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16800 | https://www.cornwallschools.com/site/handlers/filedownload.ashx?moduleinstanceid=2651&dataid=2720&FileName=El%20g_nero%20de%20los%20sustantivos.pdf | El género de los sustantivos Femenino: 1. Generalmente –a 2. Siempre femenino --Letras del alfabeto: La a, la be, la y griega --Las islas: Las Canarias, las Galápagos, las Filipinas 3. –ción, -tad, -dad, -sión, -umbre, -sis, -tud, -ie, -itis: Lección, universidad, crisis, oportunidad, costumbre, liquidación, conversación, visión, conjuntivitis, bronquitis, juventud, colisión, libertad, especie El género masculino: 1. Generalmente –o. 2. --Lagos, montañas, océanos, ríos, mares: El Mediterráneo, el Caribe, el Mar Rojo, el Mar Muerto, el Mar Negro, el Titicaca, el Michigan, el Victoria, el Hudson, el Nilo, los Cordilleras, los Andes, Los Himalayas, los Alpes --los números: El diez, el ocho, el veintiocho, el treinta y tres -- los colores: El verde, el rojo, el azul, el rosado - Los días, los meses: El martes, los martes, el agosto, el mayo, el octubre --Los árboles: El roble, el castaño, el naranjo, el cerezo, el manzano, el peral, el olivo 3. -aje: el equipaje, el pasaje, el personaje -or: el mostrador, el amor, el calor Excepciones: la flor y la labor -án: el alacrán, el tucán, el refrán -ambre: el alambre -ma, -pa, -ta: el mapa, el planeta, el problema, el diploma, el fantasma, el idioma, el tema, el sistema, el programa 4. las palabras compuestas: El cumpleaños, el sacapuntas, el rascacielos, el sacacorchos, el crucigrama, el rompecabezas, el paraguas, el parabrisas, el limpiaparabrisas, el parachoques, el parasol, el lavaplatos El/un se usa con los sustantivos femeninos que empiezan con a o ha: El alma, el arma, el ama, el alba, el arpa, el hada, el ancla, el aula, el águila, el hambre, el ave, el ala, el agua El adjetivo tiene que ser en la forma femenina: el alba bonita, mucha hambre Los sustantivos masculinos que representan a una persona y terminan en –or, -ón, -ín, -és añaden una –a en la forma femenina: El doctor la doctora El inglés la inglesa El bailarín la bailarina El campeón la campeona Los sustantivos que representan los dos géneros son: -ista, -a, -nte: El/la artista, el/la dentista, el/la estudiante, el/la cantante, el/la agente El/la atleta Pero: la presidenta Siempre iguales son: El/la testigo, el/la modelo, el/la joven El artículo fijo con las palabras: El ángel el personaje La estrella la víctima La persona el ser John Hines es una persona muy buena. Siena es el ser muy amable y bondadoso. George Clooney es la estrella del cine muy famosa. Las palabras que tienen formas diferentes en masculino y femenino: El actor la actriz El poeta la poetisa don doña el caballo la yegua el carnero la oveja el conde la condesa el duque la duquesa el emperador la emperatriz el gallo la gallina el hombre la mujer el toro la vaca el varón la hembra el yerno la nuera el rey la reina el príncipe la princesa el héroe la heroína etc. El significado que se cambia: El cometa – comet la cometa – kite El corte – cut la corte – court El capital – money la capital – capital El cura – priest la cura – cure El Papa – Pope la papa – potato El guía – guide la guía – female guide or a guidebook El frente – front la frente forehead El mañana – tomorrow la mañana – morning El orden- sequence la orden – command El pendiente – earring la pendiente -- slope Etc. Los irregulares comunes: La mano la imagen El tranvía la foto (fotografía) El día la moto (motocicleta) Los animales tienen un género predeterminado. Usen hembra o varón para distinguir: La rana La foca El elefante Etc. |
16801 | https://www.webassign.net/question_assets/buelemphys1/chapter10/section10dash5.pdf | Answer to Essential Question 10.4: Quite a number of tools and gadgets exploit torque, in the sense that they enable you to apply a small force at a relatively large distance from an axis, and the tool converts that into a large force acting at a relatively small distance from an axis. Examples include scissors, bottle openers, can openers, nutcrackers, screwdrivers, crowbars, wrenches, wheelbarrows, and bicycles.
10-5 Three Equivalent Methods of Finding Torque EXPLORATION 10.5 – Three ways to find torque A rod of length L is attached to a wall by a hinge. The rod is held in a horizontal position by a string that is tied to the wall and attached to the end of the rod, as shown in Figure 10.12.
Step 1 – In what direction is the torque applied by the string to the rod, about an axis that passes through the hinge and is perpendicular to the page? As we did in previous chapters, it’s a good idea to draw a free-body diagram of the rod (or at least part of a free-body diagram, as in Figure 10.13) to help visualize what is happening. For now the only force we’ll include on the free-body diagram is the force of tension applied by the string (we’ll go on to look at all the forces applied to the rod in Exploration 10.8). Try placing your pen over the picture of the rod. Hold the pen where the hinge is and push on the pen, at the point where the string is tied to the rod, in the direction of the force of tension. You should see the pen rotate counterclockwise. Thus, we can say that the torque applied by the string, about the axis through the hinge, is in a counterclockwise direction.
Note that we are dealing with direction for torque much as we did for angular velocity. The true direction of the torque can be found by curling your fingers on your right hand counterclockwise and placing your hand, little finger down, on the page. When you stick out your thumb it points up, out of the page. This is the true direction of the torque, but for simplicity we can state directions as either clockwise or, as in this case, counterclockwise.
Now we know the direction of the torque, relative to an axis through the hinge, applied by the string, let’s focus on determining its magnitude.
Step 2 – Measuring the distance r in Equation 10.9 along the bar, apply Equation 10.9 to find the magnitude of the torque applied by the string on the rod, with respect to the axis passing through the hinge perpendicular to the page. Finding the magnitude of the torque means identifying the three variables, r, F, and !, in Equation 10.9. In this case we can see from Figure 10.13 that the distance r is the length of the rod, L; the force is the force of tension, ; and the angle ! is the angle between the line of the force (i.e., the string) and the line the distance r is measured along (the rod), so ! is the angle in Figure 10.13. In this case, then, applying Equation 10.9 tells us that the magnitude of the torque is .
Chapter 10 – Rotation I Page 10 - 10 Figure 10.12: A rod attached to a wall at one end by a hinge, and held horizontal by a string.
Figure 10.13: A partial free-body diagram for the rod, showing the force of tension applied to the rod by the string.
Step 3 – Now, determine the torque, about the axis through the hinge that is perpendicular to the page, by first splitting the force of tension into components, and then applying Equation 10.9. Which set of axes should we use when splitting the force into components? The most sensible coordinate system is one aligned parallel to the rod and perpendicular to the rod, giving the two components shown in Figure 10.14. Because the force component that is parallel to the rod is directed at the hinge, where the axis goes through, that component gives a torque of zero (it’s like trying to open a door by pushing on the door with a force directed at the line passing through the hinges). Another way to prove that the force is zero is to apply Equation 10.9 with an angle of 180˚, which means multiplying by a factor of sin(180˚), which is zero.
The torque from the force of tension is associated entirely with the perpendicular component of the force of tension. Now, identifying the three pieces of Equation 10.9 gives a force magnitude of ; a distance measured along the rod of , and an angle of between the line of the perpendicular force component and the line we measured r along. Because , applying Equation 10.9 tells us that the magnitude of the torque from the tension, with respect to our axis through the hinge, is . This agrees with our calculation in Step 2.
Step 4 – Instead of measuring r along the rod, draw a line from the hinge that meets the string (the line of the force of tension) at a 90˚ angle. Apply Equation 10.9 to find the magnitude of the torque applied by the string on the rod, with respect to the axis passing through the hinge, by measuring r along this line. As we can see from Figure 10.15, the r in this case is not L, the length of the rod, but is instead . This result comes from applying the geometry of right-angled triangles. The magnitude of the force, F , is , the magnitude of the full force of tension, and the angle between the line we measure r along and the line of the force is 90˚. This is known as the lever-arm method of calculating torque, where the lever-arm is the perpendicular distance from the axis of rotation to the force. Applying Equation 10.9 gives the magnitude of the torque as , agreeing with the other two methods discussed above.
Key idea for torque: We can find torque in three equivalent ways. It can be found using the whole force and the most obvious distance; after splitting the force into components; or by using the lever-arm method in which the distance from the axis is measured along the line perpendicular to the force. Use whichever method is most convenient in a particular situation. Related End-of-Chapter Exercises: 7, 23, 50.
Essential Question 10.5: Torque can be calculated with respect to any axis. In Exploration 10.5, what is the torque, due to the force of tension, with respect to an axis passing through the point where the string is tied to the wall? In each case, assume the axis is perpendicular to the page.
Chapter 10 – Rotation I Page 10 - 11 Figure 10.14: Splitting the force of tension into a component parallel to the rod, and a component perpendicular to the rod.
Figure 10.15: A diagram showing the lever arm, in which the distance used to find torque is measured from the axis along a line perpendicular to the line of the force. |
16802 | https://tutorial.math.lamar.edu/Solutions/Alg/DividingPolynomials/Prob2.aspx | Algebra - Dividing Polynomials
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Section 5.1 : Dividing Polynomials
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Use long division to divide x 3+2 x 2−3 x+4 x 3+2 x 2−3 x+4 by x−7 x−7.
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Let’s first perform the long division. Just remember that we keep going until the remainder has degree that is strictly less that the degree of the polynomial we’re dividing by, x−7 x−7 in this case. The polynomial we’re dividing by has degree one and so, in this case, we’ll stop when the remainder is degree zero, i.e. a constant.
Here is the long division work for this problem.
x 2+9 x+60 x−7 x 3+2 x 2−3 x+4−(x 3−7 x 2)––––––––––––9 x 2−3 x+4−(9 x 2−63 x)–––––––––––––60 x+4−(60 x−420)–––––––––––––424 x 2+9 x+60 x−7 x 3+2 x 2−3 x+4−(x 3−7 x 2) 9 x 2−3 x+4−(9 x 2−63 x) 60 x+4−(60 x−420)_ 424 Show Step 2
We can put the answer in either of the two following forms.
x 3+2 x 2−3 x+4 x−7=x 2+9 x+60+424 x−7 x 3+2 x 2−3 x+4 x−7=x 2+9 x+60+424 x−7 x 3+2 x 2−3 x+4=(x−7)(x 2+9 x+60)+424 x 3+2 x 2−3 x+4=(x−7)(x 2+9 x+60)+424
Either answer is acceptable here although one may be more useful than the other depending on the application that is being done when you need to actually do the long division.
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16803 | https://openstax.org/books/principles-economics-3e/pages/7-4-production-in-the-long-run | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Principles of Economics 3e
7.4 Production in the Long Run
Principles of Economics 3e7.4 Production in the Long Run
Search for key terms or text.
Learning Objectives
By the end of this section, you will be able to:
Understand how long run production differs from short run production.
In the long run, all factors (including capital) are variable, so our production function is .
Consider a secretarial firm that does typing for hire using typists for labor and personal computers for capital. To start, the firm has just enough business for one typist and one PC to keep busy for a day. Say that’s five documents. Now suppose the firm receives a rush order from a good customer for 10 documents tomorrow. Ideally, the firm would like to use two typists and two PCs to produce twice their normal output of five documents. However, in the short turn, the firm has fixed capital, i.e. only one PC. The table below shows the situation:
| | | | | | | | |
--- --- --- --- |
| # Typists (L) | 1 | 2 | 3 | 4 | 5 | 6 | |
| Letters/hr (TP) | 5 | 7 | 8 | 8 | 8 | 8 | For K = 1PC |
| MP | 5 | 2 | 1 | 0 | 0 | 0 | |
Table 7.11 Short Run Production Function for Typing
In the short run, the only variable factor is labor so the only way the firm can produce more output is by hiring additional workers. What could the second worker do? What can they contribute to the firm? Perhaps they can answer the phone, which is a major impediment to completing the typing assignment. What about a third worker? Perhaps the third worker could bring coffee to the first two workers. You can see both total product and marginal product for the firm above. Now here’s something to think about: At what point (e.g., after how many workers) does diminishing marginal productivity kick in, and more importantly, why?
In this example, marginal productivity starts to decline after the second worker. This is because capital is fixed. The production process for typing works best with one worker and one PC. If you add more than one typist, you get seriously diminishing marginal productivity.
Consider the long run. Suppose the firm’s demand increases to 15 documents per day. What might the firm do to operate more efficiently? If demand has tripled, the firm could acquire two more PCs, which would give us a new short run production function as Table 7.12 below shows.
| | | | | | | | |
--- --- --- --- |
| # Typists (L) | 1 | 2 | 3 | 4 | 5 | 5 | |
| Letters/hr (TP) | 5 | 6 | 8 | 8 | 8 | 8 | For K = 1PC |
| MP | 5 | 2 | 1 | 0 | 0 | 0 | |
| Letters/hr (TP) | 5 | 10 | 15 | 17 | 18 | 18 | For K = 3PC |
| MP | 5 | 5 | 5 | 2 | 1 | 0 | |
Table 7.12 Long Run Production Function for Typing
With more capital, the firm can hire three workers before diminishing productivity comes into effect. More generally, because all factors are variable, the long run production function shows the most efficient way of producing any level of output.
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SAT Math
Course: SAT Math>Unit 4
Lesson 2: Radicals and rational exponents: foundations
Radicals and rational exponents | Lesson
Radicals and rational exponents — Basic example
Radicals and rational exponents — Harder example
Radicals and rational exponents: foundations
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Radicals and rational exponents: foundations
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Radicals and rational exponents | Lesson
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A guide to radicals and rational exponents on the digital SAT
What are radicals and rational exponents?
Exponential expressions are algebraic expressions with a coefficient, one or more variables, and one or more exponents. For example, in the expression 3 x 4:
3 is the coefficient.
x is the base.
4 is the exponent.
In 3 x 4, 3 is multiplied by x4 times:
3 x 4=3⋅(x⋅x⋅x⋅x)
An expression can also be raised to an exponent. For example, for (3 x)4, the expression 3 x is multiplied by itself 4 times:
(3 x)4=3 x⋅3 x⋅3 x⋅3 x=81 x 4
Notice how 3 x 4≠(3 x)4 !
Rational exponents refer to exponents that are/can be represented as fractions: 1 2, 3, and −2 3 are all considered rational exponents. Radicals are another way to write rational exponents. For example, x 1 2 and x are equivalent.
In this lesson, we'll:
Review the rules of exponent operations with integer exponents
Apply the rules of exponent operations to rational exponents
Make connections between equivalent rational and radical expressions
You can learn anything. Let's do this!
What are the rules of exponent operations?
Powers of products & quotients (integer exponents)
Khan Academy video wrapper
Powers of products & quotients (integer exponents)See video transcript
The rules of exponent operations
Adding and subtracting exponential expressions
When adding and subtracting exponential expressions, we're essentially combining like terms. That means we can only combine exponential expressions with both the same base and the same exponent.
a x n±b x n=(a±b)x n
Examples
2 x 2+x 2=(2+1)x 2=3 x 2
5 x 3−4 x 3=(5−4)x 3=x 3
4 x 3 and 4 x 2 cannot be combined because the two terms have the same base, but not the same exponent.
2 x 2 and 2 y 2 cannot be combined because the two terms have the same exponent, but not the same base.
Multiplying and dividing exponential expressions
When multiplying two exponential expressions with the same base, we keep the base the same, multiply the coefficients, and add the exponents. Similarly, when dividing two exponential expressions with the same base, we keep the base the same and subtract the exponents.
a x m⋅b x n=a b⋅x m+n a x m b x n=a b⋅x m−n
Examples
3 x⋅4 x 2=(3⋅4)⋅x 1+2=12 x 3
10 x 3 5 x 2=10 5⋅x 3−2=2 x
When multiplying or dividing exponential expressions with the same exponent but different bases, we multiply or divide the bases and keep the exponents the same.
x n⋅y n=(x y)n x n y n=(x y)n
Examples
3 2⋅4 2=(3⋅4)2=12 2=144
6 4 2 4=(6 2)4=3 4=81
Raising an exponential expression to an exponent and change of base
When raising an exponential expression to an exponent, raise the coefficient of the expression to the exponent, keep the base the same, and multiply the two exponents.
(a x m)n=a n⋅x m n
Examples
(x 2)3=x 2⋅3=x 6
(2 x 5)2=2 2⋅x 5⋅2=4 x 10
3⋅(x 4)3=3⋅x 4⋅3=3 x 12
When the bases are numbers, we can use a similar rule to change the base of an exponential expression.
(a b)n=a b n
This is useful for questions with multiple terms that need to be written in the same base.
Example
Rewrite 3 x⋅9 6 as a single exponential expression with a base of 3.
We need to rewrite 9 6 as an expression with a base of 3 and multiply it by 3 x. Since 9=3 2:
3 x⋅9 6=3 x⋅(3 2)6=3 x⋅3 2⋅6=3 x⋅3 12=3 x+12
Negative exponents
A base raised to a negative exponent is equivalent to 1 divided by the base raised to the
opposite
of the exponent.
x−n=1 x n
Examples
2 x−3=2⋅1 x 3=2 x 3
x 5 x 7=x 5−7=x−2=1 x 2
x−3 y−4=1 x 3÷1 y 4=1 x 3⋅y 4 1=y 4 x 3
Zero exponent
A nonzero base raised to an exponent of 0 is equal to 1.
x 0=1,x≠0
How do the rules of exponent operations apply to rational exponents?
Every rule that applies to integer exponents also applies to rational exponents.
Try it!
try: divide two rational expressions
In order to divide 12 x 5 2 by 3 x 1 2, we
operation
the coefficients and
operation
the exponents of x.
12 x 5 2 3 x 1 2=
Check Explain
Try: raise an exponential expression to an exponent
To calculate (2 y 4 3)3, we
coefficient operation
and
operation
the exponents 4 3 and 3.
(2 y 4 3)3=
Check Explain
How are radicals and fractional exponents related?
Rewriting roots as rational exponents
Khan Academy video wrapper
Rewriting roots as rational exponentsSee video transcript
Roots and rational exponents
Squares and square roots are inverse operations: they "undo" each other. For example, if we take the square root of 3 squared, we get 3 2=3.
The reason for this becomes more apparent when we rewrite square root as a fractional exponent: x=x 1 2, and 3 2=(3 2)1 2=3 1.
When rewriting a radical expression as a fractional exponent, any exponent under the radical symbol (x) becomes the numerator of the fractional exponent, and the value to the left of the radical symbol (e.g., A x 3) becomes the denominator of the fractional exponent. Square root is equivalent to A x 2.
A x m n=x m n
Examples
A x 2 3=x 2 3
16 3 2=16 3
All of the rules that apply to exponential expressions with integer exponents also apply to exponential expressions with fractional exponents. Similarly, for radical expressions:
A x n⋅A y n=A x y n A x n A y n=A x y n
When working with radical expressions with the same radical, we can choose whether to convert to fractional exponents first or multiply what's under the radical symbol first to our advantage.
Example
x 3⋅x=x 3 2⋅x 1 2=x 3 2+1 2=x 2
3⋅27=3⋅27=81=9
A 8 x 3 y 6 3=A 8 3⋅A x 3 3⋅A y 6 3=8 1 3⋅(x 3)1 3⋅(y 6)1 3=(2 3)1 3⋅(x 3)1 3⋅(y 6)1 3=2 3⋅1 3⋅x 3⋅1 3⋅y 6⋅1 3=2 1⋅x 1⋅y 2=2 x y 2
Try it!
Try: determine equivalent expressions
Determine whether each of the radical expressions below is equivalent to x 3 2 y 1 3.
| | Equivalent | Not equivalent |
--- |
| x 3⋅A y 3 | | |
| x y | | |
| x 3 y | | |
| A x 9 y 2 6 | | |
Check Explain
Your turn!
Practice: multiply rational expressions
Which of the following is equivalent to 2 x 3⋅3 x 5 ?
Choose 1 answer:
Choose 1 answer:
(Choice A) 5 x 8 A 5 x 8
(Choice B) 6 x 8 B 6 x 8
(Choice C) 6 x 15 C 6 x 15
(Choice D) 8 x 15 D 8 x 15
Check Explain
Practice: change bases
If a b 2=25 for positive integers a and b, what is one possible value of b ?
Check Explain
Practice: raise to a negative exponent
If n−1 3=x, where n>0, what is n in terms of x ?
Choose 1 answer:
Choose 1 answer:
(Choice A) −1 x 3 A −1 x 3
(Choice B) 1 x 3 B 1 x 3
(Choice C) −A x 3 C −A x 3
(Choice D) A x 3 D A x 3
Check Explain
Practice: simplify radical expressions
A 8 x 8 y 6 3 4 x 2 y 6
Which of the following is equivalent to the expression above?
Choose 1 answer:
Choose 1 answer:
(Choice A) x A x
(Choice B) x A x 2 3 y B x A x 2 3 y
(Choice C) A 2 3 x 2 C A 2 3 x 2
(Choice D) 2 x 3 D 2 x 3
Check Explain
Things to remember
Adding and subtracting exponential expressions:
a x n±b x n=(a±b)x n
Multiplying and dividing exponential expressions:
a x m⋅b x n=a b⋅x m+n a x m b x n=a b⋅x m−n x n⋅y n=(x y)n x n y n=(x y)n
Raising an exponential expression to an exponent and change of base:
(a x m)n=a n⋅x m n(a b)n=a b n
Negative exponent:
x−n=1 x n
Zero exponent:
x 0=1,x≠0
All of the rules that apply to exponential expressions with integer exponents also apply to exponential expressions with fractional exponents.
A x m n=x m n A x n⋅A y n=A x y n A x n A y n=A x y n
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First you should distribute the cube root on the top and the square root on the bottom. A cube root is the same as an exponent of 1/3, and a square root is the same as an exponent of 1/2. So you can get rid of them by raising everything in the top row to the 1/3 power and everything in the bottom row to the 1/2 power. The equation now looks like this:
On top: 8^(1/3) · x^(8/3) · y^(6/3)
On bottom: 4^(1/2) · x^(2/2) · y^(6/2)
Now you can simplify some of the fraction exponents just like you'd simplify any fraction:
On top: 8^(1/3) · x^(8/3) · y^(2)
On bottom: 4^(1/2) · x^(1) · y^(3)
8^(1/3) is the same as the cube root of 8. That equals 2. If you didn't know that, you can plug 8^(1/3) into your calculator and you'll get an answer.
4^(1/2) is the same as the square root of 4. That's 2. So now we have this:
On top: 2 · x^(8/3) · y(2)
On bottom: 2 · x^(1) · y^(3)
Now you can simplify each of the variables in the problem. You do this by dividing the top by the bottom.
2 divided by 2 is 1, so the 2's cancel out.
To divide numbers with exponents, you have to subtract the exponents. So for x, you'd do 8/3 minus 1. This is just like subtracting any other fraction: You need to get a common denominator. So 1 becomes 3/3. 8/3 minus 3/3 equals 5/3. So now you have x^(5/3) on the top.
To simplify y, you subtract: 2 minus 3. That equals -1. So you have y^1 on the bottom, or just y.
On top: x^(5/3)
On bottom: y
Now you need to simplify the x^(5/3). (I'm not sure if Khan Academy has taught this in this course yet, but the process I'm using is called "simplify radicals.") This is the same as the cube root of (x^5). Inside the radical you can write five 'x's because that's what x^5 is: x·x·x·x·x. Now, because it's a cube root on the outside, you can circle groups of three 'x's from inside. That gives you one group of three, as well as two 'x's that don't fit in the group. Now you can take out the group. You do this by getting rid of all three 'x's from the inside and putting one of them outside the radical, in front of it. The other two 'x's remaining inside (that didn't fit in a group) stay inside the radical.
On top: x ∛(xx)
On bottom: y
(Edited to add that you didn't technically NEED to simplify the 5 out of the x^(5/3), but for this specific question, all of the answer options are in terms of it being simplified rather than being left alone. After I'd gotten it down to x^(5/3), I looked at the answer options and saw that Khan Academy expected it to be simplified. So that's why I did that step.)
x times x is x^2. So you can simplify the inside of the radical and get the answer:
On top: x · ∛(x^2)
On bottom: y
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christina d. a year ago Posted a year ago. Direct link to christina d.'s post “this is easily the worst ...” more this is easily the worst unit fml bro
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LiteraryPlover a year ago Posted a year ago. Direct link to LiteraryPlover's post “It shouldn't be! Here are...” more It shouldn't be! Here are some simple tips to help you:
Product Property -> When multiplying two numbers with the same base, add the exponents.
Quotient Property -> When dividing two numbers with the same base, subtract the exponents in the numerator and denominator.
Power of a Power Property -> When an exponent is raised to the power of another exponent, multiply the
exponents together.
Zero Exponent Rule -> Any number raised to the power of zero equals one.
Negative Exponent Rule -> A number with a negative exponent can be rewritten as a quotient with a positive exponent in the denominator.
The root of a number is a factor that, when multiplied by itself a specified number of times, equals the number under the radical symbol.
Roots are indicated with radical signs. Radical signs differ slightly depending on degree and include square roots, cube roots, fourth roots, and so on
The radicand refers to the number or quantity under the radical sign.
Hope this helps!
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Lara Helme Altaee 2 years ago Posted 2 years ago. Direct link to Lara Helme Altaee's post “my DSAT is in august any ...” more my DSAT is in august any tips ? please help your sister out its my first time taking any sat and im rlly nervous. i took a practice test and got 1050 any tips?
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Jang 2 years ago Posted 2 years ago. Direct link to Jang's post “Just pick the CORRECT ans...” more Just pick the CORRECT answers on English, not the seem-like-correct ones. For math, just practice on what you are not sure before the test
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jasminecoca5 2 years ago Posted 2 years ago. Direct link to jasminecoca5's post “We call this lesson at ou...” more We call this lesson at our school surds and indices. It's pretty much the same. I've researched it. It's just a matter of naming. So for anyone here who learnt this lesson as surds and indices don't worry the content is the exact same.
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16805 | http://www.kurims.kyoto-u.ac.jp/EMIS/journals/EJC/Volume_12/PDF/v12i1r1.pdf | Sequentially perfect and uniform one-factorizations of the complete graph Jeffrey H. Dinitz Mathematics and Statistics University of Vermont, Burlington, VT, USA 05405 Jeff.Dinitz@uvm.edu Peter Dukes Mathematics and Statistics University of Victoria, Victoria, BC, Canada V8W 3P4 dukes@oddjob.math.uvic.ca Douglas R. Stinson School of Computer Science University of Waterloo, Waterloo, ON, Canada N2L 3G1 dstinson@uwaterloo.ca Submitted: Aug 23, 2004; Accepted: Oct 12, 2004; Published: Jan 7, 2005 Mathematics Subject Classifications: 05C70 Abstract In this paper, we consider a weakening of the definitions of uniform and perfect one-factorizations of the complete graph. Basically, we want to order the 2n −1 one-factors of a one-factorization of the complete graph K2n in such a way that the union of any two (cyclically) consecutive one-factors is always isomorphic to the same two-regular graph. This property is termed sequentially uniform; if this two-regular graph is a Hamiltonian cycle, then the property is termed sequentially perfect. We will discuss several methods for constructing sequentially uniform and sequentially perfect one-factorizations. In particular, we prove for any integer n ≥1 that there is a sequentially perfect one-factorization of K2n. As well, for any odd integer m ≥1, we prove that there is a sequentially uniform one-factorization of K2tm of type (4, 4, . . . , 4) for all integers t ≥2 + ⌈log2 m⌉(where type (4, 4, . . . , 4) denotes a two-regular graph consisting of disjoint cycles of length four).
1 Introduction A one-factor of a graph G is a subset of its edges which partitions the vertex set. A one-factorization of a graph G is a partition of its edges into one-factors. Any one-factorization the electronic journal of combinatorics 12 (2005), #R1 1 of the complete graph K2n has 2n−1 one-factors, each of which has n edges. For a survey of one-factorizations of the complete graph, the reader is referred to , or .
A one-factorization {F0, . . ., F2n−2} of K2n is sequentially uniform if the one-factors can be ordered (F0, . . . , F2n−2) so that the graphs with edge sets Fi ∪Fi+1 (subscripts taken modulo 2n −1) are isomorphic for all 0 ≤i ≤2n −2. Since the union of two one-factors is a 2-regular graph which is 2-edge-colorable, it is isomorphic to a disjoint union of even cycles. We say the multiset T = (k1, . . . , kr) is the type of a sequentially uniform one-factorization if Fi∪Fi+1 is isomorphic to the disjoint union of cycles of lengths k1, . . . , kr, where k1 + · · ·+ kr = 2n. When the union of every two consecutive one-factors is a Hamiltonian cycle, the one-factorization is said to be sequentially perfect.
The idea to consider orderings of the one-factors in a one-factorization of K2n is not entirely academic. In fact, an ordered one-factorization of K2n is a schedule of play for a round-robin tournament (played in 2n −1 rounds). Round-robin tournaments possessing certain desired properties have been studied (see [15, Chapter 5], or ); however, to our knowledge, round robin tournaments with this “uniform” property have not been considered previously.
The definition above is a relaxation of the definition of uniform (perfect) one-factoriz-ation of K2n, which requires that the union of any two one-factors be isomorphic (Hamil-tonian, respectively). Much work has been done on perfect one-factorizations of K2n; for a survey, see Seah . Perfect one-factorizations of K2n are known to exist whenever n or 2n −1 is prime, and when 2n = 16, 28, 36, 40, 50, 126, 170, 244, 344, 730, 1332, 1370, 1850, 2198, 3126, 6860, 12168, 16808, and 29792 (see ). Recently a few new perfect one-factorizations have been found, the smallest of which is in K530 (see [4, 9, 16]); how-ever before this, no new perfect one-factorization of K2n had been found since 1992 ().
The smallest value of 2n for which the existence of a perfect one-factorization of K2n is unknown is 2n = 52. We will show that sequentially perfect one-factorizations are much easier to produce and indeed we will produce a sequentially perfect one-factorization of K2n for all n ≥1.
Various uniform one-factorizations have been constructed from Steiner triple systems, . For instance, when n = 2m for some positive m, the so-called binary projective Steiner triple systems provide a construction of uniform one-factorizations of K2n of type (4, 4, . . . , 4). There are also sporadic examples of perfect Steiner triple systems, , which give rise to uniform one-factorizations of type (2n −4, 4). When v = 3m, uniform one-factorizations (4, 6, . . ., 6) exist (these are Steiner one-factorizations from Hall triple sys-tems) and when p is an odd prime there is a uniform one-factorization of Kps+1 of type (p + 1, 2p, . . ., 2p) which arises from the elementary abelian p-group (see ).
