equation and we need to do is you need to show that the left side of this equation is equivalent to the right side of this equation now listed on the board are some techniques that you can use to do just that one of the first things that i would look to is trying to convert everything to sine and cosine that's a good place to start it doesn't work for every problem but at least for some of the easier ones it's a a good way to get started solving it now for step two sometimes you'll see a problem where there's two terms on the left but one term on the right when you see this what you could try to do is if a and b are fractions you can try to get the common denominator of both fractions convert it into a single fraction and then go from there other times you may need to factor so that you can convert the two terms into one term now step three is basically the reverse of step two if you have one term on the left and two terms on the right to convert from one to two terms you could distribute or if you have a fraction you can split that one fraction into two smaller fractions and for step four converting division into multiplication or vice versa for that i have to show it to you you just have to see it in action number five sometimes you need to multiply by the conjugate if you see one plus sign it could be one minus sign or one plus or minus cosine let's say on the denominator of a fraction that's a good indication that you need to multiply by the conjugate and then for step six sometimes you need to factor other times you need to foil and if you've factored in algebra then you can apply those same techniques here sometimes you'll see situations where you have difference differences of perfect squares or you may have a trinomial like this where you can see that it's factorable so those are some other things that you want to keep in mind when solving these problems now before we begin working on a few practice problems you may want to take down some notes by the way i recommend writing what you see on the board right now Notes so the first thing you need to know is some identities sine squared plus cosine squared is equal to one make sure you know that one that's a very common one and then i'm sure you've seen this one 1 plus tangent squared is equal to secant squared so you want to commit these identities to memory because you're going to be using them a lot the next one 1 plus cotangent squared is equal to cosecant squared now you need to be familiar with these other identities tangent is equal to sine over cosine cotangent is the reverse it's cosine divided by sine so therefore tangent is the reciprocal of cotangent tangent is 1 over cotangent and then here are some other reciprocal identities secant is equal to 1 over cosine and cosecant is one over sine so make sure you write this write all this stuff down on a piece of paper it's going to be very helpful as we work on some problems First Example let's begin with the first problem let's say we have sine x times secant x and let's say that's equal to tangent x go ahead and verify this trigonometric identity by the way for each of these problems i recommend that you pause the video try it yourself and then you know after you finish working a problem play the video to see if you have the right answer so we're going to use step number one we're going to convert everything into sine and cosine now you could start with the left side of the equation and make it look like you like the right side of the equation or you could start with the right side and then make that equivalent to the left side or you could start with both and make make them equal to each other it really doesn't matter as long as you show that the left side is equal to the right side you'll get the right answer but for me personally i like to keep the right side the same and i like to start with the left side of the equation and convert it to the right side so you don't have to do it that way but that's the way i like to do it which that's the way i'm going to be using in this video so just to give you a heads up so right now i'm going to leave the sine function the way it is now we know that secant is one over cosine now if you want to you can write this as sine x over one sine times one is just sine and on the bottom if we multiply these two one times cosine is cosine and based on the formulas that we wrote earlier we know that sine divided by cosine is tangent so now that the left side is the same as the right side we're finished here so we've verified this particular trigonometric identity so once you show that the left side is equal to the right side you're finished Second Example now let's move on to our next example number two let's say we have tangent squared times cotangent squared and that's equal to one go ahead and verify this trigonometric identity so let's begin by converting everything into sine and cosine which is step number one so we know that tangent is sine over cosine now we have tangent squared so we have two of them so this is going to be sine over cosine squared cotangent is the reciprocal of tangent we know that cotangent is cosine over sine and this is squared as well so what i'm going to do is i'm going to distribute the exponent this is sine to the first power and it's raised to the second power so 1 times 2 is 2 so this becomes sine squared and the same is true with cosine we're going to raise cosine to the second power so this becomes cosine squared times cosine squared divided by sine squared now cosine squared divided by cosine squared that's one sine squared divided by sine squared is one so basically we get one times one which is one so that's it for number two that's all that we need to do in order to verify this particular identity Third Example now let's move on to our next example number three so for this one here's the problem cotangent times secant x times sine x is equal to one go ahead and verify it so like before we're going to use step number one we're going to try converting everything into sine and cosine so we can change cotangent into cosine over sine secant we know is one over cosine and sine we could just leave it as sine x or you can write it as sine x over one so these two will cancel sine x over sine x is one and cosine x divided by cosine is one so this works out to be something similar one times one is one and so that's it for that problem all right let's work on some harder examples each problem will progressively get harder this one is a little bit harder than the last one but not too hard so here we have cosine times secant divided by cotangent and this is equal to tangent go ahead and prove it so let's begin by converting everything on the left side into sine and cosine so cosine i'm just going to leave it the way it is i'm going to write it as cosine over 1. secant we can write that as 1 over cosine now cotangent if you want to you could convert it to cosine over sine you don't have to but for this problem actually i'm going to leave it as cotangent and here's why we know we can cancel cosine and so that's going to give us a 1 in the numerator and we know that 1 over cotangent is tangent so for this problem there's no need to convert cotangent into cosine of a side if you did convert it you can still get the right answer so it wouldn't be a bad thing you'd just be introducing an extra step so the last thing to do is replace one of a cotan with tangent and that'll be it for this problem Fifth Example now let's move on to our fifth example for this one we're going to have sine x times tangent x and this is going to be equal to 1 minus cosine squared divided by cosine x so what do you think we need to do for this problem how can we change sine times tangent into 1 minus cosine squared over cosine so sometimes what you could do is look at the right side to see where to go the right side of the equation if you're starting from the left side that is the right side of the equation can be like a guide and we only have the cosine function on the right side so that tells us we need to introduce cosine into the left side right now the only thing we could change is tangent we can convert tangent into sine and cosine so i'm going to write sine as sine over one and tangent i'm going to replace that with sine over cosine now if you don't want to rewrite what's on the right side you could just put this now let's multiply sine x times sine x is sine squared and on the bottom we have one times cosine which is cosine so the good thing is we have what is on the bottom so this part is the same what we need to do is convert sine squared into 1 minus cosine squared now how can we do that well we're familiar with this identity we know that sine squared plus cosine squared is equal to one so what i'm going to do is i'm going to subtract both sides by cosine squared so these two will cancel and i'm going to get sine squared is one minus cosine squared so knowing that all i need to do is replace sine squared with an equivalent expression 1 minus cosine squared and that's going to give me the answer and so now the identity has been verified try this one cosine squared minus sine squared let's say that's equal to 1 minus 2 sine squared so what do we need to do here so once again we're going to look at the right side of the equation as a guide on the right side there is no cosine function only sine so somehow we need to convert cosine into a sine function and we know how to do that using this identity cosine squared plus sine squared is equal to one so we need to get cosine squared by itself so for this time we need to subtract both sides by sine squared so on the left side what we have left over is cosine squared and on the right side we have one minus sine squared so let's replace cosine squared with one minus sine squared so we're going to have 1 minus sine squared minus another sine squared and that's equal to what we see on the right side now there's like an invisible one here so we have negative one minus one which is negative two so then this becomes one minus two sine squared and that's all we need to do for this problem so it helps to look at the right side which tells you what you need to do with the left side so use the right side of the equation as a guide Seventh Example all right number seven so this one is going to be a little different than the others here we have sine x times tangent x plus cosine x and this one is equal to secant x so notice that we have two terms on the left one term on the right so this is like the first term sine times tangent and here this is the second term cosine and this is the third one so we have this form a plus b equals c what do we need to do here well we can't factor out a sine or cosine or tangent because we don't have a common factor so the other thing that we could do is somehow get two fractions get common denominators and then combine it into a single fraction the question is how can we do that well let's begin by turning everything into sine and cosine that is everything on the left side so let's write sine as sine of one and tangent we're going to convert that into sine over cosine and then cosine x we're just going to write it as cosine over one we want to put everything in fraction form on the left we can multiply sine times sine which will give us sine squared on the bottom one times cosine so that's just cosine so now we have two fractions but we don't have the same denominator how can we get common denominators in order to do that we need to multiply the second fraction by the denominator of the first and whatever you do to the top you must also do to the bottom so if you multiply the numerator of this fraction by cosine you need to do the same thing with the denominator so we're going to have sine squared over cosine x plus so cosine times cosine that's cosine squared and on the bottom one times cosine we know it's cosine all right let me delete a few things just to create extra space so now that these two fractions share the same denominator what we can do is combine it into a single fraction so we can now write sine squared plus cosine squared all divided by the same common denominator of cosine now hopefully you have your list of trigonometric identities with you because we have an important identity one that you're familiar with sine squared plus cosine squared what does that equal but we know that sine squared plus cosine squared is equal to one so now this becomes one over cosine and we know that secant is one over cosine so we can replace one over cosine with secant and that's it for this problem the identity has been verified so whenever you see a situation like this where you have two terms on the left one term on the right you may need to convert it into two fractions get common denominators combine it into a single fraction and then simplify now let's move on to number eight here we have secant minus cosine x and this is equal to tangent x times sine x go ahead and try it so notice that we have two terms on the left well this time it's a minus b instead of a plus b and we only have one term on the right tangent times sine that's one term so we need to do something similar either we could try to factor or we can try to convert into fractions and get common denominators we don't have a common term to factor so we need to get common denominators so let's turn secant and cosine into fractions secant we know it's 1 over cosine and cosine we can turn that into a fraction by putting it over one so now in order to get common denominators we'll need to multiply the second fraction by cosine over cosine so this becomes one over cosine minus cosine squared over cosine so now that we have the same denominator we can combine this into a single fraction so this becomes 1 minus cosine squared over cosine x and that's equal to tangent times sine x now what should we do at this point notice that this 1 minus cosine squared is an identity we know that sine squared plus cosine squared is equal to one so if we take this term move it to the other side we're gonna get sine squared is equal to one minus cosine squared it's positive on the left side but when you move it to the other side it's going to be negative so we're going to do is we're going to replace 1 minus cosine squared with sine squared so we have sine squared over cosine and this is equal to tangent times sine now notice what we have here this is basically rule number four here we have division and here we have multiplication how do we convert division into multiplication well let me show you let's say if you have a over b you can convert that fraction into multiplication by writing it this way a over 1 times 1 over b so that's a simple way in which you can convert division into multiplication so we can write this as sine squared over one times one minus cosine now sine squared is sine times sine now what you could do is you could move this sign to this fraction there's nothing wrong with that for instance let's say if i have 4 times 4 over 1 times one over two if i were to move the four to this other fraction it would be four over one times four over two the value of the entire situation is still the same this is 16 divided by two that's eight this is still four times four which is 16 divided by two that's still eight so you can move the sign from one fraction to the other when you're multiplying it doesn't change the value of the fraction so what we now have is sine x over one times sine over cosine and that is equal to tangent times sine x sine x over one we can just leave it as sine and sine divided by cosine we know it's tangent sine times tangent is equal to tangent times sine you can reverse it three times five for instance is equal to five times three they both equal fifteen so once you get to this part that's it the problem is finished |
4724 | https://www.expii.com/t/raoults-law-definition-overview-8051 | Raoult's Law — Definition & Overview - Expii
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Raoult's Law — Definition & Overview - Expii
Raoult's Law states that the vapor pressure of a solvent in solution is equal to its mole fraction in the solution times the vapor pressure of the pure solvent.
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ChemistryColligative Properties of Solutions
Raoult's Law — Definition & Overview
Raoult's Law states that the vapor pressure of a solvent in solution is equal to its mole fraction in the solution times the vapor pressure of the pure solvent.
Raoult's Law — Definition & Overview
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Explanations (3)
Eric Sears
Video
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(Video) Raoult's Law With Example Problem
by Denovo Tutor
In this video, Denovo Tutor explains Raoult's law. He starts by explaining the vapor pressure. A liquid's vapor pressure changes if you add a solvent. That's the basis of Raoult's law. He explains the equation. Finally, he shows you an example problem. This video will help you master Raoult's law!
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Related Lessons
Lowering of Vapor Pressure — Overview & Examples
Calculating Molar Mass Using Colligative Properties — Examples
Boiling Point Elevation — Overview & Examples
Solution Stoichiometry — Overview & Examples
View All Related Lessons
Trisha
Text
1
Image 4: Raoult's Law
Can help us find the vapor pressure of an ideal solution.
Psolution = Xsolvent times Psolvent
Psolution equals the vapor pressure of solution.
Xsolvent equals the mole fraction of solvent
Psolvent equals the vapor pressure of solvent.
Image Source: By Trisha
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Eric Sears
Text
1
Our last lesson was on vapor pressure lowering. We learned that adding a solute reduces the solvent's vapor pressure. So, a solution has a lower vapor pressure than the pure solvent. But, vapor pressure is a colligative property. Colligative properties are extensive. So, the physical change only depends on the amount of solute. We find it using Raoult's law. Below we'll look at an example problem.
Raoult's Law
To find the vapor pressure, we use Raoult's law. The formula is P A=X A P o A.
P A is the new vapor pressure. We call it partial pressure.
X A is the mole fraction of the solvent solution.
Remember, its formula is X A=m o l s s o l v e n t m o l s t o t a l. The mole fraction is moles divided by moles. So, it's unitless.
P o A is the vapor pressure of the pure solvent.
To find the change in the vapor pressure, we use the formula Δ P=P A−P o A.
Raoult's Law Practice Problem
Previously, we studied freezing point depression and boiling point elevation. In our example problems, we examined the effects of ethylene glycol in car engines. Let's calculate the vapor pressure change from adding ethylene glycol to water.
Again, we'll say we mixed one gallon of antifreeze solution (about 3.8 liters). The ratio of ethylene glycol to water is 1:1. The density of ethylene glycol is 0.7857 g m l. Its molar mass is 62.1 g m o l. The vapor pressure of pure water is 3.17 k P a at 25.0 o C. Calculate the change in water's vapor pressure from adding ethylene glycol.
Step 1: Use dimensional analysis to calculate the moles of ethylene glycol and water.
1.9 L×1000 m l L×0.7857 g m l×1 m o l 62.1 g=24.0 m o l s e t h y l e n e g l y c o l
1.9 L×1000 m l L×1.0 g m l×1 m o l 18.01 g=105.5 m o l s w a t e r
A shortcut is to know that the molarconcentration of water is 55.5 m o l s L. So, 55.5 m o l L×1.9 L=105.5 m o l s.
Step 2: Calculate the mole fraction.
X=m o l s w a t e r m o l s t o t a l=105.5 105.5+24.0=0.81
Step 3: Use Raoult's law to calculate the new partial pressure of water.
P A=X A P o A=0.81×3.17 k P a=2.58 k P a
Step 4: Subtract the final and initial pressures to find the change.
Δ P=P A−P o A=2.58−3.17=−0.59 k P a
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4725 | https://www.mathworks.com/help/predmaint/ref/monotonicity.html | Skip to content
Main Content
monotonicity
Quantify monotonic trend in condition indicators
collapse all in page
Syntax
Y = monotonicity(X)
Y = monotonicity(X,lifetimeVar)
Y = monotonicity(X,lifetimeVar,dataVar)
Y = monotonicity(X,lifetimeVar,dataVar,memberVar)
___
Description
Y = monotonicity(X)
returns the monotonicity of the lifetime data X. Use
monotonicity to quantify the monotonic trend in condition indicators
as the system evolves toward failure. The values of Y range from 0 to
1, where Y is 1 if X is perfectly monotonic and 0
if X is non-monotonic.
As a system gets progressively closer to failure, a suitable condition indicator
typically has a monotonic trend. Conversely, any feature with a non-monotonic trend is a
less suitable condition indicator.
Y = monotonicity(X,lifetimeVar)
returns the monotonicity of the lifetime data X using the lifetime
variable lifetimeVar.
example
Y = monotonicity(X,lifetimeVar,dataVar)
returns the monotonicity of the lifetime data X using the data
variables specified by dataVar.
example
Y = monotonicity(X,lifetimeVar,dataVar,memberVar)
returns the monotonicity of the lifetime data X using the lifetime
variable lifetimeVar, the data variables specified by
dataVar, and the member variable
memberVar.
example
___
estimates the monotonicity with additional options specified by one or more
Name,Value pair arguments. You can use this syntax with any of the
previous input-argument combinations.
example
___ with no output arguments plots a
bar chart of ranked monotonicity values.
example
Examples
collapse all
In this example, consider the lifetime data of 10 identical machines with the following 6 potential prognostic parameters—constant, linear, quadratic, cubic, logarithmic, and periodic. The data set machineDataCellArray.mat contains C, which is a 1x10 cell array of matrices where each element of the cell array is a matrix that contains the lifetime data of a machine. For each matrix in the cell array, the first column contains the time while the other columns contain the data variables.
Load the lifetime data and visualize it against time.
Copy openExample Command
Paste command in MATLAB to download and open example files
Open in MATLAB Online
load('machineDataCellArray.mat','C')
display(C)
```
C=1×10 cell array
{219×7 double} {189×7 double} {202×7 double} {199×7 double} {229×7 double} {184×7 double} {224×7 double} {208×7 double} {181×7 double} {197×7 double}
```
Copy openExample Command
Paste command in MATLAB to download and open example files
Open in MATLAB Online
for k = 1:length(C)
plot(C{k}(:,1), C{k}(:,2:end));
hold on;
end
Observe the 6 different condition indicators—constant, linear, quadratic, cubic, logarithmic, and periodic—for all 10 machines on the plot.
Visualize the monotonicity of the potential prognostic features.
Copy openExample Command
Paste command in MATLAB to download and open example files
Open in MATLAB Online
monotonicity(C)
From the histogram plot, observe that the features Var2, Var4 and Var5 rank better than the others. Hence, these features are more appropriate for remaining useful life predictions since they are the best indicators of machine health.
In this example, consider the lifetime data of 10 identical machines with the following 6 potential prognostic parameters—constant, linear, quadratic, cubic, logarithmic, and periodic. The data set machineDataTable.mat contains T, which is a 1x10 cell array of tables where each element of the cell array contains a table of lifetime data for a machine.
Load and display the data.
Copy openExample Command
Paste command in MATLAB to download and open example files
Open in MATLAB Online
load('machineDataTable.mat','T');
display(T)
```
T=1×10 cell array
{219×7 table} {189×7 table} {202×7 table} {199×7 table} {229×7 table} {184×7 table} {224×7 table} {208×7 table} {181×7 table} {197×7 table}
```
Copy openExample Command
Paste command in MATLAB to download and open example files
Open in MATLAB Online
head(T{1},2)
```
Time Constant Linear Quadratic Cubic Logarithmic Periodic
_ _ _ __ _ _ _
0 3.2029 11.203 7.7029 3.8829 2.2517 0.2029
0.05 2.8135 10.763 7.2637 3.6006 1.8579 0.12251
```
Note that every table in the cell array contains the lifetime variable 'Time' and the data variables 'Constant', 'Linear', 'Quadratic', 'Cubic', 'Logarithmic', and 'Periodic'.
Compute monotonicity using Spearman's rank correlation method with Time as the lifetime variable.
Copy openExample Command
Paste command in MATLAB to download and open example files
Open in MATLAB Online
'Time' 'Method' 'rank'
```
Y=1×6 table
Constant Linear Quadratic Cubic Logarithmic Periodic
__ __ _ _ __ _____
0.069487 1 0.17766 0.97993 0.99957 0.059208
```
From the resulting table of monotonicity values, observe that the linear, cubic, and logarithmic features have values closer to 1. Hence, these three features are more appropriate for predicting remaining useful life since they are the best indicators of machine health.
Consider the lifetime data of 4 machines. Each machine has 4 fault codes for the potential condition indicators—voltage, current, and power. monotonicityEnsemble.zip is a collection of 4 files where every file contains a timetable of lifetime data for each machine—tbl1.mat, tbl2.mat, tbl3.mat, and tbl4.mat. You can also use files containing data for multiple machines. For each timetable, the organization of the data is as follows:
When you perform calculations on tall arrays, MATLAB® uses either a parallel pool (default if you have Parallel Computing Toolbox™) or the local MATLAB session. To run the example using the local MATLAB session, change the global execution environment by using the mapreducer function.
Copy openExample Command
Paste command in MATLAB to download and open example files
Open in MATLAB Online
mapreducer(0)
Extract the compressed files, read the data in the timetables, and create a fileEnsembleDatastore object using the timetable data. For more information on creating a file ensemble datastore, see fileEnsembleDatastore.
Copy openExample Command
Paste command in MATLAB to download and open example files
Open in MATLAB Online
unzip monotonicityEnsemble.zip;
ens = fileEnsembleDatastore(pwd,'.mat');
ens.DataVariables = {'Voltage','Current','Power','FaultCode','Machine'};
ens.ReadFcn = @readtable_data;
ens.SelectedVariables = {'Voltage','Current','Power','FaultCode','Machine'};
Visualize the monotonicity of the potential prognostic features with 'Machine' as the member variable and group the lifetime data by 'FaultCode'. Grouping the lifetime data ensures that monotonicity calculates the metric for each fault code separately.
Copy openExample Command
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Open in MATLAB Online
'MemberVariable' 'Machine' 'GroupBy' 'FaultCode'
```
Evaluating tall expression using the Local MATLAB Session:
- Pass 1 of 1: Completed in 0.43 sec
Evaluation completed in 0.81 sec
Evaluating tall expression using the Local MATLAB Session:
- Pass 1 of 1: Completed in 0.14 sec
Evaluation completed in 0.27 sec
Evaluating tall expression using the Local MATLAB Session:
- Pass 1 of 1: Completed in 0.29 sec
Evaluation completed in 0.34 sec
```
monotonicity returns a histogram plot with the features ranked by their monotonicity values. A higher monotonicity value indicates a more suitable prognostic parameter. For instance, the candidate feature Current has the highest monotonic trend for machines with FaultCode 1.
Input Arguments
collapse all
Lifetime data, specified as a cell array of matrices, cell array of tables and timetables,
fileEnsembleDatastore object, table, or timetable. Lifetime data
contains run-to-failure data of the systems being monitored. The term
lifetime here refers to the life of the machine defined in
terms of the units you use to measure system life. Units of lifetime can be quantities
such as the distance traveled (miles), fuel consumed (gallons), or time since the start
of operation (days).
If X is
a cell array of matrices or tables, the function assumes that each matrix
or table contains columns of lifetime data for a system. Each column of
every matrix or table, except the first column, contains data for a
prognostic variable. 'Var1','Var2', ... can be used to
refer to the matrix columns that contain the lifetime data. For instance,
the file machineDataCellArray.mat contains a 1-by-10 cell
array of matrices C, where each of the 10 matrices
contains data for a particular machine.
a table or timetable, the function assumes that each column, except the
first one, contains columns of lifetime data. The table variable names can
be used to refer to the columns that contain the lifetime data. If
lifetimeVar is not specified when
X is a table, then the first data column is used as
the lifetime variable.
a fileEnsembleDatastore object, specify the data
variables dataVar and member variables
memberVar to be used. If
lifetimeVar is not specified, then the first data
column is used as the lifetime variable for computation.
Each numerical member in X is of type double.
Lifetime variable, specified as a string or character vector. lifetimeVar
measures the lifetime of the systems being monitored and the lifetime data is sorted
with respect to lifetimeVar. The value of
lifetimeVar must be a valid ensemble or table variable
name.
For a cell array of matrices, the value 'Time' can be used to refer to the
first column of each matrix, which is assumed to contain the lifetime variable. For
instance, the file machineDataCellArray.mat contains the cell array
C, where the first column in each matrix contains the lifetime
variable while the other columns contain the data variables.
Data variables, specified as a string array, character vector, or cell array of character
vectors. Data variables are the main content of the members of an ensemble. Data
variables can include measured data or derived data for the analysis and development of
predictive maintenance algorithms.
If X is
a fileEnsembleDatastore object, the value of
dataVar supersedes the
DataVariables property of the ensemble.
a cell array of matrices, the value 'Time' can be used
to refer to the first column of each matrix, that is, the lifetime variable
lifetimeVar. 'Var1','Var2', ...
can be used to refer to the other matrix columns which contain the lifetime
data. For instance, the file machineDataCellArray.mat
contains the cell array C where the first column in each
matrix contains the lifetime variable. The other columns in the cell array
C contain the data variables.
a table, the table variable names can be used to refer to the columns
which contain the lifetime data.
The values of dataVar must be valid ensemble or table variable names. If
dataVar is not specified, the computation includes all data
columns except the one specified in lifetimeVar. For instance,
suppose that each entry in a cell array is a table with variables A,
B, C, and D. Setting
dataVar to ["A","D"] uses only
A and D for the computation while
C and D are ignored.
Member variable, specified as a string or character vector. Use memberVar
to specify the variable for identifying the systems or machines in lifetime data
X. For instance, in the fileEnsembleDatastore
object, the fifth column in each timetable contains numbers that identify data from a
particular machine. The column name corresponds to the member variable
memberVar.
memberVar is ignored when X is specified as a cell array of matrices or tables.
Name-Value Arguments
collapse all
Specify optional pairs of arguments as
Name1=Value1,...,NameN=ValueN, where Name is
the argument name and Value is the corresponding value.
Name-value arguments must appear after other arguments, but the order of the
pairs does not matter.
Before R2021a, use commas to separate each name and value, and enclose
Name in quotes.
Example: ...,'Method','rank'
Lifetime variable, specified as the comma-separated pair consisting of
'LifeTimeVariable' and either a string or character vector. If
'LifeTimeVariable' is not specified, then the first data column
is used.
'LifeTimeVariable' is equivalent to the input argument
lifetimeVar.
Data variables, specified as the comma-separated pair consisting of
'DataVariables' and either a string array, character vector or
cell array of character vectors.
'DataVariables' is equivalent to the input argument
dataVar.
Member variables, specified as the comma-separated pair consisting of
'MemberVariable' and either a string or character vector.
'MemberVariable' is equivalent to the input argument
memberVar.
Grouping criterion, specified as the comma-separated pair consisting of
'GroupBy' and either a string or character vector. Use
'GroupBy' to specify the variables for grouping the lifetime data
X by operating conditions.
The function computes the metric separately for each group that results from applying
the criterion, such as a fault condition, specified by 'GroupBy'. For
instance, in the fileEnsembleDatastore object ens,
the fourth column in each timetable in ens contains the variable
'FaultCode'. The metric is computed for each machine by grouping
the data by 'FaultCode'.
You can only group variables when X is defined as a
fileEnsembleDatastore object, table, timetable, or cell array of
tables or timetables.
Size of the centered moving average window for data smoothing, specified as the
comma-separated pair consisting of 'WindowSize' and either a scalar
or two-element vector. A Savitzky-Golay filter is used for data smoothing. For more
information, see smoothdata.
If 'WindowSize' is not specified, the window length is automatically determined from lifetime data X using smoothdata(X,'sgolay'). Set 'WindowSize' to 0 to turn off data smoothing.
Method to compute monotonicity, specified as the comma-separated pair consisting
of 'Method' and either 'sign' or
'rank'.
'sign', Use the signum formula.
'rank', Use Spearman's rank correlation formula.
For more information, see Algorithms.
Output Arguments
collapse all
Monotonicity of lifetime data, returned as a vector or table.
monotonicity characterizes the trend of a feature as the system
evolves toward failure. As a system gets progressively closer to failure, a suitable
condition indicator typically has a monotonic trend. Conversely, any feature with a
non-monotonic trend is a less suitable condition indicator. The values of
Y range from 0 to 1.
Y is 1 if X is perfectly
monotonic.
Y is 0 if X is perfectly
non-monotonic.
Selecting appropriate estimation parameters out of all available features is the
first step in building a reliable remaining useful life prediction engine. The
monotonicity values in Y are useful to determine which condition
indicators best track the degradation process of the systems being monitored. The higher
the monotonic trend, the more desirable the feature is for prognostics.
When 'GroupBy' is not specified, then Y is
returned as a row vector or single-row table. Conversely, when
'GroupBy' is specified, then each row in Y
corresponds to one group.
Limitations
When X is a tall table or tall timetable,
monotonicity nevertheless loads the complete array into memory using
gather. If the memory available is inadequate, then monotonicity returns an
error.
Algorithms
collapse all
Monotonicity is computed in the following two ways as specified by the
'Method' option.
When you specify 'Method' as 'sign', the
computation of monotonicity uses this formula:
monotonicity = 1MM∑j=1∣∣∣∣Nj−1∑k=1sgn(xj(k+1)−xj(k))Nj−1∣∣∣∣
where xj represents the vector of measurements of a feature on the jth system, M is the number of systems monitored, and Nj is the number of measurements on the jth system.
When you specify 'Method' as 'rank', the
computation of monotonicity uses this formula:
monotonicity = 1MM∑j=1∣corr(rank(xj),rank(tj))∣
where M is the number of systems monitored and tj is the vector of time points corresponding to the measurement vector xj.
References
Coble, J., and J. W. Hines.
"Identifying Optimal Prognostic Parameters from Data: A Genetic Algorithms Approach." In
Proceedings of the Annual Conference of the Prognostics and Health Management
Society. 2009.
Coble, J. "Merging Data Sources to
Predict Remaining Useful Life - An Automated Method to Identify Prognostics Parameters."
Ph.D. Thesis. University of Tennessee, Knoxville, TN, 2010.
Lei, Y. Intelligent Fault
Diagnosis and Remaining Useful Life Prediction of Rotating Machinery. Xi'an,
China: Xi'an Jiaotong University Press, 2017.
Lofti, S., J. B. Ali, E. Bechhoefer,
and M. Benbouzid. "Wind turbine high-speed shaft bearings health prognosis through a
spectral Kurtosis-derived indices and SVR." Applied Acoustics Vol. 120,
2017, pp. 1-8.
Version History
Introduced in R2018b
See Also
prognosability | trendability | fileEnsembleDatastore
Topics
Wind Turbine High-Speed Bearing Prognosis
Feature Selection for Remaining Useful Life Prediction
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4726 | https://swlimo.southwestlondon.icb.nhs.uk/wp-content/uploads/2021/06/Vitamin-D-Guidelines-for-Children-and-Young-People.pdf | Prepared by: The Medicines Optimisation Team, Croydon CCG Lead author: Shreeya Patel/ Hema Shah Approved by: Croydon Prescribing Committee Date approved: January 2019 Review Date: January 2021 Guidelines for the Management of Vitamin D Deficiency in Primary Care Children and Young People – January 2019 1 Table of Contents Purpose and Scope 2 Sources of vitamin D 2 Sunlight (this section should be read in conjunction with NICE guidelines 34: Sunlight exposure risks and benefits) ............................................................................................................................... 2 Diet ...................................................................................................................................................... 3 Supplements. ...................................................................................................................................... 3 General Lifestyle Advice 3 Groups at risk of Vitamin D deficiency 4 Other risk factors for Vitamin D deficiency 4 Symptoms of Vitamin D deficiency 5 Testing for Vitamin D deficiency 5 Recommended Investigations 6 Radiological assessment ..................................................................................................................... 6 Initial blood tests ................................................................................................................................ 6 Additional blood tests ......................................................................................................................... 6 Classification of vitamin D status 6 Indications for specialist referral 6 Treatment of vitamin D deficiency 7 Prescribing considerations .................................................................................................................. 7 Calcium supplementation ................................................................................................................... 8 Product selection and dosing .............................................................................................................. 9 Monitoring requirements during vitamin D treatment 10 Vitamin D toxicity 10 Contraindications to vitamin D supplementation 10 Cautions and drug interactions 10 Appendix 1: Flowchart: Investigation and Treatment of Vitamin D deficiency and insufficiency for Children and Young people 11 Appendix 2: Licensed Vitamin D preparations of colecalciferol products available (January 2019) and their suitability in certain diets (Children) 12 Appendix 3: Examples of OTC (Over the Counter) Vitamin D preparations available for children 14 Appendix 4 Healthy Start Vitamins and Drops Collection Points within Croydon 15 References 17 2 Purpose and Scope This guideline is intended to support primary care and secondary care (Croydon Health Services, CUH) clinicians with the prevention, identification and management of vitamin D deficiency within children (0-18 years). It incorporates recommendations from the Royal College of Paediatrics & Child Health Guide for Vitamin D in Childhood and National Osteoporosis Society Vitamin D and Bone Health: A Practical Clinical Guideline for Management in Children and Young People: June 2015. This guideline relates only to the management of vitamin D deficiency to promote optimal bone health and does not address the use of vitamin D for other potential indications such as auto-immune disease, cancer, mental health problems and cardiovascular disease. This guideline does not address the management of vitamin D deficiency in children who are under the care of a paediatrician, children who have severe or end-stage chronic kidney disease, severe liver disease, malabsorption syndromes, unexplained bone pain, unusual fractures and other evidence of metabolic bone disorders. Specialist advice should be sought for the management of such children. Sources of vitamin D The sources of vitamin D are diet and sunlight exposure, with sunlight being a major source in most people. Approximately 80-90% of human vitamin D supply is manufactured in skin under the stimulus of solar ultraviolet B (UVB) light, the other 10-20% comes from the diet. Sunlight (this section should be read in conjunction with NICE guidelines 34: Sunlight exposure risks and benefits) Environmental and personal factors greatly affect vitamin D production in the skin, making it difficult to recommend a one-size-fits-all level of exposure for the general population. In the UK, sunlight is strongest between 11am and 3pm between March/early April and the end of September. Between 11am and 3pm most people can make sufficient vitamin D by going out for short periods and leaving areas of skin uncovered, such as forearms, hands or lower legs. Longer periods may be needed for those with darker skin. It should be advised that skin that is not usually exposed to sunlight (for example the back, abdomen and shoulders) is particularly likely to burn, so extra care is needed. Many young people will have experienced sunburn. They can use this experience to know what their skin looks like normally, how it reacts to sunlight, how long they can be exposed without risking sunburn, and how to protect their skin accordingly. Before 11am and after 3pm it takes longer to synthesise sufficient vitamin D, however the risk of sunburn is less. Exposure to sunlight through windows is insufficient because glass blocks UVB light which is required for vitamin D synthesis. In the UK, from October to March, there is no ambient UVB sunlight of appropriate wavelength to generate skin synthesis of vitamin D. During this time the population relies on both body stores from sun exposure in the summer and dietary sources to maintain vitamin D status. 3 Diet Vitamin D2 is found in foods of non-animal origin (particularly mushrooms) and vitamin D3 in foods such as fatty fish, fish liver oil and egg yolk. Food sources which contain greater than 200 units (5micrograms) of vitamin D per portion include: 2 teaspoons of cod liver oil (also contains vitamin A which can be harmful in high doses) 70g sardines 100g tinned salmon, pilchards or tuna 110g of cooked mackerel or herring 130g cooked kipper Small amounts of vitamin D are provided by egg yolk, red meat and fortified foods, such as formula milks for infants and toddlers (plain cow’s milk is not fortified in the UK), some breakfast cereals, fat spreads (margarine) and some yogurts. Consumption of food sources alone, in the absence of skin synthesis, will not provide optimal vitamin D status. Supplements Following the review by the Scientific Advisory Committee on Nutrition (SACN) on the evidence of vitamin D and health, Public Health England (PHE) have advised that to protect bone and muscle health, everyone (children from the age of 1 years) needs vitamin D equivalent to an average daily intake of 400 units (10 micrograms). In the UK, during spring and summer, the majority of the population will get enough vitamin D through sunlight on the skin and a healthy balanced diet, however during autumn and winter; everyone will need to rely on dietary sources of vitamin D. Since it is difficult for people to meet the 400 units (10 micrograms) recommendation from consuming foods naturally containing or fortified with vitamin D, PHE advise that everyone (children from the age of 1 years) should consider taking a daily supplement containing 400 units (10 micrograms) of vitamin D during the autumn and winter months (i.e. between October to March). Those groups at high risk of vitamin D deficiency (see Table 1 below) should consider taking a daily supplement containing 400 units (10 micrograms) of vitamin D throughout the entire year. General Lifestyle Advice Appropriate lifestyle advice which encourages, ‘enjoying the sun safely whilst taking care not to burn’ (please note: exposure to sunlight through windows is insufficient because glass blocks UVB light), adequate dietary intake (including adequate calcium intake (see page 7)) and daily vitamin D supplementation where necessary, should be provided to all patients. In line with NHS England’s guidance on conditions for which over the counter items should not be routinely prescribed, South West London CCGs do not support the routine prescribing of vitamin D (colecalciferol, ergocalciferol) for maintenance or prophylaxis of vitamin D insufficiency. This includes babies and children. For more information, see the South West London Position statement on prescribing of vitamin D: vitamin%20D%20FINAL%20V2%20August%202018.pdf 4 Groups at risk of Vitamin D deficiency Public Health England recommends vitamin D supplementation in all children from birth to 4 years of age in order to prevent Vitamin D deficiency, as per Table 1. Table 1: Public Health England Recommendations on Vitamin D supplementation in the UK (Children) Groups at risk of vitamin D deficiency Breastfed infants from birth to 1 year of age Recommendation: Advise purchase of over the counter (OTC) supplements containing 340-400 units (8.5 - 10 micrograms /per day) Children aged 1 to 4 years of age Recommendation: Advise purchase of over the counter (OTC) supplements containing 400 units (10 micrograms /per day) Infants who are fed infant formula will not need to be given vitamin D supplementation until they are receiving less than 500ml of infant formula a day, as these products are fortified with vitamin D. Children aged from four weeks old until their fourth birthday may be eligible to obtain vitamins free of charge as part of the Healthy Start Scheme. Further information on the eligibility criteria can be found on the Healthy Start website at www.healthystart.nhs.uk. At the time this guideline was finalised, existing UK supplies of Healthy Start vitamin drops contain 7.5 micrograms (300 units) of vitamin D and are licensed for infants over 4 weeks old. Clinicians should note that this formulation is not fully compliant with the dose set out within the current Public Health England recommendations. However, NHS BSA have confirmed that a revised formulation of Healthy Start vitamin drops that is fully compliant with the recommended 400 units (10 micrograms of vitamin D) per dose is anticipated to be in the supply chain by July 2019. Parents/carers of children who are not eligible for Healthy Start, can purchase Healthy Start vitamin drops from some community pharmacies or be advised to buy an over the counter multivitamin preparation with similar vitamin and mineral composition to the Healthy Start vitamins, which are available at most pharmacies and larger supermarkets. Care should be taken with multivitamin preparations as Vitamin A toxicity is a concern. Other risk factors for Vitamin D deficiency Malabsorption conditions (Crohn’s disease, coeliac disease, short bowel syndrome, cystic fibrosis) Vegan or vegetarian diets or generally poor diets Pigmented skin (includes people of African, African-Caribbean and South Asian family origin). Limited sun-exposure (those who cover their skin for cultural reasons or health reasons e.g. those with skin photosensitivity or immobile/disabled patients) Obesity (BMI >98th BMI for age centile) Chronic liver or renal disease Medications that can increase metabolism or reduce absorption of vitamin D (e.g. carbamazepine, phenytoin, primidone, barbiturates, oral glucocorticoids (i.e. for 3 months or longer), rifampicin, colestyramine, colestipol, orlistat and some antiretrovirals Family members with proven vitamin D deficiency. 5 Symptoms of Vitamin D deficiency Symptoms of vitamin D deficiency are vague and it can be difficult to ascertain whether a low vitamin D is causal or a surrogate marker. Symptoms that could be attributed to vitamin D deficiency (especially if patients are in high risk groups) include: Infants Seizures Tetany Cardiomyopathy Children Aches and pains (long standing of more than 3 months) Myopathy causing delayed walking Rickets with bowed legs – wide range of related skeletal defects, swelling of costochondral junctions, Knock knees Poor growth and muscle weakness Fractures following minor trauma Adolescents Aches and pains (long standing of more than 3 months) Muscle weakness Bone changes of rickets or osteomalacia Fractures following minor trauma Abnormal Investigations Low serum calcium or phosphate, high alkaline phosphatase (greater than the local age appropriate reference range) Radiographs – showing osteopenia, rickets or pathological fractures revealed by radiographs Testing for Vitamin D deficiency It is not recommended to test vitamin D levels in patients at high risk of vitamin D deficiency, unless they show symptoms of deficiency. Efforts should be focused on giving appropriate lifestyle advice and encouraging daily supplementation of vitamin D (see page 3). Vitamin D testing should be considered in the following: Patients with disease where outcomes may be improved with vitamin D treatment o Confirmed rickets Patients with signs or symptoms that could be attributed to vitamin D deficiency (especially if patients are in high risk groups) o See “Symptoms of Vitamin D deficiency” o Other causes for symptoms should be excluded Disorders impacting on vitamin D metabolism o Malabsorption conditions (coeliac disease, cystic fibrosis). o Chronic renal or liver disease Bone diseases in children where correction of vitamin D deficiency prior to treatment would be indicated (under specialist recommendation): o Osteogenesis imperfecta o Idiopathic juvenile osteoporosis o Osteoporosis secondary to glucocorticoids, inflammatory disorders, immobility and other metabolic bone conditions. Routine vitamin D testing is not recommended to screen the normal population for deficiency. Vitamin D testing should be prioritised to those where the outcome will alter clinical management 6 Recommended Investigations Radiological assessment (if rickets suspected): To provide a definitive diagnosis of rickets -a radiological assessment. The long bone may show cupping, splaying and fraying of the metaphysis. Initial blood tests: Serum 25 vitamin D, calcium, phosphate, alkaline phosphatase (ALP) and parathyroid hormone (PTH) (if the patient has rickets or hypocalcaemia). Additional blood tests: Renal function (to exclude renal failure), liver function tests (to exclude hepatic failure), full blood count (to identify possible vitamin deficiencies), thyroid function tests, inflammatory markers (ESR & C-reactive protein), malabsorption screen, rheumatoid and other autoimmune screening. Classification of vitamin D status Table 2 outlines the classification of vitamin D status and recommended management strategies. These are broadly in line with Royal College of Paediatrics & Child Health Guide for Vitamin D in Childhood and National Osteoporosis Society Vitamin D and Bone Health: A Practical Clinical Guideline for Management in Children and Young People: June 2015. Table 2: Classification of vitamin D status, associated effects on bone health and management strategies Serum vitamin D concentration Vitamin D status Management <30 nmol/L Deficient Prescribed high-dose vitamin D followed by purchased over the counter (OTC) long term maintenance vitamin D supplementation PLUS lifestyle advice 30-50 nmol/L Insufficient Purchased over the counter (OTC) long term maintenance vitamin D supplementation PLUS lifestyle advice >50 nmol/L (Optimal vitamin D levels > 75 nmol/L) Sufficient Reassurance and lifestyle advice NB: Vitamin D toxicity is known to occur at serum 250HD values above 375nmol/L Indications for specialist referral All patients under 2 years with symptoms of vitamin D deficiency Patients with skeletal deformities/short stature or orthopaedic abnormalities related to rickets Patients with symptomatic/asymptomatic hypocalcaemia Patients with contraindications to vitamin D supplementation: severe renal impairment (CKD stage 4 or eGFR < 30mL/minute/1.73 m2), hypercalcaemia or metastatic calcification, primary hyperparathyroidism, renal stones, severe hypercalciuria and nephrocalcinosis Patients developing hypercalcaemia following supplementation Failure to respond to treatment after 3 months Patients with suspicion of, or with, malabsorption conditions Patients with suspicion of, or with, liver disease or tuberculosis Patients taking medications that can increase the risk of vitamin D deficiency (e.g. certain antiepileptics or oral corticosteroids) or risk of vitamin D toxicity (e.g. thiazide diuretics, digoxin) 7 Treatment of vitamin D deficiency Prescribing considerations Key aims for treating vitamin D deficiency are to ensure correction of vitamin D deficiency (serum vitamin D ideally >50nmol/L), reverse the clinical consequences of vitamin D deficiency in a timely manner and to avoid toxicity. The BNF for Children (BNFC) and National Osteoporosis Society Vitamin D and Bone Health: A Practical Clinical Guideline for Management in Children and Young People (NOS) recommends the following daily doses (see Table 3) for treatment of vitamin D deficiency over a period of 8-12 weeks. Whereas the Royal College of Paediatric & Child Health (RCPCH) Guide for Vitamin D in Childhood recommends the same daily doses over a period of 4-8 weeks. RCPCH Vitamin D in Childhood guidelines have been used to establish a cumulative treatment dose of Vitamin D required for each age range. This is a more conservative cumulative treatment dose than the BNFC and NOS but will better enable use of licensed colecalciferol products at licensed doses, where possible, especially given that the cumulative treatment dose of currently licensed preparations is significantly lower than that recommended in the BNFC and NOS. Table 3: Recommended doses for the treatment of vitamin D deficiency Age Daily treatment dose (Based on BNFC, NOS, and RCPCH) Cumulative Vitamin D treatment dose (Based on RCPCH) Up to 6 months 1,000 – 3,000 units daily 28,000 – 168,000 units 6 months up to 12 years 6,000 units daily 168,000 – 336,000 units 12-18 years 10,000 units daily 280,000 – 560,000 units Oral vitamin D is the preparation of choice. Whilst intramuscular administration results in 100% adherence, it has an unpredictable bioavailability, slower onset of action and is associated with the additional administration burden in comparison to oral preparation. As a fat soluble vitamin, oral vitamin D preparations should preferably be taken with a meal, to aid absorption. Colecalciferol (vitamin D3) is considered the preferred form of vitamin D for treatment, as it has been reported to raise vitamin D levels more effectively than ergocalciferol (vitamin D2) and has a longer duration of action. A colecalciferol preparation should be prescribed unless this is unacceptable to the patient. Colecalciferol (vitamin D3) is commercially synthesised from an animal source such as lanolin/wool fat from sheep’s wool. If the animal is not harmed in the manufacture of the raw vitamin D3, the product could be considered suitable for vegetarians. It is advised that the source of vitamin D3 be discussed with vegans, to enable them to make an informed choice as to whether this is acceptable to them or not. Ergocalciferol (vitamin D2) is derived from a common plant steroid, and could be used at equivalent doses (in a gelatine free product) for vegans who do not find it acceptable to take colecalciferol containing products. Prescribing of intramuscular vitamin D preparations should remain with the specialist, as this is usually reserved for those patients with compliance issues or malabsorption conditions. 8 The choice of preparation should be discussed for patients who have various dietary requirements (e.g. halal, kosher, vegan or vegetarian, soya allergies), so that an informed decision can be made. There are many licensed preparations of colecalciferol available for high dose (and maintenance vitamin D supplementation, if appropriate) which can be prescribed to meet the needs of various dietary requirements (e.g. vegetarian, allergies), see Appendix 2. Therefore, it is expected that prescribing of unlicensed preparations, will only be undertaken in exceptional circumstances to meet the specific needs of an individual. If in doubt of the suitability of a preparation, contact the Medicines Optimisation Team for further advice. Licensed preparations should be prescribed by BRAND, where possible to avoid potential for dispensing unlicensed preparations or preparations that do not meet the dietary requirements of specific patients. It is recommended that where high dose vitamin D is prescribed, this is prescribed as an acute prescription, to avoid inadvertent repeated prescribing. Calcium supplementation Many children with Vitamin D deficiency rickets have a poor dietary calcium intake. As their bones are growing, there is a greater risk of negative calcium balance. Therefore in children, consideration should always be given to the need for calcium supplementation. Consideration should be given to the patient’s dietary intake of calcium, possibly by using “calcium calculators” e.g. The recommended daily intake of calcium is: Age Recommended calcium intake Younger than 12 months 525mg (13.1 mmol) 1-3 years 350mg (8.8 mmol). 4-6 years 450mg (11.3 mmol) 7-10 years 550mg (13.8 mmol) 11-18 years (boys) 1000mg (25.0 mmol) 11-18 years (girls) 800mg (20.0 mmol) Many children with vitamin D deficiency will have a depleted calcium status and/or a poor calcium intake and may therefore benefit from advice about dietary calcium intake (see the British Dietetic Association (BDA) factsheet on calcium (available at www.bda.uk.com) for information on how the recommended daily calcium intake can be achieved). In some cases calcium supplementation may be required over the period of Vitamin D treatment. The BNF for Children provides details of recommended calcium supplementation doses. Dosing also needs to consider dietary intake and the size of the child. Alfacalcidol and calcitriol should not be used for the routine treatment of primary vitamin D deficiency, as unlike vitamin D, they carry a higher risk of toxicity and require long term monitoring. 9 Product selection and dosing Table 4: Preferred products and doses of licensed colecalciferol preparations for management of vitamin D deficiency and insufficiency in Children and Young people Vitamin D status and level (nmol/L) Age Product Dose and Frequency Deficient <30 nmol/L Up to 6 months Thorens 10,000 units/ml oral drops 2,000 units (10 drops) daily for 6 weeks followed by maintenance vitamin D supplementation 6 months up to 12 years 2,000 units (10 drops) daily for 12 weeks (NB: off license dosing frequency) followed by maintenance vitamin D supplementation 12 years up to 18 years Tablets and capsules Plenachol 20,000 unit capsules 20,000 units capsules once a week for 15 weeks (NB: off license dosing frequency) followed by maintenance vitamin D supplementation Stexerol 25,000 unit tablets (can be swallowed whole or crushed) 25,000 units tablets once a week for 12 weeks (NB: off license dosing frequency) followed by maintenance vitamin D supplementation Liquids: Prescribe for patients unable to swallow tablets and capsules Thorens 25,000 units/2.5ml oral solution 25,000 units once a week for 12 weeks (NB: off license dosing frequency) followed by maintenance vitamin D supplementation Insufficient 30-50 nmol/L 1 month up to 18 years Advise purchase of OTC vitamin D supplements to provide 400 – 600 units daily (maintenance therapy) (see Appendix 3) Maintenance Therapy After a treatment course with vitamin D, the following is recommended: 1. Lifestyle and dietary advice: Between March/early April to end of September short daily periods of sun exposure of skin (forearms, hands or lower legs), taking care not to burn. Dietary vitamin D can be acquired from oily fish, eggs and fortified cereals yogurts and margarine. 2. Maintenance Therapy: (See SWL CCG position statement on the prescribing of vitamin D) SWL CCG recommends maintenance therapy is purchased over the counter (OTC) in the first instance. Prescriptions should only be considered for patients who have undergone two or more previous treatment courses and other special groups (patients who have medical conditions that may pre-dispose them to inadequate vitamin D levels and where clinical assessment of the patient indicates that continuous treatment is justified e.g. conditions resulting in intestinal malabsorption e.g. short bowel syndrome) Should it be necessary to prescribe maintenance therapy, the following options are recommended: 1 - 6 months Thorens 10,000 unit/ml oral drops 400 units (2 drops) – 600 units (3 drops) daily 6 months up to 12 years 400 units (2 drops) – 600 units (3 drops) daily 12-18 years Tablets and capsules: Plenachol 20,000 unit capsules 20,000 unit capsule once every 6 weeks Stexerol 25,000 unit tablets 25,000 unit tablet once every 6 weeks Liquids: Thorens 10,000 units/ml oral drops 400 units (2 drops) – 600 units (3 drops) daily 3. Ensure calcium intake assessed, dietary advice provided and calcium supplementation prescribed where considered appropriate (see pages 7-8) 10 Monitoring requirements during vitamin D treatment Adjusted serum calcium should be checked 1 month after completing high dose vitamin D treatment or after starting maintenance vitamin D supplementation (or more regularly e.g. every 1-2 weeks in the first month of treatment in patients receiving calcium supplements in addition to high dose vitamin D treatment), in case hyperparathyroidism has been unmasked or to detect hypercalcaemia/ determine how long calcium supplementation is needed. Serum vitamin D levels should be repeated at the end of high dose vitamin D therapy. Serum vitamin D levels will take 3-6 months to reach steady state after high dose vitamin D therapy. Vitamin D levels should therefore be checked at least 3 months after treatment with high dose vitamin D. Patients who do not respond to treatment with high dose vitamin D may be considered for referral to secondary care. Vitamin D toxicity Vitamin D toxicity, defined as hypercalcaemia, is known to occur at serum vitamin D levels of above 375nmol/L. However, toxicity is rare and is unlikely to occur with recommended supplemental or therapeutic doses. Research has suggested vitamin D below 10,000 units/day is not usually associated with toxicity, whereas doses equal to or above 50,000 units/day for several weeks or months are frequently associated with toxicity. The Scientific Advisory Committee on Nutrition (SANC) has accepted the European Food Safety Authority recommendations of a safe upper limit of 1,000 units/day for infants up to 1 year old, 2,000units/day for children aged 1-10 years old and 4,000units/day for those older than 10 years old. Early symptoms of toxicity will include symptoms of hypercalcaemia, such as thirst, polyuria, constipation, nausea and vomiting and weight loss. Toxicity can lead to renal failure if left untreated, therefore, if toxicity is suspected, vitamin D must be withdrawn and serum calcium and renal function checked urgently. In severe cases, patients may require emergency inpatient care with rehydration therapy. Contraindications to vitamin D supplementation Vitamin D is contraindicated in patients with severe renal impairment (CKD stage 4 or eGFR <30mL/minute/1.73 m2), hypercalcaemia or metastatic calcification. Relative contraindications include primary hyperparathyroidism, renal stones, severe hypercalciuria and nephrocalcinosis. Cautions and drug interactions This list is not exhaustive. Please refer to individual product Summary of Characteristics accessible via: for further details. Effects of digoxin and other cardiac glycosides may be accentuated by vitamin D Risk of hypercalcaemia is increased in patients taking thiazide diuretics Concomitant treatment with carbamazepine, phenytoin, barbiturates, primidone, corticosteroids and some antiretroviral activation can decrease effect of vitamin D 11 Appendix 1: Flowchart: Investigation and Treatment of Vitamin D deficiency and insufficiency for Children and Young people Please refer to full guidelines for further details. Does the patient have…? A disease with outcomes that may be improved with vitamin D treatment: Confirmed rickets OR Symptoms that could be attributed to vitamin D deficiency: Infants: Seizures, Tetany, Cardiomyopathy Children: Aches and pains (long standing of more than 3 months), myopathy causing delayed walking, poor growth and muscle weakness, fractures following minor trauma Adolescents: Aches and pains (long standing of more than 3 months), muscle weakness, fractures following minor trauma OR Disorders impacting on vitamin D metabolism Malabsorption conditions (Coeliac disease, short bowel syndrome, cystic fibrosis) or chronic renal or liver disease Is the child aged between 0 - 4 years old? Breastfed infants from birth to 1 year of age Advise purchasing OTC supplements to provide 340 – 400 units of colecalciferol daily and provide lifestyle advice (see below) Infants who are fed infant formula will not need vitamin drops until they are receiving less than 500ml of infant formula a day, as these products are fortified with vitamin D. Children aged 1 to 4 years of age Advise purchasing OTC supplements to provide 400 units of colecalciferol daily and provide lifestyle advice (see below) Children aged from 4 weeks to 4 years may be eligible to obtain vitamins free of charge as part of the Healthy Start Scheme. (www.healthystart.nhs.uk). TEST for vitamin D deficiency: Serum vitamin D; ALP; Calcium; Phosphate; U&E LFTs, FBC & PTH (if indicated) Do any of the following apply? Child under 2 years old Child with symptomatic/asymptomatic hypocalcaemia Child with skeletal deformities/short stature or orthopaedic abnormalities related to rickets. Contraindications to vitamin D supplementation: severe renal impairment (CKD stage 4 or eGFR < 30ml/minute/1.73m2) Hypercalcaemia following supplementation Failure to respond to treatment after 3 months Suspicion of, or with, malabsorption conditions Suspicion of, or with, liver disease or tuberculosis Children taking medications that can increase the risk of vitamin D deficiency (e.g. certain antiepileptics or oral corticosteroids) or risk of vitamin D toxicity (e.g. thiazide diuretics, digoxin) Serum vitamin D: <30nmol/L – Deficiency Serum vitamin D: 30-50nmol/L – Insufficiency Serum vitamin D: >50nmol/L – Sufficient TREATMENT DOSING: TO BE PRESCRIBED Up to 6 months old: Thorens 10,000 unit/ml oral drops: 2,000 unit (10 drops) daily for 6 weeks 6 months up to 12 years old: Thorens 10,000 unit/ml oral drops: 2,000 units (10 drops) daily for 12 weeks 12 to 18 years old: Plenachol capsules: 20,000 unit capsules once a week for 15 weeks Stexerol 25,000 unit tablets: 25,000 unit tablets once a week for 12 weeks OR if unable to swallow capsules/tablets Thorens 25,000 unit/2.5ml oral solution: 25,000 units weekly for 12 weeks Off-license dosing frequency MAINTENANCE DOSING Promote self-care: Advise purchasing OTC supplements to provide 400 – 600 units of colecalciferol daily Croydon CCG recommends maintenance therapy is purchased over the counter (OTC) in the first instance Prescriptions for maintenance therapy should only be considered for patients who have undergone two or more previous treatment courses and in other special groups (see table on page 9 for more details) Ensure calcium intake assessed, dietary advice provided and calcium supplementation prescribed where considered appropriate MONITORING Check calcium levels 1 month after high dose Vitamin D replacement and re-check serum vitamin D levels after at least 3 months Have other causes for symptoms been excluded? Refer to Secondary Care Assess need for treatment based on serum vitamin D level REASSURANCE & LIFESTYLE ADVICE Safe sun exposure: Between March/early April to the end of September short periods of exposure of skin that are more often uncovered (such as forearms, hands or lower legs) between 11am and 3pm can achieve adequate Vitamin D levels. Dietary sources: Oily fish, egg yolks, fortified cereals, yogurts & margarines YES YES NO YES YES S NO NO Further investigation 12 Appendix 2: Licensed Vitamin D preparations of colecalciferol products available (January 2019) and their suitability in certain diets (Children) Name and Form Price Unit Price for 12-18 year treatment course (300,000 units) Suitability Preparations contains Halal Kosher Vegetarian Vegan Soya Nut Gelatine Lactose Colecalciferol 200 units Fultium-D3 2740 units/ml oral drops Dose: Please check SPC for the dose recommendations £10.70 / 25ml £42.80 Not halal certified Not kosher certified See notes Thorens 10,000 units/ml oral drops Dose: Please check SPC for the dose recommendations £5.85 / 10ml £17.55 See notes Colecalciferol 800 units Desunin 800 unit tablets Dose: Please check SPC for the dose recommendations £3.60 / 30 tablets £45.00 See notes See notes Fultium-D3 800 unit capsules Dose: Please check SPC for the dose recommendations £3.60 / 30 capsules £45.00 See notes See notes Colecalciferol 1,000 units Stexerol-D3 1,000 unit tablets Dose: Please check SPC for the dose recommendations £2.95 / 28 tablets £31.60 See notes See notes See notes See notes See notes Colecalciferol 20,000 units Aviticol 20,000 unit capsules Dose: Please check SPC for the dose recommendations £29.00 / 30 capsules £14.50 See notes Fultium-D3 20,000 unit capsules Dose: Please check SPC for the dose recommendations £17.04 / 15 capsules or £29.00 / 30 capsules £14.50 - £17.04 See notes Plenachol 20,000 unit capsules Dose: Please check SPC for the dose recommendations £9.00 / 10 capsules £13.50 See notes See notes See notes See notes See notes See notes Colecalciferol 25,000 units InVita D3 25,000 unit dose vials (SF oral solution) Dose: Please check SPC for the dose recommendations £4.45 / 3 x 1ml £17.80 Not halal certified Not kosher certified See notes Thorens 25,000 units/2.5ml oral solution Dose: Please check SPC for the dose recommendations £1.55 / 2.5ml £18.60 See notes Stexerol 25,000 unit tablets Dose: Please check SPC for the dose recommendations £17.00 / 12 tablets £17.00 See notes See notes See notes See notes See notes Only licensed in 12-18 year olds preferred products 13 Additional notes on products to discuss with patients who have specific dietary requirements Fultium-D3® (Oral drops, 800 unit & 20 000 unit capsules) Drops: The active ingredient is derived from the wool of sheep, the sheep are alive when the wool is taken for the lanolin. The drops contains no peanuts or tree nuts. Capsules: The gelatine used in the capsules is derived from beef bone and hide. Only the gelatine in the capsule is halal certified. The remaining constituents of the product have not been halal certified. The capsules are Kosher certified. Fultium-D3 capsules contains soybean oil. Note: Fultium-D3 800 unit capsules previously contained nuts (arachis oil) but are now reformulated with maize oil. Previous batches may still be in circulation so patients should consult their pharmacist when receiving them. Thorens® (25 000 unit oral solution & 10,000 unit oral drops) Thorens oral solution contains colecalciferol and refined olive oil. The colecalciferol is derived from cholesterol from wool grease (lanolin). Thorens is considered to be suitable for vegetarian patients but is not considered suitable for vegan patients. The colecalciferol in Thorens is halal and kosher-certified. Thorens can be mixed with a small amount of children's food, yogurt, milk, cheese or other dairy products. The parents should be warned not to mix Thorens into a bottle of milk or container of soft foods in case the child does not consume the whole portion, and does not receive the full dose. The parents should ensure that their child takes the entire dose. In children who are not breast-feeding, the prescribed dose should be administered with a meal. Thorens 10,000 unit oral drops comes with a dropper and has a shelf life of 6 months from first opening of bottle. Desunin® (800 unit tablets) The manufacturers have confirmed that the product is suitable for vegetarians. The source of colecalciferol (vitamin D3) is synthetically produced in a process that includes wool grease/lanolin from healthy live sheep. The manufacturer is not able to confirm whether the product is suitable for a Halal or Kosher diet. They cannot guarantee that the product is lactose free. Stexerol-D3® (25,000 unit & 1,000 unit) tablets The manufacturer states that it is suitable for vegetarians, however they have not sought any accreditation for vegetarian use. They have confirmed that the finished products do not contain excipients of animal origin; however, the active ingredient colecalciferol is derived from the lanolin contained in sheep’s wool, and therefore may not be suitable for all vegans. They state that the products do not contain ingredients sourced from nuts or soya derivatives; however, they cannot guarantee that the products are manufactured in a nut or soya free environment. The product does not contain lactose, however, they are unable to guarantee that the product has not come into contact with lactose during the manufacturing process. Aviticol® (20 000 unit capsules) The capsules are free from peanuts and peanut derivatives. However, it cannot be guaranteed that they are free from other nuts. The product may have come into contact with nut ingredients during manufacture or packaging. Plenachol® (20,000 unit capsules) The manufacturers confirm that the product is suitable for vegetarians, is Halal and Kosher certified, does not contain lactose, soya or gelatine and is free from peanuts or peanut oil; although they cannot guarantee that the product does not come into contact with any of these during manufacture. They cannot guarantee that the product is free from all nuts during each stage of manufacture. The Vitamin D3 used in the production of Plenachol is synthetically derived from the wool grease from live sheep. InVita D3® (25 000 unit oral solution) It does not use any ingredients from slaughtered animals, it does not contain gelatine or porcine sourced materials and it is alcohol-free. The vitamin D3 is sourced from lanolin from live sheep’s wool. Other The choice of product should be discussed with patients who have halal, kosher, vegan or vegetarian requirements, so that they may make an informed decision. 14 Appendix 3: Examples of OTC (Over the Counter) Vitamin D preparations available for children Product Dose Age Suitability Healthy Start drops (10ml) 5 drops (300 unit colecalciferol) once daily. Children aged up to the age of 4 years who are registered with a Croydon GP. • Suitable for vegetarians • Free from milk, egg, gluten, soya and peanut residues •Shelf life of 10 months from manufacture •Available to purchase OTC or free to eligible patients: DaliVit (25ml) 0.6ml or 14 drops once daily (400 unit colecalciferol). (This dose is unlicensed in children less than 1 year of age.) DaliVit multivitamin drops can be taken by children from 6 weeks. • Does not contain peanut oil • Suitable for vegetarians and vegans • No added colours Abidec (25ml) 0.6ml or 14 drops once daily (400 units colecalciferol) (This dose is unlicensed in children less than 1 year of age.) From birth. • Contains peanut oil • Once opened use within 4 weeks 15 Appendix 4: Healthy Start Vitamins and Drops Collection Points within Croydon Note each collection point operates individual times where members of the public can exchange their Healthy Start vouchers, please ensure you use link below for full details of collection times. Children Centres Aerodrome Children’s Centre Violet Lane (large red and cream building) Croydon, CR0 4HN Tel: 020 8688 4975 Byron Children’s Centre St. David’s (Off Stoneyfield Road), Coulsdon, Surrey, CR5 2XE Tel: 020 8763 6285 Woodside Children’s Centre Morland Road, Croydon, CR0 6NF Tel:020 8655 5655 Kensington Avenue Children’s Centre Buckingham Avenue, Thornton Heath, Surrey CR7 8AS -or- Hawthorn Avenue, Thornton Heath, Surrey, CR7 8BW Tel:020 8765 8128 Winterbourne Children’s Centre Winterbourne Road, Thornton Heath, Surrey CR7 7QT Tel: 020 8689 0978 Fairchildes Children’s Centre Fairchildes Avenue, New Addington, Croydon, CR0 0JD Tel: 01689 847136 Castle Hill Children’s Centre Castle Hill Academy, Dunley Drive,New Addington, Croydon, CR0 0RJ Purley Oaks Children’s Centre Bynes Road, Purley, South Croydon, CR2 0PR Tel: 020 8325 4517 Selhurst Children’s Centre 23 Dagnall Park Road, SE25 5PL Tel: 020 8684 3777 Shirley Children’s Centre 34 Lilac Gardens, Shirley, CR0 8NR Tel: 020 8777 2119 Woodlands Children’s Centre Woodlands Children’s Centre, Gilbert Scott Primary School, Farnborough Avenue, South Croydon, CR2 8HD Tel: 020 8916 0543 16 Health Centres Thornton Heath Health Centre 61A Gillett Road, Thornton Heath CR7 8RL Tel: 0208 664 1590 Purley War Memorial Hospital 856 Brighton Road, Purley CR8 2YL Tel: 020 8401 3000 Parkway Health Centre Parkway, Croydon, CR0 0JA Tel: 01689 808810 Woodside Clinic 3 Enmore Road South Norwood, Croydon, London SE25 5NT Tel: 020 8274 6900 Shirley Clinic 135 Shirley Road, Croydon, CR0 7LR Tel: 020 8714 2800 For more information on the full addresses, access points and opening times, see: nd%20Drops%20-%20collection%20sites.pdf 17 References 1. The National Osteoporosis Society. Vitamin D and Bone Health: A practical clinical guideline for patient management. Available at 2. The National Osteoporosis Society. Vitamin D and Bone Health: A practical clinical guideline for management in Children and Young People. Available at 3. National Institute for Health & Care Excellence. Vitamin D: supplement use in specific Vitamin D: supplement use in specific population groups PH Guidance 56; November 2014, last updated August 2017. 4. British Association of Dermatologists. Skin Cancer; Vitamin D. Available at 5. Pearce S, Cheetham D et al, Diagnosis and management of vitamin D deficiency. BMJ 2010; 340: b5664 6. Royal College of Paediatric and Child Health. Guide for Vitamin D in Childhood. October 2013. 7. Evelina London Paediatric Formulary. Monographs: Colecaciferol (Vitamin D3) 8. Public Health England. PHE publishes new advice on Vitamin D – Press Release. Available via: 9. UKMI Q&A, Which vitamin D preparations are suitable for a vegetarian or vegan diet? Prepared by UK Medicines Information (UKMi) pharmacists for NHS healthcare professionals, Date prepared 2nd May 2017, Available via www.evidence.nhs.uk 10. Committee on Toxicity. COT Statement on Vitamin D – lay summary. December 2014. Available via 11. UKMI Q&A What dose of vitamin D should be prescribed for the treatment of vitamin D deficiency? Prepared by UK Medicines Information (UKMi) pharmacists for NHS healthcare professionals, Date prepared 18th December 2017, Available via www.evidence.nhs.uk 12. NHS South West London Croydon Borough Team. Guideline for the Management of Vitamin D Deficiency in Primary Care. June 2011. 13. Medicines and Healthcare products Regulatory Agency. Drug Safety Update: Antiepileptics Adverse Effects on bone. 1st April 2009. Available via 14. Scientific Advisory Committee on Nutrition. Vitamin D and Health report. July 2016. Available via 15. National Institute for Health & Care Excellence. Sunlight exposure risks and benefits. NG 34; February 2016. 16. Lambeth CCG. A comparison of licensed Vitamin D preparations for Adults available in the UK. November 2015. Available via committee/Lambeth%20Borough%20Prescribing%20Committee/Clinical%20Guidelines/Comparison %20table%20of%20licensed%20Vitamin%20D%20preparations%20for%20adults%20Nov%202015.p df 17. Summary of Product Characteristics for Fultium-D3 Drops. 18. Drug Tariff Online. July 2018. 19. Summary of Product Characteristics for Desunin 800 IU tablets. Available via: 20. Summary of Product Characteristics for Fultium-D3 800IU capsules. Available via: 21. Summary of Product Characteristics for Plenachol 20 000IU capsules. Available via: 22. Summary of Product Characteristics for Aviticol 20 000IU capsules. Available via: 18 23. Summary of Product Characteristics for Fultium-D3 20 000IU capsules. Available via: 24. Summary of Product Characteristics for Thorens 10 000IU/ml oral solution. Available via: 25. Summary of Product Characteristics for Thorens 25 000IU/2.5ml oral solution. Available via: 26. Summary of Product Characteristics for InVita D3 25 000IU oral solution. Available via: 27. Summary of Product Characteristics for Stexerol 1,000 IU and 25 000IU tablets. Available via: 28. National Institute for Health & Care Excellence. Clinical Knowledge Summary; Vitamin D deficiency in Children. November 2016. Available via: 29. UNICEF UK.The Baby Friendly Initiative Position Statement on Vitamin D on Vitamin D supplementation for breastfed babies. Updated January 2017.Acessed via: 30. British National Formulary for Children. Available via: 31. Healthy Start website. Available from: 32. Abidec Multivitamin Drops Patient Information Leaflet. Available from: 33. Dalivit Drops. Available from: 34. SWL CCG. Position statement on the prescribing of vitamin D (colecalciferol, ergocalciferol) for the treatment, maintenance and prophylaxis of vitamin D insufficiency or deficiency, August 2018. Available from: amin%20D%20FINAL%20V2%20August%202018.pdf |
4727 | https://www.youtube.com/watch?v=NlmsQUxHuTY | Solve the Nonlinear System: A Cubic Polynomial and a Line
The Math Sorcerer
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Posted: 22 Apr 2022
Solve the Nonlinear System: A Cubic Polynomial and a Line
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Intro hello in this problem we're going to solve this non-linear system of equations so we have x cubed minus y equals 2x and y minus 5x equals negative 6. so to do this i am thinking that one way is to use substitution so we're going to take the second equation so y minus 5x equals negative 6 and we're going to solve it for y and plug it into the first equation i think that will be a good strategy so let's go ahead and write it down again and then add 5x to both sides to solve for y so this gives us Solving for y y equals 5x minus 6. so 5 x minus 6. all right so now we're going to plug this into the first equation so plug into i'm going to give these names one and two plug into one and so when we do that we get x cubed then minus and then y but y is all of this here so it's two terms so it's really important to put it in parentheses so 5x minus 6 and that's equal to 2x very nice okay let's go ahead and distribute there's really an invisible negative 1 here so this is x cubed minus negative 1 times 5x is minus 5x negative 1 times negative 6 is a positive 6 and this is equal to 2x let's go ahead and subtract 2x so minus 2x minus 2x so we have x cubed minus seven x plus six and that's equal to zero all right so now we have to Factoring factor this somehow so i am not seeing a very easy way to factor this uh in my mind uh maybe actually maybe maybe we can do this i'm thinking this this is something that might not work but i'm just thinking it might work so let's just think about it i don't know if it'll work so two numbers that multiply to six uh no this is not going to work that's not going to work because of the middle term scratch all that so let's go ahead and try something else let's try the rational roots theorem but before we do let's just see if one is a solution to this so let's check one and i say that because a lot of times in these problems um they're kind of like rigged like usually one works so let's see one cubed minus seven times one plus six equals zero so that's one minus seven plus six equals so negative 6 plus 6 equals 0. so 1 is a solution Synthetic Division so because 1 is a solution so x equals 1 is a solution to this equation that means that when we divide by x minus 1 the remainder is 0. so let's go ahead and divide by x minus 1 using synthetic division and we'll be able to fully factor this after that so when you're dividing by x minus 1 with synthetic division you start by writing down the 1 like this and then you write down the coefficients of your polynomial so the first coefficient is 1. the second coefficient is 0. there's really a 0x squared here it's invisible it's plus 0x squared so 0 the next one is negative seven and the next one is six and then you draw a line okay so one zero negative seven and six and in the synthetic division process the very first step is to bring down the one so you bring it down and one times one is one you add you get one one times one is one you add you get negative seven uh negative six six six negative six times one is negative six you add you get zero and we know that it should be zero because one is a solution so now uh we can factor this fully let me write it again x cubed minus seven x plus six so this is equal to well we know that one is a solution so x minus one is a factor and then it's times and then this gives us the other factor here so you want to start at one less because this was a cubic you want to start at [Music] x squared so it's 1 times x squared plus 1 times x minus 6. and again we were setting this equal to zero right that's the whole point was trying to solve this cubic equation um i think this will factor some more since this is x minus one x x let's see two numbers that multiply to negative six and add to one so i think three and two and let's make the negative there on the two because three x plus negative 2x gives us x so the inner and the outer always give you the middle term x times x is x squared that part's easy 3 times negative 2 is negative 6. no problem but the middle term it's going to be the inner which is 3x and the outer which is negative 2x and when you add those you get x so we get three different x values here we get 1 negative 3 and 2. but those aren't solutions right those aren't solutions to the problem they're solutions to the equation this problem is a system of linear of nonlinear Solving equations so basically now we have to find the y value so we can plug it into either of these let's go ahead and plug it into this one to find the corresponding y values so y equals 5x minus 6. i'm going to come down here and write that so y equals 5x minus 6. this is what we're going to use and let's go ahead and plug numbers in to get the y coordinates so when x is one we have y equals five times one minus six so five minus 6 so negative 1. so when x is 1 y is negative 1. so our first solution is the ordered pair negative 1 comma negative 1. when x is negative 3 we get y equals 5 times negative 3 minus 6. so that's going to be negative 15 minus 6 which is equal to negative 21. so when x is negative 3 y is negative 21 so that's going to be negative 3 negative 21 that's going to be another solution right these are these are our points on the uh xy plane where the graphs of these two functions intersect that's what it means intuitively and then when x is 2 you get y equals 5 times 2 minus 6. this is going to be 10 minus 6 which is 4. so we have 2 comma 4. and so that would be our other solution so we have three solutions there are three points of intersection on the graphs of these two functions where we have some cubic polynomial and the line and basically we're being asked where does the cubic polynomial and the line you know intersect and we found the points of intersection or equivalently solve the system of nonlinear equations same thing i hope this has been helpful good luck |
4728 | https://www.engineersedge.com/calculators/swameejain_friction_factor_15811.htm | Swamee-Jain Friction Factor Equation and Calculator
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Swamee-Jain Friction Factor Equation and Calculator
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Swamee-Jain Friction Factor Equation and Calculator
This is another explicit equation for calculating the friction factor typically used in the Darcy-Weisbach. It was first presented by P. K. Swamee and A. K. Jain in 1976 in the Journal of the Hydraulics Division of ASCE. This equation is the easiest of all explicit equations for calculating the friction factor.
The Swamee-Jain equation is expressed as
Equation 1
f = 0.25 / [ log 10 ( ε / 3.7 d + 5.74 / Re 0.9 ) ]2
Where:
f = friction factor, (dimensionless)
ε = absolute pipe roughness, (in, mm) depends on the condition of the pipe. It ranges from 0.001 to 0.01.
d = inside diameter of pipe, (in, mm)
Re = Reyonds number, (dimensionless)
Related:
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Reference:
Piping Calculations Manual,
E. Shashi Menon
SYSTEK Technologies, Inc
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4729 | https://lpsa.swarthmore.edu/LaplaceXform/FwdLaplace/LaplaceFuncs.html | Contents
The Unit Step Function
The Unit Impulse
The Exponential
The Sine
The Cosine
The Decaying Sine and Cosine
The Ramp
Composite Functions
To productively use the Laplace Transform, we need to be able to transform functions from the time domain to the Laplace domain. We can do this by applying the definition of the Laplace Transform
but this quickly becomes tedious. Our goal is to avoid having to evaluate the integral by finding the Laplace Transform of many useful functions and compiling them in a table. Thereafter the Laplace Transform of functions can almost always be looked by using the tables without any need to integrate. A table of Laplace Transform of functions is available here.
The Unit Step Function
The unit step function is defined as
Some notes about this function:
Most references use u(t) instead of γ(t). However, u(t) has some other common uses, so we will use γ(t) to avoid confusion (and because it's Laplace Transform Γ(s) looks a little like a step input).
Some references change the definition of the step so that the bottom inequality becomes a strict inequality (t>0) and either leave γ(0) undefined or say that γ(0)=�.
Although there is a discontinuity at t=0, we draw a vertical line to help guide the eye and to indicate that the blue line is a single function.
To find the Laplace Transform, we apply the definition.
so
Aside: Convergence of the Laplace Transform
Careful inspection of the evaluation of the integral performed above:
reveals a problem. The evaluation of the upper limit of the integral only goes to zero if the real part of the complex variable "s" is positive (so e-st→0 as s→∞). In this case we say that the "region of convergence" of the Laplace Transform is the right half of the s-plane (since s is a complex number, the right half of the plane corresponds to the real part of s being positive). As long as the functions we are working with have at least part of their region of convergence in common (which will be true in the types of problems we consider), the region of convergence holds no particular interest for us. Since the region of convergence will not play a part in any of the problems we will solve, it is not considered further.
The Unit Impulse
The unit impulse is discussed elsewhere, but to review. The impulse function is everywhere but at t=0, where it is infinitely large. The area of the impulse function is one. The impulse function is drawn as an arrow whose height is equal to its area.
To find the Laplace Transform, we apply the definition
Now we apply the sifting property of the impulse. Since the impulse is 0 everywhere but t=0, we can change the upper limit of the integral to 0+.
Since e-st is continuous at t=0, that is the same as saying it is constant from t=0- to t=0+. So we can replace e-st by its value evaluated at t=0.
So the Laplace Transform of the unit impulse is just one. Therefore the impulse function, which is difficult to handle in the time domain, becomes easy to handle in the Laplace domain. It will turn out that the unit impulse will be important to much of what we do.
The Exponential
Consider the causal (i.e., defined only for t>0) exponential:
Several notes about this function:
The function is 0 for t<0, and then follows an exponential trajectory thereafter.
We can either define the function piecewise (the first definition), or as an exponential multiplied by the unit step (the second definition). The second one is more compact, so we will generally use that one. Because all of our functions are equal to zero for t<0, the multiplication by γ(t) will often be implicit and we won't actually show γ(t), except in cases where it is needed for clarity.
If a<0, the function increases without bound. If a>0 the function decays to zero - decaying exponentials are much more common in the systems that we study.
To find the Laplace Transform, we apply the definition
Since γ(t) is equal to one for all positive t, we can remove it from the integral
Note that multiplication by the function γ(t) is implicit on the left hand side of the last equation. This is how we will commonly write our functions.
Note that if a=0, we get a step function and Y(s)=1/s.
The Sine
As before, start with the definition of the Laplace transform
Here it becomes useful to use Euler's identityfor the sine
so
But we've already done this integral (theexponential function, above)
Let's put this over a common denominator
The Cosine
The cosine can be found in much the same way, but using Euler's identity for the cosine.
Note that if ω=0, we get a step function and Y(s)=1/s.
The Decaying Sine and Cosine
The decaying sine or cosine is likewise handled in the same way. Consider, first, the decaying sine wave.
The rest of the derivation follows that of the sine function (i.e., put over a common denominator, and solve)
A similar procedure can be followed for the decaying cosine
The Ramp
So far (with the exception of the impulse), all the functions have been closely related to the exponential. It is also possible to find the Laplace Transform of other functions. For example, the ramp function:
We start as before
Integration by parts is useful at this point
Composite Functions
It is often quite easy to find the Laplace Transform of functions not in the table, by expressing them as a sum of functions in the table that are scaled and/or delayed. To understand this we need to use two Laplace Transform properties that are derived on the next page. We need the linearity property
and the time delay property
Example: Laplace Transform of a Rectangular Pulse
Find the Laplace Transform of the function shown:
We can compose this function in terms of two other functions
Solution:
We know the Laplace Transform of both of these functions
Let's see if we can use this information to find the Laplace Transform of the rectangular pulse. We can form the original rectangular pulse function from the step and the delayed step in two ways
The first method doesn't help us, because we have no property of the Laplace Transform the lets us deal with multiplied functions in the time domain. However, the second method can be used because it represents y(t) as the sum of functions, so we can use the linearity property. So we get
The addition of the two functions is shown graphically below
and y(t)=2.5·z(t)
Example: Laplace Transform of a Triangular Pulse
Find the Laplace Transform of the function shown:
Solution:
We need to figure out how to represent the function as the _sum_ of functions with which we are familiar. For this function, we need only ramps and steps; we apply a ramp function at each change in slope of y(t), and apply a step at each discontinuity.
Starting at t=0 we need to increase the slope of the function, so we add in a ramp with a slope of 0.5.
Starting at t=2, the slope decreases (to zero), so we need to subtract a ramp with a slope of -0.5.
Also at t=2, there is a negative discontinuity, so we need to subtract a step of height -1.
Functionally we want:
so
Graphically this is shown as:
Aside: An improper method to solve the problem
It may seem like it would be easier to define the function as a ramp (0.5·t) multiplied by a rectangular pulse (γ(t)-γ(t-1))
and this does, in fact, yield the correct function. However if we do this we end up with the product of two functions that are in the table separately
However, these two functions have different time delays and we have no way to deal with products of functions. That is why, when choosing the basic functions that make up the composite function, only addition is allowed.
Example: Laplace Transform of a Double Triangular Pulse
Find the Laplace Transform of the function shown:
Solution:
This function is more complicated than the last, but we can still create it as a sum of ramps and steps; we apply a ramp function at each change in slope of y(t), and apply a step at each discontinuity.
At t=0 there is a discontinuity, so we need to add a step of height 2.
Also at t=0, we need to decrease the slope of the function, so we add in a ramp with a slope of -4/3 (rise/run=-2/1.5).
Starting at t=1.5 we need to increase the slope of the function, so we add in a ramp with a slope of 8/3 (since the slope was -4/3 we need to increase it by 8/3 so that the resulting slope is 4/3).
Starting at t=3, the slope decreases (to zero), so we need to subtract a ramp with a slope of -4/3 (since the slope was 4/3 we need to decrease it by 4/3 so that the resulting slope is 0).
Also at t=3, there is a negative discontinuity, so we need to subtract a step of height -2.
The constituent functions are shown in the plot below. If you add them up, you get the original function (shown in black).
Example: Laplace Transform of a Gated Sine
Find the Laplace Transform of the function shown:
Solution:
This function is a little different than the previous in that it involves more than ramps and steps. The most obvious way to represent this function is as a sine wave multiplied by a rectangular pulse. But recall that we can't use generally use multiplication of functions (we have no multiplication property), only addition (by use of the linearity property). So the question becomes: what functions can we add to create the given function?
After a little thought it becomes apparent that we can take a sine wave starting at t=0, and subtract off a cosine beginning at t=2.5.
Key Concept: Finding the Laplace Transform of Composite Functions
When composing a complex function from elementary functions, it is important to only use addition. If you create a function by adding two functions, its Laplace Transform is simply the sum of the Laplace Transform of the two function. If you create a function by multiplying two functions in time, there is no easy way to find the Laplace Transform of the resulting function.
A table of Laplace Transform of functions is available here.
References |
4730 | https://www.ncbi.nlm.nih.gov/books/NBK606121/figure/article-149649.image.f18/ | [Figure, Polyarteritis Nodosa Pathology. Histopathology of...] - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Polyarteritis Nodosa Pathology. Histopathology of cutaneous polyarteritis nodosa reveals the involvement of the septa's small-to-medium–sized arteries and arterioles. Lesions show prominent fibrinoid necrosis and fibrin deposition, with the destruction of the tunica intima. Contributed by P Chu, MD
From: Dermatopathology Evaluation of Panniculitis
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4731 | https://www.sciencedirect.com/science/article/abs/pii/S0040403900808084 | Acidity constants of protonated simple carbonyl compounds: Comments on literature data and indirect estimates - ScienceDirect
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Tetrahedron Letters
Volume 29, Issue 43, 1988, Pages 5541-5544
Acidity constants of protonated simple carbonyl compounds: Comments on literature data and indirect estimates
Author links open overlay panel Jean Toullec
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The pK a values for OH-acidity of protonated simple carbonyl compounds (oxocarbenium ions) are estimated from keto-enol equilibrium constants combined within CH-acidity constants of the ions, calculated by application of the Marcus equation to the ketonisation process.
pK a values, more realistic than literature data, are estimated from rate and equilibrium constants for acid-catalysed keto-enol tautomerisation.
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A. Bagno et al.### Rev. Chem. Intermed. (1987)
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Stabilities and Reactivities of Carbocations
2010, Advances in Physical Organic Chemistry Citation Excerpt :
Comparisons of structurally related hydroxy- and methoxy-substituted cations show that hydroxy is more stabilizing by between 4 and 5 log units. This difference was recognized 20 years ago by Toullec who compared pKas for protonation of the enol of acetophenone and its methyl ether145 (–4.6 and 1.3, respectively) based on a cycle similar to that of Scheme 15, but with the enol replacing the hydrate, and a further cycle relating the enol ether to a corresponding dimethyl acetal and methoxycarbocation.146 Toullec concluded, understandably but incorrectly, that there was an error in the pKa of the ketone (over which there had been controversy at the time).147,148 Show abstract This chapter examines the various aspects of stabilities and reactivities of carbocations. The choice of an equilibrium constant for measuring the stability of a carbocation depends partly on experimental accessibility and partly on the choice of solvent. A desire to relate measurements to the majority of existing equilibrium constants implies the use of water as a solvent. The majority of equilibrium constants measured for carbocation formation refers to the ionization of alcohols or alkenes in acidic aqueous solution and corresponds to p K R or p K a. The application of kinetic methods for determining p K a and p K R for carbocations, by combining rate constants for their formation from an alcohol or alkene with a rate constant for the reverse reaction of the carbocation with water, has provided the most important development in the measurements of these equilibrium constants. Oxygen substitution has a radical effect on the stability of a carbocation, which is manifested in the chemistry of carbohydrates, benzopyran pigments, and the extensive acid-dependent reactions of carbonyl compounds. The greatest effects are from α-oxygen substituents, but the effects of substituents in the aromatic ring of benzylic carbocations are also large.
### Interplay between Substrate and Proton Donor Coordination in Reductions of Carbonyls by SmI2-Water Through Proton-Coupled Electron-Transfer
2018, Journal of the American Chemical Society
### March's Advanced Organic Chemistry: Reactions, Mechanisms, and Structure: Sixth Edition
2006, March S Advanced Organic Chemistry Reactions Mechanisms and Structure Sixth Edition
### Rate and equilibrium constants for formation and hydrolysis of 9-formylfluorene oxime: Diffusion-controlled trapping of a protonated aldehyde by hydroxylamine
2000, Canadian Journal of Chemistry
### Efficient intramolecular general acid catalysis of enol ether hydrolysis. Hydrogen-bonding stabilisation of the transition state for proton transfer to carbon
1994, Journal of the Chemical Society Perkin Transactions 2
### Basicity (pKBH+) and acidity constants (pKa) of some 3-X-, 4-X-, and dimethyl-4-X-benzoic acids
1993, Journal of the Chemical Society Perkin Transactions 2
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Copyright © 1988 Published by Elsevier Ltd.
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4732 | https://www.amp.i.kyoto-u.ac.jp/tecrep/ps_file/2017/2017-002.pdf | Sub-Homogeneous Optimization∗ Shota Yamanaka† and Nobuo Yamashita† August 29, 2017 Abstract We consider an optimization problem with sub-homogeneous functions in its objec-tive and constraint functions. Examples of such sub-homogeneous functions include the absolute value function and the p-norm function, where p is a positive real number. The problem, which is not necessarily convex, extends the absolute value optimization proposed in [O. L. Mangasarian, Absolute value programming, Com-putational Optimization and Applications 36 (2007) pp. 43–53]. In this work, we propose a dual formulation that, differently from the Lagrangian dual approach, has a closed-form and some interesting properties. In particular, we discuss the relation between the Lagrangian duality and the one proposed here, and give some suffi-cient conditions under which these dual problems coincide. Finally, we show that some well-known problems, e.g., sum of norms optimization and the group Lasso-type optimization problems, can be reformulated as sub-homogeneous optimization problems.
Keywords: Sub-homogeneous functions, duality, nonconvex optimization 1 Introduction Recently, the so-called absolute value equations (AVE) and absolute value optimization (AVO) problems have been attracted much attention. The AVE were introduced in 2004 by Rohn . Basically, if ˜ A, ˜ B are given matrices, and ˜ b is a given vector, one should find a vector x that satisfies ˜ Ax + ˜ B|x| = ˜ b, where |x| is a vector whose i-th entry is the absolute value of the i-th entry of x. It is known that AVE are equivalent to the linear complementarity problems (LCP) [10, 17, 21], which include many real-world applications.
As an extension of AVE, Mangasarian proposed in 2007 the AVO problems, which ∗This work was supported in part by a Grant-in-Aid for Scientific Research (C) (17K00032) from Japan Society for the Promotion of Science.
†Department of Applied Mathematics and Physics, Graduate School of Informatics, Kyoto University, Kyoto 606-8501, Japan (shota@amp.i.kyoto-u.ac.jp, nobuo@i.kyoto-u.ac.jp).
1 have the absolute value of variables in their objective and constraint functions. More precisely, the AVO problem considered is given by min ˜ cTx + ˜ dT|x| s.t.
˜ Ax + ˜ B|x| = ˜ b, ˜ Hx + ˜ K|x| ≥˜ p, where ˜ A, ˜ B, ˜ H, ˜ K are given matrices, and ˜ c, ˜ d,˜ b, ˜ p are vectors with appropriate dimensions.
Since AVE and LCP are equivalent, the AVO include the mathematical programs with linear complementarity constraints , which are one of the formulations of equilibrium problems.
As another application of AVO, Yamanaka and Fukushima presented facility location problems.
Since 2007, some methods for solving AVE have been presented in the literature.
For example, Rohn considered an iterative algorithm using the sign of variables for the case that ˜ A and ˜ B are square matrices. For more general ˜ A and ˜ B, Mangasarian provided a method involving successive linearization techniques. Another methods include a concave minimization approach, given by Mangasarian , and Newton-type methods, proposed by Caccetta et al. , Mangasarian , and Zhang and Wei .
Some generalizations of AVE were also proposed. For example, Hu et al. considered an AVE involving the absolute value of variables associated to the second-order cones.
Miao et al. investigated an AVE with the so-called circular cones. In both papers, quasi-Newton based algorithms were used.
As for AVO problems, Yamanaka and Fukushima proposed to use a branch-and-bound technique. In the branching procedure, two subproblems are generated by fixing the sign of a variable as nonnegative or nonpositive. In the bounding procedure, the dual information are considered. However, to the best of our knowledge, there is no other method that can find a global solution of AVO. When comparing to AVE, the research associated to AVO problems is insufficient and one of these reasons is the difficulty for obtaining feasible solutions of the problems. In fact, their constraints include AVE, which are known to be NP-hard .
Another optimization problem that is related to AVO was recently investigated by Friedlander et al. and Aravkin et al. . It is called gauge optimization, which ba-sically consists in an optimization problem with the so-called gauge function. However, differently from AVO, this problem does not consider multiple constraints, but only one gauge constraint. In [2, 9], the authors showed that the Lagrange dual of gauge optimiza-tion problems can be written in a closed-form by using the polar of the gauge functions.
In this paper, similarly to [2, 9], we introduce a generalized AVO problem, and show that it has a wider practical application comparing to AVO problems. It is also more general than gauge optimization problems, because multiple constraints can be consid-ered here. The generalization is done by replacing absolute value functions with sub-homogeneous functions.
The concept of sub-homogeneous functions varies from each literature, but here we consider the one given in , with the term Λ as the set of non-negative real numbers. So, the problem uses not only absolute value terms but also, for instance, p-norm functions with p ∈(0, ∞]. This generalized problem is referred here as 2 sub-homogeneous optimization (SHO).
Here, we introduce the SHO dual problem and compare it with the Lagrange dual.
We also show that the weak duality theorem holds, similarly to the AVO problems .
In addition, we investigate the relation between the sub-homogeneous duality and the Lagrange duality, proving that these dual problems are equivalent under some conditions.
In this case, the Lagrange dual of a sub-homogeneous problem can be written in a closed-form. We point out that the gauge functions are special cases of the sub-homogeneous functions, which are not necessarily convex, differently from the gauge. Moreover, the proposed problems here have linear and sub-homogeneous terms in their objective func-tions and constraints, which is different from the problem considered in [2, 9] that has only one gauge term.
Here, we also give some applications for the sub-homogeneous problems, which include p-order cone optimization, sum of norms optimization and group Lasso-type optimization problems, and we show that their Lagrange dual can be written in a closed-form even without convexity assumptions.
The paper is organized as follows. In Section 2, we give the definition of sub-homoge-neous functions as well as its dual, showing some of their properties. In Section 3, we define the SHO problems, and we prove that weak duality holds. In Section 4, the relation between the Lagrangian dual and the sub-homogeneous dual is discussed. We give some applications for SHO problems in Section 5. We conclude the paper in Section 6, with final remarks and some future works.
We consider the following notations throughout the paper.
Let x ∈Rn be a n-dimensional column vector, and A ∈Rn×m be a matrix with dimension n × m. We use T to denote transpose. For two vectors x and y, we denote the vector (xT, yT)T as (x, y)T for simplicity.
If x ∈Rn, then its i-th entry is denoted by xi, so x = (x1, . . . , xn)T.
Moreover, if I ⊆{1, . . . , n}, then xI corresponds to the subvector of x with entries xi, i ∈I. The notation #J denotes the number of elements of a set J. The identity matrix with dimension n is given by En ∈Rn×n. Also, we denote by ∥· ∥p and ∥· ∥∞the p-norm with p > 0 and the supremum norm, respectively. If no distinction is made for the norm, we just use the notation ∥· ∥.
2 Sub-homogeneous functions In this section, we first introduce the definitions of sub-homogeneous and vector sub-homogeneous functions. They are special cases of Λ-sub-homogeneous functions originally given in , with Λ as the set of nonnegative real numbers R+. Then, we define their dual, which will be used to describe the dual of SHO problems. Moreover, we show some properties associated to these functions.
Definition 1. (Sub-homogeneous functions) A function ψ: Rn →R is sub-homogeneous if the following inequality holds: ψ(λx) ≤λψ(x) for all x ∈Rn, λ ∈R+.
3 Definition 2. (Vector sub-homogeneous functions) A mapping Ψ: Rn →Rm is a vector sub-homogeneous function if the following property holds: Ψ(x) = ψ1(xI1) .
.
.
ψm(xIm) for all x ∈Rn, where ψi : Rni →R is a sub-homogeneous function for all i = 1, . . . , m, n = n1 +· · ·+nm, Ii ⊆{1, . . . , m} is a set of indices satisfying Ii ∩Ij = ∅, i ̸= j, and #Ii = ni, and xIi ∈Rni is a disjoint subvector of x.
The above definition basically says that Ψ is vector sub-homogeneous if its block components are all sub-homogeneous. We now introduce the dual function of ψ, which can be seen as a generalization of the dual norm. Similarly, we also define the dual of vector sub-homogeneous functions.
Definition 3. (Dual sub-homogeneous functions) Let ψ: Rn →R be a sub-homogeneous function. Then, ψ∗: Rn →R defined by ψ∗(y) := sup{xTy | ψ(x) ≤1} for all x ∈Rn is called the dual sub-homogeneous function of ψ.
Note that ψ∗is convex from definition. In fact, for all y, z ∈Rn and α ∈(0, 1), we have ψ∗(αy + (1 −α)z) = sup{xT(αy + (1 −α)z) | ψ(x) ≤1} ≤ α sup{xTy | ψ(x) ≤1} + (1 −α) sup{xTz | ψ(x) ≤1} = αψ∗(y) + (1 −α)ψ∗(z).
Definition 4. (Dual vector sub-homogeneous functions) Let Ψ: Rn →Rm be a vector sub-homogeneous function. A function Ψ∗: Rn →Rm is a dual vector sub-homogeneous function associated to Ψ if the following property holds: Ψ∗(y) = ψ∗ 1(yI1) .
.
.
ψ∗ m(yIm) , i = 1, . . . , m, for all y ∈Rn where ψ∗ i : Rni →R is the dual of sub-homogeneous function ψi for each i = 1, . . . m.
In this paper, we assume two conditions for sub-homogeneous functions.
Assumption 1. Let Ψ: Rn →Rm be a vector sub-homogeneous function as in Defini-tion 2. Then, for all i = 1, . . . , m, the sub-homogeneous function ψi satisfies the following conditions: 4 1. ψi(xIi) ≥0 for all xIi ∈Rni, 2. If xIi ̸= 0, then ψi(xIi) > 0.
From the definition of sub-homogeneous functions, we observe that ψi(0) = 0. In fact, if λ = 0 then ψi(0) ≤0. Moreover, if x = 0 and λ = 2, then ψi(0) ≥0. Moreover, the second condition of the above assumption shows that zero is the only point that satisfies ψi(x) = 0. We also observe that if ψi is taken as the usual vector norm, then it satisfies these assumptions. We now show an important property satisfied by vector sub-homogeneous functions and their dual.
Proposition 1. Let Ψ and Ψ∗be a vector sub-homogeneous function and its dual, respec-tively. Suppose that Assumption 1 holds. Then, the following inequalities hold: Ψ∗(y) ≥ 0, Ψ(x)TΨ∗(y) ≥ xTy for any x, y ∈Rn.
Proof . For simplicity, we take an arbitrary index i and denote ψi and xIi as ψ and x, respectively. From Definition 1, we have ψ(0) = 0. Using this result and Definition 3, we obtain ψ∗(y) = sup{xTy | ψ(x) ≤1} ≥0 for all y ∈Rn.
This shows that Ψ∗(y) ≥0 for all y ∈Rn from Definition 4.
If x = 0, then the second inequality of this proposition clearly holds. If x ̸= 0, then ψ(x) > 0 from Assumption 1 and so ψ ( x ψ(x) ) ≤ 1 ψ(x)ψ(x) = 1 holds once again from Definition 1. Therefore, we obtain ψ∗(y) ≥ ( x ψ(x) )T y for all y ∈Rn.
Then, for all x, y ∈Rn, we have ψ(x)ψ∗(y) ≥xTy, which indicates that Ψ(x)TΨ∗(y) = m ∑ i=1 ψIi(x)ψ∗ Ii(y) ≥ m ∑ i=1 xT IiyIi = xTy.
5 3 Sub-homogeneous optimization problems We consider the following sub-homogeneous optimization (SHO) problem: (P) min cTx + dTΨ(x) s.t.
Ax + BΨ(x) = b, Hx + KΨ(x) ≥p, where c ∈Rn, d ∈Rm, b ∈Rk, p ∈Rℓ, A ∈Rk×n, B ∈Rk×m, H ∈Rℓ×n and K ∈Rℓ×m are given constant vectors and matrices, and Ψ: Rn →Rm is a vector sub-homogeneous function satisfying Assumption 1.
Now we give the Lagrangian dual of the problem (P) as follows: (DL) sup u v≥0 ω(u, v), where ω: Rk × Rℓ→R is given by ω(u, v) := inf x L(x, u, v), (1) and L: Rn × Rk × Rℓ→R is the Lagrangian function of (P) defined by L(x, u, v) := cTx + dTΨ(x) + uT(b −Ax −BΨ(x)) + vT(p −Hx −KΨ(x)) = bTu + pTv −(ATu + HTv −c)Tx + (d −BTu −KTv)TΨ(x), with u ∈Rk and v ∈Rℓas the Lagrange multipliers associated to the equality and inequality constraints, respectively.
Notice that it is difficult to write concretely the objective function of the problem (DL) because it is, in general, not convex with respect to x.
In order to obtain a closed-form dual problem, we consider a convex relaxation of the original problem (P) and its Lagrangian dual. For simplicity, we investigate the case where Ψ(x) = |x| := (|x1|, . . . , |xn|)T, and (P) has a linear objective function and only inequality constraints. More precisely, we analyze the following problem: (Pa) min cTx s.t.
Ax + B|x| ≥b.
If we set x = x+ −x−and |x| = x+ + x−, where x+ i = max{0, xi} and x− i = max{0, −xi}, then we can write (Pa) as min [cT| −cT] [ x+ x− ] s.t.
[A| −A] [ x+ x− ] + [B|B] [ x+ x− ] ≥b, 6 which is equivalent to the following problem: min [cT| −cT] [ y1 y2 ] s.t.
[A| −A] [ y1 y2 ] + [B|B] [ y1 y2 ] ≥b, y1, y2 ≥0, yT 1 y2 = 0, where y1, y2 ∈Rn. Notice that the above problem is not convex due to the complemen-tarity constraint yT 1 y2 = 0. Therefore, we remove it from the problem and obtain the following relaxed one: min [cT| −cT] y s.t.
[A + B| −A + B] y ≥b, y ≥0, where y = (y1, y2)T. This problem is just a linear programming, then its Lagrangian dual can be written easily as max bTu s.t.
[ AT + BT −AT + BT ] u ≤ [ c −c ] , u ≥0.
Observing that the first constraint is equivalent to |ATu−c|+BTu ≤0, we finally obtain the following closed-form dual problem: (Da) max bTu s.t.
|ATu −c| + BTu ≤0, u ≥0.
In fact, the problem (Da) is the AVO dual of (Pa) proposed by Mangasarian in , and the weak duality clearly holds in this case.
Let us return to the general problem (P). Inspired by the above AVO dual problem (Da), we consider the following problem as the sub-homogeneous dual problem: (D) max bTu + pTv s.t.
Ψ∗(ATu + HTv −c) + BTu + KTv ≤d, v ≥0, where Ψ∗is the dual vector sub-homogeneous function associated to Ψ. Note that (D) is a convex optimization problem since each component ψ∗ i of Ψ∗is a convex function.
The theorem below shows that the proposed dual problem (D) is reasonable, in the sense that the weak duality holds between (P) and (D).
Theorem 2. (Weak duality) For problems (P) and (D), the following inequality holds: cTx + dTΨ(x) ≥bTu + pTv for all feasible points x ∈Rn and (u, v) ∈Rk × Rℓof (P) and (D), respectively.
7 Proof . Let x ∈Rn and (u, v) ∈Rk × Rℓbe feasible for (P) and (D), respectively. Then, we have cTx + dTΨ(x) ≥ cTx + (Ψ∗(ATu + HTv −c) + BTu + KTv)TΨ(x) = cTx + Ψ∗(ATu + HTv −c)TΨ(x) + uTBΨ(x) + vTKΨ(x), where the inequality holds from the first constraint of (D) and the nonnegativity of Ψ.
From the second inequality of Proposition 1, we also obtain: cTx + dTΨ(x) ≥ cTx + (ATu + HTv −c)Tx + uTBΨ(x) + vTKΨ(x) = uT(Ax + BΨ(x)) + vT(Hx + KΨ(x)).
Finally, the constraints of (P) gives cTx + dTΨ(x) ≥bTu + pTv, which completes the proof.
The weak duality theorem itself is a powerful theoretical result, but it does not mention how large the duality gap between (P) and (D) is. And the duality gap can be large depending on problems, then the dual problem (D) may be useless. Therefore, in the next section, we investigate the relation between the Lagrangian dual problem (DL) and the one (D) proposed here.
As a result, surprisingly, we find that (DL) and (D) are equivalent.
4 The sub-homogeneous duality and the Lagrangian duality In this section, we consider the relation between the sub-homogeneous duality and the more traditional Lagrangian duality of the problem (P), investigating conditions under which the Lagrangian dual problem (DL) and the sub-homogeneous dual problem (D) are equivalent. Recalling (1), we first show a condition that makes ω(¯ u, ¯ v), the objective function of (DL), unbounded from below for some (¯ u, ¯ v).
Lemma 3. Let ψ∗ i be the dual of the sub-homogeneous functions ψi for i = 1, . . . , m.
Suppose that Assumption 1 holds. Also, assume that there exists (¯ u, ¯ v) and an index i0 satisfying ψ∗ i0(αIi0) > βi0, where α := AT ¯ u + HT ¯ v −c ∈Rn, and β := d −BT ¯ u −KT ¯ v ∈Rm. Then, there exists a sequence {xk} such that ∥xk∥→+∞and L(xk, ¯ u, ¯ v) →−∞as k →+∞. Therefore, ω(¯ u, ¯ v) is unbounded from below.
Proof . Firstly, we denote ¯ α and ¯ α(λ) as follows: ¯ α := (αI1, αI2, . . . , αIi0, . . . , αIm) ∈Rn, ¯ α(λ) := (αI1, αI2, . . . , λˆ x, . . . , αIm) ∈Rn, 8 where λ ∈R+ and ˆ x ∈Rn is defined as the supreme point of the following problem: sup{xTαIi0 | ψi0(x) ≤1}.
From the definition of ˆ x, we obtain ψi0(ˆ x) ≤1 because the objective function xTαIi0 is linear. Then, from Definition 3, we have ˆ xTαIi0 = ψ∗ i0(αIi0) ≥ψi0(ˆ x)ψ∗ i0(αIi0).
The above equality and the definition of the Lagrangian function give L(¯ α(λ), ¯ u, ¯ v) = bT ¯ u + pT ¯ v −¯ αT ¯ α(λ) + βTΨ(¯ α(λ)) = bT ¯ u + pT ¯ v − ∑ i̸=i0 αT IiαIi −λˆ xTαIi0 + ∑ i̸=i0 βiψi(αIi) + βi0ψi0(λˆ x) = γ −λˆ xTαIi0 + βi0ψi0(λˆ x) ≤ γ −λψi0(ˆ x)ψ∗ i0(αIi0) + βi0ψi0(λˆ x), where γ := bT ¯ u + pT ¯ v −∑ i̸=i0 αT IiαIi + ∑ i̸=i0 βiψi(αIi) ∈R is constant with respect to λ.
Moreover, Definition 1 shows that L(¯ α(λ), ¯ u, ¯ v) ≤ γ −λψi0(ˆ x)ψ∗ i0(αIi0) + λβi0ψi0(ˆ x) = γ + λψi0(ˆ x)(βi0 −ψ∗ i0(αIi0)) = γ + λ(βi0 −ψ∗ i0(αIi0)).
Therefore, L(¯ α(λ), ¯ u, ¯ v) converges to minus infinity when λ increases. Finally, if we set xk = ¯ α(λk) where λk →+∞as k →+∞, then L(xk, ¯ u, ¯ v) →−∞and we complete the proof.
We now show that the sub-homogeneous dual problem (D) and the Lagrangian one (DL) are equivalent under some conditions.
Lemma 4. Suppose that Assumption 1 holds. Assume also that the sub-homogeneous dual problem (D) has a feasible solution (¯ u, ¯ v) ∈Rk × Rℓ, and that there exist x∗∈Rn satisfying the following equality: (d −BT ¯ u −KT ¯ v)TΨ(x∗) −(AT ¯ u + HT ¯ v −c)Tx∗= 0.
(2) Then, the sub-homogeneous dual problem (D) and the Lagrangian dual problem (DL) are equivalent.
Proof . From Lemma 3, the function ω is unbounded from below if there exists an index i0 such that ψ∗ i0(αIi0) > βi0, where α := AT ¯ u+HT ¯ v−c ∈Rn, and β := d−BT ¯ u−KT ¯ v ∈Rm.
Therefore, the problem (DL) is equivalent to (D′ L) sup ω(u, v) s.t.
Ψ∗(ATu + HTv −c) ≤d −BTu −KTv, v ≥0.
9 Let (¯ u, ¯ v) ∈Rk×Rℓbe the feasible solution of (D′ L). From the definition of the Lagrangian function, we obtain: L(x, ¯ u, ¯ v) = cTx + dTΨ(x) + ¯ uT(b −Ax −BΨ(x)) + ¯ vT(p −Hx −KΨ(x)) = bT ¯ u + pT ¯ v −(AT ¯ u + HT ¯ v −c)Tx + (d −BT ¯ u −KT ¯ v)TΨ(x).
Then, taking x∗∈Rn that satisfies (2), we have L(x∗, ¯ u, ¯ v) = bT ¯ u + pT ¯ v.
Notice that x∗is the solution of the problem inf x L(x, ¯ u, ¯ v), because L(x, ¯ u, ¯ v) ≥bT ¯ u + pT ¯ v holds from Proposition 1. Therefore, the problem (D′ L) can be described as follows: sup bTu + pTv s.t.
Ψ∗(ATu + HTv −c) ≤d −BTu −KTv, v ≥0, which is equivalent to the sub-homogeneous dual problem (D).
As a consequence of the above theorem, we obtain the following result.
Theorem 5. Suppose that the Lagrangian dual problem (DL) has a feasible solution.
Assume also that the vector sub-homogeneous function Ψ satisfies Assumption 1. Then, the sub-homogeneous dual problem (D) is equivalent to the Lagrangian dual problem (DL).
Proof . From Definition 1 and Assumption 1, we have Ψ(0) = 0. It means that equation (2) holds at x∗= 0. Thus, the problems (D) and (DL) are equivalent.
The above theorem shows that the Lagrangian dual problem (DL) can be written in a closed-form when the function Ψ is sub-homogeneous and satisfies Assumption 1. The paper does not show that the same property holds for the AVO problem. We now give it as a direct consequence of Theorem 5.
Corollary 1. If the dual of an AVO problem has a feasible solution, then it is equivalent to the Lagrangian dual problem (DL).
Proof . It holds from Theorem 5 and the fact that the absolute value function is sub-homogeneous and satisfies Assumption 1.
Corollary 2. If the optimal value of an AVO primal problem and its Lagrangian dual problem (DL) are equivalent, then the strong duality holds between the AVO primal and the AVO dual problem.
Proof . It holds straightforward from Corollary 1.
10 From the above result, AVO can be applied to solve 0-1 integer optimization prob-lems. To solve such problems, their Lagrangian dual are often considered, which is, in general, nondifferentiable due to the integer constraints. On the other hand, a 0-1 integer constraint, that is x ∈{0, 1}, is equivalent to |2x−1| = 1. Then, 0-1 integer optimization problems can be reduced to AVO, and we obtain their AVO dual, which are actually linear programming (LP) problems. These LP dual problems are much easier to solve compared to the nondifferentiable ones. Therefore, it might be worth considering AVO dual problems from the computational point of view.
5 Examples of sub-homogeneous optimization problems In this section, we present several applications that are formulated as SHO, and show their closed-form dual problems.
First, we observe that any p-norm function with p ∈[1, ∞) is sub-homogeneous. So, if ψ is the p-norm, then ψ∗becomes the q-norm, where 1/p + 1/q = 1. Therefore, if ψ is taken as ∥· ∥1, ∥· ∥2, ∥· ∥∞, then ψ∗becomes ∥· ∥∞, ∥· ∥2, ∥· ∥1, respectively. Moreover, in the case that p ∈(0, 1), the dual function ψ∗is equal to ∥· ∥∞for all p ∈(0, 1), which is proved in Proposition 6 of Appendix A. From the result, we can consider any p-norm functions as ψ in SHO problems. And, even if such functions are nonconvex with p ∈(0, 1), the Lagrangian dual problem can be written in a closed-form from Theorem 5.
We now show some sub-homogeneous problems using these p-norm functions. The first example is the so-called linear second-order cone optimization problem , which is one of the famous convex optimization problem.
Example 1. Let x = (x1, x2)T ∈R × Rn−1. Then, we consider the linear second-order cone optimization problem written by (P1) min cTx s.t.
Ax = b, x1 −∥x2∥2 ≥0, where c ∈Rn, A ∈Rm×n and b ∈Rm. The above problem can be written in SHO form as min cTx + 0TΨ(x) s.t.
Ax + 0Ψ(x) = b, Hx + KΨ(x) ≥0, with H = (1, 0, . . . , 0) ∈R1×n, K = (0, −1) ∈R1×2 and Ψ: Rn →R2, Ψ(x) = (|x1|, ∥x2∥2)T.
Then, recalling (D), its dual problem is given by max bTu s.t.
Ψ∗(ATu + HTv −c) + KTv ≤0, v ≥0, 11 where Ψ∗is identical to Ψ in this case. Then, from the definition of Ψ, we have max bTu s.t.
|(ATu)1 + v −c1| ≤0, ∥(ATu)2 −c2∥2 ≤v, v ≥0, with (ATu)1 as the first component of ATu, (ATu)2 is the rest of it, and c = (c1, c2)T ∈ R × Rn−1. The first constraint of the above problem shows that v = c1 −(ATu)1, and v ≥0 automatically holds from the second constraint. Then, we obtain max bTu s.t.
∥(ATu)2 −c2∥2 ≤c1 −(ATu)1 as the dual problem of (P1). In fact, the above problem is the standard dual of the linear second-order cone optimization problem .
Although we use the 2-norm in the above example, any p-norm function with p ∈(0, ∞] can be considered. In this case, if p ∈[1, ∞], then the primal and dual problems are p-order cone and q-order cone optimization problems, respectively, where 1/p + 1/q = 1 . If p ∈(0, 1), then the dual is ∞-order cone optimization problem.
In the next example, we consider a gauge optimization problem, which is also a convex problem with multiple gauge functions in its objective and constraint functions. Here, we recall that f is a gauge function if and only if it is nonnegative, convex, positively homogeneous and satisfies f(0) = 0 . For such a problem, we introduce its dual in SHO form.
Example 2. Let x ∈Rn. We consider the following problem: (P2) min s ∑ i=1 αifi(Aix −ai) s.t.
gj(Bjx −bj) ≤βj, j = 1, . . . , t, where αi, βj ∈R+, Ai ∈Rmi×n, Bj ∈Rkj×n, ai ∈Rmi and bj ∈Rkj are given for all i = 1, . . . , s and j = 1, . . . , t, and fi : Rmi →R and gj : Rkj →R are gauge functions.
Letting yi := Aix −ai and zj := Bjx −bj, (P2) can be written as min s ∑ i=1 αifi(yi) s.t.
gj(zj) ≤βj, j = 1, . . . , t, Aix −yi = ai, i = 1, . . . , s, Bjx −zj = bj, j = 1, . . . , t.
12 The above problem does not have a gauge function defined for the variable x, so we intro-duce such a gauge function x 7→ψ(x) and rewrite the problem into the following way: min 0 × ψ(x) + s ∑ i=1 αifi(yi) + 0 × t ∑ j=1 gj(zj) s.t.
0 × ψ(x) ≤0, 0 × fi(yi) ≤0, i = 1, . . . , s, gj(zj) ≤βj, j = 1, . . . , t, Aix −yi = ai, i = 1, . . . , s, Bjx −zj = bj, j = 1, . . . , t.
Note that ψ: Rn →R is a dummy gauge function with x as its domain.
Let ˆ x := (x, y1, . . . , ys, z1, . . . , zt) ∈Rn+∑s i=1 mi+∑t j=1 kj and Ψ(ˆ x) := (ψ(x), f1(y1), . . . , fs(ys), g1(z1), . . . , gt(zt))T.
Then the above problem can be rewritten as min dTΨ(ˆ x) s.t.
KΨ(ˆ x) ≤p, ˆ Aˆ x = ˆ b, where d = (0, α1, . . . , αs, 0, . . . , 0)T ∈R1+s+t, p = (0, . . . , 0, β1, . . . , βt)T ∈R1+s+t, K = [ 0 0 0 Et ] , ˆ A = A1 −Em1 .
.
.
...
0 As −Ems B1 −Ek1 .
.
.
0 ...
Bt −Ekt , and ˆ b = a1 .
.
.
as b1 .
.
.
bt .
Moreover, its sub-homogeneous dual problem is given by max ˆ bTu −pTv s.t.
Ψ∗( ˆ ATu) −KTv ≤d, v ≥0.
For simplification, let u = (u11, . . . , u1s, u21, . . . , u2t)T with u1i ∈Rmi, i = 1, . . . , s and u2j ∈Rkj, j = 1, . . . , t. Then the above problem is rewritten as (D2) max s ∑ i=1 aT i u1i + t ∑ j=1 bT j u2j − t ∑ ℓ=1 βℓv1+s+ℓ s.t.
s ∑ i=1 AT i u1i + t ∑ j=1 BT j u2j = 0, f ∗ i (−u1i) ≤αi, i = 1, . . . , s, g∗ j(−u2j) ≤v1+s+j, j = 1, . . . , t.
13 Notice that the last constraint implies v ≥0 because g∗ j is also a gauge function. Moreover, (D2) does not include the dual function ψ∗of the dummy gauge function ψ.
The next example is the group Lasso-type problems [18, 28], which is a special case of (P2) and consist in unconstrained minimizations of the sum of certain norms. Such problems have many applications, in particular they appear in compressed sensing area [7, 24], where the sparsity of solutions are important.
As an example, we consider a primal problem with p1-norm and p2-norm where p1, p2 ∈R+, which are used in the regularization terms.
Example 3. Let x ∈Rn and p1, p2 ∈R+. We consider the following problem: (P3) min ∥Ax −b∥2 + λ1 m′ ∑ i=1 ∥xIi∥p1 + λ2 m ∑ i=m′+1 ∥xIi∥p2 where λ1, λ2 ∈R+, b ∈Rm, A ∈Rm×n and 0 < m′ < m.
Notice that the first term of the objective function of group Lasso-type problems are usually the square of 2-norm functions. However, it is not sub-homogeneous, so we re-moved the square and considered just the 2-norm functions.
We obtain the above problem by setting, in (P2), s = m + 1, αi = λ1, if i = 1, . . . , m′, λ2, if i = m′ + 1, . . . , m, 1, if i = m + 1, Ai = { EIi, if i = 1, . . . , m, A, if i = m + 1, where EIi is a submatrix of En with Ej, j ∈Ii as its rows, ai = { 0, if i = 1, . . . , m, b, if i = m + 1, and fi(·) = ∥· ∥p1, if i = 1, . . . , m′, ∥· ∥p2, if i = m′ + 1, . . . , m, ∥· ∥2, if i = m + 1.
Then, recalling (P2) and (D2), the dual of (P3) can be written as max bTu1(m+1) s.t.
m ∑ i=1 ET Iiu1i + ATu1(m+1) = 0, ∥−u1i∥q1 ≤λ1, i = 1, . . . , m′, ∥−u1i∥q2 ≤λ2, i = m′ + 1, . . . , m, ∥−u1(m+1)∥2 ≤1, 14 where qi, i = 1, 2 are obtained by (3) qi = { pi pi −1, if pi > 1, ∞, if pi ∈(0, 1], from Proposition 6 of Appendix A. Notice that the first equality constraint can be rewrit-ten as u1i + (AT)Iiu1(m+1) = 0, i = 1, . . . , m.
Then, the above problem is described as max bTu s.t.
∥(AT)Iiu∥q1 ≤λ1, i = 1, . . . , m′, ∥(AT)Iiu∥q2 ≤λ2, i = m′ + 1, . . . , m, ∥−u∥2 ≤1, where we denote u1(m+1) as u for simplicity.
The next example is also a Lasso-type problem. In this case, the objective function is a gauge, because the sum of gauge functions is also gauge. In order to obtain the dual of a gauge optimization problem, the polar of the objective function should be considered [2, 9]. However, it may be difficult to obtain the polar of a sum of gauge functions. To overcome this drawback, we use here the SHO framework.
Example 4. Let x ∈Rn and p1, p2 ∈R+. We consider the following problem: (P4) min λ1∥x∥p1 + λ2∥x∥p2 s.t.
∥Ax −b∥2 ≤β, where λ1, λ2, β ∈R+, A ∈Rm×n and b ∈Rm. The above problem can be obtained if we set, in (P2), s = 2, t = 1, α1 = λ1, α2 = λ2, A1 = A2 = En, a1 = a2 = 0, B1 = A, b1 = b, f1(·) = ∥· ∥p1, f2(·) = ∥· ∥p2, g1(·) = ∥· ∥2. Then, recalling (D2), the dual of (P4) is written by max bTu21 −βv4 s.t.
u11 + u12 + ATu21 = 0, ∥−u11∥q1 ≤λ1, ∥−u12∥q2 ≤λ2, ∥−u21∥≤v4, which is finally rewritten as max bTu2 −βv s.t.
∥u1 + ATu2∥q1 ≤λ1, ∥−u1∥q2 ≤λ2, ∥−u2∥≤v, where we set u12, u21 and v4 as u1, u2 and v, respectively, and q1 and q2 are defined in (3).
15 In order to control the sparsity of the solutions of the above Lasso-type problems, we can use any combination of p-norm functions, with p ∈(0, ∞], as the regularization terms. Especially, it is reported that the p-norm functions with p ∈(0, 1) in (P3) is useful because they give sparser solutions than 1-norm functions [5, 6, 20].
We now give another example: the sum of norms optimization problems, which are generally nonconvex. Such problems have applications, for example, in facility location, where locations of new facilities should be decided by analyzing the distance between the new and the existing facilities . Moreover, the problem of the following example can be applied not only to the minimization of the distance but also maximization of it by taking the constant λi as −λi. Such a situation can be found for instance in locating obnoxious facilities in residential areas.
Example 5. Let x ∈Rn. We consider the following problem: (P5) min s ∑ i=1 λifi(Aix −ai), s.t.
Bx ≤b, where λi ∈R, Ai ∈Rmi×n, B ∈Rk×n, ai ∈Rmi and b ∈Rk are given, and fi : Rmi →R, i = 1, . . . , s are sub-homogeneous functions. We now introduce its sub-homogeneous dual by taking almost the same procedure as in Example 2. Let yi := Aix −ai, then (P5) is equivalent to min s ∑ i=1 λifi(yi) s.t.
Aix −yi = ai, i = 1, . . . s, Bx ≤b.
By introducing additional constraints, we consider the following problem: min s ∑ i=1 λifi(yi) s.t.
Aix −yi = ai, i = 1, . . . s, Bx ≤b, cifi(yi) ≤di, i = 1, . . . s, where ci and di are strictly positive constants.
Notice that the additional constraints ensure the boundedness of the each term of the objective function especially when λi is strictly negative. Without such constraints, (P5) can be unbounded depending on the linear constraint, and then its dual becomes infeasible. Note that the additional constraints do not change solutions, when we choose ci and di so that the constraint cifi(yi) ≤di will include reasonable solutions.
Let ˆ x := (x, y1, . . . , ys)T ∈Rn+∑s i=1 mi and Ψ(ˆ x) := (ψ(x), f1(y1), . . . , fs(ys))T ∈R1+s, where ψ(·) is a dummy sub-homogeneous function. Then the above problem can be de-scribed as min dTΨ(ˆ x) s.t.
ˆ Aˆ x = ˆ a, Hˆ x + KΨ(ˆ x) ≥p, 16 where d = (0, λ1, . . . , λs)T, ˆ a = (a1, . . . , as)T, p = (−b, −d1, . . . , −ds)T, ˆ A = A1 −Em1 0 .
.
.
...
As 0 −Ems , H = [ −B 0 0 0 ] , and K = 0 0 −c1 0 ...
−cs .
Then, recalling the sub-homogeneous dual (D), the dual of the above problem can be writ-ten as max ˆ aTu + pTv s.t.
Ψ∗( ˆ ATu + HTv) ≤d −KTv, v ≥0, which is rewritten by max s ∑ i=1 aT i ui −bTv1 − s ∑ i=1 dT i vi+1 s.t.
s ∑ i=1 AT i ui −BTv1 = 0, f ∗ i (−ui) ≤λi + ci, , i = 1, . . . s, v ≥0, where v = (v1, . . . , vs+1)T.
6 Conclusion In this paper, we proposed an optimization problem with sub-homogeneous functions, which we call sub-homogeneous optimization problem. We also introduced its dual prob-lem and showed the weak duality theorem between these problems. Moreover, we gave sufficient conditions for the equivalency between the proposed dual and the Lagrangian dual problems. Finally, we presented some examples of sub-homogeneous problems to show their value in real-world applications. One natural future work will be to propose methods that obtain approximate solutions of sub-homogeneous optimization problems.
We believe the theoretical results described here are essential for that.
Acknowledgements The authors are grateful to Prof. Ellen. H. Fukuda for helpful com-ments and suggestions. This work was supported in part by a Grant-in-Aid for Scientific Research (C) (17K00032) from Japan Society for the Promotion of Science.
A Appendix The following proposition shows that the dual of the p-norm function is the ∞-norm even when p is less than 1.
17 Proposition 6. Suppose that p ∈(0, 1). Then, the dual of the p-norm function is equal to the ∞-norm.
Proof . Let y ∈Rn be an arbitrary vector. If y = 0, this proposition clearly holds. If y ̸= 0, from Definition 3, we obtain ∥y∥∗ p = sup{xTy | ∥x∥p ≤1} ≤ sup{|xTy| | ∥x∥p ≤1} ≤ sup { n ∑ i=1 |xi||yi| | ∥x∥p ≤1 } ≤ max j |yj| ( sup { n ∑ i=1 |xi| | ∥x∥p ≤1 }) = max j |yj| ( sup{∥x∥1 | ∥x∥p ≤1} ) .
Since p ∈(0, 1), we note that ∥x∥1 ≤∥x∥p holds . Then, we have ∥y∥∗ p ≤max j |yj| ( sup{∥x∥p | ∥x∥p ≤1} ) = max j |yj| = ∥y∥∞.
Now, take an arbitrary i0 ∈argmax i |yi|, and define ¯ xi as follows: ¯ xi = { sign(yi), if i = i0, 0, otherwise, where sign(yi) = 1, if yi > 0, 0, if yi = 0, −1, if yi < 0.
Then, ∥¯ x∥p = 1 and we have ∥y∥∗ p = sup{xTy | ∥x∥p ≤1} ≥¯ xTy = max i |yi| = ∥y∥∞, which completes the proof.
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4733 | https://bio.libretexts.org/Bookshelves/Introductory_and_General_Biology/General_Biology_(Boundless)/19%3A_The_Evolution_of_Populations/19.03%3A_Adaptive_Evolution/19.3B%3A_Stabilizing_Directional_and_Diversifying_Selection | Skip to main content
19.3B: Stabilizing, Directional, and Diversifying Selection
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: Nov 23, 2024
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19.3A: Natural Selection and Adaptive Evolution
19.3C: Frequency-Dependent Selection
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Boundless
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
Contrast stabilizing selection, directional selection, and diversifying selection.
Stabilizing Selection
If natural selection favors an average phenotype by selecting against extreme variation, the population will undergo stabilizing selection. For example, in a population of mice that live in the woods, natural selection will tend to favor individuals that best blend in with the forest floor and are less likely to be spotted by predators. Assuming the ground is a fairly consistent shade of brown, those mice whose fur is most-closely matched to that color will most probably survive and reproduce, passing on their genes for their brown coat. Mice that carry alleles that make them slightly lighter or slightly darker will stand out against the ground and will more probably die from predation. As a result of this stabilizing selection, the population’s genetic variance will decrease.
Stabilizing selection: Stabilizing selection occurs when the population stabilizes on a particular trait value and genetic diversity decreases.
Directional Selection
When the environment changes, populations will often undergo directional selection, which selects for phenotypes at one end of the spectrum of existing variation.
A classic example of this type of selection is the evolution of the peppered moth in eighteenth- and nineteenth-century England. Prior to the Industrial Revolution, the moths were predominately light in color, which allowed them to blend in with the light-colored trees and lichens in their environment. As soot began spewing from factories, the trees darkened and the light-colored moths became easier for predatory birds to spot.
Directional selection: Directional selection occurs when a single phenotype is favored, causing the allele frequency to continuously shift in one direction.
Over time, the frequency of the melanic form of the moth increased because their darker coloration provided camouflage against the sooty tree; they had a higher survival rate in habitats affected by air pollution. Similarly, the hypothetical mouse population may evolve to take on a different coloration if their forest floor habitat changed. The result of this type of selection is a shift in the population’s genetic variance toward the new, fit phenotype.
Diversifying (or Disruptive) Selection
Sometimes natural selection can select for two or more distinct phenotypes that each have their advantages. In these cases, the intermediate phenotypes are often less fit than their extreme counterparts. Known as diversifying or disruptive selection, this is seen in many populations of animals that have multiple male mating strategies, such as lobsters. Large, dominant alpha males obtain mates by brute force, while small males can sneak in for furtive copulations with the females in an alpha male’s territory. In this case, both the alpha males and the “sneaking” males will be selected for, but medium-sized males, which cannot overtake the alpha males and are too big to sneak copulations, are selected against.
Diversifying (or disruptive) selection: Diversifying selection occurs when extreme values for a trait are favored over the intermediate values.This type of selection often drives speciation.
Diversifying selection can also occur when environmental changes favor individuals on either end of the phenotypic spectrum. Imagine a population of mice living at the beach where there is light-colored sand interspersed with patches of tall grass. In this scenario, light-colored mice that blend in with the sand would be favored, as well as dark-colored mice that can hide in the grass. Medium-colored mice, on the other hand, would not blend in with either the grass or the sand and, thus, would more probably be eaten by predators. The result of this type of selection is increased genetic variance as the population becomes more diverse.
Comparing Types of Natural Selection
Key Points
Stabilizing selection results in a decrease of a population ‘s genetic variance when natural selection favors an average phenotype and selects against extreme variations.
In directional selection, a population’s genetic variance shifts toward a new phenotype when exposed to environmental changes.
Diversifying or disruptive selection increases genetic variance when natural selection selects for two or more extreme phenotypes that each have specific advantages.
In diversifying or disruptive selection, average or intermediate phenotypes are often less fit than either extreme phenotype and are unlikely to feature prominently in a population.
Key Terms
directional selection: a mode of natural selection in which a single phenotype is favored, causing the allele frequency to continuously shift in one direction
disruptive selection: (or diversifying selection) a mode of natural selection in which extreme values for a trait are favored over intermediate values
stabilizing selection: a type of natural selection in which genetic diversity decreases as the population stabilizes on a particular trait value
19.3A: Natural Selection and Adaptive Evolution
19.3C: Frequency-Dependent Selection |
4734 | https://finto.fi/mesh/en/page/D009894 | Opportunistic Infections - MeSH - Finto
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"an infect caused by an organism which becomes pathogenic under certain conditions, e.g., during immunosuppression"; coord IM with specific name of opportunistic infect (IM) + specific infect to which it is opportunistic (IM), using /compl for all 3
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An infection caused by an organism which becomes pathogenic under certain conditions, e.g., during immunosuppression.
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4735 | https://www.youtube.com/watch?v=dAClgx0aJeg | Expand (x+1/x)^6 using Binomial Theorem
Ah Sing Math TV
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Posted: 23 Aug 2022
A simpler way of applying the Binomial Theorem.
Expand (x+1/x)^6 by using the Binomial Theorem.
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Transcript:
hi disaster you are now watching us in math tv today we would like to share how to expand this exponential by using the binomial trm we have x plus 1 over x to the power of 6 and we should always start from the integer 0 followed by the next integer 1 2 3 4 5 and 6 since the power is 6 which means that the maximum number that we can have is 6 and we know that binomial theorem uses this combination with this maximum value we have 6 0 6 c 1 6 c 2 6 c 3 6 c 4 6 c 5 and 6 c 6 next is to make use of the combination between x and 1 over x if we give all the changes to x which is x to the power of 6 and 1 over x should have zero since the total of the powers must be always equal to this number which is six so if let's say we take only five for x in other words one over x we should take one plus 5 equal to 6 if we take x to the power 4 so we should have 1 over x for 2 since 4 plus 2 equal to 6 and so for the right so we have x to the power 3 1 over x to the power 3 then we have x to the power 2 1 over x to the power of 4 and we have x to the power 1 1 over x to the power 5 and lastly x to power 0 1 over x to the power of 6 and we just have to add up we can evaluate the coefficient by using formula recall that ncr is equal to n factorial divided by r factorial times m minus r factorial which means that if you have 6 c 0 this is equal to 6 factorial divided by 0 factorial times 6 minus 0 factorial and this is equal to 1. if we have 6 c 2 this is equal to 6 factorial divided by 2 factorial and we times 6 minus 2 which is equal to 4 and we have factorial and this is equal to 6 times 5 times 4 times 3 times 2 times 1 divided by 2 times 1 times 4 times 3 times 2 times 1 and we can simplify so we cancel the similar term so the numerator is equal to 30 divided by the denominator 2 so this is equal to 50. by using the same idea we could find auditing efficiency if we don't like to calculate manually we can make use of the calculator so we press number 6 shift divide to get c followed by the number that we want for example 30 and we press equal then we should obtain the correct value we could also determine all the coefficients by using the pascal triangle start from the value one one plus nothing is one one plus 9 is what 1 plus nothing 1 1 plus 1 2 1 plus nothing 1 1 plus nothing 1 1 plus 2 3 2 plus 1 3 1 plus nothing 1 1 plus nothing 1 1 plus 3 is equal to four three plus three six three plus one four one plus nothing one one plus nothing one one plus four five four plus six ten six plus four ten four plus one five one plus nothing one one plus nothing one one plus five says five plus ten fifteen ten plus ten twenty ten plus five fifteen five plus one six one plus nothing one so this is for the power 0 1 2 3 4 5 6 and we have power 6 so which means that the efficient should be this row so we have 1 6 15 20 15 6 and 1. next let's evaluate the terms for x so we have x about 6 divided by x to the power of 0 so we have x about 6 x about 5 divided by x power 1 so simply 5 minus 1 is equal to 4 4 minus 2 equal to 2 3 minus 3 is equal to 0 and 2 minus four is equal to negative two one minus five is equal to negative four and zero minus six is negative six and this is equal to x above six plus 6 x power 4 plus 15 x squared plus 20 plus 15 divided by x squared plus 6 divided by x power 4 plus 1 over x to power six enhance with that okay that's all for this video thanks for watching have like this see you |
4736 | https://www.cod.edu/academics/tasc/mathematics-assistance/pdf/diy_trigonometry_identities.pdf | _______________ Stop by or call (630) 942-3339 Introduction: Identities are equations that are true for every value of the variable(s) in the equation. For example, 3 + 7 = 10 and 3x + 7x = 10x are identities, but 3x + 7x = 10 is not an identity since it is only true if x = 1. Since all six trigonometric functions are defined by only two independent qualities (the x- and y-coordinates of a point on the unit circle, or the lengths of the opposite and adjacent sides of a right triangle), there is much interdependence in these functions. An example of a trigonometric identity comes from the definitions of sine and cosine; sin 𝑥= 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 and cos 𝑥= 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 so sin 𝑥 cos 𝑥= 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒∙ ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡= 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡= tan 𝑥 by definition. So tan 𝑥= sin 𝑥 cos𝑥 is an identity. It is true for any angle x. To learn about other trigonometric identities, how to prove they are identities, and how to use identities to simplify expressions and evaluate trigonometric quantities, watch the following set of YouTube videos. They are followed by several practice problems for you to try, using the concepts covered in the videos, with answers and detailed solutions. Some additional resources are included for more practice at the end. 1. Co-function identities: 2. Basic Pythagorean identities:- 3. Sum and difference and double angle identities:- 4. Use of sum and difference identities to evaluate trigonometric ratios of certain angles a. Example 1:- b. Example2:- c. Example3:- d. Example4:- 5. Half angle identities:- 6. Power Reducing Identities and Product-to-Sum Identities: 7. Verifing identities: 8. More on verifying identities: 9. Solving Trigonometric equations: 10. Solving trigonometric equations: DIY: Trigonometric Identities ______________ Stop by or call (630) 942-3339 Summary of Trigonometric Identities 1. Fundamental Identities: Cofunction Identities: sin𝑥= cos(90𝑜−𝑥) cos 𝑥= sin(90° −𝑥) tan 𝑥= cot (90° −𝑥) cot 𝑥= tan (90° −𝑥) sec 𝑥= 𝑐𝑠𝑐(90° −𝑥) csc 𝑥= sec(90° −𝑥) (Note: substitute π/2 for 90⁰ if using radian measure) Quotient Identities: tan 𝑥= sin𝑥 cos 𝑥 cot 𝑥= cos 𝑥 sin𝑥 Reciprocal Identities: cot 𝑥= 1 tan 𝑥 sec 𝑥= 1 cos 𝑥 csc 𝑥= 1 sin𝑥 Even/Odd functions: sin(−𝑥) = −sin 𝑥 (odd function cos(−𝑥) = cos 𝑥 (even function) tan(−𝑥) = −tan 𝑥 (odd function) Pythagorean Identities: cos2 𝑥+ sin2 𝑥= 1 1 + tan2 𝑥= sec2 𝑥 1 + cot2 𝑥= csc2 𝑥 2. Sum Identities: sin(𝑥± 𝑦) = sin 𝑥cos 𝑦± sin𝑦cos𝑥 cos(𝑥± 𝑦) = cos 𝑥cos 𝑦 ∓sin𝑥sin 𝑦 tan(𝑥± 𝑦) = tan 𝑥 ±tan𝑦 1∓𝑡𝑎𝑛 𝑥 𝑡𝑎𝑛𝑦 3. Double Angle and Power Reducing Identities: sin2𝑥= 2 sin 𝑥cos 𝑥 cos 2𝑥= cos2 𝑥−sin2 𝑥= 2 cos2 𝑥−1 = 1 −2 sin2 𝑥 tan 2𝑥= 2 tan 𝑥 1−tan2 𝑥 sin2 𝑥= 1 2 (1 −cos 2𝑥) cos2 𝑥= 1 2 (1 + cos 2𝑥) 4. Half-Angle Identities: sin 𝑥 2 = ±√1−cos𝑥 2 cos 𝑥 2 = ±√1+cos 𝑥 2 tan 𝑥 2 = ±√1−cos 𝑥 1+cos 𝑥= sin𝑥 1+cos 𝑥= 1−cos𝑥 sin𝑥 5. Product-to-Sum and Sum-to-Product Identities: Product-to-Sum: sin𝑥cos 𝑦= 1 2 [sin(𝑥+ 𝑦) + sin (𝑥−𝑦] sin𝑥sin𝑦= 1 2 [cos(𝑥−𝑦) −cos(𝑥+ 𝑦)] cos 𝑥cos𝑦= 1 2 [cos(𝑥+ 𝑦) + cos (𝑥−𝑦)] Sum-to-Product: sin𝑥+ sin𝑦= 2 sin 𝑥+𝑦 2 cos 𝑥−𝑦 2 sin𝑥−sin 𝑦= 2 sin 𝑥−𝑦 2 cos 𝑥+𝑦 2 cos 𝑥+ cos𝑦= 2 cos 𝑥+𝑦 2 cos 𝑥−𝑦 2 cos 𝑥+ cos 𝑦= −2 sin 𝑥+𝑦 2 sin 𝑥−𝑦 2 _______________ Stop by or call (630) 942-3339 Practice problems: The following problems use the techniques demonstrated in the above videos. The 1) Verify the following identities: a) sin4x - cos4𝑥=(sin x – cos x)(sin x+ cos x) b) 𝑠ec2A - 1=tan2A c) sin θ 1+cos θ + 1+cos θ sin θ =2csc θ d) 1+tan2𝛽 1+cot2β = tan2 𝛽 2) Evaluate the following without using a calculator. Write the exact answer: a) sin2117° + cos2117° b) cos(-135°) c) sin 100° cos 10° – sin10° cos 100° d) tan5𝜋 12 + tan𝜋 4 1− tan5𝜋 12 tan𝜋 4 e) -sin 150° sin 60° +cos 150° cos 60° f) sin 75° g) 2 sin𝜋 6 cos𝜋 6 1−2sin2𝜋 6 h) 1 sec 2(𝜋 4)−1 i) cos215° - sin215° j) sin229° +2 sin 60° cos 60° + cos2 29° 3) Find the exact values for the following, using identities and values of trig functions for special angles (such as 30⁰, 45⁰, 60⁰, 90⁰, etc.) a) sin 105⁰ b) tan 225⁰ c) cos 165⁰ 4) Verify the following identities: (Note: these are more challenging than the identities in problem 1, involving using identities and algebraic techniques.) a) 1−tan 𝐴 1+tan 𝐴= cot 𝐴 − 1 cot 𝐴 + 1 b) cos6 θ+ sin6 θ=1-3 sin2θ cos2θ (hint: think of the left-hand side as the sum of cubes, then factor) c) sec 2𝑥= sec2 𝑥+sec4 𝑥 2+sec2 𝑥−sec4 𝑥 d) tan 𝐴 2 = 1−cos 𝐴 sin 𝐴 ______________ Stop by or call (630) 942-3339 e) sin 420° cos 390° + cos (-300°) sin(-330°) =1 f) sin B = 2 tan 𝐁 𝟐 1+tan 2 𝐁 𝟐 (hint: use the identity from part d) g) cos 2t = cot 𝑡−tan 𝑡 csc 𝑡sec 𝑡 h) √1−sin A 1+sin A = sec A – tan A i) ( sin α+csc α )2+ ( cos α+ sec α )2= tan2α+cot2α + 7 5) Evaluate: a) If sin x = 12 13 , find tan x and cos x c) Find sin (s+t) and cos (s+t) given that sin s = -4/5 and cos t = 12/13. Angle s is in Q III and angle t is in Q IV. What quadrant is (s+t) in? b) If cos2 A = 3 4 , find cot A and csc A 6) Find all solutions in [0,2π) : Note: these are NOT identities. They are only true for particular values of the variable. a) sin2 𝜃+ 2 cos 𝜃−2 = 0 b) cot x +tan x =2csc x c) cot2 𝐵+ (√3 + 1 √3) cot 𝐵+ 1 = 0 d) sin 5θ = cos 5𝜃 e) tan 𝛼 + sec 𝛼 = √3 f) csc x = 1 + cot x 7) Prove the following: a) sec2 x 2 = 2 1+cos 𝑥 b) - cot x 2 = sin 2x+sin x cos 2x−cos x c) sin 3t+sin 2t sin 3t−sin 2t = tan 5t 2 tant 2 d) tan ( 𝑥 2 + 𝜋 4) = sec x+ tan x Hint: Use the sum identity for tangent, then convert all terms to sines and cosines. Hint: Use Sum-to-Product Identities _______________ Stop by or call (630) 942-3339 Answers: 1. (Proofs) 2. a. 1 b. −1 √2 = −√2 2 c. 1 d. −√3 e. −√3 2 f. √2+√6 4 g. √3 h. 1 i. √3 2 j. 1 + √3 2 = 2+√3 2 3. a. √6+√2 4 b. 1 c. −√6+√2 4 4. (Proofs) 5. a. cos 𝑥= ± 5 13 , tan 𝑥= ± 12 5 b. cot 𝐴= ±√3, csc 𝐴= ±√2 c. sin(𝑠+ 𝑡) = −33 65 , cos(𝑠+ 𝑡) = −56 65 , (s+t) is in Quadrant III 6. a. 0 b. 𝜋 3 , 5𝜋 3 c. 2𝜋 3 , 5𝜋 3 , 5𝜋 6 , 11𝜋 6 d. 𝜋 20 , 9𝜋 20 , 17𝜋 20 , 25𝜋 20 = 5𝜋 4 , 33𝜋 20 , 5𝜋 20 = 𝜋 4 , 13𝜋 20 , 21𝜋 20 , 29𝜋 20 , 37𝜋 20 e. 𝜋 6 f. 𝜋 2 7. (Proofs) _______________ Stop by or call (630) 942-3339 Detailed Solutions ______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 ______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 ______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 ______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 ______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 ______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 _______________ Stop by or call (630) 942-3339 __________________ Stop by or call (630) 942-3339 Additional Resources 1. Go To 2. Under “Trigonometry” click on: a) Simple trig equations b) Simple trig equations c) Inverse trig functions d) Fundamental identities e) Equations with factoring and fundamental identities f) Sum and Difference Identities g) Multiple-Angle Identities h) Product-to-Sum Identities i) Equations and Multiple-Angle Identities 3. You can print out the worksheets and work on them. 4. For help please contact the Math Assistance Area. |
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2011 IMO Problems/Problem 4
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1 Problem
2 Solution
3 Alternative Solution
4 See Also
Problem
Let be an integer. We are given a balance and weights of weight . We are to place each of the weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed. Determine the number of ways in which this can be done.
Solution
Call our answer . We proceed to prove .
It is evident .
Now, the key observation is that smaller weights can never add up to the weight of a larger weight, ie which side is heavier is determined completely by the heaviest weight currently placed. It follows, therefore, that the number of ways to place weights on the balance according to the rule is the same no matter which distinct powers of two are the weights, as each weight completely overpowers any smaller weight and is completely overpowered by any larger weight. That is, there is the 1st heaviest weight, the 2nd heaviest, the 3rd, ..., the n-th heaviest, and each weight is negligible compared to any heavier weight. Thus, any valid placement of weights of weight can changed by replacing with the -th heaviest weight in the set , where , and vice versa, forming a relation. With this in mind, we use recursion upon the last weight placement. There are choices; namely, you can put any weight on either side except for the heaviest weight on the right. For the first weight placements, the answer reduces to . We can reduce in the same way.
Alternative Solution
We can compute the answer by conditioning on the position of the heaviest weight in the order of placement: The heaviest weight can only go to the left pan. is the number of ways to place the first weights which got placed before the heaviest weight at position . We used the fact that the number of valid ways does not depend on the actual weights because each weight is heavier than the sum of all the weights lighter than it. There are ways to select weights ( the order matters ) after the heaviest one out of other weights . This is because there are ways to select the 1st weight after the heaviest, to select the next one etc. Each of these weights can go to the left or the right pan so there are ways to create all left-right combinations. Rearange as recursion: That is: as since if you have just one weight it can only go to the left pan.
--alexander_skabelin 9:24, 13 July 2023 (EST)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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4738 | https://discourse.julialang.org/t/stuck-using-backtracking/107222 | Stuck using backtracking
Hello.
To improve my skills I am experimenting with backtracking as a potential approach to solve a verbal arithmetic (alphametics) puzzle.
As a starting point I tried to develop a simple implementation of a backtracking algorithm as described on the Wikipedia:
# just for testing purposes the backtracking should guess a sequence of numbers containing no duplicates (here:[5,3,1]`)
the procedures / functions are named (abbreviated) according to the nomenclature on the Wikipedia page
the 'reject' procedure:
"returns true only if the partial candidate [sequence] is not worth completing"
here: rejects sequences with duplicates (prunes branches)
r(v::Vector) = !allunique(v)
the 'accept' procedure:
"returns true if a candidate [sequence] is a solution"
a(v::Vector) = v == [5,3,1] # to keep it simple just a comparison of a candidate sequence against a predetermined solution
the 'first' procedure:
"generates the first extension of a candidate [sequence]"
f(v::Vector) = length(v) == 3 ? nothing : append!(v,0) # to keep it small the solution can consist of max. 3 numbers (a.k.a. 'n')
the 'next' procedure:
"generates the next alternative extension of a candidate [sequence]"
function n(v::Vector)
v[end] == 5 && return nothing # again, to keep it small the range of possible numbers is restricted to 0-5 (a.k.a. 'm'; 'first' / f() sets lower boundary)
v[end] += 1
return v
end
the main 'backtracking' procedure:
numbered @show() (1-3) to be able to follow along
function bt(v::Vector)
@show((1, v))
r(v) && return
a(v) && return "bingo"
s = f(v)
@show((2, s))
while !isnothing(s) # same results with s != nothing
bt(s)
s = n(s)
@show((3, s))
end
end
bt([])`
The procedure does not find a solution and returns nothing.
nothing
The outputs of @show show that the backtracking enters the first ‘subtree/branch’ and then terminates early before even exploring other ‘subtrees/branches’ like [1,...], [2,...], [3,...], etc or even [0,2,...], [0,3,...], etc… (seems like the while loop does not continue)
@show
[1,...]
[2,...]
[3,...]
[0,2,...]
[0,3,...]
# outputs of@show(1-3):
(1, v) = (1, Any[])
(2, s) = (2, Any)
(1, v) = (1, Any)
(2, s) = (2, Any[0, 0])
(1, v) = (1, Any[0, 0])
(3, s) = (3, Any[0, 1])
(1, v) = (1, Any[0, 1])
(2, s) = (2, Any[0, 1, 0])
(1, v) = (1, Any[0, 1, 0])
(3, s) = (3, Any[0, 1, 1])
(1, v) = (1, Any[0, 1, 1])
(3, s) = (3, Any[0, 1, 2])
(1, v) = (1, Any[0, 1, 2])
(2, s) = (2, nothing)
(3, s) = (3, Any[0, 1, 3])
(1, v) = (1, Any[0, 1, 3])
(2, s) = (2, nothing)
(3, s) = (3, Any[0, 1, 4])
(1, v) = (1, Any[0, 1, 4])
(2, s) = (2, nothing)
(3, s) = (3, Any[0, 1, 5])
(1, v) = (1, Any[0, 1, 5])
(2, s) = (2, nothing)
(3, s) = (3, nothing)
(3, s) = (3, nothing)
(3, s) = (3, nothing)
What am I doing wrong? I guess, it is probably something obvious, but even after considerable time experimenting I am not able to see it. (e.g. I do not understand why the while loop would exit early.)
I would gratefully appreciate if someone could kindly point me in the right direction.
I apologize should this be the wrong channel for this kind of questions. I consider myself still being fairly new to Julia.
Also my searches on backtracking did not return something that I would have associated with a potential (if only a partial) solution to my confusion.
Thanks for your time in advance!
v[end] == 5 && return nothing # again, to keep it small the range of possible numbers is restricted to 0-5 (a.k.a. 'm'; 'first' /f()sets lower boundary)
I think you still want to return v here, just without the last element. i.e.:
v
v[end] == 5 && (pop!(v); return v)
You might have to change the while condition too, to isempty or something like that.
while
isempty
Many thanks!
v[end] == 5 && (pop!(v); return nothing)
did the trick!
v[end] == 5 && (pop!(v); return nothing)
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4739 | https://math.stackexchange.com/questions/3286150/understanding-the-difference-between-pre-image-and-inverse | Stack Exchange Network
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Understanding the difference between pre-image and inverse
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$\begingroup$
I am a little confused as to what the difference between the pre-image and the inverse of a function are and how to find each given a particular function ( I had though they were essentially the same thing but I realise now I was mistaken in that thought ).
From Wikipedia the definitions given are :
Inverse
Let $f$ be a function whose domain is the set $X$, and whose image (range) is the set $Y$. Then $f$ is invertible if there exists a function $g$ with domain $Y$ and image $X$, with the property: $f ( x ) = y \iff g ( y ) = x . $ If $f$ is invertible, the function $g$ is unique, which means that there is exactly one function $g$ satisfying this property (no more, no less). That function $g$ is then called the inverse of $f$, and is usually denoted as $f^{1}$
Pre-image
Let $f$ be a function from $X$ to $Y$. The preimage or inverse image of a set $B\subseteq Y$ under $f$ is the subset of $X$ defined by $f ^{ 1 }[ B ] = { x \in X \mid f ( x ) \in B }$.
So using an example I want to see if I have this about right ( please correct any mistakes I make)
Example 1:
Lets say $f:\Bbb R \rightarrow \Bbb R$
$f(x)=x^2$
For this example, clearly $f$ cannot be invertible (hence no inverse) as there exists no function $g$ which will satisfy $ f ( x ) = y \iff g ( y ) = x$. (as it would only map to positive values of $\Bbb R$ (i.e. not the whole set))
The pre-image of this function I believe is related to the inverse except it does not require that we map to the whole set $\Bbb R$, but rather just a subset of it. Therefore we can find the function $f^{-1}$ in an analogous way to to finding the inverse we just have to be more considerate about what the co-domain of this function is.
So if $f(x)=x^2 \Rightarrow y=x^2 $swap variables to get $x=y^2 \Rightarrow \sqrt{x}=y=f^{-1} $
So the pre-image is the set $f ^{ 1 }[ \Bbb R_+ ] = { x \in X \mid f ( x ) \in \Bbb R_+, f^{-1}=\sqrt{x} } $
Example 2:
An example of an invertible function would be $f:\Bbb R \rightarrow \Bbb R$
$f(x)=5x$, as a function $g(x)=x/5$ has domain $\Bbb R$ and range $\Bbb R$ and satisfies $f ( x ) = y g ( y ) = x .$
The pre-image in this case will be equal to the inverse.
Could anyone please explain to me any mistakes I'm making here ?
functions
definition
inverse
inverse-function
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edited Jul 8, 2019 at 1:24
Maximilian Janisch
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asked Jul 7, 2019 at 20:52
excalibirrexcalibirr
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3
$\begingroup$ For starters, they are two very different objects. The pre-image of a function is a subset of the domain and the inverse function is a function from the range back to the domain that satisfies certain properties. A function may not be invertible but we can always talk about pre-image of a function. $\endgroup$
Anurag A
– Anurag A
2019-07-07 20:58:43 +00:00
Commented Jul 7, 2019 at 20:58
$\begingroup$ @AnuragA I know now that they are not the same thing , I was just wondering if I was correct in identifying how they are different $\endgroup$
excalibirr
– excalibirr
2019-07-07 21:00:50 +00:00
Commented Jul 7, 2019 at 21:00
$\begingroup$ No, one peaks in almost all cases of the preimage of a subset of the target space. Take, for instance, the squaring function $f(x)=x^2$, and ask for the preimage of the set of odd numbers. Then this will be the set of all $\pm\sqrt{2k+1}$, with $k$ running through all integers. If one were to speak of the preimage of a function, presumably that would be the preimage of the whole target space, which is all of the domain of definition of the function. $\endgroup$
Lubin
– Lubin
2019-07-07 21:09:58 +00:00
Commented Jul 7, 2019 at 21:09
Add a comment |
3 Answers 3
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11
$\begingroup$
The biggest difference between a preimage and the inverse function is that the preimage is a subset of the domain. The inverse (if it exists) is a function between two sets.
In that sense they are two very different animals. A set and a function are completely different objects.
So for example: The inverse of a function $f$ might be: The function $g:\mathbb R \to \mathbb R: g(x) = \sqrt{x-9}$. Whereas the preimage of a set $B$ of the function might be $[1,3.5)\cup {e, \pi^2}$.
Now $g(x) = \sqrt{x-9}$ and $[1,3.5)\cup {7, \pi^2}$ are completely different types of things.
This will be the case if $f$ is $f:\mathbb R \to \mathbb R: f(x) = x^3 + 9$ and $B= [10, 51.875) \cup {352, \pi^6 + 9}$.
The inverse $f^{-1}(x)$ (if it exist) is the function $g$ so that if $f(x) = y$ if and only if $g(y) = x$. So if $f(x) = x^3 + 9 = y$ then if such a function exists it must be that $g(y)^3 + 9 = y$ so $g(y)^3 = y-9$ and $g(y) = \sqrt{y-9}$ so $g(x) = \sqrt{x-9}$.
That's that.
The pre-image of $A= [10, 51.875) \cup {352, \pi^6 + 9}$ is the set ${x\in \mathbb R| f(x) \in [10, 51.875) \cup {352, \pi^6 + 9}}=$
${x\in \mathbb R| x^3 + 9 \in [10, 51.875) \cup {352, \pi^6 + 9}}=$
${x\in \mathbb R| x^3 \in [1, 42.875) \cup {343, \pi^6 }}=$
${x\in \mathbb R| x \in [1, 3.5) \cup {7, \pi^2 }}=$
$[1, 3.5) \cup {7, \pi^2 }}$.
And that's the other.
........
Now that's not to say the inverse of a function and the pre-image of a set under the function aren't related. They are. But they refer to different concepts. This is similar to how a rectangle and its area are related. But one is a geometric shape... the other is a positive real number. THey are two different types of animals.
....
I'll add more in an hour or so but I have to take the dog for a walk. I'll be back.
.....
It occurred to me as I was walking the dog that maybe what is confusing you is that the inverse function (if it exists) and the preimage of a set have very similar notation and the only way to tell them apart is in context.
If $f$ is invertible then the inverse function is written as $f^{-1}$ so if $f(x) = x^3 + 9$ then $f^{-1}(x) = \sqrt{x-9}$.
But the preimage of $B$ under $f$ whether $f$ is invertible or or not is writen as $f^{-1}(B)$.
So if $f(x) = x^3 + 9$ then $f^{-1}(17) = 2$ means that if you enter $17$ into the function $\sqrt{x -9}$ you get $2$. But $f^{-1}({17})={2}$ and $f^{-1}({36,17}) = {2,3}$ means that set of values that will output ${17}$ is the set ${2}$ and the set of values that will output ${36,17}$ is the set ${2,3}$.
A few things to note:
If $f$ is invertible then the preimage of a set is the same thing as the image of the set under the inverse function and that means the notation is compatible.
If $f(x) = x^3 + 9$ then $f^{-1}([1,36)) = [1,3)$ can be interpretated as both the the image of the set under the inverse function: $f^{-1}([1,36))= {f^{-1}(x) = g(x) = \sqrt{x-9}| x\in [1,36)}$
OR it can be interpreted as the preimage for $f$: $f^{-1}([1,36)) = {x\in \mathbb R| f(x) \in [1,36)}$.
but this is not the case if $f$ is not invertible.
Say $f:\mathbb R \to [-1,1]; f(x)\to \sin x$. This is not invertible.
The pre-image of$B= {\frac {\sqrt 2}2}$ is ${...-\frac {11\pi}4, -\frac {9\pi}4,-\frac{3\pi}4,-\frac \pi 4, \frac \pi 4, \frac {3\pi}4, \frac {9\pi}4, \frac {11\pi}4,....}$ this is still written as $f^{-1}( {\frac {\sqrt 2}2})$ even though there is no function $f^{-1}:[-1,1]\to \mathbb R$.
Another thing to note is that not all the elements in $B$ have to have pre-image values.
If $f= x^2+9$ then $f^{-1}({8}) = \emptyset$. This is because ${x\in \mathbb R| f(x) = x^2 + 9 \in {8}} = \emptyset$.
And some elements may have many preimages.
And $\sin^{-1}\left({\frac {\sqrt2} 2}\right)$ showed.
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edited Nov 25, 2023 at 22:57
epelaez
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answered Jul 7, 2019 at 21:38
fleabloodfleablood
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$\begingroup$ Thank you for such a brilliantly comprehensive answer :) there's literally nothing else I could think to ask . $\endgroup$
excalibirr
– excalibirr
2019-07-15 07:34:32 +00:00
Commented Jul 15, 2019 at 7:34
$\begingroup$ Actually , I did manage to think of question lol, (it's in the context of continuity in topology) say we have the function $f=x^2$ and a basis set $(-q,q), q \in \Bbb Q$ in our target. then I know that $f^{-1}((-q,q))={ x \in X| x\in (-\sqrt{q}, \sqrt{q})}$ but why isn't it $(\sqrt{-q}, \sqrt{q})$ instead ?, also say we have the set (a,b) (a $\endgroup$
excalibirr
– excalibirr
2019-07-15 09:28:03 +00:00
Commented Jul 15, 2019 at 9:28
$\begingroup$ Amazing answer! But I believe a small typo $f^{-1}({17})={3}$ should be 2 not 3. Thanks! $\endgroup$
CormJack
– CormJack
2022-12-20 12:41:29 +00:00
Commented Dec 20, 2022 at 12:41
Add a comment |
4
$\begingroup$
When we talk about pre-image then it has two components: a set and a function. So when we say $f^{-1}[B]$, then we want the pre-image of the set $B$ (a subset of the co-domain) under the function $f$. So asking about pre-image of a function is a bit ambiguous.
Let us consider $f:{1,2,3} \rightarrow {a,b,c}$ such that $f(1)=a, f(2)=a$ and $f(3)=c$. Then \begin{align} f^{-1}\left[{a}\right] & ={1,2}\ f^{-1}\left[{c}\right] & ={3}\ f^{-1}\left[{a,c}\right] & ={1,2,3}\ f^{-1}\left[{b}\right] & =\emptyset\ f^{-1}\left[{a,b,c}\right] & ={1,2,3} \end{align}
The function $f$ (as described above) is not invertible. So relating the inverse function (which doesn't exist) to the inverse image is meaningless in this case.
One can check invertibility of a function $f: A \rightarrow B$ by checking the inverse images of singleton subsets of the co-domain.
What it means is that: if we can ensure that for every $b \in B$, the inverse image set $f^{-1}\left[{b}\right]$ has exactly one element (this is to ensure both one-one and ontoness), then $f$ is invertible.
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answered Jul 7, 2019 at 21:14
Anurag AAnurag A
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So, there's no such thing as the preimage of a function. Functions can have inverses; functions do not have preimages.
An inverse is something that certain functions have, and the inverse of a function is another function.
Given a function, a preimage is something that sets have, and the preimage of a set is another set.
Specifically:
Given a function $f : A \to B$, that function may or may not have an inverse. If it does, then that inverse is a function $B \to A$.
Given a function $f : A \to B$, and a set $s$ which is a subset of $B$, that set always has a preimage under $f$. That preimage is a subset of $A$.
That might clear up some of your confusion.
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answered Jul 7, 2019 at 22:03
Sophie SwettSophie Swett
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Nicholas's birthday party will cost $12 if he invites 6 guests. If there are 9 guests, how much will Nicholas's birthday party cost? Assume the relationship is directly proportional.
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4741 | https://www.youtube.com/watch?v=F4IOxTuVMgQ | Proof of equation of a line - the slope-intercept form y =mx+c
mattam66
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Proof of equation of a line - the slope-intercept form y =mx+c
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Transcript:
in this video I'm going to prove the well-known slope intercept form of a line which is y = mx + C where m is the gradient or slope and C is the Y intercept so this is what we want to prove prove that the equation of a line with a slope a gradient of M and y intercept of C is given by this equation which is called slope intercept form y = mx + C so this is your X and Y axis so this is your y AIS and this is my this is your x axis okay and I've drawn a line an arbitrary line which makes an angle of theta with the positive one of the x-axis and uh B is uh the point where the line intersects the Y AIS so you can say this is the origin o is the origin and the coordinate of B is 0 comma C so this distance o so let's write OB is equal to C because the Y intercept is C okay so let us take an arbitrary Point anywhere on this line X say let us say p x comma Y is an arbitrary Point anywhere on the line okay so now I'm going to drop a perpendicular from p on the X AIS so this is a perpendicular from p on x axis and I'm joining uh B with the with this line so basically this is a right angle and this is also a right angle so let us name this point say m and this point is n okay so I hope you can see uh that these two lines can can you I hope you can see that line BN is parallel to xais is parallel to xais BN is parallel to x-axis and as it's parallel you can consider the line AB as a transversal so if this angle is Theta that angle you can say if angle b a o is Theta B AO is Theta then you can say this angle pbn is also Theta so this angle is also Theta I hope you can see uh it is Theta because of corresponding angles formed by parall lines and the transversal so you can say angle p b n is also Theta okay I'm writing in short this is because of corresponding angles parallel lines okay this is how you can write in short okay so what happens now now in Triangle uh in Triangle in Triangle p n b in Triangle p n b angle p n B is 90 or right angle so I can say tan Theta is opposite over hypotenuse so the opposite is PN sorry opposite over adjacent so PN over BN tan is uh to R is opposite over adjacent now this is nothing but rise over run this is rise over run so if you consider these two points say if you consider B and P uh you can cons you can say that BN is the PN is the rise PN is the rise and BN is the r uh run and rise over run is nothing but gradient and that is nothing we can denote the letter M so we can say m therefore I can say let me change color therefore m is equal to tan Theta which is equal to pn/ BN okay now coming looking back to the figure PN PN let me scroll this up slightly up PN is nothing but p m minus n m from PM if you take away n m you get PN okay so can I write this as p m minus PN no minus MN sorry p m minus MN over BN okay so let me do so what is what is PM PM is nothing but y PM is y and NM is C so you can say this is C that is if B if B is C this implies uh you can say MN is also equal to C because if B is C MN is also C and PM is y because the coordinate of p is X comma y so I'm going to write this therefore m is PM is y minus MN is minus MN is C and what is BN the distance from B to n is X if this is X sorry if this is X this distance is X so if this is X this is also X so in place of BN I can put X okay let me repeat so o o is C so NM is c om m is X so BN is also X because C is an arbitrary Point having the coordinate X comma y so the distance from o to m is X and the distance from M to P is y okay so now this this implies MX is = to y - c now multiplying or adding C to both sides mx + C is equal y therefore y = mx + C is the general form of an equation of a line which is called the slope intercept form |
4742 | https://www.storyofmathematics.com/difference-of-squares/ | JUMP TO TOPIC [show]
What is Difference of Squares?
Difference of Squares Formula
How to Factor Difference of Squares?
Practice Questions
Difference of Squares – Explanation & Examples
A quadratic equation is a second degree polynomial usually in the form of f(x) = ax2 + bx + c where a, b, c, ∈ R, and a ≠ 0. The term ‘a’ is referred to as the leading coefficient, while ‘c’ is the absolute term of f (x). Every quadratic equation has two values of the unknown variable, usually known as the roots of the equation (α, β).
What is Difference of Squares?
The difference of two squares is a theorem that tells us if a quadratic equation can be written as a product of two binomials, in which one shows the difference of the square roots and the other shows the sum of the square roots.
One thing to note about this theorem is that it does not apply to the SUM of squares.
Difference of Squares Formula
The difference of square formula is an algebraic form of the equation used to express the differences between two square values. A difference of square is expressed in the form:
a2 – b2, where both the first and last term is perfect squares. Factoring the difference of the two squares gives:
a2 – b2 = (a + b) (a – b)
This is true because, (a + b) (a – b) = a2 – ab + ab – b2 = a2 – b2
How to Factor Difference of Squares?
In this section, we will learn how to factorize algebraic expressions using the difference of square formula. To factor a difference of squares, the following steps are undertaken:
Check if the terms have the greatest common factor (GCF) and factor it out. Remember to include the GCF in your final answer.
Determine the numbers that will produce the same results and apply the formula: a2– b2 = (a + b) (a – b) or (a – b) (a + b)
Check whether you can factor the remaining terms any further.
Let’s solve a few examples by applying these steps.
Example 1
Factor 64 – x2
Since we know the square of 8 is 64, then we can rewrite the expression as;
64 – x2 = (8)2 – x2
Now, apply the formula a2 – b2 = (a + b) (a – b) to factorize expression;
= (8 + x) (8 – x).
Example 2
Factorize x 2 −16
Solution
Since x2−16 = (x) 2− (4)2, therefore apply the difference square formula a2 – b2 = (a + b) (a – b), where a and b in this case are x and 4 respectively.
Therefore, x2 – 42 = (x + 4) (x – 4)
Example 3
Factor 3a2 – 27b2
Solution
Since 3 is GCF of the terms, we factor it out.
3a2 – 27b2 = 3(a2 – 9b2)
=3[(a)2 – (3b)2]
Now apply a2 – b2 = (a + b) (a – b) to get;
= 3(a + 3b) (a – 3b)
Example 4
Factor x3 – 25x
Solution
Since the GCF = x, factor it out;
x3 – 25x = x (x2 – 25)
= x (x2 – 52)
Apply the formula a2 – b2 = (a + b) (a – b) to get;
= x (x + 5) (x – 5).
Example 5
Factor the expression (x – 2)2 – (x – 3)2
Solution
In this problem a = (x – 2) and b = (x – 3)
We now apply a2 – b2 = (a + b) (a – b)
= [(x – 2) + (x – 3)] [(x – 2) – (x – 3)]
= [x – 2 + x – 3] [x – 2 – x + 3]
Combine the like terms and simplify the expressions;
[x – 2 + x – 3] [x – 2 – x + 3] = > [2x – 5]
= [2x – 5]
Example 6
Factor the expression 25(x + y)2 – 36(x – 2y)2.
Solution
Rewrite the expression in the form a2 – b2.
25(x + y)2 – 36(x – 2y)2 => {5(x + y)}2 – {6(x – 2y)}2
Apply the formula a2 – b2 = (a + b) (a – b) to get,
= [5(x + y) + 6(x – 2y)] [5(x + y) – 6(x – 2y)]
= [5x + 5y + 6x – 12y] [5x + 5y – 6x + 12y]
Collect like terms and simplify;
= (11x – 7y) (17y – x).
Example 7
Factor 2x2– 32.
Solution
Factor out the GCF;
2x2– 32 => 2(x2– 16)
= 2(x2 – 42)
Applying the difference squares formula, we get;
= 2(x + 4) (x – 4)
Example 8
Factor 9x6 – y8
Solution
First, rewrite 9x6 – y8 in the form a2 – b2.
9x6 – y8 => (3x3)2 – (y4)2
Apply a2 – b2 = (a + b) (a – b) to get;
= (3x3 – y4) (3x3 + y4)
Example 9
Factor the expression 81a2 – (b – c)2
Solution
Rewrite 81a2 – (b – c)2 as a2 – b2
= (9a)2 – (b – c)2
By applying the formula of a2 – b2 = (a + b) (a – b) we get,
= [9a + (b – c)] [9a – (b – c)]
= [9a + b – c] [9a – b + c ]
Example 10
Factor 4x2– 25
Solution
= (2x)2– (5)2
= (2x + 5) (2x – 5
Practice Questions
Previous Lesson | Main Page | Next Lesson |
4743 | https://www.youtube.com/watch?v=6QSPvedEePI | Transmission and Reflection at a Potential Barrier, PHYS 372
Stephen Remillard
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Posted: 23 Oct 2022
The transmission and reflection coefficients for a probability wave at a step in potential are derived and explained. This is an excerpt from one of my Modern Physics lectures.
Please consider clicking on "Like" so that I know what sort of content is being appreciated. I don't ask for support for developing this content, just the satisfaction of knowing that it is appreciated.
GriffithsQuantumMechanics
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Transcript:
there are several problems in physics and Engineering where a wave encounters a sudden change in potential energy one example is a classic problem of an ultimate contact between a metal and a semiconductor an electron incident on the interface between the two encounters a sudden change in potential energy so if it's propagating in this case from left to right it's going to encounter a sudden increase in its kinetic energy because of a sudden decrease in its potential energy and that's the problem that we'll model right now the energy experienced by a wave in one dimensional space suddenly changes at x equals zero either can go up or down and so it's either a barrier in potential or a step in potential and we want to find the transmission and reflection coefficient a sketch of the potential function shows a sudden change at x equals zero it has value b0 and negative space and a value zero in positive space a wave is incident at energy e and when that wave arrives at the interface some of it has to reflect because there's been a sun change in the potential energy it's like a light wave arriving at a dielectric interface and there's a sudden change in the impedance to that wave and so some of it reflects back and some of it proceeds through the interface I'll call the negative region region 1 and the positive region is Region 2 and already down the wave functions in those two regions on the left in region one you have an incident plane wave that is a free particle and it reflects it's still a free particle so it's still a plane wave so the wave function in the negative region has two plane waves one is forward propagating that's e to the ik One X and one is reverse propagating either minus ik1x K1 is the wave number in the negative region in the positive region the wave numbers K2 and the wave function is just e to the I K2 X there's no reverse propagating wave because it's only transmitting through and heading on to the right nothing's coming from the left let's figure out these wave numbers in this two regions that come from the kinetic energy of the wave which is presumably a particle in region one the kinetic energy is the energy of the wave minus the potential energy as well as in region two it's just there's no potential energy so the kinetic energy is total energy so we can write the wave numbers in the two regions it's momentum over H bar in region one that would be the square root of 2m kinetic energy over H bar and in Region 2 is the same thing the square of 2m kinetic energy over H bar so now we have expressions for the wave numbers in each of the regions we have preliminary expressions for the wave functions with these coefficients a b and c let's proceed to find expressions for these coefficients in order to do that there are two conditions that have to be met by the wave function at the interface the first is continuity it has to have the same value just before the interface as it has just after the interface consequence of that then is that a plus B has to equal C go ahead and write out the two wave functions at x equals zero you realize you just have e to the ik 0 which is one and so ignore it and you have a plus b equals c we'll call that equation one the second condition is differentiability of the wave function at the interface now you might ask why does the wave function need to be differentiable at the interface if you look at the schroederer equation it has the second derivative of the wave function in it so if the wave function isn't differentiable whether it's the first derivative or the second derivative that means the derivative is infinite and the shorter equation is therefore undefined which would be unphysical so to have a solution to the Schrodinger equation you need to have defined derivative at the interface and so that requires differentiability across the interface so differentiate those two expressions and evaluate them at x equals zero and again you have these e to the ik zeros that you can ignore and you have a new expression which I'm going to rearrange quickly for C in terms of A and B put a box around it and call that equation two you can insert this expression for C into equation 1 above and you have a new expression that has only A and B in it rearrange this and solve it for B and you have expression for B I'll put a box around it and I'm going to insert that back into equation 2 that we just looked at but here it is as a reminder and you have a new expression for C in terms of a so we have an expression for B in terms of a and we have an expression for C in terms of a I will let you do the algebra between this line and the next to last line and convince yourself that's correct now you see we have expressions of the coefficients except that one of them still is unknown you have a b and c and we have these two equations you will never actually know all three coefficients without some initial condition let's not assume an initial condition that would allow us to solve for all three there are a lot of questions we can proceed to answer without knowing that initial condition let's find the probability current what is that it's not electric current but it's very similar to it we'll even represent it with a capital J for current density it's momentum over Mass which is the velocity of the wave times the number of particles per unit volume now if you stare at that long enough you realize that that's velocity times number per unit volume is the number of particles per unit area per unit time and it really isn't number of particles because it's a probability wave but you can go ahead and cross that out it's actually the probability so J is the probability current density which is the same Principle as the electric current density and in your quantum mechanics textbook you've got an expression for current density in terms of the wave function so go ahead and evaluate this with those expressions for the wave functions beginning with region one I'll go ahead and just plug in the complex conjugate of the wave function expression that we're using and the derivative of the wave function that expression that we're using and so on I will let you do the algebraic simplification on it and you will come up with this in terms of A and B where K1 again is the wave number in region one now there are two terms here H bar K1 over M and H bar K 1 over M and then a squared B squared and they're subtracted from each other and remember what A and B really are a is the coefficient of the incident plane wave and B this is the coefficient of the reflected plane wave so these two terms are literally the probability current density for the incident wave and the probability current density associated with the reflected wave the algebra for Region 2 is a little simpler because you don't have two plane waves to contend with I'll let you do all of that and you get probability current density is H part K Over M times the c squared and that's transmitted probability current density one thing you notice when you look at these Expressions is that J the probability current density is a real number and that's very important it would make no sense to have a complex probability to summarize the four equations then that we have the probability current density in the two regions and the coefficients B and C in terms of the coefficient a will keep a as our unknown it's the initial condition plug them in for region one you have an expression of the probability current density for region two you have an expression for the probability current density both of them in terms of this amplitude of the incoming wave a you know something interesting here they're the same when you put B in here and C into here and reduce the algebra you have the exact same expression probability current is conserved that is just the same on either side of that interface if the wave hits an interface some of it reflects some of it transmits let's find the percent that reflects the percent that transmits begin with the reflection coefficient which is the amount of reflected probability current divided by the amount of incident current density that's an intuitive statement on what reflection should be put in those expressions that we have for J reflected that's the second term of that region one current that's the divide by J instant that's the first term in that region one current density you can cancel a lot of things and all you have is B array so this B over a squared is the percentage of reflected probability refer to the expression that we have for B in terms of a and you have a more useful expression and I would remind you that these wave numbers K1 and K2 are written in terms of the potential energies in the two regions and the total energy of the particle so that's the reflection coefficient at the interface transmission as well can be found from again an intuitive statement that the transmission should be the percentage of probability as transmitted transmitted probability current density divided by the incident probability current density refer again to those expressions for J in Region 2 and the part of J in region one that corresponds to the incident current h-bars and M's cancel but you're left with a ratio of K's and a ratio of the coefficients one thing I point out is it's not just C over a that's a common mistake made with these kinds of problems is that the transmission coefficient is the ratio of probability coefficients but you need to account for the change in impedance to the wave in the two regions so it's not just C over a squared now we have that expression for C in terms of a use it and you have a transmission coefficient is this I'll let you do the algebra and again the wave numbers K1 and K2 are up above for reference this tests out fairly well you should be able to add the transmission coefficient and the reflection coefficient together and get one because the incident wave is divided into two parts the part that's reflected and the part that's transmitted the probability of being reflected plus the problem really being transmitted should add up to one add those two expressions together expand it out and you'll quickly realize it's one that's exactly what it must be so there we have the transmission and reflection coefficients for a potential barrier you could write it in terms of e energy and V zero potential energy that's just extra messy but you could do that with these Expressions up here I won't ask you to do that in the homework problem but it's not really a problem it's more like an exercise because I just walked you through the whole solution now all you have to do is make one little change have that incident wave come from the right from the positive region instead of the left from the negative region and come up with expression for the reflection and transmission coefficients okay well after this we'll be able to work on finite potential well |
4744 | https://www.hopkinsmedicine.org/health/treatment-tests-and-therapies/appendectomy | Skip to Main Content
Health
Appendectomy
What is an appendectomy?
An appendectomy is surgery to remove the appendix when it is infected. This condition is called appendicitis. Appendectomy is a common emergency surgery that cures appendicitis.
The appendix is a thin pouch that is attached to the large intestine. It sits in the lower right part of your belly. If you have appendicitis, your appendix is usually removed right away. If not treated, your appendix can burst (rupture). This is a medical emergency.
There are two types of surgery to remove the appendix. The standard method is an open appendectomy. A less invasive method is a laparoscopic appendectomy.
Open appendectomy.A cut or incision about 2 to 4 inches long is made in the lower right-hand side of your belly or abdomen. The appendix is taken out through the incision.
Laparoscopic appendectomy.This method is less invasive. That means it’s done without a large incision. Instead, 1 to 3 tiny cuts are made. A long, thin tube called a laparoscope is put into one of the incisions. It has a tiny video camera and surgical tools. The surgeon looks at a TV monitor to see inside your abdomen and guide the tools. The appendix is removed through one of the incisions.
During a laparoscopic surgery, your surgeon may decide that an open appendectomy is needed.
If your appendix has burst and infection has spread, you may need an open appendectomy. If an area of infection called an abscess has formed around the appendix, the surgeon may use antibiotics and drain the abscess before the appendectomy.
A laparoscopic appendectomy may cause less pain and scarring than an open appendectomy. For either type of surgery, the scar is often hard to see once it has healed.
Both types of surgery have low risk for complications. A laparoscopic appendectomy has a shorter hospital stay, shorter recovery time, and lower infection rates. Some studies suggest that intravenous antibiotics alone could treat appendicitis without the need for surgery. But appendectomy remains the standard of care since antibiotics alone do not always cure appendicitis.
Why might I need an appendectomy?
You may need an appendectomy to remove your appendix if you show symptoms of appendicitis.
Appendicitis is a medical emergency. It is when your appendix becomes sore, swollen, and infected.
If you have appendicitis, there is a serious risk your appendix may burst or rupture. This can happen as soon as 48 to 72 hours after you have symptoms. It can cause a severe, life-threatening infection called peritonitis in your belly.
If you have appendicitis symptoms, get medical care right away.
What are the risks of an appendectomy?
All surgery has risks. The risks of an appendectomy include:
Bleeding
Wound infection
Infection, redness, swelling inside the belly (peritonitis) that can occur if the appendix bursts during surgery
Blocked bowels
Injury to nearby organs
You may have other risks that are unique to you. Be sure to discuss any concerns with your healthcare provider before surgery.
How do I get ready for an appendectomy?
Your healthcare provider will explain the surgery to you. Ask any questions you may have.
You will be asked to sign a consent form that gives your permission to do the surgery. Read the form carefully and ask questions if anything is not clear.
Your healthcare provider will ask questions about your past health. They will also give you a physical exam. This is to make sure you are in good health before you have surgery. You may also need blood tests and other diagnostic tests.
You must not eat or drink anything for 8 hours before the surgery. This often means no food or drink after midnight. Sometimes, your healthcare provider may decide that the urgency of having the surgery is more important than when you last ate or drank.
Make sure your healthcare provider has a list of all medicines (prescribed and over-the-counter) and all herbs, vitamins, and supplements that you are taking.
You may be given a medicine to help you relax (a sedative) before the surgery.
Tell your healthcare provider if you:
Are pregnant or think you may be pregnant
Are allergic to or sensitive to latex, medicines, tape, or anesthesia medicines (local and general)
Have a history of bleeding disorders or are taking any blood-thinning (anticoagulant) medicines, aspirin, or other medicines that affect blood clotting. You may have to stop taking these medicines before surgery.
Your healthcare provider may have other instructions for you based on your medical condition.
What happens during an appendectomy?
In most cases, an appendectomy is an emergency surgery and will require a hospital stay. You will have either an open appendectomy or a laparoscopic appendectomy. This will depend on your condition and your healthcare provider’s practices.
An appendectomy is done while you are given medicines to put you into a deep sleep (under general anesthesia).
Generally, the appendectomy follows this process:
You will be asked to remove any jewelry or other objects that might get in the way during surgery.
You will be asked to remove your clothing and will be given a gown to wear.
An IV (intravenous) line will be put in your arm or hand.
You will be placed on the operating table on your back.
If there is a lot of hair at the surgical site, it may be clipped off.
A tube will be put down your throat to help you breathe while you are under anesthesia. The anesthesiologist will check your heart rate, blood pressure, breathing, and blood oxygen level during the surgery.
Open appendectomy
A cut or incision will be made in the lower right part of your belly.
Your abdominal muscles will be separated and the abdominal area will be opened.
Your appendix will be tied off with stitches and removed.
If your appendix has burst or ruptured, your abdomen will be washed out with salt water (saline).
The lining of your abdomen and your abdominal muscles will be closed with stitches. A small tube may be put in the incision to drain out fluids.
Laparoscopic appendectomy
A tiny incision will be made for the tube (laparoscope). More cuts may be made so that other tools can be used during surgery.
Carbon dioxide gas will be used to swell up your abdomen so that your appendix and other organs can be easily seen.
The laparoscope will be put in and your appendix will be found.
Your appendix will be tied off with stitches and removed through an incision.
When the surgery is done, the laparoscope and tools will be removed. The carbon dioxide will be let out through the cuts. A small tube may be placed in the cut to drain out fluids.
At the end of either method
Your appendix will be sent to a lab to be tested.
Your cuts will be closed with stitches or surgical staples.
A sterile bandage or dressing will be used to cover the wounds.
What happens after an appendectomy?
In the hospital
After surgery, you will be taken to the recovery room. Your healthcare team will watch your vital signs, such as your heart rate and breathing. Your recovery will depend on the type of surgery that was done and the type of anesthesia you had. Once your blood pressure, pulse, and breathing are stable and you are awake and alert, you will be taken to your hospital room.
A laparoscopic appendectomy may be done on an outpatient basis. In this case, you may be discharged and sent home from the recovery room.
You will have pain medicine as needed. This may be by prescription or from a nurse. Or you may give it to yourself through a device connected to your IV line.
You may have a thin plastic tube that goes through your nose into your stomach. This is used to remove stomach fluids and air that you swallow. The tube will be taken out when your bowels are working normally. You will not be able to eat or drink until the tube is removed.
You will be asked to get out of bed a few hours after a laparoscopic surgery or by the next day after an open surgery.
You may be allowed to drink liquids a few hours after surgery. You may slowly be able to add more solid foods.
You will schedule a follow-up visit with your healthcare provider. This is often 2 to 3 weeks after surgery.
At home
When you are home, you must keep the incision clean and dry. Your healthcare provider will give you instructions on how to bathe. Any stitches or surgical staples used will be removed at a follow-up office visit. If adhesive strips were used, they should be kept dry. They will often fall off in a few days.
The incision and your abdominal muscles may ache, often after long periods of standing. Take pain medicine as recommended by your healthcare provider. Aspirin or other pain medicines may raise your risk of bleeding. Only take medicines that your healthcare provider has approved.
If you had a laparoscopy, you may feel pain from the carbon dioxide gas that is still in your belly. This pain may last for a few days. You should feel a bit better each day.
Your healthcare provider will likely want you to walk and move around a bit. But don't do any tiring activity. Your healthcare provider will tell you when you can return to work and your normal activities.
Call your healthcare provider if you have any of the following:
Fever or chills
Redness, swelling, bleeding, or other drainage from the incision site
More pain around the incision site
Vomiting
Loss of appetite or inability to eat or drink anything
Constant coughing, trouble breathing, or shortness of breath
Belly pain, cramping, or swelling
No bowel movement for 2 days or longer
Watery diarrhea for more than 3 days
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4745 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5074415/ | Corrections for the Combined Effects of Decay and Dead Time in Live-Timed Counting of Short-Lived Radionuclides - PMC
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Appl Radiat Isot
. Author manuscript; available in PMC: 2017 Mar 1.
Published in final edited form as: Appl Radiat Isot. 2015 Dec 2;109:335–340. doi: 10.1016/j.apradiso.2015.11.108
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Corrections for the Combined Effects of Decay and Dead Time in Live-Timed Counting of Short-Lived Radionuclides
R Fitzgerald
R Fitzgerald
1 National Institute of Standards and Technology, 100 Bureau Drive, Gaithersburg, MD 20899 USA
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1 National Institute of Standards and Technology, 100 Bureau Drive, Gaithersburg, MD 20899 USA
1
Corresponding author. Tel.: 1 301 975 5579; ryan.fitzgerald@nist.gov
Issue date 2016 Mar.
PMC Copyright notice
PMCID: PMC5074415 NIHMSID: NIHMS819303 PMID: 26682893
The publisher's version of this article is available at Appl Radiat Isot
Abstract
Studies and calibrations of short-lived radionuclides, for example 15 O, are of particular interest in nuclear medicine. Yet counting experiments on such species are vulnerable to an error due to the combined effect of decay and dead time. Separate decay corrections and dead-time corrections do not account for this issue. Usually counting data are decay-corrected to the start time of the count period, or else instead of correcting the count rate, the mid-time of the measurement is used as the reference time. Correction factors are derived for both those methods, considering both extending and non-extending dead time. Series approximations are derived here and the accuracy of those approximations are discussed.