The remainder of this paper is organized as follows. In Section 2, we review the classical “starter” construction for one-factorizations and we show that sequentially perfect one-factorizations of K2n exist for all n. In Section 3, we summarize existence results obtained by computer for small orders. In Section 4, we investigate the construction of sequentially uniform one-factorizations from so-called quotient starters in noncyclic abelian groups.
Here we obtain interesting number-theoretic conditions that determine if the resulting one-factorizations can be ordered so that they are sequentially uniform. In Section 5, the electronic journal of combinatorics 12 (2005), #R1 2 we present a recursive product construction which yields infinite classes of sequentially uniform one-factorizations of K2tm of type (4, 4, . . ., 4), for any odd integer m.
2 Starters We describe our main tool for finding sequentially uniform one-factorizations. Let Γ be an abelian group of order 2n −1, written additively. A starter in Γ is a set of n −1 pairs S = {{x1, y1}, . . . , {xn−1, yn−1}} such that every nonzero element of Γ appears as some xi or yi, and also as some difference xj −yj or yj −xj. Let S∗= S ∪{{0, ∞}} and define x + ∞= ∞+ x = ∞for all x ∈Γ. Then {S∗+ x : x ∈Γ} forms a one-factorization of K2n (with vertex set Γ ∪{∞}).
Many of the known constructions for (uniform and perfect) one-factorizations use starters in this way. In our first lemma we note the connection between starter-induced one-factorizations and sequentially uniform one-factorizations. Clearly, the order in which the 1-factors are listed is essential to the type of a sequentially uniform one-factorization.
Thus we will sometimes refer to ordered one-factorizations in this context. Whenever we discuss sequentially uniform one-factorizations, we will always give the 1-factor ordering.
Lemma 2.1. Let S be a starter in Z2n−1 with n ≥1. Then the ordered one-factorization of K2n generated by S, namely (S∗, S∗+1, S∗+2, . . ., S∗+(2n−2)) is sequentially uniform.
Proof: For any x ∈Z2n−1, we have (S∗+ x) ∪(S∗+ (x + 1)) = x + (S∗∪(S∗+ 1)), so all unions of two consecutive one-factors in the given order are isomorphic.
2 Remark: When gcd(k, 2n −1) = 1, the ordering (S∗, S∗+ k, S∗+ 2k, . . . , S∗+ (2n −2)k) of the same one-factorization is also sequentially uniform. Note, however, that it is not necessarily of the same type as the ordered one-factorization (S∗, S∗+ 1, S∗+ 2, . . . , S∗+ (2n −2)).
The most well-known one-factorization of K2n (called GK(2n)) is generated from the patterned starter P = {{x, −x} : x ∈Z2n−1} in the cyclic group Z2n−1. It is known when 2n −1 is prime that GK(2n) is a perfect one-factorization and, in general, GK(2n) is a uniform one-factorization for all n ≥1.
The cycle lengths in P ∗∪(P ∗+ k) for k ∈Z2n−1 \ {0} are now given.
Lemma 2.2. Let P be the patterned starter in Z2n−1 with n ≥1. Let k ∈Z2n−1 \ {0} with gcd(2n −1, k) = d. Then P ∗∪(P ∗+ k) consists of a cycle of length 1 + (2n −1)/d and (d −1)/2 cycles of length 2(2n −1)/d.
Proof: The cycle through infinity is (∞, 0, 2k, −2k, 4k, −4k, . . ., −k, k), which has length 1 + (2n −1)/d. All other cycles (if any) are of the form (i, −i, 2k + i, −2k −i, 4k + i, −4k −i, . . . , −2k + i, 2k −i), for 1 ≤i < d.
2 Combining Lemmas 2.1 and 2.2 (with d = 1) we have the following result.
the electronic journal of combinatorics 12 (2005), #R1 3 Theorem 2.3. For every n ≥1 there exists a sequentially perfect one-factorization of K2n.
Contrast this with the known results for perfect one-factorizations: the sporadic small values mentioned in the Introduction, and only two infinite classes (each of density zero).
3 Small orders The one-factorizations of K4 and K6 are unique and in each case they are perfect. Hence both are sequentially perfect (the only possible type in these small cases).
The one-factorization of K8 obtained from the unique Steiner triple system of order 7 has type (4, 4) while GK(8) is a perfect one-factorization. Hence there exist sequentially uniform one-factorizations of K8 of all possible types.
We have checked all starters in Z9 by computer and report that no ordering of the translates of any of these starters yields a sequentially uniform one-factorization of K10 of type (4, 6). However, there does exist a uniform one-factorization of type (4, 6) (it is one-factorization #1 in the list of all 396 non-isomorphic one-factorizations of K10 given in [1, p. 655]). Clearly this is also sequentially uniform of type (4, 6) under any ordering of the one-factors. From Theorem 2.3 there exists a sequentially perfect one-factorization of K10. Thus sequentially uniform one-factorizations of K10 exist for both possible types.
Obviously, the ordering of the one-factors can affect the type of the 2-factors formed from consecutive 1-factors in an ordered one-factorization. Given a starter S in Z2n−1, let FS(k) denote the ordered one-factorization (S∗, S∗+ k, S∗+ 2k, . . ., S∗+ (2n −2)k) of K2n. In the following examples we discuss sequentially uniform one-factorizations in K12 and K14. In Z13 we will give one starter which induces all possible types of ordered one-factorizations when different orderings are imposed on translates of that starter.
Example 3.1. Given the following starter in Z11, S = {{1, 2}, {3, 8}, {4, 6}, {5, 9}, {7, 10}}, FS(1) is sequentially uniform of type (6, 6), FS(2) is sequentially uniform of type (4, 8) and FS(3) is sequentially uniform of type (12).
By checking all starters in Z11, we found that no ordering of any of the one-factoriz-ations formed by these starters gave a sequentially uniform one-factorization of type (4, 4, 4). However, Figure 1 provides a non-starter-induced ordered one-factorization which is sequentially uniform of this type.
In it is found that there exist exactly five nonisomorphic perfect one-factorizations of K12 and in a uniform one-factorization of type (6, 6) is given. From the enumer-ation in , it is known that there exist no other uniform one-factorizations of K12.
Hence it is noteworthy that FS(2) (defined in Example 3.1) gives a sequentially uni-form one-factorization of K12 of type (8, 4) and Figure 1 gives a sequentially uniform one-factorization of type (4, 4, 4).
the electronic journal of combinatorics 12 (2005), #R1 4 Figure 1: A sequentially uniform one-factorization of K12 with type (4, 4, 4) F0 : {{0, 1}, {2, 6}, {3, 4}, {7, 9}, {8, 10}, {5, 11}} F1 : {{0, 2}, {1, 6}, {3, 9}, {4, 7}, {5, 10}, {8, 11}} F2 : {{0, 3}, {1, 4}, {5, 8}, {6, 7}, {2, 9}, {10, 11}} F3 : {{0, 4}, {1, 3}, {2, 8}, {7, 10}, {6, 11}, {5, 9}} F4 : {{0, 5}, {1, 2}, {3, 8}, {4, 9}, {6, 10}, {7, 11}} F5 : {{0, 8}, {1, 7}, {2, 11}, {3, 5}, {4, 6}, {9, 10}} F6 : {{0, 6}, {1, 5}, {2, 10}, {4, 8}, {3, 7}, {9, 11}} F7 : {{0, 7}, {1, 10}, {2, 5}, {3, 6}, {4, 11}, {8, 9}} F8 : {{0, 9}, {1, 11}, {2, 3}, {4, 10}, {5, 6}, {7, 8}} F9 : {{0, 11}, {1, 9}, {2, 4}, {3, 10}, {6, 8}, {5, 7}} F10 : {{0, 10}, {1, 8}, {2, 7}, {3, 11}, {4, 5}, {6, 9}} Example 3.2. The following starter in Z13, S = {{1, 10}, {2, 3}, {4, 9}, {5, 7}, {6, 12}, {8, 11}}, yields sequentially uniform one-factorizations of K14 of all possible types: namely (14), (10, 4), (8, 6), and (6, 4, 4). Specifically, FS(3) is sequentially uniform of type (6, 4, 4), FS(1) is sequentially uniform of type (8, 6), FS(2) is sequentially uniform of type (10, 4) and FS(5) is sequentially uniform of type (14).
For large n, there are many more possible types than there are translates, so the starter in Example 3.2 is of particular interest. In the Appendix we give examples of sequentially uniform one-factorizations of K2n of all possible types, for 14 ≤2n ≤24.
4 Starters in non-cyclic groups Many uniform and perfect one-factorizations are known to be starter-induced over a non-cyclic group; for example, see . So it is natural to also expect sequentially uniform one-factorizations where the ordering is not cyclic. In this section we give a numerical condition that determines when certain starter-induced one-factorizations over non-cyclic groups are sequentially uniform.
Let q be an odd prime-power (not a prime) and write q = 2rt + 1, where t is odd.
In order to eliminate trivial cases, we will assume that t > 1. Suppose ω is a generator of the multiplicative group of Fq and let Q be the subgroup (of order t) generated by ω2r. Suppose the cosets of Q are Ci = ωiQ, i = 0, . . ., 2r −1. A starter S in Fq is said to be an r-quotient starter if, whenever {x, y}, {x′, y′} ∈S with x, x′ ∈Ci, it holds that y/x = y′/x′. An r-quotient starter S can be completely described by a list of quotients (a0, . . . , ar−1), such that S = {{x, aix} : (ai −1)x ∈Ci, i = 0, . . . , r −1}.
the electronic journal of combinatorics 12 (2005), #R1 5 It is not hard to see that S∗∪(S∗+x) is isomorphic to S∗∪(S∗+y) whenever x/y ∈C0∪Cr.
It follows that every 1-quotient starter yields a uniform one-factorization. We now show that, although r-quotient starters might not generate uniform one-factorizations when r > 1, , the resulting one-factorizations usually can be ordered in such a way that they are sequentially uniform.
Theorem 4.1. Suppose q = pd is an odd prime-power (with p prime and d > 1) such that q = 2rt + 1 and t > 1 is odd. Let S be any r-quotient starter in Fq. Then the one-factorization generated by S can be ordered to be sequentially uniform if and only if the multiplicative order of p modulo t is equal to d.
Proof: Let q = pd = 2rt + 1 with t odd. C0 is the multiplicative subgroup of F∗ q generated by a primitive tth root of 1 in Fq, say α. The splitting field of xt −1 over Fp is Fpe, where e is the smallest positive integer such that pe ≡1 (mod t). Hence the extension field Fp(α) = Fq if and only if the multiplicative order of p modulo t, which we denote by ordt(p), is equal to d.
Suppose that ordt(p) = d. Then Fp(α) = Fq and 1, α, . . ., αd−1 is a basis of Fq over Fp. Therefore, every element x ∈Fq can be expressed uniquely as a d-tuple (x1, . . ., xd) ∈ (Zp)d, where x = n X i=1 xiαi−1.
Now, consider the graph on vertex set (Zp)d in which two vertices are adjacent if and only if they agree in d −1 coordinates and their values in the remaining coordinate differ by 1 modulo p (this is a Cayley graph of the elementary abelian group of order pd). It is not hard to check that this graph has a hamiltonian cycle, say C = (y1, y2, . . . , ypd, y1).
The cycle C provides the desired ordering of Fq because the difference between any two consecutive elements yi and yi+1 is in C0 ∪Cr (note that one of yi −yi+1 and yi+1 −yi is a power of α and hence in C0, while the other is in Cr).
Conversely, suppose that ordt(p) = e < d. Then Fp(α) = Fpe which is a strict subfield of Fq. Clearly C0 ∪Cr ⊆Fpe. Suppose that y1, y2, . . . is an ordering of the elements of Fq such that adjacent elements always have a difference that is an element of C0 ∪Cr.
Without loss of generality we can take y1 = 0. But then every element yi is in the subfield Fpe, which is a contradiction. Hence, the desired ordering cannot exist.
2 It is interesting to note that the proof above does not depend on the structure of the starter S. Either all r-quotient starters in Fq yield sequentially uniform one-factorizations or they all do not do so.
Example 4.2. Let q = 25 so that t = 3 and r = 4. We have ordt(p) = 2 = d, so Theorem 4.1 asserts that any 4-quotient starter will yield a sequentially uniform one-factorization. In particular, if we take F25 = Z5[x]/(x2 + x + 2) then C0 contains a basis {1, α} for the field, where α = x8 = 3x + 1. The field elements can be cyclically ordered 0, 1, 2, 3, 4, 3x, . . ., 3x + 4, x, . . ., x + 4, 4x, . . ., 4x + 4, 2x, . . . , 2x + 4, 0 so that the difference of consecutive elements is either 1 or α.
the electronic journal of combinatorics 12 (2005), #R1 6 Most applications of r-quotient starters use values of r that are powers of two (see, for example, ). It is interesting to determine the conditions under which the hypotheses of Theorem 4.1 are satisfied in this case. This is done in Lemma 4.3.
Lemma 4.3. Suppose q = pd is an odd prime-power (with p prime and d > 1) such that q = 2kt + 1 and t > 1 is odd. Then one of the two following conditions hold: 1. ordt(p) = d, or 2. p = 2j −1 for some integer j (i.e., p is a Mersenne prime) and d = 2. (In this case, ordt(p) = 1 is less than d.) Proof: Suppose that pe ≡1 (mod t) for some positive integer e < d (note that e|d). Let pe = bt + 1 where b is a positive integer. Then 2kt + 1 = q = (pe)d/e = (bt + 1)d/e = cbt + 1 for some integer c. Hence, b|2k, and therefore b = 2ℓfor some positive integer ℓ≤k. So we have that pe = 2ℓt + 1.
Let ρ = pe and f = d/e. Then we have that t = ρf −1 2k = ρ −1 2ℓ .
Removing common factors, we obtain ρf−1 + ρf−2 + · · · + ρ + 1 = 2k−ℓ.
(1) Suppose that k = ℓ. Then the right side of (1) is equal to 1, so f = 1 and d = e. This contradicts the assumption that d > e. Therefore k > ℓand the right side of (1) is even.
Now, suppose that f is odd. Then the left side of (1) is odd, and we have a contra-diction. Therefore f is even, and ρ + 1 is a factor of the left side of (1). This implies that 2k−ℓ≡0 (mod ρ + 1), and hence ρ = 2j −1 for some integer j ≥2. Then, after dividing (1) by the factor ρ + 1, we obtain the following equation: ρf−2 + ρf−4 + · · · + ρ2 + 1 = 2k−ℓ−j.
(2) Suppose that j < k −ℓ. Then the right side of (2) is even and ρ2 + 1 is a factor of the left side of (2), so ρ2 = 2i −1 for some integer i ≥2. But ρ = 2j −1 where j ≥2, so ρ ≡3 (mod 4). Then ρ2 ≡1 (mod 4), which contradicts the fact that ρ2 = 2i −1 where i ≥2.
Therefore we have that j = k −ℓ. This implies that f = 2 and so d = 2e. So ρ = 2j −1 for some integer j and q = ρ2.
However, it is easy to prove that the Diophantine equation 2u −yv = 1 has no solution in positive integers with u, v > 1†. See, for example, Cassels [3, Corollary 2]. Therefore, †This result is a special case of Catalan’s Conjecture, which states that the Diophantine equation xu −yv = 1 has no solution in positive integers with u, v > 1 except for 32 −23 = 1. Catalan’s Conjecture was proven correct in 2002 by Mih˘ ailescu (see Mets¨ ankyl¨ a for a recent exposition of the proof).
the electronic journal of combinatorics 12 (2005), #R1 7 we can conclude that ρ is prime. Hence, p = 2j −1 is a Mersenne prime, e = 1 and d = 2.
In this case, we have q = p2. Then we have that q −1 = (p −1)(p + 1) = (p −1)2j ≡0 (mod t).
But t is odd, so p ≡1 (mod t). Therefore ordt(p) = 1.
2 Example 4.4. Let q = 961 = 312 so p = 31 and d = 2. Here p = 31 = 25 −1 is a Mersenne prime. We can write q = 2515 + 1, so t = 15. We see that ordt(p) = 1 < 2, as asserted by Lemma 4.3.
The following corollary is an immediate consequence of Theorem 4.1 and Lemma 4.3.
Corollary 4.5. Suppose q = pd is an odd prime-power (with p prime and d > 1) such that q = 2kt + 1 and t > 1 is odd. Let S be any 2k−1-quotient starter in Fq. Then the one-factorization generated by S can be ordered to be sequentially uniform if and only if it is not the case that p is a Mersenne prime and d = 2.
5 Product construction We now recall the usual product construction for one-factorizations, and apply it to determine another infinite class of sequentially uniform one-factorizations.
Suppose that F is a one-factor on X and G is a one-factor on Y , where |X| = 2n and |Y | = 2m. Define various one-factors of X × Y by F ∗ = {(xi, y), (x′ i, y)} : {xi, x′ i} ∈F, y ∈Y , G∗ = {(x, yj), (x, y′ j)} : x ∈X, {yj, y′ j} ∈G , FG = {(xi, yj), (x′ i, y′ j)} : {xi, x′ i} ∈F, {yj, y′ j} ∈G .
Given one-factorizations F = {F0, . . . , F2n−2} and G = {G0, . . . , G2m−2} of K2n and K2m on the points X and Y , respectively, it is easy to see that FG = FiGj : i = 0, . . . , 2n −2 and j = 0, . . ., 2m −2 [ F ∗ i : i = 0, . . . , 2n −2 [ G∗ j : j = 0, . . . , 2m −2 is a one-factorization of X × Y .
The following are easy lemmas about the cycle types of pairs of one-factors in FG.
Lemma 5.1. For any i ∈{0, . . . , 2n −2} and j ∈{0, . . . , 2m −2}, the following all have cycle type (4, 4, . . ., 4): (i) F ∗ i ∪G∗ j, (ii) FiGj ∪F ∗ i , and (iii) FiGj ∪G∗ j.
the electronic journal of combinatorics 12 (2005), #R1 8 Lemma 5.2. If (F0, F1, . . . , F2n−2) is sequentially uniform of type (4, 4, . . ., 4), then the following all have cycle type (4, 4, . . ., 4): (i) F ∗ i ∪F ∗ i+1, and (ii) FiGj ∪Fi+1Gj, for any i, j, where the subscripts i + 1 are reduced modulo 2n −1.
We can use the above results to give a product construction for sequentially uniform one-factorizations of type (4, 4, . . ., 4).
Theorem 5.3. Suppose there exists a sequentially uniform one-factorization of K2n of type (4, 4, . . ., 4). Let m ≤n. Then there is a sequentially uniform one-factorization of K4mn of type (4, 4, . . ., 4).
Proof: We use all the notation above, with (F0, F1, . . . , F2n−2) sequentially uniform of type (4, 4, . . ., 4) and G any one-factorization of K2m. The ordered one-factorization G∗ 0, F0G0, F1G0, . . . , F2n−2G0, F ∗ 2n−2, G∗ 1, F1G1, F2G1, . . ., F0G1, F ∗ 0 , G∗ 2, F2G2, F3G2, . . ., F1G2, F ∗ 1 , .
.
.
G∗ 2m−2, F2m−2G2m−2, . . . , F2m−3G2m−2, F ∗ 2m−3, F ∗ 2m−2, F ∗ 2m−1, . . . , F ∗ 2n−3 of K4mn is sequentially uniform of type (4, 4, . . ., 4) by Lemmas 5.1 and 5.2.
2 By applying the above product construction with 2n a power of 2 — for which the existence of uniform one-factorizations of type (4, 4, . . ., 4) are known — one immediately has the following corollary.
Corollary 5.4. For any odd integer m ≥1, there is a sequentially uniform one-factoriz-ation of K2tm of type (4, 4, . . ., 4) for all integers t ≥2 + ⌈log2 m⌉.
Let t0 = t0(m) denote the smallest integer such that there is a sequentially uniform one-factorization of K2tm of type (4, 4, . . ., 4) for all integers t ≥t0. Corollary 5.4 provides an explicit upper bound on t0(m); however, for a particular value of m, we might be able to give a better bound on t0(m). For example, the sequentially perfect one-factorization of K4 shows that t0(1) = 2, the sequentially uniform one-factorization of K12 of type (4, 4, 4) given in Figure 1 yields t0(3) = 2, and the sequentially uniform one-factorization of K20 of type (4, 4, 4, 4, 4) exhibited in the Appendix gives t0(5) = 2. In fact, we conjecture that t0(m) = 2 for all odd integers m ≥1.
As a final note, we observe that the existence results for sequentially uniform one-factorizations of K2tm of type (4, 4, . . ., 4) provide an interesting contrast to those for uniform one-factorizations of K2tm of type (4, 4, . . ., 4), which exist only when m = 1 (see Cameron [2, Proposition 4.3]).
the electronic journal of combinatorics 12 (2005), #R1 9 Acknowledgements The authors would like to to thank Dan Archdeacon, Cameron Stewart and Hugh Williams for useful discussions and pointers to the literature.
D. Stinson’s research is supported by the Natural Sciences and Engineering Research Council of Canada through the grant NSERC-RGPIN #203114-02. P. Dukes was sup-ported by an NSERC post-doctoral fellowship.
References L.D. Andersen. Factorizations of graphs. In The CRC Handbook of Combinatorial Designs, C.J. Colbourn and J.H. Dinitz, eds., CRC Press, Boca Raton, 1996, pp.
653–666.
P.J. Cameron. Minimal edge-colourings of complete graphs. J. London Math. Soc.
11 (1975) 337–346.
J.W.S. Cassels. On the equation ax −by = 1. Amer. J. Math. 75 (1953), 159–162.
J.H. Dinitz and P. Dukes. Some new perfect one-factorizations of the complete graph.
University of Vermont Mathematics Research Report No. 2004-01.
J.H. Dinitz, D.K. Garnick, and B.D. McKay. There are 526, 915, 620 nonisomorphic one-factorizations of K12. J. Combin. Des. 2 (1994), 273–285.
J.H. Dinitz and D.R. Stinson. Some new perfect one-factorizations from starters in finite fields. J. Graph Theory 13 (1989), 405–415.
N.J. Finizio. Tournament designs balanced with respect to several bias categories.
Bull. Inst. Combin. Appl. 9 (1993), 69–95.
M.J. Grannell, T.S. Griggs, and J.P. Murphy. Some new perfect Steiner triple sys-tems. J. Combin. Des. 7 (1999), 327–330.
V. Leck. Personal communication, 2001.
(See E. Mendelsohn and A. Rosa. One-factorizations of the complete graph — A survey.
J. Graph Theory 9 (1985), 43–65.
T. Mets¨ ankyl¨ a. Catalan’s Conjecture: Another old Diophantine problem solved. Bull.
Amer. Math. Soc. 41 (2004), 43–57.
A.P. Petrenyuk and A.Ya. Petrenyuk. Intersection of perfect 1-factorizations of com-plete graphs (Russian). Cybernetics 1980 6–8, 149.
the electronic journal of combinatorics 12 (2005), #R1 10 E. Seah. Perfect one-factorizations of the complete graph — a survey. Bull. Inst.
Combin. Appl. 1 (1991), 59–70.
W.D. Wallis. One-factorizations of the complete graph. In Contemporary Design The-ory: A Collection of Surveys, J.H. Dinitz and D.R. Stinson, eds., John Wiley & Sons, New York, 1992, pp. 593–631.
W.D. Wallis. One-factorizations. Kluwer Academic Publishers, Dordrecht, NL, 1997.
I.M. Wanless. Atomic Latin squares based on cyclotomic orthomorphisms. Submitted.
J.Z. Zhang. Some results on perfect 1-factorizations of K2n. Dianzi Keji Daxue Xue-bao 21 (1992), 434–436.
the electronic journal of combinatorics 12 (2005), #R1 11 Appendix Below is a table giving all possible types for sequentially uniform one-factorizations of K2n, 14 ≤2n ≤24. Each type is realized by the ordered 1-factorization FS(1) corresponding to the starter S = {{xi, xi + i} : i = 1, . . ., n −1} in Z2n−1.
type (x1, . . . , xn−1) (14) (1, 5, 8, 6, 12, 3) (10, 4) (1, 9, 4, 6, 3, 12) (8, 6) (1, 5, 9, 6, 3, 11) (6, 4, 4) (7, 2, 9, 1, 6, 10) (16) (1, 4, 8, 9, 7, 14, 3) (12, 4) (1, 4, 8, 9, 5, 12, 7) (10, 6) (1, 7, 11, 4, 5, 12, 6) (8, 8) (1, 3, 7, 9, 6, 8, 12) (8, 4, 4) (2, 11, 9, 4, 5, 1, 14) (6, 6, 4) (3, 11, 2, 6, 7, 8, 9) (4, 4, 4, 4) (3, 6, 11, 12, 5, 7, 2) (18) (1, 3, 7, 12, 8, 9, 4, 6) (14, 4) (1, 3, 11, 8, 4, 10, 6, 7) (12, 6) (1, 5, 8, 12, 9, 4, 13, 15) (10, 8) (1, 3, 6, 11, 8, 10, 7, 4) (10, 4, 4) (1, 13, 5, 7, 9, 4, 16, 12) (8, 6, 4) (1, 4, 12, 5, 8, 10, 7, 3) (6, 6, 6) (1, 6, 10, 5, 11, 15, 7, 12) (6, 4, 4, 4) (3, 7, 12, 2, 8, 10, 11, 14) (20) (1, 3, 6, 11, 13, 8, 10, 4, 7) (16, 4) (1, 3, 9, 13, 10, 8, 4, 18, 16) (14, 6) (1, 3, 9, 7, 13, 8, 10, 15, 16) (12, 8) (1, 3, 9, 13, 6, 10, 8, 18, 14) (12, 4, 4) (1, 5, 12, 6, 9, 17, 11, 8, 13) (10, 10) (1, 3, 12, 9, 11, 4, 7, 17, 18) (10, 6, 4) (1, 4, 10, 11, 7, 18, 9, 14, 8) (8, 8, 4) (1, 3, 12, 10, 6, 7, 16, 9, 18) (8, 6, 6) (1, 4, 5, 13, 9, 10, 11, 7, 3) (8, 4, 4, 4) (2, 5, 11, 4, 10, 12, 13, 9, 16) (6, 6, 4, 4) (2, 14, 8, 13, 7, 4, 18, 1, 15) (4, 4, 4, 4, 4) (6, 16, 17, 11, 4, 8, 3, 5, 12) type (x1, . . . , xn−1) (22) (1, 3, 4, 10, 11, 13, 8, 12, 9, 17) (18, 4) (1, 3, 4, 13, 11, 12, 8, 6, 10, 20) (16, 6) (1, 3, 4, 10, 11, 12, 13, 9, 6, 19) (14, 8) (1, 3, 4, 12, 9, 11, 13, 10, 6, 19) (14, 4, 4) (1, 4, 7, 11, 12, 14, 19, 8, 9, 3) (12, 10) (1, 3, 6, 11, 13, 10, 12, 20, 8, 4) (12, 6, 4) (1, 3, 7, 15, 11, 8, 13, 4, 9, 17) (10, 8, 4) (1, 3, 7, 11, 14, 6, 13, 9, 16, 8) (10, 6, 6) (1, 3, 7, 16, 12, 8, 6, 11, 9, 15) (10, 4, 4, 4) (1, 5, 11, 13, 19, 6, 8, 10, 16, 20) (8, 8, 6) (1, 3, 12, 6, 13, 14, 4, 9, 7, 19) (8, 6, 4, 4) (1, 3, 10, 16, 12, 9, 4, 6, 19, 8) (6, 6, 6, 4) (1, 8, 6, 11, 12, 19, 13, 18, 7, 14) (6, 4, 4, 4, 4) (1, 11, 5, 16, 9, 6, 17, 10, 19, 15) (24) (1, 3, 4, 9, 11, 15, 12, 14, 8, 10, 18) (20, 4) (1, 3, 4, 13, 10, 16, 12, 6, 11, 8, 21) (18, 6) (1, 3, 4, 10, 16, 13, 8, 9, 11, 12, 18) (16, 8) (1, 3, 4, 10, 12, 15, 11, 8, 13, 19, 9) (16, 4, 4) (1, 3, 6, 10, 16, 11, 15, 12, 4, 8, 19) (14, 10) (1, 3, 4, 13, 16, 8, 11, 12, 6, 9, 22) (14, 6, 4) (1, 3, 6, 10, 12, 16, 13, 11, 21, 8, 4) (12, 12) (1, 3, 4, 10, 12, 16, 13, 11, 6, 8, 21) (12, 8, 4) (1, 3, 4, 13, 10, 12, 9, 14, 20, 11, 8) (12, 6, 6) (1, 3, 4, 15, 9, 10, 11, 12, 13, 21, 6) (12, 4, 4, 4) (1, 3, 13, 15, 4, 6, 14, 10, 8, 20, 11) (10, 10, 4) (1, 3, 6, 15, 16, 8, 11, 12, 4, 7, 22) (10, 8, 6) (1, 3, 4, 11, 9, 16, 12, 13, 8, 10, 18) (10, 6, 4, 4) (1, 3, 4, 12, 15, 13, 10, 6, 9, 21, 11) (8, 8, 8) (1, 3, 4, 11, 16, 8, 10, 12, 13, 9, 18) (8, 8, 4, 4) (1, 3, 19, 14, 10, 7, 4, 12, 8, 6, 21) (8, 6, 6, 4) (1, 3, 9, 13, 22, 15, 7, 10, 11, 6, 8) (8, 4, 4, 4, 4) (1, 4, 16, 10, 13, 9, 5, 22, 17, 11, 20) (6, 6, 6, 6) (1, 5, 6, 12, 15, 21, 10, 11, 13, 8, 3) (6, 6, 4, 4, 4) (1, 6, 14, 7, 13, 15, 3, 12, 19, 22, 16) (4, 4, 4, 4, 4, 4) (5, 17, 9, 11, 13, 21, 1, 22, 16, 10, 3) the electronic journal of combinatorics 12 (2005), #R1 12 |
16806 | https://opencw.aprende.org/courses/electrical-engineering-and-computer-science/6-432-stochastic-processes-detection-and-estimation-spring-2004/readings/ | Readings | Stochastic Processes, Detection, and Estimation | Electrical Engineering and Computer Science | MIT OpenCourseWare
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This course has no required or recommended text. However, the optional reference text (A. Papoulis, Probability, Random Variables, and Stochastic Processes, McGraw-Hill, third ed., 1991, ISBN: 0070484686) can be used as a supplement to the course notes. This book covers many of the same topics we will, but overwhelming student opinion in the past has been that the book is one they tend to refer to much more after the course than during it for a variety of reasons. While you may well find this book a useful addition to your personal library, we will not assume you have access to the book during the term. Several other texts which may be a useful resource to you are listed below.
Reference Texts
Anderson, B. D. O., and J. B. Moore. Optimal Filtering. Prentice-Hall, 1979. ISBN: 0-13-638122-7 (Out of Print).
Very readable treatment of Kalman and Wiener filtering.