1. Introduction
Studies and calibrations of short-lived radionuclides are of particular interest in nuclear medicine. For instance, positron emission tomography (PET) imaging by 15 O, with a half-life of 122.46 s (Chisté and Bé, 2015), is the “gold standard” for assessment of blood flow in studies of tumor vascularity (Aboagye, et al., 2012).
Yet, counting measurements on short-lived nuclear states (either excited states or radionuclides) using live-timed counting systems are vulnerable to an error due to the combined effect of decay and dead time. The effect causes a bias in the apparent count rate due to the changing live-time fraction throughout the measurement. Since the count rate is decreasing during the measurement, the fractional live-time is increasing and the latter parts of the measurement contribute relatively more to the recorded count rate. If decay corrections do not account for this, then there remains a bias in the count rate at the reference time, which could be important when producing calibration sources or standards for short-lived radionuclides.
Counting experiments are subject to dead time, τ, which is the period of time following an event during which any part of the apparatus is unable to count another event. Moreover, a dead time is extending if events that are lost during that time cause further dead time, or non-extending if lost events do not cause dead time. A system could have mixed dead time, which is not considered further here, beyond noting that by designing a counting system with an imposed, extending dead-time that is longer than any intrinsic dead time of the system, one can approximate the pure extending dead-time case. The uncertainties from various dead-time, pileup, and live-time corrections are the subject of a recent review article (Pommé et al, 2015), which indeed cites the final result of this paper.
Axton and Ryves (1963) derived an approximate correction formula with leading-order expansion for the count rate at the start of the counting period for non-extending dead times. Müller (1981) derived expressions for non-extending and extending systems without live-timing. For modern live-timed systems, which may have imposed dead times, it is desirable to use the live-timed count rate determined by the instrument, rather than calculating the dead-time correction.
Derivations are presented for the actual and apparent live count rate for a decaying source over a finite measurement duration. Furthermore, correction factors are defined for both the decay-to-start method and midpoint approximation. These expressions are meant to enable a simple implementation of the corrections and aid in experimental design, thereby limiting errors caused by this combination dead-time and decay effect. Tabulated results are presented for various conditions and as well as simple approximations to the correction formulas.
Finally, the equivalence between the correction factors derived in this work and the approximations given in previous reports is demonstrated for certain cases.
2. Derivation
From the start, let us note that the background count rate, B, will be ignored in this work because it would hardly affect the final correction factors f m and f d that conclude this paper. This fact was verified by deriving the results with B included and noting the equivalence to the present results for extending dead time. For the non-extending case, the change in the correction factor due to B for initial net count rate due to the radionuclide, ρ(0), is of the order B/ρ(0). As the corrections derived here are needed only at high count rates, this small correction on the correction is not significant.
Consider a counting experiment with initial true, count rate at time t = 0 of ρ(0). If the measurement is live-timed with an extending or non-extending dead time, τ, then the observed average live-timed count rate, R̅, is the ratio of the total live counts, N L, to the total live time, T L. That is,
(1)
When ρ(t) is decreasing exponentially with decay constant, λ, as,
(2)
then one of two methods is typically used to report the experimental value for the instantaneous true count rate from the observed average rate. For a total count time, T tot, that is much lower than λ−1, the decay is nearly linear during the measurement. In that case, the midpoint method is sometimes employed. That is, R̅ is assumed to approximate the true rate at the mid-time,
(3)
A more accurate method is to average Equation (2) over the domain from t = 0 to t = T tot and then solve for ρ(0) in terms of the average rate, ρ̅, as
(4)
Ignoring the decay-dead-time effect in this case is equivalent to the assumption that,
(5)
We next calculate R̅ from ρ(t) and the resulting correction factors for the two assumptions.
For a counting system with a fixed, extending or non-extending dead time, τ, the instantaneous recorded count rate, R'(t), is given by the well-known formulae (ICRU, 1994):
(6a)
(6b)
For measurement duration T tot, the recorded live counts N L and live time T L can be calculated as:
(7)
(8a)
(8b)
The calculated value of R̅, called R̅ c, can now be written explicitly by combining equations 2, 6, 7 and 8 as:
(9a)
(9b)
Let us define x = λ T tot and k = ρ(0)τ as convenient parameters. The integrals in Equations 9 can be solved to be:
(10a)
(10b)
Here, E i(x) is the exponential integral of x, which can be evaluated numerically.
Correction factors f d and f m are now defined as the ratio of the R̅ values assumed by the decay-to-start time approximation in Equation (5) and mid-time approximation in Equation (3), respectively to the true R̅ c,
(11)
and,
(12)
In this way, the instantaneous rates determined by the usual simple methods can be corrected to give the true rates at t = 0 and as,
(13)
and
(14)
Exact forms of f d and f m are given by solving Equation (13) and Equation (14):
(15a)
(15b)
(16a)
(16b)
Series expansions, calculated to 4 th power of x, are given by the following:
(17a)
(18a)
(17b)
(18b)
Limiting the expansions to second order yields results that are accurate to about 5·10−3 for most cases of interest, as illustrated in Table 1, and should be particularly useful for experimental design, due to their simplicity.
Table 1.
Some calculated values for f d and f m (Equations 15a, 15b, 16a, and 16b) as a function of x and k. Numbers in parentheses are the errors in the last reported digits that would occur from using the series expansion to 4 th-order (bold) and 2 nd-order respectively.
| | | Extending | Non-Extending |
---: :---: |
| x | k | f d | f m | f d | f m |
| 0.5 | 0.5 | 1.0082 (1) (22) | 0.9978 (0) (22) | 1.0059 (0) (11) | 0.9955 (0) (12) |
| |
| 1.0 | 0.2 | 1.0104 (5) (63) | 0.9695 (3) (55) | 1.0092 (1) (47) | 0.9683 (1) (39) |
| |
| 0.5 | 1.0 | 1.0164 (1) (44) | 1.0059 (1) (45) | 1.0091 (0) (13) | 0.9988 (1) (12) |
| |
| 1.0 | 3.0 | 1.1561 (61) (940) | 1.1093 (102) (990) | 1.0546 (3) (79) | 1.0119 (1) (89) |
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3. Examples
For example, a scintillation detector system may have a fixed dead time of τ = 5.0·10−5 s. For a decaying source with a half-life of T 1/2 = 10 s, an initial true count rate of ρ(0) = 1·10 5 s−1, and a real count period of T tot = 5.0 s, we would have x = 0.347 and k = 5.0. Using Equations (15) and (16), then f d = 1.041 and f m = 1.036 for the extending case, and f d = 1.0081 and f m = 1.0031 for the non-extending case.
Table 1 lists some examples of f d and f m calculated using Equations (15) and (16) as well as the error from using either the 4 th-order and only the 2 nd-order expansions of Equations (17) and (18). Figure 1 and Figure 2 shows 3D plots created using the exact solutions to f d and f m,, respectively.
Figure 1.
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Plots of f d as a function of k and x for extending (left) and non-extending (right) dead-times.
Figure 2.
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Plots of f m as a function of k and x for extending (left) and non-extending (right) dead times.
3. Tabulations
Table 2 through Table 5 display values for f d and f m for pairs of x and k values. For reference, the measurement time, T tot, and true count rate, ρ(0), corresponding to x and k respectively in the case of 15 O with τ = 50 μs are listed.
Table 2.
Calculated values of f d, for the extending case using Equation (15a) as a function of x and k. Also listed are the corresponding values of the total measurement time T tot, in seconds, and of the real initial count rate of the source ρ 0, in s−1, for 15 O (λ=5.66 10−3 s−1) and τ = 5·10−5 s.
| k_x_ | Extending | 0.10 | 0.30 | 0.50 | 0.70 | 0.90 | 1.1 | 1.3 | 1.5 | 1.7 | 1.9 |
:---: ---: ---: ---: ---: ---: |
| | ρ 0_T_ tot | 18 | 53 | 88 | 123 | 159 | 194 | 229 | 265 | 300 | 335 |
| 0.2 | 4.0·10 3 | 1.000 | 1.001 | 1.003 | 1.006 | 1.009 | 1.012 | 1.015 | 1.019 | 1.022 | 1.026 |
| 0.4 | 8.0·10 3 | 1.000 | 1.003 | 1.007 | 1.012 | 1.018 | 1.024 | 1.031 | 1.038 | 1.045 | 1.052 |
| 0.6 | 1.2·10 4 | 1.001 | 1.004 | 1.010 | 1.018 | 1.027 | 1.036 | 1.047 | 1.057 | 1.068 | 1.078 |
| 0.8 | 1.6·10 4 | 1.001 | 1.005 | 1.013 | 1.024 | 1.036 | 1.049 | 1.063 | 1.077 | 1.091 | 1.105 |
| 1.0 | 2.0·10 4 | 1.001 | 1.007 | 1.016 | 1.029 | 1.044 | 1.061 | 1.078 | 1.096 | 1.114 | 1.132 |
| 1.2 | 2.4·10 4 | 1.001 | 1.008 | 1.020 | 1.035 | 1.053 | 1.073 | 1.094 | 1.116 | 1.138 | 1.159 |
| 1.4 | 2.8·10 4 | 1.001 | 1.009 | 1.023 | 1.041 | 1.062 | 1.086 | 1.110 | 1.136 | 1.161 | 1.187 |
| 1.6 | 3.2·10 4 | 1.001 | 1.010 | 1.026 | 1.047 | 1.071 | 1.098 | 1.126 | 1.155 | 1.185 | 1.214 |
| 1.8 | 3.6·10 4 | 1.001 | 1.012 | 1.030 | 1.053 | 1.080 | 1.110 | 1.142 | 1.175 | 1.208 | 1.241 |
| 2.0 | 4.0·10 4 | 1.002 | 1.013 | 1.033 | 1.059 | 1.089 | 1.122 | 1.158 | 1.194 | 1.231 | 1.269 |
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Table 5.
Calculated values of f m, for the non-extending case using Equation (16b) as a function of x and k. Also listed are the corresponding values of the total measurement time T tot, in seconds, and of the real initial count rate of the source ρ 0, in s−1, for 15 O (λ=5.66 10−3 s−1) and τ = 5·10−5 s.
| k_x_ | Non- Extending | 0.10 | 0.30 | 0.50 | 0.70 | 0.90 | 1.1 | 1.3 | 1.5 | 1.7 | 1.9 |
:---: ---: ---: ---: ---: ---: |
| | ρ 0_T_ tot | 18 | 53 | 88 | 123 | 159 | 194 | 229 | 265 | 300 | 335 |
| 0.2 | 4.0·10 3 | 1.000 | 0.997 | 0.992 | 0.985 | 0.975 | 0.962 | 0.946 | 0.928 | 0.907 | 0.884 |
| 0.4 | 8.0·10 3 | 1.000 | 0.998 | 0.995 | 0.989 | 0.981 | 0.970 | 0.956 | 0.940 | 0.922 | 0.901 |
| 0.6 | 1.2·10 4 | 1.000 | 0.999 | 0.996 | 0.992 | 0.985 | 0.976 | 0.965 | 0.951 | 0.935 | 0.916 |
| 0.8 | 1.6·10 4 | 1.000 | 0.999 | 0.998 | 0.994 | 0.989 | 0.982 | 0.973 | 0.960 | 0.946 | 0.929 |
| 1.0 | 2.0·10 4 | 1.000 | 1.000 | 0.999 | 0.997 | 0.993 | 0.987 | 0.979 | 0.969 | 0.956 | 0.940 |
| 1.2 | 2.4·10 4 | 1.000 | 1.000 | 1.000 | 0.998 | 0.996 | 0.991 | 0.985 | 0.976 | 0.964 | 0.950 |
| 1.4 | 2.8·10 4 | 1.000 | 1.000 | 1.000 | 1.000 | 0.998 | 0.995 | 0.990 | 0.982 | 0.972 | 0.960 |
| 1.6 | 3.2·10 4 | 1.000 | 1.001 | 1.001 | 1.001 | 1.000 | 0.998 | 0.994 | 0.988 | 0.979 | 0.968 |
| 1.8 | 3.6·10 4 | 1.000 | 1.001 | 1.002 | 1.002 | 1.002 | 1.001 | 0.998 | 0.993 | 0.986 | 0.976 |
| 2.0 | 4.0·10 4 | 1.000 | 1.001 | 1.002 | 1.003 | 1.004 | 1.004 | 1.002 | 0.998 | 0.991 | 0.983 |
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4. Comparisons to previous derivations
Michotte et al. (2007) have cited an internal report by B. Chauvenet giving an equation accounting for the decay-to-start-time correction for extending dead-time. His correction term (Michotte, 2010) is equivalent to our f d−1 written in the form,
(19)
Also, our exact equation for f d [Equation (17a)] for the extending case agrees with Baerg's term in square brackets in his Equation (10) (Baerg et al., 1976) to better than 2 % for x< 2 and k< 2 (f d = 1.29). Baerg’s equation can be expressed as,
(20)
Here, we have omitted his additional decay correction to an arbitrary reference time.
For non-extending dead time, Axton and Ryves (1963) derived their equation (4) for their error “k”, which is a measure of our f d−1.
(21)
where their ñ is the observed average count rate, without dead-time correction, that is, N L/T tot. To zeroeth order in x this is,
(22)
Therefore, Axton and Ryves’ formula would be equivalent to,
(23)
in agreement with the 2 nd –order expansion given in our Equation(17b).
5. Summary
Correction factors have been derived for the combined effect of decay and dead-time for live-timed systems with extending and non-extending dead times. The uncertainty in these correction factors can be determined from the uncertainty of the input parameters, k and x. In addition, if an approximate expression is used, then the uncertainty in the approximation can be included, as indicated for representative values in Table 1 or by examining the next-order term.
Two useful parameters have been defined:
x = λ T tot (radioactive decay constant times total real count time)
k = ρ(0)τ (initial true count rate times dead time)
The corrections (f d and f m) to live-timed count rates that have been already corrected for decay (f d) or assuming the mid-time approximation (f m), are given by 2 nd order expansions in x as follows:
| Extending | Non-Extending |
:---: |
| (24a) | (24b) |
| (25a) | (25b) |
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Table 3.
Calculated values of f m, for the extending case using Equation (16a) as a function of x and k. Also listed are the corresponding values of the total measurement time T tot, in seconds, and of the real initial count rate of the source ρ 0, in s−1, for 15 O (λ=5.66 10−3 s−1) and τ = 5·10−5 s.
| k_x_ | Extending | 0.10 | 0.30 | 0.50 | 0.70 | 0.90 | 1.1 | 1.3 | 1.5 | 1.7 | 1.9 |
:---: ---: ---: ---: ---: ---: |
| | ρ 0_T_ tot | 18 | 53 | 88 | 123 | 159 | 194 | 229 | 265 | 300 | 335 |
| 0.2 | 4.0·10 3 | 1.000 | 0.998 | 0.993 | 0.986 | 0.976 | 0.963 | 0.947 | 0.929 | 0.909 | 0.886 |
| 0.4 | 8.0·10 3 | 1.000 | 0.999 | 0.996 | 0.991 | 0.984 | 0.974 | 0.962 | 0.947 | 0.929 | 0.909 |
| 0.6 | 1.2·10 4 | 1.000 | 1.000 | 0.999 | 0.997 | 0.993 | 0.986 | 0.977 | 0.964 | 0.949 | 0.932 |
| 0.8 | 1.6·10 4 | 1.000 | 1.001 | 1.003 | 1.003 | 1.001 | 0.998 | 0.991 | 0.982 | 0.970 | 0.955 |
| 1.0 | 2.0·10 4 | 1.000 | 1.003 | 1.006 | 1.009 | 1.010 | 1.009 | 1.006 | 1.000 | 0.991 | 0.978 |
| 1.2 | 2.4·10 4 | 1.001 | 1.004 | 1.009 | 1.014 | 1.019 | 1.021 | 1.021 | 1.018 | 1.011 | 1.002 |
| 1.4 | 2.8·10 4 | 1.001 | 1.005 | 1.012 | 1.020 | 1.027 | 1.033 | 1.036 | 1.036 | 1.032 | 1.025 |
| 1.6 | 3.2·10 4 | 1.001 | 1.007 | 1.016 | 1.026 | 1.036 | 1.044 | 1.050 | 1.054 | 1.053 | 1.049 |
| 1.8 | 3.6·10 4 | 1.001 | 1.008 | 1.019 | 1.032 | 1.045 | 1.056 | 1.065 | 1.071 | 1.074 | 1.073 |
| 2.0 | 4.0·10 4 | 1.001 | 1.009 | 1.022 | 1.037 | 1.053 | 1.068 | 1.080 | 1.089 | 1.095 | 1.096 |
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Table 4.
Calculated values of f d, for the non-extending case using Equation (15b) as a function of x and k. Also listed are the corresponding values of the total measurement time T tot, in seconds, and of the real initial count rate of the source ρ 0, in s−1, for 15 O (λ=5.66 10−3 s−1) and τ = 5·10−5 s.
| k_x_ | Non- Extending | 0.10 | 0.30 | 0.50 | 0.70 | 0.90 | 1.1 | 1.3 | 1.5 | 1.7 | 1.9 |
:---: ---: ---: ---: ---: ---: |
| | ρ 0_T_ tot | 18 | 53 | 88 | 123 | 159 | 194 | 229 | 265 | 300 | 335 |
| 0.2 | 4.0·10 3 | 1.000 | 1.001 | 1.003 | 1.005 | 1.008 | 1.011 | 1.014 | 1.017 | 1.020 | 1.023 |
| 0.4 | 8.0·10 3 | 1.000 | 1.002 | 1.005 | 1.009 | 1.014 | 1.019 | 1.025 | 1.031 | 1.037 | 1.043 |
| 0.6 | 1.2·10 4 | 1.000 | 1.003 | 1.007 | 1.012 | 1.019 | 1.026 | 1.034 | 1.043 | 1.051 | 1.060 |
| 0.8 | 1.6·10 4 | 1.000 | 1.003 | 1.008 | 1.015 | 1.023 | 1.032 | 1.042 | 1.053 | 1.064 | 1.075 |
| 1.0 | 2.0·10 4 | 1.000 | 1.004 | 1.009 | 1.017 | 1.027 | 1.038 | 1.049 | 1.062 | 1.075 | 1.088 |
| 1.2 | 2.4·10 4 | 1.000 | 1.004 | 1.010 | 1.019 | 1.030 | 1.042 | 1.055 | 1.070 | 1.085 | 1.100 |
| 1.4 | 2.8·10 4 | 1.001 | 1.004 | 1.011 | 1.020 | 1.032 | 1.046 | 1.061 | 1.077 | 1.094 | 1.111 |
| 1.6 | 3.2·10 4 | 1.001 | 1.004 | 1.012 | 1.022 | 1.034 | 1.049 | 1.065 | 1.083 | 1.101 | 1.120 |
| 1.8 | 3.6·10 4 | 1.001 | 1.005 | 1.012 | 1.023 | 1.036 | 1.052 | 1.070 | 1.089 | 1.109 | 1.129 |
| 2.0 | 4.0·10 4 | 1.001 | 1.005 | 1.013 | 1.024 | 1.038 | 1.055 | 1.074 | 1.094 | 1.115 | 1.137 |
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Acknowledgments
The author is grateful to Dr. Carine Michotte of the BIPM who, as chair of the CCRI(II) Transfer Instrument working group, encouraged this work and provided valuable feedback on an early draft of the manuscript.
References
Aboagye EO, Gilbert FJ, Fleming IN, et al. Recommendations for measurement of tumour vascularity with positron emission tomography in early phase clinical trials. Eur. Radio. 2012;22:1465–1478. doi: 10.1007/s00330-011-2311-3. [DOI] [PubMed] [Google Scholar]
Axton EJ, Ryves TB. Dead-time corrections in the measurement of short-lived radionuclides. Int. J. Appl. Radiat. Isot. 1963;14:159–161. doi: 10.1016/0020-708x(63)90111-x. [DOI] [PubMed] [Google Scholar]
Baerg AP, Munzenmayer K, Bowes GC. Live-timed anti-coincidence counting with extending Dead-time circuitry. Metrologia. 1976;12:77–80. [Google Scholar]
Chisté V, Bé MM. Decay Data Evaluation Project; 2015. [accessed April 2015]. Table de Radionucléides 15O. [Google Scholar]
ICRU. Particle Counting in Radioactivity Measurements, ICRU Report, 52. Bathesda, MD: International Commission on Radiation Units and Measurements; 1994. pp. 56–57. [Google Scholar]
Michotte C. Report of the CCRI(II) Transfer Instrument Working Group 2005–2007. CCRI(II)/07-14. 2007 [Google Scholar]
Michotte C. Bureau International des Poids et Mesures. Sèvres, personal communication. 2010 [Google Scholar]
Müller JW. Counting statistics of short-lived nuclides. Journal of Radioanalytical Chemistry. 1981;61:345–359. [Google Scholar]
Pommé S, Fitzgerald R, Keightley J. Uncertainty of nuclear counting. Metrologia. 2015;52:S3–S17. [Google Scholar]
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1. Introduction
2. Derivation
3. Examples
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4746 | https://math.stackexchange.com/questions/34040/proof-and-purpose-of-fermats-little-theorem | Skip to main content
Proof and purpose of Fermat's Little Theorem
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What is the purpose of using Fermat's Little Theorem, and a proof of it?
group-theory
number-theory
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edited May 16, 2024 at 6:34
Bill Dubuque
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asked Apr 20, 2011 at 10:21
marymary
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Did you try en.wikipedia.org/wiki/Fermat's_little_theorem ?
– lhf
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Apr 20, 2011 at 10:37
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There are a full 6 proofs on wikipedia.
– davidlowryduda
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What do you mean by "purpose?"
– Thomas Andrews
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Below is what Weil wrote on proofs of Fermat's little theorem, from p. 56 of his Number theory: An approach through history. As for its purpose, it helps serve to reveal the multiplicative structure of Z/p, and problems in Z are often best solved by reducing them to much simpler problems in its finite images Z/p, e.g. so-called "local-global" principles. No doubt a search on "Fermat little" will uncover many applications - both here and elsewhere.
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edited Apr 20, 2011 at 19:00
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answered Apr 20, 2011 at 15:04
Bill DubuqueBill Dubuque
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where could i find Leibniz's proof using multinomial theorem ?? can someone summarize thanks
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In modular arithmetic e.g. mod m you can reduce big numbers down to numbers between 0 and m.
Fermat's little theorem lets you reduce the exponents (of numbers of the form be) down to numbers between 0 and φ(m) (without knowing anything about the base).
For example of the first, x=0,1,2 or 3 mod 4 so the number x2 must be equal to 0 or 1 mod 4. So x2 is of the form 4k or 4k+1.
For an example of the second, if p is prime φ(p)=p−1 then p2=1 mod p-1 so x(p2)=x mod p. I just made that up but you can probably see from it that you can use this for lots of things.
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edited Apr 20, 2011 at 10:40
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Surprisingly, Euler's extension of Fermat's Little Theorem plays a crucial role in modern cryptography.
– lhf
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Apr 20, 2011 at 10:38
@lhf: Actually Fermat's little theorem is enough for RSA, and Euler's theorem is not needed. The reason is that given distinct primes p,q and jk=a(p−1)(q−1)+1 we have xjk≡(xa(p−1))q−1⋅x≡x(modq) (using Fermat's little theorem if q∤x), and by symmetry xjk≡x(modp), and hence p,q∤xjk−x, and the proof is finished using Euclid's lemma. =)
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Fermat's little theorem states that, for a prime p:
ap≡a(modp)
alternatively:
ap−1≡1(modp)
The easiest proof that I've seen is as follows: Consider any integer, g, that is relatively prime to p. Let x be the product of all of the (distinct) integers (modp). i.e.:
x=1⋅2⋅3…(p−1)(modp)
Now, multiply each term on the right by g:
gp−1x=(g⋅1)(g⋅2)(g⋅3)…(g⋅(p−1))(modp)
Since multiplication is 1-1 and onto, all this does is permute the values involved in the product on the right to give back x. So:
gp−1x=x(modp)
gp−1=1(modp)
One needs to prove multiplication is 1-1 and onto (for p prime), but once that's done, the proof above is (in my opinion) straight forward and simple.
EDIT: I originally restricted the product, x, to be the product of powers of a generator, g. As lhf points out, this can be any number. I've changed the above to reflect this.
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edited Apr 21, 2011 at 1:12
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No need to use a generator. Consider the product of ax for x=1,…,p−1.
– lhf
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Apr 20, 2011 at 11:57
Wow, that's not "easy" because knowing that there is a generator for the multiplicative group is non-trivial, isn't it? And if you are gonna jump to group theory, you know that g|G|=1 for all g∈G for any group G, so you don't really need a generator.
– Thomas Andrews
Commented
Apr 20, 2011 at 19:26
@lhf, you are absolutely right. @Thomas, the above proof without the generator condition is now much simpler.
– user4143
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Apr 21, 2011 at 1:13
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We prove by fermat's little theorem by induction. Firstly, the base case tells us that
1p≡1 (mod p), which is true.
Assume true for a=k, namely that we assume kp≡k (mod p).
Then for a=k+1, we observe that (k+1)p−(kp+1)=pk + (p2) k2 + … kp−1 (pp−1).
I leave it as an exercise to check that the R.H.S. is divisible by p, and so we have the chain of congruences (using the induction hypothesis) that (k+1)p≡kp+1≡k+1 (mod p). So the statement of fermat's little theorem is true for a=1, a=k and a=k+1 and hence is true for all natural numbers k.
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Newton's Sums
Newton sums give us a clever and efficient way of finding the sums of roots of a polynomial raised to a power. They can also be used to derive several factoringidentities.
Contents
[hide]
1 Statement
2 Proof
3 Example
4 Practice
5 See Also
Statement
Consider a polynomial of degree ,
Let have roots . Define the sum:
Newton's sums tell us that,
(Define for .)
We also can write:
where denotes the -th elementary symmetric sum.
Proof
Let be the roots of a given polynomial . Then, we have that
Thus,
Multiplying each equation by , respectively,
Sum,
Therefore,
Note (Warning!): This technically only proves the statements for the cases where . For the cases where , an argument based on analyzing individual monomials in the expansion can be used (see for example.)
Example
For a more concrete example, consider the polynomial . Let the roots of be and . Find and .
Newton's Sums tell us that:
Solving, first for , and then for the other variables, yields,
Which gives us our desired solutions, and .
Practice
2019 AMC 12A Problem 17
2003 AIME II Problem 9
2008 AIME II Problem 7
2024 AIME II Problem 13
See Also
Vieta's Formulas
Newton's Inequality
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Vieta's Formulas
In algebra, Vieta's formulas are a set of results that relate the coefficients of a polynomial to its roots. In particular, it states that the elementary symmetric polynomials of its roots can be easily expressed as a ratio between two of the polynomial's coefficients.
It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in many math contests/tournaments.
Contents
1 Statement
2 Proof
3 Problems
3.1 Introductory
3.2 Intermediate
4 Advanced
5 See also
Statement
Let be any polynomial with complex coefficients with roots , and let be the elementary symmetric polynomial of the roots.
Vieta’s formulas then state that This can be compactly summarized as for some such that .
Proof
Let all terms be defined as above. By the factor theorem, . We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients.
When expanding the factorization of , each term is generated by a series of choices of whether to include or the negative root from every factor . Consider all the expanded terms of the polynomial with degree ; they are formed by multiplying a choice of negative roots, making the remaining choices in the product , and finally multiplying by the constant .
Note that adding together every multiplied choice of negative roots yields . Thus, when we expand , the coefficient of is equal to . However, we defined the coefficient of to be . Thus, , or , which completes the proof.
Problems
Here are some problems with solutions that utilize Vieta's quadratic formulas:
Introductory
2005 AMC 12B Problem 12
2007 AMC 12A Problem 21
2010 AMC 10A Problem 21
2003 AMC 10A Problem 18
2021 AMC 12A Problem 12
Intermediate
2017 AMC 12A Problem 23
2003 AIME II Problem 9
2008 AIME II Problem 7
2021 Fall AMC 12A Problem 23
2019 AIME I Problem 10
Advanced
2020 AIME I Problem 14
See also
Polynomial
Retrieved from "
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Theorems
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4749 | https://math.stackexchange.com/questions/1804628/calculate-minimal-variance | probability theory - Calculate minimal Variance - Mathematics Stack Exchange
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Calculate minimal Variance
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My task is to calculate the minimal variance. I got a result, but don't know for sure if it's correct. Maybe some of you could help me out here.
Let X X be some real-valued random variable. We know for sure that P(X=10)=P(X=20)=1 3 P(X=10)=P(X=20)=1 3. Calculate the minimal variance.
So i did the following:
E(X 2)=⋯+⋯+10 2⋅1 3+⋯+20 2⋅1 3+⋯≥10 2⋅1 3+20 2⋅1 3≈166.67 E(X 2)=⋯+⋯+10 2⋅1 3+⋯+20 2⋅1 3+⋯≥10 2⋅1 3+20 2⋅1 3≈166.67 and
E(X)2=(⋯+⋯+10⋅1 3+⋯+20⋅1 3+…)2≥(10⋅1 3+20⋅1 3)2=100 E(X)2=(⋯+⋯+10⋅1 3+⋯+20⋅1 3+…)2≥(10⋅1 3+20⋅1 3)2=100, therefore by
V a r(X)=E(X 2)−E(X)2≥166.67−100=66.67 V a r(X)=E(X 2)−E(X)2≥166.67−100=66.67
It this how it's done? I do really want to understand the concept.
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asked May 29, 2016 at 14:55
nobodynobody
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Easiest way to reduce variance is to put everything you can right at the mean. Here, for example, suppose P(X=15)=1 3 P(X=15)=1 3. Variance in that case is a lot lower than 66.67 66.67!lulu –lulu 2016-05-29 15:01:34 +00:00 Commented May 29, 2016 at 15:01
Also, note that A≥Y A≥Y and B≥Z B≥Z does not imply that A−B≥Y−Z A−B≥Y−Z.lulu –lulu 2016-05-29 15:03:10 +00:00 Commented May 29, 2016 at 15:03
Note that if Pr(X=15)=1 3 Pr(X=15)=1 3 then the variance is 50 3 50 3.André Nicolas –André Nicolas 2016-05-29 15:09:53 +00:00 Commented May 29, 2016 at 15:09
Hi @lulu, thanks i do see now that the last step is not true. (i.e. 2≥1∧4≥2 2≥1∧4≥2 does not imply −2=2−4≥1−2=−1−2=2−4≥1−2=−1) I'm really unfamiliar with this topic, how would one "put everything at the mean"? Any hint will help.nobody –nobody 2016-05-29 15:13:06 +00:00 Commented May 29, 2016 at 15:13
I gave you the actual distribution! Just take P(X=15)=1 3 P(X=15)=1 3. Then that term contributes 0 0 to the variance, and a quick computation shows that (A−μ)2+(B−μ)2(A−μ)2+(B−μ)2 is minimized when μ=A+B 2 μ=A+B 2.lulu –lulu 2016-05-29 15:19:57 +00:00 Commented May 29, 2016 at 15:19
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Geometrically, the idea is to simply put the other values of X X at the mean (15 15). In this case, we simply declare that P(X=15)=1 3 P(X=15)=1 3. Thus our candidate distribution has X=10,15,20 X=10,15,20 each with probability 1 3 1 3. To see that this gives the minimum variance:
First note: for general A,B A,B the expression (A−μ)2+(B−μ)2(A−μ)2+(B−μ)2 is minimized when μ=A+B 2 μ=A+B 2, so μ μ is the average of A,B A,B. This follows at once by differentiating the expression in μ μ.
Thus, if your distribution has any mean other that 15 15 we see that it's variance must exceed the variance of our candidate.
Likewise, if A A is any value other than 10,15,20 10,15,20 with P(X=A)>0 P(X=A)>0 the term (A−15)2(A−15)2 will contribute to the variance, and hence this distribution will have greater variance than ours. A similar argument for continuous distributions shows that if we have [a,b][a,b] containing none of 10,15,20 10,15,20 then P(X∈[a,b])=0 P(X∈[a,b])=0 and we are done.
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answered May 29, 2016 at 16:11
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4750 | http://asuonline-dev.asu.edu/math/142/pdf/sets_and_counting.pdf | MAT 142 College Mathematics Sets and Counting Terri L. Miller & Elizabeth E. K. Jones Contents Basic Sets Concepts 2 Set Operations 5 Venn Diagrams 11 Fundamental Counting Principle 17 Permutations and Combinations 19 c ⃝2020 ASU School of Mathematical & Statistical Sciences and Terri L. Miller & Elizabeth E. K. Jones 1 2 Sets and Counting Basic Sets Concepts What is a set?
A set is a collection of objects. The objects in the set are called elements of the set. A set is well-defined if there is a way to determine if an object belongs to the set or not. To indicate that we are considering a set, the objects (or the description) are put inside a pair of set braces, {}.
Example 1. Are the following sets well-defined?
(1) The set of all groups of size three that can be selected from the members of this class.
(2) The set of all books written by John Grisham.
(3) The set of great rap artists.
(4) The best fruits.
(5) The 10 top-selling recording artists of 2007.
Solution: (1) You can determine if a group has three people and whether or not those people are members of this class so this is well-defined.
(2) You can determine whether a book was written by John Grisham or not so this is also a well-defined set.
(3) A rap artist being great is a matter of opinion so there is no way to tell if a particular rap artist is in this collection, this is not well-defined.
(4) Similar to the previous set, best is an opinion, so this set is not well-defined.
(5) This is well-defined, the top selling recording artists of any particular year are a matter of record.
Set Equality.
Two sets are equal if they contain exactly the same elements.
Example 2.
(1) {1, 3, 4, 5} is equal to the set {5, 1, 4, 3} (2) The set containing the letters of the word railed is equal to the set containing the letters of the word redial.
Equivalent Sets.
Two sets are euivalent if they contain exactly the same number of elements. The elements of equivalent sets do not have to be the same.
Example 3.
(1) {1, 3, 4, 5} is equivalent to the set {a, b, c, d} (2) The set containing the letters of the word ambidextrous is equivalent to the set containing the names of the months of the year.
MAT 142 - Module Sets and Counting 3 There are three basic ways to describe a set. The first is by giving a description, as we did in Example 1 and the second is by listing the elements as we did in Example 2 (1). This second method is called roster notation. A third way is called set-builder notation. In set-builder notation a set is described by stating the properties that its elements must satisfy.
An example of set-builder notation can be seen in Example 4 below with set C.
Notation:. We usually use an upper case letter to represent a set and a lower case x to represent a generic element of a set. The symbol ∈is used to replace the words “is an element of”; the expression x ∈A would be read as x is an element of A. If two sets are equal, we use the usual equal sign: A = B.
Example 4. A = {1, 2, 3, 5} B = {m, o, a, n} C = {x|x ≥3 and x ∈R} D = {persons | the person is a registered Democrat} U = {countries | the country is a member of the United Nations} Universal Set.
In order to work with sets we need to define a Universal Set, U, which contains all possible elements of any set we wish to consider. The Universal Set is often obvious from context but on occasion needs to be explicitly stated.
For example, if we are counting objects, the Universal Set would be whole numbers. If we are spelling words, the Universal Set would be letters of the alphabet. If we are considering students enrolled in ASU math classes this semester, the Universal Set could be all ASU students enrolled this semester or it could be all ASU students enrolled from 2000 to 2005.
In this last case, the Universal Set is not so obvious and should be clearly stated.
Empty Set.
On occasion it may turn out that a set has no elements, the set is empty. Such a set is called the empty set and the notation for the empty set is either the symbol ∅or a set of braces alone, {}.
Example 5. Suppose A is the set of all integers greater than 3 and less than −1. What are the elements of A? There are no numbers that meet this condition, so A = ∅.
Subset.
A is a subset of B if every element that is in A is also in B. The notation for A is a subset of B is A ⊆B. Note: A and B can be equal.
Example 6.
A = {0, 1, 2, 3, 4, 5} B = {1, 3, 4} C = {6, 4, 3, 1} D = {0, 1, 2, 5, 3, 4} E = {} Which of the sets B, C, D, E are subsets of A?
4 Sets and Counting B ⊆A since it’s elements 1, 3, and 4 are all also in A. C is NOT a subset of A (C ̸⊆A) since there is a 6 in C and there is no 6 in A. D is a subset of A since everything that is in D is also in A; in fact D = A. Finally, E is a subset of A; this is true since any element that IS in E is also in A.
Notice that every set is a subset of itself and the empty set is a subset of every set. If A ⊆B and A ̸= B, then we say that A is a proper subset of B. The notation is only a bit different: A ⊂B. Note the lack of the “equal” part of the symbol.
Size of a set.
The cardinality of a set is the number of elements contained in the set and is denoted n(A).
Example 7. If A = {egg, milk, flour, sugar, butter}, then n(A) = 5. Note, the empty set, {}, has no elements, so n({}) = 0.
MAT 142 - Module Sets and Counting 5 Set Operations We will need to be able to do some basic operations with sets.
Set Union ∪.
The first operation we will consider is called the union of sets. This is the set that we get when we combine the elements of two sets. The union of two sets, A and B is the set containing all elements of both A and B; the notation for A union B is A ∪B. So if x is an element of A or of B or of both, then x is an element of A ∪B.
Example 8. For the sets A = {bear, camel, horse, dog, cat} and B = {lion, elephant, horse, dog}, we would get A ∪B = {bear, camel, horse, dog, cat, lion, elephant}.
Set Intersection ∩.
The next operation that we will consider is called the intersection of sets. This is the set that we get when we look at elements that the two sets have in common. The intersection of two sets, A and B is the set containing all elements that are in both A and B; the notation for A intersect B is A ∩B. So, if x is an element of A and x is an element of B, then x is an element of A ∩B.
Example 9. For the sets A = {bear, camel, horse, dog, cat} and B = {lion, elephant, horse, dog}, we would get A ∩B = {horse, dog}.
Set Complement.
Every set is a subset of some universal set. If A ⊆U then the complement of A is the set of all elements in U that are NOT in A. This is denoted: A′. Note that (A′)′ = A, i.e. the complement of the complement is the original set.
Example 10. Consider the same sets as in Example 6. It appears that the set of all integers, U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, would be a natural choice for the universe in this case. So, we would have A′ = {6, 7, 8, 9} B′ = {0, 2, 5, 6, 7, 8, 9} C′ = {0, 2, 5, 7, 8, 9} E′ = U Set Difference.
For sets A and B, the difference between sets A and B (notation A −B) is the set of elements in A that are not in B.
Example 11. Consider the sets A = {0, 1, 2, 3, 4, 5} and B = {6, 4, 3, 1}. A −B = {0, 2, 5}.
6 Sets and Counting Venn Diagrams. Pictures are your friends! It is often easier to understand relationships if we have something visual. For sets we use Venn diagrams. A Venn diagram is a drawing in which there is a rectangle to represent the universe and closed figures (usually circles) inside the rectangle to represent sets.
U A 1 One way to use the diagram is to place the elements in the diagram.
To do this, we write/draw those items that are in the set inside of the circle. Those items in the universe that are not in the set go outside of the circle.
Example 12. With the sets U = {red, orange, yellow, green, blue, indigo, violet} and A = {red, yellow, blue}, the diagram looks like U A red, yellow, blue indigo, violet orange, green Notice that you can see A′ as well. Everything in the universe not in A, A′={orange, indigo, violet, green}.
If there are more sets, there are more circles, some pictures with more sets follow.
MAT 142 - Module Sets and Counting 7 U A B U A B U A B U A B C 1 To see set unions using a Venn diagram, we would give each set a color. Then A ∪B would be anything in the diagram with any color.
Example 13. If we color the set A with blue and the set B with yellow, we see the set A ∪B as the parts of the diagram that have any color (blue, gray, yellow).
A B U To see set intersections using a Venn diagram, we would give each set a color. Then A ∩B would be anything in the diagram with both colors.
Example 14. If we color the set A with blue and the set B with yellow, we see the set A ∩B as the part of the diagram that has both blue and yellow resulting in a gray colored “football” shape.
8 Sets and Counting A B U Note: A ∩A′ = {}, the intersection of a set with its complement is the empty set.
We can extend these ideas to more than two sets. With three sets, the Venn diagram would look like In this diagram, the three sets create several “pieces” when they intersect. I have given each piece a lower case letter while the three sets are labelled with the upper case letters A, B, and C. I will describe each of the pieces in terms of the sets A, B, and C.
With three sets, the keys to completing the venn diagrams are the “triangle” pieces, t, v, w, and x. The darkest blue piece in the center, w, is the intersection of all three sets, so it is A ∩B ∩C; that is the elements in common to all three sets, A and B and C. The yellow piece t is part of the intersection of 2 of the sets, it is the elements that are in both A and B but not in C, so it is A ∩B ∩C′. Similarly v, the purple piece, is the elements that are in MAT 142 - Module Sets and Counting 9 both A and C but not in B, A ∩C ∩B′. Finally, the blue piece x is the elements that are in both B and C but not in A, A′ ∩B ∩C.
The next pieces to consider are the “football” shaped pieces formed by joining two of the triangle pieces. The piece composed of t and w is the elements in both A and B, so it is the set A ∩B. The “football” formed by v and w is the elements that are in both A and C, that is A ∩C. The last “football”, formed by x and w, is the elements in both B and C, i.e.
B ∩C.
The areas to consider are the large outer pieces of each circle. The red region marked with s is the elements that are in A but not in B or C, this is the set A ∩(B′ ∪C′). Similarly, the green region, u is the elements that are in B but not in A or C, i.e. B ∩(A′ ∪C′).
Lastly, the blue region marked with y is the elements that are in C but not in A or B, the set C ∩(A′ ∪B′).
The final region, z, outside of all the circles is the elements that are not in A nor in B nor in C. It is the set (A ∪B ∪C)′.
Properties. de Morgan’s Laws (A ∪B)′ = A′ ∩B′, and (A ∩B)′ = A′ ∪B′ I will illustrate the first law using Venn diagrams and leave the illustration of the other equation to the reader.
Consider the Venn diagram with two sets using the colors blue for A and yellow for B, from example 13, we know that A ∪B is the part containing any color.
A B U Figure 1. A ∪B So (A ∪B)′ is the white part of the current diagram that is outside of both circles. The diagram to illustrate this is given below, the set is colored in red.
Figure 1 and 2 demonstrate the left side of the equation (A ∪B)′ = A′ ∩B′. To demonstrate the right hand side of the equation, we will start with diagrams for A′ and B′.
Here we see that everything outside of A is blue and everything outside of B is green. If we lay one diagram over the other, any part containing both colors will be the intersection of those two sets, A′ ∩B′. The result will be Figure 2.
10 Sets and Counting Figure 2. (A ∪B)′ Figure 3.
A′ B′ MAT 142 - Module Sets and Counting 11 Venn Diagrams In this section, I will illustrate the use of Venn diagrams in some examples. Often we use the cardinality of a set for the Venn diagram rather than the actual objects.
Example 15. Two programs were broadcast on television at the same time; one was the Big Game and the other was Ice Stars. The Nelson Ratings Company uses boxes attached to television sets to determine what shows are actually being watched.
In its survey of 1000 homes at the midpoint of the broadcasts, their equipment showed that 153 households were watching both shows, 736 were watching the Big Game and 55 households were not watching either. How many households were watching only Ice Stars? What percentage of the households were watching only the Big Game? What percentage of the households were not watching either broadcast?
We begin by constructing a Venn diagram, we will use B for the Big Game and I for Ice Stars. Rather than entering the name of every household involved, we will put the cardinality of each set in its place within the diagram. So, since they told us that 153 households were watching both broadcasts, we know that n(B∩I) = 153, this number goes in the dark purple “football” area.
We are told that 736 were watching the Big Game, n(B) = 736, since we already have 153 in that part of B that is in common with I, the remaining part of B will have 736−153 = 583.
This tells us that 583 households were watching only the Big Game.
We are also told that 55 households were watching neither program, n((A ∪B)′) = 55, so that number goes outside of both circles.
12 Sets and Counting Finally, we know that the total of everything should be 1000, n(U) = 1000. Since only one area does not yet contain a number it must be the missing amount to add up to 1000. We add the three numbers that we have, 583 + 153 + 55 = 791, and subtract that total from 1000, 1000 −791 = 209, to get the number that were watching only Ice Stars. Filling in this number, we have a complete Venn diagram representing the survey.
Now, we have the information needed to answer any questions about the survey results. In particular, we were asked how many households were watching only Ice Stars, we found this number to be 209. We were also asked what percentage of the households were watching only the Big Game. The number watching only the game was found to be 583, so we compute the percentage, (583/1000) ∗100% = 58.3%. Finally, we were asked what percentage of households were not watching either broadcast. The number that we found were not watching either broadcast was 55, so we compute the percentage (55/1000) ∗100% = 5.5%.
Example 16. In a recent survey people were asked if the took a vacation in the summer, winter, or spring in the past year. The results were 73 took a vacation in the summer, 51 took a vacation in the winter, 27 took a vacation in the spring, and 2 had taken no vacation.
Also, 10 had taken vacations at all three times, 33 had taken both a summer and a winter MAT 142 - Module Sets and Counting 13 vacation, 18 had taken only a winter vacation, and 5 had taken both a summer and spring but not a winter vacation.
(1) How many people had been surveyed?
(2) How many people had taken vacations at exactly two times of the year?
(3) How many people had taken vacations during at most one time of the year?
(4) What percentage had taken vacations during both summer and winter but not spring?
To begin to answer these questions we will make a Venn diagram representing the infor-mation. Using S for summer, W for winter, and P for spring our diagram looks like the following.
We start by writing down all of the information given.
n(S) = 73 n(W) = 51 n(P) = 27 n((S ∪W ∪P)′) = 2 n(S ∩W ∩P) = 10 n(S ∩W) = 33 n(W ∩(S ∪P)′) = 18 n(S ∩P ∩W ′) = 5 14 Sets and Counting Look at this information to see if any of it can be entered into the diagram with no further work. It is best to start with the center if possible, and then the remainder of the “trianglar” pieces. Our information given, tells us that e = 10, h = 2. c = 18, and d = 5.
We know that the football shape that is cyan and white is supposed to have 33 people, since 10 are accounted for in the white portion, 33 −10 = 23 = b.
MAT 142 - Module Sets and Counting 15 We now have three of the four pieces that make up S and we know the total of all of the pieces of S, so we find the green piece, a, 73−(23+10+5) = 35 = a. We also know three of the four pieces that make up W, this gives us the purple piece, f, 51−(23+10+18) = 0 = f.
Now we have three of the four pieces of P and only need to find the red piece g, 27−(5+10) = 12 = g, to complete the Venn diagram and begin answering the questions.
16 Sets and Counting Now, to answer the questions (1) This is asking the size of our universal set, so we add all of the numbers in our diagram, 35 + 23 + 10 + 5 + 18 + 12 + 2 = 105, thus 105 people were surveyed.
(2) Those who have taken vacations exactly two times of the year would be the triangular pieces that are the intersection of only two of the sets, this is all of the triangular pieces except the white one in the center, 23 + 5 + 0 = 28, hence 28 people took vacations at exactly two times of the year.
(3) Those people who took vacations at most one time of the year either took a vacation during exactly one of the seasons, these are the green, blue, and red pieces, or had not taken a vacation at all, the number outside of all the circles. Adding these together, 35 + 18 + 12 + 2 = 67, we get that 67 people had taken at most one vacation.
(4) The number that took vacations during both summer and winter but not spring is the cyan section. Thus the percentage who took vacations only during those two seasons is (23/105) ∗100% = 21.905%.
MAT 142 - Module Sets and Counting 17 Fundamental Counting Principle There are various methods that can be used in counting. The first one is a pictorial method.
Tree Diagram. A tool that is very useful in counting is called a tree diagram. This is a picture that branches for each option in choice.
Example 17. If I toss a penny and a nickel, how many possible outcomes are there?
Using the symbols H for heads and T for tails, we get the following tree: H HH H 2 , T HT 5 ) H TH T 2 , T TT The first coin has a possible outcome of H or T, this is represented with the first branch. For each branch here, the second coin has the outcome of H or T, represented by the second set of branches. At the end of each branch, I have listed the result of starting at the beginning and following along the branches to the end.
There are 4 ends, so the total number of possible outcomes of tossing a penny and a nickel is 4.
A second method is making an organized list of all the possibilities. I will introduce this counting technique with an example.
Example 18. Suppose a cafeteria offers a $5 lunch special which includes one entree, a beverage, and a side. For the entree, you can choose to have either soup or a sandwich; the beverage choices are soda, lemonade, or milk; and the side choices are chips or cookies. How many different lunch combinations are there?
We will list the possibilities: soup, soda, chips soup, soda, cookies soup, lemonade, chips soup, lemonade, cookies soup, milk, chips soup, milk, cookies sandwich, soda, chips, sandwich, soda, cookies sandwich, lemonade, chips sandwich, lemonade, cookies sadwich, milk, chips sandwich, milk, cookies Notice that this task involved a sequence of choices. We had to make a sequence of three choices to complete the task. For each choice we had options. If we think of the task as a sequence of boxes to fill, we can set it up as: 18 Sets and Counting entree beverage side .
This leads us to a third method.
The Fundamental Counting Principle gives us a quicker way to count up the number of ways to complete the task. If a task requires a sequence of choices, then the number of ways to complete the task is to multiply together the number of options for each choice.
Example 19. In our previous example, we counted up the number of ways to make a lunch combination by listing them all out. With the Fundamental Counting Principle, we could have simply multiplied: 2 entree × 3 beverage × 2 side = 12.
Example 20. How many license plates can be made if each is to be three digits followed by 3 letters. The plate umber cannot begin with a 0.
We can think of this as a sequence of tasks and apply the Fundamental Counting Principle: 9 digit × 10 digit × 10 digit × 26 digit × 26 digit × 26 digit = 15, 818, 400.
MAT 142 - Module Sets and Counting 19 Permutations and Combinations Factorial. Before we can continue on to our next counting technique, we will need to learn a new idea and notation. The idea is called the factorial, it has the notation !. It is easiest to understand the idea by looking at the pattern.
0! = 1 1! = 1 2! = 2 · 1 3! = 3 · 2 · 1 4! = 4 · 3 · 2 · 1 The idea can be expressed in general as: n! = n(n −1)(n −2)(n −3) · · · 3 · 2 · 1.
We could stop the expansion process at any point and indicate the remainder of the factorial in terms of a lower factorial.
5! = 5 · 4!
= 5 · 4 · 3!
= 5 · 4 · 3 · 2!
= 5 · 4 · 3 · 2 · 1!
= 5 · 4 · 3 · 2 · 1 The ability to expand any factorial partially can be an aid in simplifying expressions involving more than one factorial.
Example 21. Compute, by expanding and simplifying, 12!
9! .
Solution: 12!
9! = 12 · 11 · 10 · 9!
9!
(by expanding) = 12 · 11 · 10 · 9!
9!
(cancelling like terms) = 12 · 11 · 10 = 1320 Next we will consider the difference in the following tasks and learn how to count them.
The first task is to take 3 books from a pile of 8 distinct books and line them up. The second task is to select 3 of the 8 distinct books.
20 Sets and Counting The first thing to notice is that we can distinguish each object from any other and that we cannot replace any book that has been used, this is called without replacement. Both of our tasks have this feature. Next, we note that the first task is to line up the books, thus order makes a difference; {math, art, english} would look different from {art, math, english}.
So for task one, ORDER MATTERS. In the second task, we simply choose a group of 3 with no arranging, so for task two, ORDER DOES NOT MATTER.
This leads us to the definitions for these situations. A permutation is an arrangement of objects. A combination is a collection of objects.
Next, we will learn how to count the number of permutations and combinations.
The number of permutations (arrangements) without replacement of r objects from a group of n distinct objects is denoted P(n, r) or nPr and is calculated with the formula: P(n, r) = n!
(n −r)!.
The number of combinations (groups) without replacement of r objects chosen from a group of n distinct objects is denoted C(n, r) or nCr and is calculated with the formula C(n, r) = n!
r!(n −r)!.
Example 22. Consider the set {a, b, 5}.
(1) How many permutations of 2 of the objects are possible?
Solution: The first solution is to simply list out the permutations: ab, a5, ba, b5, 5a, 5b to see that there are a total of 6. The second solution is to use the formula where r is 2 (the number of objects being arranged) and n is 3 (the number of objects from which we are selecting those to arrange). This gives the result: P(3, 2) = 3!
(3 −2)! = 3!
1! = 6 1 = 6.
(2) How many groups of two of the objects are there?
Solution: Again, our first solution will be to list the possibilities: {a, b}, {a, 5}, {b, 5} to see that there are only three ways to choose two of the objects. The second solution is to use the formula with r = 2 (the number of objects to be chosen) and n = 3 (the number of objects from which we are choosing). This gives the result: C(3, 2) = 3!
2!(3 −2)! = 3!
2! 1! = 6 2 · 1 = 6 2 = 3.
Example 23. A museum has 12 paintings. They have space to display 6 of them in the current exhibit. How many ways are there for the museum to choose and arrange the 6 paintings?
MAT 142 - Module Sets and Counting 21 Solution: Since we are asked how many ways there are to arrange the painting, then the order is important and we are dealing with a permutation. We want to arrange 6 of the 12 painting. Thus in the formula P(n, r) = n!
(n−r)!, n = 12 and r = 6. We then calculate P(12, 6) to find that there are 665,280 ways for the museum to arrange six of the twelve paintings.
Example 24. Katie is taking an exam containing eight questions. She is required to answer five of the questions. How many ways are there for Katie to choose the five questions?
Solution: Since it does not matter which order Katie answers the questions, this is a combi-nation problem. Thus in the formula C(n, r) = n!
r!(n−r)!, n = 8 and r = 5. We then calculate C(8, 5) to find that there are 56 ways for Katie to pick out the five questions to answer.
Example 25. An animal shelter has 30 dogs and 20 cats. They are going to be holding an adoption event where there will be room for 15 dogs and 10 cats. How many ways are there for the shelter to choose the dogs and the cats for the adoption event?
Solution: To solve this problem, we will need to use multiple counting methods. Since there are two types of animals that we will be choosing, the question is a fundamental counting principle question at its base. We will need to figure out how many ways there are to choose the 15 dogs and then how many ways there are to choose the 10 cats.
ways to choose dogs × ways to choose cats .
We now just need to figure out what the numbers are that go into each box. We will start with the ways to choose the dogs. Since the order that we choose the 15 dogs out of the 30 dogs does not matter, this is a combination questions. The same is true for choosing the cats. Thus we can fill in the boxes with the appropriate combinations for the dogs and the cats.
C(30, 15) ways to choose dogs × C(20, 10) ways to choose cats.
When we do the final calculation 155, 117, 520 ways to choose dogs × 184, 756 ways to choose cats = 28, 658, 892, 530, 000 ways to pick the animals for the event. |
4751 | https://script.byu.edu/latin-handwriting/tools/grammar/verbs | Script Tutorial
Making sense of old handwriting
Latin Verbs
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| General | Dictionaries | Nouns | Pronouns | Adjectives |
Verbs, from the Latin verbum 'word,' are the most basic part of speech, along with nouns. Verbs describe the action that is being completed and can be seen as the heart of a sentence.
Like many other parts of speech in Latin, verbs are highly inflected and do much more 'work' than verbs do in English and even the romance languages that derive from Latin. This means that there is a lot of information packed into a single word.In order to be able to organize all this information, Latin verbs are usually given in four principal parts. Each of the principal parts gives information on the nature of the verb and how to conjugate all the possible forms of the said verb. Thus a verb in the dictionary will look like this: amō, amāre, amāvī, amātum 'to love. The principal parts are as follows:
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| amō, | amāre, | amāvī, | amātum |
| The 1-person present, i.e., “I love.” | The infinitive, i.e., “to love.” Used to conjugate the present system. | The 1-person perfect, i.e., “I loved.” Used to conjugate the perfect system | The participle. Used to decline the participle system. |
The purpose of all these principal parts and inflections is to show different pieces of information or characteristics. like many languages, Latin verbs exhibit five characteristics:
Person: who is the subject doing the verb and is categorized based on the speaker's point of view: 1st person, the speaker (I, we), 2nd person, the one being spoken to (you, you all), 3rd person, the person being spoken about (he, she, it, they).
Number: how many people are doing the action, either singular or plural.
Tense: the time the action occurred. Most genealogical records are in the past or the perfect tense.
Mood: showed the way the action or mode of the verb. This includes the indicative, stating facts; the subjunctive, stating opinions, wishes, hypotheticals, or desires; and the imperative mood, orders, and commands.
Voice: shows if the subject performs the action, active voice, or if the subject receives the action, passive voice.
Verb Conjugations
All Latin verbs exhibit all five of these characteristics; furthermore, verbs are divided into different conjugations. The first two principal parts of a verb are useful for identifying what conjugation a verb belongs to. It is important to be aware of the type of conjugation a verb belongs to in order to avoid misinterpreting some of the characteristics of the said verb. There are five different conjugations that exist in Latin:
First Conjugation: are verbs like amō, amāre, amāvī, amātum, 'to love,' and are identifiable by the -āre ending in the infinitive. This conjugation is very regular throughout its principal parts.
Second Conjugation: are verbs like maneō, manēre, mānsī, mānsum, 'to remain,' and are identifiable by the -ēre ending in the infinitive. It is extremely important to recognize the penultimate long 'e' as this differentiates it from the third conjugation.
Third Conjugation: are verbs like dīcō, dīcere, dīxī, dictum, 'to say,' and are identifiable by the -ere ending in the infinitive. Notice that, unlike the second declension, the third conjugation has a short 'e.' It is the only verb conjugation to have a short penultimate vowel.
Fourth Conjugation: are verbs like audiō, audīre, audīvī, audītum, 'to hear,' and are identifiable by the -īre ending in the infinitive.
Irregulars: are verbs that do not fit in with the other conjugations. There are not many irregular verbs, but the ones that are are often very important such as sum, esse, fuī, futūrum 'to be.'
To anyone who has taken Spanish or another Romance language, these verb endings will look familiar. As Latin changed into Vulgar Latin and then into modern romance languages, the short 'e' in the third conjugation shrank into an 'i,' combined with the fourth conjugation, and resulted in the -ar, -er, -ir verbs that they are today.
Active Verbs
Most verbs in Latin are used in the Active Mood. This is where the subject performs the action; for example, this sentence, "the doctor heals the man," the verb to see hereis active because the subject, the doctor, is acting, healing in this case, on the object, the man. Within the active mood, Latin verbs conjugate to show different tenses to show the action through time. To show this, there are two systems: the present system, which uses the first principal part, and the perfect system, which uses the second principal part.
The present system uses the second principal part as its base to form the following tenses:
The present tense: "The doctor heals the man."
The imperfect tense: "The doctor was healing the man."
The future tense: "The doctor will love the man."
The other system, called the perfect system, uses the third principal part as its base to form the following tenses:
The perfect tense: "The doctor healed the man."
The pluperfect tense: "The doctor had healed the man."
The future perfect tense: "The doctor will have healed the man."
While verbs conjugate across these various tenses, the verbs reflect number and person fairly regularly. These are called personal endings. Since personal endings are present in nearly every sentence, it would be beneficial for any researcher working in Latin to memorize them. These are:
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| Singular | Plural |
| 1-Person | -ō or -m | -mus |
| 2-Person | -s | -tis |
| 3-Person | -t | -nt |
Thus if we take this with the verb, sānō, sānāre, sānāvī, sānātum 'to heal,' and conjugate it in the active present tense, it will look like this:
"Ego sānō" > I heal
"Tu sānās" > You heal
"Is/ ea/ id sānat" > He/ she/ it heals
"Nōs sānāmus" > We heal
"Vōs sānātis" > You all heal
"Eī/ eæ/ ea sānant" > They heal
Note that most records that family historians will read will be either in the imperfect, perfect, or pluperfect tenses.
Passive Verbs
Latin Verbs can also conjugate to become passive. While the subject performs the action with active verbs, with passive verbs, the object becomes the subject and receives the action. For example, the previous sentence, "the doctor heals the man," is in the active the same sentence in the passive form would be, "the man was healed by the doctor."
The passive mood conjugates in a very similar manner as the active verbs and forms the same tenses. The passive present system also uses the second principal part; thus, the only difference is the personal endings. These are:
| | |
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| Singular | Plural |
| 1-Person | -or or -r | -mur |
| 2-Person | -ris | -minī |
| 3-Person | -tur | -ntur |
Thus if we take this with the verb, sānō, sānāre, sānāvī, sānātum 'to heal,' and conjugate it in the passive present tense, it will look like this:
"Ego sānor" > I was healed
"Tu sānāris" > You were healed
"Is/ ea/ id sānātur" > He/ she/ it was healed
"Nōs sānāmur" > We were healed
"Vōs sānāminī" > You were all healed
"Eī/ eæ/ ea sānantur" > They were healed
Meanwhile, the passive perfect system is formed differently as it uses the fourth principal part combined with sum, esse, fuī, futūrum 'to be' for the perfect system. Thus the same verb conjugated in the passive perfect tense will look like this:
"Sānātus sum" > I was healed
"Sānātus es" > You were healed
"Sānātus est" > He was healed
"Sānātus sumus" > We were healed
"Sānātus estis" > You all were healed
"Sānātus sunt" > They (masculine) were healed
Since the fourth principal part is used for the perfect system, this means that the passive perfect system must also inflect gender to agree with the noun it acts on. This declension is the same as the 1 and 2 declension nouns. For example, notice how the verb declines with the words puella ('girl,' feminine), vir ('man,' masculine), and animal ('animal,' neuter):
"Puella sānāta est" > 'the girl was healed.'
"Vir sānātus est" > 'the man was healed.'
"Animal sānātum est" > 'the animal was healed.'
Another feature of passive sentences, as you may have noticed in the previous example, is that they only require a subject and a verb. For example, in vir sānātus est "the man (subject) was healed (passive verb); this means that the actual doer of the sentence never needs to be mentioned, instead focusing on the who was receiving the action. However, in English and Latin, the actor can be added in a passive sentence. For example, with the previous sentence, "The man was healed," we can add the doer to the sentence by adding by the doctor to make, "the man was healed by the doctor.
To do this in Latin, one must use what is called the ablative of agent. This is done by using the preposition ā/ab 'from, by,' plus the doer or actor of a passive sentence. Also, note that the preposition ā/ab works like the English word 'a/an,' in that it is ā when the next word starts with a consonant and ab when the next word starts with a vowel. Now with these rules, we can make sentences like the following:
"Vir sānātur ā medicō" > the man is healed by the doctor
"Carnem mandūcāta est ab animalī" > the meat was eaten by the animal
"Puellæ ab puerīs amātæ sunt" > The girls were loved by the boys
Passive constructions are very common in parish registers. Especially in entries like "he was born," "she was baptized," or "they were buried."
Deponent Verbs
Along with regular verbs that can be formed in the active and the passive, Latin has another class of verbs called deponent verbs. There are verbs that are formed and look like they are in the passive form but work like regular active verbs. For example:
"Sanctus Iacobus linguam latinam loquitur" > St. James speaks Latin."Senēs sequuntor fīliī suī" > The old men follow their children
It is important to recognize deponent verbs as they can cause confusion when translating. However, they are quite easy to form as they are conjugated like any other verb in the passive mood, but they must be read like verbs in the active voice. Furthermore, because they are only formed in the passive mood, they only have three principal parts: the first, second, and fourth. Thus in the dictionary, they look like this:
Loquor, loquī, locūtum (dep.) > 'to speak, talk'
Sequor, sequī, sēcūtum (dep.) > 'to follow'
Participles
The last major form of verbs is participles. Participles are adjectives formed by the verb. An example of this in English can be seen in the phrase "the loving woman," here, the word 'loving' is an adjective used to describe the woman, having been formed by the verb 'to love.'
In Latin, most verbs have two active participles and two passive participles. The present active and the future passive participles are formed from the first principal part, while the perfect passive and the future active paticibles are formed from the fourth principal part. If we take the verb amō, amāre, amāvī, amātum 'to love' and make a chart out of its participle forms, it would look like this:
| | |
--- |
| Active | Passive |
| Present | amāns, amantis | - Does not exist - |
| Perfect | - Does not exist - | amātus, -a, -um |
| Future | amātūrus, -a, -um | amandus, -a, -um |
The reason that there are so many forms of the participle is because they each have their own connotation or meaning. These are:
The present active: amāns, amantis = loving
The perfect passive: amātus, -a, -um = loved, having been loved
The future active: amātūrus, -a, -um = about to love, going to love
The future passive: amandus, -a, -um = to be loved, that which ought to be loved1
These concepts might be better understood if seen in their 'natural habitat.' Therefore in a sentence, these participles do the following:
If we now take these participles and attach them to a noun, we get the following:
"Fēmina amāns" > a loving woman
"Fēmina amāta > a loved woman/ a woman, having been loved,
"Fēmina amātūra" > an about-to-be-loved woman
"Fēmina amanda" > a woman to love/ a woman who ought to be loved
If we take the future passive feminine of this verb, we get the name Amanda, which literally means 'She who should be loved,' in Latin.
Illumiation 1: Master of the Codex Manesse, Codex Manesse 311r Alram con Gresten, 1305-1315, in Codex Manesse (Zürich, 1305-1315) Cod. Pal. germ. 848, fol. 311r. This image is in the public domain.
Illumiation 2: Master of the Codex Manesse, Codex Manesse 158r Der von Sachsendorf, 1305-1315, in Codex Manesse (Zürich, 1305-1315) Cod. Pal. germ. 848, fol. 158r. This image is in the public domain.
The Latin Tutorial Introduction Paleography Introduction Practical Suggestions Grammar Numbers & Calendars Common Phrases Abbreviations Given Names Genealogical Glossary Record Types Church Civil Notarial Document Gallery |
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Prime Factorization of 777
Prime Factors of
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What is the Prime Factorization of 777?
Explanation of number 777 Prime Factorization
Prime Factorization of 777 it is expressing 777 as the product of prime factors. In other words it is finding which prime numbers should be multiplied together to make 777.
Since number 777 is a Composite number (not Prime) we can do its Prime Factorization.
To get a list of all Prime Factors of 777, we have to iteratively divide 777 by the smallest prime number possible until the result equals 1.
Here is the complete solution of finding Prime Factors of 777:
The smallest Prime Number which can divide 777 without a remainder is 3. So the first calculation step would look like:
777 ÷ 3 = 259
Now we repeat this action until the result equals 1:
259 ÷ 7 = 37
37 ÷ 37 = 1
Now we have all the Prime Factors for number 777. It is: 3, 7, 37
Prime Factor Tree of 777
We may also express the prime factorization of 777 as a Factor Tree:
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Prime Factorization Table
| Number | Prime Factors |
--- |
| 762 | 2, 3, 127 |
| 763 | 7, 109 |
| 764 | 22 × 191 |
| 765 | 32 × 5 × 17 |
| 766 | 2, 383 |
| 767 | 13, 59 |
| 768 | 28 × 3 |
| 769 | 769 |
| 770 | 2, 5, 7, 11 |
| 771 | 3, 257 |
| 772 | 22 × 193 |
| 773 | 773 |
| 774 | 2 × 32 × 43 |
| 775 | 52 × 31 |
| 776 | 23 × 97 |
| 777 | 3, 7, 37 |
| 778 | 2, 389 |
| 779 | 19, 41 |
| 780 | 22 × 3 × 5 × 13 |
| 781 | 11, 71 |
| 782 | 2, 17, 23 |
| 783 | 33 × 29 |
| 784 | 24 × 72 |
| 785 | 5, 157 |
| 786 | 2, 3, 131 |
| 787 | 787 |
| 788 | 22 × 197 |
| 789 | 3, 263 |
| 790 | 2, 5, 79 |
| 791 | 7, 113 |
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4753 | https://www.expii.com/t/rational-exponents-definition-examples-4456 | Expii
Rational Exponents - Definition & Examples - Expii
Exponents can be rational, such as x^(a/b). The numerator is the power being raised. The denominator is the root being taken.
Explanations (3)
Alex Federspiel
Video
7
(Video) Simplifying Radical Expressions
by mathman1024
This video by mathman1024 works through converting fractional exponents into radicals.
Summary
There are many rules about the combination or breakdown of exponents.
amn=n√amam⋅an=am+naman=am−n(am)n=am⋅n
To review the laws of exponents, you can check out the lessons on: negative exponents, product rule, quotient rule, and power to a power.
The most relevant rule for this section is the first one with a fractional exponent. The denominator of the exponent is the degree of the radical. The numerator of the exponent is the power of the term inside of the radical.
We can combine this with the other rules to simplify radical and exponential expressions.
The first example is: √x⋅5√x.
√x⋅5√x=x12⋅x15=x12+15=x710=10√x7
The second example he does is: 4√x33√x2.
4√x33√x2=x34x23=x34−23=x112=12√x
These examples will be really helpful when solving word problems!
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Sum and Difference of Square Roots - Rules & Examples
Solving Radical Equations - Examples & Practice
Solve by Radical: xⁿ=c
Radical Equation Word Problems - Examples & Practice
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Zora Gilbert
Text
5
Fraction exponents are like an exponent and a radical
Fraction exponents tell you to take the power of a number, and then take a root. It's hard to explain in words, so here's a rule:
abc=c√ab=(c√a)b
Basically, in a fraction exponent the numerator (top) is the power, and the denominator (bottom) is the root you need to take.
Image source: By Caroline Kulczycky
Let's look at a quick example using real numbers: 2753
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Ben Ferrara
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A rational exponent can always be turned into a radical. If b to the nth root of a number is real, and m is a positive integer, then rational exponents and radicals are related like this:
b1n=n√bandbmn=n√bm=(n√b)m
The denominator of the fraction is the value of the root. For example, if the denominator is 3, you'll be taking the cube root. You may be dealing with positive or negative fractions, so be aware of that. Positive numbers are most common, as in the example above and below. Negative numbers means you'll need to follow the rules of exponent.
Let's try this with a number first.
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4754 | https://sites.math.duke.edu/first_year/teachers.only/Lab_Manual/Series_1.pdf | Duke University Series #1 181 Series #1: Limits of Partial Sums Before beginning our study of infinite sums we first need to discuss briefly the convergence of sequences of real numbers. A sequence is just a list of numbers in a given order. Here are three examples: 2, 4, 6, 8, 10, 12, . . .
1, −1, 1, −1, 1, −1, . . .
1, 1 1 2, 1 3 4, 1 7 8, 1 15 16, . . .
Sometimes we give names to the terms in a sequence by letting c1 denote the first term, c2 denote the second term, and so forth. We represent the entire sequence by {cn}∞ n=1, where n is the index which runs from 1 to ∞, indicating that the successive terms of the sequence are c1, c2, c3, . . .. In the case of the first sequence above, cn = 2n.
Definition. We say that a sequence {cn}∞ n=1 converges to a limit L if we can make cn as close to L as we like by choosing n large. We shall sometimes use the notation lim n→∞cn = L to indicate that the sequence {cn}∞ n=1 converges to a limit L.
Let’s see which of the sequences given above converge. The first sequence can’t get close to any one number because each term is larger by 2 than the preceeding term. The second sequence can’t get close to any one number because the terms oscillating between +1 and −1. The third sequence is cn = 2 − 1 2 n−1 Since ( 1 2)n gets smaller and smaller as n get larger, we see that cn approaches the limit L = 2.
Example 1. Here are some more examples of convergent sequences.
(a) The sequence cn = 1 n converges to 0.
(b) The sequence cn = 5 −42−n + 6 n converges to 5.
(c) The sequence cn = 2 + 1 n 7 −1 n2 − 5 −2−n + 6 n 2 182 Series #1 Duke University converges to −11.
(d) The sequence cn = 4 + 7n 2 + 3n is a little more difficult. Divide numerator and denominator by n to obtain cn = 4 n + 7 2 n + 3 Now we can see that as n gets large this sequence approaches 7 3.
Now we are ready to turn our attention to series. A series is an infinite sum a1 + a2 + a3 + a4 + a5 + . . .
(1) We have already seen examples of such sums in our study of probability. Since it is awkward to write out sums (whether finite or infinite), we shall use the sum-mation notation m X j=n aj = an + an+1 + . . . + am−1 + am.
Using this notation, the infinite sum in (1) would be written ∞ X j=1 aj.
There is a practical and theoretical problem with infinite sums: we cannot compute a sum by adding one term after another if there are infinitely many terms. We need another way to make sense of and to compute infinite sums. For this reason, we introduce partial sums and limits. We define the nth partial sum of this series to be the sum of the first n terms Sn = n X j=1 aj of the series. Now we can say what we mean by the infinite sum (1).
Definition. We say that the infinite sum (1) converges if the sequence of partial sums {Sn} converges to a finite limit S as n gets larger. In this case, we define ∞ X j=1 aj = S.
Duke University Series #1 183 If the sequence of partial sums does not converge to a limit, we say that the series diverges.
This definition makes sense because it says that, as we add more and more terms to the sum, if we get closer and closer to a number S, then we define this number S to be the sum of the series.
Example 2. Let r ̸= 1 and define aj = rj. This is the geometric series which we studied in the probability laboratory. The nth partial sum is Sn = 1 + r + r2 + . . . + rn−1 = 1 −rn 1 −r If |r| < 1, we can see that the sequence of partial sums Sn converges to 1 1−r.
Therefore, by definition, if |r| < 1, the series converges and 1 + r + r2 + . . . + rn−1 + . . . = 1 1 −r If r = 1, the nth partial sum is Sn = 1 + 1 + 12 + . . . + 1n−1 = n so the sequence of partial sums does not converge. Thus the geometric series does not converge if r = 1. If fact, the geometric series does not converge for any r satisfying |r| ≥1 (see problem 2).
Example 3. Consider the series ∞ X j=1 1 j(j + 1) = 1 2 + 1 6 + 1 12 + 1 20 + . . . .
(2) Since 1 j(j + 1) = 1 j − 1 j + 1, we can compute the nth partial sum explicity.
Sn = 1 1 −1 2 + 1 2 −1 3 + 1 3 −1 4 + . . . + 1 n − 1 n + 1 = 1 − 1 n + 1.
184 Series #1 Duke University Since Sn = 1 − 1 n+1 we see that the sequence of partial sums converges to 1. Thus, the series (2) converges and ∞ X j=1 1 j(j + 1) = 1.
Example 4. In Example 3 you may have been tempted to perform an infinite cancellation on the series ( 1 1 −1 2) + ( 1 2 −1 3) . . .. The following example shows that such cancellation can give incorrect results. Consider the series (5 −5 1 2) + (5 1 2 −5 1 3) + (5 1 3 −5 1 4) + . . . + (5 1 j −5 1 j+1) + . . .
The nth partial sum is Sn = 5 −5 1 n+1 which converges to 4 as n gets larger and larger. The infinite cancellation gives 5, which is incorrect.
The series in Examples 2, 3, and 4 are very special. Unfortunately, for most series there is no way to calculate a simple formula for the nth partial sum, Sn.
If we can’t calculate Sn, how are we ever going to be able to tell whether Sn converges? Good question! If we can somehow determine that a series converges, we can approximate the sum by a partial sum with high n. We shall see in the next lecture that there are some tests which one can apply to a series which guarantee that it converges if the test comes out positive. For the moment, we raise the question of how large the terms aj can be if the sum is finite. We will see that “most of the terms ” must be small in the following sense: if P∞ j=1 aj, then the sequence a1, a2, a3, . . . must converge to zero. For any positive integer m, we can write am = Sm −Sm−1 = (S −Sm−1) −(S −Sm).
As m gets larger and larger, both terms on the right get smaller and smaller (since, for a convergent series, the partial sums converge to the sum). Thus am gets smaller and smaller as m gets larger and larger.
The nth Term Test for Convergence. If a series P∞ j=1 aj converges, then the sequence of terms, {aj}, must converge to 0.
Duke University Series #1 185 Example 5. This condition can sometimes be used to show that series do not converge. Consider the series ∞ X j=1 (−1)j The jth term is aj = (−1)j. Since the sequence {aj} does not converge to 0 (it oscillates between +1 and −1), the series can not converge.
The condition that {aj} converges to 0 is a necessary condition for a series to converge. That is, if the series converges, it must be true. However, it is not a sufficient condition for convergence. That is, just because {aj} converges to 0 does not prove that the series converges. Here is an example.
Example 6 (The harmonic series). Consider the series ∞ X j=1 1 j .
Notice that aj = 1 j , so it is certainly true that the sequence {aj} converges to 0.
However, we shall see that the series does not converge. We cleverly group the terms of the series in the following way: 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 + 1 9 + 1 10 + 1 11 + 1 12 + 1 13 + 1 14 + 1 15 + 1 16 + . . .
= 1 + 1 2 + { 1 3 + 1 4} + { 1 5 + 1 6 + 1 7 + 1 8} + { 1 9 + 1 10 + 1 11 + 1 12 + 1 13 + 1 14 + 1 15 + 1 16} + . . .
≥ 1 + 1 2 + 1 2 + 1 2 + 1 2 + . . .
This shows that the sequence partial sums increases without bound, so the series does not converge. Note that the series diverges even though the sequence of terms approaches zero.
A Note on Notation. We have been using the the letter j for the index in sums: m X j=n aj = an + an+1 + . . . + am−1 + am.
If we used the letter k for the index, the sum m X k=n ak = an + an+1 + . . . + am−1 + am would still be the same. Thus, there is no difference between the series P∞ j=1 1 j(j+1) and the series P∞ k=1 1 k(k+1) except for the name given to the index.
186 Series #1 Duke University Problems.
(1) For each of the following sequences, say whether it converges and if it does say what the limit is.
(a) cn = (3 −2 n).
(b) cn = (3 −2 n)2.
(c) cn = 3n.
(d) cn = 3n 2 n.
(e) cn = 2 + (−1)n.
(f) cn = 2 + (−1)n n .
(g) cn = (2 −(−1)n n )(5 + 17 n ) −4.
(h) cn = 6−n 2+3n.
(i) cn = 6−n+3n2 2+3n+15n2 .
(2) By looking at each of the cases, r = 1, r = −1, r > 1, r < −1, show that the geometric series does not converge if |r| ≥1.
(3) Find the sum of the series ∞ X k=1 3 4 k .
(4) Find the sum of the series ∞ X k=1 7 1 k −7 1 k+1 .
(5) By using the idea in Example 3, show that ∞ X k=2 1 k2 −1 = 3 4.
(6) We shall show later in the course that the series P∞ j=1 1 j2 converges. Use your calculator to approximate the sum accurate to two decimal places. Explain how you decide when you have the required accuracy.
(7) Compute S10, S80, and S700 for the series 4 −4 3 + 4 5 −4 7 + . . .
Do you think that this series has a sum? If so, what do you think it is?
Duke University Series #1 187 (8) We showed in Example 6 that the harmonic series P∞ j=1 1 j diverges. Use your hand calculator to compute S100 and S200. Using the inequality from the di-vergence argument in Example 6, determine how large n would have to be so that Sn is greater than 10. What does this tell you about using a calculator to determine whether a series converges or diverges?
(9) Suppose that we change every other sign in the harmonic series to a minus sign obtaining the series 1 −1 2 + 1 3 −1 4 + . . .
We shall prove later in the course that this new series converges. Use partial sums to approximate the sum. How many terms do you think you need to get 2 decimal place accuracy? Why do you think that the behavior of these partial sums is so different from the partial sums of the harmonic series?
(10) Use the necessary condition to prove that the following series do not converge: (a) P∞ j=1 2j.
(b) P∞ j=1 sin (π 4 j).
(c) P∞ k=1(1 + 1 k).
Answers to Selected Problems.
1. (g) converges to 6; (h) converges to −1 3; (i) converges to 1 5.
3. 3.
7. S10 = 3.041, S80 = 3.129, S700 = 3.140. |
4755 | https://www.youtube.com/watch?v=c4m-YQJK5mY | How to simplify a rational trigonometric expression by factoring
Brian McLogan
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Posted: 22 Nov 2013
👉 Learn how to verify trigonometric identities having rational expressions. To verify trigonometric expression means to verify that the terms on the left-hand side of the equality sign is equal to the terms on the right-hand side. To verify rational trigonometric identities, it is usually more convenient to start with getting rid of the denominator(s) of the rational term(s). This can be done by multiplying both the numerator and the denominator by the conjugate of the denominator, if the denominator involves addition/subtraction or by the reciprocal of the denominator, if the denominator involves product or the expression can be converted to Pythagoras trigonometric identity if possible.
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Transcript:
welcome all right so in this case ladies and gentlemen what i have is the sine squared of x plus 2 uh sine of x plus 1 divided by sine of x plus 1. and we have this rational expression that we're going to have to somehow eliminate and a couple of ways that i've you know talked about going through this especially when we have um a a binomial in the bottom and especially when it's you know it it's a uh in terms of our trigonometric functions we can multiply by the conjugate to be able to get now a uh we to be able to now get a pythagorean identity and we could still do that but i don't really want to multiply sine of x minus one on the top and bottom just because you know that works but then i have to multiply a binomial times a trinomial and it just doesn't look like it's going to be fun so i want to kind of think backwards rather than creating more work for myself let's try to see how can i reduce the amount of work and how can i kind of maybe work backwards well you can see here i have some form of a trinomial that i'm going to be able to work with and we know that's trimming wheels raised to a second power then it has a linear term then it has a constant so what i can do in this case is try to see well how can i maybe possibly factor this so if i just looked at this in terms of variables i would have x squared plus 2x plus 1. well that's pretty basic i can see that's going to be x plus 1 times x plus 1. so that's exactly what i'm going to do but now rather than just using x i'm going to use sine of x so therefore i have this as sine of x plus 1 times sine of x plus 1 all divided by sine of x plus 1. so therefore you can see that these are now similar terms so now i can just divide out those two terms and my final solution is just going to be sine of x plus 1. so there you go ladies and gentlemen that is how you simplify that expression thanks |
4756 | https://www.who.int/publications/i/item/WHO-MHP-HPS-EML-2023.04 | AWaRe classification of antibiotics for evaluation and monitoring of use, 2023
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AWaRe classification of antibiotics for evaluation and monitoring of use, 2023
AWaRe classification of antibiotics for evaluation and monitoring of use, 2023
26 July 2023
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Overview
The AWaRe classification of antibiotics was developed in 2017 by the WHO Expert Committee on Selection and Use of Essential Medicines as a tool to support antibiotic stewardship efforts at local, national and global levels, Antibiotics are classified into three groups, Access, Watch and Reserve, taking into account the impact of different antibiotics and antibiotic classes on antimicrobial resistance, to emphasize the importance of their appropriate use. It is updated every 2 years.
The AWaRe classification is intended as a tool for monitoring antibiotic consumption, defining targets and monitoring the effects of stewardship policies that aim to optimize antibiotic use and curb antimicrobial resistance. The WHO 13 th General Programme of Work 2019–2023 includes a country-level target of at least 60% of total antibiotic consumption being Access group antibiotics.
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4757 | https://math.stackexchange.com/questions/668619/proving-a-trig-identity-frac-sina-b-sina-b-frac-tan-a-tan-b | Stack Exchange Network
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Proving a trig identity: $\frac{\sin(A + B)}{\sin(A - B)}=\frac{\tan A + \tan B}{\tan A - \tan B}$
I'm learning about trig identities, and I'm struggling to prove that two expressions are equal:
$$
\frac{\sin(A + B)}{\sin(A - B)}=\frac{\tan A + \tan B}{\tan A - \tan B}
$$
How do you go about proving this? I know about compound angles - i.e. the sine, cosine and tangent of $(A \pm B)$, but don't know how to apply it in this situation.
'
,
5 Answers 5
We start with $\frac{\tan A + \tan B}{\tan A - \tan B}$:
$$\frac{\tan A + \tan B}{\tan A - \tan B}=\frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}}=\frac{\sin A\cos B +\sin B\cos A}{\sin A\cos B -\sin B\cos A}=\frac{\sin(A+B)}{\sin(A-B)}.$$
Using the tangent addition formulas, we get
$$\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$
and
$$\tan (A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$
From this, we get
$$\frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\tan (A+B)}{\tan (A-B)} \frac{1 - \tan A \tan B}{1 + \tan A \tan B} = \frac{\tan (A+B)}{\tan (A-B)} \frac{\cos A \cos B - \sin A \sin B}{\cos A \cos B + \sin A \sin B}$$
$$= \frac{\tan (A+B)}{\tan (A-B)} \frac{\cos (A+B)}{\cos (A-B)} = \frac{\sin (A+B)}{\sin (A-B)}.$$
$$
\frac{\sin(A + B)}{\sin(A - B)}=\frac{\sin A\cos B+\sin B\cos A}{\sin A\cos B-\sin B\cos A}=\frac{\frac{\sin A\cos B+\sin B\cos A}{\cos A\cos B}}{\frac{\sin A\cos B-\sin B\cos A}{\cos A\cos B}}=\frac{\tan A + \tan B}{\tan A - \tan B}
$$
Using the fact that $e^{i x}=\cos x + i\sin x$, where $i=\sqrt{-1}$, we have
$$\frac{\sin(A+B)}{\sin(A-B)} = \frac{\Im(e^{iA}e^{iB})}{\Im(e^{iA}e^{-iB})} = \frac{\Im((\cos A+i\sin A)(\cos B+i\sin B))}{\Im((\cos A+i\sin A)(\cos B-i\sin B))},$$
where $\Im$ denotes the imaginary part. Expanding and taking the imaginary ($\Im$) parts, we get
$$\frac{\cos A\sin B+\sin A\cos B}{\sin A\cos B-\cos A\sin B}.$$
Dividing numerator and denominator by $\cos A \cos B$ gives
$$\frac{\left(\displaystyle\frac{\cos A\sin B+\sin A\cos B}{\cos A\cos B}\right)}{\left(\displaystyle\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}\right)} = \frac{\tan B+\tan A}{\tan A-\tan B},$$
as required.
A very general (but extremely useful) approach is by noting the following
$$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$
$$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$
and since $\tan(x) = \frac{\sin(x)}{\cos(x)}$
$$\tan(x) = \frac{1}{i}\frac{e^{ix} - e^{-ix}}{e^{ix} + e^{-ix}} = -i \frac{e^{ix} - e^{-ix}}{e^{ix} + e^{-ix}} $$
We note here that $i$ is the imaginary constant (basically its a number such that $i^2 = -1$)
I will leave you to go ahead and verify these formulas work for every trig Identity you have already memorized and more information is underneath:
So we wish to prove:
$$\frac{\sin(A+B)}{\sin(A-B)} = \frac{\tan(A) + \tan(B)}{\tan(A) - \tan(B)} $$
We prepare the left side (noting that both fractions take form $\frac{A}{2i}$ and therefore we can drop the $2i$ denominators
$$\frac{\sin(A+B)}{\sin(A-B)} = \frac{e^{i(A+B)} - e^{-i(A+B)}}{e^{i(A-B)} - e^{-i(A-B)}}$$
So because I'm slightly lazy (and for the sake of giving you something to practice) you need to do the exact same job with the tan(x) expression where each instance of x becomes A or B depending on whats being evaluated.
Now the goal is to systematically simplify both expressions by transforming expressions of the form $e^{-k}$ to $\frac{1}{e^k}$
Followed by taking sums and giving them common denominators $\frac{A}{C} + \frac{B}{D} = \frac{AD + BC}{CD}$
And dividing out common factors.
Its a tedious process but once done. Both expression will look exactly the same... Well you don't have to take my word for it, do it yourself and prove that it works ;)
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4758 | https://boards.straightdope.com/t/is-there-a-difference-between-naive-and-gullible/954992 | Is there a difference between naïve and gullible? - In My Humble Opinion - Straight Dope Message Board
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Is there a difference between naïve and gullible?
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Nars_GlinleyGuest
Nov 2021
I used naïve to describe someone earlier and then later in the conversation used the word gullible. Obviously the two words are similar but gullible sounds more insulting to me now that I think about it and that wasn’t my intent.
What’s the difference to you? I’m not interested in dictionary definitions but in common usage. That’s why I put this in IMHO and not FQ.
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PastTenseGuest
Nov 2021
I think that a naive person has not had the appropriate learning experiences while a gullible person has had the learning experiences–but has not learned from them.
1 Reply
kenobi_65 Corellian Nerfherder (he/him)
Nov 2021
Naive is specifically about being inexperienced and unworldly. People whom I’d classify as “naive” tend to be some combination of young, sheltered, or just lacking in a lot of experience. Gullible, on the other hand is about being overly trusting, and easily fooled/taken advantage of.
One might be gullible because one is also naive, but one can also simply be gullible by being overly trusting, wanting to think the best of people, or having a faulty “BS meter,” despite being older and experienced (and, thus, no longer truly naive).
2 Replies
Thudlow_BoinkCharter Member
Nov 2021
Nars_Glinley:
I’m not interested in dictionary definitions but in common usage.
But “gullible” isn’t in the dictionary!
4 Replies
kenobi_65 Corellian Nerfherder (he/him)
Nov 2021
Thudlow_Boink:
But “gullible” isn’t in the dictionary!
I actually was able to successfully get a college friend (an English major, at that) to grab a dictionary and look it up, to prove to me that I was wrong about that.
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Nars_GlinleyGuest
Nov 2021
Thudlow_Boink:
But “gullible” isn’t in the dictionary!.
Good grief. You had me checking google to make sure I hadn’t spelled it wrong.
I hate you.
1 Reply
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4759 | https://www.cuemath.com/calculus/differentiation-of-e-to-the-power-x/ | LearnPracticeDownload
Differentiation of e to the Power x
Differentiation of e to the power x is a process of determining the derivative of e to the power x with respect to x which is mathematically written as d(ex)/dx. An exponential function is of the form f(x) = ax, where 'a' is a real number and x is a variable. e to the power x is an exponential function with base (a) equal to the Euler's number 'e' and the differentiation of e to the power x is equal to e to the power x, that is, itself. It is written as d(ex)/dx = ex.
Let us learn about the differentiation of e to the power x and some variations of the function e to the power x. We will determine the differentiation of e to the power x using different methods including the first principle of differentiation, and derivative of exponential function along with some examples for a better understanding.
| | |
--- |
| 1. | What is the Differentiation of e to the Power x? |
| 2. | Differentiation of e to the Power x Formula |
| 3. | Differentiation of e to the Power x Using First Principle |
| 4. | Differentiation of e to the Power x Using Derivative of ax |
| 5. | FAQs on Differentiation of e to the Power x |
What is the Differentiation of e to the Power x?
The differentiation of e to the power x is equal to e to the power x because the derivative of an exponential function with base 'e' is equal to ex. Mathematically, it is denoted as d(ex)/dx = ex. e to the power x is an exponential function with a base equal to 'e', which is known as "Euler's number". It is written as f(x) = ex, where 'e' is the Euler's number and its value is approximately 2.718. The differentiation of e to the power x can be done using different methods such as the first principle of differentiation and derivative of ax.
Differentiation of e to the Power x Formula
Suppose y = ex ⇒ ln y = ln ex ⇒ ln y = x. On differentiating this with respect to x, we have (1/y) dy/dx = 1 ⇒ dy/dx = y ⇒ dy/dx = ex. If we differentiate e to the power x with respect to x, we have d(ex)/dx = ex. Hence the formula for the differentiation of e to the power x is,
Differentiation of e to the Power x Using First Principle of Derivatives
Next, we will prove that the differentiation of e to the power x is equal to ex using the first principle of differentiation. We know that for two exponential functions, if the bases are the same, then we add the powers. To prove the derivative of e to the power x, we will use the following formulas of exponential functions and derivatives:
f'(x) = lim h→0 [f(x + h) - f(x)] / h
ex + h = ex.eh
lim x→0 (ex - 1) / x = 1
Using the above formulas, we have
d(ex)/dx = lim h→0 [ex + h - ex] / h
= lim h→0 [ex.eh - ex] / h
= lim h→0 ex [eh - 1] / h
= ex lim h→0[eh - 1] / h
= ex × 1
= ex
Hence we have proved the differentiation of e to the power x to be equal to e to the power x.
Differentiation of e to the Power x Using Derivative of ax
An exponential function is of the form f(x) = ax, where 'a' is a constant (real number) and x is the variable. The derivative of exponential function f(x) = ax is f'(x) = (ln a) ax. Using this formula and substituting the value a = e in f'(x) = (ln a) ax, we get the differentiation of e to the power x which is given by f'(x) = (ln e) ex = 1 × ex = ex [Because by log rules, ln e = 1]. Hence, the derivative of e to the power x is ex.
Important Notes on Differentiation of e to the Power x:
The nth differentiation of e to the power x is equal to ex, that is, dn(ex)/dxn = ex
The derivative of the exponential function with base e is equal to ex.
The derivative of eax is aeax. Using this formula, we have the differentiation of ex to be 1.ex = ex.
☛ Related Topics:
Derivative of e2x
Derivative of Exponential Function
Derivative of log x
Read More
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SHEETS
Differentiation of e to the Power x Worksheet
Calculus Worksheet
Differentiation of e to the Power x Worksheet
Calculus Worksheet
Differentiation of e to the Power x Examples
Example 1: What is the differentiation of e to the power x to the power 2?