Drake, A.Fundamentals of Applied Probability Theory. McGraw-Hill, 1967. ISBN: 0-07-017815-1 (Out of Print).
Basic engineering text on probability theory.
Feller, W. An Introduction to Probability Theory and Its Applications. Vols. 1 and 2. Wiley, 1968. ISBN: 0-471-25708-7.
Valuable formal reference set on probability theory.
Gardner, W. A. Introduction to Random Processes: with Applications to Signals and Systems. 2nd ed. McGraw-Hill, 1990. ISBN: 0-07-022855-8.
Useful auxilliary reference for a number of topics covered in the course, particularly stochastic processes.
Gray, R. M., and L. D. Davisson. Random Processes: A Mathematical Approach for Engineers. Prentice-Hall, 1986. ISBN: 0-13-752882-5 (Out of Print).
Bridges the gap between formal mathematical texts and engineering texts on probability theory.
Helstrom, C. W. Probability and Stochastic Processes for Engineers. Macmillan, 1991. ISBN: 0-02-353571-7.
Nice undergraduate level introduction to several themes in the course.
Kay, S. M. Fundamentals of Statistical Signal Processing and Estimation Theory. Prentice-Hall, 1993. ISBN: 0-13-345711-7.
Accessible and thorough treatment of estimation theory.
Lee, E., and D. G. Messerschmitt. Digital Communication. 2nd ed. Kluwer Academic, 1994. ISBN: 0-89838-274-2.
Advanced reading on applications in communication theory.
Loève, M. Probability Theory . Vol. 1. 4th ed. Springer-Verlag, 1977. ISBN: 3-540-90210.
Formal but reasonably readable treatment of probability theory. A classic.
Naylor, A. W., and G. R. Sell. Linear Operator Theory in Engineering and Science. Springer-Verlag, 1982. ISBN: 0-387-90748-3 (Out of Print).
Accessible treatment of the mathematical foundations for vector space concepts.
Oppenheim, A. V., and R. W. Schafer. Discrete-Time Signal Processing. Prentice Hall, 1989. ISBN: 0-13-216292-X (Out of Print).
Standard text on discrete-time linear systems and signals.
Oppenheim, A. V., and A. S. Willsky. Signals and Systems. Prentice-Hall, 1982. ISBN: 0-13-814757-4.
Standard undergraduate text on signals and systems.
Parzen, E. Stochastic Processes. Holden-Day, 1962. ISBN: 0-8162-6664-6.
Classic, formal text on stochastic processes.
Scharf, L. L. Statistical Signal Processing: Detection, Estimation, and Time Series Analysis. Addison-Wesley, 1991. ISBN: 0-201-19038-9.
Slightly more advanced treatment of several detection and estimation topics.
Stark, H., and J. W. Woods. Probability, Random Processes, and Estimation Theory for Engineers. 2nd ed. Prentice-Hall, 1994. ISBN: 0-13-728791-7 (Out of Print).
A useful reference for many topics in the course, as well as the background material.
Strang, G.Linear Algebra and its Applications. 3rd ed. Harcourt Brace Jovanovich, 1988. ISBN: 0-15-551006-1.
Standard reference text on linear algebra.
Therrien, C. W. Discrete Random Signals and Statistical Signal Processing. Prentice-Hall, 1992. ISBN: 0-13-852112-3 (Out of Print).
Very accessible reference for many topics in the course.
Van, H. L. Trees, Detection, Estimation and Modulation Theory, Part I. Wiley, 1968. ISBN: 0-471-89955-0 (Out of Print).
Classic and valuable reference text on detection and estimation theory.
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Math - Algebra II: Rational Expressions and Equations Complete Unit Plan
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Farideh Dormishian - M.S.C.S
Subject
Math — Algebra, Mathematical Functions
Grade Level
Grades 10-12, Higher Education
Resource Type
Activity, Assessment, Handout, Presentation, Worksheet
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About This Lesson
The unit plan is appropriate for students in grade 11 who have passed Algebra I and Geometry. This topic is Rational Expressions & Equations and is part of a unit in Algebra II. The unit contains five lessons and it takes 10 hours of teaching. Rational expressions can be used in solving many problems involving mixtures, photography, electricity, medicine, and travel. The learning objective of the unit is for students to learn how to:
Simplify rational expressions.
Multiply and divide rational expressions
Determine the LCM of polynomials.
Add and subtract rational expressions.
Simplify complex rational expressions
Solve rational equations
Solving real-world problems.
Solve rational inequalities.
Graphing rational functions using a calculator.
Determine the vertical asymptotes and the point discontinuity for the graphs of rational functions.
Graph rational functions.
The daily lesson plan will start with a warm-up to review the previous lessons. The learning objectives will be stated using PowerPoint. A sample PowerPoint is created and you may modify it to your liking. Students will be encouraged to share any information they already know about the topic and the lesson will start by stating examples of the usage of the specific math concept in real-life. This will tap into students’ background knowledge and students will be able to know the reason they need to learn the topic. Furthermore, the lesson will have many examples and students will be engaged. In the end, the objective of the lesson will be reviewed and an exit ticket for each lesson will identify how and what students have learned. The exit-ticket is in the form of a journal. The file for the quiz, test, homework, warm-ups, how to make the journal, and a rubric for the journal are included. Since the unit will cover five lessons, student’s performance will be assessed using formative assessments and after the third lesson, a quiz can be taken and a unit test can be administered at the end of the fifth lesson.
At the end of the unit, students will be able to define, understand, and apply terms such as rational equation, rational expression, rational function, rational inequality, complex fraction, asymptote, continuity, and point of discontinuity.
Standards
CCSS.MATH.CONTENT.HSA.APR.D.7
CCSS.MATH.CONTENT.HSA.REI.B.3
CCSS.MATH.CONTENT.HSF.IF.C.7
CCSS.MATH.CONTENT.HSF.IF.C.7.D
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Standards
### CCSS.Math.Content.HSA-APR.D.7
Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.
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### CCSS.Math.Content.HSA-REI.B.3
Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
Read More About CCSS.Math.Content.HSA-REI.B.3
### CCSS.Math.Content.HSF-IF.C.7
Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases.
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### CCSS.Math.Content.HSF-IF.C.7.d
Graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing end behavior.
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### CCSS.Math.Content.HSA-REI.A.2
Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise.
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16808 | https://askfilo.com/user-question-answers-smart-solutions/the-tank-in-fig-contains-oil-and-water-as-shown-find-the-3333353138313838 | Question asked by Filo student
The tank in Fig. 3−38 contains oil and water as shown. Find the resultant force on side ABC, which is 4ft wide.F=γhc8AFAB=(0.80)(62.4)[(10)(4)]=9980lbFAn acts at a point (52)(10), or 6.67ft below point A. Water is acting on area BC, and any superimposed liquid can be converted to an equivalent depth of water. Employ an imaginary water surface (IWS) for this calculation, locating IWS by changing 10ft of oil to (0.80)(10), or 8ft of water. Thus, FBC= (62.4)(8+26)[(6)(4)]=16470Ib.y_{\text {Gp }}=\frac{-I_{x x} \sin \theta}{h_{c 8} A}=\frac{-\left[(4) (6)^3 / 12\right]\left(\sin 90^{\circ}\right)}{\left(8+\frac{5}{2}\right)[(6)(4) ]}=-0.27 \mathrm{ft}(i.e., below the centroid of BC )Fsc acts at a point \left(2+8+\frac{5}{2}+0.27\right), or 13.27ft below A.∑MA=0;(9980+16470)(hep)−(9980)(6.67)− (16470)(13.27)=0,hep=10.78ft from A. Thus, the total resultant force on side ABC is 9980+16470, or 26450Ib acting 10.78ft below A.
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| Question Text | The tank in Fig. 3−38 contains oil and water as shown. Find the resultant force on side ABC, which is 4ft wide.F=γhc8AFAB=(0.80)(62.4)[(10)(4)]=9980lbFAn acts at a point (52)(10), or 6.67ft below point A. Water is acting on area BC, and any superimposed liquid can be converted to an equivalent depth of water. Employ an imaginary water surface (IWS) for this calculation, locating IWS by changing 10ft of oil to (0.80)(10), or 8ft of water. Thus, FBC= (62.4)(8+26)[(6)(4)]=16470Ib.y_{\text {Gp }}=\frac{-I_{x x} \sin \theta}{h_{c 8} A}=\frac{-\left[(4) (6)^3 / 12\right]\left(\sin 90^{\circ}\right)}{\left(8+\frac{5}{2}\right)[(6)(4) ]}=-0.27 \mathrm{ft}(i.e., below the centroid of BC )Fsc acts at a point \left(2+8+\frac{5}{2}+0.27\right), or 13.27ft below A.∑MA=0;(9980+16470)(hep)−(9980)(6.67)− (16470)(13.27)=0,hep=10.78ft from A. Thus, the total resultant force on side ABC is 9980+16470, or 26450Ib acting 10.78ft below A. |
| Updated On | May 12, 2025 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Grade 12 |
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16809 | https://www.youtube.com/watch?v=ODisxDsqLCQ | Understanding Similar Triangles With GeoGebra
apethan
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we're going to look at similar triangles and we're going to experiment with them using geod so the way that this works you can see the links on the site what we're going to do is we're going to take two triangles and we're going to see that these two are in fact congruent let set them right on top of each other here so these are two congruent triangles and what we're going to do is scale one of them so that it grows relative to the first so the scale now if we go less than one it will shrink but if we go greater than one it will grow so let's go ahead and get it up to a scale of two so what we're looking at right now is that this larger triangle every side length is twice as long as the original triangle and if you look at it visually that seems to make a bit of sense C looks like it's the midpoint of uh D or a depending on which triangle you're looking at all the way up to F over here and B looks like it's at the midpoint of D to e and so this this uh line segment right here from E to F corresponds to B to C and that one also looks to be about twice as long so what we did here is we formed a common ratio every time we have a corresponding side on one triangle to the next triangle we have a ratio of the scale factor 2: one we grow our scale factor all the way up to three you'll have the same thing each one of these corresponding sides this side to the smaller one here the smaller side to the larger one here it's all going to be a 3:1 ratio that we're looking at one thing that you may have noticed is that even though the side lengths are changing throughout all of this the angles do not so this angle here for smaller triangle angle a or the larger triangle angle D is the exact same no matter what scale we use same thing with this second angle here angle b or angle e which correspond to each other they're always the exact same angle no matter how big the other triangle is and even though it's not marked angle C and F also given that the three angles add up to 180 must be the same so let's move over to another geogra worksheet uh looking at specifically proportions and I also have this link below if we go ahead and change one of these triangles the other triangle is going to change so that it is in fact similar the way that it does this is all the angles are maintained to be the same so angle C corresponds to angle F and those two will always have the same angle B to angle E and A to angle D but we can scale the triangle on the right by uh grabbing a hold of Point F and shrinking it down so if we shrink it down to about here to get that exactly right but what we're going to be able to do then is get the two to be equivalent we're going to get the two to be congruent triangles so you can see even though it's not perfectly exact it's very close right now what we can do we can go ahead and drag this one out and scale it and let's get this let's get this one here to some integer length so that it's easy to make some comparisons so we have three on one side we're going to go for about four again because of the decimals here it's a little bit hard to get exactly what you want but we have a three side a four side and this side we're going to oh we can get that one to about five if you remember 3 four five you're going to be looking at a right a right triangle you move this down here and if we can get this hypotenuse this longest side here to be about 10 you'll notice that each of the other ratios are exactly the same this 10 to 5 corresponds to 6 to 3 which you know 6id 3 is also two and 8.08 to 4.04 you're looking at again a ratio of 2: 1 what if we grow this a little bit if we can grow this side to 15 side EF it's a little touchy again but let's get it pretty close so roughly 15 to 5 15 divid 5 is three so this side is three times longer than its corresponding side here here you're going to have roughly 9 ID 3 again 3: 1 and roughly 12 / 4 again 3 one so whenever you're dealing with similar triangles the proportions of corresponding sides will always be the same and it says that down here AB divided de now these are both referring to lengths uh of the sides of the triangle is going to be equal to AC over DF because angle a and angle D are corresponding angles B and E are corresponding angles and C and F so what I want you to do go ahead and try both of these geogebra worksheets by moving around moving around the triangles resizing them see what happens if you rotate them so on this one you can actually rotate them and see how that affects things as you grow things the proportions may appear to be off but you have to remember what sides actually do in fact correspond and see what you can come up with we'll discuss this again later that's it |
16810 | https://en.wikipedia.org/wiki/Positive_and_negative_predictive_values | Jump to content
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Contents
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1 Definition
1.1 Positive predictive value (PPV)
1.2 Negative predictive value (NPV)
1.3 Relationship
1.4 Worked example
2 Problems
2.1 Other individual factors
2.2 Bayesian updating
2.3 Different target conditions
3 See also
4 References
Positive and negative predictive values
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From Wikipedia, the free encyclopedia
Statistical measures of whether a finding is likely to be true
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| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "Positive and negative predictive values" – news · newspapers · books · scholar · JSTOR (March 2012) (Learn how and when to remove this message) |
The positive and negative predictive values (PPV and NPV respectively) are the proportions of positive and negative results in statistics and diagnostic tests that are true positive and true negative results, respectively. The PPV and NPV describe the performance of a diagnostic test or other statistical measure. A high result can be interpreted as indicating the accuracy of such a statistic. The PPV and NPV are not intrinsic to the test (as true positive rate and true negative rate are); they depend also on the prevalence. Both PPV and NPV can be derived using Bayes' theorem.
Although sometimes used synonymously, a positive predictive value generally refers to what is established by control groups, while a post-test probability refers to a probability for an individual. Still, if the individual's pre-test probability of the target condition is the same as the prevalence in the control group used to establish the positive predictive value, the two are numerically equal.
In information retrieval, the PPV statistic is often called the precision.
Definition
[edit]
Positive predictive value (PPV)
[edit]
The positive predictive value (PPV), or precision, is defined as
where a "true positive" is the event that the test makes a positive prediction, and the subject has a positive result under the gold standard, and a "false positive" is the event that the test makes a positive prediction, and the subject has a negative result under the gold standard. The ideal value of the PPV, with a perfect test, is 1 (100%), and the worst possible value would be zero.
The PPV can also be computed from sensitivity, specificity, and the prevalence of the condition:
cf. Bayes' theorem
The complement of the PPV is the false discovery rate (FDR):
Negative predictive value (NPV)
[edit]
The negative predictive value is defined as:
where a "true negative" is the event that the test makes a negative prediction, and the subject has a negative result under the gold standard, and a "false negative" is the event that the test makes a negative prediction, and the subject has a positive result under the gold standard. With a perfect test, one which returns no false negatives, the value of the NPV is 1 (100%), and with a test which returns no true negatives the NPV value is zero.
The NPV can also be computed from sensitivity, specificity, and prevalence:
The complement of the NPV is the false omission rate (FOR):
Although sometimes used synonymously, a negative predictive value generally refers to what is established by control groups, while a negative post-test probability rather refers to a probability for an individual. Still, if the individual's pre-test probability of the target condition is the same as the prevalence in the control group used to establish the negative predictive value, then the two are numerically equal.
Relationship
[edit]
The following diagram illustrates how the positive predictive value, negative predictive value, sensitivity, and specificity are related.
| | | | |
--- --- |
| | | Predicted condition | Sources: view talk edit |
| Total population = P + N | Predicted positive | Predicted negative | Informedness, bookmaker informedness (BM) = TPR + TNR − 1 | Prevalence threshold (PT) = √TPR × FPR − FPR/TPR − FPR |
| Actual condition | Real Positive (P) [a] | True positive (TP), hit[b] | False negative (FN), miss, underestimation | True positive rate (TPR), recall, sensitivity (SEN), probability of detection, hit rate, power = TP/P = 1 − FNR | False negative rate (FNR), miss rate type II error [c] = FN/P = 1 − TPR |
| Real Negative (N)[d] | False positive (FP), false alarm, overestimation | True negative (TN), correct rejection[e] | False positive rate (FPR), probability of false alarm, fall-out type I error [f] = FP/N = 1 − TNR | True negative rate (TNR), specificity (SPC), selectivity = TN/N = 1 − FPR |
| | Prevalence = P/P + N | Positive predictive value (PPV), precision = TP/TP + FP = 1 − FDR | False omission rate (FOR) = FN/TN + FN = 1 − NPV | Positive likelihood ratio (LR+) = TPR/FPR | Negative likelihood ratio (LR−) = FNR/TNR |
| Accuracy (ACC) = TP + TN/P + N | False discovery rate (FDR) = FP/TP + FP = 1 − PPV | Negative predictive value (NPV) = TN/TN + FN = 1 − FOR | Markedness (MK), deltaP (Δp) = PPV + NPV − 1 | Diagnostic odds ratio (DOR) = LR+/LR− |
| Balanced accuracy (BA) = TPR + TNR/2 | F1 score = 2 PPV × TPR/PPV + TPR = 2 TP/2 TP + FP + FN | Fowlkes–Mallows index (FM) = √PPV × TPR | phi or Matthews correlation coefficient (MCC) = √TPR × TNR × PPV × NPV - √FNR × FPR × FOR × FDR | Threat score (TS), critical success index (CSI), Jaccard index = TP/TP + FN + FP |
^ the number of real positive cases in the data
^ A test result that correctly indicates the presence of a condition or characteristic
^ Type II error: A test result which wrongly indicates that a particular condition or attribute is absent
^ the number of real negative cases in the data
^ A test result that correctly indicates the absence of a condition or characteristic
^ Type I error: A test result which wrongly indicates that a particular condition or attribute is present
Note that the positive and negative predictive values can only be estimated using data from a cross-sectional study or other population-based study in which valid prevalence estimates may be obtained. In contrast, the sensitivity and specificity can be estimated from case-control studies.
Worked example
[edit]
Suppose the fecal occult blood (FOB) screen test is used in 2030 people to look for bowel cancer:
| | | |
---
| | Fecal occult blood screen test outcome | view talk edit |
| | Total population(pop.) = 2030 | Test outcome positive | Test outcome negative | Accuracy (ACC) = (TP + TN) / pop.= (20 + 1820) / 2030≈ 90.64% | F1 score = 2 × precision × recall/precision + recall≈ 0.174 |
| Patients withbowel cancer(as confirmedon endoscopy) | Actual conditionpositive (AP)= 30(2030 × 1.48%) | True positive (TP)= 20(2030 × 1.48% × 67%) | False negative (FN)= 10(2030 × 1.48% × (100% − 67%)) | True positive rate (TPR), recall, sensitivity = TP / AP= 20 / 30≈ 66.7% | False negative rate (FNR), miss rate = FN / AP= 10 / 30≈ 33.3% |
| Actual conditionnegative (AN)= 2000(2030 × (100% − 1.48%)) | False positive (FP)= 180(2030 × (100% − 1.48%) × (100% − 91%)) | True negative (TN)= 1820(2030 × (100% − 1.48%) × 91%) | False positive rate (FPR), fall-out, probability of false alarm = FP / AN= 180 / 2000= 9.0% | Specificity, selectivity, true negative rate (TNR) = TN / AN= 1820 / 2000= 91% |
| | Prevalence = AP / pop.= 30 / 2030≈ 1.48% | Positive predictive value (PPV), precision = TP / (TP + FP)= 20 / (20 + 180)= 10% | False omission rate (FOR) = FN / (FN + TN)= 10 / (10 + 1820)≈ 0.55% | Positive likelihood ratio (LR+) = TPR/FPR= (20 / 30) / (180 / 2000)≈ 7.41 | Negative likelihood ratio (LR−) = FNR/TNR= (10 / 30) / (1820 / 2000)≈ 0.366 |
| | False discovery rate (FDR) = FP / (TP + FP)= 180 / (20 + 180)= 90.0% | Negative predictive value (NPV) = TN / (FN + TN)= 1820 / (10 + 1820)≈ 99.45% | Diagnostic odds ratio (DOR) = LR+/LR−≈ 20.2 |
The small positive predictive value (PPV = 10%) indicates that many of the positive results from this testing procedure are false positives. Thus it will be necessary to follow up any positive result with a more reliable test to obtain a more accurate assessment as to whether cancer is present. Nevertheless, such a test may be useful if it is inexpensive and convenient. The strength of the FOB screen test is instead in its negative predictive value — which, if negative for an individual, gives us a high confidence that its negative result is true.
Problems
[edit]
Other individual factors
[edit]
Note that the PPV is not intrinsic to the test—it depends also on the prevalence. Due to the large effect of prevalence upon predictive values, a standardized approach has been proposed, where the PPV is normalized to a prevalence of 50%. PPV is directly proportional[dubious – discuss] to the prevalence of the disease or condition. In the above example, if the group of people tested had included a higher proportion of people with bowel cancer, then the PPV would probably come out higher and the NPV lower. If everybody in the group had bowel cancer, the PPV would be 100% and the NPV 0%.[citation needed]
To overcome this problem, NPV and PPV should only be used if the ratio of the number of patients in the disease group and the number of patients in the healthy control group used to establish the NPV and PPV is equivalent to the prevalence of the diseases in the studied population, or, in case two disease groups are compared, if the ratio of the number of patients in disease group 1 and the number of patients in disease group 2 is equivalent to the ratio of the prevalences of the two diseases studied. Otherwise, positive and negative likelihood ratios are more accurate than NPV and PPV, because likelihood ratios do not depend on prevalence.[citation needed]
When an individual being tested has a different pre-test probability of having a condition than the control groups used to establish the PPV and NPV, the PPV and NPV are generally distinguished from the positive and negative post-test probabilities, with the PPV and NPV referring to the ones established by the control groups, and the post-test probabilities referring to the ones for the tested individual (as estimated, for example, by likelihood ratios). Preferably, in such cases, a large group of equivalent individuals should be studied, in order to establish separate positive and negative predictive values for use of the test in such individuals.[citation needed]
Bayesian updating
[edit]
Bayes' theorem confers inherent limitations on the accuracy of screening tests as a function of disease prevalence or pre-test probability. It has been shown that a testing system can tolerate significant drops in prevalence, up to a certain well-defined point known as the prevalence threshold, below which the reliability of a positive screening test drops precipitously. That said, Balayla et al. showed that sequential testing overcomes the aforementioned Bayesian limitations and thus improves the reliability of screening tests. For a desired positive predictive value , where , that approaches some constant , the number of positive test iterations needed is:
is the desired PPV
is the number of testing iterations necessary to achieve
is the sensitivity
is the specificity
is disease prevalence
Of note, the denominator of the above equation is the natural logarithm of the positive likelihood ratio (LR+). Also, note that a critical assumption is that the tests must be independent. As described Balayla et al., repeating the same test may violate the this independence assumption and in fact "A more natural and reliable method to enhance the positive predictive value would be, when available, to use a different test with different parameters altogether after an initial positive result is obtained.".
Different target conditions
[edit]
PPV is used to indicate the probability that in case of a positive test, that the patient really has the specified disease. However, there may be more than one cause for a disease and any single potential cause may not always result in the overt disease seen in a patient. There is potential to mix up related target conditions of PPV and NPV, such as interpreting the PPV or NPV of a test as having a disease, when that PPV or NPV value actually refers only to a predisposition of having that disease.
An example is the microbiological throat swab used in patients with a sore throat. Usually publications stating PPV of a throat swab are reporting on the probability that this bacterium is present in the throat, rather than that the patient is ill from the bacteria found. If presence of this bacterium always resulted in a sore throat, then the PPV would be very useful. However the bacteria may colonise individuals in a harmless way and never result in infection or disease. Sore throats occurring in these individuals are caused by other agents such as a virus. In this situation the gold standard used in the evaluation study represents only the presence of bacteria (that might be harmless) but not a causal bacterial sore throat illness. It can be proven that this problem will affect positive predictive value far more than negative predictive value. To evaluate diagnostic tests where the gold standard looks only at potential causes of disease, one may use an extension of the predictive value termed the Etiologic Predictive Value.
See also
[edit]
Binary classification
Sensitivity and specificity
False discovery rate
Relevance (information retrieval)
Receiver-operator characteristic
Diagnostic odds ratio
Sensitivity index
References
[edit]
^ Fletcher, Robert H. Fletcher; Suzanne W. (2005). Clinical epidemiology : the essentials (4th ed.). Baltimore, Md.: Lippincott Williams & Wilkins. pp. 45. ISBN 0-7817-5215-9.{{cite book}}: CS1 maint: multiple names: authors list (link)
^ a b Altman, DG; Bland, JM (1994). "Diagnostic tests 2: Predictive values". BMJ. 309 (6947): 102. doi:10.1136/bmj.309.6947.102. PMC 2540558. PMID 8038641.
^ Fawcett, Tom (2006). "An Introduction to ROC Analysis" (PDF). Pattern Recognition Letters. 27 (8): 861–874. doi:10.1016/j.patrec.2005.10.010. S2CID 2027090.
^ Provost, Foster; Tom Fawcett (2013-08-01). "Data Science for Business: What You Need to Know about Data Mining and Data-Analytic Thinking". O'Reilly Media, Inc.
^ Powers, David M. W. (2011). "Evaluation: From Precision, Recall and F-Measure to ROC, Informedness, Markedness & Correlation". Journal of Machine Learning Technologies. 2 (1): 37–63.
^ Ting, Kai Ming (2011). Sammut, Claude; Webb, Geoffrey I. (eds.). Encyclopedia of machine learning. Springer. doi:10.1007/978-0-387-30164-8. ISBN 978-0-387-30164-8.
^ Brooks, Harold; Brown, Barb; Ebert, Beth; Ferro, Chris; Jolliffe, Ian; Koh, Tieh-Yong; Roebber, Paul; Stephenson, David (2015-01-26). "WWRP/WGNE Joint Working Group on Forecast Verification Research". Collaboration for Australian Weather and Climate Research. World Meteorological Organisation. Retrieved 2019-07-17.
^ Chicco D, Jurman G (January 2020). "The advantages of the Matthews correlation coefficient (MCC) over F1 score and accuracy in binary classification evaluation". BMC Genomics. 21 (1): 6-1–6-13. doi:10.1186/s12864-019-6413-7. PMC 6941312. PMID 31898477.
^ Chicco D, Toetsch N, Jurman G (February 2021). "The Matthews correlation coefficient (MCC) is more reliable than balanced accuracy, bookmaker informedness, and markedness in two-class confusion matrix evaluation". BioData Mining. 14 (13): 13. doi:10.1186/s13040-021-00244-z. PMC 7863449. PMID 33541410.
^ Tharwat A. (August 2018). "Classification assessment methods". Applied Computing and Informatics. 17: 168–192. doi:10.1016/j.aci.2018.08.003.
^ Heston, Thomas F. (2011). "Standardizing predictive values in diagnostic imaging research" (PDF). Journal of Magnetic Resonance Imaging. 33 (2): 505, author reply 506–7. doi:10.1002/jmri.22466. PMID 21274995.
^ a b c Jacques Balayla. Bayesian Updating and Sequential Testing: Overcoming Inferential Limitations of Screening Tests. BMC Med Inform Decis Mak 22, 6 (2022).
^ a b Gunnarsson, Ronny K.; Lanke, Jan (2002). "The predictive value of microbiologic diagnostic tests if asymptomatic carriers are present". Statistics in Medicine. 21 (12): 1773–85. doi:10.1002/sim.1119. PMID 12111911. S2CID 26163122.
^ Orda, Ulrich; Gunnarsson, Ronny K; Orda, Sabine; Fitzgerald, Mark; Rofe, Geoffry; Dargan, Anna (2016). "Etiologic predictive value of a rapid immunoassay for the detection of group A Streptococcus antigen from throat swabs in patients presenting with a sore throat" (PDF). International Journal of Infectious Diseases. 45 (April): 32–5. doi:10.1016/j.ijid.2016.02.002. PMID 26873279.
^ Gunnarsson, Ronny K. "EPV Calculator". Science Network TV.
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16811 | https://www.youtube.com/watch?v=VpX6_aL25KQ | formulas: Sn = n/2 [2a + (n - 1)d] Sn = n/2 [a1 + an]
mkarmakar
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The sum of n terms of an AP can be easily found out using a simple formula which says that, if we have an AP whose first term is a and the common difference is d, then the formula of the sum of n terms of the AP is Sn = n/2 [2a + (n-1)d].
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welcome to you all come back to derive another formula that is the formula to sum of first in terms of an AP to do so first let us consider an AP of in Terms therefore let A1 A2 A3 and so on last turn n b in AP that means AP series to mean some of these terms written this SN SN means sum of N term of an AP or of any numbers of numbers writing so A1 + A2 + A3 plus last term a n here A1 is first term D is equal to let K term minus preceding term k minus one term and N last term again right one thing known that K from the beginning plus K ter from the end is equal to first term plus last term or constant or independent of K so if we write this by this way that means reverse way a n + a n minus one and so on A2 + A1 see first term last term second first second from the beginning second from the end so this formula can be used A+ L that is equal to first term plus last term that is K term from beginning Plus prend this is a Formula or property so if we add this SN plus SN 2 SN and grouping two at a time A1 + a n plus A2 plus a n minus1 plus and so on how will be last term here a n minus 1 plus A2 plus n A1 each bracket the terms two terms continuing first term uh in this case second term from the beginning this is second term from the end so using this formula for each bracketed terms we may write A+ L plus a + L mean last term plus a + L so what we can write now see each term equal that means A+ L A+ L A Plus L how many terms or how many brackets in Brackets so we should write or we must write n into a + L what is left hand side 2 ASN we have not written here because same thing now writing or 2 SN or 2 SN is equal to this actually what we are going to find out SN in some of the first NS so SN is equal to n transpose 2 to the right by 2 a + L therefore the formula to find sum of first in terms of AP so what we will write SN sum of first in terms of an AP that is equal to n by 2 a + L so required formula is SN isal n by 2 a + L this formula may be written another way another way see how we are writing this is Formula One SN = n by 2 we know that l mean last term means n term that is equal to first term a plus n-1 d means common difference of that AP so writing second bracket a in place of L we are writing this what is that a + n -1 into D or SN is equal to n by 2 a + a 2 a + n -1 into d we may call this is second formula this is first Formula so this is another formula to find first n terms is equal to n by 2 2 A + n -1 into D this is another formula actually these two formulas are same but in different form or therefore what is the use of two formulas if first term and last term known then to find the sum of N terms we will use this formula if first term known common difference known then we can use to find sum of NS this formula as for example suppose find the sum of 5 9 13 and so on See n not given we don't know what is the nth term so we will do we will use this formula then the formula we will use this but for another problem problems are taking find the sum of 2 4 6 24 then in this case first time last of given then we will use this formula but in both cases number of terms not given so we write SN formula in this case SN formula we should better to write this formula then the formula will take the form n by 2 2 A is given 5 + n minus one n will remain same n because the value of n is not given and common five and 9 subtract 9 five from 9 that will be four so this will be actually n by 2 5 and 2 multiped 10 here we will get - 4 10 - 4 6 and 4 n so writing 4N first + 6 in this case what will be the formula SN that is equal to n by 2 first term 2 plus last term 4 24 that is equal to 24 + 2 24 + 2 26 by 2 13 n by one that is equal to 13 n in this case only n here is also only n because n is not given if n is given then you will find exact value of some of the N terms or some of the like that sum of 10 terms sum of f ter five terms and so on if any problem you may write in comment box even if you want to get any problem you may write that will be solved thank you very much for watching |
16812 | https://math.stackexchange.com/questions/2344457/notation-for-a-sub-sub-sequence | real analysis - Notation for a sub-sub-sequence - Mathematics Stack Exchange
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Notation for a sub-sub-sequence
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In Real Mathematical Analysis by Charles Pugh, sequences are denoted using the standard notation (a n)(a n), which represents (a 1,a 2,…)(a 1,a 2,…). Furthermore, sub-sequences are denoted by (a n k)(a n k) which expands to (a n 1,a n 2,…)(a n 1,a n 2,…) i.e. n k∈N n k∈N indicates the index (from the mother sequence) of the k k-th term of the sub-sequence. These two notations are thus far standard.