Solution: To determine the differentiation of e to the power x to the power 2, that is, ex2, we will use the chain rule.
We know that for y = f(g(x)), the derivative is y' = f'(g(x)) × g'(x). Let u = g(x) = x2 and f(x) = ex and y = f(g(x)) = ex2 ⇒ y = f(u) = eu
We have dy/dx = dy/du × du/dx
= eu × 2x
= 2xex2
Answer: Hence the derivative of ex2 is 2xex2
2. Example 2: Determine the differentiation of e to the power x sin x.
Solution: To evaluate the value of the derivative of exsinx, we will use product rule of differentiation.
d(exsinx)/dx = (ex)' sin x + (sin x)' ex
= ex sin x + ex cos x [Because derivative of sin x is cos x]
= ex (sin x + cos x)
Answer: Hence differentiation of e to the power x sin x is ex (sin x + cos x).
View Answer >
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Differentiation of e to the Power x Questions
Check Answer >
FAQs on Differentiation of e to the Power x
What is the Differentiation of e to the Power x in Calculus?
The differentiation of e to the power x is equal to e to the power x itself because the derivative of an exponential function with base 'e' is equal to ex. Mathematically, it is denoted as d(ex)/dx = ex.
What is the Differentiation of e to the Power Minus x?
The differentiation of e to the power minus x is equal to the negative of e to the power minus x, that is, d(e-x)/dx = -ex.
What is the Differentiation of e to the Power Sin x?
The differentiation of e to the power sin x is equal to the product of cos x and e to the power sin x, that is, d(esin x)/dx = cos x esin x.
What is the Derivative of e to the Power x log x?
The derivative of e to the power x log x is given by, d(ex ln x)/dx = ex ln x (1 + ln x). This follows from chain rule.
How do you Find the Derivative of an Exponential Function?
The derivative of exponential function f(x) = ax is f'(x) = (ln a) ax which can be calculated by using the first principle of differentiation.
What is the Formula for Exponential Differentiation?
The formula for exponential differentiation for f(x) = ax is f'(x) = (ln a) ax. If a = e, then the formula for the differentiation of e to the power is ex.
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4760 | https://www.youtube.com/watch?v=JQzP5hzjvgc | Finding The Real And Imaginary Parts To Complex Numbers Part Four - Multiply n Factors Of i
Amour Learning
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Posted: 8 Apr 2021
In this video, we simplify i^n.
This problem can be found in the free open-access textbook: "A First Course in Complex Analysis", authored by Dr. Matthias Beck, Dr. Gerald Marchesi, Dr. Dennis Pixton and Dr. Lucas Sabalka
Any feedback on how these videos can be improved would be great. I will be constantly updating these videos to ensure that everything is accurate. Thank you all for being such a great community. All content is released under Creative Commons "No Rights Reserved", so please feel free to use this video however you want.
My name is Kody Amour, and I teach free undergraduate mathematics courses on YouTube. I'm currently developing three YouTube courses simultaneously: Discrete Math, Linear Algebra, Real Analysis, Topology, Number Theory and Complex Analysis. These courses correspond to playlists that I am constructing indefinitely.
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Transcript:
hey everyone in this video we're going to be finding the real and imaginary parts of this complex number I to the N for any n that's an integer this is problem 1.2 Part D which can be found in your free online open-source complex analysis textbook I highly recommend you take a look I'll leave a link in the description below so how do we find the real and imaginary parts to this complex number well first let's identify what I to the N is the definition of i to the N is is I I I I and so on I where there are n i here now this doesn't really help me figure out what the real and imaginary part is because I would have to multiply all of these eyes together to figure that out and I don't really know how to do that considering I don't know how many eyes I have here I I have n eyes but how many eyes is that it's n what is n some integer I guess so here's what we're going to do to try to address this we're going to pick values of n specific integers for n and we're just going to look at what I raised to that power is so let's start with a table so here in the left column I'm going to pick some integers for n and on the right column we're going to try to figure out what I raised to that power is in terms of its real and imaginary parts so let's start with some easy n values zero what is i to the zeroth power well anything raised to the zeroth power any complex number raised to the zeroth power is 1 what about when n equals 1 well I to the 1 power is I so here we have a real part that is one and an imaginary part that is Z when n equals 1 we have a real part of zero and an imaginary part of one sort of swapped there when n equal 2 we have I2 which is -1 here the real part is -1 and the imaginary part is zero and then if we go to three I cubed is just i^ 2 I and if I write it like this it makes it easy to figure out what i^ 2 I is because I already know what i^ 2 is i^ 2 is -1 and so i^ 2 I is -1 I which is just netive I and so in this specific instance the real part is zero and the imaginary part is -1 so you might notice we're rotating between real and imaginary 0 1 01 and on top of that the signs are flipping it went from positive to negative let's see what happens at Nal 4 I'm going to give myself some more room here well I to 4th is just I cubed I and I can always do this because then I can just take the result from above and replace that with I cubed here and I can just keep doing that with whatever number I have minus one I can just replace with the above result and rinse and repeat so this is I I which from here we can see is i^ 2 and i^ 2 is -1 which is just 1 i^ 2 is -1 so i^ 2 is positive 1 so in this instance the real part is 1 and the imaginary part is zero which is the same as this instance here where the real part is one and the imaginary part is zero let's go to Nal 5 here I the 5th power is just I 4th power I and we do this so that we can take the result from our previous output and we can say this is 1 I substitution which is just I and in this instance we have the real part is zero and the imaginary part is one just like above we had real part here was zero and the imaginary part was one and so you notice you might notice that we're going into a cycle we're just going back and forth here we're alternating so it's 0 1 0ga 1 0 1 0ga 1 0 1 so let's try to represent this relationship with some formula now in complex analysis technically I haven't introduced the divides operation although I have done videos on the divides operation and I highly recommend you take a look at those I'm going to try to make this function definition a little bit more approachable so this might look a little sloppy but bear in mind I'm doing this so that everyone can understand what I to the N is i to the N is one of four things I to the N is one if N is a multiple of four now I should clarify when I say multiple I mean it could be 4 or 8 or 12 or 16 or even 0 or -4 or8 specifically all of the integer multiples of four and then for Nal 1 we get I as a result but that is when Nal 1 or Nal 5 or N = 9 or Nal 13 so how do we represent those numbers well if we just take that number and we subtract one then we're back to a multiple of four so is a multiple of four and then up next we have -1 because I 2 is1 if n minus 2 is a multiple of four again technically this portion is a concept that I develop in my number Theory lectures my number Theory videos and so I don't expect you to understand how we develop this portion so if I were to make a test for my complex analysis students I would not ask this question explicitly because this requires a little bit of number Theory to understand and then lastly we have negative I if nus 3 is a multiple of four and I'll just put quotation marks just to represent it's the same as above so we're not done yet I want to identify what the real and the imaginary part of I to the N is well that is dependent on what the output of I to the N is and that is dependent on which of these is divisible by four so let's go through each case if I to the N is 1 then the real part is one and the imaginary part is zero if I to the N is I then the real part is zero and the imaginary part is one if I to the N is negative 1 then the real part is negative 1 and the imaginary part is zero and if I to the N is NE I then the real part is zero and the imaginary part is NE one that's how I would answer this question thanks everyone and I'll see you in the next video |
4761 | https://en.wikipedia.org/wiki/Cellular_respiration | Jump to content
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Contents
1 Aerobic respiration
1.1 Glycolysis
1.2 Oxidative decarboxylation of pyruvate
1.3 Citric acid cycle
1.4 Oxidative phosphorylation
2 Efficiency of ATP production
3 Fermentation
4 Anaerobic respiration
5 See also
6 References
7 External links
Cellular respiration
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From Wikipedia, the free encyclopedia
Process of releasing energy from nutrients using inorganic electron acceptors
Cellular respiration is the process of oxidizing biological fuels using an inorganic electron acceptor, such as oxygen, to drive production of adenosine triphosphate (ATP), which stores chemical energy in a biologically accessible form. Cellular respiration may be described as a set of metabolic reactions and processes that take place in the cells to transfer chemical energy from nutrients to ATP, with the flow of electrons to an electron acceptor, and then release waste products.
If the electron acceptor is oxygen, the process is more specifically known as aerobic cellular respiration. If the electron acceptor is a molecule other than oxygen, this is anaerobic cellular respiration – not to be confused with fermentation, which is also an anaerobic process, but it is not respiration, as no external electron acceptor is involved.
The reactions involved in respiration are catabolic reactions, which break large molecules into smaller ones, producing ATP. Respiration is one of the key ways a cell releases chemical energy to fuel cellular activity. The overall reaction occurs in a series of biochemical steps, some of which are redox reactions. Although cellular respiration is technically a combustion reaction, it is an unusual one because of the slow, controlled release of energy from the series of reactions.
Nutrients that are commonly used by animal and plant cells in respiration include sugar, amino acids and fatty acids, and the most common oxidizing agent is molecular oxygen (O2). The chemical energy stored in ATP (the bond of its third phosphate group to the rest of the molecule can be broken, allowing more stable products to form, thereby releasing energy for use by the cell) can then be used to drive processes requiring energy, including biosynthesis, locomotion, or transportation of molecules across cell membranes.
Aerobic respiration
[edit]
Aerobic respiration requires oxygen (O2) in order to create ATP. Although carbohydrates, fats, and proteins are consumed as reactants, aerobic respiration is the preferred method of pyruvate production in glycolysis, and requires pyruvate be transported by the mitochondria in order to be oxidized by the citric acid cycle. The products of this process are carbon dioxide and water, and the energy transferred is used to make bonds between ADP and a third phosphate group to form ATP (adenosine triphosphate), by substrate-level phosphorylation, NADH and FADH2.[citation needed]
| | |
--- |
| Mass balance of the global reaction: | C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) + energy |
| ΔG = −2880 kJ per mol of C6H12O6 |
The negative ΔG indicates that the reaction is exothermic (exergonic) and can occur spontaneously.
The potential of NADH and FADH2 is converted to more ATP through an electron transport chain with oxygen and protons (hydrogen ions) as the "terminal electron acceptors". Most of the ATP produced by aerobic cellular respiration is made by oxidative phosphorylation. The energy released is used to create a chemiosmotic potential by pumping protons across a membrane. This potential is then used to drive ATP synthase and produce ATP from ADP and a phosphate group. Biology textbooks often state that 38 ATP molecules can be made per oxidized glucose molecule during cellular respiration (2 from glycolysis, 2 from the Krebs cycle, and about 34 from the electron transport system). However, this maximum yield is never quite reached because of losses due to leaky membranes as well as the cost of moving pyruvate and ADP into the mitochondrial matrix, and current estimates range around 29 to 30 ATP per glucose.
Aerobic metabolism is up to 15 times more efficient than anaerobic metabolism (which yields 2 molecules of ATP per 1 molecule of glucose). However, some anaerobic organisms, such as methanogens are able to continue with anaerobic respiration, yielding more ATP by using inorganic molecules other than oxygen as final electron acceptors in the electron transport chain. They share the initial pathway of glycolysis but aerobic metabolism continues with the Krebs cycle and oxidative phosphorylation. The post-glycolytic reactions take place in the mitochondria in eukaryotic cells, and in the cytoplasm in prokaryotic cells.
Although plants are net consumers of carbon dioxide and producers of oxygen via photosynthesis, plant respiration accounts for about half of the CO2 generated annually by terrestrial ecosystems.: 87
Glycolysis
[edit]
Main article: Glycolysis
Glycolysis is a metabolic pathway that takes place in the cytosol of cells in all living organisms. Glycolysis can be literally translated as "sugar splitting", and occurs regardless of oxygen's presence or absence. The process converts one molecule of glucose into two molecules of pyruvate (pyruvic acid), generating energy in the form of two net molecules of ATP. Four molecules of ATP per glucose are actually produced, but two are consumed as part of the preparatory phase. The initial phosphorylation of glucose is required to increase the reactivity (decrease its stability) in order for the molecule to be cleaved into two pyruvate molecules by the enzyme aldolase. During the pay-off phase of glycolysis, four phosphate groups are transferred to four ADP by substrate-level phosphorylation to make four ATP, and two NADH are also produced during the pay-off phase. The overall reaction can be expressed this way:
: Glucose + 2 NAD+ + 2 Pi + 2 ADP → 2 pyruvate + 2 NADH + 2 ATP + 2 H+ + 2 H2O + energy
Starting with glucose, 1 ATP is used to donate a phosphate to glucose to produce glucose 6-phosphate. Glycogen can be converted into glucose 6-phosphate as well with the help of glycogen phosphorylase. During energy metabolism, glucose 6-phosphate becomes fructose 6-phosphate. An additional ATP is used to phosphorylate fructose 6-phosphate into fructose 1,6-bisphosphate by the help of phosphofructokinase. Fructose 1,6-biphosphate then splits into two phosphorylated molecules with three carbon chains which later degrades into pyruvate.: 88–90
Oxidative decarboxylation of pyruvate
[edit]
Main article: Pyruvate decarboxylation
Pyruvate is oxidized to acetyl-CoA and CO2 by the pyruvate dehydrogenase complex (PDC). The PDC contains multiple copies of three enzymes and is located in the mitochondria of eukaryotic cells and in the cytosol of prokaryotes. In the conversion of pyruvate to acetyl-CoA, one molecule of NADH and one molecule of CO2 is formed.
Citric acid cycle
[edit]
Main article: Citric acid cycle
The citric acid cycle is also called the Krebs cycle or the tricarboxylic acid cycle. When oxygen is present, acetyl-CoA is produced from the pyruvate molecules created from glycolysis. Once acetyl-CoA is formed, aerobic or anaerobic respiration can occur. When oxygen is present, the mitochondria will undergo aerobic respiration which leads to the Krebs cycle. However, if oxygen is not present, fermentation of the pyruvate molecule will occur. In the presence of oxygen, when acetyl-CoA is produced, the molecule then enters the citric acid cycle (Krebs cycle) inside the mitochondrial matrix, and is oxidized to CO2 while at the same time reducing NAD to NADH. NADH can be used by the electron transport chain to create further ATP as part of oxidative phosphorylation. To fully oxidize the equivalent of one glucose molecule, two acetyl-CoA must be metabolized by the Krebs cycle. Two low-energy waste products, H2O and CO2, are created during this cycle.
The citric acid cycle is an 8-step process involving 18 different enzymes and co-enzymes. During the cycle, acetyl-CoA (2 carbons) + oxaloacetate (4 carbons) yields citrate (6 carbons), which is rearranged to a more reactive form called isocitrate (6 carbons). Isocitrate is modified to become α-ketoglutarate (5 carbons), succinyl-CoA, succinate, fumarate, malate and, finally, oxaloacetate.
The net gain from one cycle is 3 NADH and 1 FADH2 as hydrogen (proton plus electron) carrying compounds and 1 high-energy GTP, which may subsequently be used to produce ATP. Thus, the total yield from 1 glucose molecule (2 pyruvate molecules) is 6 NADH, 2 FADH2, and 2 ATP.: 90–91
Oxidative phosphorylation
[edit]
Main articles: Oxidative phosphorylation, Electron transport chain, Electrochemical gradient, and ATP synthase
In eukaryotes, oxidative phosphorylation occurs in the mitochondrial cristae. It comprises the electron transport chain that establishes a proton gradient (chemiosmotic potential) across the boundary of the inner membrane by oxidizing the NADH produced from the Krebs cycle. ATP is synthesized by the ATP synthase enzyme when the chemiosmotic gradient is used to drive the phosphorylation of ADP. The electrons are finally transferred to exogenous oxygen and, with the addition of two protons, water is formed.
Efficiency of ATP production
[edit]
The table below describes the reactions involved when one glucose molecule is fully oxidized into carbon dioxide. It is assumed that all the reduced coenzymes are oxidized by the electron transport chain and used for oxidative phosphorylation.
| Step | coenzyme yield | ATP yield | Source of ATP |
--- --- |
| Glycolysis preparatory phase | | −2 | Phosphorylation of glucose and fructose 6-phosphate uses two ATP from the cytoplasm. |
| Glycolysis pay-off phase | | 4 | Substrate-level phosphorylation |
| 2 NADH | 3 or 5 | Oxidative phosphorylation: Each NADH produces net 1.5 ATP (instead of usual 2.5) due to NADH transport over the mitochondrial membrane |
| Oxidative decarboxylation of pyruvate | 2 NADH | 5 | Oxidative phosphorylation |
| Krebs cycle | | 2 | Substrate-level phosphorylation |
| 6 NADH | 15 | Oxidative phosphorylation |
| 2 FADH2 | 3 | Oxidative phosphorylation |
| Total yield | 30 or 32 ATP | From the complete oxidation of one glucose molecule to carbon dioxide and oxidation of all the reduced coenzymes. |
Although there is a theoretical yield of 38 ATP molecules per glucose during cellular respiration, such conditions are generally not realized because of losses such as the cost of moving pyruvate (from glycolysis), phosphate, and ADP (substrates for ATP synthesis) into the mitochondria. All are actively transported using carriers that utilize the stored energy in the proton electrochemical gradient.[citation needed]
Pyruvate is taken up by a specific, low Km transporter to bring it into the mitochondrial matrix for oxidation by the pyruvate dehydrogenase complex.
The phosphate carrier (PiC) mediates the electroneutral exchange (antiport) of phosphate (H2PO−4; Pi) for OH− or symport of phosphate and protons (H+) across the inner membrane, and the driving force for moving phosphate ions into the mitochondria is the proton motive force.
The ATP-ADP translocase (also called adenine nucleotide translocase, ANT) is an antiporter and exchanges ADP and ATP across the inner membrane. The driving force is due to the ATP (−4) having a more negative charge than the ADP (−3), and thus it dissipates some of the electrical component of the proton electrochemical gradient.
The outcome of these transport processes using the proton electrochemical gradient is that more than 3 H+ are needed to make 1 ATP. Obviously, this reduces the theoretical efficiency of the whole process and the likely maximum is closer to 28–30 ATP molecules. In practice the efficiency may be even lower because the inner membrane of the mitochondria is slightly leaky to protons. Other factors may also dissipate the proton gradient creating an apparently leaky mitochondria. An uncoupling protein known as thermogenin is expressed in some cell types and is a channel that can transport protons. When this protein is active in the inner membrane it short circuits the coupling between the electron transport chain and ATP synthesis. The potential energy from the proton gradient is not used to make ATP but generates heat. This is particularly important in brown fat thermogenesis of newborn and hibernating mammals.[citation needed]
According to some newer sources, the ATP yield during aerobic respiration is not 36–38, but only about 30–32 ATP molecules / 1 molecule of glucose , because:
ATP : NADH+H+ and ATP : FADH2 ratios during the oxidative phosphorylation appear to be not 3 and 2, but 2.5 and 1.5 respectively. Unlike in the substrate-level phosphorylation, the stoichiometry here is difficult to establish.
ATP synthase produces 1 ATP / 3 H+. However the exchange of matrix ATP for cytosolic ADP and Pi (antiport with OH− or symport with H+) mediated by ATP–ADP translocase and phosphate carrier consumes 1 H+ / 1 ATP as a result of regeneration of the transmembrane potential changed during this transfer, so the net ratio is 1 ATP : 4 H+.
The mitochondrial electron transport chain proton pump transfers across the inner membrane 10 H+ / 1 NADH+H+ (4 + 2 + 4) or 6 H+ / 1 FADH2 (2 + 4).
: So the final stoichiometry is
: 1 NADH+H+ : 10 H+ : 10/4 ATP = 1 NADH+H+ : 2.5 ATP
: 1 FADH2 : 6 H+ : 6/4 ATP = 1 FADH2 : 1.5 ATP
ATP : NADH+H+ coming from glycolysis ratio during the oxidative phosphorylation is
1.5, as for FADH2, if hydrogen atoms (2H++2e−) are transferred from cytosolic NADH+H+ to mitochondrial FAD by the glycerol phosphate shuttle located in the inner mitochondrial membrane.
2.5 in case of malate-aspartate shuttle transferring hydrogen atoms from cytosolic NADH+H+ to mitochondrial NAD+
So finally we have, per molecule of glucose
Substrate-level phosphorylation: 2 ATP from glycolysis + 2 ATP (directly GTP) from Krebs cycle
Oxidative phosphorylation
2 NADH+H+ from glycolysis: 2 × 1.5 ATP (if glycerol phosphate shuttle transfers hydrogen atoms) or 2 × 2.5 ATP (malate-aspartate shuttle)
2 NADH+H+ from the oxidative decarboxylation of pyruvate and 6 from Krebs cycle: 8 × 2.5 ATP
2 FADH2 from the Krebs cycle: 2 × 1.5 ATP
Altogether this gives 4 + 3 (or 5) + 20 + 3 = 30 (or 32) ATP per molecule of glucose
These figures may still require further tweaking as new structural details become available. The above value of 3 H+ / ATP for the synthase assumes that the synthase translocates 9 protons, and produces 3 ATP, per rotation. The number of protons depends on the number of c subunits in the Fo c-ring, and it is now known that this is 10 in yeast Fo and 8 for vertebrates. Including one H+ for the transport reactions, this means that synthesis of one ATP requires 1 + 10/3 = 4.33 protons in yeast and 1 + 8/3 = 3.67 in vertebrates. This would imply that in human mitochondria the 10 protons from oxidizing NADH would produce 2.72 ATP (instead of 2.5) and the 6 protons from oxidizing succinate or ubiquinol would produce 1.64 ATP (instead of 1.5). This is consistent with experimental results within the margin of error described in a recent review.
The total ATP yield in ethanol or lactic acid fermentation is only 2 molecules coming from glycolysis, because pyruvate is not transferred to the mitochondrion and finally oxidized to the carbon dioxide (CO2), but reduced to ethanol or lactic acid in the cytoplasm.
Fermentation
[edit]
Main article: Fermentation
Without oxygen, pyruvate (pyruvic acid) is not metabolized by cellular respiration but undergoes a process of fermentation. The pyruvate is not transported into the mitochondrion but remains in the cytoplasm, where it is converted to waste products that may be removed from the cell. This serves the purpose of oxidizing the electron carriers so that they can perform glycolysis again and removing the excess pyruvate. Fermentation oxidizes NADH to NAD+ so it can be re-used in glycolysis. In the absence of oxygen, fermentation prevents the buildup of NADH in the cytoplasm and provides NAD+ for glycolysis. This waste product varies depending on the organism. In skeletal muscles, the waste product is lactic acid. This type of fermentation is called lactic acid fermentation. In strenuous exercise, when energy demands exceed energy supply, the respiratory chain cannot process all of the hydrogen atoms joined by NADH. During anaerobic glycolysis, NAD+ regenerates when pairs of hydrogen combine with pyruvate to form lactate. Lactate formation is catalyzed by lactate dehydrogenase in a reversible reaction. Lactate can also be used as an indirect precursor for liver glycogen. During recovery, when oxygen becomes available, NAD+ attaches to hydrogen from lactate to form ATP. In yeast, the waste products are ethanol and carbon dioxide. This type of fermentation is known as alcoholic or ethanol fermentation. The ATP generated in this process is made by substrate-level phosphorylation, which does not require oxygen.[citation needed]
Fermentation is less efficient at using the energy from glucose: only 2 ATP are produced per glucose, compared to the 38 ATP per glucose nominally produced by aerobic respiration. Glycolytic ATP, however, is produced more quickly. For prokaryotes to continue a rapid growth rate when they are shifted from an aerobic environment to an anaerobic environment, they must increase the rate of the glycolytic reactions. For multicellular organisms, during short bursts of strenuous activity, muscle cells use fermentation to supplement the ATP production from the slower aerobic respiration, so fermentation may be used by a cell even before the oxygen levels are depleted, as is the case in sports that do not require athletes to pace themselves, such as sprinting.[citation needed]
Anaerobic respiration
[edit]
Main article: Anaerobic respiration
Anaerobic respiration is used by microorganisms, either bacteria or archaea, in which neither oxygen (aerobic respiration) nor pyruvate derivatives (fermentation) is the final electron acceptor. Rather, an inorganic acceptor such as sulfate (SO2−4), nitrate (NO−3), or sulfur (S) is used. Such organisms could be found in unusual places such as underwater caves or near hydrothermal vents at the bottom of the ocean,: 66–68 in digestive tracts, as well as in anoxic soils or sediment in wetland ecosystems.
In July 2019, a scientific study of Kidd Mine in Canada discovered sulfur-breathing organisms which live 7900 feet (2400 meters) below the surface. These organisms are also remarkable because they consume minerals such as pyrite as their food source.
See also
[edit]
Maintenance respiration: maintenance as a functional component of cellular respiration
Microphysiometry
Pasteur point
Respirometry: research tool to explore cellular respiration
Tetrazolium chloride: cellular respiration indicator
Complex 1: NADH:ubiquinone oxidoreductes
References
[edit]
^ Bailey, Regina. "Cellular Respiration". Archived from the original on 2012-05-05.
^ a b "Metabolism Without Oxygen - OpenStax Biology 2E". openstax.org. 28 March 2018. Retrieved 2025-03-21.
^ "How much ATP is produced in aerobic respiration".
^ a b c Rich, P. R. (2003). "The molecular machinery of Keilin's respiratory chain". Biochemical Society Transactions. 31 (Pt 6): 1095–1105. doi:10.1042/BST0311095. PMID 14641005.
^ Buckley, Gabe (2017-01-12). "Krebs Cycle - Definition, Products and Location". Biology Dictionary. Retrieved 2025-01-31.
^ O'Leary, Brendan M.; Plaxton, William C. (2016). "Plant Respiration". eLS. pp. 1–11. doi:10.1002/9780470015902.a0001301.pub3. ISBN 9780470016176.
^ a b c d Mannion, A. M. (12 January 2006). Carbon and Its Domestication. Springer. ISBN 978-1-4020-3956-0.
^ Reece, Jane; Urry, Lisa; Cain, Michael; Wasserman, Steven; Minorsky, Peter; Jackson, Robert (2010). Campbell Biology Ninth Edition. Pearson Education, Inc. p. 168.
^ Chaudhry, Raheel; Varacallo, Matthew A. (2025), "Biochemistry, Glycolysis", StatPearls, Treasure Island (FL): StatPearls Publishing, PMID 29493928, retrieved 2025-01-31
^ Sapkota, Anupama (2024-10-17). "Krebs Cycle: Steps, Enzymes, Products & Diagram". microbenotes.com. Retrieved 2025-02-01.
^ a b R. Caspi (2012-11-14). "Pathway: TCA cycle III (animals)". MetaCyc Metabolic Pathway Database. Retrieved 2022-06-20.
^ a b R. Caspi (2011-12-19). "Pathway: TCA cycle I (prokaryotic)". MetaCyc Metabolic Pathway Database. Retrieved 2022-06-20.
^ Haddad, Aida; Mohiuddin, Shamim S. (2025), "Biochemistry, Citric Acid Cycle", StatPearls, Treasure Island (FL): StatPearls Publishing, PMID 31082116, retrieved 2025-02-01
^ Deshpande, Ojas A.; Mohiuddin, Shamim S. (2025), "Biochemistry, Oxidative Phosphorylation", StatPearls, Treasure Island (FL): StatPearls Publishing, PMID 31985985, retrieved 2025-02-01
^ Porter, R.; Brand, M. (1 September 1995). "Mitochondrial proton conductance and H+/O ratio are independent of electron transport rate in isolated hepatocytes". The Biochemical Journal (Free full text). 310 (Pt 2): 379–382. doi:10.1042/bj3100379. ISSN 0264-6021. PMC 1135905. PMID 7654171.
^ a b c Stryer, Lubert (1995). Biochemistry (fourth ed.). New York – Basingstoke: W. H. Freeman and Company. ISBN 978-0716720096.
^ Stock, Daniela; Leslie, Andrew G. W.; Walker, John E. (1999). "Molecular architecture of the rotary motor in ATP synthase". Science. 286 (5445): 1700–5. doi:10.1126/science.286.5445.1700. PMID 10576729.
^ Watt, Ian N.; Montgomery, Martin G.; Runswick, Michael J.; Leslie, Andrew G. W.; Walker, John E. (2010). "Bioenergetic Cost of Making an Adenosine Triphosphate Molecule in Animal Mitochondria". Proc. Natl. Acad. Sci. USA. 107 (39): 16823–16827. Bibcode:2010PNAS..10716823W. doi:10.1073/pnas.1011099107. PMC 2947889. PMID 20847295.
^ P.Hinkle (2005). "P/O ratios of mitochondrial oxidative phosphorylation". Biochimica et Biophysica Acta (BBA) - Bioenergetics. 1706 (1–2): 1–11. doi:10.1016/j.bbabio.2004.09.004. PMID 15620362.
^ Lumen Boundless Microbiology. "Anaerobic Respiration-Electron Donors and Acceptors in Anaerobic Respiration". courses.lumenlearning.org. Boundless.com. Retrieved November 19, 2020. Anaerobic respiration is the formation of ATP without oxygen. This method still incorporates the respiratory electron transport chain, but without using oxygen as the terminal electron acceptor. Instead, molecules such as sulfate (SO2−4), nitrate (NO−3), or sulfur (S) are used as electron acceptors
^ Lollar, Garnet S.; Warr, Oliver; Telling, Jon; Osburn, Magdalena R.; Sherwood Lollar, Barbara (2019). "'Follow the Water': Hydrogeochemical Constraints on Microbial Investigations 2.4 km Below Surface at the Kidd Creek Deep Fluid and Deep Life Observatory". Geomicrobiology Journal. 36 (10): 859–872. Bibcode:2019GmbJ...36..859L. doi:10.1080/01490451.2019.1641770. S2CID 199636268.
^ World's Oldest Groundwater Supports Life Through Water-Rock Chemistry Archived 2019-09-10 at the Wayback Machine, July 29, 2019, deepcarbon.net.
^ Strange life-forms found deep in a mine point to vast 'underground Galapagos' Archived 2019-09-09 at the Wayback Machine, By Corey S. Powell, Sept. 7, 2019, nbcnews.com.
External links
[edit]
Library resources about Cellular respiration
Resources in your library
A detailed description of respiration vs. fermentation
Kimball's online resource for cellular respiration
Cellular Respiration and Fermentation at Clermont College
| v t e Metabolism, catabolism, anabolism |
| General | Metabolic pathway Metabolic network Primary nutritional groups |
| Energy metabolism | | | | --- | | Aerobic respiration | Glycolysis → Pyruvate decarboxylation → Citric acid cycle → Oxidative phosphorylation (electron transport chain + ATP synthase) | | Anaerobic respiration | Electron acceptors other than oxygen | | Fermentation | Glycolysis → Substrate-level phosphorylation + ABE + Ethanol + Lactic acid | |
| Specific paths | | | | | | | | --- --- --- | | Protein metabolism | Protein synthesis Catabolism (protein→peptide→amino acid) | | | --- | | Amino acid | Amino acid synthesis Amino acid degradation (amino acid→pyruvate, acetyl CoA, or TCA intermediate) Urea cycle | | Nucleotide metabolism | Purine metabolism Nucleotide salvage Pyrimidine metabolism Purine nucleotide cycle | | | Carbohydrate metabolism(carbohydrate catabolismand anabolism) | | | | | | | | --- --- --- | | Human | | | | Glycolysis ⇄ Gluconeogenesis | | Glycogenolysis ⇄ Glycogenesis | | Pentose phosphate pathway Fructolysis + Polyol pathway Galactolysis + Leloir pathway | | Glycosylation + N-linked + O-linked | | | Nonhuman | | | | Photosynthesis Anoxygenic photosynthesis Chemosynthesis Carbon fixation DeLey-Doudoroff pathway Entner-Doudoroff pathway | | Xylose metabolism Radiotrophism | | | | Lipid metabolism (lipolysis, lipogenesis) | | | | --- | | Fatty acid metabolism | Fatty acid degradation (Beta oxidation) Fatty acid synthesis | | Other | Steroid metabolism Sphingolipid metabolism Eicosanoid metabolism Ketosis Reverse cholesterol transport | | | Other | Metal metabolism + Iron metabolism Ethanol metabolism Phospagen system (ATP-PCr) Chlororespiration | |
| v t e Metabolism map |
| | | | Carbonfixation Photo-respiration Pentosephosphatepathway Citricacid cycle Glyoxylatecycle Ureacycle Fattyacidsynthesis Fattyacidelongation Betaoxidation Peroxisomal betaoxidation Glyco-genolysis Glyco-genesis Glyco-lysis Gluconeo-genesis Pyruvatedecarb-oxylation Fermentation Keto-lysis Keto-genesis feeders togluconeo-genesis Direct / C4 / CAMcarbon intake Light reaction Oxidativephosphorylation Amino aciddeamination Citrateshuttle Lipogenesis Lipolysis Steroidogenesis MVA pathway MEP pathway Shikimatepathway Transcription &replication Translation Proteolysis Glycosyl-ation Sugaracids Double/multiplesugars & glycans Simplesugars Inositol-P Amino sugars& sialic acids Nucleotide sugars Hexose-P Triose-P Glycerol P-glycerates Pentose-P Tetrose-P Propionyl-CoA Succinate Acetyl-CoA Pentose-P P-glycerates Glyoxylate Photosystems Pyruvate Lactate Acetyl-CoA Citrate Oxalo-acetate Malate Succinyl-CoA α-Keto-glutarate Ketonebodies Respiratorychain Serine group Alanine Branched-chainamino acids Aspartategroup Homoserinegroup& lysine Glutamategroup& proline Arginine Creatine& polyamines Ketogenic &glucogenicamino acids Amino acids Shikimate Aromatic aminoacids & histidine Ascorbate(vitamin C) δ-ALA Bilepigments Hemes Cobalamins (vitamin B12) Variousvitamin Bs Calciferols(vitamin D) Retinoids(vitamin A) Quinones (vitamin K)& tocopherols (vitamin E) Cofactors Vitamins& minerals Antioxidants PRPP Glycoproteins& proteoglycans Acetyl-CoA Terpenoids& carotenoids (vitamin A) Bile acids Polyunsaturatedfatty acids Neurotransmitters& thyroid hormones | Major metabolic pathways in metro-style map. Click any text (name of pathway or metabolites) to link to the corresponding article.Single lines: pathways common to most lifeforms. Double lines: pathways not in humans (occurs in e.g. plants, fungi, prokaryotes). Orange nodes: carbohydrate metabolism. Violet nodes: photosynthesis. Red nodes: cellular respiration. Pink nodes: cell signaling. Blue nodes: amino acid metabolism. Grey nodes: vitamin and cofactor metabolism. Brown nodes: nucleotide and protein metabolism. Green nodes: lipid metabolism. |
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Lecture 13: Lagrange Multipliers
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Last time we saw things about gradients and directional derivatives. Before that we studied how to look for minima and maxima of functions of several variables.
And today we are going to look again at min/max problems but in a different setting, namely, one for variables that are not independent. And so what we will see is you may have heard of Lagrange multipliers.
And this is the one point in the term when I can shine with my French accent and say Lagrange's name properly.
OK. What are Lagrange multipliers about? Well, the goal is to minimize or maximize a function of several variables.
Let's say, for example, f of x, y, z,but where these variables are no longer independent.
They are not independent. That means that there is a relation between them. The relation is maybe some equation of the form g of x, y, z equals some constant.
You take the relation between x, y, z, you call that g and that gives you the constraint. And your goal is to minimize f only of those values of x, y, z that satisfy the constraint. What is one way to do that?
Well, one to do that, if the constraint is very simple, we can maybe solve for one of the variables.
Maybe we can solve this equation for one of the variables, plug it back into f, and then we have a usual min/max problem that we have seen how to do.
The problem is sometimes you cannot actually solve for x,y, z in here because this condition is too complicated and then we need a new method. That is what we are going to do.
Why would we care about that? Well, one example is actually in physics. Maybe you have seen in thermodynamics that you study quantities about gases,and those quantities that involve pressure,volume and temperature. And pressure,volume and temperature are not independent of each other.
I mean you know probably the equation PV = NRT.
And, of course, there you could actually solve to express things in terms of one or the other.
But sometimes it is more convenient to keep all three variables but treat them as constrained.
It is just an example of a situation where you might want to do this. Anyway, we will look mostly at particular examples, but just to point out that this is useful when you study guesses in physics.
The first observation is we cannot use our usual method of looking for critical points of f.
Because critical points of f typically will not satisfy this condition and so won't be good solutions.
We need something else. Let's look at an example,and we will see how that leads us to the method.
For example, let's say that I want to find the point closest to the origin -- -- on the hyperbola xy equals 3 in the plane. That means I have this hyperbola, and I am asking myself what is the point on it that is the closest to the origin?
I mean we can solve this by elementary geometry,we don't need actually Lagrange multipliers,but we are going to do it with Lagrange multipliers because it is a pretty good example. What does it mean?
Well, it means that we want to minimize distance to the origin.
What is the distance to the origin?
If I have a point, at coordinates (x,y) and then the distance to the origin is square root of x squared plus y squared. Well, do we really want to minimize that or can we minimize something easier?
Yeah. Maybe we can minimize the square of a distance. Let's forget this guy and instead -- Actually, we will minimize f of x,y equals x squared plus y squared,that looks better, subject to the constraint xy =3. And so we will call this thing g of x, y to illustrate the general method.
Let's look at a picture. Here you can see in yellow the hyperbola xy equals three. And we are going to look for the points that are the closest to the origin.
What can we do? Well, for example,we can plot the function x squared plus y squared,function f. That is the contour plot of f with a hyperbola on top of it. Now let's see what we can do with that. Well, let's ask ourselves,for example, if I look at points where f equals 20 now. I think I am at 20 but you cannot really see it. That is a circle with a point whose distant square is 20. Well, can I find a solution if I am on the hyperbola? Yes, there are four points at this distance. Can I do better?
Well, let's decrease for distance.
Yes, we can still find points on the hyperbola and so on.
Except if we go too low then there are no points on this circle anymore in the hyperbola. If we decrease the value of f that we want to look at that will somehow limit value beyond which we cannot go, and that is the minimum of f.
We are trying to look for the smallest value of f that will actually be realized on the hyperbola.
When does that happen? Well, I have to backtrack a little bit. It seems like the limiting case is basically here. It is when the circle is tangent to the hyperbola. That is the smallest circle that will hit the hyperbola. If I take a larger value of f,I will have solutions. If I take a smaller value of f,I will not have any solutions anymore.
So, that is the situation that we want to solve for.
How do we find that minimum? Well, a key observation that is valid on this picture, and that actually remain true in the completely general case, is that when we have a minimum the level curve of f is actually tangent to our hyperbola.
It is tangent to the set of points where x,y equals three, to the hyperbola.
Let's write that down. We observe that at the minimum the level curve of f is tangent to the hyperbola.
Remember, the hyperbola is given by the equal g equals three, so it is a level curve of g.
We have a level curve of f and a level curve of g that are tangent to each other. And I claim that is going to be the general situation that we are interested in.
How do we try to solve for points where this happens?
How do we find x, y where the level curves of f and g are tangent to each other? Let's think for a second.
If the two level curves are tangent to each other that means they have the same tangent line. That means that the normal vectors should be parallel. Let me maybe draw a picture here. This is the level curve maybe f equals something. And this is the level curve g equals constant. Here my constant is three.
Well, if I look for gradient vectors, the gradient of f will be perpendicular to the level curve of f.
The gradient of g will be perpendicular to the level curve of g. They don't have any reason to be of the same size, but they have to be parallel to each other. Of course, they could also be parallel pointing in opposite directions.
But the key point is that when this happens the gradient of f is parallel to the gradient of g.
Well, let's check that. Here is a point.
And I can plot the gradient of f in blue.
The gradient of g in yellow. And you see,in most of these places, somehow the two gradients are not really parallel. Actually, I should not be looking at random points. I should be looking only on the hyperbola. I want points on the hyperbola where the two gradients are parallel.
Well, when does that happen? Well, it looks like it will happen here. When I am at a minimum,the two gradient vectors are parallel.
It is not really proof. It is an example that seems to be convincing. So far things work pretty well.
How do we decide if two vectors are parallel?
Well, they are parallel when they are proportional to each other. You can write one of them as a constant times the other one, and that constant usually one uses the Greek letter lambda. I don't know if you have seen it before. It is the Greek letter for L.
And probably, I am sure, it is somebody's idea of paying tribute to Lagrange by putting an L in there. Lambda is just a constant.
And we are looking for a scalar lambda and points x and y where this holds. In fact,what we are doing is replacing min/max problems in two variables with a constraint between them by a set of equations involving, you will see, three variables.
We had min/max with two variables x, y,but no independent. We had a constraint g of x,y equals constant. And that becomes something new.
That becomes a system of equations where we have to solve, well, let's write down what it means for gradient f to be proportional to gradient g. That means that f sub x should be lambda times g sub x, and f sub y should be lambda times g sub y. Because the gradient vectors here are f sub x, f sub y and g sub x,g sub y. If you have a third variable z then you have also an equation f sub z equals lambda g sub z.
Now, let's see. How many unknowns do we have in these equations? Well, there is x,there is y and there is lambda. We have three unknowns and have only two equations. Something is missing.
Well, I mean x and y are not actually independent.
They are related by the equation g of x,y equals c, so we need to add the constraint g equals c.
And now we have three equations involving three variables.
Let's see how that works. Here remember we have f equals x squared y squared and g = xy. What is f sub x?
It is going to be 2x equals lambda times,what is g sub x, y.
Maybe I should write here f sub x equals lambda g sub x just to remind you. Then we have f sub y equals lambda g sub y. F sub y is 2y equals lambda times g sub y is x. And then our third equation g equals c becomes xy equals three.
So, that is what you would have to solve.
Any questions at this point? No.
Yes? How do I know the direction of a gradient? Do you mean how do I know that it is perpendicular to a level curve?
Oh, how do I know if it points in that direction on the opposite one? Well, that depends.
I mean we'd seen in last time, but the gradient is perpendicular to the level and points towards higher values of a function. So it could be -- Wait.
What did I have? It could be that my gradient vectors up there actually point in opposite directions.
It doesn't matter to me because it will still look the same in terms of the equation, just lambda will be positive or negative, depending on the case. I can handle both situations.
It's not a problem. I can allow lambda to be positive or negative. Well, in this example,it looks like lambda will be positive.
If you look at the picture on the plot.
Yes? Well, because actually they are not equal to each other. If you look at this point where the hyperbola and the circle touch each other,first of all, I don't know which circle I am going to look at. I am trying to solve,actually, for the radius of the circle.
I am trying to find what the minimum value of f is.
And, second, at that point,the value of f and the value of g are not equal.
g is equal to three because I want the hyperbola x equals three. The value of f will be the square of a distance, whatever that is.
I think it will end up being 6, but we will see.
So, you cannot really set them equal because you don't know what f is equal to in advance. Yes?
Not quite. Actually, here I am just using this idea of finding a point closest to the origin to illustrate an example of a min/max problem.
The general problem we are trying to solve is minimize f subject to g equals constant. And what we are going to do for that is we are really going to say instead let's look at places where gradient f and gradient g are parallel to each other and solve for equations of that. I think we completely lose the notion of closest point if we just look at these equations.
We don't really say anything about closest points anymore.
Of course, that is what they mean in the end.
But, in the general setting, there is no closest point involved anymore. OK.
Yes? Yes.
It is always going to be the case that,at the minimum, or at the maximum of a function subject to a constraint, the level curves of f and the level curves of g will be tangent to each other.
That is the basis for this method.
I am going to justify that soon. It could be minimum or maximum.
In three-dimensions it could even be a saddle point.
And, in fact, I should say in advance,this method will not tell us whether it is a minimum or a maximum. We do not have any way of knowing, except for testing values.
We cannot use second derivative tests or anything like that.
I will get back to that. Yes?
Yes. Here you can set y equals to favor x. Then you can minimize x squared plus nine over x squared. In general, if I am trying to solve a more complicated problem, I might not be able to solve. I am doing an example where,indeed, here you could solve and remove one variable,but you cannot always do that. And this method will still work.
The other one won't. OK.
I don't see any other questions. Are there any other questions?
No. OK.
I see a lot of students stretching and so on,so it is very confusing for me. How do we solve these equations?
Well, the answer is in general we might be in deep trouble.
There is no general method for solving the equations that you get from this method. You just have to think about them. Sometimes it will be very easy.
Sometimes it will be so hard that you cannot actually do it without the computer. Sometimes it will be just hard enough to be on Part B of this week's problem set.
I claim in this case we can actually do it without so much trouble, because actually we can think of this as a two by two linear system in x and y. Well, let me do something.
Let me rewrite the first two equations as 2x - lambda y = 0.
And lambda x - 2y = 0. And xy = 3.
That is what we want to solve. Well, I can put this into matrix form. Two minus lambda,lambda minus two times x, y equals 0,0.
Now, how do I solve a linear system matrix times x,y equals zero? Well, I always have an obvious solution. X and y both equal to zero.
Is that a good solution? No, because zero times zero is not three. We want another solution,the trivial solution. 0,0 does not solve the constraint equation xy equals three, so we want another solution. When do we have another solution? Well, when the determinant of a matrix is zero. We have other solutions that exist only if determinant of a matrix is zero.
M is this guy. Let's compute the determinant.
Well, that seems to be negative four plus lambda squared.
That is zero exactly when lambda squared equals four,which is lambda is plus or minus two.
Already you see here it is a the level of difficulty that is a little bit much for an exam but perfectly fine for a problem set or for a beautiful lecture like this one.
How do we deal with -- Well, we have two cases to look at.
Lambda equals two or lambda equals minus two.
Let's start with lambda equals two.
If I set lambda equals two, what does this equation become?
Well, it becomes x equals y. This one becomes y equals x.
Well, they seem to be the same. x equals y.
And then the equation xy equals three becomes,well, x squared equals three. I have two solutions.
One is x equals root three and, therefore, y equals root three as well, or negative root three and negative root three.
Let's look at the other case. If I set lambda equal to negative two then I get 2x equals negative 2y.
That means x equals negative y. The second one,2y equals negative 2x. That is y equals negative x.
Well, that is the same thing. And xy equals three becomes negative x squared equals three. Can we solve that?
No. There are no solutions here.
Now we have two candidate points which are these two points, root three, root three or negative root three, negative root three. OK.
Let's actually look at what we have here.
Maybe you cannot read the coordinates, but the point that I have here is indeed root three, root three.
How do we see that lambda equals two?
Well, if you look at this picture, the gradient of f,that is the blue vector, is indeed twice the yellow vector, gradient g. That is where you read the value of lambda. And we have the other solution which is somewhere here. Negative root three,negative root there. And there, again,lambda equals two. The two vectors are proportional by a factor of two. Yes?
No, solutions are not quite guaranteed to be absolute minima or maxima. They are guaranteed to be somehow critical points end of a constraint.
That means if you were able to solve and eliminate the variable that would be a critical point. When you have the same problem,as we have critical points, are they maxima or minima?
And the answer is, well, we won't know until we check. More questions?
No. Yes?
What is a Lagrange multiplier? Well, it is this number lambda that is called the multiplier here.
It is a multiplier because it is what you have to multiply gradient of g by to get gradient of f.
It multiplies.
Let's try to see why is this method valid?
Because so far I have shown you pictures and have said see they are tangent. But why is it that they have to be tangent in general? Let's think about it.
Let's say that we are at constrained min or max.
What that means is that if I move on the level g equals constant then the value of f should only increase or only decrease. But it means,in particular, to first order it will not change. At an unconstrained min or max,partial derivatives are zero. In this case,derivatives are zero only in the allowed directions.
And the allowed directions are those that stay on the levels of this g equals constant. In any direction along the level set g = c the rate of change of f must be zero.
That is what happens at minima or maxima.
Except here, of course, we look only at the allowed directions. Let's say the same thing in terms of directional derivatives.
That means for any direction that is tangent to the constraint level g equal c, we must have df over ds in the direction of u equals zero. I will draw a picture.
Let's say now I am in three variables just to give you different examples. Here I have a level surface g equals c. I am at my point.
And if I move in any direction that is on the level surface,so I move in the direction u tangent to the level surface,then the rate of change of f in that direction should be zero.
Now, remember what the formula is for this guy.
Well, we have seen that this guy is actually radiant f dot u.
That means any such vector u must be perpendicular to the gradient of f. That means that the gradient of f should be perpendicular to anything that is tangent to this level. That means the gradient of f should be perpendicular to the level set.
That is what we have shown.
But we know another vector that is also perpendicular to the level set of g. That is the gradient of g.
We conclude that the gradient of f must be parallel to the gradient of g because both are perpendicular to the level set of g. I see confused faces,so let me try to tell you again where that comes from.
We said if we had a constrained minimum or maximum,if we move in the level set of g, f doesn't change.
Well, it doesn't change to first order.
It is the same idea as when you are looking for a minimum you set the derivative equal to zero.
So the derivative in any direction, tangent to g equals c, should be the directional derivative of f,in any such direction, should be zero.
That is what we mean by critical point of f.
And so that means that any vector u, any unit vector tangent to the level set of g is going to be perpendicular to the gradient of f. That means that the gradient of f is perpendicular to the level set of g.
If you want, that means the level sets of f and g are tangent to each other. That is justifying what we have observed in the picture that the two level sets have to be tangent to each other at the prime minimum or maximum.
Does that make a little bit of sense?
Kind of. I see at least a few faces nodding so I take that to be a positive answer.
Since I have been asked by several of you,how do I know if it is a maximum or a minimum?
Well, warning, the method doesn't tell whether a solution is a minimum or a maximum.
How do we do it? Well, more bad news.
We cannot use the second derivative test.
And the reason for that is that we care actually only about these specific directions that are tangent to variable of g.
And we don't want to bother to try to define directional second derivatives. Not to mention that actually it wouldn't work. There is a criterion but it is much more complicated than that. Basically, the answer for us is that we don't have a second derivative test in this situation. What are we left with?
Well, we are just left with comparing values.
Say that in this problem you found a point where f equals three, a point where f equals nine, a point where f equals 15.
Well, then probably the minimum is the point where f equals three and the maximum is 15. Actually, in this case,where we found minima, these two points are tied for minimum. What about the maximum?
What is the maximum of f on the hyperbola?
Well, it is infinity because the point can go as far as you want from the origin. But the general idea is if we have a good reason to believe that there should be a minimum,and it's not like at infinity or something weird like that,then the minimum will be a solution of the Lagrange multiplier equations. We just look for all the solutions and then we choose the one that gives us the lowest value. Is that good enough?
Let me actually write that down.
To find the minimum or the maximum, we compare values of f at the various solutions -- -- to Lagrange multiplier equations.
I should say also that sometimes you can just conclude by thinking geometrically. In this case,when it is asking you which point is closest to the origin you can just see that your answer is the correct one.
Let's do an advanced example. Advanced means that -- Well,this one I didn't actually dare to put on top of the other problem sets. Instead, I am going to do it.
What is this going to be about? We are going to look for a surface minimizing pyramid. Let's say that we want to build a pyramid with a given triangular base -- -- and a given volume. Say that I have maybe in the x,y plane I am giving you some triangle.
And I am going to try to build a pyramid.
Of course, I can choose where to put the top of a pyramid.
This guy will end up being behind now.
And the constraint and the goal is to minimize the total surface area. The first time I taught this class, it was a few years ago, was just before they built the Stata Center. And then I used to motivate this problem by saying Frank Gehry has gone crazy and has been given a triangular plot of land he wants to put a pyramid.
There needs to be the right amount of volume so that you can put all the offices in there. And he wants it to be,actually, covered in solid gold.
And because that is expensive, the administration wants him to cut the costs a bit. And so you have to minimize the total size so that it doesn't cost too much.
We will see if MIT comes up with a triangular pyramid building. Hopefully not.
It could be our next dorm, you never know.
Anyway, it is a fine geometry problem.
Let's try to think about how we can do this.
The natural way to think about it would be -- Well,what do we have to look for first?
We have to look for the position of that top point.
Remember we know that the volume of a pyramid is one-third the area of base times height. In fact, fixing the volume,knowing that we have fixed the area of a base,means that we are fixing the height of the pyramid.
The height is completely fixed. What we have to choose just is where do we put that top point? Do we put it smack in the middle of a triangle or to a side or even anywhere we want?
Its z coordinate is fixed. Let's call h the height.
What we could do is something like this.
We say we have three points of a base.
Let's call them p1 at (x1, y1,0); p2 at (x2,y2,0); p3 at (x3, y3,0).
This point p is the unknown point at (x, y,h). We know the height.
And then we want to minimize the sum of the areas of these three triangles. One here, one here and one at the back. And areas of triangles we know how to express by using length of cross-product.
It becomes a function of x and y.
And you can try to minimize it. Actually, it doesn't quite work.
The formulas are just too complicated.
You will never get there. What happens is actually maybe we need better coordinates. Why do we need better coordinates? That is because the geometry is kind of difficult to do if you use x, y coordinates.
I mean formula for cross-product is fine,but then the length of the vector will be annoying and just doesn't look good. Instead, let's think about it differently.
I claim if we do it this way and we express the area as a function of x, y, well, actually we can't solve for a minimum. Here is another way to do it.
Well, what has worked pretty well for us so far is this geometric idea of base times height.
So let's think in terms of the heights of side triangles.
I am going to use the height of these things.
And I am going to say that the area will be the sum of three terms, which are three bases times three heights.
Let's give names to these quantities.
Actually, for that it is going to be good to have the point in the xy plane that lives directly below p.
Let's call it q. P is the point that coordinates x, y, h. And let's call q the point that is just below it and so it' coordinates are x,y, 0. Let's see.
Let me draw a map of this thing. p1, p2, p3 and I have my point q in the middle. Let's see.
To know these areas, I need to know the base.
Well, the base I can decide that I know it because it is part of my given data. I know the sides of this triangle. Let me call the lengths a1,a2, a3. I also need to know the height,so I need to know these lengths.
How do I know these lengths? Well, its distance in space,but it is a little bit annoying.
But maybe I can reduce it to a distance in the plane by looking instead at this distance here. Let me give names to the distances from q to the sides. Let's call u1,u2, u3 the distances from q to the sides.
Well, now I can claim I can find, actually,sorry. I need to draw one more thing.
I claim I have a nice formula for the area,because this is vertical and this is horizontal so this length here is u3, this length here is h.
So what is this length here? It is the square root of u3 squared plus h squared. And similarly for these other guys. They are square roots of a u squared plus h squared. The heights of the faces are square root of u1 squared times h squared.
And similarly with u2 and u3. So the total side area is going to be the area of the first faces,one-half of base times height, plus one-half of a base times a height plus one-half of the third one.
It doesn't look so much better. But, trust me,it will get better. Now, that is a function of three variables, u1, u2, u3.
And how do we relate u1, u2, u3 to each other?
They are probably not independent.
Well, let's cut this triangle here into three pieces like that. Then each piece has side --Well, let's look at it the piece of the bottom.
It has base a3, height u3. Cutting base into three tells you that the area of a base is one-half of a1,u1 plus one-half of a2, u2 plus one-half of a3,u3. And that is our constraint.
My three variables, u1, u2, u3, are constrained in this way. The sum of this figure must be the area of a base. And I want to minimize that guy.
So that is my g and that guy here is my f.
Now we try to apply our Lagrange multiplier equations.
Well, partial f of a partial u1 is -- Well,if you do the calculation, you will see it is one-half a1,u1 over square root of u1^2 plus h^2 equals lambda,what is partial g, partial a1?
That one you can do, I am sure. It is one-half a1.
Oh, these guys simplify. If you do the same with the second one -- -- things simplify again.
And the same with the third one. Well, you will get,after simplifying, u3 over square root of u3 squared plus h squared equals lambda.
Now, that means this guy equals this guy equals this guy.
They are all equal to lambda. And, if you think about it,that means that u1 = u2 = u3. See, it looked like scary equations but the solution is very simple.
What does it mean? It means that our point q should be equidistant from all three sides.
That is called the incenter. Q should be in the incenter.
The next time you have to build a golden pyramid and don't want to go broke, well, you know where to put the top.
If that was a bit fast, sorry. Anyway, it is not completely crucial. But go over it and you will see it works. Have a nice weekend.
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Topics covered: Lagrange multipliers
Instructor: Prof. Denis Auroux
Lecture 1: Dot Product
Lecture 2: Determinants
Lecture 3: Matrices
Lecture 4: Square Systems
Lecture 5: Parametric Equat...
Lecture 6: Kepler's Second Law
Lecture 7: Exam Review
Lecture 8: Partial Derivatives
Lecture 9: Max-Min and Leas...
Lecture 10: Second Derivati...
Lecture 11: Chain Rule
Lecture 12: Gradient
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Lecture 13: Lagrange Multip...
Lecture 14: Non-Independent...
Lecture 15: Partial Differe...
Lecture 16: Double Integrals
Lecture 17: Polar Coordinates
Lecture 18: Change of Varia...
Lecture 19: Vector Fields
Lecture 20: Path Independence
Lecture 21: Gradient Fields
Lecture 22: Green's Theorem
Lecture 23: Flux
Lecture 24: Simply Connecte...
Lecture 25: Triple Integrals
Lecture 26: Spherical Coord...
Lecture 27: Vector Fields i...
Lecture 28: Divergence Theorem
Lecture 29: Divergence Theo...
Lecture 30: Line Integrals
Lecture 31: Stokes' Theorem
Lecture 32: Stokes' Theorem...
Lecture 33: Maxwell's Equat...
Lecture 34: Final Review
Lecture 35: Final Review (c...
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Last time we saw things about gradients and directional derivatives. Before that we studied how to look for minima and maxima of functions of several variables.
And today we are going to look again at min/max problems but in a different setting, namely, one for variables that are not independent. And so what we will see is you may have heard of Lagrange multipliers.
And this is the one point in the term when I can shine with my French accent and say Lagrange's name properly.
OK. What are Lagrange multipliers about? Well, the goal is to minimize or maximize a function of several variables.
Let's say, for example, f of x, y, z,but where these variables are no longer independent.
They are not independent. That means that there is a relation between them. The relation is maybe some equation of the form g of x, y, z equals some constant.
You take the relation between x, y, z, you call that g and that gives you the constraint. And your goal is to minimize f only of those values of x, y, z that satisfy the constraint. What is one way to do that?
Well, one to do that, if the constraint is very simple, we can maybe solve for one of the variables.
Maybe we can solve this equation for one of the variables, plug it back into f, and then we have a usual min/max problem that we have seen how to do.
The problem is sometimes you cannot actually solve for x,y, z in here because this condition is too complicated and then we need a new method. That is what we are going to do.
Why would we care about that? Well, one example is actually in physics. Maybe you have seen in thermodynamics that you study quantities about gases,and those quantities that involve pressure,volume and temperature. And pressure,volume and temperature are not independent of each other.
I mean you know probably the equation PV = NRT.
And, of course, there you could actually solve to express things in terms of one or the other.
But sometimes it is more convenient to keep all three variables but treat them as constrained.
It is just an example of a situation where you might want to do this. Anyway, we will look mostly at particular examples, but just to point out that this is useful when you study guesses in physics.
The first observation is we cannot use our usual method of looking for critical points of f.
Because critical points of f typically will not satisfy this condition and so won't be good solutions.
We need something else. Let's look at an example,and we will see how that leads us to the method.
For example, let's say that I want to find the point closest to the origin -- -- on the hyperbola xy equals 3 in the plane. That means I have this hyperbola, and I am asking myself what is the point on it that is the closest to the origin?
I mean we can solve this by elementary geometry,we don't need actually Lagrange multipliers,but we are going to do it with Lagrange multipliers because it is a pretty good example. What does it mean?
Well, it means that we want to minimize distance to the origin.
What is the distance to the origin?
If I have a point, at coordinates (x,y) and then the distance to the origin is square root of x squared plus y squared. Well, do we really want to minimize that or can we minimize something easier?
Yeah. Maybe we can minimize the square of a distance. Let's forget this guy and instead -- Actually, we will minimize f of x,y equals x squared plus y squared,that looks better, subject to the constraint xy =3. And so we will call this thing g of x, y to illustrate the general method.
Let's look at a picture. Here you can see in yellow the hyperbola xy equals three. And we are going to look for the points that are the closest to the origin.
What can we do? Well, for example,we can plot the function x squared plus y squared,function f. That is the contour plot of f with a hyperbola on top of it. Now let's see what we can do with that. Well, let's ask ourselves,for example, if I look at points where f equals 20 now. I think I am at 20 but you cannot really see it. That is a circle with a point whose distant square is 20. Well, can I find a solution if I am on the hyperbola? Yes, there are four points at this distance. Can I do better?
Well, let's decrease for distance.
Yes, we can still find points on the hyperbola and so on.
Except if we go too low then there are no points on this circle anymore in the hyperbola. If we decrease the value of f that we want to look at that will somehow limit value beyond which we cannot go, and that is the minimum of f.
We are trying to look for the smallest value of f that will actually be realized on the hyperbola.
When does that happen? Well, I have to backtrack a little bit. It seems like the limiting case is basically here. It is when the circle is tangent to the hyperbola. That is the smallest circle that will hit the hyperbola. If I take a larger value of f,I will have solutions. If I take a smaller value of f,I will not have any solutions anymore.
So, that is the situation that we want to solve for.
How do we find that minimum? Well, a key observation that is valid on this picture, and that actually remain true in the completely general case, is that when we have a minimum the level curve of f is actually tangent to our hyperbola.
It is tangent to the set of points where x,y equals three, to the hyperbola.
Let's write that down. We observe that at the minimum the level curve of f is tangent to the hyperbola.
Remember, the hyperbola is given by the equal g equals three, so it is a level curve of g.
We have a level curve of f and a level curve of g that are tangent to each other. And I claim that is going to be the general situation that we are interested in.
How do we try to solve for points where this happens?
How do we find x, y where the level curves of f and g are tangent to each other? Let's think for a second.
If the two level curves are tangent to each other that means they have the same tangent line. That means that the normal vectors should be parallel. Let me maybe draw a picture here. This is the level curve maybe f equals something. And this is the level curve g equals constant. Here my constant is three.
Well, if I look for gradient vectors, the gradient of f will be perpendicular to the level curve of f.
The gradient of g will be perpendicular to the level curve of g. They don't have any reason to be of the same size, but they have to be parallel to each other. Of course, they could also be parallel pointing in opposite directions.
But the key point is that when this happens the gradient of f is parallel to the gradient of g.
Well, let's check that. Here is a point.
And I can plot the gradient of f in blue.
The gradient of g in yellow. And you see,in most of these places, somehow the two gradients are not really parallel. Actually, I should not be looking at random points. I should be looking only on the hyperbola. I want points on the hyperbola where the two gradients are parallel.
Well, when does that happen? Well, it looks like it will happen here. When I am at a minimum,the two gradient vectors are parallel.
It is not really proof. It is an example that seems to be convincing. So far things work pretty well.
How do we decide if two vectors are parallel?
Well, they are parallel when they are proportional to each other. You can write one of them as a constant times the other one, and that constant usually one uses the Greek letter lambda. I don't know if you have seen it before. It is the Greek letter for L.
And probably, I am sure, it is somebody's idea of paying tribute to Lagrange by putting an L in there. Lambda is just a constant.
And we are looking for a scalar lambda and points x and y where this holds. In fact,what we are doing is replacing min/max problems in two variables with a constraint between them by a set of equations involving, you will see, three variables.
We had min/max with two variables x, y,but no independent. We had a constraint g of x,y equals constant. And that becomes something new.
That becomes a system of equations where we have to solve, well, let's write down what it means for gradient f to be proportional to gradient g. That means that f sub x should be lambda times g sub x, and f sub y should be lambda times g sub y. Because the gradient vectors here are f sub x, f sub y and g sub x,g sub y. If you have a third variable z then you have also an equation f sub z equals lambda g sub z.
Now, let's see. How many unknowns do we have in these equations? Well, there is x,there is y and there is lambda. We have three unknowns and have only two equations. Something is missing.
Well, I mean x and y are not actually independent.
They are related by the equation g of x,y equals c, so we need to add the constraint g equals c.
And now we have three equations involving three variables.
Let's see how that works. Here remember we have f equals x squared y squared and g = xy. What is f sub x?
It is going to be 2x equals lambda times,what is g sub x, y.
Maybe I should write here f sub x equals lambda g sub x just to remind you. Then we have f sub y equals lambda g sub y. F sub y is 2y equals lambda times g sub y is x. And then our third equation g equals c becomes xy equals three.
So, that is what you would have to solve.
Any questions at this point? No.
Yes? How do I know the direction of a gradient? Do you mean how do I know that it is perpendicular to a level curve?
Oh, how do I know if it points in that direction on the opposite one? Well, that depends.
I mean we'd seen in last time, but the gradient is perpendicular to the level and points towards higher values of a function. So it could be -- Wait.
What did I have? It could be that my gradient vectors up there actually point in opposite directions.
It doesn't matter to me because it will still look the same in terms of the equation, just lambda will be positive or negative, depending on the case. I can handle both situations.
It's not a problem. I can allow lambda to be positive or negative. Well, in this example,it looks like lambda will be positive.
If you look at the picture on the plot.
Yes? Well, because actually they are not equal to each other. If you look at this point where the hyperbola and the circle touch each other,first of all, I don't know which circle I am going to look at. I am trying to solve,actually, for the radius of the circle.
I am trying to find what the minimum value of f is.
And, second, at that point,the value of f and the value of g are not equal.
g is equal to three because I want the hyperbola x equals three. The value of f will be the square of a distance, whatever that is.
I think it will end up being 6, but we will see.
So, you cannot really set them equal because you don't know what f is equal to in advance. Yes?
Not quite. Actually, here I am just using this idea of finding a point closest to the origin to illustrate an example of a min/max problem.
The general problem we are trying to solve is minimize f subject to g equals constant. And what we are going to do for that is we are really going to say instead let's look at places where gradient f and gradient g are parallel to each other and solve for equations of that. I think we completely lose the notion of closest point if we just look at these equations.
We don't really say anything about closest points anymore.
Of course, that is what they mean in the end.
But, in the general setting, there is no closest point involved anymore. OK.
Yes? Yes.
It is always going to be the case that,at the minimum, or at the maximum of a function subject to a constraint, the level curves of f and the level curves of g will be tangent to each other.
That is the basis for this method.
I am going to justify that soon. It could be minimum or maximum.
In three-dimensions it could even be a saddle point.
And, in fact, I should say in advance,this method will not tell us whether it is a minimum or a maximum. We do not have any way of knowing, except for testing values.
We cannot use second derivative tests or anything like that.
I will get back to that. Yes?
Yes. Here you can set y equals to favor x. Then you can minimize x squared plus nine over x squared. In general, if I am trying to solve a more complicated problem, I might not be able to solve. I am doing an example where,indeed, here you could solve and remove one variable,but you cannot always do that. And this method will still work.
The other one won't. OK.
I don't see any other questions. Are there any other questions?
No. OK.
I see a lot of students stretching and so on,so it is very confusing for me. How do we solve these equations?
Well, the answer is in general we might be in deep trouble.
There is no general method for solving the equations that you get from this method. You just have to think about them. Sometimes it will be very easy.
Sometimes it will be so hard that you cannot actually do it without the computer. Sometimes it will be just hard enough to be on Part B of this week's problem set.
I claim in this case we can actually do it without so much trouble, because actually we can think of this as a two by two linear system in x and y. Well, let me do something.
Let me rewrite the first two equations as 2x - lambda y = 0.
And lambda x - 2y = 0. And xy = 3.
That is what we want to solve. Well, I can put this into matrix form. Two minus lambda,lambda minus two times x, y equals 0,0.
Now, how do I solve a linear system matrix times x,y equals zero? Well, I always have an obvious solution. X and y both equal to zero.
Is that a good solution? No, because zero times zero is not three. We want another solution,the trivial solution. 0,0 does not solve the constraint equation xy equals three, so we want another solution. When do we have another solution? Well, when the determinant of a matrix is zero. We have other solutions that exist only if determinant of a matrix is zero.
M is this guy. Let's compute the determinant.
Well, that seems to be negative four plus lambda squared.
That is zero exactly when lambda squared equals four,which is lambda is plus or minus two.
Already you see here it is a the level of difficulty that is a little bit much for an exam but perfectly fine for a problem set or for a beautiful lecture like this one.
How do we deal with -- Well, we have two cases to look at.
Lambda equals two or lambda equals minus two.
Let's start with lambda equals two.
If I set lambda equals two, what does this equation become?
Well, it becomes x equals y. This one becomes y equals x.
Well, they seem to be the same. x equals y.
And then the equation xy equals three becomes,well, x squared equals three. I have two solutions.
One is x equals root three and, therefore, y equals root three as well, or negative root three and negative root three.
Let's look at the other case. If I set lambda equal to negative two then I get 2x equals negative 2y.
That means x equals negative y. The second one,2y equals negative 2x. That is y equals negative x.
Well, that is the same thing. And xy equals three becomes negative x squared equals three. Can we solve that?
No. There are no solutions here.
Now we have two candidate points which are these two points, root three, root three or negative root three, negative root three. OK.
Let's actually look at what we have here.
Maybe you cannot read the coordinates, but the point that I have here is indeed root three, root three.
How do we see that lambda equals two?
Well, if you look at this picture, the gradient of f,that is the blue vector, is indeed twice the yellow vector, gradient g. That is where you read the value of lambda. And we have the other solution which is somewhere here. Negative root three,negative root there. And there, again,lambda equals two. The two vectors are proportional by a factor of two. Yes?
No, solutions are not quite guaranteed to be absolute minima or maxima. They are guaranteed to be somehow critical points end of a constraint.
That means if you were able to solve and eliminate the variable that would be a critical point. When you have the same problem,as we have critical points, are they maxima or minima?
And the answer is, well, we won't know until we check. More questions?
No. Yes?
What is a Lagrange multiplier? Well, it is this number lambda that is called the multiplier here.
It is a multiplier because it is what you have to multiply gradient of g by to get gradient of f.
It multiplies.
Let's try to see why is this method valid?
Because so far I have shown you pictures and have said see they are tangent. But why is it that they have to be tangent in general? Let's think about it.
Let's say that we are at constrained min or max.
What that means is that if I move on the level g equals constant then the value of f should only increase or only decrease. But it means,in particular, to first order it will not change. At an unconstrained min or max,partial derivatives are zero. In this case,derivatives are zero only in the allowed directions.
And the allowed directions are those that stay on the levels of this g equals constant. In any direction along the level set g = c the rate of change of f must be zero.
That is what happens at minima or maxima.
Except here, of course, we look only at the allowed directions. Let's say the same thing in terms of directional derivatives.
That means for any direction that is tangent to the constraint level g equal c, we must have df over ds in the direction of u equals zero. I will draw a picture.
Let's say now I am in three variables just to give you different examples. Here I have a level surface g equals c. I am at my point.
And if I move in any direction that is on the level surface,so I move in the direction u tangent to the level surface,then the rate of change of f in that direction should be zero.
Now, remember what the formula is for this guy.
Well, we have seen that this guy is actually radiant f dot u.
That means any such vector u must be perpendicular to the gradient of f. That means that the gradient of f should be perpendicular to anything that is tangent to this level. That means the gradient of f should be perpendicular to the level set.
That is what we have shown.
But we know another vector that is also perpendicular to the level set of g. That is the gradient of g.
We conclude that the gradient of f must be parallel to the gradient of g because both are perpendicular to the level set of g. I see confused faces,so let me try to tell you again where that comes from.
We said if we had a constrained minimum or maximum,if we move in the level set of g, f doesn't change.
Well, it doesn't change to first order.
It is the same idea as when you are looking for a minimum you set the derivative equal to zero.
So the derivative in any direction, tangent to g equals c, should be the directional derivative of f,in any such direction, should be zero.
That is what we mean by critical point of f.
And so that means that any vector u, any unit vector tangent to the level set of g is going to be perpendicular to the gradient of f. That means that the gradient of f is perpendicular to the level set of g.
If you want, that means the level sets of f and g are tangent to each other. That is justifying what we have observed in the picture that the two level sets have to be tangent to each other at the prime minimum or maximum.
Does that make a little bit of sense?
Kind of. I see at least a few faces nodding so I take that to be a positive answer.
Since I have been asked by several of you,how do I know if it is a maximum or a minimum?
Well, warning, the method doesn't tell whether a solution is a minimum or a maximum.
How do we do it? Well, more bad news.
We cannot use the second derivative test.
And the reason for that is that we care actually only about these specific directions that are tangent to variable of g.
And we don't want to bother to try to define directional second derivatives. Not to mention that actually it wouldn't work. There is a criterion but it is much more complicated than that. Basically, the answer for us is that we don't have a second derivative test in this situation. What are we left with?
Well, we are just left with comparing values.
Say that in this problem you found a point where f equals three, a point where f equals nine, a point where f equals 15.
Well, then probably the minimum is the point where f equals three and the maximum is 15. Actually, in this case,where we found minima, these two points are tied for minimum. What about the maximum?
What is the maximum of f on the hyperbola?
Well, it is infinity because the point can go as far as you want from the origin. But the general idea is if we have a good reason to believe that there should be a minimum,and it's not like at infinity or something weird like that,then the minimum will be a solution of the Lagrange multiplier equations. We just look for all the solutions and then we choose the one that gives us the lowest value. Is that good enough?
Let me actually write that down.
To find the minimum or the maximum, we compare values of f at the various solutions -- -- to Lagrange multiplier equations.
I should say also that sometimes you can just conclude by thinking geometrically. In this case,when it is asking you which point is closest to the origin you can just see that your answer is the correct one.
Let's do an advanced example. Advanced means that -- Well,this one I didn't actually dare to put on top of the other problem sets. Instead, I am going to do it.
What is this going to be about? We are going to look for a surface minimizing pyramid. Let's say that we want to build a pyramid with a given triangular base -- -- and a given volume. Say that I have maybe in the x,y plane I am giving you some triangle.
And I am going to try to build a pyramid.
Of course, I can choose where to put the top of a pyramid.
This guy will end up being behind now.
And the constraint and the goal is to minimize the total surface area. The first time I taught this class, it was a few years ago, was just before they built the Stata Center. And then I used to motivate this problem by saying Frank Gehry has gone crazy and has been given a triangular plot of land he wants to put a pyramid.
There needs to be the right amount of volume so that you can put all the offices in there. And he wants it to be,actually, covered in solid gold.
And because that is expensive, the administration wants him to cut the costs a bit. And so you have to minimize the total size so that it doesn't cost too much.
We will see if MIT comes up with a triangular pyramid building. Hopefully not.
It could be our next dorm, you never know.
Anyway, it is a fine geometry problem.
Let's try to think about how we can do this.
The natural way to think about it would be -- Well,what do we have to look for first?
We have to look for the position of that top point.
Remember we know that the volume of a pyramid is one-third the area of base times height. In fact, fixing the volume,knowing that we have fixed the area of a base,means that we are fixing the height of the pyramid.
The height is completely fixed. What we have to choose just is where do we put that top point? Do we put it smack in the middle of a triangle or to a side or even anywhere we want?
Its z coordinate is fixed. Let's call h the height.
What we could do is something like this.
We say we have three points of a base.
Let's call them p1 at (x1, y1,0); p2 at (x2,y2,0); p3 at (x3, y3,0).
This point p is the unknown point at (x, y,h). We know the height.
And then we want to minimize the sum of the areas of these three triangles. One here, one here and one at the back. And areas of triangles we know how to express by using length of cross-product.
It becomes a function of x and y.
And you can try to minimize it. Actually, it doesn't quite work.
The formulas are just too complicated.
You will never get there. What happens is actually maybe we need better coordinates. Why do we need better coordinates? That is because the geometry is kind of difficult to do if you use x, y coordinates.
I mean formula for cross-product is fine,but then the length of the vector will be annoying and just doesn't look good. Instead, let's think about it differently.
I claim if we do it this way and we express the area as a function of x, y, well, actually we can't solve for a minimum. Here is another way to do it.
Well, what has worked pretty well for us so far is this geometric idea of base times height.
So let's think in terms of the heights of side triangles.
I am going to use the height of these things.
And I am going to say that the area will be the sum of three terms, which are three bases times three heights.
Let's give names to these quantities.
Actually, for that it is going to be good to have the point in the xy plane that lives directly below p.
Let's call it q. P is the point that coordinates x, y, h. And let's call q the point that is just below it and so it' coordinates are x,y, 0. Let's see.
Let me draw a map of this thing. p1, p2, p3 and I have my point q in the middle. Let's see.
To know these areas, I need to know the base.
Well, the base I can decide that I know it because it is part of my given data. I know the sides of this triangle. Let me call the lengths a1,a2, a3. I also need to know the height,so I need to know these lengths.
How do I know these lengths? Well, its distance in space,but it is a little bit annoying.
But maybe I can reduce it to a distance in the plane by looking instead at this distance here. Let me give names to the distances from q to the sides. Let's call u1,u2, u3 the distances from q to the sides.
Well, now I can claim I can find, actually,sorry. I need to draw one more thing.
I claim I have a nice formula for the area,because this is vertical and this is horizontal so this length here is u3, this length here is h.
So what is this length here? It is the square root of u3 squared plus h squared. And similarly for these other guys. They are square roots of a u squared plus h squared. The heights of the faces are square root of u1 squared times h squared.
And similarly with u2 and u3. So the total side area is going to be the area of the first faces,one-half of base times height, plus one-half of a base times a height plus one-half of the third one.
It doesn't look so much better. But, trust me,it will get better. Now, that is a function of three variables, u1, u2, u3.
And how do we relate u1, u2, u3 to each other?
They are probably not independent.
Well, let's cut this triangle here into three pieces like that. Then each piece has side --Well, let's look at it the piece of the bottom.
It has base a3, height u3. Cutting base into three tells you that the area of a base is one-half of a1,u1 plus one-half of a2, u2 plus one-half of a3,u3. And that is our constraint.
My three variables, u1, u2, u3, are constrained in this way. The sum of this figure must be the area of a base. And I want to minimize that guy.
So that is my g and that guy here is my f.
Now we try to apply our Lagrange multiplier equations.
Well, partial f of a partial u1 is -- Well,if you do the calculation, you will see it is one-half a1,u1 over square root of u1^2 plus h^2 equals lambda,what is partial g, partial a1?
That one you can do, I am sure. It is one-half a1.
Oh, these guys simplify. If you do the same with the second one -- -- things simplify again.
And the same with the third one. Well, you will get,after simplifying, u3 over square root of u3 squared plus h squared equals lambda.
Now, that means this guy equals this guy equals this guy.
They are all equal to lambda. And, if you think about it,that means that u1 = u2 = u3. See, it looked like scary equations but the solution is very simple.
What does it mean? It means that our point q should be equidistant from all three sides.
That is called the incenter. Q should be in the incenter.
The next time you have to build a golden pyramid and don't want to go broke, well, you know where to put the top.
If that was a bit fast, sorry. Anyway, it is not completely crucial. But go over it and you will see it works. Have a nice weekend.
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4763 | https://www.sparkl.me/learn/ib/chemistry-hl/ph-scale-and-calculations/revision-notes/1376 | Topic 2/3
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pH Scale and Calculations
Introduction
Key Concepts
What is the pH Scale?
Hydronium Ion Concentration
Calculating pH
Relationship Between pH and pOH
Neutralization Reactions
Buffer Solutions
Calculating [H3O+]
Applications of pH in Chemical Reactions
Indicators and pH Measurement
Strength of Acids and Bases
Advanced Concepts
pKa and pKb Values
Henderson-Hasselbalch Equation
Ionic Product of Water
Buffer Capacity
Titration Curves
Polyprotic Acids
Le Chatelier’s Principle and pH
Intermolecular Forces and pH
Interdisciplinary Connections
Complex Calculations Involving pH
Comparison Table
| Aspect | Acids | Bases |
---
| Definition | Substances that donate protons ($H^+$) in solution. | Substances that accept protons or donate hydroxide ions ($OH^-$) in solution. |
| pH Range | Less than 7 | Greater than 7 |
| Examples | Hydrochloric acid ($HCl$), sulfuric acid ($H_2SO_4$) | Sodium hydroxide ($NaOH$), ammonia ($NH_3$) |
| Conjugate Pair | Conjugate base | Conjugate acid |
| Reaction with Water | $HA \leftrightarrow H^+ + A^-$ | $B + H_2O \leftrightarrow BH^+ + OH^-$ |
| Effect on pH | Increase $[H_3O^+]$, decreasing pH | Increase $[OH^-]$, increasing pH |
Summary and Key Takeaways
Coming Soon!
Examiner Tip
Tips
Remember the pH and pOH Relationship: Always keep in mind that pH + pOH = 14. This can help you quickly find one if you know the other.
Mnemonic for Acids and Bases: Use "ACID donates H+" and "BASE accepts H+" to remember their behaviors.
Utilize the Henderson-Hasselbalch Equation: Familiarize yourself with this equation for solving buffer problems efficiently:
$$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$$
Did You Know
1. The pH Scale Origin: The pH scale was introduced in 1909 by the Danish chemist Søren Peder Lauritz Sørensen to simplify the expression of acidity and alkalinity. Its logarithmic nature means each whole number on the scale represents a tenfold change in acidity.
2. Human Body pH: Different parts of the human body maintain specific pH levels crucial for their functions. For example, human blood maintains a slightly basic pH around 7.4, which is vital for proper physiological processes.
3. Extreme pH Environments: Some organisms, known as extremophiles, thrive in environments with extreme pH levels, such as highly acidic hot springs or alkaline lakes, showcasing the versatility of life in diverse chemical conditions.
Common Mistakes
Mistake 1: Confusing pH with pOH. Students often forget that pH and pOH are related by the equation pH + pOH = 14.
Incorrect: Calculating pH and pOH independently.
Correct: Use the relationship to find one value when the other is known.
Mistake 2: Incorrectly applying the Henderson-Hasselbalch equation. Not accounting for significant figures or misidentifying the conjugate acid/base can lead to errors.
Incorrect: Using concentrations instead of ratios.
Correct: Ensure the correct acid and base forms are used in the equation.
Mistake 3: Overlooking the effect of temperature on pH. The ionic product of water ($K_w$) changes with temperature, affecting pH calculations.
Incorrect: Assuming $K_w$ is always $1.0 \times 10^{-14}$.
Correct: Use the appropriate $K_w$ value for the given temperature.
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4764 | https://ocw.mit.edu/courses/18-385j-nonlinear-dynamics-and-chaos-fall-2004/7f15d4558d3e8cab180482e489de9295_bead_on_wire.pdf | Bead moving along a thin, rigid, wire. Rodolfo R. Rosales, Department of Mathematics, Massachusetts Inst. of Technology, Cambridge, Massachusetts, MA 02139 October 17, 2004 Abstract An equation describing the motion of a bead along a rigid wire is derived. First the case with no friction is considered, and a Lagrangian formulation is used to derive the equation. Next a simple correction for the effect of friction is added to the equation. Finally, we consider the case where friction dominates over inertia, and use this to reduce the order of the system. Contents 1 Equations with no friction. 2 Principle of least action. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Equation for the motion of the bead. . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Bead on a vertical rotating hoop. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Problem 1.1: Bead on an horizontal rotating hoop.
. . . . . . . . . . . . . . . . . 4 Wire on vertical plane, rotating around a vertical plane.
. . . . . . . . . . . . . . . 5 Problem 1.2: Wires rotating with variable rates. . . . . . . . . . . . . . . . . . . . 5 Parametric instabilities.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 The Euler Lagrange equation: derivation. . . . . . . . . . . . . . . . . . . . . . . . . 6 Problem 1.3: Hoop moving up and down in a vertical plane. . . . . . . . . . . . . 6 Parametric stabilization. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Wire restricted to a fixed vertical plane.
. . . . . . . . . . . . . . . . . . . . . . . . 7 2 Add friction to the equations. 7 Equation of motion with friction included. . . . . . . . . . . . . . . . . . . . . . . . 8 Add friction to the vertical rotating hoop.
. . . . . . . . . . . . . . . . . . . . . . . 8 Wire restricted to a fixed vertical plane.
. . . . . . . . . . . . . . . . . . . . . . . . 8 Wire moving rigidly up and down. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Sliding wires and friction.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1 Rosales Bead moving along a thin, rigid, wire. 2 3 Friction dominates inertia. 10 3.1 Nondimensional equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 3.2 Limit of large friction.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 4 Problem Answers. 11 1 Equations with no friction. Consider the motion of a bead of mass M moving along a thin rigid wire, under the influence of gravity. Let (x, y, z) be a system of Cartesian coordinates, with z the vertical direction and z increasing with height. Let g be the acceleration of gravity. Describe the wire in parametric form, as follows: x = X(s, t) , y = Y (s, t) , and z = Z(s, t) , (1.1) where s is the arclength along the wire. Note that: — Because the wire is “thin”, we have approximated it as just a curve in space. — The wire is allowed to move and change shape as a function of time. — The bead will be approximated as a mass point. — We use the same length units for x , y , z, and s, so that: X 2 + Y 2 + Z2 = 1. s s s Furthermore: we will assume that the wire does not stretch as it moves. In particular, s is not only arclength, but a material coordinate along the wire (i.e. s constant denotes always the same material point on the wire). This will be important when we consider the effects of friction. It is well known that, in the presence of rigid body constraints, the best way to formulate equations in mechanics is by the use of Lagrangian mechanics — that is: by using the principle of least action. Remark 1.1 Of course, one can get the equations using the classical newtonian formulation. What complicates things is that one must consider the “unknown” forces that arise to keep the constraints valid. For example, in this case the bead experiences the force of gravity, and an unknown force from the wire — which must be exactly right to keep the bead moving along the wire (this is what determines this force). When the wire is static, calculating this force is relatively easy. However, when the wire moves, things get complicated. Z Z 3 Rosales Bead moving along a thin, rigid, wire. P Remark 1.2 (Principle of least action). In its simplest form, the principle of least action for a conservative system (no energy is dissipated) can be stated as follows: Let be a mechanical trajectory, joining the points P1 and P2 in phase space. Parameterize this trajectory by time t, with t = t1 the starting time (when the trajectory is at P1), and t = t2 the final time (when 2 is reached). Consider now all the possible trajectories (curves) in phase space, starting at P1 for t = t1, and ending at P2 for t = t2 — is, of course, one member of this set. For any each such curve the action L can be computed: t2 L = (Kinetic Energy −Potential Energy) dt. t1 Then the principle of least action says that is a stationary point for L within the set of all possible trajectories described above. We now apply this principle to our problem. Let the position of the bead along the wire be given by s = S(t) . Then: 1 2 Kinetic Energy = M x ˙ 2 + ˙ y 2 + ˙ z 2 1 2 = MS ˙ + M (Xs Xt + Ys Yt + Zs Zt) S ˙ + 1 M (X2 + Y 2 + Zt 2) 2 t t 2 1 2 = MS ˙ + S ˙ + α , (1.2) 2 Potential Energy = g M Z , (1.3) 1 where: (A) = (S, t) = M (Xs Xt + Ys Yt + Zs Zt) and α = α(S, t) = 2 M (Xt 2 + Yt 2 + Zt 2) . (B) X, Y , Z, and all their derivatives are evaluated at (S, t). (C) The dots indicate (total) derivatives with respect to time. (D) Since x = X(S, t), we have x ˙ = S ˙ Xs(S, t) + Xt(S, t) — with similar formulas for ˙ y and ˙ z. (E) We have used that s is arclength, so that X2 + Y 2 + Z2 = 1. s s s The action is then given by t2 L = L[S] = L(S ˙, S, t) dt , (1.4) t1 1 2 + ˙ where L = MS ˙ S −V , with V = V (S, t) = g M Z −α . 2 ! q 4 Rosales Bead moving along a thin, rigid, wire. Because of the principle of least action, the Euler-Lagrange equation d @L @L dt @S ˙ − @S = 0 , (1.5) must apply for the mechanical trajectory. Using the expression for L above, this yields the equation: d2S @V @ M = − (S , t) − (S , t) . (1.6) dt2 @S @t This is the equation for the motion of a bead along a rigid moving wire, under the action of gravity, when friction is neglected. Example 1.1 When the wire moves normal to itself, 0. An example is a hoop rotating around a diameter. Consider now the case with vertical rotation axis: s X = +R sin cos( t) , R s Y = +R sin sin( t) , R s Z = −R cos , R where R is the radius of the hoop and is its angular velocity. Then S S 1 = 0 and V = −gMR cos − MR2 2 sin2 , (1.7) R 2 R where we notice that V is independent of time. The equation of motion is then: S d2S 1 2S = −g sin + R 2 sin . (1.8) dt2 R 2 R S Rewrite the equation using the angle = that the bead makes with the vertical down-radius: R d2 1 = −!2 sin + 2 sin 2 = ( 2 cos − !2) sin , (1.9) dt2 2 where ! = g/R is the angular frequency for a pendulum of length R. The two terms on the right in equations (1.8 – 1.9) correspond to the effects of gravity and the centrifugal force. Problem 1.1 Consider the case of a hoop rotating around an horizontal diameter. Rosales Bead moving along a thin, rigid, wire. 5 Example 1.2 A generalization of example 1.1 is that of a wire on a vertical plane, rotating around a vertical axis. Then: X = R(s) cos( t) , Y = R(s) sin( t) , Z = Z(s) , where (R0)2 + (Z0)2 = 1, and is the rotation angular velocity. Then 1 = 0 and V = gMZ(S) − M 2 R2(S), (1.10) 2 where, again, V is independent of time. The equation of motion is then d2S + gZ0(S) − 2R(S)R0 (S) = 0. (1.11) dt2 Problem 1.2 Generalize the derivations in examples 1.1 and 1.2 to consider variable rotation rates. That is to say: in the formulas describing the wires, replace t ! Γ, where Γ = Γ(t) is some function of time. Define = (t) = d /dt, then show that the ONLY change in the governing equations — (1.8), (1.9), and (1.11) — is that is no longer a constant, and is replaced by = (t), as defined above. Example 1.3 Consider equation (1.9) for the case of a variable, and periodic, rotation rate = (t). Assume that the amplitude of remains below ! for all times: 0 2 < !2 . In this case, intuition would seem to indicate that 0 is a stable solution to the equation, with small perturbations near it obeying the linear equation: d2 + (!2 − 2) = 0. (1.12) dt2 However, it turns out that, when the period of the forcing is appropriately selected, it is possible to make the solutions of (1.12) grow in time — thus turning the solution 0 unstable.1 This phenomena is known as a parametric instability, and we will study it later — when we cover Floquet theory. When nonlinearity and dissipation are added to the system, chaotic behavior can result from it. 1This can happen even if the amplitude of is small. Z Z Z ! 6 Rosales Bead moving along a thin, rigid, wire. Remark 1.3 — Derivation of the Euler-Lagrange equation (1.5). Let βS = βS(t) be a small (but arbitrary) perturbation to the mechanical trajectory S, vanishing at t1 and t2. Let βL be the corresponding perturbation to the action: t2 βL = L[S + βS] −L[S] = L(S ˙ + βS ˙ , S + βS , t) −L(S ˙ , S , t) dt t1 t2 = L2(S ˙ , S , t) βS + L1(S ˙ , S , t) βS ˙ dt + O(βS2) t1 t2 = L2(S ˙ , S , t) − d L1(S ˙ , S , t) βS dt + O(βS2) , t1 dt @L @L where L1 = , L2 = , and we have integrated by parts in the third line. Because S is a @S ˙ @S stationary point, the O(βS) terms in βL must vanish for all possible choices of βS. It follows that L2(S ˙ , S , t) − d L1(S ˙ , S , t) = 0 , dt which is exactly the Euler-Lagrange equation (1.5). Problem 1.3 Consider the case of a hoop that is restricted to move up and down on a vertical plane. Derive the equation for the bead motion in this case. Parameterize the wire in the form s s X = R sin , Y 0, and Z = −R cos + A sin( t), (1.13) R R where R is the radius of the hoop (constant), is the (constant) angular frequency of the up-down motions of the hoop, and A is the (constant) amplitude of the hoop oscillations. Then show that the equation for the bead position, written in terms of the angle = S/R that the bead makes with the bottom point on the hoop, is given by: d2 + !2 −θ 2 sin( t) sin = 0, (1.14) dt2 where !2 = g/R, and θ = A/R For small values of , the linearization of this equation is of the same form as equation (1.12). Thus this is another example where parametric instability can occur. Furthermore, consider the linearization of this equation near the equilibrium solution ν — bead on the top end of the hoop. Then, for = ν + ω and ω small, we can write: d2ω − !2 −A 2 sin( t) ω = 0. (1.15) dt2 ! 7 Rosales Bead moving along a thin, rigid, wire. Intuitively we expect the ν solution to be unstable. However, it is possible to select the frequency and amplitude of the up-down oscillations so that this solution becomes stable. This is an example of parametric stabilization — to be covered later in the course. Example 1.4 Consider now2 a wire restricted to move on a fixed vertical plane — so that Y 0 in equation (1.1). Furthermore, assume that the wire position is given in the form z = F(x, t) . (1.16) What is then the equation of motion, with the bead position given as x = (t)? To find this equation, we could start from equation (1.6), and transform variables. This, however, is quite messy. It is much easier to re-write the Lagrangian L in (1.4) in terms of the new variables, and then use again the Euler-Lagrange equation. Clearly, we have: 1 1 1 L = M 1 + F 2 ψ ˙ 2 + M Fx Ft ψ ˙ − gMF − 2 MFt 2 = µ ψ ˙ 2 + ψ ˙ − V , (1.17) 2 x 2 where F and its derivatives are evaluated at (ψ , t). Furthermore: µ = µ(ψ, t) = M(1 + Fx 2), = (ψ , t) = MFxFt, and V = V (ψ , t) = gMF − (1/2)MFt. The Euler-Lagrange equation d @L @L − = 0 dt @ψ ˙ @ψ then yields: d dψ ! 1 dψ !2 dt µ dt = 2 µ dt − V − t . (1.18) 2 Add friction to the equations. We want now to add the effects of friction to the equation of motion for the bead. The friction laws for solid-to-solid contact are quite complicated, thus we will simplify the problem by assuming that the wire is covered by a thin layer of lubricant liquid. In this case, an adequate model is provided by: friction produces a force opposing the motion, of magnitude proportional to the velocity of the bead relative to the wire. Since s is not just arclength, but a material 2The example in problem 1.3 is of this type. Z Z ! Z Rosales Bead moving along a thin, rigid, wire. 8 coordinate along the wire, the sliding velocity of the bead along the wire is simply S ˙. Thus we modify equation (1.6) to: d2S dS @V @ M = −κ − (S , t) − (S , t) , (2.1) dt2 dt @S @t where κ is the friction coefficient — which we assume is a constant. This is the equation for a bead moving along a lubricated rigid moving wire, under the influence of gravity, with the friction force included. Example 2.1 Add friction to the rotating hoop in example 1.1. It is easy to see that equation (1.9) is modified to: d2 κ d + + (!2 − 2 cos ) sin , (2.2) dt2 M dt by the addition of friction. When the rotation rate is variable, = (t) — see problem 1.2. More generally, equation (1.11) for a rotating wire is modified to: d2S κ dS + + gZ0(S) − 2R(S)R0(S) = 0. (2.3) dt2 M dt Example 2.2 Consider the situation in example 1.4 above. To add the effects of friction when the equation of motion is written as in (1.18), we need to compute first what the velocity of the bead is relative to the wire. This is not as simple as it may appear, for equation (1.16) does not provide enough information as to the actual motion of the wire mass points — for example: the wire may have a fixed shape, with the points in the wire “sliding” along this shape.3 Thus extra information is needed. For example, let us assume that there is some material point in the wire for which x = x0 is constant as the wire deforms and moves. In this case we can proceed as follows: ds2 = dx2 + dz2 = (1 + F x 2) dx2 = ∂2 dx2 , = ≡ S = ∂(x , t) dx , = ≡ x0 dS dψ = ∂(ψ , t) + ∂t(x , t) dx , = ≡ dt dt x0 dψ F = κ ∂(ψ , t) + ∂t(x , t) dx , (2.4) dt x0 3Think of a circular wire rotating around its center. See example 2.3. q ! ! ! Rosales Bead moving along a thin, rigid, wire. 9 q where ∂ = ∂(x , t) = 1 + F 2 x (thus µ = M ∂2 , and ∂ ∂t = Fx Ft ) , and F is the magnitude of the friction force. The magnitude of the component of the friction force along the x-axis is thus given by F /∂. This component must oppose the inertial force along the x-axis, which is given ¨ by M ψ. Since µ = M ∂2, we conclude that equation (1.18) must be modified as follows: ! !2 d dψ 1 dψ µ + ∂F = µ − V − t , (2.5) dt dt 2 dt where F = F ( ˙ ψ, ψ, t) is given by equation (2.4). A particularly simple instance of this occurs in the case of a wire moving rigidly up and down. In this case we can write, for F in (1.16), z = F (x , t) = f (x) + h(t) . (2.6) Then µ = µ(ψ) = M 1 + (f 0(ψ))2 , = (ψ, t) = M f 0(ψ)h ˙ (t), ∂ = ∂(ψ) = 1 + (f 0(ψ))2 , 1 V = V (ψ, t) = gM f (ψ) − gM h(t) + M (h ˙ (t))2 , and F = κ∂ψ ˙. Thus, the equation of motion 2 can be written in the form: d dψ κ dψ 1 ¨ ∂ + ∂ + f 0(ψ) g + h(t) . (2.7) dt dt M dt ∂ Example 2.3 Sliding wires and friction. We consider two examples. First a horizontal straight wire sliding at a variable speed. Thus: X = s + u(t), and Y = Z = 0. (2.8) du dv Let v = be the wire speed, and a = be the wire acceleration. Then the formulas above equa dt dt 1 1 tion (1.6) yield: = M v, α = M v 2, and V = − M v 2 . Thus, from equation (2.1) 2 2 d2S κ dS d2ψ κ dψ + + a = 0 → ≡ + − v = 0, (2.9) dt2 M dt dt2 M dt where ψ = S + u. The second example is that of a horizontal circular wire rotating around its center. Thus X = R cos χ, Y = R sin χ, and Z = 0, (2.10) dπ where R is the radius of the hoop, χ = s/R + π, and π = π(t) indicates the wire motion. Let = dt S be the wire angular velocity. Then equation (2.1) yields, for = + π, R d2 κ d + − = 0. (2.11) dt2 M dt Rosales Bead moving along a thin, rigid, wire. 10 3 Friction dominates inertia. 3.1 Nondimensional equations. We consider equation (2.1), for a simple situation where the equation is autonomous (no time dependence). Thus, the equation takes the form d2S dS M + κ = F (S), (3.1) dt2 dt where F is the force driving the bead along the wire (produced by gravity and the wire motion). Assume now that F varies on some characteristic length scale, and has a typical size. Thus we can write S F = F0Γ (3.2) L where F0 is a constant with force dimensions, L is a constant with dimensions of length, and Γ is nondimensional and has O(1) size, with an O(1) derivative. We introduce nondimensional variables as follows: = S L and δ = F0 κL t, (3.3) κL where is the time scale that results from the balance of the applied forces and the viscous forces. F0 In terms of these variables, the equation becomes: d2 d θ + = Γ(), (3.4) dδ 2 dδ M F0 where θ = is a nondimensional parameter. Lκ2 3.2 Limit of large friction. If the inertial forces are negligible relative to the viscous forces, to be precise, if 0 < θ ≤ 1 in (3.4), the inertial terms in the equation can be neglected. This leads to: d = Γ(). (3.5) dδ We point out that it is, generally, dangerous to neglect terms that involve the highest derivative in an equation — even if the term is multiplied by a very small parameter. Rosales Bead moving along a thin, rigid, wire. 11 In the particular case here, this can be entirely justified,4 but (in general) one must be very careful. 4 Problem Answers. The problem answers will be handed out separately, with the solutions to problem sets and exams. 4This will be done later in the course. |
4765 | https://www.npr.org/2013/09/22/224946206/adjunct-professor-dies-destitute-then-sparks-debate | Accessibility links
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The Sad Death Of An Adjunct Professor Sparks A Labor Debate After 25 years of teaching French for Duquesne University in Pittsburgh, 83-year-old Margaret Mary Vojtko was let go. She died shortly after, penniless and nearly homeless. Her story has spurred sharp anger over the treatment of part-time faculty.
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Adjunct professors at Duquesne University in Pittsburgh are trying to unionize. The death of a longtime, part-time employee has put the debate in a larger forum. Ronald Woan/Flickr hide caption
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Adjunct professors at Duquesne University in Pittsburgh are trying to unionize. The death of a longtime, part-time employee has put the debate in a larger forum.
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The death of a long-time, part-time professor in Pittsburgh is gathering the attention of instructors nationwide. The trend of relying on part-time faculty has been in the works for decades, and Margaret Mary Vojtko's story is seen by some as a tragic byproduct.
Last spring, months before her death, Vojtko showed up at a meeting between adjunct professors at Duquesne University and the union officials who had been trying to organize them. The professors are trying to organize a union affiliated with the United Steelworkers.
Daniel Kovalik, senior counsel to the Steelworkers union, says Vojtko was distraught. "She had cancer; she had very high medical bills," Kovalik says.
After 25 years of teaching French at Duquesne, the university had not renewed her contract. As a part-time professor, she had been earning about $10,000 a year, and had no health insurance.
"She didn't want charity," Kovalik says. "She thought that after working 25 years for Duquesne that she was owed a living wage and some sort of retirement and benefits."
Vojtko died Sept. 1 after a heart attack at the age of 83, destitute and nearly homeless.
After her funeral, Kovalik submitted a biting op-ed piece to the Pittsburgh Post-Gazette, critical of how Duquesne had treated Vojtko. Almost immediately, a bigger debate unfolded on Facebook, Twitter and listervs.
In large part, the story struck a nerve because compensation and treatment of adjunct professors has been a simmering issue since the early 1970s, when campuses began to see a shift from full-time to part-time faculty.
Today, these itinerant teachers make up a whopping 75 percent of college instructors, with their average pay between $20,000 and $25,000 annually.
The shift toward adjunct teachers has helped institutions save lots of money. But Duquesne Provost Tim Austin says it's unfair to cast his school as "heartless and greedy."
"First of all, I don't accept that the arrangements that we make with part-timers are dictated by cost savings," Austin says.
Second, says Austin, Duquesne pays adjunct professors more than most institutions.
"The least that an adjunct professor could be paid is $3,500 for a course, $7,000 for a given semester," he says. "Whether those are appropriate in a yet larger context is ... a matter that the academic world has not yet found a decisive answer."
The answer is staring university leaders in the face, says Maria Maisto, head of New Faculty Majority, which advocates for adjunct professors: Pay college presidents and coaches less, and part-time professors more.
"If education is really at the heart of what we do, then there's absolutely no excuse for not putting the bulk of the resources into what happens in the classroom," Maisto says.
But that's not what institutions are doing, she says.
"In fact, here in Ohio, I have colleagues who have recently had to sell their plasma in order to buy groceries," she says.
Maisto says that's why so many adjunct professors identified with Vojtko's story. Still, the university's provost says the Vojtko case has been shamelessly exploited. Duquesne did reach out to help Vojtko, Austin says, and at one point even offered her temporary housing.
Kovalik says he hopes Duquesne will be "shamed" into allowing adjunct professors to unionize.
"If Margaret Mary can help in that way, she would be very proud," Kovalik says.
Duquesne officials say there are no immediate plans to allow adjunct professors to unionize, despite professors' vote to do so. |
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Last reviewed:
15 Aug 2025
Last updated:
19 May 2023
Summary
Enuresis has primarily nocturnal symptoms in children older than 5 years of age.
Differentials include diabetes, medications, emotional problems, urinary tract infection, spina bifida, seizure disorder, and neurogenic bladder.
Treatment commonly involves behavioral changes, alarm therapy, or desmopressin.
Emotional support and encouragement is vital to management.
Definition
Enuresis is defined as normal micturition that occurs at an inappropriate or socially unacceptable time or place. As recommended by the International Children's Continence Society, in this topic "enuresis" is reserved for micturition during sleep, or bedwetting. Daytime wetting is called "incontinence".
History and exam
Key diagnostic factors
increased fluid intake at night
urinary frequency
constipation
caffeine and other bladder irritants
urinary urgency
Full details
Other diagnostic factors
abnormal voiding habits
abnormal breathing pattern at night
Full details
Risk factors
genetic predisposition
constipation
upper airway obstruction/sleep-disordered breathing
ADHD
psychological disorders
male sex
Full details
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Diagnostic tests
1st tests to order
urinalysis
Full details
Tests to consider
urinary tract ultrasound
Full details
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Treatment algorithm
ACUTE
age <7 years
age ≥7 years
ONGOING
recurrence
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Contributors
Authors
Erin C. Grantham, MD
Pediatric Urologist
Department of Urology
Billings Clinic
Billings
MT
Disclosures
ECG declares that she has no competing interests.
Acknowledgements
Dr Erin C. Grantham would like to gratefully acknowledge Dr Duncan T. Wilcox and Dr Nicholas G. Cost, the previous contributors to this topic. Dr Erin C. Grantham would also like to acknowledge Jillian Hatfield for her contribution to this topic.
Disclosures
DTW and NGC declare that they have no competing interests.
Peer reviewers
Prasad Godbole, FRCS, FRCS (Paed), FEAPU
Consultant Paediatric Urologist
Paediatric Surgical Unit
Sheffield Children's NHS Foundation Trust
Western Bank
Sheffield
UK
Disclosures
PG declares that he has no competing interests.
Elizabeth Jackson, MD
Associate Professor
Pediatric Nephrology
Cincinnati Children's Hospital Medical Center
Cincinnati
OH
Disclosures
EJ declares that she has no competing interests.
Peer reviewer acknowledgements
BMJ Best Practice topics are updated on a rolling basis in line with developments in evidence and guidance. The peer reviewers listed here have reviewed the content at least once during the history of the topic.
Disclosures
Peer reviewer affiliations and disclosures pertain to the time of the review.
References
Our in-house evidence and editorial teams collaborate with international expert contributors and peer reviewers to ensure that we provide access to the most clinically relevant information possible.
Key articles
American Psychiatric Association. Diagnostic and statistical manual of mental disorders, 5th ed. Text Revision, (DSM-5-TR). Washington, DC: American Psychiatric Publishing; 2022.
European Association of Urology. Guidelines on paediatric urology. 2023 [internet publication].
Full text
European Association of Urology. Guidelines on paediatric urology. 2019 [internet publication].
Full text
Nevéus T, Eggert P, Evans J, et al. Evaluation of and treatment for monosymptomatic enuresis: a standardization document from the International Children's Continence Society. J Urol. 2010 Feb;183(2):441-7.
Full text
Abstract
Nevéus T, Eggert P, Evans J, et al. Evaluation of and treatment for monosymptomatic enuresis: a standardization document from the International Children's Continence Society. J Urol. 2010 Feb;183(2):441-7.
Abstract
Caldwell PH, Codarini M, Stewart F, et al. Alarm interventions for nocturnal enuresis in children. Cochrane Database Syst Rev. 2020 May 4;5(5):CD002911.
Full text
Abstract
Longstaffe S, Moffat M, Whalen J. Behavioral and self-esteem changes after six months of enuresis treatment: a randomized, controlled trial. Pediatrics. 2000;105:935-940.
Abstract
Reference articles
A full list of sources referenced in this topic is available to users with access to all of BMJ Best Practice.
Differentials
Congenital abnormality of the urinary tract (e.g., ectopic ureter, ureterocele, and urethral valves)
Constipation
Diabetes
More Differentials
#### Guidelines
Guidelines on paediatric urology
Diagnosis and management of nocturia
More Guidelines
#### Patient information
Bedwetting
More Patient information
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4767 | https://byjus.com/maths/trigonometry/ | Trigonometry is one of the important branches in the history of mathematics that deals with the study of the relationship between the sides and angles of a right-angled triangle. This concept is given by the Greek mathematician Hipparchus. In this article, we are going to learn the basics of trigonometry such as trigonometry functions, ratios, trigonometry table, formulas and many solved examples.
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| Table of contents: Trigonometry Definition Trigonometric Ratios (Sin, Cos, Tan) Six Trigonometric Functions Even and Odd Trigonometric Functions Trigonometric Angles Table Unit Circle Formulas Identities Euler’s Formula Basics of Trigonometry Examples Applications Solved Problems FAQs |
What is Trigonometry?
Trigonometry is one of the most important branches in mathematics that finds huge application in diverse fields. The branch called “Trigonometry” basically deals with the study of the relationship between the sides and angles of the right-angle triangle. Hence, it helps to find the missing or unknown angles or sides of a right triangle using the trigonometric formulas, functions or trigonometric identities. In trigonometry, the angles can be either measured in degrees or radians. Some of the most commonly used trigonometric angles for calculations are 0°, 30°, 45°, 60° and 90°.
Trigonometry is further classified into two sub-branches. The two different types of trigonometry are:
Plane Trigonometry
Spherical Trigonometry
In this article, let us discuss the six important trigonometric functions, ratios, trigonometry table, formulas and identities which helps to find the missing angles or sides of a right triangle.
Trigonometry Ratios-Sine, Cosine, Tangent
The trigonometric ratios of a triangle are also called the trigonometric functions. Sine, cosine, and tangent are 3 important trigonometric functions and are abbreviated as sin, cos and tan. Let us see how are these ratios or functions, evaluated in case of a right-angled triangle.
Consider a right-angled triangle, where the longest side is called the hypotenuse, and the sides opposite to the hypotenuse are referred to as the adjacent and opposite sides.
Six Important Trigonometric Functions
The six important trigonometric functions (trigonometric ratios) are calculated using the below formulas and considering the above figure. It is necessary to get knowledge about the sides of the right triangle because it defines the set of important trigonometric functions.
| | | |
---
| Functions | Abbreviation | Relationship to sides of a right triangle |
| Sine Function | sin | Opposite side/ Hypotenuse |
| Tangent Function | tan | Opposite side / Adjacent side |
| Cosine Function | cos | Adjacent side / Hypotenuse |
| Cosecant Function | cosec | Hypotenuse / Opposite side |
| Secant Function | sec | Hypotenuse / Adjacent side |
| Cotangent Function | cot | Adjacent side / Opposite side |
Even and Odd Trigonometric Functions
The trigonometric function can be described as being even or odd.
Odd trigonometric functions: A trigonometric function is said to be an odd function if f(-x) = -f(x) and symmetric with respect to the origin.
Even trigonometric functions: A trigonometric function is said to be an even function, if f(-x) = f(x) and symmetric to the y-axis.
We know that
Sin (-x) = – Sin x
Cos (-x) = Cos x
Tan (-x) = -Tan x
Csc (-x) = – Csc x
Sec (-x) = Sec x
Cot (-x) = -Cot x
Therefore, cosine and secant are the even trigonometric functions, whereas sine, tangent, cosecant and cotangent are the odd trigonometric functions. If we know the even and odd trigonometric functions, it helps us to simplify the trigonometric expression when the variable inside the trigonometric function is negative.
Trigonometry Angles
The trigonometry angles which are commonly used in trigonometry problems are 0°, 30°, 45°, 60° and 90°. The trigonometric ratios such as sine, cosine and tangent of these angles are easy to memorize. We will also show the table where all the ratios and their respective angle’s values are mentioned. To find these angles we have to draw a right-angled triangle, in which one of the acute angles will be the corresponding trigonometry angle. These angles will be defined with respect to the ratio associated with it.
For example, in a right-angled triangle,
Sin θ = Perpendicular/Hypotenuse
or θ = sin-1 (P/H)
θ = cos-1 (Base/Hypotenuse)
θ = tan-1 (Perpendicular/Base)
Trigonometry Table
Check the table for common angles which are used to solve many trigonometric problems involving trigonometric ratios.
| | | | | | |
--- --- --- |
| Angles | 0° | 30° | 45° | 60° | 90° |
| Sin θ | 0 | ½ | 1/√2 | √3/2 | 1 |
| Cos θ | 1 | √3/2 | 1/√2 | ½ | 0 |
| Tan θ | 0 | 1/√3 | 1 | √3 | ∞ |
| Cosec θ | ∞ | 2 | √2 | 2/√3 | 1 |
| Sec θ | 1 | 2/√3 | √2 | 2 | ∞ |
| Cot θ | ∞ | √3 | 1 | 1/√3 | 0 |
In the same way, we can find the trigonometric ratio values for angles beyond 90 degrees, such as 180°, 270° and 360°.
Unit Circle
The concept of unit circle helps us to measure the angles of cos, sin and tan directly since the centre of the circle is located at the origin and radius is 1. Consider theta be an angle then,
Suppose the length of the perpendicular is y and of base is x. The length of the hypotenuse is equal to the radius of the unit circle, which is 1. Therefore, we can write the trigonometry ratios as;
| | |
--- |
| Sin θ | y/1 = y |
| Cos θ | x/1 = x |
| Tan θ | y/x |
List of Trigonometry Formulas
The Trigonometric formulas or Identities are the equations which are true in the case of Right-Angled Triangles. Some of the special trigonometric identities are given below –
Pythagorean Identities
sin²θ + cos²θ = 1
tan2θ + 1 = sec2θ
cot2θ + 1 = cosec2θ
sin 2θ = 2 sin θ cos θ
cos 2θ = cos²θ – sin²θ
tan 2θ = 2 tan θ / (1 – tan²θ)
cot 2θ = (cot²θ – 1) / 2 cot θ
Sum and Difference identities-
For angles u and v, we have the following relationships:
sin(u + v) = sin(u)cos(v) + cos(u)sin(v)
cos(u + v) = cos(u)cos(v) – sin(u)sin(v)
(\begin{array}{l}tan(u+v) = \frac{tan(u)\ +\ tan(v)}{1-tan(u)\ tan(v)}\end{array} )
sin(u – v) = sin(u)cos(v) – cos(u)sin(v)
cos(u – v) = cos(u)cos(v) + sin(u)sin(v)
(\begin{array}{l}tan(u-v) = \frac{tan(u)\ -\ tan(v)}{1+tan(u)\ tan(v)}\end{array} )
If A, B and C are angles and a, b and c are the sides of a triangle, then,
Sine Laws
a/sinA = b/sinB = c/sinC
Cosine Laws
c2= a2+ b2– 2ab cos C
a2= b2+ c2– 2bc cos A
b2= a2+ c2– 2ac cos B
| |
| Trigonometry Table Trigonometry For Class 10 Trigonometry For Class 11 Trigonometry Formulas for Class 10 Trigonometry Formulas for Class 11 Trigonometry Formulas for Class 12 |
Trigonometry Identities
The three important trigonometric identities are:
sin²θ + cos²θ = 1
tan²θ + 1 = sec²θ
cot²θ + 1 = cosec²θ
Euler’s Formula for trigonometry
As per the euler’s formula,
eix = cos x + i sin x
Where x is the angle and i is the imaginary number.
(\begin{array}{l}\sin x=\frac{e^{i x}-e^{-i x}}{2 i}\ \cos x=\frac{e^{i x}+e^{-i x}}{2}\ \tan x=\frac{\left(e^{i x}-e^{-i x}\right)}{i\left(e^{i x}+e^{-i x}\right)}\end{array} )
Trigonometry Basics
The three basic functions in trigonometry are sine, cosine and tangent. Based on these three functions the other three functions that are cotangent, secant and cosecant are derived.
All the trigonometrical concepts are based on these functions. Hence, to understand trigonometry further we need to learn these functions and their respective formulas at first.
If θ is the angle in a right-angled triangle, then
Sin θ = Perpendicular/Hypotenuse
Cos θ = Base/Hypotenuse
Tan θ = Perpendicular/Base
Perpendicular is the side opposite to the angle θ.
The base is the adjacent side to the angle θ.
The hypotenuse is the side opposite to the right angle
The other three functions i.e. cot, sec and cosec depend on tan, cos and sin respectively, such as:
Cot θ = 1/tan θ
Sec θ = 1/cos θ
Cosec θ = 1/sin θ
Hence,
Cot θ = Base/Perpendicular
Sec θ = Hypotenuse/Base
Cosec θ = Hypotenuse/Perpendicular
Trigonometry Examples
There are many real-life examples where trigonometry is used broadly.
If we have been given with height of the building and the angle formed when an object is seen from the top of the building, then the distance between object and bottom of the building can be determined by using the tangent function, such as tan of angle is equal to the ratio of the height of the building and the distance. Let us say the angle is ∝, then
Tan ∝ = Height/Distance between object & building
Distance = Height/Tan ∝
Let us assume that height is 20m and the angle formed is 45 degrees, then
Distance = 20/Tan 45°
Since, tan 45° = 1
So, Distance = 20 m
Applications of Trigonometry
Its applications are in various fields like oceanography, seismology, meteorology, physical sciences, astronomy, acoustics, navigation, electronics, etc.
It is also helpful to measure the height of the mountain, find the distance of long rivers, etc.
Video Lesson on Applications of Trigonometry
68,824
Trigonometry Problems and Solutions
Example 1: Two friends, Rakesh and Vishal started climbing a pyramid-shaped hill. Rakesh climbs 315 m and finds that the angle of depression is 72.3 degrees from his starting point. How high is he from the ground?
Solution: Let m is the height above the ground.
To find: Value of m
To solve m, use the sine ratio.
Sin 72.3° = m/315
0.953 = m/315
m= 315 x 0.953
m=300.195 mtr
The man is 300.195 mtr above the ground.
Example 2: A man is observing a pole of height 55 foot. According to his measurement, pole cast a 23 feet long shadow. Can you help him to know the angle of elevation of the sun from the tip of shadow?
Solution:
Let x be the angle of elevation of the sun, then
tan x = 55/23 = 2.391
x = tan-1(2.391)
or x = 67.30 degrees
Trigonometry Questions
Practise these questions given here to get a deep knowledge of Trigonometry. Use the formulas and table given in this article wherever necessary.
Q.1: In △ABC, right-angled at B, AB=22 cm and BC=17 cm. Find:
(a) sin A Cos B
(b) tan A tan B
Q.2: If 12cot θ= 15, then find sec θ.
Q.3: In Δ PQR, right-angled at Q, PR + QR = 30 cm and PQ = 10 cm. Determine the values of sin P,cos Pand tan P.
Q.4: If sec 4θ = cosec (θ- 300), where 4θ is an acute angle, find the value of A.
Frequently Asked Questions on Trigonometry
Q1
What do you Mean by Trigonometry?
Trigonometry is one of the branches of mathematics which deals with the relationship between the sides of a triangle (right triangle) with its angles. There are 6 trigonometric functions for which the relation between sides and angles are defined. Learn more about trigonometry now by visiting BYJU’S.
Q2
What are the six basic Trigonometric Functions?
There are 6 trigonometric functions which are:
Sine function
Cosine function
Tan function
Sec function
Cot function
Cosec function
Q3
What is the formula for six trigonometry functions?
The formula for six trigonometry functions are:
Sine A = Opposite side/Hypotenuse
Cos A = Adjacent side / Hypotenuse
Tan A = Opposite side / Adjacent side
Cot A = Adjacent side / Opposite side
Sec A = Hypotenuse / Adjacent side
Cosec A = Hypotenuse / Opposite side
Q4
What is the primary function of trigonometry?
The three primary functions of trigonometry are Sine function, Cosine function and Tangent Function.
Q5
Who is the founder of trigonometry?
A greek astronomer, geographer and mathematician, Hipparchus discovered the concept of trigonometry.
Q6
What are the Applications of Trigonometry in Real Life?
One of the most important real-life applications of trigonometry is in the calculation of height and distance. Some of the sectors where the concepts of trigonometry are extensively used are aviation department, navigation, criminology, marine biology, etc. Learn more about the applications of trigonometry here.
Learn about Trigonometry in a simple manner with detailed information, along with step by step solutions to all questions, only at BYJU’S. Download the app to get personalised videos.
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4768 | https://www.thyroid.org/patient-thyroid-information/ct-for-patients/vol-2-issue-4/vol-2-issue-4-p-5-6/ | Vol 2 Issue 4 p.5-6 | American Thyroid Association
Vol 2 Issue 4 p.5-6 | American Thyroid Association
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Home » Patients Portal » Clinical Thyroidology for the Public » Vol 2 Issue 4 » Vol 2 Issue 4 p.5-6
CLINICAL THYROIDOLOGY FOR PATIENTS
A publication of the American Thyroid Association
Summaries for Patients from Clinical Thyroidology (July 2009)
Table of Contents| PDF File for Saving and Printing
ANTI-THYROID MEDICATION
ABBREVIATIONS & DEFINITIONS
Hyperthyroidism: a condition where the thyroid gland is overactive and produces too much thyroid hormone. Hyperthyroidism may be treated with antithyroid medications (Methimazole, Propylthiouracil), radioactive iodine or surgery.
Graves’ disease: the most common cause of hyperthyroidism in the United States.
Agranulocytosis: a marked decrease in the white blood cell count that causes a patient to be more likely to develop an infection. This is commonly associated with a fever and/or a sore throat.
White blood cells: the infection-fighting cells of the blood.
Methimazole: an antithyroid medication that blocks the thyroid from making thyroid hormone. Methimazole is used to treat hyperthyroidism, especially when it is caused by Graves’ disease.
What is the study about?
Methimazole is an antithyroid medication that blocks the thyroid from making thyroid hormone. It is used to treat hyperthyroidism. The most common cause of hyperthyroidism is Graves’ disease, an autoimmune disease where the body makes antibodies that turn on the thyroid gland. Methimazole is usually a well-tolerated and safe drug, but on rare occasions, serious side effects can occur. The most serious side effect is called agranulocytosis, which may occur in 1 in 500 – 1000 patients. This is when the number of infection-fighting white blood cells in the blood decreases and cause the patient to be more likely to get an infection. The most common symptoms of this rare side effect are fever and/or a sore throat. For this reason, any patient on Methimazole is told to stop the drug and call their doctor should either of these symptoms occur. In the vast majority of patients with agranulocytosis, the white blood cell count returns to normal within 7–10 days. Some studies suggest that agranulocytosis is more likely to occur at higher rather than lower doses of Methimazole. This study attempts to see if agranulocytosis is more common in patients on 30 mg of Methimazole as compared to 15 mg of the drug.
The full article title: Takata, K., Kubota S, Fukata S., Kudo T, Nishihara E, Ito M, Amino N, Miyauchi A. Methimazole-induced agranulocytosis in patients with Graves’ disease is more frequent with an initial dose of 30 mg daily than with 15 mg daily. Thyroid 2009; 19: 559–63.
What was the aim of the study?
The aim of this study was to see if the rare side effect of agranulocytosis is more common in patients on 30 mg of Methimazole as compared to 15 mg of the drug.
Who was studied?
This study looked at 6,658 patients with Graves’ disease treated with Methimazole at Kuma hospital in Japan between 1991–2005.
How was the study done?
The records of the study patients diagnosed with Graves’ and treated with Methimazole were reviewed. If a patient developed a fever, Methimazole was discontinued and a white blood cell count was performed. Agranulocytosis was diagnosed if the white blood cell count was <1000 (normal range usually 5000–10,000).
What were the results of the study?
In all, 28 of the 6,658 patients (0.4%) were diagnosed with agranulocytosis. A total of 17 patients of the 2087 on 30 mg of Methimazole developed agranulocytosis (0.8%). A total of 6 patients of the 2739 on 15 mg of Methimazole developed agraqnulocytosis (0.3%). Dividing the groups more broadly, a total of 22 of the 3174 patients (0.7%) treated with over 20 mg of Methimazole developed agranulocytoisis as compared to 6 of the 3484 patients (0.2%) on less than 15 mg of Methimazole. No patient on less than 10 mg of Methimazole developed agranulocytosis.
How does this compare with other studies?
There are few studies on this topic. However, those studies do show that agranulocytosis is related to the dose of methimazole.
What are the implications of this study?
Agranulocytosis is a very rare side effect of the antithyroid drug Methimazole which is commonly used to treat hyperthyroidism caused by Graves’ disease. This rare side effect is more likely in patients treated with a daily dose of 30 mg than with 15 mg of Methimazole.
— Heather Hofflich, MD
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4769 | https://www.sciencedirect.com/science/article/pii/S2211034819304407 | Differential diagnosis of multiple sclerosis and other inflammatory CNS diseases - ScienceDirect
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1. Introduction
2. Multiple sclerosis
3. Neuromyelitis optica spectrum disorders
4. Neuropsychiatric systemic lupus erythematosus
5. Antiphospholipid syndrome
6. Primary central nervous system vasculitis
7. ANCA-vasculitis
8. Neuro-Behҫet disease
9. Neurosarcoidosis
10. Sjogren syndrome
11. Practical considerations
12. Conclusion
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Multiple Sclerosis and Related Disorders
Volume 37, January 2020, 101452
Review article
Differential diagnosis of multiple sclerosis and other inflammatory CNS diseases
Author links open overlay panel Paula Wildner, Mariusz Stasiołek, Mariola Matysiak
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•MRI remains the cornerstone of differential diagnosis of MS.
•Excessive reliance on radiological features may generate diagnostic errors.
•Inaccurate diagnosis of MS may impose potentially pernicious consequences.
•Rheumatic diseases are an underrated cause of white matter lesions on brain MRI.
•Further search for highly specific biomarkers seems to be of primary importance.
Abstract
Multiple sclerosis (MS) is the most common acquired demyelinating disorder of the central nervous system (CNS). Diagnosing MS can be very challenging owing to its variable clinical features and lack of specific tests. Magnetic resonance imaging (MRI) is a key measure in this process. Although white matter lesions on brain MRI are regarded as a hallmark of MS, they are a common radiological finding and their pattern may overlap in particular CNS inflammatory diseases. The increasing availability of therapies for MS and the knowledge of benefits associated with an early treatment underscore the importance of precise and quick diagnosis. Despite an extensive research, currently no fully specific diagnostic test is available to distinguish between CNS inflammatory disorders. In this review, we discuss characteristic findings and distinctive features of CNS inflammatory disorders, with particular focus on rheumatic diseases.
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Keywords
Multiple sclerosis
Rheumatic diseases
White matter lesions
CNS inflammatory disorder
Magnetic resonance imaging
Abbreviations
AC-13
apolipoprotein A1 C-terminal fragment
aCL
anticardiolipin antibodies
ADEM
acute disseminated encephalomyelitis
ANA
antinuclear antibodies
ANCA
antineutrophil cytoplasmic antibodies
anti-dsDNA
anti-double stranded DNA
anti-P
anti-ribosomal P
anti-Sm
anti-Smith
anti-β 2 GP1
anti-beta2glycopretein 1
aPL
antiphospholipid antibodies
AQP4
aquaporin-4
BBB
blood-brain barrier
BD
Behҫet disease
BLyS
B-lymphocyte stimulator
CIS
clinically isolated syndrome
CNS
central nervous system
CRP
C-reactive protein
CSF
cerebrospinal fluid
DIS
dissemination in space
DIT
dissemination in time
EAN
European Academy of Neurology
ECTRIMS
European Committee of Treatment and Research in Multiple Sclerosis
EGPA
eosinophilic granulomatosis with polyangitis
ESR
erythrocyte sedimentation rate
GABAR
gamma-aminobutyric acid receptor
GFAP
glial fibrillary acid protein
GPA
granulomatosis with polyangitis
HLA
human leukocyte antigen
ICAM
Intercellular Adhesion Molecule
IgG
immunoglobulin G
IL
interleukin
INF
interferon
LAC
lupus anticoagulant
LETM
longitudinally extensive transverse myelitis
MAGNIMS
Magnetic Resonance Imaging in Multiple Sclerosis
MOG
myelin oligodendrocytes glycoprotein
MPA
microscopic polyangitis
MPO
myeloperoxidase
MRI
magnetic resonance imaging
MS
multiple sclerosis
NBD
Neuro-Behҫet disease
NMDA
N-methyl-D-aspartate
NMO
neuromyelitis optica
NMOSD
neuromyelitis optica spectrum disorders
NPSLE
neuropsychiatric systemic lupus erythematosus
NR2
anti-N-methyl-D-aspartate receptor subtype 2 receptors
OCBs
oligoclonal bands
OCT
optical coherence tomography
PCNSV
primary central nervous system vasculitis
PPMS
primary progressive multiple sclerosis
Q alb
abumin quotient
RRMS
relapsing-remitting multiple sclerosis
SLE
systemic lupus erythematosus
SPMS
secondary progressive multiple sclerosis
STM
short transverse myelitis
TIA
transient ischemic attack
VCAM
vascular cell adhesion molecule
VEGF
vascular endothelial growth factor
WML
white matter lesions
1. Introduction
White matter lesions (WML) on brain magnetic resonance imaging (MRI) are common findings and in high percentage of cases they imply further diagnosis of multiple sclerosis (MS) and related demyelinating disorders. The reported prevalence of MS is increasing (Browne et al., 2014), which is at least partially a consequence of growing consciousness of MS in the society and also a wider availability of diagnostic tools, especially MRI. This situation, together with an inappropriate use of the diagnostic criteria, results in a considerable proportion of misdiagnoses. Since early implementation of treatment is crucial for its efficacy, precise diagnosis seems to be essential. Recent studies have emphasized the issue of erroneous diagnosis of MS and potential pernicious consequences which it entails, including aggressive immunosuppressive therapy (Solomon et al., 2016; Siva,2018 Feb; Solomon et al., 2012; Solomon and Weinshenker,2013). Studies have revealed that the most frequent disorder misdiagnosed as MS is migraine alone or in combination with other diagnosis, and it accounts for 22% of incorrectly diagnosed patients. Other conditions commonly misdiagnosed as MS are fibromyalgia, nonspecific white matter lesions, psychiatric disorders and neuromyelitis optica spectrum disorders (NMOSD). White matter lesions due to vasculitis also should not be disregarded (Solomon et al., 2016). Additionally, acute disseminated encephalomyelitis (ADEM) has to be considered as a cause of the first demyelinating episode, especially in the pediatric population. Differential diagnosis of MS involves also a broad spectrum of disorders eg. genetic metabolic disorders, such as Alexander disease, X-linked adrenoleukodystrophy, metachromatic leukodystrophy, Canavan disease, infectious diseases, mostly opportunistic such as toxoplasmosis, progressive multifocal leukoencephalopathy, neuroborreliosis and lymphoproliferative diseases (Siva,2018 Feb). On the other hand, delayed MS diagnosis in people who obtain treatment for another disorder even for years is an equally important issue. Although MRI is currently the most valuable tool in differential diagnosis of MS, excessive reliance on radiological features is often responsible for diagnostic errors. Because the issue of MS misdiagnosis is of such a paramount importance, further search for specific biomarkers seems mandatory. In this review, we shall discuss MS and other demyelinating autoimmune disorders of the central nervous system (CNS), with a particular focus on rheumatic diseases, and highlight the distinctive features which may facilitate the proper diagnosis.
2. Multiple sclerosis
MS is presumed to be the most common acquired demyelinating disorder of the central nervous system. MS is assumed to affect over 2.5 million people globally, and is regarded as one of the primary causes of disability, especially among young adults (Pelletier and Hafler,2012). The prevalence of MS is highly heterogeneous. The incidence of MS differs between populations and reaches even 10 new cases per 100 000 (Kingwell et al., 2015; Grytten et al., 2016). Although a meta-analysis, including 59 countries, revealed a statistically significant latitudinal gradient for MS prevalence (Simpson et al., 2011), other studies have shown no remarkable latitudinal or longitudinal gradient in some regions (Poppe et al., 2008; Melcon et al., 2008). The risk of developing MS is determined by genetic as well as environmental factors. Current data on genetic, epigenetic, and transcriptome studies in monozygotic twins has proven discordance for MS (Mumford et al., 1994; Baranzini et al., 2010).
Even though demyelination is regarded as the pathologic hallmark of MS, studies indicate that axonal injury is playing a key role in the persistence of neurological deficits (van Waesberghe et al., 1999; Fisher et al., 2007). The exact etiology and pathomechanisms of MS remain not fully known. However, the leading role of autoimmune and neurodegenerative processes has long been postulated. The cascade of MS pathology comprises demyelination, oligodendrocyte loss, neuronal loss, axonal damage, astrogliosis, and progressive failure of remyelination (Zeydan and Kantarci,2018; Lassmann et al., 2012). It is assumed that inflammatory reaction associated with different cell types including T cells, B cells, activated microglia, macrophages and their products results in demyelination (Lassmann,2011), whereas various inflammation induced and/or mediated processes e.g. oxidative stress leading to mitochondrial injury perpetuate neurodegeneration (Lassmann,2011). All the processes coincide at every stage of the disease, though their intensity varies in time and the progressive phase of MS becomes clinically apparent when the axonal damage threshold is surpassed (Confavreux et al., 2003). The present disease course classification comprises the relapsing-remitting MS (RRMS) and progressive MS, which is divided into primary progressive MS (PPMS) and secondary progressive MS (SPMS) (Lublin et al., 2014; Miller and Leary,2007). Each phenotype is further defined by the status of disease activity and progression (Lublin et al., 2014).
Current 2017 McDonald diagnostic criteria for multiple sclerosis include clinical, imaging and laboratory findings (Thompson et al., 2018). MRI is of utmost importance in the diagnosis of MS. Magnetic Resonance Imaging in Multiple Sclerosis (MAGNIMS) network and the Consortium of Multiple Sclerosis Centers have given recommendations on the use of MRI in MS diagnosis (Wattjes et al., 2015; Traboulsee et al., 2016). Brain and spinal cord MRI may replace clinical features to meet the dissemination in space (DIS) or dissemination in time (DIT) criteria. Typical MS MRI findings are T2-hyperintense lesions in four areas of the CNS: periventricular, cortical or juxtacortical, infratentorial and spinal cord (Thompson et al., 2018). Besides the diagnostic criteria of MS, MRI is highly advantageous in detecting findings suggestive of disorders other than MS. Awareness of such features seems to be essential for understanding the concept of ‘no better explanation’ (Charil et al., Oct). Table 1 shows common MS MRI findings and “red flags”- suggestive of other inflammatory CNS diseases. Fig.1 (a-f) demonstrates different types of white matter lesions. Although cerebrospinal fluid (CSF) examination is not obligatory to set the diagnosis of MS, 2017 revisions of McDonald criteria has emphasized the significance of the presence of CSF-specific oligoclonal bands (OCBs), which may substitute MRI criterion for DIT in patients with clinically isolated syndrome (CIS) (Thompson et al., 2018). Moreover, the presence of CSFoligoclonal bands was shown to represent an independent predictor of a second attack in patients with CIS (Andreadou et al., 2013; Dobson et al., 2013). On the other hand, such CSF findings as elevated protein concentration above 100 mg/dL, pleocytosis over 50 cells/mm 3 or presence of neutrophils, eosinophils or atypical cells are uncommon in MS, and should imply consideration of other diagnosis (Stangel et al., 2013a). Until now, no fully specific biomarker of MS is available, which makes careful exclusion of other possible causes of observed clinical symptoms mandatory (Miller et al., 2008). It is worth to mention, that neurofilament light chain (NfL) levels in serum of MS patients are associated with clinical and MRI-related measures of disease activity and neuroaxonal damage and have prognostic, but not diagnostic value (Kuhle et al., 2019). The European Committee of Treatment and Research in Multiple Sclerosis (ECTRIMS) and the European Academy of Neurology (EAN) have issued practical guidelines on the pharmacological MS treatment (Montalban et al., 2018). Despite the knowledge of a few prognostic factors (Tintore et al., 2015; Jokubaitis et al., 2016), there is still an unmet need for biomarkers, which would facilitate the stratification of patients into groups suited for particular treatment approach.
Table 1. MRI ‘red flags’ (Jog and James,2017; Ungprasert and Matteson,2017; Maggi et al., Feb; Chen et al., Sep; Ramanathan et al., 2016a; Charil et al., Oct).
| MRI finding | typical for MS | ‘red flags’ |
---
| shape and size of lesions | ovoid, perivenular, Dawson's fingers | massive, confluent |
| U-fibers involvement | yes | no |
| lesion morphology | regular, distinct border | fluffy, indistinct border |
| gadolinium enhancement | homogeneous; ring, open ring; pattern changing in successive MRIs | heterogeneous enhancement; cloud-like, pencil-thin; pattern persistent over months |
| location | periventricular, juxtacortical/ asymmetrical | peripheric, cortical/ symmetrical |
| mass effect | rare, eg. in tumefactive MS | prominent in granulomatous diseases |
| optic neuritis | | more than half of the length of the optic nerve, comprising intracranial segment, typically extending to the chiasm and optic tract |
| corpus callosum | small ovoid lesions, atrophy in later phase | ependymal surface involvement; “arch bridge” appearance - NMOSD |
| posterior fossa-brainstem lesions | small | upward/downward extension of brainstem lesion - NBD |
| spinal cord lesions | short (less than 3 vertebral segments); <50% of the cross-sectional area; peripheral | longitudinally extensive (at least 3 vertebral segments); >50% of the cord area; central gray matter |
| meningeal involvement | no | yes, leptomeningeal and pachymeningeal |
| infarcts or hemorrhages | no | yes, especially multiple |
| calcifications | no | yes |
| typical sign | central vein sign | Bagel sign - NBD; punctuate enhancement and swelling of basal ganglia – NPSLE; diencephalon involvement – NMOSD; area postrema involvement - NMOSD; bright spotty lesions in the spinal cord - NMOSD; “string of pearls” (clusters of snowball-like callosal lesions) – Susac syndrome |
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Fig. 1. Different types of white matter lesions in CNS inflammatory diseases. (a). Typical MRI findings in MS: T2-hyperintense periventricular and juxtacortical lesions. (b). T1 scan showing ring and open ring gadolinium enhancement pattern of lesions present in Fig.1a. (c). Subcortical FLAIR-hyperintense white matter lesions in a patient with APS secondary to SLE. (d). Cortical-subcortical malacia after an ischaemic stroke in the same patient with APS secondary to SLE presented in fig. 1c. (e). T2 dark fluid-hyperintense lesion in area postrema in a patient with SLE which needs to be differentiated from NMOSD. (f). Diffuse T2-hyperintense lesion at the level Th1 to Th4 affecting white as well as gray matter of spinal cord in NMO patient. Small T2-hyperintense lesions visible also at the level of Th8 and Th10.
3. Neuromyelitis optica spectrum disorders
NMOSD, including neuromyelitis optica (NMO) or Devic's syndrome, comprise a group of the central nervous system (CNS) inflammatory conditions, affecting preferentially optic nerves and spinal cord (Bruscolini et al., 2018). NMO is a relatively uncommon disease, with the incidence and prevalence ranging from 0.05–0.4 and 0.52–4.4 per 100,000, respectively. NMO primarily affects young adults, with a mean age 32.6–45.7 years. A significant predominance in women is observed, who constitute from 68% to 88% of the affected population (Pandit et al., 2015). The core clinical characteristics of NMOSD are optic neuritis, acute myelitis, area postrema syndrome, acute brainstem syndrome, symptomatic narcolepsy or acute diencephalic clinical syndrome with NMOSD-typical diencephalic MRI lesions and symptomatic cerebral syndrome with NMOSD-typical brain lesions (Wingerchuk et al., 2015). Acute optic neuritis, in particular bilateral or rapidly sequential, and longitudinally extensive transverse myelitis (LETM), are the most usual manifestations of NMO (Wingerchuk et al., 1999). Thus, the most common symptoms involve ocular pain with a severe visual acuity loss, symmetric paraplegia, sensory loss and bladder dysfunction (Wingerchuk et al., 2007). Other alarm signs are nausea/vomiting, intractable hiccups, diplopia and nystagmus, hearing and balance disorder, narcolepsy, obesity or even acute respiratory failure (Bruscolini et al., 2018; Sahraian et al., 2013).
Current criteria divide NMOSD into two subtypes according to the seropositivity of aquaporin-4 immunoglobulin G (AQP4-IgG): NMOSD with AQP4-IgG (NMOSD- AQP4-IgG) and NMOSD without AQP4-IgG or with unknown AQP4-IgG status (Wingerchuk et al., 2015). The diagnosis of NMOSD is based on clinical manifestations, serological testing and MRI. In the presence of AQP4-IgG, only one of the core clinical characteristics is sufficient to make the diagnosis, provided that alternative diagnoses have been excluded. In patients with a negative AQP4-IgG test or when the test is unavailable, at least two core characteristics of NMOSD have to be present (one of them obligatory being optic neuritis, acute myelitis with LETM or area postrema syndrome) alongside with supportive MRI features and an exclusion of alternative diagnoses (Wingerchuk et al., 2015). AQP4 is the key target in NMO pathogenesis. AQP4 is an astrocyte water channel protein which promotes the movement of water across cell membranes in response to osmotic gradients (Papadopoulos and Verkman,2012). AQP4 is primarily expressed in astrocyte foot processes, in particular within the optic nerves, spinal cord, periventricular areas, hypothalamus, subpial regions in cerebellar hemispheres, brainstem and area postrema (Zekeridou and Lennon,2015). The clinical course of NMOSD seems to vary between AQP4-IgG-seropositive and seronegative patients. In the AQP4-IgG-seropositive group, more severe and frequent relapses, higher female to male ratio, more unfavorable outcomes and more often coexisting autoimmune disorders should be expected (Jarius et al., 2012). In addition to AQP4, a few other immune pathogenic targets have been revealed in the context of NMOSD, mainly myelin oligodendrocytes glycoprotein (MOG), but also glial fibrillary acid protein (GFAP), S100 protein, metalloprotease-9, VEGF A, ICAM-1, VCAM-1 (Lennon et al., 2004; Jarius et al., 2014). The available information regarding the MOG-IgG-positive patients with NMOSD indicate that the disease usually starts at a younger age, the female to male ratio is lower, is more often monophasic and with a less severe outcome than in AQP4-IgG-positive NMOSD (Sato et al., 2014; van Pelt et al., 2016; Jurynczyk et al., 2017). The recently described concept of MOG antibody disease comprises a group of autoimmune disorders with a predilection for optic nerve and spinal cord involvement (Denève et al., 2019; Lana-Peixoto and Talim,2019).
MRI is of great importance in the process of NMOSD diagnosis, especially among the AQP4-negative patients. Optic neuritis in NMOSD is commonly bilateral and longitudinally extensive, involving over one half of the optic nerve, most typically in the posterior segment prolonging to the chiasm and optic tract (Ramanathan et al., 2016a). On the contrary, optic neuritis associated with MOG antibody localizes typically in the anterior part of the nerve, is stretched, oedematous, with more prominent inflammation than in AQP-4-IgG positive patients (Carra-Dalliere et al., 2018). Characteristic MRI feature of MOG antibody disease is enhancement of the peri‑optic nerve sheath, partly extending into the surrounding orbital fat (Ramanathan et al., 2016b). LETM, as the manifestation of NMOSD, extends for at least three adjacent vertebral segments and involves primarily the central gray matter of the spinal cord and over 50% of the cord area, constituting transversally extensive lesions (Pekcevik et al., 2016a; Dumrikarnlert et al., 2017). Lesions within the cervical spine, typically spread to the area postrema (Kim et al., 2015). Bright spotty lesions with a strong hyperintensity on axial T2-weighted images, with higher signal intensity than the surrounding cerebrospinal fluid, without flow void effects are suggested as one of the most characteristic CNS MRI features in NMOSD (Yonezu et al., 2014). Cord swelling and irregular enhancement on T1-weighted images can be observed in acute phase. The ring enhancement pattern in NMOSD LETM has a lens-shaped appearance on sagittal scans (Zalewski et al., 2017). Besides LETM, the most typical NMOSD lesion, short transverse myelitis (STM), previously considered contradictory to NMOSD, is the first sign of 14,5% of all NMO myelitis (Flanagan et al., 2015). However, STM in NMO is generally located in the central gray matter and is often longer than the spinal lesion in MS (Huh et al., 2017). Interestingly, conus involvement is highly specific for MOG antibody disease (Jarius et al., 2018). Accurate diagnosis of myelopathy is crucial as irreversible neurological deterioration may develop promptly without adequate treatment. Moreover, results of a recently published retrospective analysis suggest that the majority of patients diagnosed with idiopathic transverse myelitis have other specific myelopathy (Zalewski et al., 2018). Brain lesions are not an uncommon phenomenon in NMOSD, as they can be detected in 25–59% of patients (Wang et al., 2011; Chan et al., 2011). Most importantly, 16% of these lesions are found to meet the Barkhof MRI criteria for MS (Wingerchuk et al., 2015). The prevailing brain NMO lesions are confluent hyperintensities on FLAIR/T2-weighted images scattered asymmetrically in periependymal areas. The ependymal surface of corpus callosum, diencephalon and brainstem, involving the area postrema, are the key sites of NMO brain lesions (Wingerchuk et al., 2015). Cortical lesions are incompatible with NMO, and thus they are considered a ‘red flag’ implying the need for other diagnosis. Nevertheless, cortical involvement, particularly in the subpial layer associated with leptomeningeal enhancement, has been reported (Pekcevik et al., 2016b). The most frequent patterns of enhancement are cloud-like enhancement and the periependymal linear pattern, or pencil-thin enhancement, along the ventricular surface system. When existing together, these two patterns might present a flame-like impression. Ring and ‘open ring’ enhancement indicate rather the diagnosis of MS (Pekcevik et al., 2016b). However, MRI of patients with acute optic neuritis may be normal. In these cases, visual evoked potentials and optical coherence tomography (OCT) are substantial. OCT is a valuable tool in the assessment of optic nerve fibers damage, and it was suggested that this parameter may potentially serve as a marker of disease progression and treatment response (Bennett et al., 2015). Importantly, OCT in NMOSD shows a more apparent than in MS retinal nerve fiber layer and ganglion cells layer thinning (Bennett et al., 2015). Unlike in MS, CSF tests in NMOSD more often show elevated pleocytosis (in 35%), especially with predominant neutrophils or eosinophils (Wingerchuk et al., 2015). Another differentiating factor between NMOSD and MS is an increased CSF level of IL-6 in the former disease (Uzawa et al., 2014). Although CSF oligoclonal bands might be transiently detectable during exacerbations, their absence is considered as supportive evidence for NMOSD (Wingerchuk et al., 2015). The proper differentiation between MS and NMOSD is crucial for therapy planning. Several maintenance therapies approved for MS have been found ineffective or even aggravating the severity of clinical symptoms in NMOSD (Uzawa et al., 2014). Furthermore, recently published results of clinical trials have proved effectiveness of therapeutic approaches based on the pathomechanism of NMOSD such as intervention in the cascade of complement (Pittock et al., 2019)
4. Neuropsychiatric systemic lupus erythematosus
Systemic lupus erythematosus (SLE) is a chronic systemic autoimmune disease displaying a wide variety of clinical manifestations (Jafri et al., 2017). Central nervous system involvement in SLE is associated with a more severe course of the disease and a worse prognosis (Cervera et al., 1999). The reported prevalence of neuropsychiatric SLE (NPSLE) differs considerably, ranging from 14% to 75% of all SLE cases, most probably due to the vast variety of symptoms and changing rules of assignment to the primary disease (Unterman et al., 2011). The American College of Rheumatology defined 19 manifestations of NPSLE, which affect central, peripheral or autonomic nervous system. Among the CNS manifestations of SLE, cerebrovascular disease and epilepsy are considered the most clinically important. Demyelinating syndrome, aseptic meningitis and psychiatric disorders also need to be mentioned as significant medical challenges (Hanly,2004; Hanly,2014a). NPSLE manifestations can be categorized as diffuse or focal, depending on the prevailing pathomechanism. Diffuse manifestations include headaches, affective disorders, psychosis, cognitive dysfunction and acute confusional state (Jafri et al., 2017). Crucial focal manifestations of NPSLE are cerebrovascular disease and epilepsy. Cerebrovascular events are reported in 5% to 18% of SLE patients. The increased risk of stroke is contributed to procoagulant factors such as antiphospholipid antibodies (aPL), endothelial activation and vasculitis, as well as accelerated atherosclerosis (Hanly et al., Oct). Seizures affect 8% to 18% of SLE patients (Muscal and Brey,2010) and are associated with the presence to anti-Smith antibodies (Mikdashi et al., 2005) and aPL antibodies in serum (Herranz et al., 1994). Another focal NPSLE manifestation is transverse myelitis, which mimics NMOSD. There is a strong association between transverse myelitis and the presence of aPL antibodies in patients with SLE (Kovacs et al., 2000a). Histopathological studies have shown that the predominant process in NPSLE is a noninflammatory microangiopathy with concomitant brain microinfarction. Other reported pathological findings have been glial hyperplasia and diffuse neuronal and axonal loss (Sibbitt et al., 2010). It is also assumed that NPSLE may be a result of the blood-brain barrier (BBB) disruption, resulting from complex immune disturbances (Hanly,2014b). Among them, antibody production and immune complex formation are of primary importance in the pathogenesis of SLE. Until now, as many as 180 different autoantibodies have been detected in SLE patients (Yaniv et al., 2015). In routine clinical practice, usually only anti-double stranded DNA (anti-dsDNA) and aPL are evaluated. Other important antibodies associated with NPSLE are anti-ribosomal P (anti-P), anti-GABAR B1b and anti-GABAR B2, anti-neuronal and anti-endothelial antibodies (Clark et al., 2017). Anti-dsDNA antibodies are highly specific (92% to 96%) and of moderate sensitivity (57% to 67%) for SLE (Cozzani et al., 2014). High levels of anti-dsDNA antibodies are found in 70% of patients presenting NPSLE symptoms (Joseph et al., 2007). A subset of anti-dsDNA antibodies cross-react with anti-N-methyl-D-aspartate receptor subtype 2 receptors (NR2). Anti-NR2 antibodies in the CSF can be identified in one third of all SLE patients, which is a higher prevalence than in any other autoimmune disease (Gono et al., 2011). Bound to active NMDA receptors, the anti-NR2 antibodies induce excess calcium influx leading to neuronal dysfunction and cell death (Faust et al., 2010). Antiphospholipid antibodies comprise anticardiolipin antibodies (aCL), lupus anticoagulant (LAC) and anti-beta2glycoprotein 1 (anti-β 2 GP1) antibodies. Among focal NPSLE manifestations, aPL are highly prevalent in patients with transverse myelitis, which involves especially the thoracic cord (Kovacs et al., 2000b). aCL antibodies are the most sensitive and the least specific finding. Importantly, aCL antibodies can be temporarily induced by many drugs and infections. LAC has a strong correlation with cerebrovascular disease, particularly sinus thrombosis (Hanly et al., 2011). Anti-β 2 GP1 antibodies are more specific for thrombotic events than aCL (Bertolaccini and Khamashta,2006). Anti-P antibodies are detected in 15% to 20% of SLE patients (Zandman-Goddard et al., 2007) and are considered to have a strong association with psychosis and depression (Hirohata et al., 2007). Besides the abovementioned antibodies, serum analysis may reveal an elevated erythrocyte sedimentation rate (ESR) with normal C-reactive protein levels (Joseph et al., 2007) and decreased concentrations of complement – finding typical of active disease (Magro-Checa et al., 2016). Cerebrospinal fluid analysis should be performed in each case of suspected NPSLE. The usual mild non-specific abnormalities observed in NPSLE are: slightly raised white cell count (typically lymphocytosis), elevated protein and IgG index with a decreased or normal glucose level (Joseph et al., 2007). It might be useful to assess BBB integrity. Since albumin is not generated intrathecally, the relatively high leakage across the BBB may be demonstrated by the assessment of albumin concentration gradient between CSF and plasma - albumin quotient (Q alb) (Abbott et al., 2003). CSF oligoclonal bands can be found in 15% to 85% of NPSLE cases (Clark et al., 2017), which may indicate considerable heterogeneity of investigated SLE patient populations. CNS MRI is the key measure in diagnosis of NPSLE. The primary MRI manifestations of SLE are large infarcts and multifocal white matter T2-hyperintense lesions. White matter T2-hyperintense lesions are the most common radiological finding in NPSLE and it seems crucial to differentiate them from MS lesions. In contrary to MS, SLE white matter lesions more often exhibit a vascular distribution (Jennekens and Kater,2002). Another distinction from MS is the bilateral involvement of basal ganglia, with swelling and punctuate enhancement. Calcifications in the basal ganglia, dentate nucleus, centrum ovale and corticosubcortical junctions can also be observed (Raymond et al., 1996). Leptomeningeal enhancement may be also present in CNS MRI of SLE patients. Spinal cord lesions are often longitudinally extensive (Jog and James,2017). Importantly, MRI findings in NPSLE may in some cases fulfill the radiologic diagnostic criteria for MS, which underlines the need for development of specific biomarkers. Given the abundant psychiatric manifestations of SLE, neuropsychological assessment should be performed. Treatment strategies of NPSLE depend on the prevailing manifestations and severity of the disease. Interestingly, it is suggested that the optimal treatment window in NPSLE occurs much earlier than the average time of diagnosis. Thus, the therapy should be introduced without delay, in patients with unspecific clinical syndrome and multifocal brain and spinal cord lesions. Unfortunately, no specific test allows early and definite diagnosis of NPSLE. Thus, further research is required to facilitate the diagnostic strategy.
5. Antiphospholipid syndrome
As mentioned above, aPL antibodies are of paramount importance in the context of the pathogenesis of neuropsychiatric lupus manifestations. Antiphospholipid syndrome (APS) is an acquired systemic disorder related to the circulating aPL antibodies. Current classification criteria for APS include fulfillment of at least one of clinical criteria such as vascular thrombosis or particular pregnancy related conditions and at least one of the laboratory criteria which is a positive test for aPL antibodies: LAC, aCL or anti-β 2 GP1, present on 2 or more occasions not less than 12 weeks apart (Miyakis et al., 2006). APS might be primary or secondary to other autoimmune disease, with SLE being the most common cause of secondary APS (Graf,2013). Besides vascular thrombosis and pregnancy morbidity, APS comprises a wide spectrum of symptoms, including hematological, cutaneous, renal, cardiac and pulmonary involvement (Negrini et al., 2017). Hematological manifestations of APS are aPL-associated idiopathic thrombocytopenic purpura and autoimmune hemolytic anemia (Cervera et al., 2002). Livedo reticularis is the hallmark of APS (Cervera et al., 2002) and has been proposed as an independent risk factor of arterial thrombosis (Frances et al., 2005). APS nephropathy can be acute or chronic and in secondary APS may coexist with immune-complex-mediated lupus nephritis (Pons-Estel and Cervera,2014; Alchi et al., 2010). Other systemic manifestations of APS involve Libman-Sacks endocarditis, intraalveolar hemorrhage, acute respiratory distress syndrome and fibrosing alveolitis (Hojnik et al., 1996; Kanakis et al., 2013; Barnini et al., 2012). The most frequently affected arterial system in APS is the cerebral circulation and consequently the most prevalent neurological complication of APS is a cerebrovascular event (Rodrigues et al., 2010). Ischemic stroke may be related to vessels of any size and located in any area of the brain, without typical predilection (Tanne and Hassin-Baer,2001). It is estimated that at least 20% of stroke cases under the age of 45 can be attributed to APS (Hughes,2003). In particular, in patients with secondary APS, LAC is a remarkably stronger predictor of CNS ischemic events than the aCL antibodies (de Amorim et al., 2017; Petri et al., 1987). Cognitive symptoms are another important manifestation of APS. It has been demonstrated that between 42% and 80% of APS patients exhibit global cognitive impairment (Yelnik et al., 2016; Brey et al., 2011). Studies have shown a positive correlation between cognitive dysfunction and aCL antibodies (Jacobson et al., 1999; Menon et al., 1999; Hanly et al., 1999). Epilepsy is considered as one of the main CNS manifestations of APS. It is more frequent in secondary APS, in which it affects 13,7% of patients (Shoenfeld et al., 2004). Although epileptogenic foci secondary to ischemic events are considered as the most important cause of seizures in APS, autoimmune process is presumed to be also substantially involved. It needs to be indicated that clinical features of APS might resemble MS. Possible MS-like manifestations of APS encompass e.g. optic neuritis or transverse myelitis with CNS MRI revealing MS-type white matter lesions. It may be particularly misleading, as scattered hyperintensive subcortical white matter lesions are the most common MRI finding in APS (Zhu et al., 2014). Taking into consideration the lack of fully specific diagnostic tests, differential diagnosis in such cases may prove to be complex and difficult (Rodrigues et al., 2010). Similarly to NPSLE, treatment approach in APS needs to be adjusted to the prevailing manifestations and severity of symptoms.
6. Primary central nervous system vasculitis
Primary central nervous system vasculitis (PCNSV), also known as primary angitis of the CNS, is an insufficiently investigated inflammatory condition which affects vessels of brain, spinal cord and meninges (Mandal and Chung,2017). Due to an ample spectrum of clinical manifestations and absence of any specific diagnostic test, PCNSV remains a commonly overlooked disorder. Although rare, PCNSV should be taken into consideration in each case of suspected CNS vasculitis when secondary causes e.g. infectious, connective tissue diseases, or systemic vasculitis, cannot be confirmed. Unlike in secondary vasculitis, men are more susceptible to PCNSV than women, with the male to female ratio estimated at 2:1, and a mean age of onset being 50 years (Salvarani et al., 2007). Patients with PCNSV may exhibit multiple symptoms. The disease usually starts with a protracted prodromal period, lasting weeks or months, during which headaches and mild cognitive disturbances often appear (Salvarani et al., 2015). After prodromal phase, diverse neurological manifestations can emerge, such as visual disturbances, cranial neuropathies, stroke with focal deficits, cerebellar syndrome, epilepsy (Boysson et al., 2014). Constitutional symptoms, such as fever and weight loss, are usually absent, and they imply rather the diagnosis of secondary vasculitis (Salvarani et al., 2015). PCNSV is a highly heterogeneous disease, which may affect vessels of various size, mainly medium and small. Also the pathological findings differ considerably. Several PCNSV subtypes have been characterized, with granulomatous angitis of the CNS being the most common (Salvarani et al., 2007). The clinical course of granulomatous angitis is indolent. The prodromal period with headaches and cognitive impairment lasts from 3 to 6 months, and is typically followed by focal deficits or epilepsy (Suri et al., 2014). As granulomatous angitis typically involves small vessels, angiography usually shows no abnormalities (Suri et al., 2014). Currently, there is no specific laboratory test available for PCNSV. Immunological tests, ESR and CRP should be assessed to exclude secondary vasculitis. CSF examination should be performed in each case of suspected PCNSV, mainly to differentiate with infections and neoplastic causes. In majority of patients with PCNSV some nonspecific CSF abnormality is found, usually mild lymphocytic pleocytosis or elevated protein level, with normal glucose (Salvarani et al., 2015). Brain MRI is an obligatory step in the diagnosis of PCNSV and it typically shows multifocal, often bilateral, infarcts in numerous vascular territories, with parenchymal enhancement (Boulouis et al., 2017; Goertz et al., 2010). Conventional cerebral angiography remains the basic tool to evaluate cases of highly probable PCNSV. Characteristic picture of “beading”, which is formed by alternating segments of stenotic and dilated vessels, is a cornerstone of diagnosis (Edgell et al., 2016). Other possible findings include focal occlusion, collateral circulation, microaneurysms. Magnetic resonance angiography, in particular high resolution multicontrast wall and lumen imaging, with the use of the 3T scanner or stronger, can also prove beneficial (Cosottini et al., 2013). Gold standard test for PCNSV diagnosis is brain and leptomeningeal biopsy. The biopsied sample needs to contain leptomeninges, cortex and subcortical white matter from radiologically altered location (Hajj-Ali et al., 2011; Miller et al., 2009).
7. ANCA-vasculitis
The antineutrophil cytoplasmic antibodies (ANCA)-associated CNS vasculitis is a group of primary systemic vasculitic diseases comprising granulomatosis with polyangitis (GPA), microscopic polyangitis (MPA) and eosinophilic granulomatosis with polyangitis (EGPA) (Graf,2017 Nov). GPA is an idiopathic necrotizing small-vessel disease related to ANCA specific for proteinase 3 (PR3) (Graf,2017 Nov). The most common neurological manifestation of GPA, observed in 60% of cases, is peripheral neuropathy. CNS is affected in 4% to 11% of GPA patients (Dutra et al., Feb). Three different types of manifestations can be distinguished including: cerebral vasculitis and two predominantly granulomatous phenotypes - chronic hypertrophic pachymeningitis and pituitary disease (Graf,2017 Nov). The vascular phenotype can lead to ischemic as well as hemorrhagic complications. The imaging results of GPA ischemic lesions are nonspecific. Brain MRI might reveal both extensive cerebral infarctions and diffuse white matter lesions as seen in small-vessel vasculopathy (De Luna et al., 2015). The size of involved vessels is often below the level of resolution of MRI or conventional angiography (Ghinoi et al., 2010). GPA affects more frequently pachymeninges than leptomeninges and intracranial dura matter is more often involved than spinal (Holle and Gross,2011). Chronic hypertrophic pachymeningitis can manifest clinically with variety of symptoms such as headache, cranial neuropathy, ataxia, myelopathy, epilepsy and thickening of the dura on MRI scans (Shimojima et al., 2017). MPA is associated with ANCA directed against myeloperoxidase (MPO) and affects small- to medium-sized vessels (Graf,2017 Nov). Although MPA mostly exhibits as glomerulonephritis and diffuse alveolar hemorrhage, CNS may be involved as well (Duvuru and Stone,2013). CNS vasculitis in the course of EGPA is a rare phenomenon and usually constitutes the final phase of the disease. The primary EGPA manifestations are asthma and nasal polyps or rhinosinusitis, followed by an eosinophilic phase characterized by peripheral eosinophilia and organ involvement (Greco et al., 2015). EGPA affects the CNS in 6% to 10% cases and causes encephalopathy, ischemic infarcts and hemorrhages (Murthy et al., 2013).
8. Neuro-Behҫet disease
Behҫet disease (BD) is a chronic relapsing inflammatory vascular disease of not fully known etiopathogenesis. CNS presentation is described in 5% to 10% of BD cases (Siva and Saip,2009). Several genetic, environmental and microbiological factors have been postulated as risk factors for BD (Onder and Gürer,2001). Among genetic factors, a significant relationship has been found between human leukocyte antigen (HLA)-B51 and BD (Demirseren et al., 2014). HLA-B51 gene is presumed to be the strongest risk factor for BD (Kirino et al., 2013). It has been also shown that HLA-B51 is associated with a lower frequency of neurological involvement (Hamzaoui et al., Jun). However, this correlation may be more complicated and HLA-B51 subtype specific, since in one study an increased frequency of HLA-B5103 has been observed in BD patients with CNS involvement (Demirseren et al., 2014). As there is no pathognomonic test for BD, the diagnosis is based on clinical criteria. To meet current International Criteria for Behҫet Disease (ICBD), a patient needs to score at least 4 points, 2 points being ascribed for any of ocular lesions, oral aphthosis, genital aphthosis and 1 point for any of skin lesions, neurological manifestations, and vascular manifestations. An extra point may be assigned for positive pathergy test (International Team for the Revision of the International Criteria for Behçet's Disease(ITR-ICBD) 2014). There are two major forms of Neuro-Behҫet disease (NBD). A vascular-inflammatory disease, also called parenchymal NBD, often mimics MS clinical picture. The other major NBD presentation - non-parenchymal NBD encompasses an isolated cerebral venous sinus thrombosis and intracranial hypertension (Siva and Saip,2009; Al-Araji and Kidd,2009). Parenchymal NBD is usually more severe than non-parenchymal NBD, and it comprises such manifestations as meningoencephalitis involving brainstem, cranial nerve palsies and epilepsy (Al-Araji and Kidd,2009). In contrary to MS, NBD affects men more often than women. Optic neuritis, spinal cord involvement and sensory symptoms are rare in NBD and thus should always warrant careful differential diagnostics with other possible causes e.g. MS (Kocer et al., 1999). MRI is the gold-standard in NBD diagnosis. Typical CNS MRI finding in NBD is a large lesion in brainstem, without apparent border, with a tendency to spread to diencephalon and basal ganglia, less frequently caudally (Kocer et al., 1999). Heterogeneous contrast enhancement may be observed in acute phase, resolving usually incompletely in subsequent imaging. Typically, brainstem lesions on MRI are much smaller in MS. As mentioned above, spinal cord involvement is uncommon in NBD, but when it occurs, the lesions are longitudinally extensive, resembling lesions characteristic of NMOSD (Uygunoglu et al., 2016). The most specific spinal MRI feature of NBD, described till now, is the “bagel sign” pattern, visible in axial T2WIs, and defined as a central spinal cord lesion with hypointense core and hyperintense rim, with or without contrast enhancement (Uygunoglu et al., 2017). CSF OCBs positivity in NBD patients is significantly less common than in MS, observed in only up to 15% of cases (Siva,2018 Feb).
9. Neurosarcoidosis
Another inflammatory disease exhibiting multiple brain lesions on MRI is neurosarcoidosis, a chronic granulomatous disorder of unknown origin. Nervous system involvement is reported in 3% to 10% of patients with sarcoidosis, but such prevalence is presumed to be underestimated (Ungprasert and Matteson,2017). Although isolated neurosarcoidosis is a rare condition, in about 70% to 80% of patients neurological symptoms are one of the earliest disease manifestations (Ferriby et al., 2001). The most prevalent neurological complication of sarcoidosis is cranial neuropathy, with a particular predilection for cranial nerves VII, II and III. Facial nerve palsy may be bilateral and is probably a result of epi‑ and perineural inflammation and external compression by granuloma (Carlson et al., 2014). Optic neuritis, often bilateral, has been observed as an initial disease presentation in even 35% of neurosarcoidosis cases (Pawate et al., 2009a). Vestibulocochlear nerve injury is considered to result from granulomatous meningitis (Kane,1976). Brain parenchymal disease constitutes another significant issue and is presumed to be a result of chronic inflammation and consecutive atherosclerosis (Pawate et al., 2009b). It can manifest as headache, cognitive impairment, affective disorders and epilepsy (Ungprasert et al., 2017). Sarcoidal granulomas may infiltrate the pituitary gland and hypothalamus, causing endocrinological disturbances (Stern et al., 1985), or spinal cord with a predilection for thoracic and cervical segments (Cohen-Aubert et al., 2010). The most prevalent MRI findings in neurosarcoidosis are multiple nonenhancing periventricular white matter lesions, which imitate MS or vasculitis (Smith et al., 2004). CSF findings are similar to those seen in SLE, including hypoglycorrhachia (Terushkin et al., 2010). Increased level of CSF angiotensin-converting enzyme is insensitive and not sufficiently specific, as it may be observed in other pathologic conditions i.e. MS (Nozaki and Judson,2012).
10. Sjogren syndrome
Primary Sjogren syndrome belongs to autoimmune diseases, which may affect CNS. The pathological process in primary Sjogren syndrome is associated with the presence of antinuclear antibodies, especially anty-Ro/SSA and anty-La/SSB. Although the main affected organs are salivary and lacrimal glands, neurologic symptoms may also appear (Delalande et al., 2004; Massara et al., 2010). Peripheral neuropathies, mainly distal sensory and sensorimotor, are the most common neurological manifestations in primary Sjogren syndrome (Pavlakis et al., 2012; Brito-Zeron et al., 2013). However, CNS involvement might be the first presentation of primary Sjogren syndrome and comprises diverse neurological syndromes such as multiple sclerosis-like disease, encephalopathy, psychiatric disorders, NMOSD – like disease, cerebellar syndromes and movement disorders (Margaretten,2017). The underlying pathology is mainly mononuclear inflammatory vasculopathy (Alexander,1993). However, small vessel vasculitis can be observed in a remarkable percentage of primary Sjogren syndrome patients with the use of cerebral angiography. It has been found that such angiographic findings are associated with anti-Ro/SSA antibodies seropositivity (Alexander et al., 1994). CSF analysis might reveal increased lymphocytosis, elevated IgG index and protein concentration as well as oligoclonal bands (Delalande et al., 2004).
MRI abnormalities in primary Sjogren syndrome are nonspecific. In majority of patients with focal CNS involvement, diffuse hyperintense T2-weighted lesions are observed (Manthorpe et al., 1992). Lesions are often located in areas typically observed in MS (Massara et al., 2010). In contrary to other autoimmune diseases, primary Sjogren syndrome itself is not related to an elevated risk of ischemic stroke. However, aPL antibodies and LAC, associated with increased risk of thrombotic events, are commonly detected in this condition (Pasoto et al., 2012).
11. Practical considerations
Differential diagnosis of disseminated white matter lesions is very complex and apart from MS, primarily systemic diseases including connective tissue disorders, diverse vasculopathies e.g. genetic and infectious diseases have to be taken into account. Symptoms such as impaired consciousness, rapidly progressive cognitive deficits, aphasia and epilepsy should make us consider diagnoses other than MS. A fulminant course of the disease is also rare in MS. Extrapyramidal symptoms should imply rather a suspicion of a hereditary or metabolic disorder. Apparent symmetricity of symptoms and lack of involvement of the optic nerves also raise the probability of disease other than MS. In clinical practice, an absolute lack of response to intravenous methylprednisolone treatment is not typical of MS (Siva,2018 Feb). Despite extensive search for biomarkers, until now no laboratory test has proven 100% specific of MS. Nonetheless, such findings as an elevated ESR and lymphopenia indicate the need for search for a systemic disease. On the other hand, positive ANA tests have been observed in 22.5% to 30.4% of MS patients without any systemic autoimmune disease. However, in most cases of such nonspecific autoantibodies presence, ANA titers are usually not higher than 1:320 (Barned et al., 1995). One study in MS population has shown ANA positivity in as much as 51.0% and a positivity for either aCL or anti-β 2 GP1 antibodies was observed in 32.6% of participants (Roussel et al., 2000). Although such nonspecific prevalence of these autoantibodies is considerably high, the possibility of coexisting autoimmune disease should not be neglected, especially in NMOSD (Siva,2018 Feb). Taking into consideration the dynamic development of the idea of NMO spectrum and most recently MOG spectrum disorders, other antibodies may prove to be useful in the diagnosis of atypical cases, including AQP4-IgG, anti-MOG, anti-NMDAr antibodies and panels for rheumatologic, infectious and paraneoplastic disorders (Siva,2018 Feb). According to the latest revisions of McDonald criteria for MS, CSF examination and especially OCBs detection should be a standard diagnostic step. Although OCBs are not specific and are often present in other disorders e.g. connective tissue diseases, their presence is reported in approximately 90% of MS patients. Consequently, lack of CSF OCBs should raise the suspicion of disorder other than MS (Stangel et al., 2013b). The pattern of OCBs is also a valuable information (Gastaldi et al., 2016). Vast majority of MS patients show type 2 pattern, that is CSF OCBs positivity with no corresponding abnormality in serum. In some cases of MS type 3 OCBs can be detected, which means the IgG bands are present both in the CSF and serum with additional bands detected in the CSF. Identical CSF and serum OCBs constitute type 4 positivity and are usually detected in systemic inflammatory diseases or infections (Deisenhammer et al., 2006). Moreover, CSF pleocytosis higher than 20–30 cells/µL and protein level increased over 60 mg/dL should raise clinician's vigilance (Stangel et al., 2013b; Zettl and Tumani,2005). Table 2 presents serological/CSF findings characteristic of all abovementioned CNS inflammatory disorders.
Table 2. Serum/CSF typical findings in CNS inflammatory disorders (Jog and James,2017; Clark et al., 2017; Bruscolini et al., 2018; Graf,2013; Mandal and Chung,2017; Siva,2018 Feb; Dutra et al., Feb; Stangel et al., 2013b; Gastaldi et al., 2016; Nozaki and Judson,2012; Massara et al., 2010).
| Disease | Typical serological/CSF finding | Comment |
---
| MS | CSF IgG OCBs | OCBs positivity in 90% of MS usually type 2, in some cases type 3 |
| | CSF pleocytosis <30 cells/µL CSF protein <60 mg/Dl | |
| NMOSD | AQP4-IgGs MOG-IgGs | sensitivity 73%, specificity 91%, present in 68%−91% present in 10%−25% |
| NPSLE | ANA, eg. - anti-dsDNA - anti-Sm - anti-P -anti-NR2 - antineuronal - anti-Ro/SSA Anti-GABAR B1b IgGs Anti-GABAR B2 IgGs Nitrated nucleosome levels Increased erythrocyte C4d levels Increased B-cell C4d levels Decreased C3 and C4 levels Increased INF-γ, IL-5, IL-6 Increased BLyS (B-lymphocyte stimulator) | titer 1:80 high specificity high specificity associated with psychosis and depression associated with cognitive dysfunction typical of NPSLE and not SLE Sjogren syndrome, vasculitis exclusive to SLE, present in 15% exclusive to SLE, present in 15% serum titers twice as high in NPSLE flares |
| APS | Anticardiolipin antibody Anti-β−2-glycoprotein I antibody Lupus anticoagulant | Antiphospholipid antibodies should be tested on at least 2 occasions more than 12 weeks and less than 5 years apart |
| PCNSV | none | Extremely elevated erythrocyte sedimentation rate argues against PCNSV |
| EGPA | ANCA, mainly MPO-ANCA | active EGPA |
| | Eosinophilia >1500 cells/μl | active EGPA |
| | Increased eotaxin-3 (CCL26) Increased CCL17 | |
| MPA | MPO-ANCA | non-specific |
| | AC-13 (apolipoprotein A1 C-terminal fragment) | activity marker |
| GPA | ANCA, mainly PR3-ANCA | |
| | Increased High-mobility group box 1 protein Increased serum S100A8/A9 levels | activity marker |
| Behçet's disease | less than 15% positivity for CSF OCBs, prominent pleocytosis and protein level | non-specific |
| Neurosarcoidosis | increased level of CSF angiotensin-converting enzyme | insensitive, not fully specific |
| Sjögren's syndrome | ANA, particularly to Ro/SSA and La/SSB | |
12. Conclusion
Despite the growing availability of tools, diagnosis of CNS inflammatory disorders remains complex and often extends over time. Although MRI is the cornerstone of this process, in some cases it may be a source of confusion, as radiological findings in certain conditions can overlap. Taking into consideration the abovementioned data, the need for specific biomarkers seems to be of primary importance. It appears that serological markers deserve particular attention. Unlike CSF, peripheral blood samples are easily obtainable in routine ambulatory care on numerous time points, and thus they would not only facilitate proper diagnosis, but also enable non-invasive monitoring of treatment response. Nevertheless, a thorough combination of clinical examination, radiological assessment, laboratory tests and often an interdisciplinary approach remains currently mandatory to provide a proper diagnosis.
Declaration of Competing Interest
None of the authors has any potential financial conflict of interest related to this manuscript.
Our manuscript has not, in whole or in part, been published previously and is not under consideration for publication elsewhere.
Acknowledgments
This work has not been supported by any funding sources.
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4770 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9121095/ | Leaf Venation Architecture in Relation to Leaf Size Across Leaf Habits and Vein Types in Subtropical Woody Plants - PMC
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Front Plant Sci
. 2022 May 6;13:873036. doi: 10.3389/fpls.2022.873036
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Leaf Venation Architecture in Relation to Leaf Size Across Leaf Habits and Vein Types in Subtropical Woody Plants
Guoquan Peng
Guoquan Peng
1 College of Chemistry and Life Sciences, Zhejiang Normal University, Jinhua, China
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1, Yingjie Xiong
Yingjie Xiong
1 College of Chemistry and Life Sciences, Zhejiang Normal University, Jinhua, China
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1, Mengqi Yin
Mengqi Yin
1 College of Chemistry and Life Sciences, Zhejiang Normal University, Jinhua, China
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Xiaolin Wang
1 College of Chemistry and Life Sciences, Zhejiang Normal University, Jinhua, China
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Wei Zhou
1 College of Chemistry and Life Sciences, Zhejiang Normal University, Jinhua, China
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1 College of Chemistry and Life Sciences, Zhejiang Normal University, Jinhua, China
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2 School of Biology and Ecology, University of Maine, Orono, ME, United States
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1 College of Chemistry and Life Sciences, Zhejiang Normal University, Jinhua, China
2 School of Biology and Ecology, University of Maine, Orono, ME, United States
Edited by: Dongliang Xiong, Huazhong Agricultural University, China
Reviewed by: Yajun Chen, Xishuangbanna Tropical Botanical Garden (CAS), China; Jun Sun, Fujian Normal University, China
✉
Correspondence: Dongmei Yang, yangdm@zjnu.cn
This article was submitted to Plant Physiology, a section of the journal Frontiers in Plant Science
Received 2022 Feb 10; Accepted 2022 Mar 21; Collection date 2022.
Copyright © 2022 Peng, Xiong, Yin, Wang, Zhou, Cheng, Zhang and Yang.
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PMCID: PMC9121095 PMID: 35599892
Abstract
Leaves are enormously diverse in their size and venation architecture, both of which are core determinants of plant adaptation to environments. Leaf size is an important determinant of leaf function and ecological strategy, while leaf venation, the main structure for support and transport, determines the growth, development, and performance of a leaf. The scaling relationship between venation architecture and leaf size has been explored, but the relationship within a community and its potential variations among species with different vein types and leaf habits have not been investigated. Here, we measured vein traits and leaf size across 39 broad-leaved woody species within a subtropical forest community in China and analyzed the scaling relationship using ordinary least squares and standard major axis method. Then, we compared our results with the global dataset. The major vein density, and the ratio of major (1° and 2°) to minor (3° and higher) vein density both geometrically declined with leaf size across different vein types and leaf habits. Further, palmate-veined species have higher major vein density and a higher ratio of major to minor vein density at the given leaf size than pinnate-veined species, while evergreen and deciduous species showed no difference. These robust trends were confirmed by reanalyzing the global dataset using the same major vein classification as ours. We also found a tradeoff between the cell wall mass per vein length of the major vein and the major vein density. These vein scaling relationships have important implications on the optimization of leaf size, niche differentiation of coexisting species, plant drought tolerance, and species distribution.
Keywords: leaf habit, leaf size, leaf vein type, scaling relationship, subtropical forest, vein density, vein distribution
Introduction
The leaf is the main organ of photosynthesis in higher plants and a critical component in the plant water transport system, which accounts for 30% or more of whole-plant hydraulic resistance (Sack and Holbrook, 2006). Leaf size is an important determinant of plant physiological function and ecological strategy. It reflects the efficiency of light interception and the ability of carbon capture in plants (Parkhurst and Loucks, 1972; Givnish and Vermeij, 1976). Leaf size also shapes the tradeoff between carbon assimilation and water use efficiency, which is crucial for leaf temperature regulation under different climatic conditions (Michaletz et al., 2014; Fauset et al., 2018; Li et al., 2020). Leaf venation is the main structure for physical support and water/nutrient transport in the leaf, which has an important role in maintaining the growth and development of a leaf. It also transports photosynthate and signal molecules from the mesophyll to the rest of the plant. Thus, leaf venation is strongly related to the leaf hydraulic conductance, gas exchange rates, and plant performance (Niklas, 1999; Sack and Holbrook, 2006; Sack et al., 2012, 2013). The leaf hydraulic conductance (K leaf) is determined by the conductance of a series of the xylem (K x) and outside-xylem pathways (K ox). The vein density is a determinant of both K x and K ox because higher densities provide more numerous xylem flow pathways that are parallel per leaf area and shorten pathways for water movement outside the xylem (Cochard et al., 2004; Sack and Frole, 2006; Brodribb et al., 2007; McKown et al., 2010). The higher vein densities and conductivities are expected to be adaptations to higher-resource conditions (Sack et al., 2005; McKown et al., 2010), and large leaves are predominant in moister and/or shaded habitats (Givnish, 1987; Fonseca et al., 2000). Smaller leaves and higher major vein densities are more frequent in dry habitats (Givnish, 1987; Ackerly, 2004; Scoffoni et al., 2011). In addition, the development of the algorithm for vein formation during leaf expansion also provides the basis correlation for vein trait and leaf size (Sack and Scoffoni, 2013). Hence, leaf venation has an important role in the optimization of leaf size. In addition, the variation of leaf size would be closely related to leaf venation architecture.
The scaling of vein traits with leaf size across species has been explored by several studies, but not systematically (except for Sack et al., 2012, 2013). Hence, the conclusions are not consistent. Previous studies with fewer species (≤10) found a negative correlation between major vein density and leaf size, while no relationship between minor vein density and leaf size was found in most studies (Sack et al., 2008; Dunbar-Co et al., 2009; Scoffoni et al., 2011). However, Walls (2011) showed a weak relationship (R 2 = 0.11). Then, Price et al. (2012) showed that vein density was independent of leaf size by using an automated analysis of low-resolution images, but did not distinguish vein orders in 339 species collected from the National Cleared Leaf Collection at the Museum of Natural History, Smithsonian Institution. Another study analyzed 485 globally distributed species with new, high-resolution measurements of vein systems (Sack et al., 2012). They found that larger leaves had major veins with larger diameters, but lower major vein density. Meanwhile, minor vein traits were independent of leaf size, and total leaf vein density was not related to leaf size for both palmate-veined (multiple 1° veins) and pinnate-veined (with a single 1° veins) species. These inconsistent conclusions were not only because of the different image resolutions of leaf venation, but also the different classification standards of major veins and minor veins. In Sack et al. (2012), the major vein included 1°, 2°, and 3° veins, while the minor veins were defined as all more higher-order veins (Sack et al., 2012). In contrast, Price et al. (2012) did not distinguish vein orders. However, the major vein might also be defined based on different formation timing and the gene expression during development (Haritatos et al., 2000; Evert, 2006). Based on the newly developed synthetic model for the development of vein hierarchy, the formation of 1° and 2° veins coincides with the first slow phases, while most 3° and higher-order veins form during the rapid expansion phase (Sack et al., 2012). Hence, it could be reasonable to define the major vein as the sum of 1° and 2° veins, and the minor veins as the 3° and higher-order veins. This classification is the same as in Ellis et al. (2009) and Walls (2011), which will be used in this study. To the best of our knowledge, although the scaling relationship between leaf venation architecture and leaf size has been studied within the given genera and families and at the global level, it has not been investigated within a community sharing similar environmental conditions. It is not clear if this general scaling relationship is conserved across species that coexist in a community.
The major and minor veins differ in the timing of development and xylem and phloem formation (Sack et al., 2012; Sack and Scoffoni, 2013). The variation in the ratio of major to minor vein density could affect leaf hydraulic conductance and vulnerability to cavitation (Scoffoni et al., 2011). Hence, the ratio of major to minor vein density should be considered because even within species with the same total vein density and leaf size, leaf vein distribution could be different. Additionally, the major vein continually thickens with the increase of the leaf size, while the diameter of the minor vein quickly reaches the maximum and is kept constant with the further increase in leaf size (Sack et al., 2012). Thus, the construction cost of extending the major vein and minor vein should be different. Also, the cell wall mass of veins should be considered when studying the scaling relationship between vein architecture and leaf size. These will be helpful for us to understand the covariant relationship of leaf venation structure with leaf size.
In addition, species with different vein types (palmate- vs. pinnate-veined) have different numbers of midribs, resulting in different major vein densities and tolerance to vein damages (Sack et al., 2008; Scoffoni et al., 2011). Species with different leaf habits (i.e., evergreen versus deciduous species) also present significant differences in leaf size, leaf mass per area (LMA), vessel size, photosynthesis, and stress tolerance (Cavender-Bares and Holbrook, 2001; Wright et al., 2004; Yang et al., 2008). Therefore, leaf habits and leaf vein types might also change the general scaling relationship.
Here, we measured leaf size, leaf vein length, and leaf vein cell wall dry mass across 39 broad-leaved woody species within a subtropical forest community in Tiantong National Forest Park of China. These species belong to 27 genera in 18 families, including different leaf habits and leaf vein types, with leaf sizes ranging from 7.68 to 196.00 cm 2. We aimed to (1) test whether the general scaling relationship of leaf vein density with leaf size holds across species within a community, (2) determine whether the ratio of major to minor vein density and vein cell wall mass per length are also related to leaf size, and (3) test whether the scaling exponents are consistent between different leaf habits and different leaf vein-type species. Additionally, we compiled the global dataset from Sack et al. (2012), redefined the major vein and minor vein as we did in this study, and reanalyzed and compared the scaling relationship between our data and data from Sack et al. (2012). This comprehensive study on the scaling trends of leaf venation with leaf size would provide a fundamental understanding of the adaptive significance of leaf size and venation and their ecological strategies.
Materials and Methods
Study Sites and Leaf Materials
Leaf materials were collected from 39 broad-leaved woody species (including 8 palmate-veined deciduous, 7 pinnate-veined deciduous, and 24 pinnate-veined evergreen species) in an evergreen broad-leaved forest of Tiantong National Forest Park, China (29°48′ N, 121°47′ E) on August 2019 when leaf expansion was completed. In all collected species, there were only a few that were not native species, but have been planted for many years in the park. The study site has a subtropical monsoon climate. The mean annual temperature and precipitation from 2012 to 2017 was approximately 16.6°C and 1824.4 mm, respectively. The meteorological data were from the Zhejiang Tiantong Forest Ecosystem National Observation and Research Station.
For each species, four to five healthy adult individuals within a similar environment were selected, and three to five random branches with tips at the outer edge of the middle layer of the plant crown were chosen. Healthy, undamaged, and fully developed current-year leaves located on the third or fourth leaf position were sampled for venation architecture and leaf traits measurements.
Vein Systems Analysis
One leaf per branch from three branches on each of five individuals was collected. Therefore, a total of 15 leaves per species were used for vein system measurements. The leaf vein orders were classified and divided into major (including 1° and 2° veins) and minor veins (3° veins and higher-order veins) according to Ellis et al. (2009), Walls (2011). Firstly, cleaned fresh leaves were scanned using a scanner (LiDE 300, Canon, Vietnam), and leaf area and the length of 1° and 2° veins in each leaf were measured by ImageJ 18.0 (National Institutes of Health1). Secondly, all leaves were chemically cleared with 5% NaOH solution and boiled in a water bath with a constant temperature for 20–30 min. Then, 1-cm 2-sized samples (avoiding 1° and 2° veins as much as possible) were cut from symmetrical locations of the tip, middle, and bottom of the leaves (6 samples per leaf, 15 leaves per species, see Supplementary Figure 1) and bleached in 5% NaClO solution, then stained with an alcoholic solution of toluidine blue (3%) overnight. The above protocols were similar to other studies (Scoffoni et al., 2011; Sack et al., 2012; Petruzzellis et al., 2019). Finally, three images of each sample (a total of 18 images per leaf) were obtained with a digital camera (Leica, DFC7000 T, Germany) mounted on an optical microscope (Leica, DM6B Wetzlar, Germany) at 40 × magnification. Then, minor vein length was measured using the phenoVein software with manual correction (Bühler et al., 2015). The major and minor vein densities were calculated by dividing leaf area with vein length in major and minor veins, respectively. The ratio of major to minor vein density was obtained by the major vein length divided by the minor vein length.