Finally, the confusion arises with the notation of sub-sub-sequences, which are denoted by (a n k(l))(a n k(l)). The expansion seems to be on the l l variable, so that (a n k(l))=(a n k(1),a n k(2),…)(a n k(l))=(a n k(1),a n k(2),…). If this expansion is correct, what is n k(1)n k(1) supposed to indicate? Is k k now a function, where n k(i)∈N n k(i)∈N denotes the index of the i i-th term in the sub-sub-sequence (from the mother sequence, or grandmother sequence)? Is this notation standard?
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asked Jul 2, 2017 at 23:06
jIIjII
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3 (1)(1) I find it oddly progressive that you imagine sequences as matrilineal. (2)(2) Your interpretation seems correct, but I wouldn't worry too much about the notation. I've seen triply-subsequential indices used maybe once thus far in my mathematical experience.Chris –Chris 2017-07-02 23:15:23 +00:00 Commented Jul 2, 2017 at 23:15
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The values n 1,n 2,...n 1,n 2,... represent the values of a strictly increasing f:N→N,f:N→N, with n j=f(j).n j=f(j). The values k 1,k 2,...k 1,k 2,... also represent the values of a strictly increasing k:N→N.k:N→N.
So a n k(j)=a f(k(j)).a n k(j)=a f(k(j)).
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answered Jul 3, 2017 at 6:16
DanielWainfleetDanielWainfleet
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It is common denote subsequence of an arbitrary sequence (a n)(a n) by (a n(k))(a n(k)), where n n is now a monotonic function of k k.
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edited Jul 3, 2017 at 2:45
answered Jul 2, 2017 at 23:22
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I prefer to write (a f(n))(a f(n)) to avoid using the same letter n n in a formula for two different things. If you ever saw a presentation by Prof. Stavo Todorcevic you might think there was a shortage of letters in the alphabet. Even pros had trouble with it.DanielWainfleet –DanielWainfleet 2017-11-30 20:02:53 +00:00 Commented Nov 30, 2017 at 20:02
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16813 | https://www.facebook.com/groups/557157681859053/permalink/1110262599881889/?comment_id=1145131349728347 | Riddles and answers | I am something, we are two brothers moving in the same direction but we can never meet | Facebook
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16814 | https://pubmed.ncbi.nlm.nih.gov/7733230/ | Physiological roles and properties of potassium channels in arterial smooth muscle - PubMed
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. 1995 Apr;268(4 Pt 1):C799-822.
doi: 10.1152/ajpcell.1995.268.4.C799.
Physiological roles and properties of potassium channels in arterial smooth muscle
M T Nelson1,J M Quayle
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1 Department of Pharmacology, University of Vermont, Colchester 05446, USA.
PMID: 7733230
DOI: 10.1152/ajpcell.1995.268.4.C799
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Physiological roles and properties of potassium channels in arterial smooth muscle
M T Nelson et al. Am J Physiol.1995 Apr.
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. 1995 Apr;268(4 Pt 1):C799-822.
doi: 10.1152/ajpcell.1995.268.4.C799.
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M T Nelson1,J M Quayle
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DOI: 10.1152/ajpcell.1995.268.4.C799
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This review examines the properties and roles of the four types of K+ channels that have been identified in the cell membrane of arterial smooth muscle cells. 1) Voltage-dependent K+ (KV) channels increase their activity with membrane depolarization and are important regulators of smooth muscle membrane potential in response to depolarizing stimuli. 2) Ca(2+)-activated K+ (KCa) channels respond to changes in intracellular Ca2+ to regulate membrane potential and play an important role in the control of myogenic tone in small arteries. 3) Inward rectifier K+ (KIR) channels regulate membrane potential in smooth muscle cells from several types of resistance arteries and may be responsible for external K(+)-induced dilations. 4) ATP-sensitive K+ (KATP) channels respond to changes in cellular metabolism and are targets of a variety of vasodilating stimuli. The main conclusions of this review are: 1) regulation of arterial smooth muscle membrane potential through activation or inhibition of K+ channel activity provides an important mechanism to dilate or constrict arteries; 2) KV, KCa, KIR, and KATP channels serve unique functions in the regulation of arterial smooth muscle membrane potential; and 3) K+ channels integrate a variety of vasoactive signals to dilate or constrict arteries through regulation of the membrane potential in arterial smooth muscle.
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16815 | https://web.evanchen.cc/handouts/cmplx/en-cmplx.pdf | Bashing Geometry with Complex Numbers Evan Chen《陳誼廷》 29 August 2015 This is a (quick) English translation of the complex numbers note I wrote for Taiwan IMO 2014 training. Incidentally I was also working on an airplane.
1 The Complex Plane Let C and R denote the set of complex and real numbers, respectively.
Each z ∈C can be expressed as z = a + bi = r (cos θ + i sin θ) = reiθ where a, b, r, θ ∈R and 0 ≤θ < 2π. We write |z| = r = √ a2 + b2 and arg z = θ.
More importantly, each z is associated with a conjugate z = a −bi. It satisfies the properties w ± z = w ± z w · z = w · z w/z = w/z |z|2 = z · z Note that z ∈R ⇐ ⇒z = z and z ∈iR ⇐ ⇒z + z = 0.
Im Re 0 z = 3 + 4i z = 3 −4i −1 −2i |z| = 5 θ Figure 1: Points z = 3 + 4i and −1 −2i; z = 3 −4i is the conjugate.
We represent every point in the plane by a complex number. In particular, we’ll use a capital letter (like Z) to denote the point associated to a complex number (like z).
1 Evan Chen《陳誼廷》— 29 August 2015 Bashing Geometry with Complex Numbers Complex numbers add in the same way as vectors. The multiplication is more interest-ing: for each z1, z2 ∈C we have |z1z2| = |z1| |z2| and arg z1z2 = arg z1 + arg z2.
This multiplication lets us capture a geometric structure. For example, for any points Z and W we can express rotation of Z at W by 90◦as z 7→i(z −w) + w.
Im Re 0 z w i(z −w) + w z −w i(z −w) Im Re 0 z = 3 + 4i iz = −4 + 3i Figure 2: z 7→i(z −w) + w.
2 Elementary Propositions First, some fundamental formulas: Proposition 1. Let A, B, C, D be pairwise distinct points. Then AB ⊥CD if and only if d−c b−a ∈iR; i.e.
d −c b −a + d −c b −a = 0.
Proof. It’s equivalent to d−c b−a ∈iR ⇐ ⇒arg d−c b−a ≡±90◦⇐ ⇒AB ⊥CD.
Proposition 2. Let A, B, C be pairwise distinct points. Then A, B, C are collinear if and only if c−a c−b ∈R; i.e.
c −a c −b = c −a c −b .
Proof. Similar to the previous one.
Proposition 3. Let A, B, C, D be pairwise distinct points. Then A, B, C, D are concyclic if and only if c −a c −b : d −a d −b ∈R.
Proof. It’s not hard to see that arg c−a c−b = ∠ACB and arg d−a d−b = ∠ADB. (Here angles are directed).
2 Evan Chen《陳誼廷》— 29 August 2015 Bashing Geometry with Complex Numbers Im Re Im Re 0 a b c d 0 b −a d −c Figure 3: AB ⊥CD ⇐ ⇒ d−c b−a ∈iR.
Now, let’s state a more commonly used formula.
Lemma 4 (Reflection About a Segment). Let W be the reflection of Z across AB. Then w = (a −b)z + ab −ab a −b .
Of course, it then follows that the foot from Z to AB is exactly 1 2(w + z).
Im Re 0 1 a b z w Im Re 0 1 b −a z −a w −a Im Re 0 1 z−a b−a w−a b−a Figure 4: The reflection of Z across AB.
Proof. According to Figure 4 we obtain w −a b −a = z −a b −a = z −a b −a .
From this we derive w = (a−b)z+ab−ab a−b .
Here are two more formulas.
Theorem 5 (Complex Shoelace). Let A, B, C be points. Then △ABC has signed area i 4 a a 1 b b 1 c c 1 .
In particular, A, B, C are collinear if and only if this determinant vanishes.
3 Evan Chen《陳誼廷》— 29 August 2015 Bashing Geometry with Complex Numbers Proof. Cartesian coordinates.
Often, Theorem 5 is easier to use than Proposition 2.
Actually, we can even write down the formula for an arbitrary intersection of lines.
Proposition 6. Let A, B, C, D be points. Then lines AB and CD intersect at (¯ ab −a¯ b)(c −d) −(a −b)(¯ cd −c ¯ d) (¯ a −¯ b)(c −d) −(a −b)(¯ c −¯ d) .
But unless d = 0 or a, b, c, d are on the unit circle, this formula is often too messy to use.
3 The Unit Circle, and Triangle Centers On the complex plane, the unit circle is of critical importance. Indeed if |z| = 1 we have z = 1 z .
Using the above, we can derive the following lemmas.
Lemma 7. If |a| = |b| = 1 and z ∈C, then the reflection of Z across AB is a + b −abz, and the foot from Z to AB is 1 2 (z + a + b −abz) .
Lemma 8. If A, B, C, D lie on the unit circle then the intersection of AB and CD is given by ab(c + d) −cd(a + b) ab −cd .
These are much easier to work with than the corresponding formulas in general. We can also obtain the triangle centers immediately: Theorem 9. Let ABC be a triangle center, and assume that the circumcircle of ABC coincides with the unit circle of the complex plane. Then the circumcenter, centroid, and orthocenter of ABC are given by 0, 1 3(a + b + c), a + b + c, respectively.
Observe that the Euler line follows from this.
Proof. The results for the circumcenter and centroid are immediate. Let h = a + b + c.
By symmetry it suffices to prove AH ⊥BC. We may set z = h −a b −c = b + c b −c.
Then z = b + c b −c = b + c b −c = 1 b + 1 c 1 b −1 c = c + b c −b = −z so z ∈iR as desired.
We can actually even get the formula for the incenter.
Theorem 10. Let triangle ABC have incenter I and circumcircle Γ. Lines AI, BI, CI meet Γ again at D, E, F. If Γ is the unit circle of the complex plane then there exists x, y, z ∈C satisfying a = x2, b = y2, c = z2 and d = −yz, e = −zx, f = −xy.
Note that |x| = |y| = |z| = 1. Moreover, the incenter I is given by −(xy + yz + zx).
Proof. Show that I is the orthocenter of △DEF.
4 Evan Chen《陳誼廷》— 29 August 2015 Bashing Geometry with Complex Numbers 4 Some Other Lemmas Lemma 11. Let A, B be on the unit circle and select P so that PA, PB are tangents.
Then p = 2 a + b = 2ab a + b.
Proof. Let M be the midpoint of AB and set O = 0. One can show OM · OP = 1 and that O, M, P are collinear; the result follows from this.
a b 2ab a+b Figure 5: Two tangents. p = 2 a+b.
Lemma 12. For any x, y, z, the circumcenter of △XY Z is given by x x¯ x 1 y y¯ y 1 z z¯ z 1 ÷ x ¯ x 1 y ¯ y 1 z ¯ z 1 .
This formula is often easier to apply if we shift z to the point 0 first, then shift back afterwards.
5 Examples Example 13 (MOP 2006). Let H be the orthocenter of triangle ABC. Let D, E, F lie on the circumcircle of ABC such that AD ∥BE ∥CF. Let S, T, U respectively denote the reflections of D, E, F across BC, CA, AB. Prove that points S, T, U, H are concyclic.
Proof. Let (ABC) be the unit circle and h = a + b + c. WLOG, AD, BE, CF are perpendicular to the real axis (rotate appropriately); thus d = a and so on. Thus s = b + c −bcd = b + c −abc and so on; we now have s −t s −u = b −a c −a and h −t h −u = b + abc c + abc.
Compute s −t s −u : h −t h −u = (b −a)(c + abc) (c −a)(b + abc) = 1 b −1 a 1 c + 1 abc 1 c −1 a 1 b + 1 abc = ⇒s −t s −u : h −t h −u ∈R as desired.
5 Evan Chen《陳誼廷》— 29 August 2015 Bashing Geometry with Complex Numbers A B C I E F D G H Q M Example 14 (Taiwan TST 2014). In △ABC with incenter I, the incircle is tangent to CA, AB at E, F. The reflections of E, F across I are G, H. Let Q be the intersection of GH and BC, and let M be the midpoint of BC. Prove that IQ and IM are perpendicular.
Solution. Let D be the foot from I to BC, and set (DEF) as the unit circle. (This lets us exploit the results of Section 3.) Thus |d| = |e| = |f| = 1, and moreover g = −e, h = −f. Let x = d = 1 d and define y, z similarly. Then b = 2 d + f = 2 x + z .
Similarly, c = 2 x+y, so m = 1 2(b + c) = 1 x + y + 1 x + z = 2x + y + z (x + y)(x + z).
Next, we have Q = DD ∩GH, which implies q = dd(g + h) −gh(d + d) d2 −gh = 1 x2 −1 y −1 z −1 yz 2 x 1 x2 −1 yz = 2x + y + z x2 −yz .
so m/q = x2 −yz (x + y)(x + z).
Now, m/q = 1 x2 −1 yz 1 x + 1 y 1 x + 1 z = yz −x2 (x + y)(x + z) = −m/q thus m/q ∈iR, as desired.
Example 15 (USAMO 2012). Let P be a point in the plane of △ABC, and γ a line through P. Let A′, B′, C′ be the points where the reflections of lines PA, PB, PC with respect to γ intersect lines BC, AC, AB respectively. Prove that A′, B′, C′ are collinear.
Solution. Let p = 0 and set γ as the real line. Then A′ is the intersection of bc and p¯ a.
So, using Proposition 6 we get a′ = ¯ a(¯ bc −b¯ c) (¯ b −¯ c)¯ a −(b −c)a.
6 Evan Chen《陳誼廷》— 29 August 2015 Bashing Geometry with Complex Numbers A B C P A′ Note that ¯ a′ = a(b¯ c −¯ bc) (b −c)a −(¯ b −¯ c)¯ a.
Thus by Theorem 5, it suffices to prove 0 = ¯ a(¯ bc−b¯ c) (¯ b−¯ c)¯ a−(b−c)a a(b¯ c−¯ bc) (b−c)a−(¯ b−¯ c)¯ a 1 ¯ b(¯ ca−c¯ a) (¯ c−¯ a)¯ b−(c−a)b b(c¯ a−¯ ca) (c−a)b−(¯ c−¯ a)¯ b 1 ¯ c(¯ ab−a¯ b) (¯ a−¯ b)¯ c−(a−b)c c(a¯ b−¯ ab) (a−b)c−(¯ a−¯ b)¯ c 1 .
This is equivalent to 0 = ¯ a(¯ bc −b¯ c) a(¯ bc −b¯ c) (¯ b −¯ c)¯ a −(b −c)a ¯ b(¯ ca −c¯ a) b(¯ ca −c¯ a) (¯ c −¯ a)¯ b −(c −a)b ¯ c(¯ ab −a¯ b) c(¯ ab −a¯ b) (¯ a −¯ b)¯ c −(a −b)c .
Evaluating the determinant gives X cyc ((¯ b −¯ c)¯ a −(b −c)a) · − b ¯ b c ¯ c · (¯ ca −c¯ a) ¯ ab −a¯ b or, noting the determinant is b¯ c −¯ bc and factoring it out, (¯ bc −c¯ b)(¯ ca −c¯ a)(¯ ab −a¯ b) X cyc ab −ac + ¯ c¯ a −¯ b¯ a = 0.
Example 16 (Taiwan TST Quiz 2014). Let I and O be the incenter and circumcenter of ABC. A line ℓis drawn parallel to BC and tangent to the incircle of ABC. Let X, Y be on ℓso that I, O, X are collinear and ∠XIY = 90◦. Show that A, X, O, Y are concyclic.
Solution. Let X′ and Y ′ respectively denote the reflections of X and Y across I. Note that X, Y lie on BC. Also, let P, Q be the intersections of IY with the circumcircle.
Of course, (ABC) is the unit circle. Let j be the complex number corresponding to I (to avoid confusion with i = √−1). Thus, x′ = bc −bc (j −0) − j0 −j0 (b −c) (b −c)(j −0) −(b −c)(j −0) = j · c2−b2 bc j · c−b bc −(b −c)j = j(b + c) j + bcj .
We seek y′ now. Consider the quadratic equation in z given by z −j j + 1 z −j j = 0 ⇐ ⇒z2 −2jz + j/j = 0.
7 Evan Chen《陳誼廷》— 29 August 2015 Bashing Geometry with Complex Numbers A B C O I X Y P Q Y ′ X′ Its zeros in z are p and q, which implies that p + q = 2j and pq = j/j (by Vieta!). From this we can compute y′ = pq(b + c) −bc(p + q) pq −bc = j(b + c) −2bcjj j −bcj = j(b + c) −2bcjj j −bcj .
which gives x = 2j −x′ = j(2j −b −c + 2bcj) j + bcj and y = 2j −y′ = j(2j −b −c) j −bcj .
From this we can obtain y −x = j · (2j −b −c)(j + bcj) −(2j −b −c + 2bcj)(j −bcj) (j −bcj)(j + bcj) = j · 2bcj(2j −b −c) −2bcj(j −bcj) (j −bcj)(j + bcj) = j · 2bcj j −b −c + bcj (j −bcj)(j + bcj) X = y −x x = 2bcj j −b −c + bcj (j −bcj)(2j −b −c + 2bcj) A = y −a a = j(2j −b −c −a) + abcj a(j −bcj) We need to prove X/A = X/A. Now set a = x2, b = y2, c = z2, j = −(xy + yz + zx), j = −x+y+z xyz (this is a different x, y than the points X and Y .) So, the above rewrites as X = 2yz x (x + y + z)( yz x (x + y + z) + y2 + z2 + xy + yz + zx) −yz x (x + y + z) + xy + yz + zx y2 + z2 + 2(xy + yz + zx) + 2 yz x (x + y + z) = 2yz(x + y + z) 2xyz + P sym x2y (y + z)(x2 −yz) (x(y + z)(2x + y + z) + 2yz(x + y + z)) 8 Evan Chen《陳誼廷》— 29 August 2015 Bashing Geometry with Complex Numbers = 2yz(x + y + z)(x + y)(x + z) (x2 −yz) ((x2 + yz)(y + z) + (xy + yz + zx)(x + y + z)) and A = (xy + yz + zx)(x + y + z)2 −xyz(x + y + z) x2(−(xy + yz + zx) + yz x (x + y + z)) = (x + y + z)(x + y)(y + z)(z + x) x(yz −x2)(y + z) thus X/A = −2xyz (x2 + yz)(y + z) + (x + y + z)(xy + yz + zx) = −2 xyz ( 1 x2 + 1 yz)( 1 y + 1 z) + ( 1 x + 1 y + 1 z)( 1 xy + 1 yz + 1 zx) = X/A.
6 Practice Problems 1. Let ABCD be cyclic. Let HA, HB, HC, HD denote the orthocenters of BCD, CDA, DAB, ABC. Show that AHA, BHB, CHC, DHD are concurrent.
2. (China TST 2011) Let Γ be the circumcircle of a triangle ABC. Assume AA′, BB′, CC′ are diameters of Γ. Let P be a point inside ABC and let D, E, F be the feet from P to BC, CA, AB. Let X be the reflection of A′ across D; define Y and Z similarly. Prove that △XY Z ∼△ABC.
3. In circumscribed quadrilateral ABCD with incircle ω, Prove that the midpoint of AC and the midpoint of BD are collinear with the center of ω.
4. (Simson Line) Let ABC be a triangle and P a point on its circumcircle.
(a) Let D, E, F be the feet from P to BC, CA, AB. Show that D, E, F are collinear.
(b) Moreover, prove that the line through these points bisects PH, where H is the orthocenter of ABC.
5. (PUMaC Finals) Let γ and I be the incircle and incenter of triangle ABC. Let D, E, F be the tangency points of γ to BC, CA, AB and let D′ be the reflection of D about I. Assume EF intersects the tangents to γ at D and D′ at points P and Q. Show that ∠DAD′ + ∠PIQ = 180◦.
6. (Schiffler Point) Let triangle ABC have incenter I. Prove that the Euler lines of △AIB, △BIC, △CIA, △ABC are concurrent.
7. (USA TST 2014) Let ABCD be a cyclic quadrilateral and let E, F, G, H be the midpoints of AB, BC, CD, DA. Call W, X, Y , Z the orthocenters of AHE, BEF, CFG, DGH. Prove that ABCD and WXY Z have the same area.
8. (Iran 2004) Let O be the circumcenter of ABC. A line ℓthrough O cuts AB and AC at points X and Y . Let M and N be the midpoints of BY , CX. Show that ∠MON = ∠BAC.
9. (APMO 2010) Let ABC be an acute triangle, where AB > BC and AC > BC.
Denote by O and H the circumcenter and orthocenter. The circumcircle of AHC intersects AB again at M; the circumcircle of AHB intersects AC again at N.
Prove that the circumcenter of triangle MNH lies on line OH.
9 Evan Chen《陳誼廷》— 29 August 2015 Bashing Geometry with Complex Numbers 10. (Iran 2013) Let ABC be acute, and M the midpoint of minor arc d BC. Let N be on the circumcircle of ABC such that AN ⊥BC, and let K, L lie on AB, AC so that OK ∥MB, OL ∥MC. (Here O is the circumcenter of ABC). Prove that NK = NL.
11. (MOP 2006) Cyclic quadrilateral ABCD has circumcenter O. Let P be a point in the plane and let O1, O2, O3, O4 be the circumcenters of PAB, PBC, PCD, PDA. Show that the midpoints of O1O3, O2O4, OP are concurrent.
10 |
16816 | https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation | Contents
Cauchy's functional equation
Cauchy's functional equation is the functional equation:
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
.
{\displaystyle f(x+y)=f(x)+f(y).\ }
A function
f
{\displaystyle f}
that solves this equation is called an additive function. Over the rational numbers, it can be shown using elementary algebra that there is a single family of solutions, namely
f
:
x
↦
c
x
{\displaystyle f\colon x\mapsto cx}
for any rational constant
c
.
{\displaystyle c.}
Over the real numbers, the family of linear maps
f
:
x
↦
c
x
,
{\displaystyle f:x\mapsto cx,}
now with
c
{\displaystyle c}
an arbitrary real constant, is likewise a family of solutions; however there can exist other solutions not of this form that are extremely complicated. However, any of a number of regularity conditions, some of them quite weak, will preclude the existence of these pathological solutions. For example, an additive function
f
:
R
→
R
{\displaystyle f\colon \mathbb {R} \to \mathbb {R} }
is linear if:
On the other hand, if no further conditions are imposed on
f
,
{\displaystyle f,}
then (assuming the axiom of choice) there are infinitely many other functions that satisfy the equation. This was proved in 1905 by Georg Hamel using Hamel bases. Such functions are sometimes called Hamel functions.
The fifth problem on Hilbert's list is a generalisation of this equation. Functions where there exists a real number
c
{\displaystyle c}
such that
f
(
c
x
)
≠
c
f
(
x
)
{\displaystyle f(cx)\neq cf(x)}
are known as Cauchy-Hamel functions and are used in Dehn-Hadwiger invariants which are used in the extension of Hilbert's third problem from 3D to higher dimensions.
This equation is sometimes referred to as Cauchy's additive functional equation to distinguish it from the other functional equations introduced by Cauchy in 1821, the exponential functional equation
f
(
x
+
y
)
=
f
(
x
)
f
(
y
)
,
{\displaystyle f(x+y)=f(x)f(y),}
the logarithmic functional equation
f
(
x
y
)
=
f
(
x
)
+
f
(
y
)
,
{\displaystyle f(xy)=f(x)+f(y),}
and the multiplicative functional equation
f
(
x
y
)
=
f
(
x
)
f
(
y
)
.
{\displaystyle f(xy)=f(x)f(y).}
Solutions over the rational numbers
A simple argument, involving only elementary algebra, demonstrates that the set of additive maps
f
:
V
→
W
{\displaystyle f\colon V\to W}
, where
V
,
W
{\displaystyle V,W}
are vector spaces over an extension field of
Q
{\displaystyle \mathbb {Q} }
, is identical to the set of
Q
{\displaystyle \mathbb {Q} }
-linear maps from
V
{\displaystyle V}
to
W
{\displaystyle W}
.
Theorem: Let
f
:
V
→
W
{\displaystyle f\colon V\to W}
be an additive function. Then
f
{\displaystyle f}
is
Q
{\displaystyle \mathbb {Q} }
-linear.
Proof: We want to prove that any solution
f
:
V
→
W
{\displaystyle f\colon V\to W}
to Cauchy’s functional equation,
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
{\displaystyle f(x+y)=f(x)+f(y)}
, satisfies
f
(
q
v
)
=
q
f
(
v
)
{\displaystyle f(qv)=qf(v)}
for any
q
∈
Q
{\displaystyle q\in \mathbb {Q} }
and
v
∈
V
{\displaystyle v\in V}
. Let
v
∈
V
{\displaystyle v\in V}
.
First note
f
(
0
)
=
f
(
0
+
0
)
=
f
(
0
)
+
f
(
0
)
{\displaystyle f(0)=f(0+0)=f(0)+f(0)}
, hence
f
(
0
)
=
0
{\displaystyle f(0)=0}
, and therewith
0
f
(
0
)
=
f
(
v
+
(
−
v
)
)
=
f
(
v
)
+
f
(
−
v
)
{\displaystyle 0=f(0)=f(v+(-v))=f(v)+f(-v)}
from which follows
f
(
−
v
)
=
−
f
(
v
)
{\displaystyle f(-v)=-f(v)}
.
Via induction,
f
(
m
v
)
=
m
f
(
v
)
{\displaystyle f(mv)=mf(v)}
is proved for any
m
∈
N
∪
{
0
}
{\displaystyle m\in \mathbb {N} \cup {0}}
.
For any negative integer
m
∈
Z
{\displaystyle m\in \mathbb {Z} }
we know
−
m
∈
N
{\displaystyle -m\in \mathbb {N} }
, therefore
f
(
m
v
)
=
f
(
(
−
m
)
(
−
v
)
)
=
(
−
m
)
f
(
−
v
)
=
(
−
m
)
(
−
f
(
v
)
)
=
m
f
(
v
)
{\displaystyle f(mv)=f((-m)(-v))=(-m)f(-v)=(-m)(-f(v))=mf(v)}
. Thus far we have proved
Let
n
∈
N
{\displaystyle n\in \mathbb {N} }
, then
f
(
v
)
=
f
(
n
n
−
1
v
)
=
n
f
(
n
−
1
v
)
{\displaystyle f(v)=f(nn^{-1}v)=nf(n^{-1}v)}
and hence
f
(
n
−
1
v
)
=
n
−
1
f
(
v
)
.
{\displaystyle f(n^{-1}v)=n^{-1}f(v).}
Finally, any
q
∈
Q
{\displaystyle q\in \mathbb {Q} }
has a representation
q
m
n
{\displaystyle q={\frac {m}{n}}}
with
m
∈
Z
{\displaystyle m\in \mathbb {Z} }
and
n
∈
N
{\displaystyle n\in \mathbb {N} }
, so, putting things together,
Properties of nonlinear solutions over the real numbers
We prove below that any other solutions must be highly pathological functions.
In particular, it is shown that any other solution must have the property that its graph
{
(
x
,
f
(
x
)
)
|
x
∈
R
}
{\displaystyle {(x,f(x))\vert x\in \mathbb {R} }}
is dense in
R
2
,
{\displaystyle \mathbb {R} ^{2},}
that is, that any disk in the plane (however small) contains a point from the graph.
From this it is easy to prove the various conditions given in the introductory paragraph.
Lemma—Let
t
0
{\displaystyle t>0}
. If
f
{\displaystyle f}
satisfies the Cauchy functional equation on the interval
[
0
,
t
]
{\displaystyle [0,t]}
, but is not linear, then its graph is dense on the strip
[
0
,
t
]
×
R
{\displaystyle [0,t]\times \mathbb {R} }
.
WLOG, scale
f
{\displaystyle f}
on the x-axis and y-axis, so that
f
{\displaystyle f}
satisfies the Cauchy functional equation on
[
0
,
1
]
{\displaystyle [0,1]}
, and
f
(
1
)
=
1
{\displaystyle f(1)=1}
.
It suffices to show that the graph of
f
{\displaystyle f}
is dense in
(
0
,
1
)
×
R
{\displaystyle (0,1)\times \mathbb {R} }
, which is dense in
[
0
,
1
]
×
R
{\displaystyle [0,1]\times \mathbb {R} }
.
Since
f
{\displaystyle f}
is not linear, we have
f
(
a
)
≠
a
{\displaystyle f(a)\neq a}
for some
a
∈
(
0
,
1
)
{\displaystyle a\in (0,1)}
.
Claim: The lattice defined by
L
:=
{
(
r
1
+
r
2
a
,
r
1
+
r
2
f
(
a
)
)
:
r
1
,
r
2
∈
Q
}
{\displaystyle L:={(r_{1}+r_{2}a,r_{1}+r_{2}f(a)):r_{1},r_{2}\in \mathbb {Q} }}
is dense in
R
2
{\displaystyle \mathbb {R} ^{2}}
.
Consider the linear transformation
A
:
R
2
→
R
2
{\displaystyle A:\mathbb {R} ^{2}\to \mathbb {R} ^{2}}
defined by
A
(
x
,
y
)
=
[
1
a
1
f
(
a
)
]
[
x
y
]
{\displaystyle A(x,y)={\begin{bmatrix}1&a\1&f(a)\end{bmatrix}}{\begin{bmatrix}x\y\end{bmatrix}}}
With this transformation, we have
L
A
(
Q
2
)
{\displaystyle L=A(\mathbb {Q} ^{2})}
.