Leaf Vein Cell Wall Mass Analysis
One leaf on a sun-exposed branch from each of the same five individuals measured for vein architecture was collected for leaf vein cell wall mass measurements (cell wall mass denotes cell wall dry mass except where we specifically mentioned fresh mass). After scanning and measuring the leaf area, the leaves were chemically cleared with the same protocols as described above for vein systems. Then, the leaves were washed with deionized water and dried with absorbent paper. Then, 1°, 2°, and minor veins were separated from the cleared leaves with scissors and tweezers before the fresh mass was measured with an analytical balance (±0.1 mg, Mettler Toledo, XPE 205, Switzerland). For the leaf vein cell wall extraction, we used the same protocol as Wang et al. (2017). Briefly, the weighted fresh vein was transferred to a centrifuge tube that was injected with 75% ethanol, and the samples were ground to homogenate with a grinder (70 Hz, 2 min). Then, the homogenates were washed into a 50 ml centrifuge tube with 75% ethanol and kept in an ice-cold bath for 20 min. Later, the homogenates were centrifuged at 1,000 RPM for 10 min, and sediments were washed according to a sequence of 1:7 (vein fresh mass/volume) of ice-cold acetone, methanol-chloroform mixture (1:1, V/V), and methanol. The supernatant in each washing was discarded, and the final deposit was dried in an oven at 70°C for 48 h. The dry mass of the powder is the vein cell wall mass, hence the cell wall mass of 1°, 2°, and minor veins were separately obtained. Except for vein traits mentioned above, the other current-year leaves located on the third or fourth leaf positions of selected branches were collected. Then, the leaf area and leaf mass were obtained for calculation of (LMA).
Ideally, the vein cell wall mass and vein length should be measured at the same leaf area to get the vein cell mass per vein length. However, we used different leaves with different leaf areas for these two measurements. Hence, the vein cell wall mass was converted proportionally to the corresponding vein cell wall mass in the leaf area that was used for measuring vein length for each branch in each species. Based on the LMA values, the same proportional conversion was performed for obtaining the leaf mass of the same leaf area that was used for vein length measurements. In this way, all traits were obtained with the same leaf area that was measured for vein length in each species. For ensuring that the vein cell wall mass conversion was correct, we tested the scaling relationship of vein cell wall mass with leaf area across species. Since the scaling exponents were not different between original vein cell wall mass measurements and converted data with the same leaf area in the vein length measurements, we confirmed that the conversion was correct. The leaf area of each species in the two measurements was similar. Particularly, the points were on the 1:1 line, and the slope did not deviate from 1.0 (R 2 = 0.93).
Data Analyses
Preliminary regression analyses showed that all bivariate relationships were log-log linear. Thus, log 10-transformed data were used in all statistical analyses for scaling relationships. The bivariate relationship was described by the equation y = ax b, where x and y were two traits and a and b represented the intercept and slope of the linear relationship. We followed the other published papers on vein architecture and leaf trait studies in using ordinary least squares (OLS) or standard major axis (SMA) linear regression for fitting the data. They mainly depended on the calculation of the given trait and thus its relative level of measurement error (Niklas, 1994; Smith, 2009; Scoffoni et al., 2011; Price et al., 2012; Sack et al., 2012). The SMA is preferred for allometric scaling analyses, particularly when the measurement error in both variables is proportional and when there is no dependent variable (Warton et al., 2006). Furthermore, when measurement error in the ordinate variable is significantly higher than in the abscissa variable, the results analyzed by SMA are not as accurate as that in the OLS (Kimura, 1992; Sack et al., 2012). Therefore, for the scaling relationship of vein traits with leaf size, we used OLS, and alternatively, we used SMA. We noted that for most of the relationships, the directions of the relationships were not different regardless of using SMA or OLS, and the scaling exponents were similar when the correlation coefficients were high. The analysis of scaling relationships was conducted using Standardised Major Axis Tests and Routines (SMATR) (Falster et al., 2006), and confidence intervals for individual regression slopes were calculated following Pitman (1939). The common slops were obtained where homogeneity of slopes was demonstrated based on the methods of Warton and Weber (2002). Then, the differences in elevation of regression lines (y-intercept) were tested as in standard analysis of covariance (ANCOVA) (Wright and Westoby, 2002; Westoby and Wright, 2003; Yang et al., 2009).
Phylogenetically independent contrasts (PIC) were also performed for analyzing the correlation between functional traits throughout their phylogeny. The phylogenetic tree was constructed using PHYLOMATIC, which is based on the Angiosperm Phylogeny Group III classification of angiosperms (APG III2). The PIC analysis was conducted using the “ape” package (Paradis et al., 2004) in R 4.1.0 version.
To test trait differences between species groups of this study, the t-test was performed. The partial correlation analysis was performed for the intercorrelated relationship among the major vein density, the ratio of major to minor vein density, and leaf size across species, testing the relationship between two variables while holding the third variable constant (Scoffoni et al., 2011).
The global dataset (from Sack et al., 2012) was reanalyzed based on the same definition of major and minor veins as ours, so that the data are comparable. There were 485 species in the global dataset, including data from previous literature (36 species) and original data from their study, which had 410 of 449 species collected from the Daniel I. Axelrod cleared leaf image collection (Museum of Paleontology of the University of California, Berkeley, California). We extracted species in two ways. One contained detailed data of major and minor veins (63 species, called “Sack’s data”) to compare with our data. The other included all species from Axelrod, which accounted for most of the global dataset, including 1° and 2° but not more than 3° veins. Thus, we can get major vein data (401species) from Axelrod (called “Axelrod’s data”) to compare with ours or Sack’s data. In addition, because there was no significant difference in the y-intercept of major vein density and the ratio of major to minor vein density vs. leaf area between evergreen and deciduous species within pinnate-veined species in our results, only venation type species were separately analyzed when comparing with Sack’s data or Axelrod’s data.
Results
Relationships Between Leaf Vein Density and Leaf Size
In the current study, the leaf size ranged from 7.69 to 196.00 cm 2, with a considerable range in total vein density from 52.04 to 122.54 cm–1. Detailed functional traits can be found in Supplementary Table 1.
A strong negative correlation between the major vein density and leaf area in palmate-veined deciduous, pinnate-veined deciduous, and pinnate-veined evergreen species groups was found (all R 2> 0.824), with a common slope of –0.525 [95% confidence interval (CI) –0.576, –0.450, p = 0.158], which did not differ significantly from –0.5 (Figure 1A and Table 1). The same result was found when leaf size was represented by lamina mass with a common slope of –0.453 (95% CI –0.561, –0.328, p = 0.517) (Table 1). These results indicated that the major vein density geometrically declined with leaf size. However, palmate-veined deciduous species had a significantly higher y-intercept than both the pinnate-veined deciduous and evergreen species (p< 0.01, Figure 1A and Table 1), suggesting that the palmate-veined species have a greater major vein density at the given leaf area than in the pinnate-veined species. However, deciduous and evergreen species showed no difference within the same pinnated-veined vein type. By contrast, the density of minor veins was independent of leaf size, as was the total vein density (Figure 1B and Table 1). The correlation between the major vein density and leaf size was also significant when expressed as correlated evolutionary divergences (Table 2).
FIGURE 1.
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The relationships of the major vein density (A), the minor vein density (B), the major vein length (C), and minor vein length (D) with leaf area across different group species. P almate D, plamate-veined deciduous species; P innate D, pinnate-veined deciduous species; and P innate E, pinnate-veined evergreen species.
TABLE 1.
Parameters for the scaling relationships of vein traits and leaf traits.
x y b-Value (95% CIs)
a-value
Method P almate D P innate D P innate E P almate D P innate D P innate E
Leaf area Vein length, cm
(cm 2)Major vein OLS 0.461(0.412, 0.545)1.217a 1.134b 1.140b
R 2: 0.969 0.887 0.879
Minor vein OLS 0.970(0.856, 1.053)2.032a 1.926b 1.923b
R 2: 0.990 0.952 0.827
Total vein OLS 0.958(0.846, 1.039)2.064a 1.958b 1.955b
R 2: 0.990 0.952 0.830
Vein density, cm cm–2
Major vein OLS–0.525(–0.576, –0.450)1.212a 1.122b 1.132b
R 2: 0.979 0.935 0.824
Minor vein OLS ns ns ns–––
R 2: 0.007 0.337 0.085
Total vein OLS ns ns ns–––
R 2: 0.003 0.387 0.103
Vein cell wall mass per length, g cm–1
Major vein OLS 0.583(0.437, 0.721)–4.968a–4.702b–4.614b
R 2: 0.672 0.887 0.554
Minor vein OLS ns ns ns–––
R 2: 0.001 0.390 0.021
Total vein OLS ns ns ns–––
R 2: 0.001 0.397 0.049
MVD/M i VD OLS–0.454(–0.554, –0.298)–0.892a–0.828a–0.845a
R 2: 0.950 0.592 0.496
Lamina mass MVD OLS–0.453(–0.561, –0.328)0.158a 0.010a 0.156a
(g)R 2: 0.764 0.734 0.635
MVD/M i VD OLS–0.391(–0.509, –0.230)–1.802a–1.825a–1.688b
R 2: 0.682 0.843 0.376
MVD MVW/MVL SMA–1.295(–1.708, –1.048)–3.575a–3.422ab–3.282b
R 2: 0.638 0.814 0.501
M i VD M i VW/M i VL SMA ns ns ns–––
R 2: 0.191 0.003 0.017
MVD MVD/M i VD SMA 0.975 (0.820,1.130)–1.967a–1.865b–1.868b
R 2: 0.965 0.872 0.377
Leaf area Lamina mass SMA 0.960 (0.839,1.103)–1.964ab–2.100a–1.880b
R 2: 0.721 0.862 0.872
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CIs, confidence intervals; OLS, ordinary linear regression; SMA, standard major axis; P almate D, plamate-veined deciduous species; P innate D, pinnate-veined deciduous species; and P innate E, pinnate-veined evergreen species; MVD, major vein density, cm cm–2; M i VD, minor vein density, cm cm–2; MVL, major vein length, cm; M i VL, minor vein length, cm; MVD/M i VD, the ratio of major to minor vein density; MVM/MVL, the cell wall mass per vein length of the major vein, g cm–1; M i VM/M i VL, the cell wall mass per vein length of the minor vein, g cm–1. b-value is the common slope among different species groups by ordinary linar regression (OLS) or standard major axis (SMA) method; a-value is the y-intercept based on the common slope; the small letter after the a-value is the significance test, different letters mean the significant difference; ns represented the scaling relationship was not significant. R 2 is the absolute coefficient of scaling relationship between two traits.
TABLE 2.
The regression slopes between functional traits (log-log transformed data) of correlated evolutionary divergences for 39 subtropical woody plants in Tiantong National Forest Park, China.
Traits (x-axis–y-axis)Slopes R 2 P-value
LA–MVD–0.428 0.914<0.001
LA–MVL 0.574 0.953<0.001
LA–M i VL 0.994 0.960<0.001
LA–MVM/MVL 0.433 0.638<0.001
LA–MVD/M i VD–0.427 0.835<0.001
LM–MVD–0.432 0.701<0.001
LM–MVD/M i VD–0.431 0.641<0.001
MVD–MVW/MVL–1.035 0.732<0.001
MVD–MVD/M i VD 0.942 0.814<0.001
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The OLS method on log-transformed variables was applied. All the regression lines were highly significant (p < 0.001). LA, leaf area; LM, lamina mass. For other abbreviations, see Table 1.
Vein length was significantly and positively related to leaf area within each species group (all R 2> 0.827). The slope of vein length vs. leaf area was not different among species groups, with the common slope 0.461 (95% CI 0.412, 0.545, p = 0.072) and 0.970 (95% CI 0.856, 1.053, p = 0.202) for major and minor veins, respectively (Figures 1C,D and Table 1). The palmate-veined deciduous species were significantly greater in the vein length than two pinnate-veined group species at the same leaf area (p< 0.01, y-intercept in Figures 1C,D and Table 1), indicating that the major and minor vein lengths were both significantly larger in palmate-veined species than in pinnate-veined species. The result of the PIC analysis also showed a positive correlation between correlated evolutionary divergences (Table 2).
Relationships Between Vein Cell Wall Mass per Length and Leaf Size
The major vein cell wall mass per length was significantly and positively related to leaf area in three groups (all R 2> 0.554), with a common slope of 0.583 (95% CI 0.437, 0.721, p = 0.919), which did not significantly deviate from 0.5 (Figure 2A and Table 1). This was consistent with the positive relationship between correlated evolutionary divergences (Table 2). The difference in y-intercept was found to be significant between palmate-veined deciduous species and pinnate-veined (evergreen and deciduous) species (p< 0.001), but not significant between two pinnated-veined species groups (p> 0.05, Table 1). In contrast, minor vein cell wall mass per length was independent of leaf size (Figure 2B and Table 1).
FIGURE 2.
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The relationships of the cell wall mass per length of the major vein (A), and that of the minor vein (B) with leaf area; the relationship of the cell wall mass per length of the major vein with the major vein density (C). For abbreviations, see Figure 1.
In addition, there was a significantly negative correlation between the major vein cell wall mass per vein length and the major vein density within each species group (all R 2> 0.501), with the common slope of –1.295 (95% CI –1.708, –1.048, p = 0.355), which marginally but significantly deviated from –1.0 (Figure 2C and Table 1). This suggested a tradeoff between the major vein density and the major vein cell wall biomass investment per unit vein length. The relationship was also strong when expressed as correlated evolutionary divergences (Table 2). However, there was no significant correlation between the minor vein cell wall mass per vein length and minor vein density in each species group (all p> 0.05).
The Scaling Relationship of the Ratio of Major to Minor Vein Density With Leaf Size
A significant scaling relationship was found between the ratio of major to minor vein density and leaf area within each species group (all R 2> 0.496). Neither the slopes nor the intercepts of the ratio of major to minor vein density vs. leaf area relationships differed among the species groups, with a common slope of –0.454 (95% CI –0.554, –0.298, p = 0.064) that did not significantly deviate from –0.5 (Figure 3A and Table 1). A similar scaling relationship was also found when the leaf size was represented by lamina mass (Table 1). These results indicated that the leaf vein density distribution was tightly correlated with leaf size. Furthermore, the ratio of major to minor vein density was significantly related to the major vein density in each species group, with the common scaling exponent of 0.975 (95% CI 0.820, 1.130, p = 0.472) not being different from 1.0. Thus, these two venation traits have an isometrical relationship (Figure 3B and Table 1).
FIGURE 3.
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The relationships of the ratio of major to minor vein density with leaf area (A) and the major vein density (B). For abbreviations, see Figure 1.
In addition, the relationships between the ratio of major to minor vein density and leaf size, and between the ratio of major to minor vein density and major vein density were also highly significant when expressed as correlated evolutionary divergences (Table 2).
The Comparisons Between Our Results and the Global Dataset
The major vein density negatively scaled to leaf area in both this study and Sack’s data (both R 2> 0.60). Neither the slopes nor the intercepts of relationships differed between these two groups, with common slopes of –0.532 (95% CI –0.594, –0.454) and –0.496 (95% CI –0.553, –0.439) for palmate- and pinnated-veined species, respectively, which did not significantly deviate from –0.5 (Figures 4A,B and Table 3). However, the scaling exponent in the Axelrod’s data (401 species) was significantly larger (less negative) than for this study and Sack’s data, with exponents of –0.408 (95% CI –0.440, –0.375) and –0.427 (95% CI –0.447, –0.407) for palmate- and pinnated-veined species, respectively, both being significantly deviated from –0.5 (Table 3). Consequently, when we plot the scaling relationship for Axelrod’s species, our data, and Sack’s data, there was no common slope among them in both venation type species (Figure 5A and Table 3).
FIGURE 4.
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The relationships of major vein density of palmate-veined species (A), major vein density of pinnate-veined species (B), the ratio of major to minor vein density of palmate-veined species (C), and the ratio of major to minor vein density of pinnate-veined species (D) with leaf area.
TABLE 3.
Parameters for the scaling of vein traits with leaf size analyzed by OLS method in this study, Sack’s data, and Axelrod’s data (Sack et al., 2012).
Vein type Traits Data source N R 2 b 1-value b 2-value a 2-value Tests for differences between groups
Slope (95% CIs)Common slope
(95% CIs)P-value
Palmate MVD Sack’s 34 0.607–0.478(–0.616, –0.339)–0.532 (–0.594, –0.454)1.196 b 2: 0.434
(cm cm–2)This study 8 0.979–0.544(–0.624, –0.464)1.224 a 2: 0.211
MVD/M i VD Sack’s 34 0.391–0.402(–0.583, –0.221)–0.509 (–0.602, –0.378)–0.765 b 2: 0.172
This study 8 0.950–0.545(–0.670, –0.420)–0.792 a 2: 0.382
MVD Axelrod’s 67 0.906–0.408 (–0.440, –0.375)––b 2: 0.016
This study 8 0.979–0.544 (–0.624, –0.464)–
Sack’s 34 0.607–0.478 (–0.616, –0.339)
Pinnate MVD Sack’s 29 0.740–0.442(–0.545, –0.339)–0.496 (–0.553, –0.439)1.137 b 2: 0.209
This study 31 0.897–0.519(–0.586, –0.453)1.087 a 2: 0.069
MVD/M i VD Sack’s 29 0.513–0.509(–0.704, –0.313)–0.437 (–0.523, –0.347)–0.738 b 2: 0.389
This study 30 0.727–0.415(–0.514, –0.317)–0.876 a 2: 0.006
MVD Axelrod’s 334 0.839–0.427 (–0.447, –0.407)––b 2: 0.042
This study 31 0.897–0.519 (–0.586, –0.453)–
Sack’s 29 0.740–0.442 (–0.545, –0.339)
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CIs, confidence intervals; OLS, ordinary linear regression; MVD, major vein density, cm cm–2; MVD/M i VD, the ratio of major to minor vein density. Parameters for the scaling of vein traits, that is, of b value in the equation log (trait) = a + b log (leaf size), across species in different data sources. b 1-value is the slope of each data group, b 2-value is the common slope of two or three groups. a 2-value is the y-intercept based on the common slope. P-value is the test between groups for common slope b 2-value and y-intercept a 2-value, respectively.
FIGURE 5.
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The comparisons of the scaling relationship of major vein density with leaf area (A), and the independence of total vein density from leaf area (B) in our study (39 woody species within subtropical forestry community), Sack’s data (63 comprehensive species), and Axelrod’s data. The red line is the common slope of this study and Sack’s data, while the blue line is the slope of Axelrod’s data. Inset in panel (A) is the same plot but with raw data.
The ratio of major to minor vein density was significantly and negatively correlated with leaf area in both this study and Sack’s data (both R 2> 0.39), with common slopes of –0.509 (95% CI –0.602, –0.378) and –0.437 (95% CI –0.523, –0.347) for palmate- and pinnated-veined species, respectively, which did not differ from –0.5 (Figures 4C,D and Table 3). The y-intercept was also not different between this study and Sack’s data in the palmate-veined species. However, the Sack’s data had a significantly higher ratio of major to minor vein density than our data at a given leaf area in the pinnate-veined species. In addition, similar to Sack’s global dataset, our data showed a considerable range in total vein density. The two datasets, both used high-resolution images, did not differ significantly (Figure 5B, p = 0.231 of t-test, Supplementary Table 1).
Discussion
Our results showed an extremely strong and consistent scaling relationship of leaf venation architecture with leaf size across species with different leaf habits (evergreen and deciduous) and leaf vein types (palmate-veined and pinnate-veined) within a community. The scaling relationship between leaf vein density and leaf size across species was the same as what was found in the global dataset (Sack et al., 2012). Additionally, we found that the leaf vein distribution (the ratio of major to minor vein density) and the major vein cell wall mass per length were significantly correlated with leaf size across species. Indeed, it is intriguing when different leaf habits and leaf vein types species have the same scaling relationships between leaf vein traits and leaf size. This pattern suggests that these leaf vein traits are of importance for leaf size and are governed by the law of physics or physiological requirements. These relationships are core discoveries in leaf structure and function, which have important ecological and biogeographic implications. These results also provide a new way to understand the optimization of leaf size.
The Scaling Relationship of Leaf Vein Density With Leaf Size
We found a strong negative correlation of major vein density with leaf size across 39 woody broad-leaved species within a community. However, minor vein density was independent of leaf size (Figure 1). Since total vein density was mainly determined by the minor vein length per area, which accounted for > 97% of the total vein length in this study (Supplementary Table 1), the total vein density was also not related to leaf size. This was consistent with findings from the previous global-scale dataset and small groups with no more than 10 species in previous studies (Sack et al., 2008, 2012; Dunbar-Co et al., 2009; Scoffoni et al., 2011).
These relationships were robust across different leaf habits and leaf vein types, with the common slope of major vein density with leaf size not different from –0.5 (Figure 1 and Table 1). The consistent results when leaf size was represented with leaf area and lamina mass was because of the isometric relationship between leaf area and lamina mass (b = 0.96, 95% CI 0.84, 1.10, Table 1). This robust geometrical scaling of the major vein density with leaf size could be directly demonstrated from the scaling relationship between vein length and leaf area in the current study. The significant relationship between major vein length (MVL) and leaf area (A) among all groups showed a scaling slope not significantly different from 0.5 (b = 0.46; Figure 1C and Table 1), i.e., MVL scaling with A was described as MVL∝ A 0.46, and the leaf area scaled with leaf area as A∝A 1. Therefore, the major vein density was determined by MVD = MVL/A ∝A–0.54, and the scaling exponent was not different from –0.5 (Figure 1 and Table 1). Additionally, considering the geometric dimensions, geometric scaling predictions have been derived (Niklas, 1994) by treating each fundamental trait as an area (A) as a two-dimensional variable and length (L) as a one-dimensional variable. Hence, vein density is an L/A∝A–0.5. Thus, we can say that the major vein density geometrically declined with leaf size. However, the scaling of minor vein length (M _i_ VL) with leaf area was described as M _i_ VL∝ A 0.97, which was not different from 1.0 (Table 1). Thus, the minor vein density was determined by M _i_ VD = M _i_ VL/A ∝ A–0.03, suggesting that M _i_ VD did not change with leaf size. These trends can also be explained by the development mechanism of venation according to the synthetic model (Sack et al., 2012). With the leaf development, the 1° and 2° veins are formed during a “slow” limited expansion phase due to cell proliferation, and the vein density peaks as procambium forms and declines as leaves are pushed apart during subsequent rapid expansion. Thus, the major vein density would geometrically decline with increasing leaf size. In contrast, the 3° and other higher-order veins are principally formed during a “rapid” dramatic expansion phase mainly because of cell expansion, although cell divisions continue. Thus, the minor vein density stabilizes as their initiation, and is maintained during leaf expansion (Sack et al., 2012). The declining trend of major vein density with leaf area was consistent among different leaf vein types and leaf habit groups. However, the palmate-veined species had higher major vein density than pinnate-veined species at a given leaf area (Figure 1A), which can be mainly ascribed from palmate-veined species having more midrib compared with pinnate-veined species (Sack et al., 2008; Scoffoni et al., 2011). Additionally, major vein length in palmate-veined species was significantly higher than in the pinnate-veined species at the same leaf area (Figure 1C and Table 1), which also contributed to the above difference. Even though evergreen species normally have smaller leaf area, photosynthesis efficiency, and higher LMA than for deciduous species (Zhang et al., 2013; Qi et al., 2021), major vein length and the major vein density were both insignificantly different between deciduous and evergreen species within the same leaf vein type in the current study (p> 0.05). Therefore, the y-intercept of scaling relationship in leaf vein density with leaf size was only impacted by vein type, not by leaf habit. Also, in the future, the evergreen and deciduous species could be combined within the same leaf vein type to study the scaling relationship of leaf venation architecture with leaf size.
The exponent of the geometrically scaling between major vein density and leaf area (no different from –0.5) in this study was different from that in the global dataset, which has the major vein density conservatively declined with leaf size (b = –0.341, 95% CI –0.360, –0.322) (Sack et al., 2012). This difference is not caused by different definitions of the major vein, but by the different range of leaf size of the studied species. When we reanalyzed the scaling relationship from the global dataset by recalculating major vein density based on the same definition of ours, the scaling exponent was still significantly larger than –0.5 (common slope of palmate- and pinnate-veined species, b = –0.405, 95% CI –0.383, –0.427), in agreement with original results reported in Sack et al. (2012). However, considering the data source, we found that the scaling exponent of Axelrod’s data (401 species) was significantly larger than –0.5 (Figure 5A and Table 3), while the scaling exponent of Sack’s data (63 species including major and minor veins) was not significantly different from –0.5, in agreement with our results (Figures 4A,B and Table 3). The conservative scaling exponent (>–0.5) found in the original global dataset was mainly determined by the Axelrod’s data because Axelrod’s species accounted for 85% of total species. The significant difference in scaling exponent between Axelrod’s data and ours or Sack’s data might attribute to the source of species collected by Axelrod or their venation treatment. We noted that the leaf area of the species collected by Axelrod ranged from 0.16 to 57.67 cm 2, compared to 7.69 to 196.00 cm 2 in our study and 3.77 to 279.65 cm 2 in Sack’s 63 species. Thus, most species collected by Axelrod were smaller than ours or Sack’s, which could be related to the environment or climate condition. Therefore, we could say that within the same subtropical broad-leaved forest community, the major vein density was geometrically declined with leaf size, and this scaling exponent is consistent with another dataset that includes big leaves, but not with that including only small leaf species (Axelrod). The conserved scaling relationship between major vein density and leaf size may have important implications for understanding leaf size evolution, biogeography, physiological adaptation, and paleobiology. Further research work could be done to test whether the scaling exponent is related to the climate by studying different communities with different climate/environments. This will provide a new understanding of the species distribution based on the relationship between leaf venation structure and leaf size.
The Scaling Relationships of the Vein Cell Wall Dry Mass per Length With Leaf Size and Vein Density
Despite that the total dry mass of the cell wall of veins increased with leaf size across species in this study (in each group, R 2> 0.50, p< 0.001), the cell wall mass per vein length was not fixed with the leaf growth, and the major and minor veins have different patterns. There was a significantly positive correlation between the cell wall mass per length and leaf size in major veins (Figure 2A and Table 1), but not in minor veins (Figure 2B and Table 1). As minor veins usually account for most of the total vein length, total vein cell wall mass per length was not significantly related to leaf size (Table 1). The above difference between major and minor veins could be caused by different diameter growth patterns of them during leaf development. The 1° and 2° veins have a prolonged diameter growth with the leaf development, and the power scaling exponent of vein diameter vs. leaf area was 0.452 (95% CI 0.426, 0.480) and 0.368 (95% CI 0.344, 0.394) for 1° and 2° veins, respectively (Sack et al., 2012). The thickening of major veins will lead to the increase of major vein volume per leaf area, resulting in the increase of cell wall biomass investment in the major vein (Niinemets et al., 2007; Niklas et al., 2007). However, minor veins (3° and higher-order) rapidly reach maximum diameter, and there is no correlation with leaf area (Sack et al., 2012). Hence, with the increase of leaf size, the cost of lengthening the major vein will increase, while the cost of that for the minor vein is relatively stable.
The results showed that the cell wall mass per unit length of major vein allometrically scaled with leaf area, with the scaling exponent not significantly different with 0.5 (Figure 2A and Table 1) across leaf habits and leaf vein types, indicating that the increase of cell wall mass per unit length of major vein could not keep up with the increase in leaf area. In other words, the leaf area that could be obtained by investing in the biomass of major vein cell wall per unit length increased with leaf area, i.e., “increasing returns.” Based on this allometrically relationship, it will be beneficial for the plant to increase the cell wall mass per unit length of the major vein. However, it is impossible for plants to infinitely increase their investment in the major vein cell wall mass per length, because the major vein density would decline with the increase of cell wall mass per unit vein length of major veins (Figure 2C and Table 1). Therefore, there is a trade-off between the length growth and thickness growth of the major vein at a given biomass investment for the major vein. For a leaf, the greatest mechanical stress occurs along its longitudinal axis (Anita et al., 2001), and the mechanical reinforcement is determined by its low-order veins (Kull and Herbig, 1995). With the increase of leaf size, the larger major veins are required to provide stronger mechanical support, and increasing the diameter of the major vein increases the hydraulic conductivity within the leaf, which is the premises to improve the water transport efficiency of the whole plant (McKown et al., 2010). Consequently, at the same major vein cell wall mass investment, plants will prefer increasing vein diameter but not extending the vein length, leading to a tradeoff between the major vein cell wall mass per length and major vein density. This might be one of the reasons for the continuous increase in thickness of the major vein, while the major vein density peaks as procambium forms during leaf growth.
The trade-off of cell wall mass per unit length of the major vein and major vein density was of great significance for the optimization of leaf size. Although a thicker major vein is more conducive to support a larger leaf size, it will limit the length of the major vein and shorten the water transport distance within the major vein, which would impact the whole leaf hydraulic conductance. In order to maintain the water transport efficiency, it is impossible to increase leaf area by infinitely increasing the cell wall mass per unit length of the major vein. Rather, plants reach a reasonable leaf size due to this trade-off.
The Ratio of Major to Minor Vein Density Scales With Leaf Size
In this study, the ratio of major to minor vein density was significantly scaled with leaf size among different species groups, which was consistent with a 10 species study (Scoffoni et al., 2011). This could be because the major and minor veins have different functions. For example, major veins (primary and secondary veins) act as the support and distribution network for leaves (Roth-Nebelsick et al., 2001; Ellis et al., 2009), while minor veins act as the sites of exchange between the mesophyll and the vascular system (Haritatos et al., 2000; Sack and Holbrook, 2006). Hence, the different distribution patterns of major and minor veins within a leaf would be preferred by different leaf sizes for adaption to the specific environment. In addition, this negative correlation could be ascribed to the slower speed of the increase in major vein length with leaf area compared to that of the minor vein. The scaling exponents were not far from 0.5 and 1.0 for major vein length and minor vein length, respectively (Figures 1C,D and Table 1). Thus, the ratio of major to minor vein density scaled with leaf area with an exponent not different from –0.5 (Figure 3 and Table 1). The y-intercept of this scaling relationship was not different between different leaf vein types or leaf habitats, although at the same leaf area major vein length was higher in palmate-veined species than in pinnate-veined species (Figure 2 and Table 1). However, the same trend was found in the minor vein length vs. leaf area. Therefore, the y-intercept difference was offset during the major vein length divided by the minor vein length to analyze the scaling relationship between the ratio of major to minor vein density and leaf area. The above results demonstrated that the scaling relationship between the ratio of major to minor vein density and leaf area was robust, approximately –0.5, regardless of palmate- or pinnate-veined species and evergreen or deciduous species.
The ratio of major to minor vein density was scaled with leaf area in our study and Sack’s data (63 species). This stable scaling relationship indicated that the ratio of major to minor vein density was another key venation trait linked with leaf area that maintained the same scaling exponent –0.5, which could be used for explaining the evolution of leaf size and adaptation to the environment. Also, the ratio of major to minor vein density was isometrically related to the major vein density [SMA results: b = 0.975 (95% CI 0.820, 1.130)] in our data (Table 1). However, within the subtropical community, we found that the evergreen species have higher mean values in the ratio of major to minor vein density and smaller leaf area compared to deciduous species within the same pinnate-veined species (Figure 3, both p< 0.05 by t-test). These trends are beneficial for species with small leaf areas because the higher the ratio of major to minor vein density, the smaller leaf area and more tolerance to leaf xylem cavitation (Scoffoni et al., 2011). This can be extended to the global scale species, of which small leaf species can survive unfavorable situations with a higher ratio of major to minor vein density.
The dramatic linkage between venation architecture (including major vein density and the ratio of major to minor vein density) and leaf size in the subtropical forest community or at the global scale (Figures 1, 3, 4) provides a hydraulic mechanism for explaining the ecological or biogeographical distribution of leaf size. Small leaves are predominant in drier and more exposed habitats (Givnish, 1987; Peppe et al., 2011; Sack et al., 2012), while large leaves in moister and/or shaded habitats (Givnish, 1987; Fonseca et al., 2000). A spatially explicit model showed that the greatest impact for the increase of K leaf in the reticulate hierarchy system was from increases in major vein conductivity and in the minor vein density (McKown et al., 2010). Hence, the major veins normally have long and wide conduits (Choat et al., 2005), which in turn have higher vulnerability to cavitation (Choat et al., 2005; Blackman et al., 2010; Scoffoni et al., 2011). Because of this, higher major vein density in small leaves provides redundant hydraulic “superhighways,” i.e., pathways around embolized major veins (Sack et al., 2008, 2012; Scoffoni et al., 2011). Both a high major vein density and a high ratio of major to minor density could reduce hydraulic vulnerability (Scoffoni et al., 2011). This study showed that small leaves generally have higher major vein density and ratio of major to minor vein density across leaf habits and leaf vein types (Figures 1, 3). Therefore, the hydraulic mechanism in providing the benefit of small leaves in dry and exposed habitats are the tight scaling relationships between leaf major vein density and leaf area, and between the ratio of major to minor vein density and leaf area. With a partial correlation analysis, the relationship of the ratio of major to minor vein density with leaf area was still significant after partialing out major vein density (r = –0.585, p< 0.01), and the correlation of major vein density with leaf area was also significant after partialing out the ratio of major to minor vein density (r = –0.778, p< 0.01) for pooled data of our study. These results indicated that the major vein density and the ratio of major to minor vein density both played a key role in leaf size distribution. Small leaves would have a lower hydraulic vulnerability that is preferred in dry habitats (Ackerly et al., 2002; Bragg and Westoby, 2002; McDonald et al., 2003; Sack et al., 2012). This provides an explanation for the fact that leaf size declines as annual temperature and precipitation decrease (McDonald et al., 2003).
In conclusion, we found strong correlations of the major vein density and the ratio of major to minor vein density with leaf size, and the isometrical relationship between the major vein density and the ratio of major to minor vein density across 39 species within a subtropical forest. These findings were confirmed by reanalyzing the global dataset. However, these relationships were not found in small leaf species collected by Axelrod, thereby asking for further studies to test this scaling exponent in different ecosystems with leaf size ranges. Interestingly, our results also demonstrated that these trends were robust in different vein types and leaf habits. However, palmate-veined species have higher major vein density and ratio of major to minor vein density at the given leaf size than pinnate-veined species, which was mainly due to a more uniform distribution of large veins in palmate-veined leaves (Niinemets et al., 2007). In contrast, evergreen and deciduous species have similar venation architecture at a certain leaf area. The linkages of the major vein density, the ratio of major to minor vein density with leaf size, and the negative relationship between the major vein density and the cell wall mass per vein length of major vein could have important implications in the optimization of leaf size, niche differentiation of coexisting species, plant drought tolerance, and species distribution.
Data Availability Statement
The original contributions presented in the study are included in the article/Supplementary Material, further inquiries can be directed to the corresponding author/s.
Author Contributions
DY and GP conceived the research plans, supervised the experiments, analyzed the data, and wrote the manuscript. GP performed most of the experiments. YX, MY, and XW performed the leaf vein length measurements. WZ and ZC performed the leaf vein cell wall mas measruments. Y-JZ helped to revise the manuscript. All authors contributed to the article and approved the submitted version.
Conflict of Interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Publisher’s Note
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.
Acknowledgments
We thank Kailu Wei, Sili Zhu, Zhiwen Song, Ningjing Zheng, Xiangxiang Tang, Jiahe Xu, and Hongru Geng for their field and laboratory assistance with vein traits measurements and samples collections. We also thank Yaxin Li for helping draw the illustration of sampling positions for minor veins within a leaf, and thank workers of Zhejiang Tiantong Forest Ecosystem National Observation and Research Station.
Footnotes
1
2
Funding
This research was supported by grants from the National Natural Science Foundation of China (No. 31770647), the Zhejiang Provincial Natural Science Foundation of China (LY19C150007), and the Ten Thousand Talents Program of Zhejiang Province (2019R52014).
Supplementary Material
The Supplementary Material for this article can be found online at:
Click here for additional data file. (18.1KB, XLSX)
Click here for additional data file. (269KB, pdf)
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4771 | https://www.aetna.com/content/dam/aetna/pdfs/aetnacom/individuals-families/document-library/cardiovascular-diseases.pdf | The following article features consumer information from Columbia University College of Dental Medicine. All information is intended for your general knowledge only and isn't a substitute for medical advice or treatment for specific medical conditions. Talk to your doctor about what’s best for you. Cardiovascular diseases Some cardiovascular (heart and artery) diseases may require changes in your dental treatment and how you receive dental care. Recent research has linked periodontal disease with the risk of coronary artery disease and stroke. Periodontal disease has also been associated with preterm birth, low birth weight and intrauterine growth restriction, diabetes, osteoporosis, respiratory disease and cancer. Treatment of periodontal disease can reduce overall inflammation in the body and of course will make your teeth and gums healthier. However, at present there is limited evidence that periodontal treatment can prevent heart disease, heart attack or stroke. If you have a one of these conditions, make sure that your dentist always has an up-to-date list of all the medicines you take. The list should be neatly written or typed. It should include prescription drugs, and over-the-counter medicines such as antacids. It also should include vitamins, herbal pills and other nutritional supplements. Your list should provide the name of each drug, the dosage, how often you take it and when your physician prescribed it. Put the date that you made the list at the top of the page. This will let the dentist know that it is a current list. Periodontal disease and cardiovascular conditions Periodontal disease can affect your overall health. Over time, it may increase the risk for heart disease and stroke. Several studies have shown that people with periodontal disease may be more likely to have coronary artery disease than people with healthy mouths. Right now, scientists have two possible explanations for this association. One is that the bacteria that cause periodontal disease can release toxins into or travel through the bloodstream and help to form fatty plaques in the arteries. These plaque deposits can lead to serious problems, such as blood clots, which can block blood flow. The other explanation is that these bacteria cause the liver to make high levels of certain proteins, which inflame the blood vessels. The inflammatory response, that these bacteria can cause, makes periodontal disease a risk factor for cardiovascular events such as stroke, congestive heart failure, myocardial infarction (heart attack), peripheral artery disease and atherosclerosis. Symptoms of periodontal disease include: • Persistent bad breath • Red, swollen or tender gums • Gums that bleed when you brush your teeth • Gums that have pulled away from the teeth 2291605-01-01 (06/23) • Loose teeth • A change in the way your teeth come together when you bite down If you have symptoms of periodontal disease, see your dentist soon for treatment. Heart disease and dental treatment Patients with certain heart conditions have a higher risk of endocarditis. This is an infection of the heart. It can be life threatening. It happens when bacteria in the bloodstream attach to damaged heart valves or other damaged heart tissue. People with certain heart conditions may need antibiotics before they have certain types of dental procedures. Make sure to inform your dentist of any heart issues. In 2017, the American Heart Association and the American College of Cardiology published a focused update to their 2014 guidelines for the use of antibiotics prior to dental treatment. The new policy advises antibiotics for fewer conditions than the old policy did. Pre-treatment with antibiotics is still recommended for people who fall into certain categories: • People who have had infective endocarditis in the past • People with artificial cardiac valves • People who have had transplants and later developed heart valve problems People with a prosthetic material used for valve repair Pre-treatment with antibiotics also is recommended for people with certain heart conditions that were present at birth: • Cyanotic heart disease that has not been repaired or was repaired incompletely. This includes people with shunts and conduits. • A heart defect that was completely repaired with a prosthetic material or device. In this case, antibiotics are advised only for the first six months after the procedure. • Any repaired heart defect that still has some defect at or next to the site of a prosthetic patch or device Taking antibiotics before dental treatment is no longer advised for people with: • Acquired heart valve dysfunction (for example, rheumatic heart disease) • Mitral valve prolapse • Bicuspid valve disease • Calcified aortic stenosis • Congenital heart conditions, such as ventricular septal defect, atrial septal defect, and hypertrophic cardiomyopathy The American Heart Association guidelines recommend pre-treatment antibiotics for dental procedures that involve an incision or manipulation of the gums or the tissues around a tooth root. 2291605-01-01 (06/23) Antibiotics are not required for the following: • Routine anesthetic injections through non-infected tissue • X-rays • Placement of dentures • Placement or adjustment of removable orthodontic appliances • Placement of the bracket part of braces (not bands) • The natural loss of baby teeth in children • Bleeding from trauma to the lips or mouth The most important dental factors in the prevention of infective endocarditis are good oral hygiene at home, and regular dental cleanings. Heart attack (myocardial infarction) Oral effects: A heart attack can sometimes feel like pain that starts in the chest and spreads to the lower jaw. Other times it may be pain that starts in the jaw or in the left arm or shoulder. At the dentist: You should wait at least six weeks after a heart attack to have most dental treatments. Your dentist should have oxygen and nitroglycerin available during your appointment. Your dentist and physician should discuss your condition before dental treatment. Some medicines you take can change the way your dentist treats you. For example, if you are taking blood thinning drugs (anticoagulants), your blood is less likely to clot. You may need to stop taking your blood thinning medicines before some dental procedures. Do not stop taking any medicines until you have spoken to your physician. This is something your dentist will discuss with you and your physician. Let your dentist know the medicines you take, and their doses. You may need to take blood tests before some dental procedures, such as deep scaling (cleaning) of your gums, gum surgery (periodontal surgery) or extractions, to measure bleeding and clotting times. High blood pressure (hypertension) Oral effects: Some drugs that treat high blood pressure (anti-hypertensive medicines) cause dry mouth (xerostomia) or an altered sense of taste (dysgeusia). Others may make you more likely to faint when you are raised from the relatively flat position in the dentist's chair to a sitting or standing position quickly. This reaction is called orthostatic hypotension. Gum overgrowth is a possible side effect of some drugs that treat high blood pressure. These include calcium channel blockers. It can begin as soon as one month after you start drug therapy. Some people's gums become so large that they have difficulty chewing. In some cases, surgery is needed to remove part of the overgrown gum tissue. 2291605-01-01 (06/23) At the dentist: If you have high blood pressure, your dentist should check your blood pressure at each visit. Your dentist can decide whether it's OK for you to have non-emergency dental treatment. It will depend on: • How high your blood pressure is • How well your blood pressure is controlled • Whether you have other medical conditions The first time you visit the dental office after being diagnosed with high blood pressure, your dentist may take your blood pressure two or three times. This is to establish a "baseline" blood pressure. This way, the dentist will know if your blood pressure changes in response to treatment or a medicine. Most people with high blood pressure can safely take anti-anxiety drugs — such as nitrous oxide or diazepam (Valium) — for dental procedures. They can also safely receive local anesthetics even if they contain epinephrine. If you have concerns about these drugs, talk to your dentist, physician or both. Some people taking calcium channel blockers may notice gum overgrowth (gingival hyperplasia). Your dentist will give you detailed oral hygiene instructions and may ask you to visit more often for professional cleanings. Remember that your daily tooth brushing and flossing at home is very important. If you stop taking the drugs, and only do so after consultation with your physician, your gums recede somewhat. However, this may take several months. Some people's gums do not return to normal on their own. Gum surgery may be necessary. Make sure your dentist knows which drugs you are taking for your high blood pressure. Before a dental visit, take your medicines as you normally do. Coronary artery bypass graft (CABG) Oral effects: There are no oral effects of this procedure. At the dentist: For the first couple of weeks after surgery, you may feel severe pain when reclining in the dental chair. This is a side effect of the surgery. Work with your dentist to find a comfortable position in the chair. Unless they need dental treatment within a few weeks after the surgery, people who have had CABG generally do not require antibiotics before a dental procedure. If you have had this surgery, speak to your physician before having any dental treatment within the next six months. Angina Oral effects: Angina is pain that starts in the chest. Sometimes it spreads to your lower jaw. Some people with angina take drugs called calcium channel blockers. These drugs can cause gum overgrowth. This can happen as soon as one month after you start these drugs. Some people's gums become so large that they have problems chewing. People who have this problem will most likely need surgery on their gums (periodontal surgery). 2291605-01-01 (06/23) At the dentist: People with stable angina can be treated like any other patients, with a few differences. Your dentist should have oxygen and nitroglycerin available during your visit. Your dentist should talk to your physician before your appointment. People with unstable angina should not receive non-emergency dental care. If you need emergency dental care, your heart should be continuously monitored. Stress can trigger angina attacks. If being in the dental chair increases your anxiety, speak with your dentist about ways to reduce this stress. If you feel any chest discomfort, tell your dentist or the dental staff right away. High cholesterol (hyperlipidemia) Oral effects: People with high cholesterol have too much fat in their blood. There are no oral effects of high cholesterol. At the dentist: Some drugs used to treat high cholesterol can make you feel faint after you get up from the dental chair. Oral side effects of these drugs include dry mouth (xerostomia), fruit-like breath odor and joint pain. High cholesterol puts you at risk of hardening of the arteries, which can lead to a heart attack or stroke. Your dentist should know about your condition and the drugs you are taking. Statins that are used to reduce treat high cholesterol can also have positive oral effects because of their anti-inflammatory and antioxidant properties and their positive effect on wound healing. Stroke Oral effects: Stroke can cause many long-term effects. These include: • Paralysis • Difficulty speaking and swallowing • Increased or decreased sensitivity to pain • Blurred vision • Poor memory • Personality changes (anxiety, depression) In some people, a stroke paralyzes one side of the body. If this happens to you, a family member or caregiver may need to help you with activities of daily living, including your dental care. Special toothbrushes and floss holders also are available. If you wear dentures, they may need to be adjusted. If your face or tongue is paralyzed, you may not be able to rinse your mouth. You may also not realize when you have food left in your mouth. You may bite your lip or tongue and not realize it. To keep your teeth and gums healthy, your dentist may suggest that you use a fluoride gel or saliva substitute. At the dentist: Some stroke survivors take blood thinners. Tell your dentist about all the medicines you take. You may need to stop taking your blood-thinning medicines before some dental procedures. Do not stop taking any medicines until you have spoken to your physician. This is something your dentist will discuss with you and your physician. 2291605-01-01 (06/23) Usually, routine dental treatment is safe. Bring a copy of your most recent blood tests to your dentist at every visit. Congestive heart failure Oral effects: Many of the medicines used to treat congestive heart failure (CHF) cause dry mouth. The medical term for dry mouth is "xerostomia." At the dentist: If you are being treated for CHF and have no complications, side effects or physical limitations, there are usually no special changes needed for dental treatment. However, the dentist may make some changes, depending on the medicines you take and your overall health. If you have more severe heart failure, you should not lie down in the dental chair too far. The fluid build-up in your lungs may affect your breathing. You also should not sit up or lie down very quickly. These changes can make you dizzy and light-headed. Your dentist can confirm how serious your CHF is by talking with your physician or cardiologist. Some people with severe CHF may need to have their dental treatment in a hospital setting. This includes people whose disease is considered class III or IV under the New York Heart Association functional classification system. Pacemaker/defibrillator implantation Oral effects: There are no specific oral effects caused by having a pacemaker or defibrillator. At the dentist: If you have a pacemaker or defibrillator you should confirm that there are no interactions between electromagnetic devices in your dentist's office and your pacemaker or defibrillator. Certain machines that a dentist or dental hygienist may use could potentially interact and cause a problem with a pacemaker or defibrillator. Examples include machines used for ultrasound or electrosurgery. The chance of any interaction is very small. You or your dentist should be able to find out about interactions from your physician or from the device’s manufacturer. Talk with your physician about possible interactions before visiting the dental office. If there is a chance of interaction, your dentist can take precautions to prevent it. You should avoid elective dental care within the first few weeks after receiving your pacemaker. If you must receive dental care within that time, your dentist and physician should decide if you need to take antibiotics before treatment. Last updated: 06-01-23 2291605-01-01 (06/23) References 1. Liccardo D, Cannavo A, Spagnuolo G, Ferrara N, Cittadini A, Rengo C, Rengo G. Periodontal Disease: A Risk Factor for Diabetes and Cardiovascular Disease. Int J Mol Sci. 2019 Mar 20;20(6):1414. doi: 10.3390/ijms20061414. PMID: 30897827; PMCID: PMC6470716. 2. Sanz M, Del Castillo AM, Jepsen S, Gonzalez-Juanatey JR, D'Aiuto F, Bouchard P, Chapple I, Dietrich T, Gotsman I, Graziani F, Herrera D, Loos B, Madianos P, Michel JB, Perel P, Pieske B, Shapira L, Shechter M, Tonetti M, Vlachopoulos C, Wimmer G. Periodontitis and Cardiovascular Diseases. Consensus Report. Glob Heart. 2020 Feb 3;15(1):1. doi: 10.5334/gh.400. PMID: 32489774; PMCID: PMC7218770. 3. ada.org/en/member-center/oral-health-topics/antibiotic-prophylaxis (accessed 03-31-21) 4. Samulak-Zielińska R, Dembowska E, Lizakowski P. Dental treatment of post-myocardial infarction patients: A review of the literature. Dent Med Probl. 2019 Jul-Sep;56(3):291-298. doi: 10.17219/dmp/109232. PMID: 31577073. 5. Southerland JH, Gill DG, Gangula PR, Halpern LR, Cardona CY, Mouton CP. Dental management in patients with hypertension: challenges and solutions. Clin Cosmet Investig Dent. 2016 Oct 17;8:111-120. doi: 10.2147/CCIDE.S99446. PMID: 27799823; PMCID: PMC5074706. 6. Tom J. Management of Patients With Cardiovascular Implantable Electronic Devices in Dental, Oral, and Maxillofacial Surgery. Anesth Prog. 2016 Summer;63(2):95-104. doi: 10.2344/0003-3006-63.2.95. PMID: 27269668; PMCID: PMC4896049. 7. perio.org/consumer/gum-disease-and-other-diseases (accessed 03-31-21) 8. Albert DA, Sadowsky D, Papapanou P, Conicella ML, Ward A. An examination of periodontal treatment and per member per month (PMPM) medical costs in an insured population. BMC Health Serv Res. 2006 Aug 16;6:103. doi: 10.1186/1472-6963-6-103. PMID: 16914052; PMCID: PMC1574303. 9. Lauritano D, Martinelli M, Baj A, Beltramini G, Candotto V, Ruggiero F, Palmieri A. Drug-induced gingival hyperplasia: An in vitro study using amlodipine and human gingival fibroblasts. Int J Immunopathol Pharmacol. 2019 Jan-Dec;33:2058738419827746. doi: 10.1177/2058738419827746. PMID: 31663449; PMCID: PMC6822186. 10. Tahamtan S, Shirban F, Bagherniya M, Johnston TP, Sahebkar A. The effects of statins on dental and oral health: a review of preclinical and clinical studies. J Transl Med. 2020 Apr 6;18(1):155. doi: 10.1186/s12967-020-02326-8. PMID: 32252793; PMCID: PMC7132955. Aetna is the brand name used for products and services provided by one or more of the Aetna group of companies, including Aetna Life Insurance Company and its affiliates (Aetna). All information is intended for your general knowledge only and is not a substitute for medical advice or treatment for specific medical conditions. You should seek prompt medical care for any specific health issues and consult your physician before starting a new fitness regimen. Use of this online service is subject to the disclaimer and the terms and conditions. External website links provided on this site are meant for convenience and for informational purposes only; they do not constitute an endorsement. External links open in a different window. 2291605-01-01 (06/23) |
4772 | https://www.khanacademy.org/v/sn1-mechanism-carbocation-rearrangement | Published Time: Tue, 15 Jul 2025 20:16:58 GMT
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Organic chemistry
Course: Organic chemistry>Unit 5
Lesson 5: Sn1 and Sn2
Identifying nucleophilic and electrophilic centers
Curly arrow conventions in organic chemistry
Intro to organic mechanisms
Alkyl halide nomenclature and classification
Sn1 mechanism: kinetics and substrate
Sn1 mechanism: stereochemistry
Carbocation stability and rearrangement introduction
Carbocation rearrangement practice
Sn1 mechanism: carbocation rearrangement
Sn1 carbocation rearrangement (advanced)
Sn2 mechanism: kinetics and substrate
Sn2 mechanism: stereospecificity
Sn1 and Sn2: leaving group
Sn1 vs Sn2: Solvent effects
Sn1 vs Sn2: Summary
Science>
Organic chemistry>
Substitution and elimination reactions>
Sn1 and Sn2
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Sn1 mechanism: carbocation rearrangement
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Examples of Sn1 reactions involving carbocation rearrangements.
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Kim 7 years ago Posted 7 years ago. Direct link to Kim's post “I know this is somewhat s...” more I know this is somewhat stupid, but how do you know when to perform a hydride shift and a methyl shift when you're doing an Sn1 reaction. For instance if my teacher gave me a molecule and said predict the Sn1 reaction, is there any hints in the molecule that would tell me that I should let the molecule undergo a hydroxide or methyl shift?
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anthonygeorge617 2 years ago Posted 2 years ago. Direct link to anthonygeorge617's post “At 3:15, why would one of...” more At 3:15 , why would one of the extra ethanol molecules take the hydrogen bonded to the Oxygen that is now a part of the initial reactant? I understand it's to get to the final product shown, but is there a reason this is done to help with stability?
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Richard 2 years ago Posted 2 years ago. Direct link to Richard's post “The last step is a simple...” more The last step is a simple acid-base reaction where the ethanol is acting as a base and the substrate is acting as an acid. Ethanol is a weak base similar in strength to water. Likewise, the protonated substrate is a weak acid. So yeah if you put an acid and a base in the same solution, they’re going to do acid-base chemistry.
Hope that helps.
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Sidharth Gat 7 years ago Posted 7 years ago. Direct link to Sidharth Gat's post “Hydroxide taking Proton f...” more Hydroxide taking Proton from hydronium in first step makes sense but water taking back Proton in last step doesn't?
Or
If 1st step can occur last step can't or if last step can occur the 1st can't?
Please explain if I'm getting wrong somewhere (as to why this reaction should reach completion through way it's shown).
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kumar.aks99 8 years ago Posted 8 years ago. Direct link to kumar.aks99's post “At 3:45, why is hydroxide...” more At 3:45 , why is hydroxide a bad leaving group. Is it because you're in a polar aprotic solvent and hydroxide is a very strong base?
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heybati.shayan 2 years ago Posted 2 years ago. Direct link to heybati.shayan's post “At 6:22, how do you know ...” more At 6:22 , how do you know to do a hydride shift versus methyl shift?
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PineappleSoup 2 years ago Posted 2 years ago. Direct link to PineappleSoup's post “If we had done a methyl s...” more If we had done a methyl shift, the resulting carbocation would still be a secondary carbocation, so there would be no change in stability. The hydride shift allows the formation of a more stable tertiary carbocation, which is why it is favored.
When trying to decide whether to use a hydride shift vs. a methyl shift, think about whether the resulting carbocation will be secondary or tertiary. A methyl shift from a quaternary carbon will result in a tertiary carbocation, and a hydride shift from a tertiary carbon will result in a tertiary carbocation.
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athena 5 months ago Posted 5 months ago. Direct link to athena's post “Why isn't the I- ion we h...” more Why isn't the I- ion we have from the beginning bonding with the H+ at the end to create HI instead of another Ethanol molecule needing to step in at 3:09 ?
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Jonathon 8 years ago Posted 8 years ago. Direct link to Jonathon's post “In the first example, we ...” more In the first example, we have protonated Ethanol and Iodide ions. Is Iodic Acid the ultimate byproduct of the reaction? Is there an equilibrium for this reaction or does the HI bubble out?
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Video transcript
[Lecturer] Since the SN1 mechanism involves the formation of a carbocation a rearrangement is possible. So let's look at this SN1 reaction. On the left is our alkyl halide, ethanol is our solvent and on the right is our product. The first step should be loss of a leaving group. So these electrons come off on to the iodine to form the iodide ion. We are taking a bond away from this carbon in red so the carbon in red is gonna get a plus one formal charge. So let's draw this in. We are gonna form a carbocation and the carbon in red is this carbon so this carbon gets a plus one formal charge. If we look at this carbocation it is secondary. The carbon in red is directly bonded to two other carbons so this is a secondary carbocation. If you think about the video on carbocations and rearrangements. We can have a rearrangement here to form a more stable carbocation. A methyl shift will make sense. So this methyl group is gonna shift over to this carbon and let's draw what we would make now. So we would have a methyl group now that's moved over so the carbon in red is this carbon. And we are taking a bond away from the carbon that I just circled in green so the carbon I circled in green is this carbon and that carbon gets a plus one formal charge now. So this is our carbocation. What kind of carbocation is it? Well if you look at the carbon that I circled in green that's directly bonded to one, two, three other carbons so it is a tertiary carbocation, which we know is more stable than a secondary carbocation so a methyl shift increases the stability going from a secondary to a tertiary carbocation. This carbocation is our electrophile. Now it is time for our nucleophile to attack our electrophile. Our nucleophile is our solvent ethanol. So this is a solvolysis reaction. So if I draw in an ethanol molecule here and put in two lone pairs of electrons on the oxygen, so one of those lone pairs is gonna form a bond with the carbon that I circled in green. So the nucleophile attacks the electrophile and we form a bond between the oxygen and the carbon. So if I sketch this in here, now we would have a bond to this carbon. This oxygen is still bonded to a hydrogen, still bonded to the ethyl and lets say that the lone pair of electrons form the bond. Let's make them these electrons in magenta. So those electrons in magenta on the oxygen form this bond. We still have a lone pair of electrons left on the oxygen so here they are. Here they are right here. And that's a plus one formal charge on the oxygen. If we compare this to our final product, notice all we have to do is deprotonate. So the last step of the mechanism is just a proton transfer and acid base reaction. Another molecule of ethanol could come along and serve as the base. So if I draw on my lone pairs of electrons on the oxygen one of those lone pairs could pick up this proton, leave these electrons behind on this oxygen and finally give us our product so you could draw your product a few different ways. But that is our SN1 mechanism with the carbocation rearrangement. Let's do another carbocation rearrangement in an SN1 mechanisms but this time we are starting with this alcohol. In the previous example, the first step was loss of a leaving group. But if we show the electrons going on to the oxygen now to form the hydroxide ion, that doesn't work because the hydroxide ion is a bad leaving group. So we actually have to protonate the alcohol first to form a good leaving group. So the first step of this mechanism is a proton transfer. The alcohol is gonna function as a base and HC3O plus is gonna donate a proton. So the hydronium ion is gonna act as our asset so that's HC3O plus and our alcohol acts as our base. Lone pair of electrons on the oxygen pick up a proton from hydronium. So our first step is a proton transfer. And that gives us this oxygen here with a plus one formal charge. So still one lone pair of electrons, a plus one formal charge on that oxygen and let's say that this lone pair of electrons picked up this proton to form this bond. Now we are ready for loss of a leaving group because when these electrons come off on to this oxygen now that gives us water and water is a good leaving group. So first step is proton transfer. Second step is loss of a leaving group and we are taking a bond away from this carbon in red. So that carbon in red gets a plus one formal charge. So let me draw this in here. So the carbon in red is this carbon. So let me write a plus one formal charge on this carbon. What kind of carbocation is that? The carbon in red is directly bonded to two other carbons. So this is a secondary carbocation and we think about the possibility of a rearrangement. So in the carbocation and rearrangements video, another shift that we did was a hydride shift. So on this carbon in magenta there's still a hydrogen and a hydride shift would give us a more stable carbocation. Remember a hydride shift think about this hydrogen and these two electrons here. Shifting over to this carbon in red. So let's draw in what we would have now. So now if you think about that carbon in red, we already had a hydrogen bonded to it. So let me go ahead and highlight this carbon as being the carbon in red. We already had a hydrogen bonded to it in the example, in the carbocation to the left I should say. So let me draw in that original hydrogen. With the hydride shift, we are adding in another hydrogen here to this carbon in red. So there is no more formal charge on the carbon in red. The formal charge moves to this carbon, the one I circled in green. It is losing a bond so that carbon that I circled in green now gets a plus one formal charge. What kind of carbocation did we form. Well the carbon in green is directly bonded to one, two, three, other carbons. So this is a tertiary carbocation. A hydride shift gives us a more stable carbocation. Now we formed our electrophile and our nucleophile will come along in the next step. And our nucleophile should be methanol. So let me draw in a molecule of methanol here. So put in a hydrogen and two lone pairs of electrons on the oxygen. So our nucleophile will attack our electrophile so lone pair of electrons on this oxygen are gonna form a bond with this carbon that I circled in green. So let's draw the result of our nucleophilic attack. So now we are forming a bond between the carbon in green and this oxygen here. So let me sketch in the rest of this. And then we still have a methyl on this oxygen. We still have a hydrogen on this oxygen and let me highlight those electrons. So our electrons, let's make them blue. So these electrons in blue here on this oxygen form a bond between the oxygen and that carbon. We still have a lone pair of electrons on this oxygen which gives the oxygen a plus one formal charge. And also notice what I did with this group right here. So these two carbons I just drew them going down this way. This free rotation around this bond right here so you can draw it however you want. I just drew it to look a little bit more like our product here. When we compare our two, the last step must be a proton transfer. All we have to do is take a proton away and we have our final product. I have a couple of choices here as my base. I could use either the water molecule that we lost back in this step or I could use another molecule of methanol as our base. It doesn't really matter what we use here. I'll just use the water molecule. Either water or methanol acts as a base in the last step of our mechanism. And takes this proton leaving these electrons behind and forming our final product.
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4773 | https://www.quora.com/Why-is-the-unit-for-acceleration-m-s-2 | Why is the unit for acceleration m/s^2? - Quora
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Why is the unit for acceleration m/s^2?
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Chris Harrington
Bachelor of Fine Arts from Academy of Art University (Graduated 2011) · Author has 20.3K answers and 46.9M answer views
·Updated May 10
Originally Answered: Why is acceleration m/s2? ·
It doesn't have to be meters, but using metric units is easier and requires less conversion. It could be feet per second² for the USA-ans.
So acceleration is a measurement of how fast you are picking up speed. That means, if you start from zero and pick up speed, you are going to have more and more speed over time.
The phrase “m/s²” means “meters per second squared,” or more accurately, “meters per second, per second.” (One second, per second is shortened to seconds².) For example, at zero seconds, you're not moving. Then in the next second, you are going one meter per second. Then in the next s
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It doesn't have to be meters, but using metric units is easier and requires less conversion. It could be feet per second² for the USA-ans.
So acceleration is a measurement of how fast you are picking up speed. That means, if you start from zero and pick up speed, you are going to have more and more speed over time.
The phrase “m/s²” means “meters per second squared,” or more accurately, “meters per second, per second.” (One second, per second is shortened to seconds².) For example, at zero seconds, you're not moving. Then in the next second, you are going one meter per second. Then in the next second, you are going two meters per second. Then in the third second, three meters per second. The amount of your speed increases by one meter per second, and it does that every second. So your acceleration is 1m/s², or “one meter per second, per second.”
That's what acceleration in “m/s²” means. It means that your speed, (given in m/s,) increases by the given amount every second.
OP: “Why is acceleration m/s2?”
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More answers below
What is the unit of acceleration?
What does the unit s^2/m^2 mean?
What is the SI unit for acceleration?
Why is velocity measured in m/s and acceleration measured in m/s^2?
Why does a unit of time occur twice in a unit of acceleration?
Cody Reisdorf
Math-Phys B.Sc.—nescient swot · Author has 629 answers and 1.1M answer views
·8y
Originally Answered: Why is the acceleration of gravity have m/s^2 instead of m^s? ·
Because acceleration is a change in velocity—velocity is a vector quantity, meaning it has a magnitude and a direction, so a change in the magnitude, or the direction (or both), implies (or requires) an acceleration of some sort.
The magnitude part of velocity is speed, which is distance divided by time, like miles per hour, or kilometers per hour, or meters per second (m/s). Since an acceleration is a change in velocity, it must be some amount of distance divided by time, divided by time again.
For example, if your in a car driving 60 miles per hour (mph), and you step on the accelerator (calle
Continue Reading
Because acceleration is a change in velocity—velocity is a vector quantity, meaning it has a magnitude and a direction, so a change in the magnitude, or the direction (or both), implies (or requires) an acceleration of some sort.
The magnitude part of velocity is speed, which is distance divided by time, like miles per hour, or kilometers per hour, or meters per second (m/s). Since an acceleration is a change in velocity, it must be some amount of distance divided by time, divided by time again.
For example, if your in a car driving 60 miles per hour (mph), and you step on the accelerator (called the gas peddle here in America—what do they call it in countries where they call fuel petrol instead of gas?), and you “speed up” to 80 mph, then you’ve changed your velocity by 20 mph—but depending on far down you pushed the accelerator peddle, you may have taken a short amount of time to gain that extra speed, or a long amount of time. If you took exactly five minutes, then the acceleration was 20 miles per hour per five minutes. (We could divide the 20 by five and say 4 miles per hour per minute—if we want we could use other conversions to get the acceleration in miles per hour per second, or miles per minute per minute, or meters per minute per minute, etc., for any units we want.)
Meters per second per second is m s s 1=m s∗s=m s 2 m s s 1=m s∗s=m s 2
I’m not really sure what m s m s would even mean really—I’m not aware of any circumstance in which units of distance raised to the power of time has any useful interpretation or application. (That doesn’t mean there isn’t a useful meaning, but there’s a good chance there isn’t a useful meaning.)
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Science Hidden Facts
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Shiva Kumar
· Author has 1.1K answers and 2.1M answer views
·Jan 11, 2022
Why is the unit for acceleration m/s²?
Linear motion of a body is described by velocity and acceleration.
Velocity is the time rate of displacement. It's SI unit is m/s. While,
Acceleration is the time rate of velocity. In other words,
Velocity per second.
(m/s)/s
That's, meter per second or second
It's SI unit is m/s².
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Acceleration measures how quickly velocity changes with time. Velocity has units of metres per second (m/s) because it expresses distance travelled per unit time. Acceleration is the rate of change of that velocity with respect to time, so its units are “velocity per time.”
Step-by-step:
Velocity = distance / time → units: m / s.
Acceleration = change in velocity / time → units: (m / s) / s.
Dividing by seconds once more gives m / s^2 (metres per second per second).
Interpretations and examples:
m/s^2 means the velocity increases by that many metres per second every second. If acceleration = 2 m/s^2
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Acceleration measures how quickly velocity changes with time. Velocity has units of metres per second (m/s) because it expresses distance travelled per unit time. Acceleration is the rate of change of that velocity with respect to time, so its units are “velocity per time.”
Step-by-step:
Velocity = distance / time → units: m / s.
Acceleration = change in velocity / time → units: (m / s) / s.
Dividing by seconds once more gives m / s^2 (metres per second per second).
Interpretations and examples:
m/s^2 means the velocity increases by that many metres per second every second. If acceleration = 2 m/s^2, speed rises by 2 m/s after one second, 4 m/s after two seconds, etc.
Negative m/s^2 indicates a reduction in velocity (deceleration or acceleration opposite to motion).
Dimensional form: [a] = L T^−2 (length per time squared), used in physics to check equations (e.g., F = ma → [F] = M L T^−2).
Alternate notation: a = m·s^−2 or “metres per second per second” — all equivalent ways to express the same physical quantity.
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Can you explain what the unit of acceleration (m/s^2) really means?
What is a unit of acceleration and velocity?
How can we practically explain acceleration having s^-2 in its unit?
Why is s^2 used in the case of acceleration's unit?
What is the reason for measuring acceleration in m/s^2 in physics instead of other units such as m/s^-1?
Sergio N. Bordalo
Knows Portuguese · Author has 882 answers and 925.4K answer views
·4y
Originally Answered: Why is velocity measured in m/s and acceleration measured in m/s^2? ·
That’s a very intelligent question.
Acceleration is the CHANGE in velocity per time.
Suppose a spaceship was initially travelling at 1000 m/s. - - - ->
And after one hour it’s speed has increased to 4600 m/s. - - - - - - - - - - - >
So, the speed changed by 3600 m/s in ONE HOUR. It’s acceleration was 3600 m/s per hour….
…Or 1 m/s per second (the speed increased by 1 m/s at every second that passed).
Mathematically we have
a = (1 m/s)/s = 1 m/s².
You see, it’s completely natural! It’s not made up.
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Gaurav Singh
Studying at The College and University Experience · Author has 252 answers and 1.7M answer views
·Updated 7y
Originally Answered: What is the logic (not the math) behind the term (s^2) in the unit of acceleration (m/s^2)? ·
Acceleration relates the time it takes to change your speed which is already defined as the time it takes to change your location. So acceleration is measured in distance units over time x time.
Explanation:
We have already discovered that when something moves, it changes its location. It takes some time to complete that movement, so the change in location over the time is defined as speed, or its rate of change. If the thing is moving in a particular direction, the speed can then be defined as velocity.
Velocity is the rate or speed an object is moving from A to B over a measurable time.
It is un
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Acceleration relates the time it takes to change your speed which is already defined as the time it takes to change your location. So acceleration is measured in distance units over time x time.
Explanation:
We have already discovered that when something moves, it changes its location. It takes some time to complete that movement, so the change in location over the time is defined as speed, or its rate of change. If the thing is moving in a particular direction, the speed can then be defined as velocity.
Velocity is the rate or speed an object is moving from A to B over a measurable time.
It is unusual to maintain a constant velocity in a given direction for very long; at some point the speed will increase or decrease, or the direction of motion will change. All of these changes are a form of acceleration. And all of these changes take place over time.
Acceleration is the rate or speed at which an object is increasing or decreasing its velocity over a measurable time.
We can think of acceleration as doing two things at once. We are still moving across a distance over a time, but we are also increasing how fast we are doing it. We are multi-tasking to arrive sooner, so we have to multiply the time x time to calculate the correct numerical value for our acceleration.
Edit : look into comments, there's additional detailed explanation.
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James Garry
R&D director (2016–present) · Author has 2.5K answers and 1.5M answer views
·5y
Originally Answered: Why are the units of acceleration meters per second squared? ·
As long as you measure a speed, and the rate of change of that speed, you’re welcome to use whatever units you like.
Don’t like seconds? The hour is perfectly fine - and perhaps the thing is accelerating so slowly that a mm per hour per hour makes more sense.
Or are you confused about the ‘squared’ part?
Well, you’re measuring the rate at which speed is changing - right?
And speed has units of length per unit time.
So speed could be measured in yards per day, or metres per second. Both are valid.
And now you want to know how fast that speed is changing - how much extra speed is gained in a period of
Continue Reading
As long as you measure a speed, and the rate of change of that speed, you’re welcome to use whatever units you like.
Don’t like seconds? The hour is perfectly fine - and perhaps the thing is accelerating so slowly that a mm per hour per hour makes more sense.
Or are you confused about the ‘squared’ part?
Well, you’re measuring the rate at which speed is changing - right?
And speed has units of length per unit time.
So speed could be measured in yards per day, or metres per second. Both are valid.
And now you want to know how fast that speed is changing - how much extra speed is gained in a period of time.
So you could write the acceleration of a slow thing as 1 mm per hour, per day.
In other words, your plant’s root is speeding up, each day, by 1mm per hour.
Day 0 = at rest
Day 1 = moving at 1mm per hour
Day 2 = moving at 2mm per hour
Day 3 = moving at 3mm per hour
Note that we would write this acceleration as 1mm per hour per day
Which is awkward.
Far simpler to write it as xxx mm per hour per hour
And if you do that you notice that you can write “per hour squared” instead of “per hour per hour”.
1/(a^2) = 1/a 1/a
Right?
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Leonard Baczek
SM from Massachusetts Institute of Technology (Graduated 1970) · Author has 887 answers and 774.8K answer views
·3y
Originally Answered: How do I derive the unit of acceleration? Why do we multiply m/s by s? ·
We do not multiply m/s by s ; we divide it. Here is why. Acceleration is how fast a velocity changes with each unit of time. If a velocity is 5 meters per second, written 5 m/s, then an object travels 5 meters in each second. Suppose that the same object speeds up by 5 meters per second with each passing second. At time 0, the velocity is 0 m/s. At time t=1sec the object has sped up (accelerated) by 5 m/s to a velocity of 5 m/s. During the 2nd second, the velocity increases again by 5 m/s to a velocity of 10 m/s and so on. The acceleration is said to be a change in velocity of 5 m/s with each
Continue Reading
We do not multiply m/s by s ; we divide it. Here is why. Acceleration is how fast a velocity changes with each unit of time. If a velocity is 5 meters per second, written 5 m/s, then an object travels 5 meters in each second. Suppose that the same object speeds up by 5 meters per second with each passing second. At time 0, the velocity is 0 m/s. At time t=1sec the object has sped up (accelerated) by 5 m/s to a velocity of 5 m/s. During the 2nd second, the velocity increases again by 5 m/s to a velocity of 10 m/s and so on. The acceleration is said to be a change in velocity of 5 m/s with each passing second or 5 m/s per second written 5 m/s/s. The double division is just like dividing fractions, flipping the denominator and multiplying. The result is an acceleration of 5 m/s^2 which is just a more advanced way of writing m/s/s. Hope this helps.
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Loring Chien
Using measurement tools for 50 years · Author has 67.9K answers and 250M answer views
·9y
Originally Answered: Why is velocity measured in m/s and acceleration measured in m/s^2? ·
In calculus the velocity is the derivative of position with respect to time.
And acceleration is the derivative of velocity with respect to time.
Knowledge of simple differentiation will show that a simple derivative is the base unit divided by time.
Hence the units meters, meters per second and meters per second^2 for position, velocity, and acceleration.
Simply put, acceleration is the change in velocity per unit time. If a car goes from 15 m/s to 20 m/s in a second then the acceleration is said to be 5 m/s in a second or 5 m/s^2. This makes a lot of assumptions like steady acceleration but is
Continue Reading
In calculus the velocity is the derivative of position with respect to time.
And acceleration is the derivative of velocity with respect to time.
Knowledge of simple differentiation will show that a simple derivative is the base unit divided by time.
Hence the units meters, meters per second and meters per second^2 for position, velocity, and acceleration.
Simply put, acceleration is the change in velocity per unit time. If a car goes from 15 m/s to 20 m/s in a second then the acceleration is said to be 5 m/s in a second or 5 m/s^2. This makes a lot of assumptions like steady acceleration but is a clear illustration.
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Anupam Kumar
Studied at Career Launcher, Patna · Author has 420 answers and 1.3M answer views
·5y
Originally Answered: How do I derive the unit of acceleration? Why do we multiply m/s by s? ·
SI unit of velocity = m/s
SI unit of time = s
We know that acceleration is the rate of change of velocity.
Acceleration = change in velocity ÷ time taken
Hence, SI unit of acceleration = SI unit of velocity ÷ SI unit of time
= (m/s)÷s(m/s)÷s
= m/s 2 m/s 2
A⚡K
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C Stuart Hardwick
Scifi author and science nerd. · Author has 13.6K answers and 204.4M answer views
·9y
Originally Answered: Why is velocity measured in m/s and acceleration measured in m/s^2? ·
For the very simple reason that acceleration is the change in speed over time, so it must be expressed in a speed over time. In science, this is most often meters per second per second. On average, and object falling near the surface of the earth, ignoring wind resistance, increases its speed of descent at a rate of 9.8 meters per scond, for each second it falls. So after falling for two seconds, if would be traveling at 19.6 meters per second.
The way I wrote it out is terribly wordy, so instead we say, "the gravitational accelleration is 9.8 meters per second per second, or 9.8mps^2.
I can als
Continue Reading
For the very simple reason that acceleration is the change in speed over time, so it must be expressed in a speed over time. In science, this is most often meters per second per second. On average, and object falling near the surface of the earth, ignoring wind resistance, increases its speed of descent at a rate of 9.8 meters per scond, for each second it falls. So after falling for two seconds, if would be traveling at 19.6 meters per second.
The way I wrote it out is terribly wordy, so instead we say, "the gravitational accelleration is 9.8 meters per second per second, or 9.8mps^2.
I can also express acceleration in other units, with varying results. The Tesla Motors S P85D can do 0 to 60 in 3.1 seconds, which is 26.8 meters per second in 3.1 seconds or 8.7mps^2. But we don't measure the acceleration of cars in mps/s because no car ever made can sustain it's take-off acceleration for more than a few seconds, the it's acceleration varies greatly over than time, so the "0 to 60" time is a more meaningful unit of acceleration for cars.
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John Nygate
Artist · Author has 592 answers and 900.3K answer views
·9y
Originally Answered: Why is velocity measured in m/s and acceleration measured in m/s^2? ·
Well, say at time 0 you are travelling at 3m/s. A second later you are travelling at 5m/s. So in one second you have have increased you velocity by 2m/s. So assuming constant acceleration we say the acceleration was 2m/s per second or 2m/s/s and we write this as 2m/s^2.
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Victor Mazmanian
Former Associate Prof. Of Physics (Retired) at United States Air Force Academy · Author has 1.5K answers and 4.7M answer views
·7y
Originally Answered: How do I derive the unit of acceleration? Why do we multiply m/s by s? ·
Both responses provided are sufficient as an answer. My answer deals with understanding the significance of the unit itself.
All of the given/acceptable units for acceleration [m/s/s; m/s^2; ms^-2] mean the following: there is a chunk of speed (measured in m/s) that is CHANGING (it could be increase or decrease) per second. So the units above may be read as: “so much change in speed every second”. So when we say g = 9.81 m/s/s; we are really saying: the acceleration of an object in free-fall is an increase of a chunk of 9.81 m/s speed EVERY SECOND.
A favorite method of mine is to illustrate “g”
Continue Reading
Both responses provided are sufficient as an answer. My answer deals with understanding the significance of the unit itself.
All of the given/acceptable units for acceleration [m/s/s; m/s^2; ms^-2] mean the following: there is a chunk of speed (measured in m/s) that is CHANGING (it could be increase or decrease) per second. So the units above may be read as: “so much change in speed every second”. So when we say g = 9.81 m/s/s; we are really saying: the acceleration of an object in free-fall is an increase of a chunk of 9.81 m/s speed EVERY SECOND.
A favorite method of mine is to illustrate “g” in non-standard units (considered a huge no-no by purists; but I beg to differ because it is very instructive): I present “g” not as 9.81 m/s/s but as its weird BE-MKS mixture of: 22 mph EVERY SECOND. What this says is: in free-fall, objects will gain speed. They will ACCELERATE. After one second of fall from rest, the object’s speed will be 22 mph. After two seconds, it would be 44 mph. After three seconds, it would be 66 mph; etc. Try this: what would its speed be after 2.5 seconds? Answer: 55 mph!
If the object was thrown down with an initial speed, say 15 mph, then after one second it would be traveling at 37 mph (gaining 22 mph); and after two seconds, it would be traveling at 59 mph, etc. Try this: what would its speed be after 3.5 seconds? Answer: 92 mph!
If the object is a projectile in a vertically upward motion; then the acceleration would be “- 22 mph every second”. Suppose we project it at 88 mph upwards. We can immediately deduce that it would reach its peak in 4 seconds! Wasn’t that easy?
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Related questions
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4774 | https://people.hsc.edu/drjclassics/latin/general_info_about_grammar/verbs.shtm | How to Translate a Verb
DR. J'S ILLUSTRATED GUIDE TO THE CLASSICAL WORLD site indexsites of Greece | sites of Italy | other sites | Myth | Romans in... lectures | texts | Latin | other materials (classics +) | Dr. J's Dossier Dr J's Audio-Visual Resources for Classics Back to Latin Instruction Home Paradigms Explanations Vocabulary Web ResourcesHOW TO TRANSLATE A VERB IN A MAIN CLAUSE ex: Hominem ad me ducent. 1. Identify the verb (ducent). 2. Identify the conjugation of the verb based on its principal parts (duco, ducere, duxi, ductus...a short -ere ending on the infinitive identifies this verb as 3rd conjugation) 3. Determine the tense based on rules for that particular conjugation (in the 3rd conjugation, an "e" indicates future) 4. Determine the person and number of the verb (3rd person plural). 5. Translate the verb. Ducent = "they will lead" 6. SOV the sentence: O prep phrase S/V Hominemad meducent 7. Translate the sentence: They will lead the man to me. copyright 2001 Janice Siegel, All Rights Reserved send comments to: Janice Siegel (jfsiege@ilstu.edu) date this page was edited last: 06/29/2005 the URL of this page: |
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Markup and commission word problems
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| Markup and commission word problems |
| |
| Description |
| Exercise Name: | Markup and commission word problems |
| Math Missions: | 7th grade (U.S.) Math Mission, Arithmetic essentials Math Mission, Pre-algebra Math Mission, Mathematics I Math Mission |
| Types of Problems: | 2 |
The Markup and commission word problems exercise appears under the 7th grade (U.S.) Math Mission, Arithmetic essentials Math Mission, Pre-algebra Math Mission and Mathematics I Math Mission. This exercise concentrates on business applications of percentage, specifically markup and commission.
Types of Problems[]
There are two types of problems in this exercise:
Find the commission: This problem provides a sales situation where an employee is partially paid on commission. The problem asks the user to provide the amount of commission a person would earn based on an amount of sales.
Find the commission
Find the markup: This problem provides a circumstance where an owner is supposed to markup a price in order to realize a profit. Users are asked to find the markup amount, and the final markup price on different problems.
Find the markup
Strategies[]
Knowledge of percentage as related to business would be an advantage here, but is not necessary. Knowing how to work word problems involving percentage is really the most important thing.
Research has witnessed instances where the markup problems asks for the markup and the markup price, but no such distinction on the commission problems yet.
Real-life Applications[]
Fractions, decimals, and percents are used in many sports such as basketball to determine how many field goals the player has made over how many they missed.
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Heat is a type of energy and like any other energy it flows between two different potentials. Heat flows from a body with a higher amount of heat energy to a body with a low amount of it. Just like any other energy transfer heat transfer also has a rate of transfer. If the transfer of heat energy between two objects is considered, the rate of heat transfer per unit area is known as the heat flux. It is defined as the amount of heat energy transferred per unit time per unit area to or fro from a surface. It is measured in the units of Watt per meter squared. This Physics article will explain the heat flux formula along with some solved examples.
What is Heat Flux?
Heat flux is the rate of heat transfer per unit area. It describes the amount of thermal energy that passes through a surface per unit time, typically expressed in units of watts per square meter (W/m²).
Heat flux can be either convective or radiative. Convective heat flux is the transfer of heat due to the movement of fluids, such as air or water, over a surface. Radiative heat flux is the transfer of heat due to the emission and absorption of electromagnetic radiation, such as infrared radiation.
Heat Flux Units
The S.l. unit of heat flux is watts per square meter (W/m²). Other commonly used heat flux units include calories per square centimeter per second (cal/cm²·s) and BTUs per hour per square foot (BTU/h·ft²).
Heat Flux Formula
The formula for heat flux is mathematically described as
(JH_{c}=\lambda \frac{dT}{dZ})
(JH_{c}) = conductive heat flux
T = Temperature
(\lambda = thermal conductivity constant
The formula for heat flow rate is mathematically stated as
(Q=-k(\frac{A}{l})(\Delta T))
Where,
Q = Heat transfer per unit time
k = Thermal conductivity of the object
A = Cross-sectional area of the face where the heat is applied.
l = Length of the object
(\Delta T) = Temperature difference between the object.
Also learn about thermodynamic process here.
Measurement of Heat Flux
The most popular, though frequently impracticable, approach is to measure the temperature differential across a piece of material whose thermal conductivity is already known. Since the thermal resistance of the material being tested is often unknown, this method is not proper and very difficult to perform.
Heat flux sensors are devices that are used to measure heat flux directly. These sensors work by measuring the temperature difference across a thin surface exposed to the heat source. The heat flux is calculated by measuring the temperature gradient and the thermal conductivity of the material.
Heat Flux Formula Solved Examples
Example 1. One face of a copper plate is 1 m thick and maintained at 60°C, and the other face is maintained at 20°C. Calculate the heat flux given that the thermal conductivity of copper is 385.
Solution 1: Since the formula for heat flux is (JH_{c}=\lambda \frac{dT}{dZ})
dT = 60 -20 = 40°C
dZ = 1 m
(\lambda) = 385
Thus, (JH_{c}=\lambda \frac{dT}{dZ})
(JH_{c}=385 \times \frac{40}{1})
(\phi_{q} = 15400 W/m^2)
Example 2. Heat has to flow between a copper plate of thickness 0.08 m across two temperatures of 160 °C and 40°C. Calculate the heat flux. The thermal conductivity of glass is 386 W/mK
Solution 2: Since the formula for heat flux is (JH_{c}=\lambda \frac{dT}{dZ})
dT = 160 -40 = 120°C
dZ = 0.08 m
(\lambda) = 386 W/mK
Thus, (JH_{c}=386 \times \frac{120}{0.08})
(\phi_{q} = 386 \times \frac{120}{0.08})
(\phi_{q} = 579000 W/m^2)
Example 3. An iron plate of thickness 0.12 m was maintained across two bodies of temperatures of 170 °C and 30 °C respectively. Calculate the heat flux. The thermal conductivity of glass is 80 W/mK
Solution 3: Since the formula for heat flux is (JH_{c}=\lambda \frac{dT}{dZ})
Since,
(dT = 170 - 30 = 140^o C)
dZ = 0.12 m
(\lambda) = 80 W/mK
Thus, (JH_{c}=\lambda \frac{dT}{dZ})
(JH_{c}=80 \times \frac{140}{0.12})
(\phi_{q} = 93333.33 \frac{W}{m^2})
Example 4. A gold plate of thickness 0.05 m was maintained across two bodies of temperatures of 190 °C and 50 °C respectively. Calculate the heat flux. The thermal conductivity of glass is 310 W/mK
Solution 4: Since the formula for heat flux is (JH_{c}=\lambda \frac{dT}{dZ})
Since,
(dT = 190 - 50 = 140^o C)
dZ = 0.05 m
(\lambda) = 310 W/mK
Thus, (JH_{c}=\lambda \frac{dT}{dZ})
(JH_{c}=310 \times \frac{140}{0.05})
(\phi_{q} = 868000 \frac{W}{m^2})
Hope this article about heat flux formulas was able to get the concept of this topic in motion. There are many such interesting topics and their real-life applications to learn about, just download the Testbook app and start browsing to get insights on them which can clear all your concepts regarding them.
More Articles for Physics Formulas
Heat Release Rate Formula
Impulse Formula
Induced Voltage Formula
Resistivity Formula
Surface Charge Density Formula
Radioactive Decay Formula
Torque Formula
Average Speed Formula
Beam Deflection Formula
Gravitational Acceleration Formula
Heat Flux Formula FAQs
How do you calculate heat flux?
Heat flux is calculated using the formula (JH_{c}=\lambda \frac{dT}{dZ}).
What is heat flux?
Heat Flux is defined as the rate of heat energy transfer per unit area to or from a surface.
What is the unit of heat flux?
The unit of heat flux are in (W/m^2).
Is heat flux the same as heat rate?
No, heat flux is the amount of heat transferred per unit time per unit area, whereas heat rate is the amount of heat transferred per unit time.
Is heat flux in watts?
Heat flux is given in units of (W/m^2).
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4778 | https://byjus.com/biology/photosynthesis/ | Photosynthesis is a process by which phototrophs convert light energy into chemical energy, which is later used to fuel cellular activities. The chemical energy is stored in the form of sugars, which are created from water and carbon dioxide.
Table of Contents
What is Photosynthesis?
Site of photosynthesis
Factors
Equation
Structure
Process
Importance
Photosynthesis definition states that the process exclusively takes place in the chloroplasts through photosynthetic pigments such as chlorophyll a, chlorophyll b, carotene and xanthophyll. All green plants and a few other autotrophic organisms utilize photosynthesis to synthesize nutrients by using carbon dioxide, water and sunlight. The by-product of the photosynthesis process is oxygen.Let us have a detailed look at the process, reaction and importance of photosynthesis.
What Is Photosynthesis in Biology?
The word “photosynthesis” is derived from the Greek words phōs(pronounced: “fos”) and σύνθεσις (pronounced: “synthesis“)Phōs means “light” and σύνθεσιςmeans, “combining together.” This means “combining together with the help of light.”
Photosynthesis also applies to other organisms besides green plants. These include several prokaryotes such as cyanobacteria, purple bacteria and green sulfur bacteria. These organisms exhibit photosynthesis just like green plants.The glucose produced during photosynthesis is then used to fuel various cellular activities. The by-product of this physio-chemical process is oxygen.
A visual representation of the photosynthesis reaction
Photosynthesis is also used by algae to convert solar energy into chemical energy. Oxygen is liberated as a by-product and light is considered as a major factor to complete the process of photosynthesis.
Photosynthesis occurs when plants use light energy to convert carbon dioxide and water into glucose and oxygen. Leaves contain microscopic cellular organelles known as chloroplasts.
Each chloroplast contains a green-coloured pigment called chlorophyll. Light energy is absorbed by chlorophyll molecules whereas carbon dioxide and oxygen enter through the tiny pores of stomata located in the epidermis of leaves.
Another by-product of photosynthesis is sugars such as glucose and fructose.
These sugars are then sent to the roots, stems, leaves, fruits, flowers and seeds. In other words, these sugars are used by the plants as an energy source, which helps them to grow. These sugar molecules then combine with each other to form more complex carbohydrates like cellulose and starch. The cellulose is considered as the structural material that is used in plant cell walls.
Also Read:Photosynthesis in Higher plants
Where Does This Process Occur?
Chloroplasts are the sites of photosynthesis in plants and blue-green algae. All green parts of a plant, including the green stems, green leaves, and sepals – floral parts comprise of chloroplasts – green colour plastids. These cell organelles are present only in plant cells and are located within the mesophyll cells of leaves.
| |
| Factors Affecting Photosynthesis Photosynthesis process requires several factors such as: Light Intensity: Increased light intensity results in a higher rate of photosynthesis. On the other hand, low light intensity results in a lower rate of photosynthesis. The concentration of CO2: Higher concentration of carbon dioxide helps in increasing the rate of photosynthesis. Usually, carbon dioxide in the range of 300 – 400 PPM is adequate for photosynthesis. Temperature: For efficient execution of photosynthesis, it is important to have a temperature range between 25° to 35° C. Water: As water is an important factor in photosynthesis, its deficiency can lead to problems in the intake of carbon dioxide. The scarcity of water leads to the refusal of stomatal opening to retain the amount of water they have stored inside. Pollution: Industrial pollutants and other particulates may settle on the leaf surface. This can block the pores of stomata which makes it difficult to take in carbon dioxide. |
Also Read:Photosynthesis Early Experiments
Photosynthesis Equation
Photosynthesis reaction involves two reactants, carbon dioxide and water. These two reactants yield two products, namely, oxygen and glucose. Hence, the photosynthesis reaction is considered to be an endothermic reaction. Following is the photosynthesis formula:
| |
| 6CO2 + 6H2O —> C6H12O6 + 6O2 |
Unlike plants, certain bacteria that perform photosynthesis do not produce oxygen as the by-product of photosynthesis. Such bacteria are called anoxygenic photosynthetic bacteria. The bacteria that do produce oxygen as a by-product of photosynthesis are called oxygenic photosynthetic bacteria.
| |
| Photosynthetic Pigments There are four different types of pigments present in leaves: 1. Chlorophyll a 2. Chlorophyll b 3. Xanthophylls 4. Carotenoids |
Structure Of Chlorophyll
The structure of Chlorophyll consists of 4 nitrogen atoms that surround a magnesium atom. A hydrocarbon tail is also present. Pictured above is chlorophyll-f,which is more effective in near-infrared light than chlorophyll-a
Chlorophyll is a green pigment found in the chloroplasts of the plant celland in the mesosomes of cyanobacteria. This green colour pigment plays a vital role in the process of photosynthesis by permitting plants to absorb energy from sunlight. Chlorophyll is a mixture of chlorophyll-a and chlorophyll-b.Besides green plants, other organisms that perform photosynthesis contain various other forms of chlorophyll such as chlorophyll-c1, chlorophyll-c2, chlorophyll-d and chlorophyll-f.
Also Read: Biological Pigments
Process Of Photosynthesis
At the cellular level, the photosynthesis process takes place in cell organelles called chloroplasts. These organelles contain a green-coloured pigment called chlorophyll, which is responsible for the characteristic green colouration of the leaves.
As already stated, photosynthesis occurs in the leaves and the specialized cell organelles responsible for this process is called the chloroplast. Structurally, a leaf comprises a petiole, epidermis and a lamina. The lamina is used for absorption of sunlight and carbon dioxide during photosynthesis.
Structure of Chloroplast. Note the presence of the thylakoid
“Photosynthesis Steps:”
During the process of photosynthesis, carbon dioxide enters through the stomata, water is absorbed by the root hairs from the soil and is carried to the leaves through the xylem vessels. Chlorophyll absorbs the light energy from the sun to split water molecules into hydrogen and oxygen.
The hydrogen from water molecules and carbon dioxide absorbed from the air are used in the production of glucose. Furthermore, oxygen is liberated out into the atmosphere through the leaves as a waste product.
Glucose is a source of food for plants that provide energy for growth and development, while the rest is stored in the roots, leaves and fruits, for their later use.
Pigments are other fundamental cellular components of photosynthesis. They are the molecules that impart colour and they absorb light at some specific wavelength and reflect back the unabsorbed light. All green plants mainly contain chlorophyll a, chlorophyll b and carotenoids which are present in the thylakoids of chloroplasts. It is primarily used to capture light energy. Chlorophyll-a is the main pigment.
The process of photosynthesis occurs in two stages:
Light-dependent reaction or light reaction
Light independent reaction or dark reaction
Stages of Photosynthesis in Plants depicting the two phases – Light reaction and Dark reaction
Light Reaction of Photosynthesis (or) Light-dependent Reaction
Photosynthesis begins with the light reaction which is carried out only during the day in the presence of sunlight. In plants, the light-dependent reaction takes place in the thylakoid membranes of chloroplasts.
The Grana, membrane-bound sacs like structures present inside the thylakoid functions by gathering light and is called photosystems.
These photosystems have large complexes of pigment and proteins molecules present within the plant cells, which play the primary role during the process of light reactions of photosynthesis.
There are two types of photosystems: photosystem I and photosystem II.
Under the light-dependent reactions, the light energy is converted to ATP and NADPH, which are used in the second phase of photosynthesis.
During the light reactions, ATP and NADPH are generated by two electron-transport chains, water is used and oxygen is produced.
The chemical equation in the light reaction of photosynthesis can be reduced to:
2H2O + 2NADP+ + 3ADP + 3Pi → O2 + 2NADPH + 3ATP
Dark Reaction of Photosynthesis (or) Light-independent Reaction
Dark reaction is also called carbon-fixing reaction.
It is a light-independent process in which sugar molecules are formed from the water and carbon dioxide molecules.
The dark reaction occurs in the stroma of the chloroplast where they utilize the NADPH and ATP products of the light reaction.
Plants capture the carbon dioxide from the atmosphere through stomata and proceed to the Calvin photosynthesis cycle.
In the Calvin cycle, the ATP and NADPH formed during light reaction drive the reaction and convert 6 molecules of carbon dioxide into one sugar molecule or glucose.
The chemical equation for the dark reaction can be reduced to:
3CO2 + 6 NADPH + 5H2O + 9ATP → G3P + 2H+ + 6 NADP+ + 9 ADP + 8 Pi
G3P – glyceraldehyde-3-phosphate
Calvin photosynthesis Cycle (Dark Reaction)
Also Read:Cyclic And Non-Cyclic Photophosphorylation
Importance of Photosynthesis
Photosynthesis is essential for the existence of all life on earth. It serves a crucial role in the food chain – the plants create their food using this process, thereby, forming the primary producers.
Photosynthesis is also responsible for the production of oxygen – which is needed by most organisms for their survival.
Frequently Asked Questions
Q1
1. What is Photosynthesis? Explain the process of photosynthesis.
Photosynthesis is a biological process utilized by all green plants to synthesize their own nutrients. The process of photosynthesis requires solar energy, water and carbon dioxide. The by-product of this process is oxygen.
Q2
2. What is the significance of Photosynthesis?
During photosynthesis, oxygen gas is liberated out into the environment and is utilized by humans, animals and other living species during the process of respiration.
Q3
3. List out the factors influencing Photosynthesis.
There are several factors that affect the rate of photosynthesis. Light intensity, water, soil pH, carbon dioxide concentration, temperature and other climatic conditions are the main factors affecting the rate of photosynthesis.
Q4
4. What are the different stages of Photosynthesis?
Photosynthesis takes place in two stages, namely light-dependent reactions and light-independent reactions. Light-dependent reactions are also called light reactions and occur during the day time. Light-independent reaction is also called the dark reaction or the Calvin cycle.
Q5
5. What is the Calvin Cycle?
The Calvin cycle is also called the light-independent reaction. The complete process of the Calvin cycle takes place in the stroma of the chloroplasts.
Q6
6. Write down the Photosynthesis Equation.
6CO2 + 6H2O —> C6H12O6 + 6O2
Register at BYJU’S Biology to explore more photosynthesis notes or notes for other related concepts. Important MCQs on Photosynthesis
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Maroti ingole September 26, 2019 at 2:12 pm
Nice
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very useful
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It’s very helpful ☺️
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Please What Is Meant By 300-400 PPM
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nazia July 7, 2020 at 6:02 pm
PPM stands for Parts-Per-Million. It corresponds to saying that 300 PPM of carbon dioxide indicates that if one million gas molecules are counted, 300 out of them would be carbon dioxide. The remaining nine hundred ninety-nine thousand seven hundred are other gas molecules.
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4779 | https://www.varsitytutors.com/hotmath/hotmath_help/topics/graphing-quadratic-equations-using-the-axis-of-symmetry | Graphing Quadratic Equations using the Axis of Symmetry
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Graphing Quadratic Equations using the Axis of Symmetry
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Graphing Quadratic Equations using the Axis of Symmetry
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Graphing Quadratic Equations using the Axis of Symmetry
Study Guide
Key Definition
A quadratic equation is in the form y=a x 2+b x+c, where a≠0. The axis of symmetry is x=−b 2 a.
Important Notes
•The vertex of the parabola is at (−b 2 a,f(−b 2 a)), where the y-coordinate is found by substituting x = \frac{-b}{2a} into the quadratic
•The axis of symmetry is a vertical line
•The parabola is symmetric with respect to this line
•The x-coordinate of the vertex is −b 2 a
•Substitute x-values to find corresponding y-values
Mathematical Notation
represents addition
− represents subtraction
× or ∗ represents multiplication
÷ or / represents division
a b represents a fraction
Remember to use proper notation when solving problems
Why It Works
The formula x=−b 2 a gives the x-value of the vertex, which is the highest or lowest point of the parabola.
Remember
The axis of symmetry helps in plotting parabolas symmetrically.
Quick Reference
Axis of Symmetry:x=−b 2 a
Vertex:(−b 2 a,y)
Understanding Graphing Quadratic Equations using the Axis of Symmetry
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Beginner Explanation
The axis of symmetry is the line x=−b 2 a. It divides the parabola into two mirror images.
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Practice Problems
Test your understanding with practice problems
1
Quick Quiz
Single Choice Quiz
Beginner
What is the axis of symmetry for y=x 2−7 x+2?
A
x=7 2
B
x=−7 2
C
x=7 4
D
x=7 3
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2
Real-World Problem
Question Exercise
Intermediate
Teenager Scenario
A skateboarder jumps off a ramp, and the path is modeled by y=−x 2+6 x+1. Find the highest point of the jump.
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3
Thinking Challenge
Thinking Exercise
Intermediate
Think About This
A parabola opens upwards. If the axis of symmetry is x=4, what can you deduce about the coefficients of the quadratic equation?
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4
Challenge Quiz
Single Choice Quiz
Advanced
Find the vertex of the parabolay=2 x 2−8 x+3.
A
(2,−1)
B
(2,−5)
C
(2,−9)
D
(2,0)
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4780 | https://oeis.org/A005843 | 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120
OFFSET
0,2
COMMENTS
-2, -4, -6, -8, -10, -12, -14, ... are the trivial zeros of the Riemann zeta function. - Vivek Suri (vsuri(AT)jhu.edu), Jan 24 2008
If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-2) is the number of 2-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007
Omitting the initial zero gives the number of prime divisors with multiplicity of product of terms of n-th row of A077553. - Ray Chandler, Aug 21 2003
The number of hydrogen atoms in straight-chain (C(n)H(2n+2)), branched (C(n)H(2n+2), n > 3), and cyclic, n-carbon alkanes (C(n)H(2n), n > 2). - Paul Muljadi, Feb 18 2010
For n >= 1; a(n) = the smallest numbers m with the number of steps n of iterations of {r - (smallest prime divisor of r)} needed to reach 0 starting at r = m. See A175126 and A175127. A175126(a(n)) = A175126(A175127(n)) = n. Example (a(4)=8): 8-2=6, 6-2=4, 4-2=2, 2-2=0; iterations has 4 steps and number 8 is the smallest number with such result. - Jaroslav Krizek, Feb 15 2010
For n >= 1, a(n) = numbers k such that arithmetic mean of the first k positive integers is not integer. A040001(a(n)) > 1. See A145051 and A040001. - Jaroslav Krizek, May 28 2010
a(k) is the (Moore lower bound on and the) order of the (k,4)-cage: the smallest k-regular graph having girth four: the complete bipartite graph with k vertices in each part. - Jason Kimberley, Oct 30 2011
Let n be the number of pancakes that have to be divided equally between n+1 children. a(n) is the minimal number of radial cuts needed to accomplish the task. - Ivan N. Ianakiev, Sep 18 2013
For n > 0, a(n) is the largest number k such that (k!-n)/(k-n) is an integer. - Derek Orr, Jul 02 2014
a(n) when n > 2 is also the number of permutations simultaneously avoiding 213, 231 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl Aug 07 2014
It appears that for n > 2, a(n) = A020482(n) + A002373(n), where all sequences are infinite. This is consistent with Goldbach's conjecture, which states that every even number > 2 can be expressed as the sum of two prime numbers. - Bob Selcoe, Mar 08 2015
Number of partitions of 4n into exactly 2 parts. - Colin Barker, Mar 23 2015
Number of neighbors in von Neumann neighborhood. - Dmitry Zaitsev, Nov 30 2015
Unique solution b( ) of the complementary equation a(n) = a(n-1)^2 - a(n-2)b(n-1), where a(0) = 1, a(1) = 3, and a( ) and b( ) are increasing complementary sequences. - Clark Kimberling, Nov 21 2017
Also the maximum number of non-attacking bishops on an (n+1) X (n+1) board (n>0). (Cf. A000027 for rooks and queens (n>3), A008794 for kings or A030978 for knights.) - Martin Renner, Jan 26 2020
Integer k is even positive iff phi(2k) > phi(k), where phi is Euler's totient (A000010) [see reference De Koninck & Mercier]. - Bernard Schott, Dec 10 2020
Number of 3-permutations of n elements avoiding the patterns 132, 213, 312 and also number of 3-permutations avoiding the patterns 213, 231, 321. See Bonichon and Sun. - Michel Marcus, Aug 20 2022
a(n) gives the y-value of the integral solution (x,y) of the Pellian equation x^2 - (n^2 + 1)y^2 = 1. The x-value is given by 2n^2 + 1 (see Tattersall). - Stefano Spezia, Jul 24 2025
REFERENCES
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 28.
J.-M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 529a pp. 71 and 257, Ellipses, 2004, Paris.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 256.
LINKS
Eric Weisstein's World of Mathematics, Even Number
FORMULA
G.f.: 2x/(1-x)^2.
G.f. with interpolated zeros: 2x^2/((1-x)^2 (1+x)^2); e.g.f. with interpolated zeros: xsinh(x). - Geoffrey Critzer, Aug 25 2012
a(0) = 0, a(1) = 2, a(n) = 2a(n-1) - a(n-2). - Jaume Oliver Lafont, May 07 2008
a(n) = Sum_{k=1..n} floor(6n/4^k + 1/2). - Vladimir Shevelev, Jun 04 2009
Digit sequence 22 read in base n-1. - Jason Kimberley, Oct 30 2011
a(n) = 3a(n-1) - 3a(n-2) + a(n-3). - Vincenzo Librandi, Dec 23 2011
a(n) = 2n = Product_{k=1..2n-1} 2sin(Pik/(2n)), n >= 0 (undefined product := 1). See an Oct 09 2013 formula contribution in A000027 with a reference. - Wolfdieter Lang, Oct 10 2013
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/2 = (1/2)A002162 = (1/10)A016655. (End)
Sum_{n>=1} 1/a(n)^2 = Pi^2/24 = A222171.
Sum_{n>=1} (-1)^(n+1)/a(n)^2 = Pi^2/48 = A245058. (End)
EXAMPLE
G.f. = 2x + 4x^2 + 6x^3 + 8x^4 + 10x^5 + 12x^6 + 14x^7 + 16x^8 + ...
MAPLE
MATHEMATICA
PROG
(Magma) [ 2n : n in [0..100]];
(R) seq(0, 200, 2)
(Haskell)
a005843 = ( 2)
(Python) def a(n): return 2n # Martin Gergov, Oct 20 2022
CROSSREFS
Cf. A000027, A002061, A005408, A001358, A077553, A077554, A077555, A002024, A087112, A157888, A157889, A140811, A157872, A157909, A157910, A165900.
Moore lower bound on the order of a (k,g) cage: A198300 (square); rows: A000027 (k=2), A027383 (k=3), A062318 (k=4), A061547 (k=5), A198306 (k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10), A094626 (k=11); columns: A020725 (g=3), this sequence (g=4), A002522 (g=5), A051890 (g=6), A188377 (g=7). - Jason Kimberley, Oct 30 2011
Cf. A231200 (boustrophedon transform).
KEYWORD
nonn,easy,core,nice
AUTHOR
STATUS
approved |
4781 | https://users.renyi.hu/~p_erdos/1978-12.pdf | INTERSECTION PROPERTIES OF SYSTEMS OF FINITE SETS By M . DEZA, P . ERDOS, and P . FRANKL [Received 30 June 1976] Abstract Let X be a finite set of cardinality n .
If L = {ll , .
. ., l,} is a set of non-negative integers with 1 1 < 1 2 < . . . < lr , and k is a natural number, then by an (n, L, k)-system we mean a collection of k-element subsets of X such that the intersection of any two different sets has cardinality belonging to L . We prove that if d is an (n, L, k)-system, with I ~V I > enr -1 (c = e(k) is a constant depending on k), then (i) there exists an l1-element subset D of X such that D is contained in every member of Q, (11) (12 - 11)1(13 - 12)1 . . .I(lr - lr-1)I (k-l r), (iii) ni=1 (?,-li)/(k- li) > ICI (for n > n,(k)) .
Parts of the results are generalized for the following cases : (a) we consider t-wise intersections, where t > 2 ; (b) the condition I A I = k is replaced by I A I E K where K is a set of integers ; (c) the intersection condition is replaced by the following : among q + 1 different members A 1 , . . ., Aq+1 there are always two, A i , Aj , such that I A i n A, I E L .
We consider some related problems .
An open question : let L' - L ; do there exist an (n, L, k)-system of maximal cardinality (s2/) and an (n, L', k)-system of maximal cardinality (,') such that d ::) a'?
1 . Introduction Throughout this paper lower case latin letters denote integers, capital letters stand for sets, and capital script letters for families of sets .
Let L = {l l , . . ., l,}, where 1 < 12 < . . . < l r , and K be sets of integers .
By an (n, L, K)-system we mean a family Q of subsets of a set X, with I X I = n, such that for A t , A 2 E d we have I A1 1, A2 I E K, I A 1 n A 2 I E L .
If K = {k} then the notation (n, L, k)-system is applied, too .
A family B = {B 1, B 2 , . . ., B.} of sets is called a 0-system of cardinality c if there exists a set D - Bi , with i = 1, . . ., c, such that the sets B1\D, . . .,B,\D are pairwise disjoint .
D is called the kernel of the 0-system .
Proc. London Math . Soc. (3) 36 (1978) 369-384 5388.3 .36 Z 370 M . DEZA, P . ERDŐS, AND P . FRANKL THEOREM 1. (Erdős and Rado ) .
There exists a function T,(k) such that any fancily of 9,(k) distinct k-element sets contains a A-system of cardinality c .
An old conjecture of Rado and the second author is that there exists an absolute constant c' such that 9,(k) < (cc')k .
The best existing upper bound (of order about ckk !) is due to Spencer .
THEOREM 2 (Erdős, Ko, and Rado ) . If is an (n, {l, l + 1, . . ., k - 1}, k)-system of maximal cardinality, then for n , no (k, l) there exists a set D of cardinality l such that for every A E a'l, D A holds .
In particular, for l = 1, no (/,, l) = 2k + 1 is the best possible value for n o(k,1) .
(For l , 2 the best existing upper bound on n o(k,1) is due to Frankl .) THEOREM 3 (Deza ) .
An (n, {l}, k)-system of cardinality more than k 2 -k+1 is a A-system .
The object of this paper is to generalize Theorems 2 and 3 for (n, L, K)-systems. In the proofs heavy use is made of Theorem 1 .
The next four theorems express properties of (n, L, 7c)-systems .
Throughout the paper we assume that n > n o(k, s) for e > 0. Let us set c(k,L) = max(k-h+1, 1 22 -12 +1)+s . sI is an (n, L, k)-system .
THEOREM 4. If s1 c(k, L) 11i_ 2 (n - li)/(k - l i ) then there exists a set D of cardinality 11 such that D A for every A E sV .
THEOREM 5 .
If , k22r-lnr-1 then (12 - 11) 1(1 3 -12)1 . . . I (lr - lr-1) I (k - I,)-TimOREM 6 .
I~I 5 i=1k-l i .
The following result is a generalization of Theorems 4, 5, and 6 for (n, L, K)-systems . Let K = {k l , . . ., k,}, with kl < . . . < k, Let us define Ko = K n {0, . . . . h}, Ki = {li + 1, . . .,1i+1} n K, for i = l, . . ., r -1, and Kr = Kn{lr+1, . . .,k3} .
Let us set k2 = min{k I k c- Ki}, for i = 0, . . ., r .
THEOREM 7 .
Let sl be an (n, L, K)-system .
(i) If I sa'l I > Icsc(ks , L) YV=2 (n - l i)/(k - l i ) then there exists a set D of cardinality 11 such that D A for every A E Q/ .
INTERSECTIONS OF SYSTEMS OF FINITE SETS 371 (ü) If I (z/ I > k,S32r-inr-1 then there exists a k e Kr such that ( 1 2 -11) 1(13-12) 1- 1 ('r - lr-1) I I~EE~ n -l1 i < i=o j=1 ki -lj 1 j kj .
where s i = 0 if Ki = 0, s i = 1 otherwise .
The next theorem is a common generalization of Theorems 4 and 6 and a theorem of Hajnal and Rothschild .
THEOREM 8 .
Let sl be a family of k-element subsets of the n-element set X such that whenever A 1 , . . ., Aq+1 are q+ 1 different sets belonging to we can find two of them A i , A j such that I Aí n Aj I E L (q > 1 is fixed) .
(i) There exists a constant c = c(k, q) such that r n-1 • IVI > (q-1) H k-l i +cnr-1 implies the existence of sets D 1, D2 , . . ., D s such that for every A E a there exists an i, with 1 < i < s, satisfying Di c A, I Dl I = . . . = J DJI = 11 .
Further, if qi denotes the maximum number of sets A,, . . ., Aq, such that, for 1 < j < qi, D i - Aj but for i' i, Di Aj , and I Aj, n A j2 1 0 L for 1 < .%1 < ,l2 < qi, then Es =1 qi = q • (ü) Ic~/ < q1I n-1i +0(nY-1) (n > no(k,« .
i=1 k -li In the next theorem we generalize Theorems 4, 5, and 6 for the case of t-wise intersections .
THEOREM 9 .
Let ,71 be a family of k-subsets of X. Suppose that, for any t different members A 1 , . . ., A i of (l, I A1 n . . . n A, I c L . Then (i) there exists a constant c = c(Ie, t) such that I , d I > cnr- 1 implies the existence of an l1-element set D such that D A for every AEI, (ü) I I > cnr-1 implies that (l2 -l1 )1 . . .1(lr-l.r-1)I (k - 1r), (iii) IdI < (t-1)IIi=1(n-li)/(k- l i ) (n > no(k,t)) .
First versions of Theorems 4, 5, 6, and 7 were announced in , the case where I L I = 2 was considered in .
372 M . DEZA, P . ERDÜS, AND P . FRANKL 2. The proof of Theorems 4, 5, and 6 In the case where r = 1 the statements of the theorems follow from Theorem 3 .
Now suppose r > 2 . We apply induction on k . The case where k = 1 is trivial .
Let us first consider the case where 11 = 0. Then the statement of Theorem 4 is evident . Let x be an arbitrary element of X . Let us define a x = {A{x} I x E A} . Then ax is an (n-1,{1,- 1, . . ., lr -1},k- 1)-system .
Hence, by the induction hypothesis, r (n- 1)-(li -1) r n-l i Iáx 1 -(k-1)-(li-1 )-11 k-li ( 1 ) Counting the number of pairs (x, A), for x e A e c/, in two different ways we obtain kIW1= E I~x1 .
(2) XEX From (1) and (2) it follows that IX I r n-li -r 14 n-l i ICI < k n k-li -k-l i ' which proves Theorem 6 for this case .
Now we wish to prove Theorem 5 . So we may suppose that I I > k22r-1nr-1 .
Let us set d = k 22r-2nr-2 and mil° _ c/~ . If Q/j is defined and there exists an element x e X such that 0 < I c_ ,/x I < d then define ~dj+1 =cY;{AEdiIxeA} .
After finitely many steps the procedure stops, that is, we obtain a family mil' in which every element of X has either degree 0 or degree more than d, and I ,Ql' I > l c7/1 -nd > k22r-2,r-1 .
Let X' be the set of elements of X which have non-zero degree in (V' . If x c- X' then cV' is an (n -1, {12 - l, . . . , lr -1}, k -1)-system, and I ax I > d = k22r-2nr-2 whence by the induction hypothesis there exists a set Dx - X \ {x}, with I Dx I = 12 -1, such that Dx c A for every A e q/x .
We assert that for any y c- Dx , DY _ (Dx \ {y}) u {x} .
Suppose that for some y it does not hold. As any member A of mIX contains y so it has to contain D„ as well and consequently A 2 ((Dx u D,) \ {x}) .
I (Dx u D„) \ {x} I > 12 , which implies that any two elements of sVx intersect in at least 12 elements .
Hence ax is an INTERSECTIONS OF SYSTEMS OF FINITE SETS 373 (n-1, {13 -1, . . ., lr -1}, k-1)-system . So Theorem 6 implies that r n-1 • I ,4x I'< H k_h< nr-2 ' i=3 i a contradiction. So we have proved that, for x y, Dx u {x} and Dy u {y} coincide or they are disjoint .
Consequently the sets Dx u {x}, for x c- X', form a partition of the set X' such that any member A of sl' is the union of some of them . Hence 12 I k . We assert that for any 3 < i <, r there are two sets A, B E V' such that J A n B J = li. Indeed, otherwise a' is an k)-system, so by Theorem 6 1 <1 n-li I I' l , n r-1 .7oi k-l, ' a contradiction . Now if I A n B I = li then that 1, I li follows from the fact that A and B, whence A n B too, are the unions of some of the pairwise disjoint l 2 -element sets Dx u {x} .
In particular it follows that 12 = (12 - 11) 1 (13 -12) • Applying the induction hypothesis to q/x we obtain that ((13 -1 ) -( 12 -1 ))1((14 - 1) - ( 13 - 1 ))I . . .I((k-1)-(1r-1)), that is, ( 1 3 -12)1( 14 -13_101 . . .
I (k-l r ), which finishes the proof for the case where 11 = 0 .
Now we need a lemma .
LEMMA 1 .
(i) Let the sets A,, . . ., A C forma 0-system with kernel D, where I D I = l1 , and c ,> k - l 1 + 2 .
Then for any set B, with I B I ,< k, I B n A i I >, 1 1 for i = 1, . . ., c implies B D .
(ü) Let the sets Fi1, F2, . . ., Fi form a 0-system with kernel Ei , where Fi J = k for i = l j = l, . . ., t .
Suppose that the sets E i form a 0-system with kernel D, where I D 1 = l, and that t > (s-1) (k-1) .
Then there are indices 1 < ji <, t for i = 1, . . .,s such that the sets Fis form a 0-system with kernel D .
Proof.
Let us set I Bn D I = l' < 11 .
Then I B n A i 1 11 implies that I B n (A i\D) I > l1 - l' for i = 1, . . ., e. As the sets A i \ D are pairwise disjoint we obtain k = CBI l'+c(l1-l') >, l'+(k-11+2)(l1-l') or equivalently (k - 11 ) , (k - 11 + 1)(1 1 - l'), which yields 1 1 = t' as desired.
Now we prove (ü) .
Suppose that for i = 1, . . ., s' we have chosen indices 1 < ji 5 t such that the sets F s form a A-system with kernel D .
Now we wish to choose the index j = js+1 in such a way that Fs . n Fis = D 374 M . DEZA, P . ERDŐS, AND P . FRANKL for i = 1, . . . , s' and Fs .+, n Ei = D for i = s'+ 2, . . ., s . An index j does not satisfy the conditions if and only if Fs •+,\D is not disjoint from the set H S = (U (F2\D)) u ( . U (EZ\D)) x>s'+1 As the sets Fs.+, \ Es •+ , are pairwise disjoint, for j = 1, . . ., t, and IHS I = s'(k-l)+~ =s+2l~Ei\DI < (s-1) (k-1) < t, the appropriate choice of Fs •+ 1 is always possible, which proves the lemma .
Now we turn to the proof of Theorem 4 for the case where l, > 0 . If we can find k-1,+2 sets A„ . . . , Ak_a1+2 belonging to Q/ which form a A-system with kernel D, where I D = 11, then it follows that D c A for every A e q/, since I A n A Z 11, for i = 1, . . . , k -11 + 2, and from the lemma .
So we may assume that such a A-system does not exist . Now let us choose a set D2 of cardinality 12 such that D2 is the kernel of a A-system formed by k2 members of s4l (A1 1 , . . ., A,• '), and let us define 'd1= {Ae_c/ID1(-A} .
Now we choose a set D2 of cardinality 12 which is the kernel of a A-system formed by k 2 different members of t and define 2 = {A E ~2I D2 A}, and so on. After a finite number of steps, say q 2 , we cannot find a set Die+1 of cardinality 12 which is the kernel of a A-system formed by P different members of X1 42 Now we choose a set D3 of cardinality 1, which is the kernel of a A-system formed by k 3 different elements of 'C~/2Z and define 'CV1 = {A E ~QZ I D 3 A} ; after say q 3 steps we cannot find such a Dq3+ , .
Then we look for an l4 -element set which is the kernel of a A-system formed by k4 members of 'q/q3 , and so on. At last we obtain a family dr, which does not contain any A-system with kernel D1 , where ID, I = h, and of cardinality ki (j = 1, . . ., r) . As -qh is an (n, L, k) -system it means that SlI Q, does not contain any A-system of cardinality at least kr, implying that IV4 .I < Tkr(k) • (3) Now we assert that q; < <Pk;_1(l,) (j = 3, . . ., r) (4 ) and that q2 < c(k,L) • (5 ) If it is not true then we could find among the kernels of cardinality h a A-system of cardinality k 2 and kernel D i , with I D i I = li , for some 1 < i < j, or, for j = 2, a A-system of cardinality k - l, + 2 and with a kernel D, of cardinality l, .
contains a 0-system consisting of ki sets and having a kernel Di , with I Di I = li , for 1 < i < j, or for j = 2 that cl contains a A-system of cardinality k -11 + 2 and with kernel Dl , with I Dl I = 1 1 .
The first possibility contradicts the choice of qj_, while the second one is contrary to our assumptions. So (4) and (5) are proved .
If 1 < u < qj then define _ ,V(j, u) _ {A\Di I A E &Y, Dú A} .
Then ,n/ (j, u) is an (n-lj , {0,1j+1 -lj , . . ., l,.-lj}, k-lj)-system .
Hence by the induction hypothesis r (n - l •) -(li-lj) -r n-l • I ~(j u)I .
(k-h)-(li-lj) II k-li .
Consequently, r n-l • ; \ _u10 Qj I % 11 (6) i=j k - l i for j = 2, . . ., r, where VQ l = sl . From (3), (4), (5), and (6) we obtain r I~I= ZI vi-1 \ q,I+I q > I j=2 r n-l .
r+1 r n-1 .
c(k,L) fl ~+ ~Tkj- ' (l ;) II z (7) j=2 IC - l i j=s i= ; k - li In (7) we use the conventions that 1r+1 = k and that the empty product is l .
From (7) we obtain I,d I < (c(k,L)+o(1)) rl n-l i, i=2 k - 1i ' which is a contradiction as n > n,(k) .
Now the proof of Theorem 4 is finished. So in proving Theorems 5 and 6 we may suppose that there exists a set D, with I D 1 = 1, such that D si A for every A E V .
Let us define l(D) _ {A\D ;A E } .
It follows then that /(D) is an (n - h, {0, 1, - lr - h}, k - l, )-system. We know that k - l, < k as h > 0. Hence both Theorem 5 and Theorem 6 follow from the induction hypothesis .
Equality in the estimation of Theorem 6 (briefly 'equality') is realizable by the hyperplane-family of any perfect matroid-design (ef . ) of rank I L I + 1, such that for any j-flat Fj we have I F'LI I = k , I I +' JLJ+1 I = n, I P I = lj+1, 0 < j < I L I . For example, in the case when L is an arithmetic progression with difference d = l 2 -11, we may obtain equality by an (l2 - h)-inflation of an S(I L I , lc/d, n/d,) if this Steiner-system exists. The affine and projective geometries provide other examples when equality INTERSECTIONS OF SYSTEMS OF FINITE SETS 375 Applying Lemma 1 we obtain that, for the corresponding j, u,i 376 M . DEZA, P . ERDÖS, AND P . FRANKL is possible . (The collection of all the j-flats Fj with I P I = k for some 0 < i < I L I gives equality in the estimation (iii) of Theorem 7 .) In the case where L = {0, 1, 3} the equality implies the existence of an S(2, 3, k), whence 2 1 (k -1), 6 1 (k -1)k .
In the first case, Ic = 5, equality is not possible, moreover it can be proved that no (n, L, 5)-system has more than 2n 11 / 4 =o(njLj) elements though (1-0)1 (3-1)1(5-3) .
The collec-tion of the 2-dimensional subspaces of PG(s, 2), AG(s, 3), respectively, provide equality for the cases where k = 7, 9. The first open problem is to decide whether there are infinitely many values of n for which we can have equality in the case where k = 13 .
REMARK 1 (on Theorem 4) . Without changing the argument we can prove the following : if k' >_ k and 1,2/1 > c(k , , L) ÍI n-li j=2 k-l i then there are sets A,, ._ Ah,-11+2 E which form a 0-system with a kernel of cardinality 11 .
REMARK 2 (on Theorem 5) . For the case where L = {0,1} in it was shown that I q/ I > n implies 11 Ic and this estimation is the best possible (this is a generalization of the Fisher-Majumdar inequality ) .
REMARK 3 (on Theorem 6). In it was shown by Ray-Chaudhuri and Wilson that I a S (L) I I for any (n, L, k)-system (this is another generalization of the Fisher-Majumdar inequality [151) . This estimation does not depend on k, but it is weaker than Theorem 6 for I a I > C(k)n 1L 1-1 .
Its proof (using, a propos, linear independence of certain systems of vectors) will be interesting to extend for the cases of Theorems 7, 8, and 9 .
3. The proof of Theorem 7 We apply induction on r .
If n/' denotes {A E c A I > 1,1 then it follows from the induction hypothesis that n-1 • i=u j= 1 ki - j when r > 2, while the same inequality holds trivially for r = 1 as well .
Hence r i \ ,P/'I%i i-r11 n - 1 • j=1 r - tj INTERSECTIONS OF SYSTEMS OF FINITE SETS 377 First we prove (i) .
As IKr 5 k,-l,. 5 1c,-r and c(k s,L) >, 1, there exists a k E Kr such that {A c d i I AI = k} _ ~(k)I > c(Ic„L) 11 -l2 % c(k,>L) II n-li i=2 k r - li i=2 k - li Hence by Remark 1 there exist k,,-11 +2 elements A,, Ak e-h+2 C - 2/(k) such that for D = A, n A 2 , I D I = h and the sets A, \ D, . . ., Akg_i1+2\ D are pairwise disjoint . So by Lemma 1, for every A E V, A D and hence (i) holds .
Now let us prove (ü) . Now we can find a k e K,, for which IV(k) I > k,,2 2r-lnr-1 holds. As a(k) is an (n, L, k)-system, Theorem 5 implies that (ü) is true .
To prove (iii) observe first that, for x E X \ D, dx = {A \ D I x c A E a'} satisfies the hypothesis of the theorem with n' = n-11 , K' _ {k-hl k E Kr }, L={12-1P-Jr-111, so by the induction hypothesis it follows that r n-1 • I ~' I s 11 1 x j_2 kr - , Counting the number of pairs (x, A), where x c A, x c X \ D, and A E a', in two different ways we obtain Z I'Q/xl %I(kr - 11) xcX\D and consequently r n-l .
(n-11)11k l % I~'I(kr - 1 7=2 7=2k,'-7 and (iii) follows .
From the estimation (iii) of Theorem 7 it follows that in the case where L = [l, lc -1], K = [g, h], and n > n o(k), any (n, L, K)-system satisfies k n-1 I~I < iZ (2-l)' which generalizes Theorem 2 of Hilton for the case where l > 1 .
4. The proof of Theorem 8 We apply double induction on k, q .
Let us first consider the case where h = 0. In this case (i) holds automatically. To prove (ü) observe that if we define Q x = {A \ {x} I A c 9/, x c- A}, then x satisfies the hypothesis of the theorem with n' = n-1, L' _ {1 2 -1, . . ., lr k' = k-1, and q' = q .
Hence by the induction hypothesis IslxI qfV=2 (n-li)j(k-li), and this 378 M . DEZA, P . ERDŐS, AND P . FRANKL equality holds for the case where L = {l,} too. As J d I - k = ExE x it follows that ICI < glj2-,(n-li)/(k-li) .
Now suppose that li > 0. Let us choose a set Dl , with I D, = 1,, such that there exist A,', . . ., AJ.q E satisfying A? n AJ = D, for 1 < i < j S kq .
Then let us set c/1 = {A E -d I A 2 D l} . Now we choose D2 in the same way and define 5V2 , and so on . After a finite number, say p, of steps ~4p does not contain any 0-system of cardinality kq and with kernel D, where I D I = l, . We assert that p < q . Otherwise we have at least q+ 1 0-systems Ail, A2, . . ., Ak q with kernels D i , where I D i I = 1,, for i = l, . . ., q+ l . As the sets Ai \ D l , . . . , Akq \ D, are pairwise disj oint and we can find an index j, such that (A, 1 ,\ D,) n Di = 0 for i = l, . . ., q + 1 .
If we have chosen A!,-, A~ then we want to choose A~+1 in such a way that (A~ +i \D,,+,) n (Ai \ Di ) = 0 = (A~+1 \D .,+,) n Di , for 1 < i < s, s+2<i'<q+1 .
As s U (A,;\Di) i=1 q+1 U Di i=2 q+1 U Di, i'=s+2 ,<l,q<kq, S s(k-11)+(q-s)l, < qk and the sets As+1 \Ds+1 are pairwise disjoint, for j = 1, . . ., kq, such a choice of As+i is possible . But if 1 < s < s' S q+ 1, then b+1 Ale n Ass, g DS n DS., which implies that I A?s n A,: ' l 0 L, a contradiction .
Now we want to show that J a/p I = 0(nr-1 ) .
We proceed in essentially the same way as in the proof of Theorem 4 for the case where l, > 0, so the proof will only be sketched .
Let us choose D2 , with I Dl = 12, in such a way that there exist A„ . . ., AV ,, belonging to Q/ p which form a 0-system with kernel D2 1 1 Now define 'q/1 = {A E 2/P A Dl} .
Then choose D2, and so on .
When there are no more l 2-element sets which are kernels of a 0-system of cardinality qk 2 then try to find an l3-set which is the kernel of a 0-system of cardinality qk 3 , and so on .
By Lemma 1, among the A-systems of kernel l i there are no gk2-1 which form a 0-system, whence their number is less than If Dí is an li-element set, then sYD .. _ {A \ Dí I A E V, Dí - A} satisfies the hypo-thesis of the theorem with n' = n - li , k' = k - l i, L' _ {0,1i+1 - li , . . ., lr - li}, and q' = q .
INTERSECTIONS OF SYSTEMS OF FINITE SETS 379 The induction hypothesis yields r n l • _ I~D(I < (q + 1) k - h - 0(nr-1 ) for i 2, j= i j from which it follows that I q/p I = 0(nr- 1 ) .
If E is a set of cardinality more than h then the family 'VE ={AE IESA} satisfies the assumptions of the theorem with n' = n - I E ~, k' = k - I E ~, L' _ {l 2 - I E I , . . ., lr - I E I } n {0,1, 1,-, 1.1, and q' = q.
Hence it follows by induction that I aE I = 0(nr-1 ) .
Let us set '~qi = {A E I D2 A, Dj $ A for j i} . Now it follows that 2, \U-qj j=1 = 0(nr-1) as this family can be written as the union of the families D1,2 = Di, u D ie , for 1 < i1 < i2 < p, and I Di. u Die I > 1, Let c' be a sufficiently large constant and let us set qi -1 = [{I 9iI -c'nr-1}/~,jj n-lj +c o(k,q)nr-1 ~] =1 k - l lll ( j ([x] is the greatest integer not exceeding x) .
As I V 15 I aP I + I V \ Up 1 Ij I + Ep1 I -qj I, it follows for c > c o (c', k, q) that ~p1 qj >- q .
Let qi denote the greatest integer such that there exist Ai, . . ., AQ ,, c- 9i satisfying I Ail n A' 10L for 1 j1 < j2 , qi.
As -4i = {B \ Di I B E .qi} satisfies the assumptions of the theorem with n' = n - 1 1 , k' = k - 11 , L' _ {0, 12 - 113 . . . , lr - l1}, and q' = q, by induction we obtain qi , q i .
If q i = qj for i = 1, . . ., p and Zp 1 qj = q then we are done. So we may suppose that either qj > q or, for some 1 5 j < p, q,' > qj . In the latter case we may assume that ql > q1 .
Hence in any case qi+E 2qj > q .
Let us choose qi sets A1 , . . ., AQí e -1, such that I A jl n A j2 I O L for 1 ~ < j 1 < j 2 S qi .
Suppose that AQí+,, . . ., AQí+q2+ . ..+q,_1 are defined already .
Let us set 9i = 2i \ {B E Ri I there exists j, 1 5 j 5 q1 + q2 + . . . + q i- 1 , such that I B n A j I > l1 } .
It can be seen as above that I-qiI = I ° iI +0(nr-1 ) .
Hence by the induction hypothesis we can find q j sets AQí +q2+. • .+Qi-i+1, . . ., Aqí +q2+ . ..+qt-i+qt such that the cardinality of the intersection of any two different sets ,QID12 , where 380 M . DEZA, P . ERDÖS, AND P . FRANKL among them does not belong to L .
Moreover, if 1 < j, < qí+g2+ . . . +qi-1 < ja < qí+q2+ . . . +qi then I A,, n A j2 < 1 1 , implying that I A il n Ah JO L .
Finally we obtain qi + Ef=2 qi > q sets A1, . . . , A ql +q2+ . ..+qy such that lA 7i n A j2 IOL for 1 < j 1 < j 2 < q+1, a contradiction . So necessarily q~ = qi, q = 1 qi, and by the induction hypothesis yielding r n-l • ~ ~ q H k-h +-cO(k, q)nr-1 for an appropriate choice of c 0 (lc, q), which proves (ü) .
To finish the proof of (i) we have to show that .2/p = 0. Suppose it is not the case and let A 0 E 21p .
We define A i recurrently . If, for some i > 0, A 0, A,, .- Aql, Aql+1, . . ., Aql+ . . .+q,_1 are defined then first define 194, . _ Ri \ {B E RL I there exists j, 0 < j < ql + . . . + qi 1 , such that I B n A~ 'then 19 9 _ _ qi I + 0(nr-1 ) .
So by the induction hypothesis we can define ql+q2+ . ..+qi-,+1, • • •' Aql+ . . .+q;-,+qi such that, for q 1 + . . . + qi-, + 1 < j 1 < j 2 < q l + . . . + qi, A il n A j2 L .
But, as ~ 1 qi = q, this means that in the end we find q+ 1 sets A 0 , . . ., A q E -Ql such that J A,, n Aj2 J O L for 0 < jl < j2 , < q, and this final contradiction concludes the proof of the theorem .
REMARK 4. We conjecture that the assumptions of Theorem S imply that r n - l i I~I < q k-li ' that is, we may omit the last term in (ü) .
If this conjecture is true then it is the best possible in certain cases .
Let k > 2r and X = {1, . . ., n} . Let =1 , . . . I TiT q be q random permutations of X and let be an (n, L, k)-system of cardinality 11 á_ 1 (n - l i )/(k - li ) if such a system exists . If i e X then =(i) is the image of i by -r .
Further, set =(A) _ {z7(a) I a E A} for A - X and z7( Z) _ {-u(A) J A E Ql} .
Then zs1 ( Rl) l(d) are (n, L, k)-systems, and if n > n 0 (E) it can be easily seen that they are pairwise disjoint with probability not less than 1-E .
So for r n-l • 1Wi1 = 1 Wi1 < qi rl ' - + co(k -11,gi)nr-1 - 1 k-h INTERSECTIONS OF SYSTEMS OF FINITE SETS 381 an appropriate choice of zrl , . . .' ur, q the family -4 ='C-J(V) uzu-'(V) U . . . U'A q ('-Q) satisfies the assumptions of Theorem 8 and has cardinality r n-li q 11 k-li REMARK 5 . In the case where L = {11 ,11 + 1, . . ., k- 1} Theorem 8 yields that for a system 5V, of maximum cardinality there are q different 1,-sets Dl , .
. ., Dq such that every element of d contains at least one of the Di's .
The maximality of V implies that V = {A c X I there exists i, 1 < i < q, such that Di - A}, and the sets D 1 , .
.
.
' D q are pairwise disjoint, that is, Theorem 8 is indeed a generalization of the Hajnal-Rothschild theorem .
5. The proof of Theorem 9 We proceed in essentially the same way as with the proof of Theorems 4, 5, and 6 . Therefore the proof is only briefly sketched . We apply induction on k ; the case where k = 1 is trivial .
(a) h = 0. In this case (i) holds automatically with D = 0 . In proving (ü) we may suppose that r > 2 as otherwise we have nothing to prove .
Choosing the constant e(k, t) in such a way that it satisfies c(k, t) > 2c(k -1, t) we may, as in the proof of Theorem 5, successively omit the elements of X which are contained in at most c(k-1, t)nr -2 members of -q/ . Finally, we obtain a family Q/' which consists of subsets of a set X' - X, where every element of X' has degree greater than c(k-1, t)nr-2 and V'J > e(k - 1,t)nr-1 Now using the induction hypothesis we obtain that for every x e X' there exists a set Dx such that I D x I = 12 -1, for x 0 Dx , and A c Q/', for x c A' imply Dx A .
It follows, as in the proof of Theorem 5, that the sets {x} u D x form a partition of X' and, by the induction hypothesis, I V' I > e(k-1, t)nr-1 implies that for any l E L there exist A,_ ., A t E mil' such that JA 1 n . . .nA,J=l .
But A, n . . . n A, is the disjoint union of some of the l 2-element sets x u D x , and it follows that 121 l i and 1, 1 k . The property 11 1141 . . .
I l r I k follows from the induction hypothesis applied to one of the families dx = {A \ x Jx E A E mil'}, for x E X' .
382 M . DEZA, P . ERDÜS, AND P . FRANKL Now we prove (iii). Let x c X . If r > 1 then we can use the induction hypothesis for the family ql x = {A\ x I x e A E -d} and obtain 9~l r n-li x1 (t-1) H k-li If r = 1, that is, L = {0}, then it is obvious that l cv' x ( < t - l . Counting the number of the incident pairs (x, A), where x c- A E sal, in two different ways we obtain that I lx I = k j .Q1 I, yielding la I S (t - 1) lI n-lip i=~ k -1j as desired .
(b) h > 0. First we prove (i) .
If there are k + t sets A,, . . .,Ak+t C_ 'd which form a 0-system with kernel D, where I D I < h, then by the assump-tions of the theorem I D I = h and it follows that D c A for every A E '~V .
So, in the case where r = 1, the assertion follows from Theorem 1 for c = Pt+k(k) • Now we do the same thing as in the proof of Theorem 3 . We select all the l2-element subsets of X which are kernels of a A-system of eardinality (k+t) 2 , consisting of members of z,// .
As we may suppose that no (Ic+t)-element 0-system with an l i-element kernel exists, Lemma 1 yields that there are at most Pk+t(l2) such l2-element sets. Then we omit all the sets containing some of these l2-element sets and look for 0-systems of eardinality (k+t) 3 and with a kernel of ear-dinality 1, and so on .
Finally, using the fact that by the induction assumption a given li-element subset of X is contained in at most (t-1)Ij,r=i(n-lj)/(k-lj) members of jI, we obtain that ~r n-l • 99k+t(l2) _ (t-1) 11 k h +O(nr-2 ), 7=2 j which is a contradiction, for c sufficiently large .
To finish the proof of (ü) and (iii) it is sufficient to apply the induction hypothesis to the system 5V D _ {A\ D I A E _4} .
REMARK 6 . It is possible to prove Theorem 9 when the condition I A i = k is replaced by I A, n . . . n A 1_, I S k for any t -1 different members of a .
REMARK 7 . Let us introduce the following two functions fk,t(n) _ 9k,1(n) _ max{ 1 21 satisfies the assumptions of the theorem with L = {l}, and i n ., E , A i < ii ; max{ szl I satisfies the assumptions of the theorem with L = {0, l} and lX k} .
INTERSECTIONS OF SYSTEMS OF FINITE SETS 383 We know only that fk,,(n) ~< ck (ck = k2 -k+ 1 in the case where t = 2) and 9k i (n) <- ckn K' = i in the case where t = 2), where ek , ck are constants depending only on k .
REMARK S . It is proved in that in the case where L = {1, 2, . . ., k -1} (iii) holds already for k >, (t-1)t-tn .
It would be interesting to obtain better bounds for the general case as well .
6. Concluding remarks (1) Let L' - L. Is it then true that there exist an (n, L, k)-system and an (n, L', k)-system Q/', both of maximum cardinality, such that a' 91(n > no(k))?
It is easy to prove that this is true whenever In the case where L = {0, 2, 3, . . ., k- I} and L' _ {2, 3, . . ., k-1} this is equivalent to a conjecture of Sós and the second author stating that for k > 4 an (n, {0, 2, 3, . . ., k -1}, k)-system has cardinality at most n-2 (k-2 In the case where k = 3 it is not true, which shows that the answer is negative in the case where L' _ {2} and L = {0, 2} .
(2) In the case where L = {1, 2, . . ., k-1} a theorem of Hilton and Milner gives that Theorem 4 holds already for I_Q/1' (k- 1) - (n kkl l) +l, and this bound is the best possible . It would be interesting to obtain best-possible bounds in the general case, too. The third author can prove that in the case where L = {l, l+ 1, . . ., k-1} the optimal bound is n-l n-k-1 (k-l) - ( k-l ) +l for k > ko(l), n > no(k) .
(3) Let s be a positive integer . Let B be an m x k matrix with entries 0, l, . . ., s. Suppose that any two rows of B coincide in at least l positions .
The authors can prove that, for s > so(l), m < (s+ 1)k-a They conjecture that if s = k-1 and every row of B is a permutation of {0, 1, . . ., k- l } then is < (k-l)! for k > k,(l) .
This was proved by Deza and Frankl in for the following cases : l = 1, k arbitrary ; l = 2, k = q ; l = 3, k = q+ 1 where q is the power of a prime .
(4) It is possible to generalize Theorems 7 and 9 simultaneously, that is, for families of sets / such that, for i t < i2 < . . . < it , ; A il I e K, lA il n . . .nA i,IEL .
Such families are called quasi-block-designs by Sós in where the problem of studying these objects was raised .
384 INTERSECTIONS OF SYSTEMS OF FINITE SETS REFERENCES 1 . M. DEza,Solution d'un probl6me de Erdös-Lovast', J. Combinatorial Theory Ser. B 16 (1974) 166-67 .
2 . and P . ERDŐS, On intersection properties of the systems of finite sets', Notices Amer . 3lath . Soc . 22-6 (1975) A-657 .
3 .Matrices dont deux lignes quelconques coincident daps un nombre donne de positions communes', J . Combinatorial Theory Ser . A. 20 (1976) 306-18 .
4 . P. ERDŐS, and N. M. SiNCiii, Combinatorial problems on subsets and their intersections', Advances in Alath ., to appear .
5 . and P. FRANKL,Maximum number of permutations with given maximal or minimal distance', J . Combinatorial Theory Ser. A 22 (1977) .
6 . Pavage généralis© parfait', Coll .
Int .
C.N.R.S .
260-ProWntes combinatoires, to appear .
7. P. ERDŐS and R . RaDo,Intersection theorems for systems of sets', J . London Math . Soc . 35 (1960) 85-90 .
8 . CHAO Ko, and R. RADo, Intersection theorems for systems of finite sets', Quart . J . 11lath . Oxford (2) 12 (1961) 313-20 .
9. P. FRANKL,Sperner systems satisfying an additional condition', J . Combinatorial Theory Ser . A 20 (1976) 1-11 .
10 . - The theorem of Erdös-Ko-Rado holds for n >- Ck(r+ 1)', to appear .
11 . A. HAJNAL and B. RoTHSCHILD,A generalisation of the Erdös-Ko-Rado theorem on finite set systems', J . Combinatorial Theory Ser. A 15 (1973) 359-62 .
12 . A . J. W. HILTON and E . C . MILNER, Some intersection theorems for systems of finite sets', Quart . J. Math . Oxford (2) 18 (1967) 369-84 .
13 .Analogues of a theorem of Erdös, Ko and Rado on a family of finite sets', ibid . 25 (1974) 19-28 .
14 . P. YOUNG, V . MuRTY, and J . EDMONDS, Equicardinal matroids and matroid-designs', Proceedings of the second Chapel Hill conference on combinatorial mathematics, Chapel Hill, N .C. (1970) 498-541 .
15 . K. N. MAJUMDAR,On some theorems in combinatorics relating to incomplete block designs', Ann . lllath . Statist . 24 (1953) 52-75 .
16 . D. K. RAY-CHAUDHURI and R . M. WILsoN, On t-designs', Osaka J . Math .
12 (1975) 737-44 .
17 . V. T . Sós,Some remarks on the connection of graph theory, finite geometry and block designs', to appear .
18 .
J. SPENCER, `Intersection theorems for systems of sets' , to appear .
M. DEZA P. ERDOS Centre National de la Recherche The Hungarian Academy Scientifique of Sciences Paris Budapest P. FRANKL Eötvös L. University Budapest |
4782 | https://www.structuralbasics.com/simply-supported-beam-moment-shear/ | Simply supported beam: Moment and Shear hand calculation - Structural Basics
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Simply supported beam: Moment and Shear hand calculation
ByLaurin ErnstUpdated December 30, 2022
As we used FE programs to calculate the bending moments, shear forces and deflections of structures in last tutorials, we are going a step back now to the very basics of structural engineering and do hand calculations.
After starting this new series of posts with the cantilever beam, we continue our journey with the probably most used static system – the simply supported beam.
We’ll show, step-by-step, how the probably most used formula in structural engineering
𝑀=𝑞⋅𝑙 2/8
is derived and how to calculate the bending moments and shear forces for different loading situations.
Here is a quick overview of what we cover in this post
🙋♀️ What is a simply supported beam?
👆 The static system of the simply supported beam
🏢 The simply supported beam applied on real structures
⬇️ 2 Examples of loading situations
🧮 Hand calculation of bending moment and shear forces – simply supported beam
Not much more talk, let’s get started.
🙋♀️ What is a simply supported beam?
A simply supported beam is a static system acting as a beam element in bending and shear – in some situations also compression or tension due to axial forces. It’s characterized by having two supports, a roller and a pinned support. Those supports allow for rotation.
### 💡 Aroller supporttakes up only vertical forces, while thepinned supporttakes up vertical and horizontal forces.
Pinned and Roller support.
We are surrounded by simply supported beams in our daily lives.
Here are 7 examples of simply supported beams in real life
One span precast concrete beam
One span precast concrete slab
One span bridge deck supported by bearings
Timber flat roof with secondary beams
One span steel beam
Rafter beams supported by 2 Purlin beams
Benches supported on both ends (Example in the following)
👆 The static system of the simply supported beam
The simply supported beam is in most cases a horizontal beam having a roller and a pinned support on the ends. The beam can take normal and shear forces as well as bending moments.
Let’s have a look at the static system.
Static system of a simply supported beam
It can be seen from the picture that the pinned support (a) takes up
a vertical reaction force 𝑉 𝑎 and
a horizontal reaction force 𝐻 𝑎
The roller support (b) takes up
a vertical reaction force 𝑉 𝑏
🏢 The simply supported beam applied on real structures
Understanding the static system of a structure is probably one of the hardest parts about statics and structural engineering in the beginning.
From my experience it’s also poorly taught at university and there is very little information about it online as well, isn’t it?
So let’s look at some applied examples.
The secondary beams of a flat roof. Note that the primary beams can also be simply supported. In our example only the secondary beams are simply supported.
Secondary beam as simply supported beam.
The wooden beams/panels of this bench are simply supported.
Example of a simply supported beam. Bench.
⬇️ 2 Examples of loading situations
We wanna keep it a bit joyful and realistic in this blog, right? So let’s look at loading scenarios we all can relate to.
🧍 Me sitting and meditating on a bench – A Point load
Let’s look at the example where a person – in this case me😊 – sits on bench and meditates🧘♂️. Let’s assume i gained a little bit of weight compared to the last tutorial and now i weigh 76 kg.
Me meditating on a simply supported bench.
This weight of 76 kg can directly be translated into 0.745 kN.
Now because I sit in a specific point with my butt, the load is concentrated and therefore the 0.745 kN can be approximated as a point load on the simply supported beam(s).
### ❗ Watch out: Translating the self-weight to a point load is a rough approximation and simplification. In reality, the self-weight creates (most likely) 2 unevenly distributed line/area loads. Structural engineering is based on many approximations to simplify the calculations. Another good example of that are the wind and snow loads, which are calculated in area loads (𝑘𝑁/𝑚2) in Codes. Wind is quite complex and dependent on many factors and geometries, which can’t be included in normal engineering design due to economic reasons. Therefore, Codes try to simplify these complex wind situations with formulas based on experiments which are applicable to many different design situations.
❗So be always aware of your approximations and consider if they are applicable for what you want to achieve.
As we want to demonstrate a “real-life” example of a point load, this point load is not 100% realistic.
However, there are almost no “100%-correct” point loads and a person sitting on a bench comes close enough to a point load applied to a bench for dimensioning the thickness and width of the timber beam.
Alright, let’s get back to the beam model.
The 3 beams of the bench equal 1 beam when we calculate moments and shear forces with this simple approach.
We can therefore apply the Point load on the static system of the simply supported beam.
Point load applied on bench.
🏋️ The self-weight of a timber beam – Line load
So, the timber beam has a self-weight, right? I guess everyone can image that when you try to lift it up.
In structural engineering, we also call this dead load, and it’s always equally/uniformly distributed for horizontal elements.
For the case of the timber secondary beam, the self-weight per 𝑚 is calculated as
D e n s i t y 𝜌⋅C r o s s-s e c t i o n a l A r e a A
The density of timber beams is found either from manufacturer tables online or Standards.
The cross-sectional area of the timber beam is taken from the previous article, where we designed the flat roof and its beams.
𝜌=3 8 0k g/m 3
𝐴=ℎ⋅𝑤=2 8 8 0 0m m 2
3 8 0k g/m 3⋅2 8 8 0 0m m 2=0.1 1 k N/m
Now, it always helps to translate kN in non-engineering language. 0.1 1 kN/m is also 10.9 kg per 𝑚.
That’s roughly as much as a French bulldog dog – but per meter and per beam! So in case of our timber beam where we have a span of 5𝑚 the dead load of the wood is equal to 5 dogs🐕.
The dead load of the secondary wood beam is roughly as much as 5 french bulldogs.
This line load (kN/m) – or those 5 dogs – can now also be applied to the simply supported beam.
Line load applied on simply supported beam.
🧮 Hand calculation of bending moment and shear forces – simply supported beam
Now in order to design the thickness and material properties of the bench, the secondary beam – or any other simply supported structure the forces and moments acting in the beam need to be calculated.
In general, statically determinate structures such as the cantilever or the simply supported beams need to fullfill three equilibrium conditions:
Horizontal equilibrium ∑𝐻=0: The sum of all horizontal loads and reactions is 0.
Vertical equilibrium ∑𝑉=0: The sum of all vertical loads and reactions is 0.
Moment equilibrium ∑𝑀=0: The sum of all moments is 0.
Let’s continue with the 2 loading situations that we are now familiar with – point and line load.
🔎 Bending moment and shear forces of a beam – Point load on bench
The first thing we always calculate in determinate structures are the reaction forces/moment. In our case that is 𝑉 𝑎,𝐻 𝑎 at support (a) and 𝑉 𝑏 at support (b) due to the equilibrium conditions.
Point load applied on simply supported beam.
∑𝐻=0:𝐻 𝑎=0
∑𝑉=0:𝑉 𝑎+𝑉 𝑏–0.7 4 5 k N=0= ->𝑉 𝑎=𝑉 𝑏=0.7 4 5 2 k N=0.3 7 2 5k N
∑𝑀=0:𝑀 𝑎=0
Calculation of the shear and moment distribution along the beam due to the reaction forces. The parameter x is introduced as the length between point a and any point on the beam.
Parameter x
The shear forces and bending moments can be calculated in dependence of x. Let’s make a first cut at a point between the support and the point load 0<x<1.0m.
Force equilibrium at point 0 < x < 1.0m
As for the reaction force calculation, the equilibrium conditions are used to calculate the moment and shear forces at point x
∑𝐻=0:𝐻 𝑎=0
∑𝑉=0:0.3 7 2 5k N–𝑉 𝑥=0 ->𝑉 𝑥=0.3 7 2 5k N
∑𝑀=0:𝑀 𝑥–0.3 7 2 5 k N⋅𝑥=0 ->𝑀 𝑥=0.3 7 2 5 k N⋅𝑥
As we can see the shear force is constant and not dependent on the parameter x. Let’s set x = 1.0m and see what results we get for the bending moment:
𝑀 1.0𝑚=0.3 7 2 5 k N⋅1.0𝑚=0.3 7 2 5k N m
In dependence of x and the Point load Q = 0.745kN a general formula for the bending moment of a simply supported beam for 0<x<l/2 can be formulated as:
𝑀 𝑥=1/2⋅𝑄⋅𝑥
You might have already come across the formula when we set x=l/2
𝑀 𝑚 𝑎 𝑥=1/2⋅𝑄⋅𝑙/2=1/4⋅𝑄𝑙
4. Cut at a point between the point load and the endpoint 1.0m<x<2.0m
Force equilibrium at point 1.0m < x < 2.0m
The equilibrium conditions lead to
∑𝐻=0:𝐻 𝑎=0
∑𝑉=0:0.3 7 2 5𝑘𝑁–𝑉 𝑥–0.7 4 5𝑘𝑁=0 ->𝑉 𝑥=−0.3 7 2 5𝑘𝑁
∑𝑀=0:𝑀 𝑥–0.3 7 2 5𝑘𝑁⋅𝑥+0.7 4 5𝑘𝑁⋅(𝑥−1.0𝑚)=0
𝑀 𝑥=0.3 7 2 5𝑘𝑁⋅𝑥–0.7 4 5𝑘𝑁⋅𝑥+0.7 4 5𝑘𝑁𝑚=−0.3 7 2 5𝑘𝑁⋅𝑥+0.7 4 5𝑘𝑁𝑚
### 💡 Tip: The calculation of the bending moment distribution at 1.0 < x < 2.0m can be calculated much quicker when the left side of the beam is cut (see next picture).
Another possibility of equilibrium for 2.0m > x > 1.0m
5. Bending moment and shear force diagrams
The diagrams can be plotted by a tool like Excel using the formulas from above or drawn by hand when one is aware of the geometrical shape of the distribution.
Shear force diagram – simply supported beam
Point load shear force distribution. Simply supported beam.
Bending moment diagram – simply supported beam
Point load bending moment distribution. Simply supported beam.
👨🏫 Bending moment and shear forces of a beam – Line load on the secondary roof beam
1.As for the Point load, we first calculate the reaction forces 𝑉 𝑎,𝐻 𝑎and moment𝑀 𝑎 in the determinate structure – simply supported beam – due to the equilibrium conditions.
Line load applied on cantilever beam.
∑𝐻=0:𝐻 𝑎=0
∑𝑉=0:𝑉 𝑎+𝑉 𝑏–0.1 1 k N/m⋅5.0𝑚=0 ->𝑉 𝑎=0.5⋅0.1 1 k N/m⋅5m=0.2 7 5k N
∑𝑀=0:𝑀 𝑎=𝑀 𝑏=0
Calculation of the shear and moment distribution along the beam due to the reaction forces. The parameter x is introduced as the length between point a and any point on the beam.
Parameter x
The shear forces and bending moments can be calculated in dependence of x. Let’s make a cut at a point between the support a and support b 0<x<5.0m.
Force equilibrium at point 0 < x < 5.0m
As for the reaction force calculation, the equilibrium conditions are used to calculate the moment and shear forces at point x.
∑𝐻=0:𝐻 𝑎=0
∑𝑉=0:𝑉 𝑥+0.1 1 k N/m⋅𝑥–0.2 7 5 k N=0 ->𝑉 𝑥=0.2 7 5k N–0.1 1 k N/m⋅𝑥
∑𝑀=0:𝑀 𝑥–0.2 7 5 k N⋅𝑥+0.1 1 k N/m⋅𝑥 2 2=0 ->𝑀 𝑥=0.2 7 5 k N⋅𝑥–0.1 1 k N/m⋅𝑥 2 2
### ❗ Do you still follow us? Or was the moment calculation a bit too quick? Let us know in the comments below.
Compared to the distribution due to a point load, the shear force distribution is now linear and dependent on the parameter x.
From the moment formulation, we can now derive the famous formula for the maximum bending moment of a simply supported beam due to a line load. Let’s set x = l/2=2.5m.
Formula for maximum bending moment in simply supported beam 𝑞𝑙 2/8
In order to get the formula we change the load and reaction values to variables. The line load 0.11kN/m is used as q and the reaction force 𝑉 𝑎 equals ql/2.
𝑀 𝑥=𝑞⋅𝑙/2⋅𝑥–𝑞⋅𝑥 2 2
𝑀 𝑙/2=𝑞⋅𝑙/2⋅𝑙/2–𝑞⋅(𝑙/2)2 2=𝑞⋅𝑙 2/4–𝑞⋅𝑙 2 8=𝑞⋅𝑙 2/8
𝑀 𝑙/2=0.1 1 k N/m⋅(5m)2/8=0.3 4k N m
Formula for maximum shear force in simply supported beam 𝑞𝑙/2
As for the bending moment we change the load and reaction values to variables. The line load 0.11kN/m is used as q and the reaction force 𝑉 𝑎 equals ql/2.
𝑉 𝑥=𝑞⋅𝑙/2–𝑞⋅𝑥
Now, the maximum shear force acts right next to the supports, therefore x is set to 0, or l
𝑉 0=𝑞⋅𝑙/2–0
𝑉 0=0.1 1 k N/m⋅5m/2=0.2 7 5k N
4. Bending moment and shear force diagrams
The diagrams can be plotted by a tool like Excel using the formulas from above or drawn by hand when one is aware of the geometrical shape of the distribution.
Shear force diagram – simply supported beam
Shear force diagram. Simply supported beam. Line load. Linear shape.
Bending moment diagram – simply supported beam
Bending moment diagram. Simply supported beam. Line load. Parabolic shape.
Once the forces and moments are calculated for different load cases like
snow load
wind load
dead load
live load
and loads are combined in load combinations, the design of the element needs to be done.
In our case that is the timber design of the secondary beam, meaning that we pick the timber type, calculate the required cross-sectional dimensions and much more.
We actually calculated the flat roof in full scale in this article.
Now, I would like to hear from you: What simply supported structures do you know from the “real world”? Did I forget some? Let us know in the comments below📝
Post Tags: #Internal force calculation
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Class 10 (Old)
Course: Class 10 (Old)>Unit 12
Lesson 1: Combination of solids
Area of combination of solids
Problem types: surface area of combination of solids
Volume of combination of solids
Problem types: volume of combination of solids
Combination of solids (basic)
Area of combination of solids (intermediate)
Area of combination of solids (advanced)
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Class 10 (Old)>
Surface areas and volumes>
Combination of solids
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Volume of combination of solids
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Let's learn how to find the volume of a complicated solid by breaking it down into cylinders, cones, hemispheres, and cuboids.Created by Aanand Srinivas.
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rajvardhan07 2 years ago Posted 2 years ago. Direct link to rajvardhan07's post “Can't we just add and tak...” more Can't we just add and take the quantities out which are common ?
Like here :
1/3pir^3+pir^2h+2/3pir^3
We can take pir^2 common here right ?
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let's learn how to find the volume of solids that don't look like the typical cylinder or cone or a cuboid no things that we've already learnt let's look at something that's slightly more complicated than that what if we had to find the volume of something like this say a pencil here are some lengths what methods can you think of of finding the volume of this pencil now one method that strikes me immediately is hey I need to find volume right but volume is just the amount of substance that's contained in this so what I can do is take this and immerse it basically dip it in water and look at the amount of water that gets displaced you know it's basically pours out if the container was completely full the good thing Archimedes tells us is that the volume of that water that spills out and this volume volume of the pencil that we want will be exactly the same and then I can what I can do is take that water and what does the good thing about water it has no shape it will take whatever shape but I wanted to take pulled in some beaker and I can measure the volume that's probably what I would do if I really had to do this in some for some reason the other method because this pencil has this beautiful property that it looks like hey there's a cone here that I can see there's a cylinder I can see here and there's a half ball or a hemisphere I can see because this pencil looks happens to have these things within it we can also apply this other method where I can break this pencil down into a cone a cylinder and a hemisphere and this is the kind of problem we will see in this chapter solids which we can break down into things that are familiar to us and just add and the good thing about volume is that you always just have to add the amount of substance just adds so you can pause the video right now and break it down find each of the volumes individually and add I'm going to start so the first thing I have is a cylinder is Lynda so a cylinder has some radius like this if I were to cut this out over here and then it'll look something like this obviously it'll be longer I'm just showing it's small over here and then this will be the circular part this will be the height do you remember the formula for volume of a cylinder it's PI R square H i R square H and you can ask me okay is that an intuition that helps me remember this better the way I remember it is the area of the circle is PI R squared you're stacking this PI R squared H times because this is H and this area is is R I'm sorry this length is R so this circle the area is PI R squared and H times you keep it you get PI R Squared's now for the cone if I were to draw the cone over here so there is my cone and once again oh my god that circles probably the worst one I've drawn so this looks better yeah this cone what is the volume of this cone the formula for it again you don't have to remember it because as you solve more problems just keep looking it up in the book and eventually you will not be able to forget it so it's 1 by 3 PI R square it's now one way I remember it is to notice that it's 1/3 of a cylinder of the same height and the same radius right if I had a cylinder of radius R and a height H so in other words if I had a silicon sorry a cone like this that'll be one-third of the area of that cylinder that's just a way to remember it the way to actually derive this and to see why it works sort of prove it happens with me in class 12 if you learn integration you can derive these quantities using integration the third thing we have your is half a ball which is like a hemisphere so let's let's draw the hemisphere over here I'd say we again have a circle and we have a hemisphere there it is and the volume of a hemisphere the more popular formula remember is the volume of an entire sphere so volume of a sphere is actually 4 by 3 PI R cubed 4 by 3 PI R cubed so the volume of a hemisphere will be 2 by 3 PI R cube half of that and the good thing about having a volume is that you actually don't have this concept of total volume surface volume and all that volume is just the amount of matter no matter what you do it's the it's the same thing it's just the amount of matter so you just have to add it these are the only formula you have to remember for volume apart from of course Q now the cuboid volume is actually very intuitive its length into breadth into height now why that's intuitive is because very similar argument to this you take a rectangle of length into breadth area and then multiply it keep it stuck at each times so L into B into H very similar to PI R square H now the question boils down to just finding the right numbers for each of these and adding them up you can go ahead and do it I'm gonna do it now so it's 1 by 3 into PI I take this 22 by 7 into R squared now once again R is not 2 centimeters here it's the width like we saw on surface area you have to make sure that you take half of this so it's one centimeter and I'm actually going to move this a little bit to the left let's move this so I have space to do cylinder and there it is so R is one centimeter and you have R squared so one centimeter actually put one to one centimeter squared the reason I keep the unit's here is to not make mistakes it's over here is for the cone what is it it's two centimeters given over here so it's directly given to centimeters notice that for the surface area you actually care about this length the slant height but for the volume you care about the height directly in case you wanted the slant height you can actually find it using Pythagoras theorem this is 2 centimeters this is one centimeters so you can find this root of 2 square plus 1 square which is root 5 but that's not needed for us that was actually why did I even say it I don't think I should have said it it's not necessary but yeah what about a cylinder over here this does cylinder over here so you have 22 by 7 let's use blue 22 by 7 multiplied by R squared and as once again just 1 centimeter into 1 centimeter square multiplied by 8 which is 10 centimeters 10 centimeters so we can calculate this later what about the hemisphere that's going to be equal to 2 by 3 2 over 3 into 22 by 7 into R cubed an R in this case is 1 centimeter cube notice that in all these cases you finally get an answer that's in centimeter cube which is basically cubic centimeters it's called it's the unit for volume and this also helps you check that your units are fine so let's do it now what do I have I have 22 into 244 divided by 21 centimeter cube over here I owe here I have 220 by 7 220 let's use blue 220 by 7 centimeter cube and over here I have 44 by 21 again 44 by 21 coincidentally the same volume as this part of the pencil coincidentally so there you have it you can actually find the total volume you just have to add this and this and this the answer may or may not look pretty so you can go ahead and do it 44 plus if I just multiply this with a 3 and a 3 over here I will be able to add them all because they have the same denominator so at 660 plus 88 660 plus 100 760 minus 12 750 748 so 748 divided by 21 centimeter cube I'm gonna leave this one here because I know you can find what this number should should be and once again notice that the purpose of this video is to not just find the volume of this pencils but a notice that this will be the type or this will be the pattern that will happen that that you'll follow for all the questions regarding volume of solids like this you will first break it down into pieces that for which you know the formula and then you'll remember the formulae that you need to remember actually these are the only things you need to remember for a cone it's 1 by 3 PI R square H for a cylinder it's PI R square H for a sphere it's 4 by 3 PI R cube which means for half a sphere it's 2 by 3 PI R cube apart from this like I said the only other one is a cuboid which is length into breadth into which L in to blend into breadth into height that's it there's nothing new that you need to know after this that you need to solve problems in this chapter
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4784 | https://www.khanacademy.org/math/10th-grade-philippines/x9bab43254af22e3c:3rd-quarter/x9bab43254af22e3c:combinations/e/combinations_1 | Combinations (practice) | 3rd quarter | Khan Academy
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10th grade Math (MELCS)
Course: 10th grade Math (MELCS)>Unit 3
Lesson 2: Combinations
Intro to combinations
Combination formula
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Combination example: 9 card hands
Combinations
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Combinations
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CCSS.Math: HSS.CP.B.9
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4785 | https://cse232-msu.github.io/CSE232/labs/lab13.html | Lab - Linked List
HomeSyllabusLecturesVSCode Setup
Piazza
Lab - Linked List
There is no Unix tutorial this week
Coding Assignment
Today, we’re going to practice pointer manipulation in the context of a singly-linked list. You can find more information regarding linked lists here.
Background
Download the starter code provided here.
In a nutshell: a singly-linked list is a data structure for implementing a generic array of elements, where each node has data, and a pointer to the next node. The list structure typically has pointers to the list’s first node and last node. A singly-linked list’s first node is typically called the head, and the last node the tail.
Linked above is a header file containing the definition for a singly-linked list class named
plaintext
SingleLink
, and a definition for a linked list node class named
plaintext
Node
.
Below is the interface for the two classes as it stands right now:
```
struct Node {
public:
int data_;
Node next_;
Node() : data_(0), next_(nullptr) {};
Node(int d) : data_(d), next_(nullptr) {};
};
```
plaintext
Node
has two private members,
plaintext
Node next_
and
plaintext
int data_
.
plaintext
next_
is a pointer to another instance of type
plaintext
Node
; the “chain” that makes up the structure of the entire list.
plaintext
data_
is the value held by the
plaintext
Node
. For the purposes of today’s lab, it’ll only be allowed to hold integers.
It has two constructors, a default that initializes
plaintext
data_
to 0, and a one-parameter variant that initializes
plaintext
data_
to the incoming argument.
```
class SingleLink {
private:
Node head_;
Node tail_;
public:
SingleLink();
SingleLink(int dat);
void append_back(int);
friend std::ostream& operator<<(std::ostream &out, SingleLink &s);
bool del(int val);
Node& operator;
// Rule of three stuff
~SingleLink();
SingleLink(const SingleLink &);
SingleLink& operator=(SingleLink);
};
```
plaintext
SingleLink
has two private members,
plaintext
Node head_
and
plaintext
Node tail_
, which are pointers to the first and last node of the list, respectively.
If the list is ever empty,
plaintext
head_
and
plaintext
tail_
should point to
plaintext
nullptr
.
If the list ever has one
plaintext
Node
, both the
plaintext
head_
and
plaintext
tail_
should point to that one
plaintext
Node
.
plaintext
SingleLink
has two constructors; a default with no arguments, and a one-parameter variant that should add a
plaintext
Node
with data,
plaintext
dat
, to the list.
If the one-parameter variant is invoked, the head and tail pointers will need to be adjusted to point to this new
plaintext
Node
.
What’s missing?
Pretty much everything. But, this is an opportunity to get more practice with pointers, and get a feel for how to program something you’ll be becoming much more familiar with in CSE 331. You probably won’t finish all of it, but work through from the beginning and see how far you can get!
Program Specifications
Download and extract the starter code’s .zip file into your working directory (see top of Background). Next, create a ‘singlelink.cpp’ implementation file. You are only going to be modifying the implementation file and main.cpp. If you submit to Coding Rooms, only the implementation file is needed for submission.
None of the Rule of Three functions have been implemented. If you have the time, it would be good if you implemented those.
Before jumping into the methods, I would make the default and one-parameter constructors.
Listed below are the methods you need to implement. I recommend designing these methods in the order they appear down this list.
void append_back(int dat)
Method function; creates a new
plaintext
Node
instance with
plaintext
data_=dat
and appends it to the end of the list.
Make sure you use dynamic allocation (the
plaintext
new
keyword) so that the
plaintext
Node
you create isn’t deleted when it falls out of scope.
Also make sure that you’re re-routing the
plaintext
head_
and
plaintext
tail_
pointers correctly (what happens when this is the first
plaintext
Node
being appended vs. when it’s the second
plaintext
Node
being appended?)
ostream & operator<<(ostream &out, SingleLink &s)
Friend function; pushes the
plaintext
data_
member of each
plaintext
Node
instance in the list to the output stream,
plaintext
out
, and returns
plaintext
ostream &
.
I recommend modifying ‘main.cpp’ to show that your method works before moving on. Call the
plaintext
append_back()
method with, say, integers
plaintext
{1, 2, 4, 8}
, and then use
plaintext
cout
to print them.
The next task on our list is the
plaintext
del()
method (we can’t call it
plaintext
delete()
, since
plaintext
delete
is a keyword in C++ ☹️).
plaintext
del()
should remove a particular value from the list. Before you write any code, identify any edge cases your list might have to deal with. Type them as comments under the
plaintext
del()
method.
⭐ Show the TA your working
plaintext
append_back()
and
plaintext
operator<<()
functions. And, show your TA all of the cases
plaintext
del()
must account for before moving on.
bool del(int val)
Searches through the list for the first
plaintext
Node
that has the same
plaintext
data_
value as
plaintext
val
. If found, deletes the
plaintext
Node
and returns
plaintext
true
, otherwise returns
plaintext
false
.
Again, I recommend modifying ‘main.cpp’ for testing.
Node & operator
This method is an override for the
plaintext
[]
operator. On a call, such as
plaintext
sl
, the argument, 3, is assigned to the parameter,
plaintext
index
. The return value is a reference to a
plaintext
Node
so that the
plaintext
Node
can be modified (i.e., can show up on either side of an assignment operator).
You’ll have to search the list (starting from the
plaintext
head_
pointer) for the
plaintext
index
-th
plaintext
Node
. Then, return a reference to that
plaintext
Node
, or throw an
plaintext
out_of_range
exception if you’ve traversed to the end of the list.
⭐ Show the TA your completed
plaintext
SingleLink
class. Include example functionality for all of the methods.
What if I didn’t finish?
This lab is designed to prepare you for later CSE courses, especially CSE 331, Algorithms and Data Structures. If you didn’t finish the lab (asynchronously or synchronously), we strongly recommend completing it own your own. As incentive, there will be at least one final exam question on the material from this lab specifically.
The lab has 120 points of tests on it, but for normal lab credit, you need to earn only 100 points. This scoring system means that the final four test cases can be failed without penalty.
Honors Material
For this one lab, there will not be separate honor’s material. The baseline project is sufficiently challenging and Honors students are expected to complete the full project, earning all 120 points worth of test cases. |
4786 | https://study.com/academy/lesson/the-sum-of-the-first-n-terms-of-an-arithmetic-sequence.html | Sum of Arithmetic Sequence | Formula & Examples - Lesson | Study.com
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Copyright
Math Courses / Algebra II: High School Course
Sum of Arithmetic Sequence | Formula & Examples
Lesson
Transcript
Joanna Tatomir, Yuanxin (Amy) Yang Alcocer, Matthew Bergstresser
Author Joanna Tatomir
Joanna holds a PhD in Biology from the University of Michigan and is currently working towards a degree in Veterinary Medicine at Michigan State University. She has taught a combination of ESL and STEM courses to secondary and university students.
View bio
Instructor Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has been teaching math for over 9 years. Amy has worked with students at all levels from those with special needs to those that are gifted.
View bio
Expert Contributor Matthew Bergstresser
Matthew has a Master of Arts degree in Physics Education. He has taught high school chemistry and physics for 14 years.
View bio
Learn what an arithmetic sequence is and explore different examples of an arithmetic sequence. Understand how to find the sum of an arithmetic sequence. Updated: 11/21/2023
Table of Contents
What is an Arithmetic Sequence?
Arithmetic Sequence Examples
Sum of an Arithmetic Sequence
How to Find the Sum of an Arithmetic Sequence
Lesson Summary
Show
FAQs
Activities
Arithmetic Sequence
Some numbers have patterns to them such as 2, 4, 6, 8, 10. The pattern here is that each successive number is two (2) more than the previous number. Say we wanted to sum all of the numbers in this pattern from 2 to 200. That would take a long time to list out the values. Luckily there is an equation to simplify this. Let's practice using this equation.
Problems for Additional Practice
Add up the first four terms of the arithmetic sequence {2, 4, 6, 8, 10}.
Add up the first ten terms of the arithmetic sequence {100, 200, 300, ...}.
Add up the first 1,000 terms of the arithmetic sequence {3, 6, 9, ...}.
A bank wants to do a promotion. They have provided a puzzle and if it is solved correctly the winner gets the money in a savings account in their bank. The puzzle is is this. A triangle has 20 rows. The first row is $1. Each successive row has one more dollar. How much money is in the triangle?
Solutions
For each solution we will use the arithmetic sum equation (n/2)(2a + (n - 1)d), where a is the first term, d is the common difference between the terms and n is how many terms to sum.
(4/2)(2(2) + (4 - 1)(2)) = 2(4 + (3)(2)) = 2(4 + 6) = 2(10) = 20
To check ourselves we can add 2 + 4 + 6 + 8 and we get 20.
(10/2)(2(100) + (10 - 1)(100)) = 5(200 + (9)(100)) = 5(200 + 900) = 5(1100) = 5,500
(1,000/2)(2(3) + (1,000 - 1)(3)) = 500(6 + (999)(3)) = 500(5 + 2,997) = 500(3,002) = 1,501,000
(50/2)(2(1) + (50 - 1)(1)) = 25(2 + 49) = (25)(51) = $1,275
What is an arithmetic sequence?
An arithmetic sequence is a series of numbers that is defined by a constant value between adjacent terms. The constant value can be found by taking the difference between any two adjacent terms.
What formula is used in arithmetic sequence?
An arithmetic sequence refers to a series of numbers separated by a constant difference between adjacent terms. The formula used to solve the sum of an arithmetic sequence is: n/22a + (n-1)d, where n = the number of terms to be added, a = the first term, and d = the constant value.
What is an arithmetic sequence and give examples?
Arithmetic sequences are series of numbers, like 2, 4, 6, 8, ..., that have a constant difference between neighboring terms. For example, in this series, the constant value is 2 (4 -2 = 2 or 8 - 6 = 2).
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Table of Contents
What is an Arithmetic Sequence?
Arithmetic Sequence Examples
Sum of an Arithmetic Sequence
How to Find the Sum of an Arithmetic Sequence
Lesson Summary
Show
What is an Arithmetic Sequence? -------------------------------
In mathematics, many people are already familiar with common operations such as addition, subtraction, multiplication, and division. Throughout primary and secondary school, students become well-versed with adding numbers together or subtracting them from one another.
Sometimes, students are even asked to add a whole sequence of numbers together. Some terms to learn for this lesson include:
Sequence: a series of numbers placed in a specific order.
Arithmetic sequence: a special type of sequence. An arithmetic sequence, also known as an arithmetic progression or an arithmetic series, is a set of numbers in which the difference between each term is a constant value. In other words, the same value can be added to the previous number for an infinite amount of time.
Arithmetic series: another name for an arithmetic sequence.
It's a relatively straightforward process to add all of the terms together in a short series to derive the sum of an arithmetic sequence. However, what if a student is asked to add the first 20 or first 125 terms of an arithmetic sequence? What if the difference between each term is an especially large value? Is there an easier way to solve this type of problem?
Using the general formula for the nth term of an arithmetic sequence, an individual can easily find the sum of an arithmetic series for a specific number of terms.
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0:02 An Arithmetic Sequence
0:59 The Sum Formula
2:00 Using the Formula
3:49 Another Example
5:16 Lesson Summary
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Arithmetic Sequence Examples ----------------------------
An arithmetic sequence can be expressed in terms of the following expression: a,(a+d),(a+2 d),(a+3 d), ... where a is the first term and d is the constant difference between values.
Using this expression, some arithmetic sequence examples include:
1, 5, 9, 13, 17, 21, 25, 29, 33, ...
The constant value can be derived by taking the difference between any two adjacent terms. For example, 9 - 5 = 4 and 33 - 29 = 4. Therefore, 4 represents the constant value.
Using the expression above: 1, (1 + 4) , (1 + 2(4)), (1 + 3(4)), (1 + 4(4)),...
2, 4, 6, 8, 10, 12, 14, 16, 18,...
In this simple arithmetic series, the constant value between adjacent terms is 2.
Using the expression above: 2, (2 + 2), (2 + 2(2)), (2 + 3(2)), (2 + 4(2)),...
1, 8, 15, 22, 29, 36, 43, 50, ...
The constant value can be derived by taking the difference between any two adjacent terms. In this sequence, 50 - 43 = 7 and 22 - 15 = 7. Therefore, the constant value is 7.
Using the expression above: 1, (1 + 7), (1 + 2(7)), (1 + 3(7)), (1 + 4(7)),...
5, 15, 25, 35, 45, 55, 65, 75, ...
Taking the difference between any two adjacent terms, the constant value is 10. For example, 15 - 5 = 10 and 55 - 45 = 10.
Using the expression above: 5, (5 + 10), (5 + 2(10)), (5 + 3(10)), (5 + 4(10)),...
12, 24, 36, 48, 60, 72, 84, 96, ...
The constant value for this sequence is 12. This is derived by subtracting any two adjacent terms, such as 24 - 12 = 12 or 84 - 72 = 12.
Using the expression above: 12, (12 + 12), (12 + 2(12)), (12 + 3(12)), (12 + 4(12)),...
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Sum of an Arithmetic Sequence -----------------------------
Calculating the sum of the first few terms of an arithmetic series is relatively simple. These values just simply need to be added together. However, the addition of the first 100 terms, for example, would prove more problematic. Rather than manually adding all of these terms together, mathematicians have developed the arithmetic sequence formula, which allows an individual to easily calculate the sum of an arithmetic sequence.
The formula for the sum of an arithmetic sequence is: S n=n 2[2 a+(n−1)d], where:
n = the number of terms to be added
a = the first term in the sequence
d = the constant value between terms
For example, adding the first four terms of the arithmetic sequence 2, 4, 6, 8,... can be manually calculated by adding the terms together: 2 + 4 + 6 + 8 = 20. Using the formula:
S n=n 2[2 a+(n−1)d]
n = 4
a = 2
d = 2
S n=4 2[2(2)+(4−1)(2)]
S n=2[4+6]
S n=2(10)
S n=20
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How to Find the Sum of an Arithmetic Sequence ---------------------------------------------
Using both methods shown in the previous section, the sum of an arithmetic sequence can be calculated. However, when adding larger numbers of terms, the arithmetic sequence formula is especially useful. Some steps for how to find the sum of an arithmetic sequence include:
Find the first term of the sequence. This value will be a.
Find the constant value by subtracting any two adjacent terms. This value will be d.
Determine the number of terms to be added. This value will be n.
Then fill in these values into the formula and solve for S n.
Using the arithmetic sequence examples above, these steps will be used to solve some sample problems.
Example 1
What is the sum of the first 20 terms of the sequence: 5, 15, 25, ...?
The first term of the sequence is 5: a = 5.
The difference between 5 and 15 is 10 (15 - 5 = 10): d = 10.
The first 20 terms are added together: n = 20
S n=n 2[2 a+(n−1)d]
S n=20 2[2(5)+(20−1)(10)]
S n=10[10+190]
S n=10
S n=2000
Example 2
What is the sum of the first 15 terms of this sequence: 1, 8, 64, ...? The first term is 1, for a = 1. However, when looking at the difference between the adjacent terms, there is no constant value: 8 - 1 = 7 and 64 - 8 = 56. This does not represent an arithmetic sequence.
Example 3
What is the sum of the first 15 terms of this sequence: 1, 8, 15, ...?
The first term of the sequence is 1: a = 1
The difference between adjacent terms is 7 (8 - 1 = 7 and 15 - 8 = 7): d = 7
The first 15 terms are added together: n = 15
S n=n 2[2 a+(n−1)d]
S n=15 2[2(1)+(15−1)(7)]
S n=(7.5)[2+(14)(7)]
S n=(7.5)(100)
S n=750
Example 4
What is the sum of the first 14 terms of this sequence: 1, 5, 9, ...?
The first term of the sequence is 1: a = 1
The difference between adjacent terms is 4 (5 - 1 = 4 and 9 - 5 = 4): d = 4
The first 14 terms are added together: n = 14
S n=n 2[2 a+(n−1)d]
S n=14 2[2(1)+(14−1)(4)]
S n=(7)[2+(13)(4)]
S n=(7)(54)
S n=378
Example 5
What is the sum of the first 21 terms of this sequence: 2, 4, 6, 8, ...?
The first term of the sequence is 1: a = 2
The difference between adjacent terms is 2 (4 - 2 = 2 and 8 - 6 = 2): d = 2
The first 21 terms are added together: n = 21
S n=n 2[2 a+(n−1)d]
S n=21 2[2(2)+(21−1)(2)]
S n=(10.5)[4+(20)(2)]
S n=(10.5)(44)
S n=462
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Lesson Summary --------------
A sequence in mathematics refers to a series of numbers placed in a specific order. An arithmetic sequence, also known as an arithmetic series, is a special type of sequence in which a set of numbers has a constant value between each term. An example of a simple arithmetic sequence is 2, 4, 6, 8, ..., where 2 is the constant value between adjacent terms. The general formula for an arithmetic sequence of numbers is: a,(a+d),(a+2 d),(a+3 d), ..., where a is the first term and d is the constant difference between values.
The sum of an arithmetic sequence can be easily calculated using the following formula: S n=n 2[2 a+(n−1)d], where n is the number of terms to be added, a is the first term in the sequence, and d is the constant value between terms. Use the following steps to solve the formula:
The first term of the sequence will be a.
The constant value (determined by subtracting any two adjacent terms) will be d.
The number of terms to be added will be n.
Then fill in these values into the formula and solve for S n.
Following these recommendations will enable any individual to quickly and easily solve the sum of an arithmetic sequence.
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Video Transcript
An Arithmetic Sequence
Our video lesson begins with a quick overview of what an arithmetic sequence is. It is a string of numbers where the difference between each pair of successive numbers is the same. So, for example, our number line is an arithmetic sequence where the difference is 1 between each pair of the numbers. This difference between the pairs of numbers is referred to as the common difference. You can also think of an arithmetic sequence as telling you how much you have of something as it grows. Say you have a savings account at the bank. If you start with 10 dollars and you continue to put 10 dollars into the account every day, your account balance can be written as an arithmetic sequence where the difference between each pair of numbers is 10. We write our arithmetic sequence between curly brackets and with commas between the numbers. Writing our sequence for the bank, we have {10, 20, 30, 40, . . .}. We use the three dots at the end to show that the sequence goes on indefinitely.
The Sum Formula
Sometimes, we want to add up the numbers in our sequence. For example, if our arithmetic sequence shows how many strawberries we are able to harvest each day, then adding up our sequence will tell us how many total strawberries we will have picked. To do this, we have a formula that will help us. The formula says that the sum of the first n terms of our arithmetic sequence is equal to n divided by 2 times the sum of twice the beginning term, a, and the product of d, the common difference, and n minus 1. The n stands for the number of terms we are adding together.
Arithmetic sum formula
The part of this formula that we will use is the part after the equals sign. The part before the equals sign just tells you that you are going to add each term. What this first part does is gives you a way to calculate each term in your sequence and then to sum it up. But since we are looking for what this sum equals, we only need to concern ourselves with the part after the equal sign.
Using the Formula
To use this formula, we label our n, a, and d in our problem and then we plug the appropriate numbers in and evaluate. Let's see how this is done. Say we are the strawberry farmers, and we picked 20 strawberries the first day. This is strawberry season, so every day more and more strawberries are ripening. Each day, we are able to pick 5 more strawberries than the previous day. We can write this as an arithmetic sequence where the common difference is 5. We write {20, 25, 30, . . .}.
To find out how many strawberries we are able to pick after 30 days, we need to use our sum formula. We label our n as 30, our a as 20 (since that is our beginning number), and d as 5 (since that is our common difference). Now that everything is labeled, we can plug in these numbers and evaluate to get our answer. We plug our numbers into the part of the formula that is behind the equals sign. We get 30 divided by 2 times the sum of 2 times 20 and 30 minus 1 times 5.
We begin evaluating this expression by first subtracting the 30 minus 1 to get 29. We then multiply this 29 with the 5 to get 145. We then multiply the 2 with the 20 to get 40. We add the 40 to the 145 to get 185. Now we can divide the 30 by the 2 to get 15. Multiplying the 15 with the 185, we get 2775. So, after 30 days, we will have picked 2775 strawberries. That's a lot of strawberries! Just think of how many strawberry shortcakes you can make with that!
Example calculation
Another Example
Now it's your turn. See if you can find the sum of this sequence as you follow along. Our sequence is {5, 7, 9, 11, . . .}. First, we need to make sure that this is an arithmetic sequence. We check to see what the difference is between our numbers. Is this difference the same between each pair of numbers? Yes, it is. We have an arithmetic sequence with a common difference of 2 and a beginning number of 5.
We want to find the sum of the first 10 terms of this sequence. How do we do this? Yes, we will use the sum formula. We will label our n, a, and d so we can plug these numbers into the formula to evaluate. We have our n as 10, our a as 5, and our d as 2. Plugging these in, we get 10 divided by 2 times 2 times 5 plus 10 minus 1 times 2. Evaluating, we get 10 divided by 2 times 2 times 5 plus 9 times 2. This becomes 10 divided by 2 times 2 times 5 plus 18. Now we have 10 divided by 2 times 10 plus 18. Next, we have 10 divided by 2 times 28. 10 divided by 2 becomes 5, so we have 5 times 28. This becomes our answer of 140.
Example calculation
Lesson Summary
We are done! Let's review now. An arithmetic sequence is a string of numbers where the difference between each pair of successive numbers is the same. This difference is also referred to as the common difference. The formula to find the sum of the first n terms of our sequence is n divided by 2 times the sum of twice the beginning term, a, and the product of d, the common difference, and n minus 1. The n stands for the number of terms we are adding together. To use this formula, we label our n, a, and d so we can plug them in and evaluate to get our answer.