Since
det
A
=
f
(
a
)
−
a
≠
0
{\displaystyle \det A=f(a)-a\neq 0}
, the transformation is invertible, thus it is bicontinuous. Since
Q
2
{\displaystyle \mathbb {Q} ^{2}}
is dense in
R
2
{\displaystyle \mathbb {R} ^{2}}
, so is
L
{\displaystyle L}
.
◻
{\displaystyle \square }
Claim: if
r
1
,
r
2
∈
Q
{\displaystyle r_{1},r_{2}\in \mathbb {Q} }
, and
r
1
+
r
2
a
∈
(
0
,
1
)
{\displaystyle r_{1}+r_{2}a\in (0,1)}
, then
f
(
r
1
+
r
2
a
)
=
r
1
+
r
2
f
(
a
)
{\displaystyle f(r_{1}+r_{2}a)=r_{1}+r_{2}f(a)}
.
If
r
1
,
r
2
≥
0
{\displaystyle r_{1},r_{2}\geq 0}
, then it is true by additivity. If
r
1
,
r
2
<
0
{\displaystyle r_{1},r_{2}<0}
, then
r
1
+
r
2
a
<
0
{\displaystyle r_{1}+r_{2}a<0}
, contradiction.
If
r
1
≥
0
,
r
2
<
0
{\displaystyle r_{1}\geq 0,r_{2}<0}
, then since
r
1
+
r
2
a
0
{\displaystyle r_{1}+r_{2}a>0}
, we have
r
1
0
{\displaystyle r_{1}>0}
. Let
k
{\displaystyle k}
be a positive integer large enough such that
r
1
k
,
−
r
2
a
k
∈
(
0
,
1
)
{\displaystyle {\frac {r_{1}}{k}},{\frac {-r_{2}a}{k}}\in (0,1)}
. Then we have by additivity:
f
(
r
1
k
+
r
2
a
k
)
+
f
(
−
r
2
a
k
)
=
f
(
r
1
k
)
{\displaystyle f\left({\frac {r_{1}}{k}}+{\frac {r_{2}a}{k}}\right)+f\left({\frac {-r_{2}a}{k}}\right)=f\left({\frac {r_{1}}{k}}\right)}
That is,
1
k
f
(
r
1
+
r
2
a
)
+
−
r
2
k
f
(
a
)
=
r
1
k
{\displaystyle {\frac {1}{k}}f\left(r_{1}+r_{2}a\right)+{\frac {-r_{2}}{k}}f\left(a\right)={\frac {r_{1}}{k}}}
◻
{\displaystyle \square }
Thus, the graph of
f
{\displaystyle f}
contains
L
∩
(
(
0
,
1
)
×
R
)
{\displaystyle L\cap ((0,1)\times \mathbb {R} )}
, which is dense in
(
0
,
1
)
×
R
{\displaystyle (0,1)\times \mathbb {R} }
.
Existence of nonlinear solutions over the real numbers
The linearity proof given above also applies to
f
:
α
Q
→
R
,
{\displaystyle f\colon \alpha \mathbb {Q} \to \mathbb {R} ,}
where
α
Q
{\displaystyle \alpha \mathbb {Q} }
is a scaled copy of the rationals. This shows that only linear solutions are permitted when the domain of
f
{\displaystyle f}
is restricted to such sets. Thus, in general, we have
f
(
α
q
)
=
f
(
α
)
q
{\displaystyle f(\alpha q)=f(\alpha )q}
for all
α
∈
R
{\displaystyle \alpha \in \mathbb {R} }
and
q
∈
Q
.
{\displaystyle q\in \mathbb {Q} .}
However, as we will demonstrate below, highly pathological solutions can be found for functions
f
:
R
→
R
{\displaystyle f\colon \mathbb {R} \to \mathbb {R} }
based on these linear solutions, by viewing the reals as a vector space over the field of rational numbers. Note, however, that this method is nonconstructive, relying as it does on the existence of a (Hamel) basis for any vector space, a statement proved using Zorn's lemma. (In fact, the existence of a basis for every vector space is logically equivalent to the axiom of choice.) There exist models such as the Solovay model where all sets of reals are measurable which are consistent with ZF + DC, and therein all solutions are linear.
To show that solutions other than the ones defined by
f
(
x
)
=
f
(
1
)
x
{\displaystyle f(x)=f(1)x}
exist, we first note that because every vector space has a basis, there is a basis for
R
{\displaystyle \mathbb {R} }
over the field
Q
,
{\displaystyle \mathbb {Q} ,}
i.e. a set
B
⊂
R
{\displaystyle {\mathcal {B}}\subset \mathbb {R} }
with the property that any
x
∈
R
{\displaystyle x\in \mathbb {R} }
can be expressed uniquely as
x
∑
i
∈
I
λ
i
x
i
,
{\textstyle x=\sum {i\in I}{\lambda {i}x_{i}},}
where
{
x
i
}
i
∈
I
{\displaystyle {x_{i}}_{i\in I}}
is a finite subset of
B
,
{\displaystyle {\mathcal {B}},}
and each
λ
i
{\displaystyle \lambda _{i}}
is in
Q
.
{\displaystyle \mathbb {Q} .}
We note that because no explicit basis for
R
{\displaystyle \mathbb {R} }
over
Q
{\displaystyle \mathbb {Q} }
can be written down, the pathological solutions defined below likewise cannot be expressed explicitly.
As argued above, the restriction of
f
{\displaystyle f}
to
x
i
Q
{\displaystyle x_{i}\mathbb {Q} }
must be a linear map for each
x
i
∈
B
.
{\displaystyle x_{i}\in {\mathcal {B}}.}
Moreover, because
x
i
q
↦
f
(
x
i
)
q
{\displaystyle x_{i}q\mapsto f(x_{i})q}
for
q
∈
Q
,
{\displaystyle q\in \mathbb {Q} ,}
it is clear that
f
(
x
i
)
x
i
{\displaystyle f(x_{i}) \over x_{i}}
is the constant of proportionality. In other words,
f
:
x
i
Q
→
R
{\displaystyle f\colon x_{i}\mathbb {Q} \to \mathbb {R} }
is the map
ξ
↦
[
f
(
x
i
)
/
x
i
]
ξ
.
{\displaystyle \xi \mapsto [f(x_{i})/x_{i}]\xi .}
Since any
x
∈
R
{\displaystyle x\in \mathbb {R} }
can be expressed as a unique (finite) linear combination of the
x
i
{\displaystyle x_{i}}
s, and
f
:
R
→
R
{\displaystyle f\colon \mathbb {R} \to \mathbb {R} }
is additive,
f
(
x
)
{\displaystyle f(x)}
is well-defined for all
x
∈
R
{\displaystyle x\in \mathbb {R} }
and is given by:
f
(
x
)
=
f
(
∑
i
∈
I
λ
i
x
i
)
=
∑
i
∈
I
f
(
x
i
λ
i
)
∑
i
∈
I
f
(
x
i
)
λ
i
.
{\displaystyle f(x)=f{\Big (}\sum {i\in I}\lambda {i}x_{i}{\Big )}=\sum {i\in I}f(x{i}\lambda {i})=\sum {i\in I}f(x_{i})\lambda _{i}.}
It is easy to check that
f
{\displaystyle f}
is a solution to Cauchy's functional equation given a definition of
f
{\displaystyle f}
on the basis elements,
f
:
B
→
R
.
{\displaystyle f\colon {\mathcal {B}}\to \mathbb {R} .}
Moreover, it is clear that every solution is of this form. In particular, the solutions of the functional equation are linear if and only if
f
(
x
i
)
x
i
{\displaystyle f(x_{i}) \over x_{i}}
is constant over all
x
i
∈
B
.
{\displaystyle x_{i}\in {\mathcal {B}}.}
Thus, in a sense, despite the inability to exhibit a nonlinear solution, "most" (in the sense of cardinality) solutions to the Cauchy functional equation are actually nonlinear and pathological.
See also
References
External links |
16817 | https://www.plannedparenthood.org/uploads/filer_public/b2/62/b2625720-f60f-4f9f-8235-be87e43e3da1/cdc_info_2.pdf | CS246652B National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention Chlamydia – CDC Fact Sheet Chlamydia is a common sexually transmitted disease (STD) that can be easily cured. If left untreated, chlamydia can make it difficult for a woman to get pregnant.
What is chlamydia?
Chlamydia is a common STD that can infect both men and women. It can cause serious, permanent damage to a woman’s reproductive system, making it difficult or impossible for her to get pregnant later on. Chlamydia can also cause a potentially fatal ectopic pregnancy (pregnancy that occurs outside the womb).
How is chlamydia spread?
You can get chlamydia by having anal, vaginal, or oral sex with someone who has chlamydia. If your sex partner is male you can still get chlamydia even if he does not ejaculate (cum). If you’ve had chlamydia and were treated in the past, you can still get infected again if you have unprotected sex with someone who has chlamydia.
If you are pregnant, you can give chlamydia to your baby during childbirth.
How can I reduce my risk of getting chlamydia?
The only way to avoid STDs is to not have vaginal, anal, or oral sex.
• • • Being in a long-term mutually monogamous relationship with a partner who has been tested and has negative STD test results; • • Using latex condoms the right way every time you have sex.
Am I at risk for chlamydia?
Anyone who has sex can get chlamydia through unprotected anal, vaginal, or oral sex. However, sexually active young people are at a higher risk of getting chlamydia. This is due to behaviors and biological factors common among young people. Gay, bisexual, and other men who have sex with men are also at risk since chlamydia can be spread through oral and anal sex.
Have an honest and open talk with your health care provider and ask whether you should be tested for chlamydia or other STDs. If you are a sexually active woman younger than 25 years, or an older woman with risk factors such as new or multiple sex partners, or a sex partner who has a sexually transmitted infection, you should get a test for chlamydia every year. Gay, bisexual, and men who have sex with men; as well as pregnant women should also be tested for chlamydia. I’m pregnant. How does chlamydia affect my baby?
If you are pregnant and have chlamydia, you can pass the infection to your baby during delivery. This could cause an eye infection or pneumonia in your newborn. Having chlamydia may also make it more likely to deliver your baby too early. If you are pregnant, you should be tested for chlamydia at your first prenatal visit. Testing and treatment are the best ways to prevent health problems. Page 2 of 2 How do I know if I have chlamydia?
Most people who have chlamydia have no symptoms. If you do have symptoms, they may not appear until several weeks after you have sex with an infected partner. Even when chlamydia causes no symptoms, it can damage your reproductive system.
Women with symptoms may notice • • An abnormal vaginal discharge; • • A burning sensation when urinating.
Symptoms in men can include • • A discharge from their penis; • • A burning sensation when urinating; • • Pain and swelling in one or both testicles (although this is less common).
Men and women can also get infected with chlamydia in their rectum, either by having receptive anal sex, or by spread from another infected site (such as the vagina). While these infections often cause no symptoms, they can cause • • Rectal pain; • • Discharge; • • Bleeding.
You should be examined by your doctor if you notice any of these symptoms or if your partner has an STD or symptoms of an STD, such as an unusual sore, a smelly discharge, burning when urinating, or bleeding between periods.
How will my doctor know if I have chlamydia?
There are laboratory tests to diagnose chlamydia. Your health care provider may ask you to provide a urine sample or may use (or ask you to use) a cotton swab to get a sample from your vagina to test for chlamydia. Can chlamydia be cured?
Yes, chlamydia can be cured with the right treatment. It is important that you take all of the medication your doctor prescribes to cure your infection. When taken properly it will stop the infection and could decrease your chances of having complications later on. Medication for chlamydia should not be shared with anyone. Repeat infection with chlamydia is common. You should be tested again about three months after you are treated, even if your sex partner(s) was treated. What happens if I don’t get treated?
The initial damage that chlamydia causes often goes unnoticed. However, chlamydia can lead to serious health problems.
If you are a woman, untreated chlamydia can spread to your uterus and fallopian tubes (tubes that carry fertilized eggs from the ovaries to the uterus), causing pelvic inflammatory disease (PID). PID often has no symptoms, however some women may have abdominal and pelvic pain. Even if it doesn’t cause symptoms initially, PID can cause permanent damage to your reproductive system and lead to long-term pelvic pain, inability to get pregnant, and potentially deadly ectopic pregnancy (pregnancy outside the uterus).
Men rarely have health problems linked to chlamydia. Infection sometimes spreads to the tube that carries sperm from the testicles, causing pain and fever. Rarely, chlamydia can prevent a man from being able to have children.
Untreated chlamydia may also increase your chances of getting or giving HIV – the virus that causes AIDS.
I was treated for chlamydia. When can I have sex again?
You should not have sex again until you and your sex partner(s) have completed treatment. If your doctor prescribes a single dose of medication, you should wait seven days after taking the medicine before having sex. If your doctor prescribes a medicine for you to take for seven days, you should wait until you have taken all of the doses before having sex.
Where can I get more information?
Division of STD Prevention (DSTDP) Centers for Disease Control and Prevention www.cdc.gov/std CDC-INFO Contact Center 1-800-CDC-INFO (1-800-232-4636) Contact dcs/ContactUs/Form CDC National Prevention Information Network (NPIN) P.O. Box 6003 Rockville, MD 20849-6003 E-mail: npin-info@cdc.gov American Sexual Health Association (ASHA) stdsstis/ P.O. Box 13827 Research Triangle Park, NC 27709-3827 1-800-783-9877 Last reviewed: January 23, 2014 CS246652C National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention Gonorrhea – CDC Fact Sheet Anyone who is sexually active can get gonorrhea. Gonorrhea can cause very serious complications when not treated, but can be cured with the right medication.
What is gonorrhea?
Gonorrhea is a sexually transmitted disease (STD) that can infect both men and women. It can cause infections in the genitals, rectum, and throat. It is a very common infection, especially among young people ages 15-24 years.
How is gonorrhea spread?
You can get gonorrhea by having anal, vaginal, or oral sex with someone who has gonorrhea. A pregnant woman with gonorrhea can give the infection to her baby during childbirth.
How can I reduce my risk of getting gonorrhea?
The only way to avoid STDs is to not have vaginal, anal, or oral sex.
• • • Being in a long-term mutually monogamous relationship with a partner who has been tested and has negative STD test results; • • Using latex condoms and dental dams the right way every time you have sex.
Am I at risk for gonorrhea?
Any sexually active person can get gonorrhea through unprotected anal, vaginal, or oral sex. If you are sexually active, have an honest and open talk with your health care provider and ask whether you should be tested for gonorrhea or other STDs. If you are a sexually active man who is gay, bisexual, or who has sex with men, you should be tested for gonorrhea every year. If you are a sexually active women younger than 25 years or an older women with risk factors such as new or multiple sex partners, or a sex partner who has a sexually transmitted infection, you should be tested for gonorrhea every year. I’m pregnant. How does gonorrhea affect my baby?
If you are pregnant and have gonorrhea, you can give the infection to your baby during delivery. This can cause serious health problems for your baby. If you are pregnant, it is important that you talk to your health care provider so that you get the correct examination, testing, and treatment, as necessary. Treating gonorrhea as soon as possible will make health complications for your baby less likely.
How do I know if I have gonorrhea?
Some men with gonorrhea may have no symptoms at all. However, men who do have symptoms, may have: • • A burning sensation when urinating; • • A white, yellow, or green discharge from the penis; • • Painful or swollen testicles (although this is less common).
Most women with gonorrhea do not have any symptoms. Even when a woman has symptoms, they are often mild and can be mistaken for Page 2 of 2 a bladder or vaginal infection. Women with gonorrhea are at risk of developing serious complications from the infection, even if they don’t have any symptoms.
Symptoms in women can include: • • Painful or burning sensation when urinating: • • Increased vaginal discharge: • • Vaginal bleeding between periods.
Rectal infections may either cause no symptoms or cause symptoms in both men and women that may include: • • Discharge; • • Anal itching; • • Soreness; • • Bleeding; • • Painful bowel movements.
You should be examined by your doctor if you notice any of these symptoms or if your partner has an STD or symptoms of an STD, such as an unusual sore, a smelly discharge, burning when urinating, or bleeding between periods.
How will my doctor know if I have gonorrhea?
Most of the time, urine can be used to test for gonorrhea. However, if you have had oral and/or anal sex, swabs may be used to collect samples from your throat and/or rectum. In some cases, a swab may be used to collect a sample from a man’s urethra (urine canal) or a woman’s cervix (opening to the womb). Can gonorrhea be cured?
Yes, gonorrhea can be cured with the right treatment. It is important that you take all of the medication your doctor prescribes to cure your infection. Medication for gonorrhea should not be shared with anyone. Although medication will stop the infection, it will not undo any permanent damage caused by the disease. It is becoming harder to treat some gonorrhea, as drug-resistant strains of gonorrhea are increasing. If your symptoms continue for more than a few days after receiving treatment, you should return to a health care provider to be checked again.
I was treated for gonorrhea. When can I have sex again?
You should wait seven days after finishing all medications before having sex. To avoid getting infected with gonorrhea again or spreading gonorrhea to your partner(s), you and your sex partner(s) should avoid having sex until you have each completed treatment. If you’ve had gonorrhea and took medicine in the past, you can still get infected again if you have unprotected sex with a person who has gonorrhea. What happens if I don’t get treated?
Untreated gonorrhea can cause serious and permanent health problems in both women and men.
In women, untreated gonorrhea can cause pelvic inflammatory disease (PID). Some of the complications of PID are • • Formation of scar tissue that blocks fallopian tubes; • • Ectopic pregnancy (pregnancy outside the womb); • • Infertility (inability to get pregnant); • • Long-term pelvic/abdominal pain.
In men, gonorrhea can cause a painful condition in the tubes attached to the testicles. In rare cases, this may cause a man to be sterile, or prevent him from being able to father a child.
Rarely, untreated gonorrhea can also spread to your blood or joints. This condition can be life-threatening. Untreated gonorrhea may also increase your chances of getting or giving HIV – the virus that causes AIDS.
Where can I get more information?
Division of STD Prevention (DSTDP) Centers for Disease Control and Prevention www.cdc.gov/std CDC-INFO Contact Center 1-800-CDC-INFO (1-800-232-4636) Contact dcs/ContactUs/Form Last reviewed: January 29, 2014 CS246943A National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention Genital Herpes – CDC Fact Sheet Herpes is a common sexually transmitted disease (STD) that any sexually active person can get. Most people with the virus don’t have symptoms. It is important to know that even without signs of the disease, it can still spread to sexual partners. What is genital herpes?
Genital herpes is an STD caused by two types of viruses. The viruses are called herpes simplex type 1 and herpes simplex type 2.
How common is genital herpes?
Genital herpes is common in the United States. In the United States, about one out of every six people aged 14 to 49 years have genital herpes.
How is genital herpes spread?
You can get herpes by having oral, vaginal, or anal sex with someone who has the disease. Fluids found in a herpes sore carry the virus, and contact with those fluids can cause infection. You can also get herpes from an infected sex partner who does not have a visible sore or who may not know he or she is infected because the virus can be released through your skin and spread the infection to your sex partner(s). How can I reduce my risk of getting herpes?
The only way to avoid STDs is to not have vaginal, anal, or oral sex.• If you are sexually active, you can do the following things to lower your chances of getting herpes: • • Being in a long-term mutually monogamous relationship with a partner who has been tested and has negative STD test results; • • Using latex condoms the right way every time you have sex.
Herpes symptoms can occur in both male and female genital areas that are covered by a latex condom. However, outbreaks can also occur in areas that are not covered by a condom so condoms may not fully protect you from getting herpes.
I’m pregnant. How could genital herpes affect my baby?
If you are pregnant and have genital herpes, it is even more important for you to go to prenatal care visits. You need to tell your doctor if you have ever had symptoms of, been exposed to, or been diagnosed with genital herpes. Sometimes genital herpes infection can lead to miscarriage. It can also make it more likely for you to deliver your baby too early. Herpes infection can be passed from you to your unborn child and cause a potentially deadly infection (neonatal herpes). It is important that you avoid getting herpes during pregnancy.
If you are pregnant and have genital herpes, you may be offered herpes medicine towards the end of your pregnancy to reduce the risk of having any symptoms and passing the disease to your baby. At the time of delivery your doctor should carefully examine you for symptoms. If you have herpes symptoms at delivery, a ‘C-section’ is usually performed.
How do I know if I have genital herpes?
Most people who have herpes have no, or very mild symptoms. You may not notice mild symptoms or you may mistake them for another skin This fact sheet is being updated. View the online version for the most up-to-date information about genital herpes Page 2 of 2 Last reviewed: January 23, 2014 condition, such as a pimple or ingrown hair. Because of this, most people who have herpes do not know it. Genital herpes sores usually appear as one or more blisters on or around the genitals, rectum or mouth. The blisters break and leave painful sores that may take weeks to heal. These symptoms are sometimes called “having an outbreak.” The first time someone has an outbreak they may also have flu-like symptoms such as fever, body aches, or swollen glands.
Repeat outbreaks of genital herpes are common, especially during the first year after infection. Repeat outbreaks are usually shorter and less severe than the first outbreak. Although the infection can stay in the body for the rest of your life, the number of outbreaks tends to decrease over a period of years.
You should be examined by your doctor if you notice any of these symptoms or if your partner has an STD or symptoms of an STD, such as an unusual sore, a smelly discharge, burning when urinating, or, for women specifically, bleeding between periods.
How will my doctor know if I have herpes?
Often times, your healthcare provider can diagnose genital herpes by simply looking at your symptoms. Providers can also take a sample from the sore(s) and test it. Have an honest and open talk with your health care provider and ask whether you should be tested for herpes or other STDs. Can herpes be cured?
There is no cure for herpes. However, there are medicines that can prevent or shorten outbreaks. One of these herpes medicines can be taken daily, and makes it less likely that you will pass the infection on to your sex partner(s).
What happens if I don’t get treated?
Genital herpes can cause painful genital sores and can be severe in people with suppressed immune systems. If you touch your sores or the fluids from the sores, you may transfer herpes to another part of your body, such as your eyes. Do not touch the sores or fluids to avoid spreading herpes to another part of your body. If you touch the sores or fluids, immediately wash your hands thoroughly to help avoid spreading your infection.
Some people who get genital herpes have concerns about how it will impact their overall health, sex life, and relationships. It is best for you to talk to a health care provider about those concerns, but it also is important to recognize that while herpes is not curable, it can be managed. Since a genital herpes diagnosis may affect how you will feel about current or future sexual relationships, it is important to understand how to talk to sexual partners about STDs. You can find one resource here: GYT Campaign, If you are pregnant, there can be problems for you and your unborn child. See “I’m pregnant. How could genital herpes affect my baby?” above for information about this.
Can I still have sex if I have herpes?
Where can I get more information?
Division of STD Prevention (DSTDP) Centers for Disease Control and Prevention www.cdc.gov/std Personal health inquiries and information about STDs: CDC-INFO Contact Center 1-800-CDC-INFO (1-800-232-4636) Contact dcs/ContactUs/Form Resources: CDC National Prevention Information Network (NPIN) P.O. Box 6003 Rockville, MD 20849-6003 E-mail: npin-info@cdc.gov American Sexual Health Association (ASHA) /stdsstis/ P.O. Box 13827 Research Triangle Park, NC 27709-3827 1-800-783-9877 If you have herpes, you should tell your sex partner(s) and let him or her know that you do and the risk involved. Using condoms may help lower this risk but it will not get rid of the risk completely. Having sores or other symptoms of herpes can increase your risk of spreading the disease. Even if you do not have any symptoms, you can still infect your sex partners. What is the link between genital herpes and HIV?
Genital herpes can cause sores or breaks in the skin or lining of the mouth, vagina, and rectum. The genital sores caused by herpes can bleed easily. When the sores come into contact with the mouth, vagina, or rectum during sex, they increase the risk of giving or getting HIV if you or your partner has HIV.
CS246943A National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention Syphilis – CDC Fact Sheet Syphilis is a sexually transmitted disease (STD) that can have very serious complications when left untreated, but it is simple to cure with the right treatment.
What is syphilis?
Syphilis is a sexually transmitted infection that can cause serious health problems if it is not treated. Syphilis is divided into stages (primary, secondary, latent, and tertiary), and there are different signs and symptoms associated with each stage.
How is syphilis spread?
You can get syphilis by direct contact with a syphilis sore during vaginal, anal, or oral sex. Sores can be found on or around the penis, vagina, or anus, or in the rectum, on the lips, or in the mouth. Syphilis can also be spread from an infected mother to her unborn baby.
What does syphilis look like?
Syphilis is divided into stages (primary, secondary, latent, and tertiary), and there are different signs and symptoms associated with each stage. A person with primary syphilis generally has a sore or sores at the original site of infection. These sores usually occur on or around the genitals, around the anus or in the rectum, or in or around the mouth. These sores are usually (but not always) firm, round, and painless. Symptoms of secondary syphilis include skin rash, swollen lymph nodes, and fever. The signs and symptoms of primary and secondary syphilis can be mild, and they might not be noticed. During the latent stage, there are no signs or symptoms. Tertiary syphilis is associated with severe medical problems and is usually diagnosed by a doctor with the help of multiple tests. It can affect the heart, brain, and other organs of the body. How can I reduce my risk of getting syphilis?
The only way to avoid STDs is to not have vaginal, anal, or oral sex.
If you are sexually active, you can do the following things to lower your chances of getting syphilis: • • Being in a long-term mutually monogamous relationship with a partner who has been tested for syphilis and does not have syphilis; Page 2 of 2 • • Using latex condoms, the right way, ( condomeffectiveness/male-condom-use.html) every time you have sex. Condoms prevent transmission of syphilis by preventing contact with a sore. Sometimes sores occur in areas not covered by a condom. Contact with these sores can still transmit syphilis.
Am I at risk for syphilis?
Any sexually active person can get syphilis through unprotected vaginal, anal, or oral sex. Have an honest and open talk with your health care provider and ask whether you should be tested for syphilis or other STDs. All pregnant women should be tested for syphilis at their first prenatal visit. In addition, you should get tested regularly for syphilis if you are sexually active and are a man who has sex with men, are living with HIV, or have partner(s) who have tested positive for syphilis.
I’m pregnant. How does syphilis affect my baby?
If you are pregnant and have syphilis, you can give the infection to your unborn baby. Having syphilis can lead to a low birth weight baby. It can also make it more likely you will deliver your baby too early or stillborn (a baby born dead). To protect your baby, you should be tested for syphilis at least once during your pregnancy and receive immediate treatment if you test positive.
An infected baby may be born without signs or symptoms of disease. However, if not treated immediately, the baby may develop serious problems within a few weeks. Untreated babies can have health problems such as cataracts, deafness, or seizures, and can die.
What are the signs and symptoms of syphilis?
Symptoms of syphilis in adults vary by stage: Primary Stage During the first (primary) stage of syphilis, you may notice a single sore or multiple sores. The sore is the location where syphilis entered your body. Sores are usually (but not always) firm, round, and painless. Because the sore is painless, it can easily go unnoticed. The sore usually lasts 3 to 6 weeks and heals regardless of whether or not you receive treatment. Even though the sore goes away, you must still receive treatment so your infection does not move to the secondary stage.
Example of a primary syphilis sore. Secondary rash from syphilis on palms of hands. Secondary rash from syphilis on torso. Secondary Stage During the secondary stage, you may have skin rashes and/or sores in your mouth, vagina, or anus (also called mucous membrane lesions). This stage usually starts with a rash on one or more areas of your body. The rash can show up when your primary sore is healing or several weeks after the sore has healed. The rash can look like rough, red, or reddish brown spots on the palms of your hands and/or the bottoms of your feet. The rash usually won’t itch and it is sometimes so faint that you won’t notice it. Other symptoms you may have can include fever, swollen lymph glands, sore throat, patchy hair loss, headaches, weight loss, muscle aches, and fatigue (feeling very tired). The symptoms from this stage will go away whether or not you receive treatment. Without the right treatment, your infection will move to the latent and possibly tertiary stages of syphilis.
Last reviewed: February 2017 Latent Stage The latent stage of syphilis is a period of time when there are no visible signs or symptoms of syphilis. If you do not receive treatment, you can continue to have syphilis in your body for years without any signs or symptoms. Tertiary Stage Most people with untreated syphilis do not develop tertiary syphilis. However, when it does happen it can affect many different organ systems, including the heart and blood vessels, and the brain and nervous system. Tertiary syphilis is very serious and would occur 10–30 years after your infection began. In tertiary syphilis, the disease damages your internal organs and can result in death.
Neurosyphilis and Ocular Syphilis Without treatment, syphilis can spread to the brain and nervous system (neurosyphilis) or to the eye (ocular syphilis). This can happen during any of the stages described above. Symptoms of neurosyphilis include severe headache, difficulty coordinating muscle movements, paralysis (not able to move certain parts of your body), numbness, and dementia (mental disorder). Symptoms of ocular syphilis include changes in your vision and even blindness.
How will I or my doctor know if I have syphilis?
Most of the time, a blood test can be used to test for syphilis. Some health care providers will diagnose syphilis by testing fluid from a syphilis sore.
Can syphilis be cured?
Yes, syphilis can be cured with the right antibiotics from your health care provider. However, treatment might not undo any damage that the infection has already done.
I’ve been treated. Can I get syphilis again?
Having syphilis once does not protect you from getting it again. Even after you’ve been successfully treated, you can still be re-infected. Only laboratory tests can confirm whether you have syphilis. Follow-up testing by your health care provider is recommended to make sure that your treatment was successful.
Because syphilis sores can be hidden in the vagina, anus, under the foreskin of the penis, or in the mouth, it may not be obvious that a sex partner has syphilis. Unless you know that your sex partner(s) has been tested and treated, you may be at risk of getting syphilis again from an infected sex partner.
Where can I get more information?
Syphilis and MSM - Fact Sheet stdfact-msm-syphilis.htm Congenital Syphilis - Fact Sheet stdfact-congenital-syphilis.htm STDs during Pregnancy - Fact Sheet pregnancy/stdfact-pregnancy.
htm) STD information and referrals to STD Clinics CDC-INFO Contact Center 1-800-CDC-INFO (1-800-232-4636) TTY: (888) 232-6348 Contact CDC-INFO ContactUs/Form Darkfield micrograph of Treponema pallidum. National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention CS233825A What is trichomoniasis?
Trichomoniasis (or “trich”) is a very common sexually transmitted disease (STD) that is caused by infection with a protozoan parasite called Trichomonas vaginalis. Although symptoms of the disease vary, most women and men who have the parasite cannot tell they are infected.
How common is trichomoniasis?
Trichomoniasis is considered the most common curable STD. In the United States, an estimated 3.7 million people have the infection, but only about 30% develop any symptoms of trichomoniasis. Infection is more common in women than in men, and older women are more likely than younger women to have been infected.
How do people get trichomoniasis?
The parasite is passed from an infected person to an uninfected person during sex. In women, the most commonly infected part of the body is the lower genital tract (vulva, vagina, cervix, or urethra), and in men, the most commonly infected body part is the inside of the penis (urethra). During sex, the parasite is usually transmitted from a penis to a vagina, or from a vagina to a penis, but it can also be passed from a vagina to another vagina. It is not common for the parasite to infect other body parts, like the hands, mouth, or anus. It is unclear why some people with the infection get symptoms while others do not, but it probably depends on factors like the person’s age and overall health. Infected people without symptoms can still pass the infection on to others. What are the signs and symptoms of trichomoniasis?