Learning Outcomes
Once you've completed this lesson, you'll be able to:
Define arithmetic sequence and common difference
Identify the arithmetic sum formula
Explain how to calculate an arithmetic sum using this formula
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Types of Numbers & Its Classifications 6:56
Graphing Rational Numbers on a Number Line | Chart & Examples 5:02
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The Order of Real Numbers: Inequalities 4:36
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4787 | https://proofwiki.org/wiki/Existence_of_Lowest_Common_Multiple | Existence of Lowest Common Multiple - ProofWiki
Existence of Lowest Common Multiple
From ProofWiki
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Contents
1 Theorem
2 Proof 1
3 Proof 2
4 Proof 3
5 Sources
Theorem
Let a,b∈Z:a b≠0 a,b∈Z:a b≠0.
The lowest common multiple of a a and b b, denoted lcm{a,b}lcm{a,b}, always exists.
Proof 1
We prove its existence thus:
a b≠0⟹|a b|≠0 a b≠0⟹|a b|≠0
Also |a b|=±a b=a(±b)=(±a)b|a b|=±a b=a(±b)=(±a)b.
So the lowest common multiple definitely exists, and we can say that:
0<lcm{a,b}≤|a b|0<lcm{a,b}≤|a b|
Now we prove it is the lowest.
That is:
a∖n∧b∖n⟹lcm{a,b}∖n a∖n∧b∖n⟹lcm{a,b}∖n
Let a,b∈Z:a b≠0,m=lcm{a,b}a,b∈Z:a b≠0,m=lcm{a,b}.
Let n∈Z:a∖n∧b∖n n∈Z:a∖n∧b∖n.
We have:
n=x 1 a=y 1 b n=x 1 a=y 1 b m=x 2 a=y 2 b m=x 2 a=y 2 b
As m>0 m>0, we have:
n n==m q+r:0≤r<|m|=m m q+r:0≤r<|m|=m
⇝⇝r r==n−m q n−m q
==1×n+(−q)×m 1×n+(−q)×m
⇝⇝r r==x 1 a+(−q)x 2 a x 1 a+(−q)x 2 a
==y 1 b+(−q)y 2 b y 1 b+(−q)y 2 b
⇝⇝a a∖∖r r
∧∧b b∖∖r r
Since r<m r<m, and m m is the smallest positivecommon multiple of a a and b b, it follows that r=0 r=0.
So:
∀n∈Z:a∖n∧b∖n:lcm{a,b}∖n∀n∈Z:a∖n∧b∖n:lcm{a,b}∖n
That is, lcm{a,b}lcm{a,b}divides any common multiple of a a and b b.
■◼
Proof 2
Either a a and b b are coprime or they are not.
Let:
a⊥b a⊥b
where a⊥b a⊥b denotes that a a and b b are coprime.
Let a b=c a b=c.
Then:
a∖c,b∖c a∖c,b∖c
where a∖c a∖c denotes that a a is a divisor of c c.
Suppose both a∖d,b∖d a∖d,b∖d for some d∈N>0:d0:d<c.
Then:
∃e∈N>0:a e=d∃e∈N>0:a e=d∃f∈N>0:b f=d∃f∈N>0:b f=d
Therefore:
a e=b f a e=b f
and from Proposition 19 19 of Book VII VII: Relation of Ratios to Products:
a:b=f:e a:b=f:e
But a a and b b are coprime.
From:
Proposition 21 21: Coprime Numbers form Fraction in Lowest Terms
and:
Proposition 20 20: Ratios of Fractions in Lowest Terms
it follows that b∖e b∖e
Since:
a b=c a b=c
and:
a e=d a e=d
it follows from Proposition 17 17: Multiples of Ratios of Numbers that:
b:e=c:d b:e=c:d
But b∖e b∖e and therefore:
c∖d c∖d
But c>d c>d which is impossible.
Therefore a a and b b are both the divisor of no number less than c c.
Now suppose a a and b b are not coprime.
Let f f and e e be the least numbers of those which have the same ratio with a a and b b.
Then from Proposition 19 19: Relation of Ratios to Products:
a e=b f a e=b f
Let a e=c a e=c.
Then b f=c b f=c.
Hence:
a∖c a∖c b∖c b∖c
Suppose a a and b b are both the divisor of some number d d which is less than c c.
Let:
a g=d a g=d
and:
b h=d b h=d
Therefore:
a g=b h a g=b h
and so by Proposition 19 19: Relation of Ratios to Products:
a:b=f:e a:b=f:e
Also:
f:e=h:g f:e=h:g
But f,e f,e are the least such.
From Proposition 20 20: Ratios of Fractions in Lowest Terms:
e∖g e∖g
Since a e=c a e=c and a g=d a g=d, from Proposition 17 17: Multiples of Ratios of Numbers:
e:g=c:d e:g=c:d
But:
e∖g e∖g
So c∖d c∖d
But c>d c>d which is impossible.
Therefore a a and b b are both the divisor of no number less than c c.
■◼
Proof 3
Note that as Integer Divides Zero, both a a and b b are divisors of zero.
Thus by definition 0 0 is a common multiple of a a and b b.
Non-trivial common multiples of a a and b b exist.
Indeed, a b a b and −(a b)−(a b) are common multiples of a a and b b.
Either a b a b or −(a b)−(a b) is strictly positive.
Let S S denote the set of strictly positivecommon multiples of a a and b b.
By the Well-Ordering Principle, S S contains a smallest element.
This can then be referred to as the lowest common multiple of a a and b b.
■◼
Sources
1971:Allan Clark: Elements of Abstract Algebra... (previous)... (next): Chapter 1 1: Properties of the Natural Numbers: §23 γ§23 γ
1980:David M. Burton: Elementary Number Theory(revised ed.)... (previous)... (next): Chapter 2 2: Divisibility Theory in the Integers: 2.3 2.3 The Euclidean Algorithm
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Chinese Remainder Theorem
The Chinese Remainder Theorem is a number theoretic result.
Contents
1 Statement
2 Proof
3 Applicability
3.1 Solving a system of congruences using CRT
4 Extended version of the theorem
5 See Also
5.1 External links
Statement
Let be relatively prime to . Then each residue class mod is equal to the intersection of a unique residue class mod and a unique residue class mod , and the intersection of each residue class mod with a residue class mod is a residue class mod .
This means that if we have we can deduce that and
In other words, suppose you wish to find the least number which leaves a remainder of
such that are all pairwise coprime. Let , and . If the numbers satisfy , for every , then a solution for is
Proof
If , then and differ by a multiple of , so and . This is the first part of the theorem. The converse follows because and must differ by a multiple of and , and . This is the second part of the theorem.
Applicability
Much like the Fundamental Theorem of Arithmetic, many people seem to take this theorem for granted before they consciously turn their attention to it. Its ubiquity derives from the fact that many results can be easily proven mod (a power of a prime), and can then be generalized to mod using the Chinese Remainder Theorem. For instance, Fermat's Little Theorem may be generalized to the Fermat-Euler Theorem in this manner.
General Case: the proof of the general case follows by induction to the above result (k-1) times.
Solving a system of congruences using CRT
In order to solve a system of n congruences, it is typical to solve the first two, then combine that with the third, and so on. So, it suffices to know how solve a system of 2 congruences.
Let the system be (where and are relatively coprime):
Then if we find one value such that satisfies the system, then the solution set consists of . To find such , set . Then, find that satisfy the equality. This is usually easier than brute forcing for .
Let's take an example: First simplify the second equation to using modular inverses. So we have: Then let . A clear solution for this is . Then, is one solution to the system, so is the set of all solutions.
If and are not relatively prime, then let . We split the system as follows: Then, we must check that . If so, simply ignore the 3rd congruence. Now, we have: Now we have a system of 3 congruences, which we can solve for. If is not , then repeat the decomposition. Essentially, decompose until we get a system of pairwise relatively prime congruences. Then solve.
Extended version of the theorem
Suppose one tried to divide a group of objects into , and parts instead and found , and objects left over, respectively. Any number with remainder mod must be odd and any number with remainder mod must be even. Thus, the number of objects must be odd and even simultaneously, which is a contradiction. Thus, there must be restrictions on the numbers to ensure that at least one solution exists. It follows that:
The solution exists if and only if for all where stands for the greatest common divisor. Moreover, in the case when the problem is solvable, any two solutions differ by some common multiple of . (the extended version).
See Also
Modular arithmetic/Introduction
Chicken McNugget Theorem
External links
Here is an AoPS thread in which the Chinese Remainder Theorem is discussed and implemented.
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4789 | https://chemrevise.org/wp-content/uploads/2018/04/12-cie-transition-metals.pdf | N Goalby chemrevise.org 1 12. Transition Metals General properties of transition metals transition metal characteristics of elements Sc Cu arise from one or more incomplete d orbitals in ions Sc 1s22s22p63s23p6 4s23d1 Ti 1s22s22p63s23p6 4s23d2 V 1s22s22p63s23p6 4s23d3 Cr 1s22s22p63s23p6 4s13d5 Mn 1s22s22p63s23p6 4s23d5 Fe 1s22s22p63s23p6 4s23d6 Co 1s22s22p63s23p6 4s23d7 Ni 1s22s22p63s23p6 4s23d8 Cu 1s22s22p63s23p6 4s13d10 Zn 1s22s22p63s23p6 4s23d10 Sc 3+ [Ar] 4s03d0 Ti 3+ [Ar] 4s03d1 V 3+ [Ar] 4s03d2 Cr 3+ [Ar] 4s03d3 Mn 2+ [Ar] 4s03d5 Fe 3+ [Ar] 4s03d5 Co 2+ [Ar] 4s03d7 Ni 2+ [Ar] 4s03d8 Cu 2+ [Ar] 4s03d9 Zn 2+ [Ar] 4s03d10 When forming ions lose 4s before 3d Why is Zn not a transition metal?
Zn can only form a +2 ion. In this ion the Zn2+ has a complete d orbital and so does not meet the criteria of having an incomplete d orbital in one of its compounds.
these characteristics include •complex formation, •formation of coloured ions, •variable oxidation state •catalytic activity.
Why is Sc not a transition metal?
Sc can only form a +3 ion. In this ion the Sc3+ has an empty d orbital and so does not meet the criteria of having an incomplete d orbital in one of its ions.
4 of the 5 d orbitals have this shape. Each one is arranged in a different geometry The 5th d orbital has a different shape Physical properties Melting point and Density: The transition metals have high melting points. They are higher than group 2 element calcium.
This is because they have stronger attractions between cations and electrons (as more delocalised electrons) Transition metals also have higher densities than calcium. This is because they have greater Ar’s and smaller atomic radii.
Variable oxidation states Transition elements show variable oxidation states When transition metals form ions they lose the 4s electrons before the 3d General trends •Relative stability of +2 state with respect to +3 state increases across the period •Compounds with high oxidation states tend to be oxidising agents e.g MnO4 -•Compounds with low oxidation states are often reducing agents e.g V2+ & Fe2+ Transition metals form various oxidation states. They are able to donate and receive electrons and are able to oxidize and reduce. This is because they have many electrons of similar energy in their valence- shell orbital . The 4s + 3d orbitals have similar energies. The energy differences between the oxidation states are small.
Sc Ti V Cr Mn Fe Co Ni Cu Zn +3 +4 +5 +6 +7 +6 +5 +4 +3 +2 +3 +4 +5 +6 +5 +4 +3 +2 +2 +3 +4 +5 +4 +3 +2 +1 +1 +2 +3 +4 +3 +2 +1 +1 +2 +3 +2 +1 +1 +2 +1 +1 N Goalby chemrevise.org 2 Complex formation complex :is a central metal ion surrounded by ligands.
ligand.: An atom, ion or molecule which can donate a lone electron pair.
Co-ordinate bonding is involved in complex formation.
Co-ordinate bonding is when the shared pair of electrons in the covalent bond come from only one of the bonding atoms.
Co-ordination number: The number of co-ordinate bonds formed to a central metal ion.
Cu OH2 OH2 O H2 O H2 OH2 OH2 2+ Ligands can be monodentate (e.g. H2O, NH3 and Cl- ) which can form one coordinate bond per ligand or bidentate (e.g. NH2CH2CH2NH2 and ethanedioate ion C2O4 2- ) which have two atoms with lone pairs and can form two coordinate bonds per ligand or multidentate (e.g. EDTA4- which can form six coordinate bonds per ligand).
Reaction of copper and cobalt complexes Metal aqua ions are formed in aqueous solution.
[Cu(H2O)6]2+ is a blue solution [Co(H2O)6]2+ is a pink solution Reaction with limited amount of OH- and limited NH3 The bases OH- and ammonia when added to aqueous complexes in limited amounts form the hydroxide precipitates. They form in deprotonation acid base reactions Cu(OH)2(H2O)4 (s) blue ppt, Co(OH)2(H2O)4 (s) blue ppt, [Cu(H2O)6]2+ (aq) + 2OH-(aq) Cu(H2O)4(OH)2 (s) + 2H2O (l) Here the NH3 and OH- ions are acting as Bronsted-Lowry bases accepting a proton [Co(H2O)6]2+ (aq) + 2OH-(aq) Co(H2O)4(OH)2 (s) + 2H2O (l) [Cu(H2O)6]2+ (aq) + 2NH3 (aq) Cu(H2O)4(OH)2 (s) + 2NH4 + (aq) [Cu(H2O)6]2+ (aq) + 2NH3 (aq) Cu(H2O)4(OH)2 (s) + 2NH4 + (aq) Ligand Exchange Reactions If excess ammonia is added to cobalt and copper aqueous ions a ligand exchange reaction occurs The ligands NH3 and H2O are similar in size and are uncharged.
Exchange of the ligands NH3 and H2O occurs without change of co-ordination number (eg Co2+ and Cu2+).
[Co(H2O)6]2+ (aq) + 6NH3 (aq)[Co(NH3)6]2+ (aq) + 6H2O (l) This substitution may, however, be incomplete as in the case with Cu.
[Cu(H2O)6]2+ (aq) + 4NH3 (aq)[Cu(NH3)4(H2O)2]2+ (aq)+ 4H2O (l) A pale yellow solution A deep blue solution Reactions with Chloride Ions Addition of a high concentration of chloride ions (from conc HCl or saturated NaCl) to an aqueous ion leads to a ligand exchange reaction.
The Cl- ligand is larger than the uncharged H2O and NH3ligands so therefore ligand exchange can involve a change of co-ordination number.
Addition of conc HCl to aqueous ions of Cu and Co lead to a change in coordination number from 6 to 4.
[CuCl4]2- yellow/green solution [CoCl4]2- blue solution [Cu(H2O)6]2+ + 4Cl- [CuCl4]2- + 6H2O These are tetrahedral in shape [Co(H2O)6]2+ + 4Cl- [CoCl4]2- + 6H2O Be careful: If solid copper chloride (or any other metal) is dissolved in water it forms the aqueous [Cu(H2O)6]2+ complex and not the chloride [CuCl4]2- complex.
Cu(OH)2(H2O)4(s) + 4NH3 (aq)[Cu(NH3)4(H2O)2]2+ (aq)+ 2H2O (l)+ 2OH-(aq) The equation can also be written from the hydroxide precipitate Ligands can be bidentate (e.g. NH2CH2CH2NH2 and ethanedioate ion C2O4 2- ) which have two atoms with lone pairs and can form two coordinate bonds per ligand A complex with Ethane-1-2-diamine bidentate ligands e.g. [Cr(NH2CH2CH2NH2)3]3+ It has a coordination number of 6 Cu(H2O)6 2+ + EDTA4-[Cu(EDTA)]2- + 6H2O Equations to show formation of mutidentate complexes CH2 CH2 N N CH2 CH2 CH2 CH2 C C C C O O O -O -O O-O-O The EDTA4- anion has the formula with six donor sites(4O and 2N) and forms a 1:1 complex with metal(II) ions 3-C C O O C C O O C C O O Cr O O O O O O 3+ H2C NH2 NH2 CH2 H2C NH2 NH2 CH2 CH2 NH2 NH2 Cr CH2 A complex with bidentate ethanedioate ligands e.g. [Cr(C2O4)3]3-Learn the two bidentate ligands mentioned above but it is not necessary to remember the structure of EDTA.
There are 3 bidentate ligands in this complex each bonding in twice to the metal ion.
3 N Goalby chemrevise.org Ligands can be multidentate (e.g. EDTA4- which can form six coordinate bonds per ligand).
Bidentate Ligands Cu(H2O)6 2+ + 3NH2CH2CH2NH2 [Cu(NH2CH2CH2NH2)3]2+ + 6H2O Cu(H2O)6 2+ + 3C2O4 2-[Cu(C2O4)3]4- + 6H2O Equations to show formation of bidentate complexes Octahedral shape with 90o bond angles Ethane-1-2-diamine is a common bidentate ligand.
Ethane-1-2-diamine Ethanedioate Partial substitution of ethanedioate ions may occur when a dilute aqueous solution containing ethanedioate ions is added to a solution containing aqueous copper(II) ions. In this reaction four water molecules are replaced and a new complex is formed.
2-OH2 O O C C O O Cu O O C C O O OH2 Cu(H2O)6 2+ + 2C2O4 2-[Cu(C2O4)2(H2O)2]2- + 4H2O Multidentate Ligands C2O4 2-4 Manganate Redox Titration The redox titration between Fe2+ with MnO4 – (purple) is a very common exercise. This titration is self indicating because of the significant colour change from reactant to product.
MnO4 -(aq) + 8H+ (aq) + 5Fe2+ (aq)Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq) Purple colourless Choosing correct acid for manganate titrations.
The acid is needed to supply the 8H+ ions. Some acids are not suitable as they set up alternative redox reactions and hence make the titration readings inaccurate.
Only use dilute sulphuric acid for manganate titrations.
Insufficient volumes of sulphuric acid will mean the solution is not acidic enough and MnO2 will be produced instead of Mn2+.
MnO4 -(aq) + 4H+(aq) + 3e- MnO2(s) + 2H2O The brown MnO2 will mask the colour change and lead to a greater (inaccurate) volume of Manganate being used in the titration.
Using a weak acid like ethanoic acid would have the same effect as it cannot supply the large amount of hydrogen ions needed (8H+).
It cannot be conc HCl as the Cl- ions would be oxidised to Cl2 by MnO4 - as the Eo MnO4 -/Mn2+ > Eo Cl2/Cl-MnO4 -(aq) + 8H+(aq) + 5e– Mn2+ (aq) + 4H2O(l) E+1.51V Cl2(aq) +2e– 2Cl–(aq) E +1.36V This would lead to a greater volume of manganate being used and poisonous Cl2 being produced.
It cannot be nitric acid as it is an oxidising agent. It oxidises Fe2+ to Fe3+ as Eo NO3 -/HNO2> EoFe3+/Fe2+ NO3 - (aq) + 3H+(aq) + 2e– HNO2(aq) + H2O(l) Eo +0.94V Fe3+ (aq)+e– Fe2+ (aq) Eo+0.77 V This would lead to a smaller volume of manganate being used.
The purple colour of manganate can make it difficult to see the bottom of meniscus in the burette.
If the manganate is in the burette then the end point of the titration will be the first permanent pink colour.
Colourless purple N Goalby chemrevise.org Be able to perform calculations for these titrations and for others when the reductant and its oxidation product are given.
A 2.41g nail made from an alloy containing iron is dissolved in 100cm3 acid. The solution formed contains Fe(II) ions.
10cm3 portions of this solution are titrated with potassium manganate (VII) solution of 0.02M. 9.80cm3 of KMnO4 were needed to react with the solution containing the iron.
What is the percentage of Iron by mass in the nail?
Manganate titration example MnO4 -(aq) + 8H+ (aq) + 5Fe2+ Mn2+ (aq) + 4H2O + 5Fe3+ Step1 : find moles of KMnO4 moles = conc x vol 0.02 x 9.8/1000 = 1.96x10-4 mol Step 2 : using balanced equation find moles Fe2+ in 10cm3 = moles of KMnO4 x 5 = 9.8x10-4 mol Step 3 : find moles Fe2+ in 100cm3 = 9.8x10-4 mol x 10 = 9.8x10-3 mol Step 4 : find mass of Fe in 9.8x10-3 mol mass= moles x RAM = 9.8x10-3 x 55.8 = 0.547g Step 5 : find % mass %mass = 0.547/2.41 x100 = 22.6% 5 Ox C2O4 2- 2CO2 + 2e-Red MnO4 -(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O Overall 2MnO4 -(aq) + 16H+(aq) + 5C2O4 2-(aq) 10CO2(g) + 2Mn2+(aq) + 8H2O(l) With ethanedioate With Iron (II) ethanedioate both the Fe2+ and the C2O4 2- react with the MnO4 -1MnO4 - reacts with 5Fe2+ and 2 MnO4 - reacts with 5C2O4 2-MnO4 -(aq) + 8H+(aq) + 5Fe2+ Mn2+(aq) + 4H2O + 5Fe3+ 2MnO4 -(aq) + 16H+(aq) + 5C2O4 2- 10CO2 + 2Mn2+ (aq) + 8H2O So overall 3MnO4 -(aq) + 24H+(aq) + 5FeC2O4 10CO2 + 3Mn2+ (aq) + 5Fe3+ + 12H2O So overall the ratio is 3 MnO4 - to 5 FeC2O4 The reaction between MnO4 - and C2O4 2- is slow to begin with (as the reaction is between two negative ions). To do as a titration the conical flask can be heated to 60o C to speed up the initial reaction.
N Goalby chemrevise.org A 1.412 g sample of impure FeC2O4.2H2O was dissolved in an excess of dilute sulphuric acid and made up to 250 cm3 of solution. 25.0 cm3 of this solution decolourised 23.45 cm3 of a 0.0189 mol dm–3 solution of potassium manganate(VII).
What is the percentage by mass of FeC2O4.2H2O in the original sample?
Step1 : find moles of KMnO4 moles = conc x vol 0.0189 x 23.45/1000 = 4.43x10-4 mol Step 2 : using balanced equation find moles FeC2O4.2H2O in 25cm3 = moles of KMnO4 x 5/3 (see above for ratio) = 7.39x10-4 mol Step 3 : find moles FeC2O4.2H2O in 250 cm3 = 7.39x10-4 mol x 10 = 7.39x10-3 mol Step 4 : find mass of FeC2O4.2H2O in 7.39x10-3 mol mass= moles x Mr = 7.39x10-3 x 179.8 = 1.33g Step 5 ; find % mass %mass = 1.33/1.412 x100 = 94.1% Reducing Chromium Cr3+ (green) and then Cr2+ (blue) are formed by reduction of Cr2O7 2-(orange) by the strong reducing agent zinc in (HCl) acid solution. Fe2+ is a less strong reducing agent and will only reduce the dichromate to Cr3+ Cr2O7 2- + 14H+ + 3Zn 2Cr3+ + 7H2O + 3 Zn2+ Cr2O7 2- + 14H+ + 4Zn 2Cr2+ + 7H2O + 4 Zn2+ Cr2O7 2- + 14H+ + 6Fe2+ 2Cr3+ + 7H2O + 6 Fe3+ Orange green The Fe2+ and Cr2O7 2- in acid solution reaction can be used as a quantitative redox titration. This does not need an indicator Keeping the zinc/dichromate under a hydrogen atmosphere is needed to reduce it to Cr2+, because O2 in air will oxidise Cr2+ up to Cr3+ Cr2O7 2- + 14H+ + 6e- 2Cr3+ + 7H2O E +1.33V Cr3+(aq) + e– Cr2+(aq) E -0.41V Zn2+ + 2e- Zn E=-0.76V Fe3+(aq) + e– Fe2+(aq) E +0.75V Redox Potentials As zinc has a more negative electrode potential than all the chromium half equations, zinc will reduce chromium down to Cr2+ The electrode potential of Iron(II) is in between the two chromium half equations, so Fe2+ will reduce chromium down to Cr3+ N Goalby chemrevise.org 6 Shapes of complex ions transition metal ions commonly form octahedral complexes with small ligands (e.g. H2O and NH3).
transition metal ions commonly form tetrahedral complexes with larger ligands (e.g.Cl- ).
square planar complexes are also formed, e.g.
cisplatin Ag+ commonly forms linear complexes e.g. [Ag(NH3)2]+ used as Tollen’s Reagent [Co(NH3)6]2 [Cu(H2O)6]2+ [CoCl4]2-Isomerism in complex ions Complexes can show two types of stereoisomerism: cis-trans isomerism and optical isomerism Ni NH3 Cl H3N Cl Ni Cl NH3 H3N Cl Cis-[Cr(H2O )4Cl2]+ trans-Ni(NH3)2Cl2 Cis-trans isomerism in square planar complexes Ag H3N NH3 + Pt Cl Cl H3N H3N Ni NH3 NH3 H3N H3N NH3 NH3 2+ 2-Cu Cl Cl Cl Cl Complexes with 3 bidentate ligands can form two optical isomers (non-superimposable mirror images).
2+ CH2 CH2 NH2 NH2 CH2 CH2 N H2 NH2 C H2 C H2 NH2 NH2 Ni -Optical isomerism in octahedral complexes Cis-trans isomerism in octahedral complexes cis–trans isomerism is a special case of E–Z isomerism Cl H Cr 2+ OH2 OH2 O H2 OH2 ClH Cl H Cr 2+ OH2 OH2 O H2 OH2 ClH Cis-Ni(NH3)2Cl2 trans-[Cr(H2O)4Cl2]+ + + 2+ H2C NH2 NH2 CH2 H2C NH2 NH2 CH2 CH2 NH2 NH2 Ni CH2 N Goalby chemrevise.org 7 The Pt(II) complex cisplatin is used as an anticancer drug.
Pt 2+ Cl -Cl -N H3 N H3 Pt2+ NH3 Cl-Cl-N H3 cisplatin transplatin The cisplatin version only works as two chloride ions are displaced and the molecule joins on to the DNA. In doing this it stops the replication of cancerous cells.
Cisplatin Cisplatin can also prevent the replication of healthy cells by bonding on to healthy DNA which may lead to unwanted side effects like hair loss. Society needs to assess the balance between the benefits and the adverse effects of drugs, such as the anticancer drug cisplatin.
Cisplatin prevents DNA replication in cancer cells by a ligand replacement reaction with DNA in which a dative covalent bond is formed between platinum and a nitrogen atom on guanine Pt NH3 NH3 N NH NH N NH2 O The N and O atoms marked in red can’t bond to cis-platin as they are involved in the bonding within the DNA Formation of coloured ions Colour changes arise from changes in 1.
oxidation state, 2.
co-ordination number 3.
ligand.
Degenerate d orbitals split into two energy levels octahedral and tetrahedral complexes Colour arises from electronic transitions from the ground state to excited states: between different d orbitals.
A portion of visible light is absorbed to promote d electrons to higher energy levels. The light that is not absorbed is transmitted to give the substance colour.
Changing colour Changing a ligand or changing the coordination number will alter the energy split between the d- orbitals, changing E and hence change the frequency of light absorbed.
Ligands cause the 5 d orbitals to split into two energy levels.
Co(H2O)6 2+ + 4Cl- [CoCl4]2- + 6H2O pink blue [Co(NH3)6]2+ (aq) [Co(NH3)6]3+ (aq) +e-yellow brown Compounds without colour Scandium is a member of the d block. Its ion (Sc3+) hasn't got any d electrons left to move around. So there is not an energy transfer equal to that of visible light.
In the case of Zn2+ ions and Cu+ ions the d shell is full e.g.3d10 so there is no space for electrons to transfer.
Therefore there is not an energy transfer equal to that of visible light.
O2 In this equation only oxidation state is changing.
Co(H2O)6 2+ + 6 NH3 Co(NH3)6 2+ + 6H2O Pink yellow brown In this equation both ligand and co-ordination number are changing.
In this equation only the ligand is changing.
How colour arises 8 Equation to learn!
This equation links the colour and frequency of the light absorbed with the energy difference between the split d orbitals.
E = hv.
v = frequency of light absorbed (unit s-1 or Hz) h= Planck’s constant 6.63 × 10–34 (J s) E = energy difference between split orbitals (J) A solution will appear blue if it absorbs orange light. The energy split in the d orbitals E will be equal to the frequency of orange light(5 x1014 s-1) x Planck’s constant E in a blue solution = hv = 6.63 × 10–34 x 5 x1014 = 3.32 × 10–19 J N Goalby chemrevise.org Average energy of degenerate d orbitals in field of ligands dz 2 dx 2 - y 2 dxy dxz dyz E Octahedral complex ion Absorption of visible light is used in spectrometry to determine the concentration of coloured ions.
method •Add an appropriate ligand to intensify colour •Make up solutions of known concentration •Measure absorption or transmission •Plot graph of absorption vs concentration •Measure absorption of unknown and compare If visible light of increasing frequency is passed through a sample of a coloured complex ion, some of the light is absorbed.
The amount of light absorbed is proportional to the concentration of the absorbing species (and to the distance travelled through the solution).
Some complexes have only pale colours and do not absorb light strongly. In these cases a suitable ligand is added to intensify the colour.
Spectrophotometry Spectrometers contain a coloured filter. The colour of the filter is chosen to allow the wavelengths of light through that would be most strongly absorbed by the coloured solution.
Degenerate means all orbitals have the same energy N Goalby chemrevise.org 9 The substitution of monodentate ligand with a bidentate or a multidentate ligand leads to a more stable complex. This is called the chelate effect Stability of complexes [Cu(H2O)6]2+ (aq)+ EDTA4-(aq) [Cu (EDTA)]2-(aq) + 6H2O (l) The copper complex ion has changed from having unidentate ligands to a multidentate ligand.
In this reaction there is an increase in the entropy because there are more moles of products than reactants (from 2 to 7), creating more disorder.
This chelate effect can be explained in terms of a positive entropy change in these reactions as more molecules of products than reactants Free energy ΔG will be negative as ΔS is positive and ΔH is small.
The enthalpy change is small as there are similar numbers of bonds in both complexes.
The stability of the EDTA complexes has many applications. It can be added to rivers to remove poisonous heavy metal ions as the EDTA complexes are not toxic. It is in many shampoos to remove calcium ions present in hard water, so helping lathering.
[Co(NH3)6]2+ + 3NH2CH2CH2NH2 [Co(NH2CH2CH2NH2)3]2+ + 6NH3 This reaction has an increase in entropy because of the increase in moles from 4 to 7 in the reaction. ΔS is positive.
Its enthalpy change ΔH is close to zero as the number of dative covalent and type (N to metal coordinate bond) are the same so the energy required to break and make bonds will be the same.
Therefore Free energy ΔG will be negative and the complex formed is stable.
Stability constants Kstab [Cu(H2O)6]2+ + 4Cl- ⇌[CuCl4]2- + 6H2O A ligand exchange reaction is considered as equilibria then an equilibrium expression can be written [CuCl4 2-aq] .
[Cu(H2O)6 aq 2+] [Cl-aq]4 Kstab = H2O is not included in the expression because it is concentration is assumed to be constant.
The value of K is called the stability constant for complex ions. The larger the stability constant the more stable the complex ion.
Ligand Log Kstab Cl-5.62 NH3 13.1 Edta4-18.8 A complex ion with a small stability constant will not displace the ligand from a more stable complex ion with a larger stability constant.
eg adding ammonia to a complex ion of copper and edta will not result in a colour change |
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Electric Field
Electric Field
Chapter 1 - Fundamentals
What is an electric field?
An electric field is a special state that exists in the space surrounding an electrically charged particle. This special state affects all charged particles placed in the electric field. The true nature of electric fields, as well as the true nature of an electric charge is still unknown to scientists, but the effects of an electric field can be measured and predicted using known equations.
Just like a magnet creates an invisible magnetic field around it, which can be detected by placing a second magnet in its field and measuring the attractive or repulsive force acting on the magnets, electric charges create an electric field which can be detected by using a test charge. When a test charge is placed inside an electric field, an attractive or repulsive force acts upon it. This force is called the Coulomb force. In fact, magnetic and electric fields are not entirely separate phenomena. A magnetic field that changes with time creates - or “induces an electric field, while a moving electric field induces a magnetic field as a direct consequence of the movement. Because these two fields are so tightly connected, the magnetic and electric fields are combined into one, unified, electromagnetic field.
Electric field definition
The electric field can be defined as a vector field which describes the relationship between the charge of a test particle introduced in the field and the force exerted upon this charged test particle.
Where E is the electric field, F is the force exerted on the test particle introduced into the field and q is the charge of the test particle. The unit for electric field is volts per meter [V·m-1] or newtons per coulomb [N·C-1].
The application of electric field in capacitors
Electromagnetism is a science which studies static and dynamic charges, electric and magnetic fields and their various effects. Capacitors are devices which store electrical potential energy using an electric field. As such, capacitors are governed by the rules of electromagnetism. This article will define and outline some of the terms which are needed to understand the workings of capacitors. In this article, it will be considered that the electric field is uniform in all points in space.
Electric potential energy
Electric potential energy is the potential energy of a charged particle in an electric field which results from the Coulomb force acting on the particle. It is defined as the negative of the amount of work needed to bring the particle from a reference point (often infinitely far away) to the point in space where the electric potential energy is measured. The unit for electric potential energy is a joule [J], the same unit used for the amount of work in physics.
Electric potential
Electric potential, also called the electric field potential, is the amount of electric potential energy that a charged particle would have at a certain point in space. The voltage, also called potential difference between two points in space is the difference of the electric potentials of those two points. The unit used for electric potential is a volt [V], named after the italian physicist Alessandro Volta. The same unit is used for voltage. The electric potential between two points in an uniform field is the negative of the field intensity difference between those two points.
Electric field strength
In a simple parallel-plate capacitor, a voltage applied between two conductive plates creates a uniform electric field between those plates. The electric field strength in a capacitor is directly proportional to the voltage applied and inversely proportional to the distance between the plates. This factor limits the maximum rated voltage of a capacitor, since the electric field strength must not exceed the breakdown field strength of the dielectric used in the capacitor. If the breakdown voltage is exceeded, an electrical arc is generated between the plates. This electric arc can destroy some types of capacitors instantly. The standard unit used for electric field strength is volts per meter [V·m-1].
Capacitance
Capacitance represents the ability of a body to store electrical charge. This ability is used in capacitors to store electrical energy by sustaining an electric field. When voltage is applied to a capacitor, a certain amount of positive electric charge (+q) accumulates on one plate of the capacitor, while an equal amount of negative electric charge (-q) accumulates on the other plate of the capacitor. It is defined as:
where C is the capacitance, q is the amount of charge accumulated on the plates and V is the voltage across the two plates of the capacitor.
Capacitance is a function of the capacitor’s geometry. Factors such as the area of the plates, the distance between the plates and the dielectric constant of the dielectric used in the construction of the capacitor all influence the resulting capacitance. In a simple parallel plate, the electrical capacitance is directly proportional to the area of the plates and the dielectric constant, while it is inversely proportional to the distance between the plates. The unit used for capacitance is a Farad [F], named after Michael Faraday, who was the pioneer of research in electricity and magnetism.
Energy stored on a capacitor
Capacitors are devices which are used to store electrical energy in a circuit. The energy supplied to the capacitor is stored in the form of an electric field which is created between the plates of a capacitor. When the voltage is applied across a capacitor, a certain amount of charge accumulates on the plates. The energy stored on the capacitor is:
where W is the energy stored, C is the capacitance and V is the voltage applied across the capacitor.
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4791 | https://www.studocu.com/en-za/messages/question/3491891/find-z02-the-20-point-of-the-standard-normal-distribution | [Solved] Find z02 the 20 point of the Standard Normal Distribution - Intro to statistics (STA1000S) - Studocu
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University of Cape Town
Intro to statistics
Question
Find z02 the 20 point of the Standard Normal Distribution
University of Cape Town
Intro to statistics
Question
JS
### Jemealiane
2 years ago
Find z^(0,2), the 20% point of the Standard Normal Distribution
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To find the 20% point of the Standard Normal Distribution, we need to find the z-score that corresponds to a cumulative probability of 0.20.
The Standard Normal Distribution has a ...
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4792 | https://artofproblemsolving.com/wiki/index.php/Parity?srsltid=AfmBOopuH9-1aBvWIpkRTbRfgc8uFB86FaBIpd_6vjcssyUajTGwJS2V | Art of Problem Solving
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Parity
Parity refers to whether a number is even or odd.
While this may seem highly basic, checking the parity of numbers is often an useful tactic for solving problems, especially with proof by contradictions and casework.
This concept begins with integers. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without remainder; an odd number is an integer that is not evenly divisible by 2. (The old-fashioned term "evenly divisible" is now almost always shortened to "divisible".) A formal definition of an even number is that it is an integer of the form n = 2k, where k is an integer; it can then be shown that an odd number is an integer of the form n = 2k + 1.
This only applies to integers, not fractions or decimals.
Contents
[hide]
1 Problems
2 Solution 1
3 Solution 2
3.1 Introductory
3.2 Intermediate
3.2.1 Problem
3.2.2 Solution
3.2.3 Problem 2
3.2.4 Solution
Problems
Example from 1997 AJHSME:
Problem: Ten balls numbered to are in a jar. Jack reaches into the jar and randomly removes one of the balls. Then Jill reaches into the jar and randomly removes a different ball. The probability that the sum of the two numbers on the balls removed is even is
Solution 1
For the sum of the two numbers removed to be even, they must be of the same parity. There are five even values and five odd values.
No matter what Jack chooses, the number of numbers with the same parity is four. There are nine numbers total, so the probability Jill chooses a number with the same parity as Jack's is .
Solution 2
We find that it is only possible for the sum to be even if the numbers added are both even or odd. We will get an odd number when we add an even and odd. We can use complementary counting to help solve the problem. There are a total of possibilities since Jack can choose numbers and Jill can pick . There are possibilities for the two numbers to be different since Jack can pick any of the numbers and Jill has to pick from numbers in the set with a different parity than the one that Jack picks. So the probability that the sum will be odd is . Subtracting this by one gets the answer (edited by qkddud)
Introductory
Many AMC 8 problems fit this category, help us out by putting problems here!
Intermediate
2000 AIME II Problem 2:
Problem
A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola ?
Solution
Note that and have the same parities, so both must be even. We first give a factor of to both and . We have left. Since there are factors of , and since both and can be negative, this gives us lattice points.
2008 AIME I Problem 3
Problem 2
There exist unique positive integers and that satisfy the equation . Find .
Solution
Completing the square, . Thus by difference of squares.
Since is even, one of the factors is even. A parity check shows that if one of them is even, then both must be even. Since , the factors must be and . Since , we have and ; the latter equation implies that .
Indeed, by solving, we find is the unique solution.
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4793 | https://www.youtube.com/watch?v=IRy-bpILPOE | cotA+cotB+cotC = sqrt(3), then triangle ABC must be eqiilateral
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Question: If Cot A + Cot B + Cot C = √3. Then prove that triangle ABC must be equilateral
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[Music] [Applause] [Music] okay this is a pretty complicated problem I admit to have seen it before having seen it before but let's go through the solution that I am familiar with we're told that we have a triangle where the cotangent of angle a plus the cotangent of angle B plus the cotangent of angle C equals radical three prove that the triangle ABC must be equilateral which in effect is the same as saying that angle a equals angle B equals angle C equals 60 degrees and if all angles are 60 degrees then the cotangent of those angles must also be equal so in effect we want to end up somehow with the fact that the cotangent of a equals the cotangent of B equals the cotangent C and then we can deduce from there that it must be an equilateral triangle okay so that's where we're headed now part one is to establish an identity that it's used in the actual proof and that we're gonna refer to angle a B and C instead of the measure of angle a I'm just going to use the letters a B and C to stand for the measures of the angles so in any triangle if you add two of the angles together the third angle has to be 180 minus that angle and if we take the cotangent of those equal angles [Music] those cotangent will be equal and this is a trig identity the cotangent of the sum of two angles and this is also a trig identity and if we kind of simplify this we multiply both sides by this denominator we get to here distribute the cotangent of C and then get the cotangent terms on the left side and the the one on the right side we're going to use this I this quote identity when we do the actual proof and just to convince yourself that that identity is always true Here I am on desmos where I've labeled some angles a B and C and I've typed in that identity and you can see it's one and it doesn't matter what I do to these angles that value is always equal to one and here's the cotangent of 180 minus C is always equal to the opposite the cotangent of Anglesey okay so let's go to board two and we'll talk through this ugly-looking proof this is the given information what we're going to do is square both sides and when we square both sides getting from here to here I've just made a note over here what how you could do this you could group the first two terms and treat this as the square of I binomial so we'd square the first term plus twice the product of these two terms that's cotangent C times the cotangent of eighty plus the cotangent of B plus the second term squared is this guy out here okay and then you could simplify this term by squaring that binomial and then group the terms accordingly and this is what you end up with but what I have done is remove from here to here I should say we're replacing this expression right here from part one with the number one because that expression is always equal to one subtract three from both sides you get a minus one over here and now here's the really tricky part we have to multiply both sides by two and this minus one that becomes a two but then I kind of picture that's hard to see this is the number one here negative two is the same as negative two times one and what do I replace that one with we're back to using that identity that we established in part one on tab one and now when we group the terms I wrote two cotangent squared a as cotangent squared a plus cotangent squared a and the same for these terms so that I could use each of these in separate expressions so cotangent squared a minus two cotangent a cotangent B plus cotangent squared B that gives me one group and by grouping these terms we end up with something like this and each of those terms and parentheses and parentheses are actually perfect square trinomials so I can write that first one is cotangent a minus cotangent B the quantity squared similarly whoops this should have been a C and cotangent B minus cotangent C squared and now the idea is that if these terms separated by the plus signs are going to have a sum of zero then each in turn must be equal to zero because no matter what's in here if I square it that we're going to either get a positive number or zero and the only way we're going to add three positive together to get zero is if in fact they're all zero so cotangent a minus cotangent B the quantity squared has to equal zero within if I take the square root of both sides that's what I get and then the cotangent of a must equal the cotangent of B similarly we get the cotangent of B must equal the cotangent of C and since we're dealing with angles of a triangle these angles I have to all be equal so angle a equals angle B equals angle C and therefore equilateral this problem is sometimes presented as an if and only if problem which means if it's equilateral then this statement is true also well that's really easy to prove because each angle is 60 and the cotangent of 60 added together three times will give you radical three so Q edie we're done [Music] you |
4794 | https://literarydevices.net/synecdoche/ | Synecdoche - Examples and Definition of Synecdoche
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Synecdoche
Definition of Synecdoche
Synecdoche is a figure of speech in which a part of something is used to signify the whole, or vice-versa. In fact, it’s derived from the Greek word synekdoche: “simultaneous meaning.” As a literary device, synecdoche allows for a smaller component of something to stand in for the larger whole, in a rhetorical manner. Synecdoche can work in the opposite direction as well, in which the larger whole stands in for a smaller component of something. Synecdoche is a helpful device for writers to express a word or idea in a different way by using an aspect of that word or idea. This allows for variation of expression and produces an effect for the reader.
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Writing
For example, a common synecdoche for proposing marriage is to ask for a person’s “hand.” This is a figure of speech in the sense that asking for someone’s hand is for effect, not intended literally. The “hand” in this example of synecdoche is the part that signifies the whole person receiving the marriage proposal and reflects the symbolic placement of a wedding ring. As a literary device, synecdoche is a means for writers to avoid overusing words or phrases and create an artistic form of expression.
Common Examples of Synecdoche
Here are some examples of synecdoche that may be found in everyday expression:
The White House (signifies the U.S. president or executive branch)
Wearing heels (signifies high-heeled shoes)
Green thumb (signifies person who is good at gardening)
The Pentagon (signifies U.S. military leaders)
England (signifies Great Britain)
Boots on the ground (signifies soldiers)
Paper or plastic (signifies type of shopping bag)
Stars and stripes (signifies U.S. flag)
Suits (signifies people in business)
Wheels (signifies a vehicle)
Pearly gates (signifies Heaven)
Behind bars (signifies being in jail)
Threads (signifies clothing)
Tickle the ivories (signifies playing piano keys)
Twinkle toes (signifies a dancer)
Examples of Synecdoche as Character Names
Fictional characters often feature synecdoche in their names to indicate an aspect or part of them that signifies their nature as a whole. These names can be meant in a genuine or ironic way. Here are some examples:
The Brain (Alan from children’s television series “Arthur”)
Fang (Hagrid’s dog from “Harry Potter” series)
Red (Ellis Boyd Redding from “The Shawshank Redemption”)
Hot Lips (Margaret Houlihan from “MASH”)
Whiskers (cat from “Toy Story”)
Spot (dog from “Dick and Jane”)
Stickers (nickname for Lightning McQueen in “Cars”)
Blackbeard (pirate)
Freckles (title character of Gene Stratton-Porter novel)
Two-Face (villain in DC Comics)
Famous Examples of Synecdoche
Think you haven’t heard of any famous synecdoche? Here are some well-known and recognizable examples of this figure of speech:
All hands on deck
Faces in the crowd
Kansas City scored the winning touchdown
Lend me your ears
Keep the change
Have a nice day
You have my heart
Hit the sheets
I know those voices
Keep your eyes up here
Difference Between Synecdoche and Metonymy
Synecdoche and metonymy are often confused. As literary devices, they are similar but distinct from each other. Synecdoche, as a figure of speech, must indicate a relationship in which a part signifies the whole of an entity. Metonymy is also a figure of speech in which one word is used to replace another. However, in metonymy, the words are closely linked rather than one word being a smaller part of the whole word or idea that it represents.
Here is an illustrative example of the difference between synecdoche and metonymy:
“I am at the peak of my career.” This is an example of synecdoche. “Peak” is used here to indicate the highest point of the speaker’s career path. The “peak” is a smaller part of the speaker’s job experience as a whole.
“I have a mountain of work to do.” This is an example of metonymy. “Mountain” is used here as a figure of speech that would be related or closely linked to a “pile” of paperwork. Though the word “mountain” is different than “pile,” they are both associated with one another in terms of meaning.
Both synecdoche and metonymy emphasize relationships between words and ideas. They serve to establish connections for readers as a means of developing greater understanding of concepts and expression through language.
Writing Synecdoche
Overall, as a literary device, synecdoche functions as a means of expressing a “whole” entity or idea in a rhetorical way by utilizing a part of it. This is effective for readers in that synecdoche allows them to think of an object or idea in a different way, in terms of the representation of its parts. Therefore, this can enhance the meaning and understanding of an entity for the reader when synecdoche is properly used.
It’s important that writers understand the distinction between the two basic categories of synecdoche:
Microcosm: synecdoche in which a smaller part of something stands in to refer to the larger whole. For example, “blue hair” is a figure of speech that refers to an older woman. Therefore, a smaller part of a woman’s body is used to represent her as a whole in terms of age.
Macrocosm: synecdoche in which a larger entity is used to refer to a smaller part within it. For example, “Ivy League” is a figure of speech that refers to a group of prestigious colleges in the northeastern United States. Therefore, if someone is called an “Ivy League” student, this indicates that the student attends one of these specific universities.
Here are some ways that writers benefit from incorporating synecdoche into their work:
Create Connections
When writers incorporate synecdoche into their work, it is a useful literary device for creating connections for the reader. For synecdoche to be effective, both as microcosm and macrocosm, the reader must be able to connect the significance of the smaller part to the larger whole. In other words, a writer cannot just choose any part of something and create synecdoche. There must be meaning to the part as it relates to the whole in order for the reader to understand.
Enhance Expression
Synecdoche allows writers to vary and enhance their expression. Such figures of speech can emphasize the way that a part of something represents the whole. Writers can also utilize synecdoche to enhance description and create imagery for the reader. As a result, this literary device allows writers to express ideas in unique ways to create deeper meaning for their readers.
Difference Between Wholes and Parts in Synecdoche
In a synecdoche, whole means when something is big and represents many things for example the battalion excelled in the regiment which means the battalion is a part and the regiment is a whole. In case, it is reversed, the role of the regiment becomes important as it would represent the part that is the battalion.
Use of Synecdoche in Sentences
He is so much fed up with the world that he has stopped meeting even his friends.
The factory has fired all the hands including the main executive body.
america does not want its boots on the ground in its war against a big country such as China or Iran.
More wheels are prying on the roads than the last year.
Some people have suggested another war in the offing during their interview with the New York Times.
Examples of Synecdoche in Literature
Synecdoche is an effective literary device in terms of substituting part of something as a representation of its whole. Here are some examples of synecdoche and the way it adds to the significance of well-known literary works:
Example 1:The Great Gatsby – F. Scott Fitzgerald
It was the kind of voice that the ear follows up and down, as if each speech is an arrangement of notes that will never be played again.
In this quote from Fitzgerald’s novel, the narrator, Nick Carraway, is describing the allure of Daisy’s voice. Fitzgerald incorporates synecdoche in Nick’s description with “the ear” that follows the sound of Daisy’s speech. In this case, Nick means the ear in a rhetorical manner, since there isn’t an actual ear that is literally following the “up and down” of the voice.
However, by using this literary device, Fitzgerald conveys to the reader one of the central themes in the novel. The ear signifies the whole person whose attention and focus is captivated by the sound of Daisy’s voice. Daisy holds several characters “captive” in the novel through her charm and beauty. Yet, these characteristics are simply parts of her. The “whole” of Daisy is hidden by the “parts” she reveals on the surface. Therefore, like synecdoche, her character is both represented by her parts and responded to by others through their parts, such as the ear.
Example 2:I heard a Fly buzz–when I died (Emily Dickinson)
I heard a Fly buzz – when I died –
The Stillness in the Room
Was like the Stillness in the Air –
Between the Heaves of Storm –
The Eyes around – had wrung them dry –
And Breaths were gathering firm
For that last Onset – when the King
Be witnessed – in the Room –
Dickinson utilizes synecdoche to begin her second stanza with “The Eyes around.” These eyes signify the mourners at the poet’s deathbed. By describing the mourners as “Eyes,” Dickinson is able to emphasize the imagery of crying and tears, as well as the notion of voyeurism with respect to the poet’s death. This literary device enhances the vision of the scene in the poem for the reader. In addition, it creates a connection between the poet, mourners, and reader such that the end of the poem is impactful. The final line of the poem is “I could not see to see,” underscored by the use of “Eyes” as a figure of speech for all those witnessing the poet’s death: the mourners, the poet herself, and the reader. No matter how many eyes, none of the witnesses are able to actually “see.”
Example 3: Hamlet – William Shakespeare
‘Tis given out that, sleeping in my orchard,
A serpent stung me; so the whole ear of Denmark
Is by a forged process of my death
Rankly abused: but know, thou noble youth,
The serpent that did sting thy father’s life
Now wears his crown.
In this scene, the Ghost of Hamlet’s father is lamenting his death at the hand of his brother Claudius and the resulting consequences. Shakespeare utilizes synecdoche in his phrase “the whole ear of Denmark” to emphasize the implications of Claudius’s treachery and the impact on the kingdom. Of course, Denmark does not have a “whole ear.” Figuratively, the former King is referring to the lie that people in the kingdom have heard about his death. This is an effective literary device, in that the “ear of Denmark” signifies that the population has collectively heard, and therefore believes, the false story.
In addition, this synecdoche represents a part and the whole Hamlet himself. With his father’s murder, Hamlet is the true king of Denmark, and using his “ear” to listen to the Ghost’s story is a symbolic way of delivering the truth to the kingdom as well.
Example 4: The Scarlet Letter – Nathaniel Hawthorne
They were her countrywomen; and the beef and ale of their native land, with a moral diet not a whit more refined, entered largely into their composition. The bright morning sun, therefore, shone on broad shoulders and well-developed busts, and on round and ruddy cheeks, that had ripened in the far-off island, and had hardly yet grown paler or thinner in the atmosphere of New England.
Here the beef and ale mean course women of the rural areas of New England. Therefore, they represent a part of the community which means they have represented the whole as its part. In this connection, it shows the characters of these women which is a good use of synecdoche.
But the point which drew all eyes, and, as it were, transfigured the wearer—so that both men and women, who had been familiarly acquainted with Hester Prynne, were now impressed as if they beheld her for the first time—was that SCARLET LETTER, so fantastically embroidered and illuminated upon her bosom. It had the effect of a spell, taking her out of the ordinary relations with humanity, and inclosing her in a sphere by herself.
There are two important synecdoches. The first one is all eyes which represent women and the second is the scarlet letter that represents individuals forming humanity. Whereas the first is the whole the second is the part.
Example 5: Beowulf translated – Seamus Heaney
No ring-whorled prow
could up then
and away on the sea.
Wind and water
raged with storms,
wave and shingle
were shackled in ice
until another year
appeared in the yard.
These lines from beowulf show the use of synecdoche through prow that is the representation of the whole ship. It is part of a ship but its use shows as if it is a ship. This is an excellent use of synecdoche.
He had been poorly regarded
for a long time, was taken by the Geats
for less than he was worth: and their lord too
had never much esteemed him in the mead-hall.
They firmly believed that he lacked force,
that the prince was a weakling; but presently
every affront to his deserving was reversed.
Here the mead-hall is another synecdoche in beowulf. It shows the people that means it is a part but represents the whole. This is an excellent use of synecdoche.
Synonyms of Synecdoche
Synecdoche is a figure of speech that doesn’t have exact synonyms, some other literary devices come close to it in meanings such as allegory, alliteration, allusion, analogy, or even anaphora. Even a trope or image could be its nearest synonyms.
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4795 | https://www.thoughtco.com/element-groups-vs-periods-608798 | Skip to content
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Key Takeaways
Groups are vertical columns of elements with the same number of valence electrons and similar properties.
Periods are horizontal rows of elements that share the same highest unexcited electron energy level.
The atomic number increases as you move down a group or across a period.
Groups and periods are two ways of categorizing elements in the periodic table. Periods are horizontal rows (across) the periodic table, while groups are vertical columns (down) the table. Atomic number increases as you move down a group or across a period.
Element Groups
Elements in a group share a common number of valence electrons. For example, all of the elements in the alkaline earth group have a valence of two. Elements belonging to a group typically share several common properties.
The groups in the periodic table go by a variety of different names:
| IUPAC Name | Common Name | Family | Old IUPAC | CAS | notes |
| Group 1 | alkali metals | lithium family | IA | IA | excluding hydrogen |
| Group 2 | alkaline earth metals | beryllium family | IIA | IIA | |
| Group 3 | | scandium family | IIIA | IIIB | |
| Group 4 | | titanium family | IVA | IVB | |
| Group 5 | | vanadium family | VA | VB | |
| Group 6 | | chromium family | VIA | VIB | |
| Group 7 | | manganese family | VIIA | VIIB | |
| Group 8 | | iron family | VIII | VIIIB | |
| Group 9 | | cobalt family | VIII | VIIIB | |
| Group 10 | | nickel family | VIII | VIIIB | |
| Group 11 | coinage metals | copper family | IB | IB | |
| Group 12 | volatile metals | zinc family | IIB | IIB | |
| Group 13 | icoasagens | boron family | IIIB | IIIA | |
| Group 14 | tetrels, crystallogens | carbon family | IVB | IVA | tetrels from the Greek tetra for four |
| Group 15 | pentels, pnictogens | nitrogen family | VB | VA | pentels from the Greek penta for five |
| Group 16 | chalcogens | oxygen family | VIB | VIA | |
| Group 17 | halogens | fluorine family | VIIB | VIIA | |
| Group 18 | noble gases, aerogens | helium family or neon family | Group 0 | VIIIA | |
Another way to group elements is based on their shared properties (in some cases, these groupings do not correspond to the columns in the periodic table). Such groups include alkali metals, alkaline earth metals, transition metals (including rare earth elements or lanthanides and also actinides), basic metals, metalloids or semimetals, nonmetals, halogens, and noble gases. Within this classification system, hydrogen is a nonmetal. The nonmetals, halogens, and noble gases are all types of nonmetallic elements. The metalloids have intermediate properties. All of the other elements are metallic.
Element Periods
Elements in a period share the highest unexcited electron energy level. There are more elements in some periods than others because the number of elements is determined by the number of electrons allowed in each energy sub-level.
There are seven periods for naturally occurring elements:
Period 1: H, He (does not follow the octet rule)
Period 2: Li, Be, B, C, N, O, F, Ne (involves s and p orbitals)
Period 3: Na, Mg, Al, Si, P, S, Cl, Ar (all have at least 1 stable isotope)
Period 4: K, Ca, Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn, Ga, Ge, As, Se, Br, Kr (first period with d-block elements)
Period 5: Rb, Sr, Y, Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag, Cd, In, Sn, Sn, Te, I, Xe (same number of elements as period 4, same general structure, and includes first exclusively radioactive element, Tc)
Period 6: Cs, Ba, La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu, Hf, Ta, W, Re, Os, Ir, Pt, Au, Hg, Tl, Pb, Bi, Po, At, Rn (first period with f-block elements)
Period 7: Fr, Ra, Ac, Th, Pa, U, Np, Pu, Am, Cm, Bk, Cf, Es, Fm, Md, No, Lr, Rd, Db, Sg, Bh, Hs, Mt, Ds, Rg, Cn, Uut, Fl, Uup, Lv, Uus, Uuo (all elements are radioactive; contains heaviest natural elements)
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Helmenstine, Anne Marie, Ph.D. "The Difference Between an Element Group and Period." ThoughtCo, Jun. 9, 2025, thoughtco.com/element-groups-vs-periods-608798.
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Helmenstine, Anne Marie, Ph.D. "The Difference Between an Element Group and Period." ThoughtCo. (accessed August 31, 2025).
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4796 | https://mathcurve.com/courbes2d.gb/hyperbole/hyperboleequilatere.shtml | Rectangular hyperbola
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RECTANGULAR HYPERBOLA
Reduced Cartesian equation:. In the frame turned by an eight of a turn:.
a = semi-major axis;
foci: F(, 0) and F'(-, 0) .
directrices: lines with equation x = and x = -.
Cartesian parametrization: (with; form used in the following).
or (with).
Polar equation: (it is therefore a special case of sinusoidal spiral).
General polar equation of a rectangular hyperbola passing by O:(complex parametrization:)
Curvilinear abscissa:.
Radius of curvature:.
Bifocal equation:.
Tripolar equation : MF.MF' = OM 2, O midpoint of [FF'].
Polar equation (pole F, axis Fx):.
The rectangular hyperbola is the hyperbola for which the axes (or asymptotes) are perpendicular, or with eccentricity. It is to general hyperbolas what the circle is to ellipses.
Here are various geometrical definitions:
1) Conic definition:
The hyperbola is the section of a rectangular cone of revolution (angle at the vertex equal to 90°) by a plane strictly parallel to the axis of the cone.
2) Bifocal definition:
The rectangular hyperbola is the locus of a point M such that the difference of the distances to two fixed points F and F' is equal to times the distance between these two points (see the bifocal equation).
3) Definition by focus and directrix:
The rectangular hyperbola is the locus of the point M such that where H is the projection of M on the directrix (D).
4) Cissoidal definition:
Given two perpendicular lines and a point A outside of these lines, the locus of the points M such that, where P and Q are the two intersection points between the two lines and a variable line passing by A, is the hyperbola passing by A and whose asymptotes are the two initial lines. (we derive easily from this that the hyperbola is the cissoid of the two perpendicular lines).
5) a) Angular definition (special case of stelloid):
Given two different points A and B, the locus of the points M such that the bisectors of the lines (MA) and (MB) have constant directions is the rectangular hyperbola passing by A and B whose asymptotes pass by the middle of [AB] and are parallel to these constant directions.
Mechanical interpretation: a rope is attached at an end to a fixed point A, passes by a pulley B and is maintained by hand at the other end. A bucket hangs from the rope thanks to a pulley between A and B.
The bucket describes a portion of rectangular hyperbola.
See Roguet p 161 (1842)
Curve of the bucket: compare to the curve of the tightrope walker.
b) Angular definition equivalent to the previous one .
Given three distinct points A, B, C the locus of points M verifying (AC, AM) = (BM, BC) (oriented angles of lines) is the unique equilateral hyperbola of diameter [AB] (i.e. centered at middle of (A, B)), possibly degenerate, passing through C. The asymptotes are parallel to the bisectors of (AC) and (BC).
(1) also written: (AB, AM) – (BM, AB) = (AB, AC) – (BC, AB), we can give a third angular definition:
given two distinct points A,B, and a real the locus of the points M verifying measure(AB, AM) – measure(BM, AB) =modis a rectangular hyperbola of diameter [AB].
For example, measure(AB, AM) – measure(BM, AB) =modgives the rectangular hyperbola of vertices A and B.
The previous definition allows the construction opposite [J. Lemaire, hyperbole équilatère, p. 1] of the rectangular hyperbola of vertices A and B from the circle of diameter [AB] .
The point P traveling through the circle, the angles verify and we have, in the case of the figure where, for the triangle (AMB) :. The point M here describes a branch of the hyperbola, the other branch being obtained for.
Analytically, with A(a, 0) and B(-a, 0), the transformation is defined by (harmonic homology of center B and axis x = a), therefore the circleis indeed transformed into the hyperbola.
6) Strophoidal definition:
The rectangular hyperbola is the strophoid of a line (D) with respect to a point O outside (D) and a point A located at infinity on a perpendicular line.
7) Definition as the envelope of a triangle with constant area.
The rectangular hyperbola is the envelope of the segment line [AB] the ends of which move on the orthogonal axes Ox and Oy in such a way that the oriented triangle OAB has a constant area.
The contact point in the middle of [AB].
Therefore, if we consider a cubical container, full of a coloured liquid, an edge of which remains horizontal, the trace left by the liquid on the edges when the container is pivoted is delimited by a hyperbola.
This property still holds in the case where the lines are not orthogonal; the envelope is then any hyperbola.
The 3D generalisation of this problem gives the surface xyz= constant.
Figure showing the four rectangular hyperbolas and, along with the circle.
See also the conical catenary, the Gabriel's horn and the hyperboloids.
next curveprevious curve2D curves3D curvessurfacesfractalspolyhedra
© Robert FERRÉOL 2024 |
4797 | https://www.tutorialspoint.com/cplusplus-perimeter-and-area-of-varignon-s-parallelogram | C++ Perimeter and Area of Varignon’s Parallelogram
Varignon’s Parallelogram is formed by joining midpoints of each side of a quadrilateral.
Let’s say we have a quadrilateral ABCD. The midpoints of each side are P, Q, R, and S. If we connect all the midpoints, it will always form a parallelogram PQRS known as Varignon’s Parallelogram.
In this tutorial, we will discuss how to find the perimeter and area of Varignon’s Parallelogram with the given two diagonals and the area of a quadrilateral, for example −
Approach to Find the Solution
In triangle P and Q are midpoints of AB, AC respectively,
By midpoint theorem, PQ = (1/2)AC
Similarly applying theorem to triangle ADC, RS = (1/2)BD,
So PQ=RS=(1/2)AC and PS=QR=(1/2)BD
The perimeter of PQRS = AC + BD(sum of diagonals)
EF=GH=(1/2)AC and EH=FG=(1/2)BD
For the Area of PQRS, we divide the figure into four triangles, and areas of four triangles are,
A1=(1/4)area of BAD
similarly, A2=(1/4)area of ABC
A3=(1/4)area of BCD
A4=(1/4)area of ACD.
A1 + A2 + A3 + A4 = (1/4)(area of triangles ACD+ABC+BCD+BAD)
= (¼) 2 area of ABCD
= (½) area of quad ABCD
Now A1 + A2 + A3 + A4 = (½) area of quad ABCD
It means A5 = (½) area of quad ABCD
So Area of parallelogram PQRS = (½) area of quad ABCD
Now we can find the perimeter and area of PQRS by just applying the formula using C++.
Example
C++ Code for the Above Approach
Output
Conclusion
In this tutorial, we discussed Varignon's Parallelogram and how to find the area and perimeter. We discussed the derivation of perimeter and area of a parallelogram using the midpoint theorem. We also discussed the C++ program for this problem which we can do with programming languages like C, Java, Python, etc. We hope you find this tutorial helpful.
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4798 | https://chemistry.stackexchange.com/questions/10453/understanding-beta-decay | physical chemistry - Understanding Beta Decay - Chemistry Stack Exchange
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Understanding Beta Decay
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Reading Wikipedia on "Beta Decay" they list the equation of Carbon-14 decaying into N-14. see equation
I presume the 14 refers to total protons and neutrons. The 6 in C, I believe means six protons. In N, I believe the 7 refers to protons. Where did the seventh proton come from the decay of C-14 into N-14? I understand neutrons can decay into protons, but the math doesn't work for me.
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edited Jan 13, 2016 at 15:45
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For all radioactive decay (or other nuclear reaction) of a nuclide into other nuclides, the atomic number Z Z and mass number A A need to be conserved.
X A 1 Z 1 X 2 A 1 2 Z 1 X⟶X A 2 Z 2 X 2 A 2 2 Z 2 Q+X A 3 Z 3 X 2 A 3 2 Z 3 R X Z 1 A 1 X 2 Z 1 2 A 1 X⟶X Z 2 A 2 X 2 Z 2 2 A 2 Q+X Z 3 A 3 X 2 Z 3 2 A 3 R
Z 1=Z 2+Z 3 Z 1=Z 2+Z 3
A 1=A 2+A 3 A 1=A 2+A 3
Additionally, the charges must be conserved. If X A 3 Z 3 X 2 A 3 2 Z 3 R X Z 3 A 3 X 2 Z 3 2 A 3 R is an alpha particle X 2 2 X 2 2 2 2 α X 2+X 2 2 X 2 2 2 2 α X 2+ or X 2 2 X 2 2 2 2 H e X 2+X 2 2 X 2 2 2 2 H e X 2+, then:
X A Z X 2 A 2 Z X⟶X A−2 Z−2 X 2 A−2 2 Z−2 Q X 2−+X 2 2 X 2 2 2 2 H e X 2+X Z A X 2 Z 2 A X⟶X Z−2 A−2 X 2 Z−2 2 A−2 Q X 2−+X 2 2 X 2 2 2 2 H e X 2+
Neutrons decay into protons and electrons (beta particles). Both particles are needed for the math to work. And in order for the math to work, you need to know that the atomic number Z Z is 0 0 for a neutron (no protons), and 1 1 for a proton (1 proton, or the H X+H X+ ion). The electron does not have an atomic number. In fact, it is not that Z Z and charges must be conserved separately. The sum of atomic number and charge is conserved.
X 1 0 X 2 1 2 0 n⟶X 1 1 X 2 1 2 1 p X++X 0 0 X 2 0 2 0 e X−X 0 1 X 2 0 2 1 n⟶X 1 1 X 2 1 2 1 p X++X 0 0 X 2 0 2 0 e X−
For the decay of X 14 6 X 2 14 2 6 C X 6 14 X 2 6 2 14 C into X 14 7 X 2 14 2 7 N X 7 14 X 2 7 2 14 N, we first have a neutron decay into a proton and an electron, which is captured by the newly formed X 14 7 X 2 14 2 7 N X+X 7 14 X 2 7 2 14 N X+. The nitrogen nuclide is initially a cation because carbon only had six electrons, and nitrogen needs seven.
X 14 6 X 2 14 2 6 C⟶X 14 7 X 2 14 2 7 N X++e X−⟶X 14 7 X 2 14 2 7 N X 6 14 X 2 6 2 14 C⟶X 7 14 X 2 7 2 14 N X++e X−⟶X 7 14 X 2 7 2 14 N
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answered May 4, 2014 at 12:51
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4 If the emitted electron is "captured by the newly formed nitrogen", what does a Geiger counter detect?DJohnM –DJohnM 2014-05-06 21:41:47 +00:00 Commented May 6, 2014 at 21:41
Something is really wrong here. I mean wikipedia is talking about N and e- and not N+ and e-. To me your solution is logical because the charges are conserved, but everybody else appear to have the "wrong" equation... :D inf3rno –inf3rno 2016-01-12 05:14:22 +00:00 Commented Jan 12, 2016 at 5:14
Asked it again from physicists: physics.stackexchange.com/questions/228929/…inf3rno –inf3rno 2016-01-12 06:03:59 +00:00 Commented Jan 12, 2016 at 6:03
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By beta decay in the nucleus of the atom a neutron decays into a proton, an electron, and an electron antineutrino and a lot of energy. It will lose a little mass, since E=m c 2 E=m c 2, so the decay energy is coming from the mass we lose. The electron (or beta particle) has a high energy and leaves the atom.
X 1 0 X 2 1 2 0 n⟶X 1 1 X 2 1 2 1 p X++X 0 0 X 2 0 2 0 e X−+ν e¯¯¯¯¯X 0 1 X 2 0 2 1 n⟶X 1 1 X 2 1 2 1 p X++X 0 0 X 2 0 2 0 e X−+ν e¯
By the decay of radiocarbon 8(n)+6(p)=14 8(n)+6(p)=14 the product will be a nitrogen 7(n)+7(p)=14 7(n)+7(p)=14, which means we will have one less neutron and one more proton in the nucleus.
X 14 6 X 2 14 2 6 C⟶X 14 7 X 2 14 2 7 N X++X 0 0 X 2 0 2 0 e X−+ν e¯¯¯¯¯X 6 14 X 2 6 2 14 C⟶X 7 14 X 2 7 2 14 N X++X 0 0 X 2 0 2 0 e X−+ν e¯
The total charge remains zero, since we will have one more electron (beta particle) and one more proton in the system.
According to wikipedia the decay energy of radiocarbon is E d=0.156476 M e V=2.50702189⋅10−17 k J E d=0.156476 M e V=2.50702189⋅10−17 k J. For a mole atoms this means E d⋅N A=1.509764⋅10 7 k J/m o l E d⋅N A=1.509764⋅10 7 k J/m o l. The first ionization energy of nitrogen is 1.402⋅10 3 k J/m o l 1.402⋅10 3 k J/m o l, so the nitrogen cation won't be able to recapture the beta particle, since it will carry too much energy. I am not certain about how big fraction of the decay energy the electron will carry, but certainly much more than the first ionization energy. Since the nitrogen has a relative high electronegativity (3.1), it is likely, that it will get an electron from the environment from an electron donor, for example hydrogen. Radiocarbon labelled biomolecules are likely to fall apart by nuclear transmutations, since the bond energies in these molecules are around 10 2−10 3 k J/m o l 10 2−10 3 k J/m o l, so a single decay can break 10 4−10 5 10 4−10 5 bonds. That's why beta radiations are dangerous.
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edited Jan 13, 2016 at 11:34
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Charge and energy are always conserved. Mass is a form of energy, and what we call 'mass number' - the sum of neutrons + protons - is a close approximation to mass (because electrons are so light, having just over a two-thousandth of the mass of a proton or a neutron).
The conversion of a neutron into a proton when an electron is released takes care of charge conservation, because the charge on the proton is exactly the opposite of the charge on the electron.
The energy needed to create the electron in beta decay comes from the reduced binding energy of the nucleons in N-14 compared with those in C-14.
Does that make sense of the maths for you?
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answered May 4, 2014 at 19:04
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4799 | https://www.endocrinologyadvisor.com/ddi/addisons-disease/ | Addison’s Disease
Etiology & Risk Factors for Addison’s Disease
Presentation of Addison’s Disease
Physical Examination Findings
Diagnostic Workup of Addison’s Disease
Addison’s Disease Differential Diagnosis
Management of Addison’s Disease
Side Effects, Adverse Events, and Drug Interactions
Addison’s Disease Guidelines
Frequently Asked Patient Questions
Addison’s disease — also referred to as primary adrenal insufficiency — occurs when the adrenal glands do not produce enough cortisol and/or aldosterone, hormones necessary for regulation of metabolism, blood pressure, immune system functions, the stress response, and other essential physiologic activities.1
Patients with Addison’s disease experience weakness, fatigue, anorexia, nausea/vomiting, weight loss, hyperpigmentation, electrolyte disturbances, and hypotension.2 The most severe manifestation of Addison’s disease is adrenal crisis, which is a potentially life-threatening emergency unless it is promptly recognized and treated.3 This article describes the causes, presentation, diagnosis, and treatment of Addison’s disease.
Epidemiology
In the general population, Addison’s disease occurs in 1 in 5000 to 1 in 7000 individuals.4 Estimates of the prevalence vary by country; the differences may be due to genetic variation, environmental factors, and differences in each country’s diagnostic capabilities and medical database quality.4 Recent evidence suggests the prevalence of Addison’s disease is increasing, particularly in women, perhaps due to an increased predisposition to autoimmune disease.3,5 Compared with similarly aged healthy women, women with Addison’s disease tend to have lower fertility and parity.4
Adrenal crises are relatively rare, with an estimated incidence of 5 to 10 cases per 100 patient-years.3 Approximately 40% of patients with Addison’s disease experience at least 1 adrenal crisis during their lifetime, and 20% experience more than 1.3
Read about other adrenal conditions such as Adrenal Carcinoma
Etiology & Risk Factors for Addison’s Disease
In the past, Addison’s disease was predominantly caused by adrenal damage from tuberculosis, but as the treatment of tuberculous has improved, this has become less common.6 Today most cases of Addison’s disease in developed countries are the result of autoimmune disorders.6
Three general mechanisms are responsible for the hormonal insufficiency caused by intrinsic adrenal disease2:
Congenital adrenal dysgenesis;
Defective adrenal steroidogenesis; and
Autoimmune destruction of adrenocortical steroid-producing cells (autoimmune adrenalitis).
Congenital adrenal hyperplasia is responsible for most pediatric cases of Addison’s disease.7 In this condition, cortisol deficiency leads to increased levels of adrenocorticotropic hormone (ACTH), which regulates cortisol production. Subsequent development of adrenocortical hyperplasia and accumulation of the upstream precursor steroids precipitate the cascade of associated symptoms.7
Defective adrenal steroidogenesis is characterized by dysregulated production of glucocorticoids and/or mineralocorticoids, resulting in a host of downstream pathologic manifestations.8
The most common cause of Addison’s disease is autoimmune adrenalitis, which appears to result from a complex intertwining of genetic, immunologic, and environmental factors.2
Some evidence suggests that immune checkpoint inhibitors, which block checkpoint proteins from binding with their partner proteins and thereby enable T cells to kill cancer cells, can trigger autoimmune adrenalitis.9 Other medications that might cause Addison’s disease include10:
Mitotane (used to treat adrenocortical carcinoma and refractory Cushing syndrome);
Aminoglutethimide (used to treat breast cancer);
Trilostane (used to treat Cushing syndrome);
Etomidate (an intravenous anesthetic); and
Ketoconazole and fluconazole (antifungal medications).
Another potential cause of Addison’s disease is infections such as HIV, fungal adrenalitis, meningococcal sepsis, and African trypanosomiasis.10 Some evidence suggests viral infections might generate an autoimmune response by triggering inflammation in the adrenals, and that rare variants of innate immune genes may interfere with the body’s natural ability to clear these infections.11 Cancers of the lung, stomach, breast, and colon also might cause Addison’s disease.10
Prognosis
The availability of corticosteroid replacement therapy and advances in diagnostic testing have improved the prognosis of patients with Addison’s disease.5 The prognosis depends on the etiology, the time frame in which Addison’s disease is diagnosed and treated, and any challenges that arise from comorbid conditions. Because in its early stages Addison disease is generally characterized by nonspecific signs and symptoms, the correct diagnosis is often overlooked or delayed.5 The symptoms of Addison’s disease often progress very slowly, which means the disease might not be identified until a severe or acute stressor transforms the condition into a life-threatening adrenal crisis. Patients with Addison disease have lower self-reported quality of life and increased mortality.5
Presentation of Addison’s Disease
Multiple factors affect the clinical presentation of Addison’s disease, including the patient’s age, the extent of the cortisol and/or aldosterone deficiency, and the underlying etiology.10 Individuals with Addison’s disease typically present with extreme fatigue or severe weakness, along with unexplained dehydration, weight loss, fever, and abdominal pain.5
Patients with Addison’s disease may present with pronounced hyperpigmentation of the face, neck, back of the hand, and normally hyperpigmented areas such as the areola, scrotum, and labia.4 Patients may also present with nausea, vomiting, diarrhea, depression, joint pain, irregular or no menstrual periods, loss of interest in sex, and craving for salt.6,10 Hypotension is potentially indicative of Addison’s disease.10
Physical Examination Findings
During the physical examination, in addition to hyperpigmentation, clinicians should look for signs and symptoms that may indicate an adrenal crisis, such as the following3,12:
Hypotension;
Tenderness/pain in the abdomen;
Fever;
Reduced consciousness/delirium/confusion;
Severe weakness;
Nausea/vomiting;
Syncope;
Mood disturbances/behavior changes;
Decreased motivation;
Anorexia; and
Constipation.
Diagnostic Workup of Addison’s Disease
Addison’s disease should be suspected all acutely or chronically ill patients who present with general fatigue or severe weakness as well as unexplained dehydration, hypotension, weight loss, fever, abdominal pain, and hyperpigmentation.5 The index of suspicion should be high in a patient taking any medications that interfere with cortisol, particularly anticonvulsants (such as carbamazepine or phenytoin), antifungals (ketoconazole), medications used to treat cancer, (abiraterone, mitotane, immune checkpoint inhibitors), or certain over-the-counter drugs (St John’s Wort).5
If a physical examination yields evidence of Addison’s disease, laboratory testing is indicated. Measurement of cortisol and ACTH is often the initial test.5 Addison’s disease is likely if cortisol is <5 µg/dL (138 nmol/L) with a concomitant ACTH that is more than 2-fold above the upper limit of the normal range.5
Other laboratory testing findings that suggest Addison’s disease include the following10:
Hyponatremia;
Hyperkalemia;
Hypoglycemia;
Ketosis;
Eosinophilia and lymphocytosis; and
High plasma renin activity.
Dynamic testing of adrenocortical function with a corticotropin stimulation test (also called cosyntropin test, ACTH test, or Synacthen test) is the best established and validated method of diagnosing or ruling out Addison’s disease.5 In this test, patients receive 250 µg of intravenous tetracosactide, and the cortisol level is determined immediately before and 30 and 60 minutes after the injection. Although levels may differ significantly based on the method used for detection, a common cut-off for the diagnosis of Addison’s disease is a peak cortisol concentration of less than 500 nmol/L (18 µg/dL).5
Practitioners should exercise caution in ruling out adrenal insufficiency because untreated symptoms can snowball into a potentially fatal adrenal crisis, which is often brought on by a gastrointestinal infection and marked by severe hypotension, lethargy, abdominal pain, and vomiting.4
Addison’s disease can be challenging to diagnose in children, especially infants. While hyperkalemia, hyponatremia, and hypoglycemia are distinct signs of adrenal insufficiency in infants, ketosis is not always present, and hyperpigmentation is not usually seen.10 Severely ill children may present with normal serum potassium and sodium levels due to intense vomiting, which can cause potassium loss and dehydration.13
Addison’s disease also may be difficult to recognize in a patient who is pregnant because many of the clinical features of adrenal insufficiency are also present during normal pregnancies, including weakness, lightheadedness, syncope, nausea, vomiting, hyponatremia, and increased pigmentation.14
Addison’s Disease Differential Diagnosis
Determining the etiology of a patient’s symptoms is essential to the differential diagnosis. For adults, 21-hydroxylase antibodies are used to test for autoimmune destruction of the adrenal cortex.10 In children, the baseline serum 17-hydroxyprogesterone level is measured to test for congenital adrenal hyperplasia.10 As part of the differential diagnosis, clinicians should look for signs and symptoms of the following conditions15:
Infections (such as tuberculosis, HIV, cytomegalovirus, and fungal infection);
Tumors;
Hemorrhage caused by use of heparin or low-molecular-weight heparin;
Sepsis;
Adrenalectomy;
Genetic mutations (such as adrenoleukodystrophy and familial glucocorticoid deficiency; and
Medications that increase cortisol metabolism, inhibit gene transcription, or alter tissue resistance to glucocorticoids.
Addison’s disease can occur as part of a syndrome characterized by a group of autoimmune disorders, including the following10:
Type 1 autoimmune polyglandular syndrome (APS-1), in which patients present with chronic mucocutaneous candidiasis, hypoparathyroidism, and other autoimmune diseases; and
Type 2 autoimmune polyglandular syndrome (APS-2), in which patients present with autoimmune thyroid disease or type 1 diabetes.
Patients with Addison’s disease may have multiple comorbid conditions, including Hashimoto thyroiditis and Graves disease, type 1 diabetes, celiac disease, and autoimmune gastritis with vitamin B12 deficiency.13 Patients are sometimes misdiagnosed with Addison’s disease because they have other conditions that mimic its symptomatology, particularly gastroenteritis and syndrome of inappropriate secretion of antidiuretic hormone (SIADH).16
In a patient with suspected Addison’s disease, secondary and tertiary adrenal insufficiency should also be considered. In secondary adrenal insufficiency, the pituitary does not produce enough ACTH, which leads to low cortisol production.1 Secondary adrenal insufficiency has many potential causes but most commonly occurs when long-term treatment with a potent glucocorticoid is tapered too rapidly or suddenly stopped.10
In tertiary adrenal insufficiency, the hypothalamus does not produce enough cortisol-releasing hormone, which signals the pituitary to make ACTH.1 Mostly commonly, thisis a consequence of chronic corticosteroid use or illicit use of opiates.13 Other potential causes include inflammatory disorders (such as meningitis and encephalitis) , trauma, radiation therapy, surgery, tumors, and infiltrative diseases (such as sarcoidosis, histiocytosis, and hemochromatosis).13
Some evidence suggests that traumatic brain injury can lead to deficiency in corticotropin secretion, causing adrenal insufficiency and life-threatening hyponatremia and hypotension.17
Management of Addison’s Disease
The mainstay of treatment of Addison’s disease is glucocorticoid and mineralocorticoid replacement therapy.5,13 Patients typically require replacement of both glucocorticoid (hydrocortisone or prednisone) and mineralocorticoid (fludrocortisone) hormones.13
The Table outlines recommended oral doses of glucocorticoid medications; these dosages should be adjusted based on individual patient factors and the need to avoid adrenal suppression and complications.5,13
Table. Recommended Doses of Glucocorticoids for Treating Addison’s Disease5,13
| | | |
---
| Medication | Dosage Recommendation | Considerations/Cautions |
| Hydrocortisone | 15 to 25 mg/d in 2 to 3 doses | Highest dose should be administered in the morning. Last dose should be given 4 to 6 hours prior to bedtime, to avoid insomnia and nocturnal insulin resistance For pregnant women, consider increasing the dose by approximately 30% in last trimester to compensate for the rise in natural cortisol levels associated with pregnancy |
| Prednisolone | 3 to 5 mg/d as single dose or in 2 divided doses | Divided dosing may be preferable for patients who requiring simplified therapeutic regimen to improve comfort and compliance |
In patients with aldosterone deficiency, treatment consists of fludrocortisone at a dosage of 0.05 to 0.2 mg/d, depending on the patient’s age and condition, with dose increases based on blood pressure and electrolyte status (especially during the third trimester of pregnancy).5,10,13
Patients should be encouraged to increase their sodium intake and closely monitor their blood pressure.15 For women who experience chronic low libido and depressive symptoms despite glucose and mineralocorticoid therapy, clinicians might consider supplementing treatment with dehydroepiandrosterone.5
High doses of corticosteroids are linked to a higher risk of osteoporosis.18
Therefore, in addition to an increased sodium intake for those receiving aldosterone replacement, patients may benefit from taking calcium and vitamin D supplements.18
Side Effects, Adverse Events, and Drug Interactions
Although glucocorticoid replacement is the gold standard treatment for Addison’s disease, there are some shortcomings to this approach. Both increased exposure to cortisol and insufficient application of cortisol coverage during infections and other stress-related events may result in increased morbidity and a decreased life expectancy.19 This may reflecting a failure of glucocorticoid treatment to mimic the natural circadian rhythm of cortisol release and to individualize cortisol exposure.19
Because stress can aggravate the symptoms of Addison’s disease, some evidence suggests that glucocorticoid doses should be increased substantially during stressful situations and conditions, such as high body temperature, trauma, surgery, or during any intercurrent illness.13,15,19,20 However, high doses of glucocorticoids, are associated with serious adverse events, including an increased risk of cardiovascular problems, impaired glucose tolerance, and decreased bone formation.20
Complications
Adrenal crisis is a potentially life-threatening complication of Addison’s disease. Clinicians should be alert to conditions or diseases that might cause adrenal crisis in their patients with Addison’s disease. Adrenal crisis is most often precipitated by gastrointestinal infections and other infections with fever, but also can be triggered by surgery, strenuous physical activity, sudden discontinuation of glucocorticoid therapy, and psychic distress.4 Clinicians should always take into account the pernicious role an acute stressor such as a gastrointestinal infection can play in precipitating adrenal crisis.
The most important consideration is to address suspected cases of adrenal crisis as a medical emergency.21 If an adult patient exhibit severe weakness, acute abdominal pain, nausea, vomiting, syncope, and marked confusion, initiate immediate treatment with an intravenous (IV) bolus of hydrocortisone 100 mg followed by 200 to 300 mg every 24 hours as a continuous infusion in 5% glucose saline or bolus 50 mg IV every 6 hours.21 If hydrocortisone is unavailable, use IV methylprednisolone 20 mg/12 hours or dexamethasone 4 mg every 12 hours.21 Infuse 2 to 3 liters of saline solution (0.9%) in the first 24 hours.21 Take ongoing measures to control water balance and manage electrolytes (including sodium and potassium) to avoid complications.21
Addison’s Disease Guidelines
In 2016, the Endocrine Society published a clinical practice guideline for the diagnosis and treatment of primary adrenal insufficiency.22 That guideline includes the following recommendations22:
Adoption of a low diagnostic (and therapeutic) threshold for primary adrenal insufficiency in acutely ill patients, as well as those with predisposing factors;
Use of a short corticotropin test (250 μg) as the gold standard diagnostic tool or, if unavailable, measurement of morning plasma ACTH and cortisol levels;
Diagnosis of the underlying cause should include a validated assay of autoantibodies against 21-hydroxylase;
Treatment of adults with once-daily oral fludrocortisone (median, 0.1 mg/d) and hydrocortisone (15 to 25 mg/d) or cortisone acetate replacement (20 to 35 mg daily) administered in 2 to 3 doses;
Patient education about stress dosing and the need for a steroid card as well as glucocorticoid preparation for parenteral emergency administration; and
Follow-up focused on monitoring corticosteroid dosing and associated autoimmune diseases.
Frequently Asked Patient Questions
What is Addison’s disease?
Addison’s disease, otherwise known as primary adrenal insufficiency, is caused by autoimmune destruction of the adrenal cortex. This results in decreased synthesis of adrenocortical hormones leading to reduced levels of circulating cortisol, aldosterone, and adrenal androgens.
What are the typical clinical features of Addison’s disease?
The loss of cortisol in Addison’s disease produces hypoglycemia, anorexia, nausea and vomiting, weight loss, and weakness. The loss of aldosterone produces hyperkalemia, metabolic acidosis, and hypotension. Additional clinical manifestations of Addison’s disease include hyperpigmentation of the face, neck, back of the hand, areola, scrotum, and labia.23
Adrenocorticotropic hormone (ACTH) regulates cortisol production. Therefore, increased levels of ACTH results in hyperpigmentation in patients with Addison’s disease.23
How is Addison’s disease diagnosed?
A diagnosis of Addison’s disease should be suspected of all patients who present with general fatigue or severe weakness, unexplained dehydration, hypotension, weight loss, fever, abdominal pain, and hyperpigmentation. Suspicion of Addison’s disease should be increased in patients taking medications that interfere with cortisol such as carbamazepine, phenytoin, ketoconazole, abiraterone, mitotane, immune checkpoint inhibitors, or St John’s Wort. Addison’s disease is likely if cortisol is <5 µg/dL (138 nmol/L) with a concomitant ACTH that is more than 2-fold above the upper limit of the normal range. Dynamic testing of adrenocortical function with a corticotropin stimulation test is the ideal diagnostic test for Addison’s disease.
Is Addison’s disease life threatening?
Addison’s disease is not typically thought to be life threatening. But there is evidence that traumatic brain injury can lead to deficiency in the secretion of corticotropin, resulting in adrenal insufficiency (Addison’s disease) and life-threatening hyponatremia and hypotension. Given the slow progression of the symptoms of Addison’s disease, the disease may not be identified until a severe or acute stressor causes a life-threatening adrenal crisis.
How is Addison’s disease treated?
The treatment of Addison’s disease includes glucocorticoid (hydrocortisone or prednisone) and mineralocorticoid (fludrocortisone) hormone replacement therapy.
References
National Institute of Diabetes and Digestive and Kidney Diseases. Definition & facts of adrenal insufficiency & Addison’s disease. Updated September 2018. Accessed March 26, 2024.
Yeh MW,Livhits M, Duh Q-Y. The adrenal glands. In: Townsend CM, Beauchamp D, Evers M, Mattox KL, eds. Sabiston Textbook of Surgery. 21st ed. Elsevier; 2022:964-997.
Punati R, Ong RKS, Bornstein S. Acute adrenal insufficiency. In: Shifrin, A, ed. Endocrine Emergencies. 1st ed. Elsevier; 2022:154-165.
Saverino S, Falorni A. Autoimmune Addison’s disease. Best Pract Res Clin Endocrinol Metab. 2020;34(1):101379. doi:10.1016/j.beem.2020.101379
Barthel A, Benker G, Berens K, et al. An update on Addison’s disease. Exp Clin Endocrinol Diabetes. 2019;127(2-03):165-175. doi:10.1055/a-0804-2715
National Institute of Diabetes and Digestive and Kidney Diseases. Symptoms & causes of adrenal insufficiency & Addison’s disease. Updated September 2018. Accessed March 26, 2024.
Güran T. Latest insights on the etiology and management of primary adrenal insufficiency in children. J Clin Res Pediatr Endocrinol. 2017;9(Suppl 2):9-22. doi:10.4274/jcrpe.2017.S002
Flück CE. MECHANISMS IN ENDOCRINOLOGY: Update on pathogenesis of primary adrenal insufficiency: Beyond steroid enzyme deficiency and autoimmune adrenal destruction. Eur J Endocrinol. 2017;177(3):R99-R111. doi:10.1530/EJE-17-0128
Schonfeld SJ, Tucker MA, Engels EA, et al. Immune-related adverse events after immune checkpoint inhibitors for melanoma among older adults. JAMA Netw Open. 2022;5(3):e223461. doi:10.1001/jamanetworkopen.2022.3461
White PC. Adrenocortical insufficiency. In: Kliegman RM, St Geme JW, Blum NJ, Shah SS, Tasker RC, Wilson KM, eds. Nelson Textbook of Pediatrics. 21st ed. Elsevier; 2020:2959-2970.e1.
Aslaksen S, Wolff AB, Vigeland MD, et al. Identification and characterization of rare toll-like receptor 3 variants in patients with autoimmune Addison’s disease. J Transl Autoimmun. 2019;1:100005. doi:10.1016/j.jtauto.2019.100005
National Center for Advancing Translational Sciences. Genetic and Rare Diseases Information Center. Addison’s disease. Updated February 2023. Accessed March 26, 2024.
Husebye ES, Pearce SH, Krone NP, Kämpe O. Adrenal insufficiency. Lancet. 2021;397(10274):613-629. doi:10.1016/S0140-6736(21)00136-7
Huang W, Molitch ME. Pituitary and adrenal disorders in pregnancy. In: Landon MB, Galan HL, Jauniaux ERM, et al. Gabbe’s Obstetrics: Normal and Problem Pregnancies. 8th ed. Elsevier, 2021:945-953.e3.
Little JW, Miller CS, Rhodus NL. Adrenal insufficiency. In: Little JW, Miller CS, Rhodus NL, eds. Little and Falace’s Dental Management of the Medically Compromised Patient. 9th ed. Elsevier; 2018;15:255-267.
Pintaldi S, Lora A, Vecchiato K, Taddio A, Barbi E. SIADH versus adrenal insufficiency: A life-threatening misdiagnosis. Ital J Pediatr. 2019;45(1):23. doi:10.1186/s13052-019-0614-1
Sabet N, Soltani Z, Khaksari M. Multipotential and systemic effects of traumatic brain injury. J Neuroimmunol. 2021;357:577619. doi:10.1016/j.jneuroim.2021.577619
National Institute of Diabetes and Digestive and Kidney Diseases. Eating, diet, & nutrition for adrenal insufficiency & Addison’s disease. Updated September 2018. Accessed March 26, 2024.
Johannsson G, Falorni A, Skrtic S, et al. Adrenal insufficiency: Review of clinical outcomes with current glucocorticoid replacement therapy. Clin Endocrinol (Oxf). 2015;82(1):2-11. doi:10.1111/cen.12603
Øksnes M, Ross R, Løvås K. Optimal glucocorticoid replacement in adrenal insufficiency. Best Pract Res Clin Endocrinol Metab. 2015;29(1):3-15. doi:10.1016/j.beem.2014.09.009
Araujo Castro M, Currás Freixes M, de Miguel Novoa P, Gracia Gimeno P, Álvarez Escolá C, Hanzu FA. SEEN guidelines for the management and prevention of acute adrenal insufficiency. Endocrinol Diabetes Nutr (Engl Ed). 2020;67(1):53-60. doi:10.1016/j.endinu.2019.01.004
Bornstein SR, Allolio B, Arlt W, et al. Diagnosis and treatment of primary adrenal insufficiency: An Endocrine Society clinical practice guideline. J Clin Endocrinol Metab. 2016;101(2):364-389. doi:10.1210/jc.2015-1710
Endocrine physiology. In: Costanzo, LS. Physiology. 6th ed. Elsevier, Inc; 2018:395-460.
Updated: 6/12/2024
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