About 70% of infected people do not have any signs or symptoms. When trichomoniasis does cause symptoms, they can range from mild irritation to severe inflammation. Some people with symptoms get them within 5 to 28 days after being infected, but others do not develop symptoms until much later. Symptoms can come and go.
Men with trichomoniasis may feel itching or irritation inside the penis, burning after urination or ejaculation, or some discharge from the penis. Women with trichomoniasis may notice itching, burning, redness or soreness of the genitals, discomfort with urination, or a change in their vaginal discharge (i.e., thin discharge or increased volume) with an unusual smell (i.e., fishy odor) that can be clear, white, yellowish, or greenish. Having trichomoniasis can make it feel unpleasant to have sex. Without treatment, the infection can last for months or even years.
What are the complications of trichomoniasis?
Trichomoniasis can increase the risk of getting or spreading other sexually transmitted infections. For example, trichomoniasis can cause genital inflammation that makes it easier to get infected with the HIV virus, or to pass the HIV virus on to a sex partner. How does trichomoniasis affect a pregnant woman and her baby? Pregnant women with trichomoniasis are more likely to have their babies too early (preterm delivery). Also, babies born to infected mothers are more likely to have a low birth weight (less than 5.5 pounds).
Trichomoniasis - CDC Fact Sheet Two Trichomonas vaginalis parasites, magnified (seen under a microscope) How is trichomoniasis diagnosed?
It is not possible to diagnose trichomoniasis based on symptoms alone. For both men and women, your health care provider can examine you and get a laboratory test to diagnose trichomoniasis. What is the treatment for trichomoniasis?
Trichomoniasis can be treated with medication (either metronidazole or tinidazole). It is safe for pregnant women to take this medication. It is not recommended to drink alcohol within 24 hours after taking this medication. People who have been treated for trichomoniasis can get it again. About 1 in 5 people get infected again within 3 months after receiving treatment. To avoid getting reinfected, make sure that all of your sex partners get treated and wait 7-10 days after you and your partner have been treated to have sex again. Get checked again if your symptoms come back. How can trichomoniasis be prevented?
CDC National Prevention Information (NPIN) P.O. Box 6003 Rockville, MD 20849-6003 E-mail: npin-info@cdc.gov npin.cdc.gov/disease/stds American Sexual Health Association (ASHA) P. O. Box 13827 Resear ch Triangle Park, NC 27709-3827 1-800-783-9877 stdsstis/ Resources Where can I get more information?
Division of STD Prevention (DSTDP) Centers for Disease Control and Prevention www.cdc.gov/std CDC-INFO Contact Center 1-800-CDC-INFO (1-800-232-4636) Contact: dcs/ContactUs/Form The only way to avoid STDs is to not have vaginal, anal, or oral sex. If you are sexually active, you can do the following things to lower your chances of getting trichomoniasis: • Being in a long-term mutually monogamous relationship with a partner who has been tested and has negative STD test results; • Using latex condoms the right way every time you have sex. This can lower your chances of getting trichomoniasis. But the parasite can infect areas that are not covered by a condom - so condoms may not fully protect you from getting trichomoniasis.
Another approach is to talk about the potential risk of infection before you have sex with a new partner, so that you can make informed choices about the level of risk you are comfortable taking with your sex life. If you or someone you know has questions about trichomoniasis or any other STD, especially with symptoms like unusual discharge, burning during urination, or a sore in the genital area, check with a health care provider. CS246943B National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention Information for Teens and Young Adults: Staying Healthy and Preventing STDs If you choose to have sex, know how to protect yourself against sexually transmitted diseases (STDs). What are sexually transmitted diseases (STDs)?
STDs are diseases that are passed from one person to another through sexual contact. These include chlamydia, gonorrhea, genital herpes, human papillomavirus (HPV), syphilis, and HIV. Many of these STDs do not show symptoms for a long time, but they can still be harmful and passed on during sex.
How are STDs spread?
You can get an STD by having sex (vaginal, anal or oral) with someone who has an STD. Anyone who is sexually active can get an STD. You don’t even have to “go all the way” (have anal or vaginal sex) to get an STD, since some STDs, like herpes and HPV, are spread by skin-to-skin contact.
How common are STDs?
STDs are common, especially among young people. There are about 20 million new cases of STDs each year in the United States, and about half of these are in people between the ages of 15 and 24. Young people are at greater risk of getting an STD for several reasons: • • Young women’s bodies are biologically more susceptible to STDs.
• • Some young people do not get the recommended STD tests.
• • Many young people are hesitant to talk openly and honestly with a doctor or nurse about their sex lives.
• • Not having insurance or transportation can make it more difficult for young people to access STD testing.
• • Some young people have more than one sex partner.
What can I do to protect myself?
• • The surest way to protect yourself against STDs is to not have sex. That means not having any vaginal, anal, or oral sex (“abstinence”). There are many things to consider before having sex, and it’s okay to say “no” if you don’t want to have sex.
• • If you do decide to have sex, you and your partner should get tested beforehand and make sure that you and your partner use a condom— every time you have oral, anal, or vaginal sex, from start to finish. Know where to get condoms and how to use them correctly. It is not safe to stop using condoms unless you’ve both been tested, know your status, and are in a mutually monogamous relationship.
• • Mutual monogamy means that you and your partner both agree to only have sexual contact with each other. This can help protect against STDs, as long as you’ve both been tested and know you’re STD-free.
• • Before you have sex, talk with your partner about how you will prevent STDs and pregnancy. If you think you’re ready to have sex, you need to be ready to protect your body and your future. You should also talk to your partner ahead of time about what you will and will not do sexually. Your partner should always respect your right to say no to anything that doesn’t feel right.
Page 2 of 2 Last reviewed: May 22, 2014 • • Make sure you get the health care you need. Ask a doctor or nurse about STD testing and about vaccines against HPV and hepatitis B.
• • Girls and young women may have extra needs to protect their reproductive health. Talk to your doctor or nurse about regular cervical cancer screening and chlamydia testing. You may also want to discuss unintended pregnancy and birth control.
• • Avoid using alcohol and drugs. If you use alcohol and drugs, you are more likely to take risks, like not using a condom or having sex with someone you normally wouldn’t have sex with.
If I get an STD, how will I know?
Many STDs don’t cause any symptoms that you would notice, so the only way to know for sure if you have an STD is to get tested. You can get an STD from having sex with someone who has no symptoms. Just like you, that person might not even know he or she has an STD.
Where can I get tested?
There are places that offer teen-friendly, confidential, and free STD tests. This means that no one has to find out you’ve been tested. Visit FindSTDTest.org to find an STD testing location near you.
Can STDs be treated?
Your doctor can prescribe medicines to cure some STDs, like chlamydia and gonorrhea. Other STDs, like herpes, can’t be cured, but you can take medicine to help with the symptoms.
If you are ever treated for an STD, be sure to finish all of your medicine, even if you feel better before you finish it all. Ask the doctor or nurse about testing and treatment for your partner, too. You and your partner should avoid having sex until you’ve both been treated. Otherwise, you may continue to pass the STD back and forth. It is possible to get an STD again (after you’ve been treated), if you have sex with someone who has an STD.
What happens if I don’t treat an STD?
Some curable STDs can be dangerous if they aren’t treated. For example, if left untreated, chlamydia and gonorrhea can make it difficult—or even impossible—for a woman to get pregnant. You also increase your chances of getting HIV if you have an untreated STD. Some STDs, like HIV, can be fatal if left untreated.
What if my partner or I have an incurable STD?
Some STDs- like herpes and HIV- aren’t curable, but a doctor can prescribe medicine to treat the symptoms.
If you are living with an STD, it’s important to tell your partner before you have sex. Although it may be uncomfortable to talk about your STD, open and honest conversation can help your partner make informed decisions to protect his or her health.
If I have questions, who can answer them?
If you have questions, talk to a parent or other trusted adult. Don’t be afraid to be open and honest with them about your concerns. If you’re ever confused or need advice, they’re the first place to start. Remember, they were young once, too.
Talking about sex with a parent or another adult doesn’t need to be a one-time conversation. It’s best to leave the door open for conversations in the future.
It’s also important to talk honestly with a doctor or nurse. Ask which STD tests and vaccines they recommend for you.
Where can I get more information?
CDC How You Can Prevent Sexually Transmitted Diseases prevention/ Teen Pregnancy TeenPregnancy/Teens.html CDC-INFO Contact Center 1-800-CDC-INFO (1-800-232-4636) Contact dcs/RequestForm.aspx HealthFinder.gov STD Testing: Conversation Starters HealthTopics/Category/health-conditions-and-diseases/ hiv-and-other-stds/std-testing-conversation-starters American Sexual Health Association Sexual Health and You teens/index.html Teens and Young Adults sexual-health/teens-and-young-adults/ References Centers for Disease Control and Prevention. Incidence, Prevalence, and Cost of Sexually Transmitted Infections in the United States, Accessed October 14, 2014.
CS225600A If you are pregnant, you can become infected with the same sexually transmitted diseases (STDs) as women who are not pregnant. Pregnant women should ask their doctors about getting tested for STDs, since some doctors do not routinely perform these tests).
A critical component of appropriate prenatal care is ensuring that pregnant patients are tested for STDs. Test your pregnant patients for STDs starting early in their pregnancy and repeat close to delivery, as needed. To ensure that the correct tests are being performed, we encourage you to have open, honest conversations with your pregnant patients and, when possible, their sex partners about symptoms they have experienced or are currently experiencing and any high-risk sexual behaviors in which they engage. The following sections provide details on the effects of specific STDs during a woman’s pregnancy with links to web pages with additional information I’m pregnant. Can I get an STD? Yes, you can. Women who are pregnant can become infected with the same STDs as women who are not pregnant. Pregnancy does not provide women or their babies any additional protection against STDs. Many STDs are ‘silent,’ or have no symptoms, so you may not know if you are infected. If you are pregnant, you should be tested for STDs, including HIV (the virus that causes AIDS), as a part of your medical care during pregnancy. The results of an STD can be more serious, even life-threatening, for you and your baby if you become infected while pregnant. It is important that you are aware of the harmful effects of STDs and how to protect yourself and your unborn baby against infection. If you are diagnosed with an STD while pregnant, your sex partner(s) should also be tested and treated. How can STDs affect me and my unborn baby? STDs can complicate your pregnancy and may have serious effects on both you and your developing baby. Some of these problems may be seen at birth; others may not be discovered until months or years later. In addition, it is well known that infection with an STD can make it easier for a person to get infected with HIV. Most of these problems can be prevented if you receive regular medical care during pregnancy. This includes tests for STDs starting early in pregnancy and repeated close to delivery, as needed.
Should I be tested for STDs during my pregnancy?
Yes. Testing and treating pregnant women for STDs is a vital way to prevent serious health complications to both mother and baby that may otherwise happen with infection. The sooner you begin receiving medical care during pregnancy, the better the health outcomes will be for you and your unborn baby. The Centers for Disease Control and Prevention’s 2015 STD Treatment National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention STDs during Pregnancy - CDC Fact Sheet Page 2 of 2 Guidelines recommend screening pregnant women for STDs. The CDC screening recommendations that your health care provider should follow are incorporated into the table on the STDs during Pregnancy – Detailed CDC Fact Sheet
Be sure to ask your doctor about getting tested for STDs. It is also important that you have an open, honest conversation with your provider and discuss any symptoms you are experiencing and any high-risk sexual behavior that you engage in, since some doctors do not routinely perform these tests. Even if you have been tested in the past, you should be tested again when you become pregnant.
Can I get treated for an STD while I’m pregnant?
It depends. STDs, such as chlamydia, gonorrhea, syphilis, trichomoniasis and BV can all be treated and cured with antibiotics that are safe to take during pregnancy. STDs that are caused by viruses, like genital herpes, hepatitis B, or HIV cannot be cured. However, in some cases these infections can be treated with antiviral medications or other preventive measures to reduce the risk of passing the infection to your baby. If you are pregnant or considering pregnancy, you should be tested so you can take steps to protect yourself and your baby.
How can I reduce my risk of getting an STD while pregnant?
The only way to avoid STDs is to not have vaginal, anal, or oral sex.
If you are sexually active, you can do the following things to lower your chances of getting chlamydia: • • Being in a long-term mutually monogamous relationship with a partner who has been tested and has negative STD test results; • • Using latex condoms the right way every time you have sex.
Related Content Congenital Syphilis Fact Sheet stdfact-congenital-syphilis.htm Pregnancy and HIV, Viral Hepatitis, and STD Prevention pregnancy/default.htm Sexually Transmitted Diseases - Information from CDC April 2016 CS247394B/ Versión en español aprobada por CDC Multilingual Services - Order # 246656 National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention Infección por clamidia: Hoja informativa de los CDC La infección por clamidia es una enfermedad de transmisión sexual (ETS) común que es fácil de curar. Si se deja sin curar, puede hacer más difícil que una mujer quede embarazada.
¿Qué es la infección por clamidia?
La infección por clamidia es una ETS común que puede infectar tanto a los hombres como a las mujeres. Puede causar daños graves y permanentes en el aparato reproductor de una mujer y hacer más difícil o imposible que quede embarazada en el futuro. La infección por clamidia también puede provocar un embarazo ectópico (embarazo que ocurre fuera del útero) que puede ser mortal.
¿Cómo se propaga la infección por clamidia?
Usted puede contraer la infección por clamidia al tener relaciones sexuales anales, vaginales u orales con una persona que tenga esta infección. Si su pareja sexual es hombre, usted puede contraer la infección por clamidia aunque él no eyacule (acabe). Si ya ha tenido la infección por clamidia y recibió tratamiento en el pasado, usted puede todavía volver a infectarse si tiene relaciones sexuales sin protección con una persona infectada. Si está embarazada, usted puede transmitírsela a su bebé durante el parto.
¿Cómo puedo evitar contraer la infección por clamidia?
Usted puede protegerse de contraer la infección por clamidia si: • • no tiene relaciones sexuales; • • tiene una relación mutuamente monógama a largo plazo con una pareja a quien se le hayan realizado pruebas y haya tenido resultados negativos para las ETS; • • usa condones de látex y diques dentales en forma correcta cada vez que tiene relaciones sexuales. ¿Tengo riesgo de contraer la infección por clamidia?
Cualquier persona que tenga relaciones sexuales puede contraer la infección por clamidia mediante relaciones sexuales anales, vaginales u orales sin protección. No obstante, las personas jóvenes sexualmente activas tienen mayor riesgo de contraer esta infección. Esto se debe a factores conductuales y biológicos comunes entre las personas jóvenes. Los homosexuales, bisexuales y otros hombres que tienen relaciones sexuales con hombres también corren riesgo debido a que la infección por clamidia puede propagarse mediante las relaciones sexuales orales y anales. Hable con su proveedor de atención médica de manera honesta y abierta y pregúntele si debe hacerse la prueba de detección de la clamidia o de otras ETS. Si es una mujer sexualmente activa menor de 25 años, o una mujer mayor con factores de riesgo —como el tener una nueva pareja sexual o múltiples parejas sexuales, o una pareja sexual con una infección de transmisión sexual—, debe hacerse una prueba de detección de la clamidia todos los años. Estoy embarazada. ¿Cómo afecta a mi bebé la infección por clamidia?
Si está embarazada y tiene la infección por clamidia, puede transmitírsela a su bebé durante el parto. Esto podría causar una infección en los ojos o neumonía en el recién nacido. Tener la infección por clamidia puede también aumentar su probabilidad de dar a luz a su bebé de manera prematura. Si está embarazada, usted debe hacerse la prueba de detección de la clamidia en su primera visita prenatal. Las pruebas y los tratamientos son las mejores maneras de prevenir problemas de salud. Page 2 of 2 Esta página fue revisada el 23 de enero de 2014 ¿Cómo sé si tengo la infección por clamidia?
La mayoría de las personas que tienen la infección por clamidia no presentan síntomas. Si usted presenta síntomas, es posible que no aparezcan por varias semanas después de que haya tenido relaciones sexuales con una persona infectada. Incluso cuando no causa síntomas, la infección por clamidia puede dañar su aparato reproductor. Las mujeres con síntomas podrían notar los siguientes: • • secreción vaginal anormal; • • sensación de ardor al orinar. Los síntomas en los hombres pueden ser los siguientes: • • secreción del pene; • • sensación de ardor al orinar; • • dolor e inflamación de uno o ambos testículos (aunque esto es menos común). Los hombres y las mujeres también pueden infectarse por clamidia en el recto, ya sea mediante las relaciones sexuales anales receptivas o la propagación desde otra parte infectada (como la vagina). Aunque por lo general estas infecciones no causan síntomas, pueden provocar: • • dolor en el recto; • • secreciones; • • sangrado.
Debe hacerse revisar por un médico si nota cualquiera de estos síntomas o si su pareja tiene una ETS o síntomas de una ETS, como dolor inusual, secreción con olor, ardor al orinar o sangrado entre periodos. ¿Cómo sabrá mi médico si tengo la infección por clamidia?
Existen pruebas de laboratorio para diagnosticar la infección por clamidia. Es posible que su proveedor de atención médica le pida una muestra de orina o use (o le pida que use) un hisopo para obtener una muestra de las secreciones de su vagina para hacerle una prueba para detectar la clamidia. ¿La infección por clamidia se puede curar?
Sí, la infección por clamidia se puede curar con el tratamiento correcto. Es importante que tome todos los medicamentos que su médico le recete para curar su infección. Cuando se toman de manera adecuada, detienen la infección y pueden disminuir su probabilidad de tener complicaciones en el futuro. Los medicamentos contra la infección por clamidia no se deben compartir con nadie. La recurrencia de la infección por clamidia es común. Debe volver a hacerse la prueba unos tres meses después del tratamiento, incluso si su pareja sexual o parejas sexuales recibieron tratamiento. ¿Dónde puedo obtener más información?
División de Prevención de Enfermedades de Transmisión Sexual (DSTDP) Centros para el Control y la Prevención de Enfermedades default.htm Centro de información de los CDC 1-800-CDC-INFO (1-800-232-4636) Comuníquese con CDC–INFO espanol ¿Qué pasa si no recibo tratamiento?
A menudo, el daño que inicialmente causa la clamidia pasa desapercibido. Sin embargo, la infección por clamidia puede causar problemas de salud graves.
Si usted es mujer, la infección por clamidia que no se trata puede propagarse al útero y a las trompas de Falopio (los conductos que transportan los óvulos fecundados desde los ovarios hasta el útero), y causar enfermedad inflamatoria pélvica (EIP). La enfermedad inflamatoria pélvica por lo general no presenta síntomas; sin embargo, algunas mujeres pueden tener dolor abdominal y pélvico. Aun cuando no cause síntomas iniciales, la enfermedad inflamatoria pélvica puede causar daño permanente al aparato reproductor y dolor pélvico crónico, imposibilidad de quedar embarazada y embarazo ectópico (embarazo fuera del útero) que puede causar la muerte. Los hombres raramente tienen problemas de salud asociados a la infección por clamidia. En ocasiones, la infección se propaga al conducto que transporta el semen desde los testículos, lo cual causa dolor y fiebre. La clamidia, en muy pocos casos, puede causar que un hombre no pueda tener hijos. Una infección por clamidia que no se trate también puede aumentar su probabilidad de contraer o transmitir el VIH, el virus que causa el SIDA.
Recibí tratamiento contra la infección por clamidia. ¿Cuándo puedo tener relaciones sexuales nuevamente? Usted no debe tener relaciones sexuales de nuevo hasta que usted y su pareja sexual o sus parejas sexuales hayan completado el tratamiento. Si su médico le recetó un medicamento de una sola dosis, deberá esperar siete días después de haberlo tomado, antes de volver a tener relaciones sexuales. Si su médico le recetó un medicamento que debe tomar durante siete días, deberá esperar a terminar todas las dosis antes de tener relaciones sexuales. CS246652C/Versión en español aprobada por CDC Multilingual Services – Order # 246656 National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention Gonorrea: Hoja informativa de los CDC Cualquier persona que tenga relaciones sexuales puede contraer gonorrea. La gonorrea puede causar complicaciones muy graves cuando no se trata, pero se puede curar con los medicamentos correctos. ¿Qué es la gonorrea?
La gonorrea es una enfermedad de transmisión sexual (ETS) que puede infectar tanto a los hombres como a las mujeres. Puede causar infecciones en los genitales, el recto y la garganta. Es una infección muy común, especialmente en las personas jóvenes de 15 a 24 años. ¿Cómo se transmite la gonorrea? Usted puede contraer gonorrea al tener relaciones sexuales anales, vaginales y orales con una persona que tenga esta enfermedad. Una mujer embarazada con gonorrea puede transmitírsela a su bebé durante el parto. ¿Cómo puedo evitar contraer gonorrea?
Usted puede evitar contraer gonorrea si: • • no tiene relaciones sexuales; • • tiene una relación mutuamente monógama a largo plazo con una pareja a quien se le hayan realizado pruebas y haya tenido resultados negativos para las ETS; • • usa condones de látex y diques dentales en forma correcta cada vez que tiene relaciones sexuales.
¿Tengo riesgo de contraer gonorrea?
Cualquier persona que tenga relaciones sexuales puede contraer gonorrea mediante las relaciones sexuales anales, vaginales u orales sin protección. Si usted es sexualmente activo, hable con su proveedor de atención médica de manera honesta y abierta, y pregúntele si debe hacerse la prueba de detección de la gonorrea o de otras ETS. Si es una mujer sexualmente activa menor de 25 años, o una mujer mayor con factores de riesgo —como el tener una nueva pareja sexual o múltiples parejas sexuales, o una pareja sexual con una infección de transmisión sexual—, debe hacerse una prueba de detección de la gonorrea todos los años. Si usted es un hombre sexualmente activo, homosexual, bisexual o tiene relaciones con hombres y es sexualmente activo, debe hacerse la prueba de detección de la gonorrea anualmente. Estoy embarazada. ¿Cómo afecta a mi bebé la gonorrea?
Si está embarazada y tiene gonorrea, puede transmitirle la infección a su bebé durante el parto. Esto puede causarle problemas graves de salud a su bebé. Si está embarazada, es importante que hable con su proveedor de atención médica para que le hagan los exámenes físicos y las pruebas adecuadas y reciba el tratamiento correcto, según sea necesario. Tratar la gonorrea lo antes posible disminuirá las probabilidades de que su bebé tenga complicaciones de salud. ¿Cómo sé si tengo gonorrea?
Es posible que algunos hombres con gonorrea no presenten ningún síntoma. Sin embargo, los hombres que presentan síntomas pueden tener: • • sensación de ardor al orinar; • • secreción de color blanco, amarillo o verde del pene; • • dolor o inflamación en los testículos (aunque esto es menos común).
La mayoría de las mujeres con gonorrea no tienen síntomas. Incluso cuando tienen síntomas, por lo general, son leves y se pueden confundir con los síntomas de una infección vaginal o de la vejiga. Las mujeres con gonorrea corren el riesgo de tener complicaciones graves por la infección, aun cuando no presenten ningún síntoma. Page 2 of 2 Esta página fue revisada el 29 de enero de 2014 Los síntomas en las mujeres pueden ser los siguientes: • • dolor o sensación de ardor al orinar; • • aumento de la secreción vaginal; • • sangrado vaginal entre periodos. Las infecciones del recto pueden no causar síntomas tanto en los hombres como en las mujeres o pueden causarles los siguientes: • • secreciones; • • picazón anal; • • dolores; • • sangrado; • • dolor al defecar. Debe hacerse revisar por un médico si nota cualquiera de estos síntomas o si su pareja tiene una ETS o síntomas de una ETS, como dolor inusual, secreción con olor, ardor al orinar o sangrado entre periodos.
¿Cómo sabrá mi médico si tengo gonorrea?
En las mayoría de los casos, se puede utilizar una muestra de orina para detectar la gonorrea. Sin embargo, si usted ha tenido relaciones sexuales orales o anales, se puede usar un hisopo para obtener muestras de la garganta o del recto. En algunos casos, se deben tomar muestras de la uretra del hombre (canal urinario) o del cuello uterino de la mujer (la abertura de la matriz) con un hisopo. ¿Se puede curar la gonorrea?
Sí, la gonorrea se puede curar con el tratamiento correcto. Es importante que tome todos los medicamentos que su médico le recete para curar su infección. Los medicamentos contra la gonorrea no se deben compartir con nadie. Si bien los medicamentos detendrán la infección, no repararán ninguna lesión permanente que haya causado la enfermedad. Es cada vez más difícil tratar algunos casos de gonorrea debido a que las cepas de gonorrea resistentes a los medicamentos están aumentando. Si sus síntomas continúan por más de unos días después del tratamiento, debe regresar a su proveedor de atención médica para que le hagan otro chequeo. Recibí tratamiento contra la gonorrea. ¿Cuándo puedo tener relaciones sexuales nuevamente?
Debe esperar siete días después de terminar todos los medicamentos antes de tener relaciones sexuales. Para evitar que se infecte de gonorrea nuevamente o que se la transmita a su pareja sexual o sus parejas sexuales, debe evitar tener relaciones sexuales hasta que cada persona haya completado el tratamiento. Si usted ya ha tenido gonorrea y tomó medicamentos en el pasado, todavía se puede infectar nuevamente si tiene relaciones sexuales sin protección con una persona que tenga gonorrea. ¿Qué pasa si no recibo tratamiento?
Cuando la gonorrea no se trata, puede ocasionar problemas de salud graves y permanentes tanto en los hombres como en las mujeres.
En las mujeres, la gonorrea sin tratar puede causar la enfermedad inflamatoria pélvica (EIP). Algunas de las complicaciones de la EIP son las siguientes: • • formación de tejido cicatricial que obstruye las trompas de Falopio; • • embarazo ectópico (embarazo afuera del útero); • • infertilidad (incapacidad para quedar embarazada); • • dolor pélvico o abdominal crónico.
En los hombres, la gonorrea puede causar una afección dolorosa en los conductos de los testículos. En casos muy poco comunes, esto puede causarle a un hombre infertilidad o hacer que no pueda tener hijos. La gonorrea que no se trata puede también rara vez propagarse a la sangre o las articulaciones. Esta afección puede ser mortal. La gonorrea que no se trata también puede aumentar sus probabilidades de contraer o transmitir el VIH, el virus que causa el sida. ¿Dónde puedo obtener más información?
División de Prevención de Enfermedades de Transmisión Sexual (DSTDP)Centros para el Control y la Prevención de Enfermedades default.htm Centro de información de los CDC 1-800-CDC-INFO (1-800-232-4636) Comuníquese con CDC–INFO CS249164A National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention Herpes genital: Hoja informativa de los CDC El herpes es una enfermedad de transmisión sexual (ETS) que cualquier persona sexualmente activa puede contraer. La mayoría de las personas con el virus no tiene síntomas. Es importante saber que aún sin presentar los signos de la enfermedad, se puede contagiar a una pareja sexual. ¿Qué es el herpes genital?
El herpes genital es una ETS causada por dos tipos de virus. Estos virus se llaman herpes simple del tipo 1 y herpes simple del tipo 2. ¿Qué tan común es el herpes genital?
El herpes genital es común en los Estados Unidos. En los Estados Unidos, aproximadamente una de cada seis personas entre 14 y 49 años tiene herpes genital. ¿Cómo se propaga el herpes genital? Usted puede contraer herpes al tener relaciones sexuales orales, vaginales o anales con una persona que tenga la enfermedad. El líquido que se encuentra en la llaga del herpes contiene el virus y el contacto con ese líquido puede causar la infección. Usted también puede contraer el herpes de una pareja sexual infectada que no tenga llagas visibles o que no sepa que está infectada, ya que el virus puede liberarse mediante la piel y propagar la infección a su pareja sexual o parejas sexuales. ¿Cómo puedo evitar contraer el herpes?
Usted puede protegerse contra el contagio del herpes si: • • no tiene relaciones sexuales; • • tiene una relación mutuamente monógama a largo plazo con una pareja a quien se le hayan realizado pruebas y haya tenido resultados negativos para las ETS; • • usa condones de látex y diques dentales en forma correcta cada vez que tiene relaciones sexuales.
Los síntomas del herpes pueden presentarse en las áreas genitales del hombre y de la mujer que se hayan cubierto con un condón de látex. Sin embargo, los brotes también pueden ocurrir en áreas que no se hayan cubierto por un condón, por lo tanto, es posible que los condones no lo protejan completamente del herpes. Estoy embarazada. ¿Cómo podría el herpes genital afectar a mi bebé?
Si está embarazada y tiene herpes genital, es aún más importante que usted vaya a sus citas de atención médica prenatales. Debe informarle a su médico si alguna vez ha tenido síntomas, ha estado expuesta o ha recibido un diagnóstico de herpes genital. Algunas veces la infección por herpes genital puede provocar abortos espontáneos. También puede hacer que tenga mayor probabilidad de que su bebé nazca mucho antes. Usted puede pasarle la infección por herpes a su bebé en gestación y puede causarle una infección potencialmente mortal (herpes en el neonato). Es importante que usted evite contraer el herpes durante el embarazo.
Si está embarazada y tiene herpes genital, es posible que le ofrezcan medicamentos para el herpes hacia el final de su embarazo para reducir su riesgo de tener síntomas y de pasarle la enfermedad a su bebé. En el momento del parto, su médico debe examinarla atentamente para determinar si hay síntomas presentes. Si tiene síntomas del herpes durante el parto, por lo general se realiza una cesárea. ¿Cómo sé si tengo herpes genital?
La mayoría de las personas que tiene herpes no presenta síntomas o si los presenta son muy leves. Es posible que no se dé cuenta de los síntomas leves o que los Page 2 of 2 Esta página fue revisada el 23 de enero de 2014 Versión en español aprobada por CDC Multilingual Services – Order # 248503 confunda con otra afección de la piel como un grano o pelo encarnado. Es por esto que la mayoría de las personas que tienen herpes no lo saben. Las llagas del herpes genital, por lo general, se ven como una o más ampollas en los genitales, el recto o la boca. Las ampollas se abren y dejan llagas dolorosas que pueden tardar semanas en curarse. A estos síntomas a veces se los llaman “brotes” . La primera vez que una persona tiene un brote es probable que también presente síntomas similares a los de la influenza (gripe) como fiebre, dolores corporales e inflamación de glándulas. Es común que los brotes de herpes genital se repitan, en especial durante el primer año después de la infección. Los siguientes brotes generalmente duran menos tiempo y son menos graves que el primer brote. Aunque la infección puede permanecer en el cuerpo por el resto de su vida, la cantidad de brotes tiende a disminuir con los años.
Debe hacerse revisar por un médico, si nota cualquiera de estos síntomas o si su pareja tiene una ETS o síntomas de una ETS, como una llaga inusual, secreción con olor, ardor al orinar o específicamente en las mujeres, sangrado entre periodos.
¿Cómo sabrá mi médico si tengo herpes?
Muchas veces su proveedor de atención médica puede diagnosticar herpes genital simplemente con mirar los síntomas. Los proveedores de atención médica también pueden tomar una muestra de la llaga y hacerle una prueba. Hable con su proveedor de atención médica de manera honesta y abierta y pregúntele si debe hacerse la prueba de detección del herpes o de otras ETS.
¿Se puede curar el herpes?
No existe una cura para el herpes. No obstante, existen medicamentos que pueden prevenir o disminuir la duración de los brotes. Uno de estos medicamentos para el herpes puede tomarse todos los días y reduce la probabilidad de que usted le pase la infección a su pareja sexual o parejas sexuales. ¿Qué pasa si no recibo tratamiento?
El herpes genital puede causar llagas genitales dolorosas y puede ser grave en personas con el sistema inmunitario deprimido. Si se toca las llagas o toca el líquido de estas, puede pasar el herpes a otras partes del su cuerpo, como a los ojos. No se toque las llagas ni toque el líquido para evitar propagar el herpes a otra parte del cuerpo. Si se toca las llagas o toca el líquido, lávese bien las manos inmediatamente para evitar propagar su infección. Algunas personas que contraen el herpes genital sienten preocupación sobre cómo afectará su salud general, su vida sexual y sus relaciones. Es aconsejable que hable con un proveedor de atención médica acerca de estas preocupaciones, pero también es importante saber que aunque el herpes no tenga cura es una afección controlable. Como el diagnóstico del herpes genital puede afectar cómo se siente sobre las relaciones sexuales existentes o futuras, es importante saber cómo hablar con sus parejas sexuales sobre las ETS. Aquí puede encontrar un recurso: Campaña GYT Si está embarazada, pueden producirse problemas para usted y su bebé en gestación. Vea la información anterior sobre el tema “Estoy embarazada. ¿Cómo podría el herpes genital afectar a mi bebé?” ¿Puedo aún tener relaciones sexuales si tengo herpes? Si tiene herpes, debe decirle a su pareja sexual y dejarle saber que tiene esta afección y el riesgo que implica. Usar condones puede ayudar a disminuir este riesgo, pero el riesgo no desaparecerá por completo. No tener otras llagas u otros síntomas de herpes también puede disminuir el riesgo, pero no completamente. Incluso si no tiene ningún síntoma puede aún infectar a sus parejas sexuales. ¿Dónde puedo obtener más información?
División para la Prevención de ETS (DSTDP) spanish/ Centros para el Control y la Prevención de Enfermedades Para preguntas personales sobre salud e información acerca de las ETS: Centro de información de los CDC 1-800-CDC-INFO (1-800-232-4636) Comuníquese con CDC–INFO espanol Recursos: Quiero Saber (ASHA) ets.html P. O. Box 13827 Research Triangle Park, NC 27709-3827 1-800-783-9877 ¿Cuál es la relación entre el herpes genital y el VIH?
El herpes genital puede causar llagas o cortes en la piel o en el recubrimiento interno de la boca, la vagina y el recto. Las llagas genitales causadas por el herpes pueden sangrar fácilmente. Cuando las llagas entran en contacto con la boca, la vagina o el recto durante las relaciones sexuales aumentan el riesgo de transmitir o contraer el VIH si su pareja sexual o parejas sexuales tienen el VIH. ¿Cuál es la relación entre el herpes genital y el herpes oral (herpes labial en la boca)? El herpes oral (como el herpes labial o herpes bucal en la boca o su alrededor) es, por lo general, causado por el virus del herpes simple del tipo 1 (VHS-1). La mayoría de las personas se infectan por el VHS-1 durante la infancia a través de un contacto no sexual. Por ejemplo, las personas se pueden infectar con el beso de un pariente o amigo que tenga herpes oral. Más de la mitad de la población de los EE. UU. tiene el VHS-1, aunque no muestren signos ni síntomas. El VHS-1 también puede propagarse de la boca a los genitales a través de las relaciones sexuales orales. Esta es la razón por la cual algunos casos de herpes genital son causados por el VHS-1.
CS249164D National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention Sífilis: Hoja informativa de los CDC La sífilis es una enfermedad de transmisión sexual (ETS) que puede tener complicaciones muy graves cuando se deja sin tratar, pero es fácil de curar con el tratamiento adecuado.
¿Qué es la sífilis?
La sífilis es una ETS que puede causar complicaciones a largo plazo o la muerte, si no se trata de manera adecuada. Los síntomas en los adultos se dividen en fases. Estas fases son sífilis primaria, secundaria, latente y avanzada. ¿Cómo se propaga la sífilis?
Usted puede contraer sífilis mediante el contacto directo con una llaga de sífilis durante las relaciones sexuales anales, vaginales u orales. Las llagas se pueden encontrar en el pene, la vagina, el ano, el recto o los labios y la boca. La sífilis también puede propagarse de una madre infectada a su bebé en gestación. ¿Cómo se ve la sífilis?
A la sífilis se la llama “la gran imitadora” porque tiene muchísimos síntomas posibles y muchos de estos se parecen a los síntomas de otras enfermedades. La llaga de sífilis que aparece justo después de infectarse por primera vez no produce dolor y puede confundirse con un pelo encarnado, una cortadura con un cierre u otro golpe que no parece dañino. El sarpullido que aparece en el cuerpo durante la segunda fase de la sífilis y que no produce picazón se puede producir en las palmas de las manos y las plantas de los pies, por todo el cuerpo o solo en algunas partes. La sífilis también puede afectar los ojos y causar ceguera permanente. Esto se llama sífilis ocular. Usted podría estar infectado por la sífilis y tener síntomas muy leves o no presentar ningún síntoma. ¿Cómo puedo evitar contraer la sífilis?
La única manera de evitar las ETS es no tener relaciones sexuales vaginales, anales ni orales.
Si usted es sexualmente activo, puede hacer las siguientes cosas para disminuir las probabilidades de contraer la sífilis: • • Tener una relación mutuamente monógama a largo plazo con una persona que se haya hecho pruebas y haya tenido resultados negativos para las ETS.
• • Usar condones de látex de manera correcta cada vez que tenga relaciones sexuales. Los condones previenen la transmisión de la sífilis al evitar el contacto con las llagas, pero a veces, las llagas pueden estar en áreas que el condón no cubre. La sífilis todavía se puede transmitir al tener contacto con estas llagas.• ¿Tengo riesgo de sífilis?
Cualquier persona sexualmente activa puede contraer sífilis mediante las relaciones sexuales anales, vaginales u orales sin protección. Hable con su proveedor de atención médica de manera honesta y abierta y pregúntele si debe hacerse la prueba de detección de sífilis o de otras ETS. Usted debe hacerse la prueba de sífilis con regularidad si está embarazada, es un hombre que tiene relaciones sexuales con hombres, tiene la infección por el VIH o una pareja que tuvo un resultado positivo a la prueba de sífilis. Estoy embarazada. ¿Cómo afecta la sífilis a mi bebé?
Si está embarazada y tiene sífilis, puede transmitirle la infección a su bebé en gestación. Tener sífilis puede causar que su bebé nazca con bajo peso. También puede hacer que tenga mayor probabilidad de que su bebé nazca mucho antes o de tener un mortinato (un bebé que nace muerto). Para proteger a su bebé usted debe hacerse la prueba de sífilis durante el embarazo y en el momento del parto, si el resultado es positivo debe recibir tratamiento de inmediato.
Ejemplo de una úlcera de sífilis primaria.
Page 2 of 2 Esta página fue revisada el 29 de enero de 2014 Versión en español aprobada por CDC Multilingual Services – Order # 248503 Los bebés infectados pueden nacer sin los signos o síntomas de la enfermedad. Sin embargo, si no es sometido a tratamiento de inmediato, el bebé puede presentar graves problemas al cabo de unas cuantas semanas. Los bebés que no reciben tratamiento pueden tener muchos problemas de salud como cataratas, sordera o convulsiones y pueden morir. ¿Cómo sé si tengo sífilis?
Los síntomas en los adultos se dividen en fases.
Fase primaria: Durante la primera fase (primaria) de la sífilis, es posible que note una única llaga, pero que haya muchas. La llaga aparece en el sitio por donde la sífilis entró al cuerpo. Por lo general, la llaga es firme, redonda y no causa dolor. Debido a que la llaga no causa dolor es posible que pase desapercibida. Las llagas duran de 3 a 6 semanas y se curan independientemente de que reciba tratamiento o no. Aunque las llagas desaparezcan, usted aún debe recibir tratamiento para que su infección no pase a la fase secundaria. Fase secundaria Durante la fase secundaria, es posible que tenga erupciones en la piel o llagas en la boca, la vagina o el ano (también llamadas lesiones de la membrana mucosa). Esta fase suele comenzar con la aparición de una erupción en una o más áreas del cuerpo. Las erupciones pueden aparecer cuando la llaga primaria se está curando o varias semanas después de que se haya curado. Esta erupción puede tomar el aspecto de puntos duros, de color rojo o marrón rojizo en la palma de las manos o en la planta de los pies. La erupción por lo general no pica y a veces es tan poco visible que es posible que ni se dé cuenta de que la tiene. Otros síntomas que es posible que tenga pueden incluir fiebre, inflamación de las glándulas linfáticas, dolor de garganta, pérdida parcial del cabello, dolores de cabeza, pérdida de peso, dolor muscular y fatiga (sentirse muy cansado). Los síntomas de esta fase desaparecerán reciba o no tratamiento. Sin el tratamiento adecuado, la infección progresará a una fase latente y posiblemente a las fases más avanzadas de la enfermedad.
Fases latente y avanzada: La fase latente de la sífilis comienza cuando todos los síntomas que tuvo antes desaparecen. Si no recibió tratamiento, usted puede seguir teniendo sífilis en su cuerpo por años sin presentar ningún signo o síntoma. La mayoría de las personas con sífilis sin tratar no evolucionan a la fase avanzada de esta enfermedad. Sin embargo, cuando esto sucede es muy grave y ocurriría entre 10 a 30 años desde que comenzó su infección. Los síntomas de la fase avanzada de sífilis incluyen dificultad para coordinar los movimientos musculares, parálisis (no poder mover ciertas partes del cuerpo), entumecimiento, ceguera y demencia (trastorno mental). En las fases avanzadas de la sífilis, la enfermedad daña sus órganos internos y puede causar la muerte. En una infección de sífilis, un caso “temprano” es cuando un paciente ha estado infectado por un año o menos, por ejemplo la fase primaria y secundaria de la sífilis. Las personas que tienen infecciones de sífilis “tempranas” pueden propagar la infección más fácilmente a sus parejas sexuales. La mayoría de los casos de sífilis temprano ocurren actualmente entre los hombres que tienen sexo con hombres, aunque las mujeres y los bebés en gestación también presentan riesgo de infección. ¿Cómo sabrá mi médico si tengo sífilis?
En la mayoría de los casos, se puede realizar un análisis de sangre para detectar la sífilis. Algunos proveedores de atención médica diagnosticarán sífilis al analizar el líquido de una llaga de sífilis. ¿Se puede curar la sífilis? Sí, la sífilis se puede curar con los antibióticos correctos que le recetará un proveedor de atención médica. Sin embargo, el tratamiento no revertirá ningún daño que la infección haya ya causado. He recibido tratamiento. ¿Puedo contraer sífilis nuevamente?
El hecho de que haya tenido sífilis una vez no lo protege de tenerla de nuevo. Aún después de haber sido tratado de manera exitosa, usted puede volver a infectarse. Solamente las pruebas de laboratorio pueden confirmar si tiene sífilis. Se recomiendan las pruebas de seguimiento por un proveedor de atención médica para asegurarse de que su tratamiento haya sido eficaz. Debido a que las llagas de la sífilis se pueden ocultar en la vagina, el ano, debajo de la piel que recubre el pene o la boca, es posible que no sea evidente si una pareja sexual tiene sífilis. A menos que sepa que sus parejas sexuales han sido evaluadas y tratadas, puede estar en riesgo de contraer sífilis otra vez de una pareja que no haya recibido tratamiento.
¿Dónde puedo obtener más información?
Enfermedades de transmisión sexual: Página principal Sífilis: Página sobre sífilis Sífilis y HSH: Hoja informativa. STDFact-MSM-Syphilis-s.htm ETS y embarazo: Hoja informativa STDFact-Pregnancy-s.htm Centro de información de los CDC 1-800-CDC-INFO (1-800-232-4636) Comuníquese con CDC–INFO Ejemplos de una erupción secundaria en las palmas de las manos (arriba).
Ejemplos de una erupción secundaria generalizada en el cuerpo (abajo).
Microscopía de campo oscuro de Treponema pallidum. Haga clic aquí para ampliar la imagen.
National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention CS233825B ¿Qué es la tricomoniasis? La tricomoniasis (o “tric”) es una enfermedad de transmisión sexual (ETS) muy común causada por la infección transmitida por el parásito protozoario llamado Trichomonas vaginalis. Los síntomas de la enfermedad pueden variar, y la mayoría de hombres y mujeres que tienen el parásito no saben que están infectados.
¿Qué tan frecuente es la tricomoniasis? La tricomoniasis es considerada la enfermedad de transmisión sexual curable más común. En los Estados Unidos, se calcula que 3.7 millones de personas tienen esa infección, pero solo alrededor de un 30% presenta algún síntoma. Es más frecuente en las mujeres que en los hombres y las mayores son más propensas que las jóvenes a tener la infección.
¿Cómo se contrae la tricomoniasis? Una persona infectada puede transmitirle el parásito a otra persona que no tenga la infección durante las relaciones sexuales. En las mujeres, el área del cuerpo infectada con más frecuencia es la parte baja del aparato genital (la vulva, la vagina o la uretra) y en los hombres es la parte interna del pene (uretra). Durante las relaciones sexuales, el parásito por lo general se transmite del pene a la vagina o de la vagina al pene, pero también se puede transmitir de una vagina a otra. No es frecuente que el parásito infecte otras partes del cuerpo, como las manos, la boca o el ano. No está claro por qué algunas personas con la infección presentan síntomas y otras no, pero probablemente depende de factores como la edad de la persona y su salud en general. Las personas infectadas que no tengan síntomas de todos modos pueden transmitirles la infección a otras. ¿Cuáles son los signos y síntomas de la tricomoniasis? Alrededor del 70% de las personas infectadas no presentan signos ni síntomas. Cuando la tricomoniasis causa síntomas, pueden variar entre irritación leve e inflamación grave. Algunas personas presentan los síntomas durante los 5 a 28 días después de haberse infectado, pero otras los presentan mucho más tarde. Los síntomas pueden aparecer y desaparecer.
Los hombres con tricomoniasis pueden sentir picazón o irritación dentro del pene, ardor después de orinar o eyacular, o pueden tener alguna secreción del pene. Las mujeres con tricomoniasis pueden notar picazón, ardor, enrojecimiento o dolor en los genitales, molestia al orinar, o una secreción clara con un olor inusual que puede ser transparente, blanca, amarillenta o verdosa. Tener tricomoniasis puede provocar molestias al tener relaciones sexuales. Si no se trata, la infección puede durar meses y hasta años. ¿Cuáles son las complicaciones de la tricomoniasis? La tricomoniasis puede aumentar el riesgo de contraer o propagar otras infecciones de transmisión sexual. Por ejemplo, puede causar inflamación genital que hace más fácil infectarse con el virus del VIH o transmitírselo a una pareja sexual. ¿Qué efectos tiene la tricomoniasis en una mujer embarazada y en su bebé? Las mujeres embarazadas que tienen tricomoniasis son más propensas a tener sus bebés antes de tiempo (parto prematuro). Además, los bebés nacidos de madres infectadas tienen más probabilidades de tener bajo peso al nacer, según los parámetros oficiales (menos de 5.5 libras).
Tricomoniasis - Hoja informativa de los CDC Dos parásitos Trichomonas vaginalis, amplificados (observados a través del microscopio) ¿Cómo se diagnostica la tricomoniasis? Es imposible diagnosticar la tricomoniasis basándose únicamente en los síntomas. Tanto a los hombres como a las mujeres, el médico de atención primaria u otro proveedor de atención médica tiene que hacerles un examen y una prueba de laboratorio para diagnosticar la tricomoniasis. ¿Cuál es el tratamiento de la tricomoniasis? La tricomoniasis se puede curar con una sola dosis de un antibiótico recetado (puede ser metronidazol o tinidazol), en pastillas que se pueden tomar por la boca. Las mujeres embarazadas pueden tomar este medicamento. Algunas personas que consuman alcohol durante las 24 horas después de tomar este tipo de antibiótico pueden tener efectos secundarios molestos. Las personas que hayan sido tratadas por tricomoniasis pueden contraerla de nuevo. Aproximadamente 1 de cada 5 personas se infectan otra vez dentro de los 3 meses después del tratamiento. Para evitarlo, asegúrese de que todas sus parejas sexuales también reciban tratamiento y espere para tener relaciones sexuales nuevamente hasta que todos sus síntomas hayan desaparecido (alrededor de una semana). Hágase examinar otra vez si le vuelven los síntomas. ¿Cómo se puede prevenir la tricomoniasis? Usar condones de látex correctamente todas las veces que tenga relaciones sexuales le ayudará a reducir el riesgo de contraer o transmitir la tricomoniasis. Sin embargo, los condones no cubren toda el área y es posible contraer o transmitir esta infección incluso cuando se utiliza uno.
La única manera segura de prevenir las infecciones de transmisión sexual es evitar por completo las relaciones sexuales. Otra manera de abordarlo es hablar acerca de esta clase de infecciones antes de tener relaciones sexuales con una nueva pareja, para tomar decisiones fundamentadas acerca del nivel de riesgo con que la persona se siente cómoda en su vida sexual. Si usted o alguna persona que conozca tiene preguntas acerca de la tricomoniasis o cualquier otra enfermedad de transmisión sexual, especialmente con síntomas como una secreción inusual, ardor al orinar o una úlcera en el área genital, consulte a un proveedor de atención médica para obtener respuestas. Quiero Saber American Sexual Health Association (ASHA) P. O. Box 13827 Resear ch Triangle P ark, NC 27709-3827 1-800-783-9877 www.quierosaber.org/ets.html Recursos ¿Dónde puedo obtener más información? División de Prevención de Enfermedades de Transmisión Sexual (DSTDP) www.cdc.gov/std/spanish/ Centros para el Control y la Prevención de Enfermedades Centro de información de los CDC 1-800-CDC-INFO (1-800-232-4636) Comuníquese con CDC–INFO CS265655A / Versión en español aprobada por CDC Multilingual Services – Order # 265202 National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention Las enfermedades de transmisión sexual (ETS) durante el embarazo: Hoja informativa de los CDC Una mujer embarazada que tiene una ETS puede infectar a su bebé antes, durante y después del parto. Una mujer embarazada que tiene una ETS puede infectar a su bebé antes, durante y después del parto. Si está embarazada puede infectarse de las mismas enfermedades de transmisión sexual (ETS) que las mujeres que no están embarazadas. Las mujeres embarazadas deben pedirle al médico que les haga pruebas de detección de ETS, ya que algunos doctores no lo hacen de manera habitual.
Un componente fundamental de la atención prenatal adecuada es garantizar que se les hagan pruebas de detección de ETS a las pacientes embarazadas. Hágales las pruebas de detección de ETS a sus pacientes embarazadas en etapas tempranas del embarazo y repítalas al acercarse el parto, si es necesario. Para garantizar que se realicen las pruebas de detección correctas, lo animamos a que tenga conversaciones abiertas y sinceras con sus pacientes embarazadas y, si es posible, con sus parejas sexuales sobre los síntomas que hayan tenido o tengan y sobre cualquier comportamiento sexual de alto riesgo que puedan tener.
Las siguientes secciones proporcionan detalles sobre los efectos de ETS específicas durante el embarazo con enlaces a páginas web con información adicional.
Estoy embarazada. ¿Puedo contraer una ETS? Sí, sí puede. Las mujeres embarazadas pueden infectarse con las mismas enfermedades de transmisión sexual que las mujeres que no están embarazadas. El embarazo no ofrece a las mujeres ni a sus bebés ninguna protección adicional contra las ETS. Muchas son “silenciosas” , o no tienen síntomas, por lo que usted podría no saber si está infectada. Si está embarazada, debe hacerse pruebas de detección de ETS, incluido el VIH (el virus que causa el sida), como parte de la atención médica de rutina durante el embarazo. Las consecuencias de una ETS para usted y su bebé pueden ser más graves y hasta mortales, si se infecta durante el embarazo. Es importante que usted conozca los efectos dañinos de las ETS y que sepa cómo protegerse y proteger a su bebé contra las infecciones. Si le diagnostican una ETS durante el embarazo, también le deben hacer pruebas de detección y dar tratamiento a su pareja sexual. ¿Cómo me afectarán o afectarán a mi bebé en gestación las ETS? Las ETS pueden causarle complicaciones en el embarazo y tener graves efectos en usted y su bebé en gestación. Algunos de estos problemas se pueden notar al momento del nacimiento, mientras que otros no se descubrirán sino hasta meses o años después. Además, se sabe que la infección por una enfermedad de transmisión sexual, puede hacer más fácil que una persona se infecte con el VIH. La mayoría de estos problemas pueden prevenirse si usted recibe atención médica de rutina durante el embarazo. Esto incluye hacer pruebas de detección de ETS en etapas tempranas del embarazo y repetirlas cerca del momento del parto, si es necesario.
Page 2 of 2 ¿Debo hacerme pruebas de detección de ETS durante mi embarazo?
Sí. Las pruebas de detección y el tratamiento de las enfermedades de transmisión sexual en las mujeres embarazadas es una forma vital de prevenir graves complicaciones tanto para la salud de la madre como la del bebé, que de otra forma se presentarían por la infección. Mientras más pronto usted reciba atención médica durante el embarazo, mejores serán los resultados para su salud y la de su bebé en gestación. Las directrices de los Centros para el Control y la Prevención de Enfermedades del 2015 para el tratamiento de las ETS recomiendan las pruebas de detección de ETS para las mujeres embarazadas. Las recomendaciones de los CDC sobre las pruebas de detección que su proveedor de atención médica debe seguir están incorporadas en la tabla sobre ETS durante el embarazo: Hoja informativa detallada de los CDC (en inglés).
Asegúrese de preguntarle a su médico sobre hacerse las pruebas de detección de ETS. También es importante que usted tenga una conversación abierta y sincera con su proveedor y hablen de cualquier síntoma que tenga y todas las conductas sexuales de alto riesgo en las que participe, ya que algunos doctores no realizan estas pruebas de manera rutinaria. Aunque usted se haya hecho pruebas en el pasado, debería volvérselas a hacer si queda embarazada.
¿Puedo recibir tratamiento para una ETS mientras estoy embarazada?
Esto depende de varios factores. Las enfermedades de transmisión sexual como la clamidia, gonorrea, sífilis, tricomoniasis y vaginosis bacteriana pueden tratarse y curarse con antibióticos que se pueden tomar en forma segura durante el embarazo. Las ETS causadas por virus, como el del herpes genital, la hepatitis B o el VIH, no se pueden curar. Sin embargo, en algunos casos estas infecciones se pueden tratar con medicamentos antivirales u otras medidas preventivas para reducir el riesgo de transmisión de la infección al bebé. Si está embarazada o planea quedar embarazada, debe hacerse las pruebas de detección para que pueda tomar medidas para protegerse y proteger a su bebé.
¿Cómo puedo reducir mi riesgo de contraer una ETS mientras estoy embarazada?
La única manera de evitar las ETS es no tener relaciones sexuales vaginales, anales ni orales.
Si usted es sexualmente activa, puede hacer las siguientes cosas para disminuir las probabilidades de contraer clamidia: • • Tener una relación mutuamente monógama a largo plazo con una pareja que se haya hecho pruebas de ETS y haya obtenido resultados negativos.
• • Usar condones de látex de manera correcta cada vez que tenga relaciones sexuales Related Content ¿Dónde puedo obtener más información? División de Prevención de Enfermedades de Transmisión Sexual (DSTDP) default.htm Centros para el Control y la Prevención de Enfermedades Centro de información de los CDC 1-800-CDC-INFO (1-800-232-4636) Comuníquese con CDC–INFO Quiero Saber (from ASHA)
html 1-800-783-9877 May 2016 |
16818 | https://math.jhu.edu/~brown/courses/s12/Lectures/Week7.pdf | 110.109 CALCULUS II Week 7 Lecture Notes: March 12 - March 16 Improper Integrals (cont’d.) Going back to the definition of an improper integral, we can add a second part: Definition 1. For f(x) continuous for all x ≥a, the improper integral ∞ a f(x) dx := lim b→∞ b a f(x) dx, provided the limit exists. If the limit exists, we say the improper integral converges. Else, we say it diverges.
For f(x) continuous for all x ≤b, the improper integral b −∞ f(x) dx := lim a→−∞ b a f(x) dx, provided the limit exists.
Example 2. Calculate ∞ 0 te−t dt, if it exists (Note that this is much like Example 7.8.2 in the book, but going the other way). Here the function is continuous everywhere, but the integral is improper. Hence we need to appeal to the limit: ∞ 0 te−t dt = lim b→∞ b 0 te−t dt.
Using the technique of Integration by Parts, where f(t) = t, and g′(t) = e−t, we get f ′(t) = 1, and g(t) = −e−t, and lim b→∞ b 0 te−t dt = lim b→∞ −te−t b 0 − b 0 −e−t dt = lim b→∞ −te−t −e−t dt b 0 = lim b→∞ −be−b −e−b −(0 −1) = lim b→∞ −b + 1 eb + 1 = 1.
How would you show the last step? Try L’Hospital’s Rule. Hence, this integral converges and its value is 1.
Date: March 20, 2012.
1 2 Example 3. Calculate 0 −∞ sin x dx, if it exists. Here the function is continuous everywhere, and the integral is again improper. Appealing to the limit: 0 −∞ sin x dx = lim a→−∞ 0 a sin x dx = lim a→−∞(−cos x) 0 a = lim a→−∞(−cos 0 + cos a) .
But this limit does not exist for a different reason; the cosine function does not have a limit at infinity (it does not have a horizontal asymptote!). Hence the improper integral diverges.
Exercise 1. Show, for p ∈R, the improper integral ∞ 1 1 xp dx converges if p > 1 and diverges is p ≤1.
And finally, an integral is also improper if both of its limits “run offthe page”. Well, to handle this case, remember that we can always break up an interval given by the limits into two pieces, evaluate the definite integral on both pieces, and then add the results. This is important for the following definition.
Definition 4. Suppose f(x) is continuous on R. Then for ANY choice of c ∈R, the improper integral ∞ −∞ f(x) dx = c −∞ f(x) dx + ∞ c f(x) dx, provided that BOTH of the improper integrals on the right-hand-side exist.
Example 5. Calculate ∞ −∞ 3xe−x2 dx, if is exists. Here, the integrand is continuous on the entire real line (so the integral will exist on ANY finite interval), and the integral is improper (both limits need to be addressed). Choose our intermediate value to be 0 for symmetry (though this does not matter), and ∞ −∞ 3xe−x2 dx = 0 −∞ 3xe−x2 dx + ∞ 0 3xe−x2 dx = lim a→−∞ 0 a 3xe−x2 dx + lim b→∞ b 0 3xe−x2 dx.
Here, a straightforward substitution of u = x2, du = 2x dx is very helpful. Working just with the antiderivative for a minute, we get 3xe−x2 dx = 3 2e−u du = −3 2e−u + C = −3 2e−x2 + C.
3 Back to our improper integral, we have ∞ −∞ 3xe−x2 dx = lim a→−∞ 0 a 3xe−x2 dx + lim b→∞ b 0 3xe−x2 dx = lim a→−∞ ⎛ ⎝ −3 2e−x2 0 a ⎞ ⎠+ lim b→∞ ⎛ ⎝ −3 2e−x2 b 0 ⎞ ⎠ = lim a→−∞ −3 2 + 3 2e−a2 + lim b→∞ −3 2e−b2 + 3 2 = 3 2 + 3 2 = 3.
The improper integral converges to 3.
Next, I defined another way for an integral to be improper. Go back to the example of an integral where the integrands is f(x) = 1 x2, but this time choose to integrate on the interval [b, 1], where 0 < b ≤1. For a choice of b here, we get 1 b 1 x2 dx = −1 x 1 b = −1 + 1 b > 0.
We know it is greater than 0 since the entire graph of f(x) > 0, and also for the last expression 1 b > 1 (why?) Question 6. What if we pushed b all the way back to 0?
Again, we would produce a problem with the limits of the integral. And again, the problem would involve an interpretation of the definite integral as the area of an unbounded region.
However, this time, the region is unbounded in the vertical direction due to the vertical asymptote at x = 0. To deal with this situation, we employ the same trick as before, noting that the definite integral is perfectly well-defined for all positive values of b: define 1 0 1 x2 dx = lim b→0+ 1 b 1 x2 dx = lim b→0+ −1 x 1 b = lim b→0+ −1 + 1 b = lim b→0+ 1 −b b .
Finishing this calculation means evaluating the limit at the end of the equations above. We will get ∞, and can interpret this as the area between f(x) and the x-axis grows without bound as we push the lower limit back to 0. This is an example of another type of improper integral, where the integrand is not defined at one or both of the limits.
Example 7. Do the same thing by replacing the integrand with the new function g(x) = 1 3 √x.
Note that the graph look remarkably similar, and there is again a vertical asymptote of g(x) at x = 0. However, this time, we get 1 0 1 3 √x dx = lim b→0+ 1 b 1 3 √x dx = lim b→0+ 3 2x 2 3 1 b = lim b→0+ 3 2 −3 2 3 √ b2 = 3 2.
4 Here the resulting limit does exist (the function 3 √ x2 is continuous from the right at x = 0).
The interpretation is that the area of the unbounded region between the curve and the x-axis is finite and is 3 2.
Exercise 2. Show, for p ∈R, the improper integral 1 0 1 xp dx converges if p < 1 and diverges is p ≥1.
Exercise 3. Show, for ANY p ∈R, the improper integral ∞ 0 1 xp dx diverges.
Example 8. How about h(x) = (x+1)(x−2) x−2 on the interval [1, 3]?
Here both limits of the integral of h(x) would be fine. However, there is a point inside the interval (namely x = 2) where the function is not defined. We cannot simply ignore this point. We can, however, adjust the calculation to accommodate it knowing the Sum Law for Integrals (Section 5.2, page 374). We can write 3 1 (x + 1)(x −2) x −2 dx = 2 1 (x + 1)(x −2) x −2 dx + 3 2 (x + 1)(x −2) x −2 dx, knowing that both of the integrals on the right-hand side are improper. Keep in mind that they are improper because one of their limits is outside of the domain of the function, and not because there is an asymptote there (there isn’t in this case). In this calculation, we get 3 1 (x + 1)(x −2) x −2 dx = 2 1 (x + 1)(x −2) x −2 dx + 3 2 (x + 1)(x −2) x −2 dx = lim c→2− c 1 (x + 1)(x −2) x −2 dx + lim c→2+ 3 c (x + 1)(x −2) x −2 dx = lim c→2− c 1 (x + 1) dx + lim c→2+ 3 c (x + 1) dx.
Notice here two things: (1) We set up the limits only from one side. In each case, we only need to evaluate the limit from the side within the interval of integration.
This is very important. And (2), since we are now offthe “bad” point at x = 2, the function h(x) is equivalent to x + 1, and we can simplify the integrand. Continuing the calculation 3 1 (x + 1)(x −2) x −2 dx = lim c→2− c 1 (x + 1) dx + lim c→2+ 3 c (x + 1) dx = lim c→2− x2 2 + x c 1 + lim c→2+ x2 2 + x 3 c = lim c→2− c2 + 2c 2 −3 2 + lim c→2+ 15 2 −c2 + 2c 2 .
5 Both of these limits exist as the expressions are perfectly continuous at x = 2. To finish, 3 1 (x + 1)(x −2) x −2 dx = lim c→2− c2 + 2c 2 −3 2 + lim c→2+ 15 2 −c2 + 2c 2 = 8 2 −3 2 + 15 2 −8 2 = 12 2 = 6.
Hence the improper integral exists and its value is 6.
So let’s now define exactly what I mean by an improper (definite) integral of this kind, and the plan for evaluating it, if possible.
Definition 9. If f(x) is continuous on [a, b) and discontinuous at x = b, then b a f(x) dx = lim c→b− c a f(x) dx, provided the limit exists.
Definition 10. If f(x) is continuous on (a, b] and discontinuous at x = a, then b a f(x) dx = lim c→a+ b c f(x) dx, provided the limit exists.
Note that, like before, if the limit exists in either case, we say that the improper integral converges. If the limit does not exist, then the integral diverges.
Definition 11. If f(x) is continuous on [a, b] except at the point a < c < b, then b a f(x) dx = c a f(x) dx + b c f(x) dx, where at least one of the integrals on the right-hand side are improper.
We will continue next time.
Lecture 2: Improper Integrals (cont’d) and Sequences Example 12. Calculate 2π 0 f(x) dx, if possible, for the piecewise-defined function f(x) = x if x ∈(−∞, π) cos(x −π) if x ∈[π, ∞) .
6 Here we not that the function is discontinuous at x = π, and use the last definition to write 2π 0 f(x) dx = π 0 x dx + 2π π cos(x −π) dx = lim c→π− c 0 x dx + 2π π cos(x −π) dx since only one of the integrals is actually improper (f(x) is continuous from the right at x = π. Hence the last integral on the right is okay). Continuing, we get 2π 0 f(x) dx = lim c→π− c 0 x dx + 2π π cos(x −π) dx = lim c→π− x2 2 c 0 + sin(x −π) 2π π = lim c→π− c2 2 = π2 2 .
Remark 13. You may have noticed (like in Examples 8 and 12) that when there is no asymptote involved, that the limit exists and hence the improper integral converges. This is NOT always the case, and it is always necessary to evaluate the limit explicitly than to make an assumption.
Example 14. Calculate 3 0 g(x) dx, if possible, for the piecewise-defined function g(x) = 1 if x ≤2 1 (x−2)2 if x > 2 .
Note that here, as in the last example, the domain of g(x) is all real numbers. However, there is a vertical asymptote for g(x) at x = 2.
We start be separating out the discontinuity: 3 0 g(x) dx = 2 0 dx + 3 2 1 (x −2)2 dx = 2 0 dx + lim c→2+ 3 c 1 (x −2)2 dx 7 since g(x) is continuous from the right at x = 2. Continuing, we get 3 0 g(x) dx = 2 0 dx + lim c→2+ 3 c 1 (x −2)2 dx = x 2 0 + lim c→2+ − 1 (x −2) 3 c = 2 + lim c→2+ −1 + 1 c −2 = 2 + lim c→2+ 3 −c c −2 = ∞.
This integral diverges even though part of it is fine.
Example 15. For a > 0, calculate a −a 1 5 √x dx, if possible.
This integrand is an odd function. Hence it is symmetric with respect to the origin. Recall the the integral of an odd continuous function on the interval [−a, a] should be 0. However, since the integral is improper (asymptote at x = 0), we do not know if the integral even converges. To check, we need to actually do the calculation.
Here we get a −a 1 5 √x dx = 0 −a 1 5 √x dx + a 0 1 5 √x dx = lim c→0− c −a 1 5 √x dx + lim c→0+ a c 1 5 √x dx = lim c→0− 5 4 5 √ x4 c −a + lim c→0+ 5 4 5 √ x4 a c = lim c→0− 5 4 5 √ c4 −5 4 5 (−a)4 + lim c→0+ 5 4 5 √ a4 −5 4 5 √ c4 = −5 4 5 √ a4 + 5 4 5 √ a4 = 0.
The thing here is that both of the limits actually do exist, and are the same magnitude yet opposite signs. Hence the origin integral converges for any value of a > 0, and the integral is 0.
Finding the value of a definite integral which is improper means evaluating a limit as well as using the Fundamental Theorem of calculus (using the antiderivative evaluated at the limits). However, many time, the actual value of the integral is not nearly as important as knowing whether the integral converges or not. This is particularly true when the integral is difficult or impossible to solve analytically. Fortunately, and like other areas of calculus, integrals behave well when compared to each other. We start with a major theorem: 8 Theorem 16. Suppose f(x), g(x) are continuous with f(x) ≥g(x) ≥0 for all x ≥a.
Then if ∞ a f(x) dx converges, then ∞ a g(x) dx converges. And if ∞ a g(x) dx diverges, then ∞ a f(x) dx diverges.
Notes: • The improper integrals above represent unbounded areas between their respective curves and the x-axis. The integral of the larger function must be bigger than that of the smaller (there is more area). Thus if the integral of the larger is finite, the area of the smaller must also be finite. Also, if the area of the smaller is not finite, then neither can the area of the larger.
• This is also true is the two continuous functions cross each other a number of times, as long as at some point they stop crossing and one dominates the other in the “tail”.
• really, this is ONLY an existence theorem, and says nothing about the actual value of either of the integrals, if they exist.
I then did example 9 from this section in detail. Also pay attention to Example 10.
1. Lecture 3: Sequences On to sequences.
Definition 17. A sequence of numbers is simply a infinite list of numbers, denoted by a variable and a subscript that takes values in the natural numbers N = {1, 2, 3, . . .}, a1, a2, . . . , or {an} or {an}n∈N or {an}∞ n=1 .
Notes: • Sometimes, the subscript will take values other than the natural numbers. For ex-ample, it may start at 0, or may incorporate some of the negative integers. it will be obvious, however, from the context, what the range of values of the subscript will be in the application.
• Many times, the sequence follows a pattern which may be set out according to a rule (like a function) based on the subscript: an = 1 n, or bn = (−1)n 2n , or cn = sin 2πn.
These sequences are actually functions of their subscript, and all can be written as an = f(n). The first and last look like the integer points of a continuous function. For example, an = f(n), where f(x) = 1 x. This would be harder to do for the sequence {bn}.
9 • Graphing a sequence is like graphing y = f(x), except that the graph is a discrete set of points in the plane with coordinates (n, an).
Definition 18. A sequence {an} has a limit L and we write lim n→∞{an} = L, or lim n→∞an = L, or {an} − →L if, for every ϵ > 0, there is a corresponding N where, if n > N, then |an −L| < ϵ.
Here, the last inequality is not difficult to see. |an −L| < ϵ is the same as −ϵ < an−L < ϵ, which is the same as L −ϵ < an < L + ϵ. This is simply a small band around a number L of width ϵ is either direction (for a total width of 2ϵ). The definition says that given a sequence, the number L will be the limit of the sequence provide no matter how small we make the band, eventually the sequence will enter that band and never leave again. This is exactly the same as what you studied for function way back in Calculus I.
Really, having a limit or not is the ONLY interesting thing about sequences. And it is an important quality to have. So having the means to determine if a sequence has a limit and to be able to find it if it does have one are paramount. There are many ways to discover limits of sequences.
Use the function. Suppose an = f(n) for some function f(x) which is continuous on [1, ∞). If f(x) has a horizontal asymptote at ∞, then the sequence has no choice but to follow the function, and the sequence converges. In other words: Proposition 19. For an = f(n), where f(x) is continuous on [1, ∞), if lim x→∞f(x) = L, then lim n→∞an = L.
Example 20. Let an = 2 −1 n. Since f(x) = 2 −1 x = 2x−1 x is continuous on [1, ∞), and lim x→∞f(x) = 2, it follows that lim n→∞an = 2 also.
Note: The converse is definitely NOT true. Just because a sequence defined by an = f(n) converges, it does not follow that the original function has a horizontal asymptote at infinity!
Example 21. Let f(x) = cos 2πx. f(x) is continuous for all reals. And f(x) does not have a horizontal asymptote at infinity. This means that lim x→∞f(x) does not exist. It is the cosine function. But an = f(n) = cos 2πn = 1 for all n ∈N. This sequence does converge (to 1, that is). In fact, the sequence is just the peaks of all of the humps that is the graph of the cosine function.
Also, when using the continuous function f(x) to evaluate the limit of a sequence an = f(n), one can use all of the tools of calculus. But ONLY on the continuous function!
10 Example 22. Does an = n2e−3n have a limit? Again, associate to this sequence the con-tinuous function f(x) = x2e−3x = n2 e3x. Does f(x) have a horizontal asymptote? Since f(x) is not only continuous but differentiable, we can use all of our calculus tools to study this function. Evaluate lim x→∞ x2 e3x. Here, if we push out x to infinity in the expression (in a sense, evaluating the limits of the top and bottom of the fraction), we get the indeterminate form ∞ ∞. Thus we can apply L’Hospital’s Rule so that lim x→∞ x2 e3x = lim x→∞ d dxx2 d dxe3x = lim x→∞ 2x 3e3x.
Repeating the idea, we see again that we would get the indeterminate form ∞ ∞. Applying L’Hospital’s Rule a second time gets us lim x→∞ x2 e3x = lim x→∞ d2 dx2x2 d2 dx2e3x = lim x→∞ 2 9e3x = 0.
Hence an does indeed have a limit.
And now a practical word.
One CANNOT use something like L’Hospital’s Rule on the sequence an. The sequence is NOT differentiable.
More notes: • if a limit of a sequence exists, we say that the sequence converges. if the limit does not exist, then the sequence diverges.
• Limits of sequences behave exactly like limits of functions and expression that you learned in an early chapter of the book. You should pay attention to all of the rules on page 678.
Example 23. Product Rule for Limits. Remember the Product rule for Limits: Suppose {an}, and {bn} are two convergent sequences (their limits exist!). Then lim n→∞anbn = lim n→∞an lim n→∞bn .
But caution: This ONLY works when both {an}, and {bn} are convergent.
For example, let {an} = {n}, and {bn} = 2 n . Then 2 = lim n→∞n · 2 n ̸= lim n→∞n lim n→∞ 2 n since the second limit on the right is 0, while the first is ∞(it does not exist). Hence the product rule FAILS when one of the individual sequences does not have a limit.
work with these limit rules and get very comfortable with them....
11 Squeezing sequences into limits. It should be quite clear that if one convergent sequence is term-by-term larger than another convergent sequence, then the limit of the first will be bigger then the limit of the second. We can take this one step further by saying that if one sequence lives term-by-term within two other convergent sequences, than the limit of the “sandwiched” sequence, if it exists, will live between the other two. The following theorem follows this theme to a very nice conclusion: 10 20 30 40 50 0.15 0.10 0.05 0.05 0.10 0.15 Theorem 24. Suppose an ≤cn ≤bn for n > n0, and lim n→∞an = lim n→∞bn = L. Then limn→∞cn = L.
Example 25. Show the sequence {cn} = sin n n converges to 0. This should not be hard to see as the numerator never ventures far from 0, while the denominator gets large as the sequence progresses.
However, to actually show the result takes a bit of work. One can use the corresponding function f(x) = sin x x and try to find an asymptote at infinity, as above. But L’Hospital’s Rule doesn’t apply here (why not?).
Instead, we will employ the Squeezing Theorem.
To start, notice that the numerator satisfies −1 ≤sin n ≤1 (one of the nice things about this trig function).
Divide the inequality by n to get −1 n ≤sin n n ≤1 n.
Let an = −1 n and bn = 1 n, and we have an ≤cn ≤bn just as in the theorem (choose n0 = 1 but any choice of n0 will do). And since lim n→∞an = lim n→∞bn = 0, the theorem concludes that also limn→∞cn = 0. Hence sin n n − →0. See the picture, where an is in blue on the bottom, and bn is in red on top. Both are squeezing cn in black (the more random-looking points in the middle).
We will continue next time.... |
16819 | https://www.albert.io/blog/understanding-unit-rates-with-fractions-from-production-to-speed/ | Unit Rates with Fractions: From Production to Speed | Albert Blog & Resources
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Unit Rates with Fractions: From Production to Speed
The Albert Team
Last Updated On: November 25, 2024
Understanding unit rates with fractions is essential for mastering various math concepts, and it frequently appears in standardized tests such as the ACT® WorkKeys Applied Math Test. Unit rates with fractions play a vital role in our daily lives, aiding in the comprehension of speed (measured in miles or kilometers per hour) and production rates (reflecting output over time), as well as assisting in determining prices per ounce or per mile. This article aims to simplify the process of calculating unit rates with fractions through clear, step-by-step methods and applications.
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What We Review
Toggle - [x]
What is a Rate, and How Does it Differ from a Ratio?
The Importance of Unit Rates with Fractions
Step-by-Step Guide to Calculate Unit Rates with Fractions
Understanding Speed as a Rate with Fractions
Proportionality and Simplifying Fractions in Rates
Real-World Applications of Rates with Fractions
Tips for Solving Word Problems Involving Rates with Fractions
Conclusion: The Relevance of Mastering Rates with Fractions
Sharpen Your Skills for ACT® WorkKeys Applied Math
Need help preparing for the ACT® WorkKeys Applied Math Test?
What is a Rate, and How Does it Differ from a Ratio?
A rate is a way to compare two different amounts that are measured in different units. Rates are often used to answer questions such as “How much?” or “How fast?” For instance, when we talk about driving 60 miles in one hour, we are expressing a rate. In this case, we refer to this as 60 miles per hour, which indicates how distance is being covered relative to time.
Conversely, a ratio is used to compare quantities measured in the same unit, illustrating their relationship. For example, if a recipe calls for 3 cups of flour and 2 cups of sugar, we can express this relationship as a ratio of 3:2. This ratio shows that for every 3 cups of flour, there are 2 cups of sugar in the recipe. Unlike rates, which involve different measurements, ratios focus solely on comparable units.
The Importance of Unit Rates with Fractions
Unit rates with fractions are useful because they enable us to compare various situations in a straightforward manner. For example, when we want to buy groceries, understanding the cost per item can help us determine which product offers the best value for money. Similarly, when assessing the fuel efficiency of a vehicle, knowing how many miles a car can travel per gallon of gas helps us gauge its performance.
Step-by-Step Guide to Calculate Unit Rates with Fractions
To calculate unit rates with fractions, follow these simple steps.
Identify the given rate: This could refer to any quantity you can measure, like speed in miles per hour or how many items are made in a day.
Express the rate as a fraction: Use fractions to represent the rate, such as (3 4)( \frac{3}{4} )(4 3) hours per task.
Divide to find the unit rate: Divide the numerator (top number) by the denominator (bottom number). Simplify the fraction to express it as a “per one” unit rate.
Practicing these steps helps you become more confident and faster at solving unit rate problems. Being able to quickly change complex rates into simple ones makes it easier to deal with real-world situations.
Example: Calculating Production Rates
Imagine a factory that can make 90 items in 3.5 hours. To find out how many items it produces in one hour, we look at it as (90 items 3.5 hours)( \frac{90\text{ items}}{3.5\text{ hours}} )(3.5 hours 9 0 items).
Now, let’s simplify this. When you divide 90 by 3.5, you get about 25.7 items each hour.
This rate shows how well the factory is running.
Understanding these rates is useful for managers. If on another day, the factory produced 150 items in an 8-hour shift, its production rate was 18.75 items per hour. This is much less productive than 25.7 items per hour. If production consistently drops on certain days, it might indicate issues such as equipment malfunctions or insufficient staffing.
Practice Question: Production Scenario
You are a baker. Over the past 3 days, you have baked 120 loaves of bread. If you work 8 hours each day, how many loaves of bread do you bake per hour at your current baking rate?
Solution:
First, calculate the total number of hours you worked over the past 3 days:
3 days×8 hours/day=24 hours 3 \text{ days} \times 8 \text{ hours/day} = 24 \text{ hours} 3 days×8 hours/day=2 4 hours
Next, to find the rate of loaves baked per hour, divide the total number of loaves by the total number of hours worked:
120 loaves 24 hours=5 loaves/hour \frac{120 \text{ loaves}}{24 \text{ hours}} = 5 \text{ loaves/hour} 2 4 hours 1 2 0 loaves=5 loaves/hour
Thus, at your current baking rate, you bake 5 loaves of bread per hour.
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Understanding Speed as a Rate with Fractions
Speed is often expressed as a rate involving distance and time. It indicates how fast an object moves over a period. Fractions make it easier to represent these rates precisely.
Using fractions can simplify complex speed calculations. For instance, knowing the exact distance traveled per time unit is useful. Converting speeds into unit rates makes comparisons between different speeds more straightforward.
Example: Determining Speed Rates
A car travels 50 miles in 2.5 hours. To find the speed, we can express this as (50 miles 2.5 hour)( \frac{50\text{ miles}}{2.5\text{ hour}} )(2.5 hour 5 0 miles).
Now, let’s simplify the fraction by dividing 50 by 2.5.
When we divide 50 by 2.5, we get a speed of 20 miles per hour. This tells us how fast the car is going during the trip.
Practice Question: Speed Scenario
A cyclist covers 36 miles in 1.5 hours. What is the cyclist’s speed in miles per hour? Follow the steps to find the rate and verify the answer for accuracy.
To find the cyclist’s speed in miles per hour, we need the distance traveled and the time taken. The cyclist traveled 36 miles in 1.5 hours. We use the speed formula:
Speed=Distance Time \text{Speed} = \frac{\text{Distance}}{\text{Time}} Speed=Time Distance
Substituting the values:
Speed=36 miles 1.5 hours \text{Speed} = \frac{36 \text{ miles}}{1.5 \text{ hours}} Speed=1.5 hours 3 6 miles
Calculating this gives:
36 1.5=24 \frac{36}{1.5} = 24 1.5 3 6=2 4
Thus, the cyclist’s speed is 24 miles per hour.
To confirm, we multiply the speed (24 mph) by the time (1.5 hours):
24 mph×1.5 hours=36 miles 24 \text{ mph} \times 1.5 \text{ hours} = 36 \text{ miles} 2 4 mph×1.5 hours=3 6 miles
Since this matches the original distance, our calculation is correct. The cyclist’s speed is indeed 24 miles per hour.
Proportionality and Simplifying Fractions in Rates
Understanding proportionality is important when dealing with rates and fractions. Proportionality means that two ratios are equal, which indicates a constant relationship. This idea is essential for comparing different rates.
Simplifying fractions can make calculations simpler. When you reduce a fraction to its simplest form, it helps you understand the rate better. Simplified rates are also easier to compare with other rates.
For example, if you have a rate of (12 6)( \frac{12}{6} )(6 1 2), simplifying it to 2 is helpful. This means you have 2 items for each unit. This simplification makes it easier to grasp how the rate works.
Real-World Applications of Rates with Fractions
We often come across rates that include fractions in our daily lives. For example, when we cook, we use unit rates to determine how much of each ingredient we need. This includes measuring things like how many cups of flour to use for a recipe.
Another example is when we travel. When planning a trip, it’s important to know how many miles your vehicle can go on one gallon of gas. This miles-per-gallon information helps you manage travel costs and stick to your budget.
Tips for Solving Word Problems Involving Rates with Fractions
To solve word problems, it’s important to have a clear plan. Start by reading the problem carefully. Make sure you understand all parts of the situation. Look for key details and numbers before thinking of solutions. This step is crucial for effective problem-solving.
Here are some helpful tips:
Check Units: Make sure all units in the problem match and make sense together. Different units can cause confusion, so it’s essential to ensure they align.
Simplify Fractions: Always use simplified fractions when solving. This makes calculations clearer and easier to follow. Simplifying can make your work lighter and more straightforward.
Double-Check: After you finish your calculations, review them. This helps catch mistakes and confirms that your answers are correct.
By following these steps and tips, you’ll find that even tough problems can be easier to manage. This will help you feel more confident in solving them.
Conclusion: The Relevance of Mastering Rates with Fractions
Grasping rates that involve fractions is an important skill useful in many everyday situations. These math skills help you solve different problems you might face and prepare you for more advanced math challenges later on. Knowing about rates is especially important for doing well on tests like the ACT® WorkKeys Applied Math Test and similar exams. By taking advantage of this learning opportunity, you can improve your math abilities and gain confidence in handling numbers and solving equations.
Sharpen Your Skills for ACT® WorkKeys Applied Math
Are you preparing for the ACT® WorkKeys Applied Math test? We’ve got you covered! Try our review articles designed to help you confidently tackle real-world math problems. You’ll find everything you need to succeed, from quick tips to detailed strategies. Start exploring now!
Ratios, Proportions, and Rates: Solving Real-World Problems
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16820 | https://math.stackexchange.com/questions/2290879/geodesic-curvature-projection-to-tangent-space | differential geometry - Geodesic curvature - projection to tangent space - Mathematics Stack Exchange
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Geodesic curvature - projection to tangent space
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Wikipedia article on geodesic curvature says the following :
Consider a curve γ γ in a manifold M¯M¯, parametrized by arclength, with unit tangent vector T=d γ/d s T=d γ/d s. Its curvature is the norm of the covariant derivative of T:k=||D T/d s||T:k=||D T/d s||. If γ γ lies on M M, the geodesic curvature is the norm of the projection of the covariant derivative D T/d s D T/d s on the tangent space to the submanifold and the normal curvature is the norm of the projection of the covariant derivative D T/d s D T/d s on the normal bundle at the point considered. If the ambient manifold is the euclidean space R n R n, then the covariant derivative D T/d s D T/d s is just the usual derivative d T/d s d T/d s.
I am studying geodesic/normal curvature of curves on surfaces in R 3 R 3.
Let σ:U→R 3 σ:U→R 3 be a parametrized surface, γ:I→R 3 γ:I→R 3 be a parametrized curve on σ σ i.e., γ=σ∘μ γ=σ∘μ where μ:I→U μ:I→U given by μ(s)=(x 1(s),x 2(s))μ(s)=(x 1(s),x 2(s)) is a regular smooth parametrized curve.
For this, we have defined t^(s)=γ′(s),m^(s)=N(μ(s))t^(s)=γ′(s),m^(s)=N(μ(s)) and u^(s)=m^(s)×t^(s)u^(s)=m^(s)×t^(s) where
N^(u,v)=σ u×σ v||σ u×σ v||N^(u,v)=σ u×σ v||σ u×σ v||
We have then seen that {t^(s),m^(s),u^(s)}{t^(s),m^(s),u^(s)} forms a coordinate system and any vector perpendicular to one of these three is in the span of remaining two vectors. As γ′′(s)=t^′(s)γ″(s)=t^′(s) is perpendicular to t^(s)t^(s) we need to have γ′′(s)γ″(s) as linear combination of m^(s)m^(s) and u^(s)u^(s) so there exists k g(s),k n(s)∈R k g(s),k n(s)∈R such that
γ′′(s)=k g(s)u^(s)+k n(s)m^(s).γ″(s)=k g(s)u^(s)+k n(s)m^(s).
k g(s)k g(s) defined as above is called the geodesic curvature of γ γ at s s and k n(s)k n(s) defined as above is called the normal curvature of γ γ at s s.
Now, I would like to relate this to what wikipedia article says about geodesic curvature. Projection of the covariant derivative γ′′(s)γ″(s) on the normal bundle i.e., ||γ′′(s)⋅N^(μ(s))||=||γ′′(s)⋅m^(s)||=k n(s)||γ″(s)⋅N^(μ(s))||=||γ″(s)⋅m^(s)||=k n(s) is the normal curvature. Projection of the covariant derivative γ′′(s)γ″(s) on the tangent space ? It is not really clear which tangent space are they talking about. Any explanation on this is useful.
differential-geometry
curvature
geodesic
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asked May 21, 2017 at 18:11
user312648 user312648
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Since your curve has unit speed (i.e., is parametrized by arc length), the equation
d T d s=γ′′(s)=k g(s)u^(s)+k n(s)m^(s)d T d s=γ″(s)=k g(s)u^(s)+k n(s)m^(s)
is the decomposition of the accereration into tangential and normal components at γ(s)γ(s). That is, the projection of γ′′(s)γ″(s) on the tangent plane T γ(s)M T γ(s)M is γ′′(s)γ″(s) minus the normal component:
γ′′(s)−k n(s)m^(s)=k g(s)u^(s).γ″(s)−k n(s)m^(s)=k g(s)u^(s).
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answered May 22, 2017 at 1:13
Andrew D. HwangAndrew D. Hwang
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Thanks for your answer. Any reference regarding this would be helpful. So, u^(s)u^(s) is the tangent vector(space) they are talking about? But u^(s)=m^(s)×t^(s)u^(s)=m^(s)×t^(s).. Please say more about this.user312648 –user312648 2017-05-22 05:50:52 +00:00 Commented May 22, 2017 at 5:50
The "tangent space" is the tangent plane T γ(s)M T γ(s)M, spanned by t^t^ and u^u^. (The acceleration γ′′(s)γ″(s) has zero tangential component because the curve is unit-speed: 1=∥γ′∥2=γ′⋅γ′1=‖γ′‖2=γ′⋅γ′, so the product rule gives 0=2 γ′⋅γ′′0=2 γ′⋅γ″.) Not sure what type of reference you're seeking; everything in this answer comes straight from your post...?Andrew D. Hwang –Andrew D. Hwang 2017-05-22 10:02:26 +00:00 Commented May 22, 2017 at 10:02
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16821 | https://classics.osu.edu/Undergraduate-Studies/Latin-Program/Grammar/mood/subjunctive/Dependent-subjunctives/sequence-tenses | Sequence of Tenses | Department of Classics
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Present Indicative
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Subjunctive
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Infinitive
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Subject Accusative of the Infinitive
Participles
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Comparison with Ablative or quam
Quam
Origin of Quam
Indirect Discourse
Infinitive in Indirect Discourse
Literary Latin
The Origins of the Indirect Statement
Subordinate Clauses in Indirect Discourse
Sequence of Tense
English into Latin
Converting a Latin statement
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Catiline
Oratio in Catilnam Prima
I. 1-3
I.1-3 Notes
II.4-6
II.4-6 Notes
III. 6-8
III.6-8 Notes
IV. 8-10
IV.8-10 Notes
V. 10-13
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VI. 13-16
VI.13-16 Notes
VII.16-18
VII.16-18 Notes
VIII.19-21
VIII.19-21 Notes
IX.22-24
IX.22-24 Notes
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Subjunctive
Dependent Subjunctives
Sequence of Tenses
Sequence of Tenses
In the discussion of indirect statement in Unit 25, there was a certain emphasis on following a sequence of tenses when translating the accusative infinitive structure. Similarly in subjunctive constructions, both in Latin and English, a system of sequence of tenses exists when a speaker or writer proceeds from a main clause into a subordinate clause. Understanding of these sequences is imperative in order to successfully compose Latin sentences.
Latin has two sequences:
Primary
Secondary/
Historical
For most scenarios in Latin, the rules are such: A primary tense main verb is followed by a primary tense subjunctive subordinate verb
A secondary tense main verb is followed by a
secondarytense subjunctive subordinate verb.
Each of these sequences is futher sub-divided into two categories based on the time relationship between the main clause and the subordinate clause. The action of the subordinate clause can either be contemporary to the action of the main verb or occur prior to that action.
Observe the chart below:
| Sequence | Main verb | Subordinate Subjunctive |
---
| Primary | present future perfect future perfect | present (time same/time contemporary) perfect (time prior/time after) |
| Secondary/Historical | imperfect perfect pluperfect | imperfect (time same/time contemporary) pluperfect (time prior/time after) |
Note that the perfect tense may be considered a tense in primary sequence (a present perfect: I have done this) or a secondary tense (a simple past: I did this).
Resources
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Grammar
Paradigms
Present Indicative
Imperfect Indicative
Future Indicative
Perfect Indicative
Pluperfect Indicative
Future Perfect Indicative
Present Subjunctive
Imperfect Subjunctive
Perfect Subjunctive
Pluperfect Subjunctive
Present Infinitive
Perfect Infinitive
Future Infinitive
Irregular: Sum
Irregular: Fero
Irregular: Fio
Personal Endings
Verb Stems
Cases
History
English Case
Instrumental Case
Object Case
English "of"
Nouns to Adjectives
Latin Case
Nominative
Genitive
Dative
Dative of Agent
Accusative
The Small Island
Go Home
Ablative
Ablative Absolute
Ablative of Comparison
Tense
Imperfect
Imperfective Aspect
Perfect
Perfect Tense
Pluperfect Tense
Future Perfect
Mood
Conditions
Conditional Terms
Conditions in Latin
Should/Would Conditions
Table of Indicative Conditions in Latin
Traditional Conditions in Latin
Table of Traditional Conditions in Latin
Subjunctive
Independent Subjunctive
Dependent Subjunctives
Characterizing Clauses in English
Sequence of Tenses
Subjunctive Questions
Infinitive
Complementary Infinitive
Subject Accusative of the Infinitive
Participles
Comparison
Comparitive
Superlative
Comparison with Ablative or quam
Quam
Origin of Quam
Indirect Discourse
Infinitive in Indirect Discourse
Literary Latin
The Origins of the Indirect Statement
Subordinate Clauses in Indirect Discourse
Sequence of Tense
English into Latin
Converting a Latin statement
Indirect Questions
Catiline
Oratio in Catilnam Prima
I. 1-3
I.1-3 Notes
II.4-6
II.4-6 Notes
III. 6-8
III.6-8 Notes
IV. 8-10
IV.8-10 Notes
V. 10-13
V.10-13 Notes
VI. 13-16
VI.13-16 Notes
VII.16-18
VII.16-18 Notes
VIII.19-21
VIII.19-21 Notes
IX.22-24
IX.22-24 Notes
X.25-27
X.25-27 Notes
XI.27-29
XI.27-29 Notes
XII.29-30
XII.29-30 Notes
XIII.31-33
XIII.31-33 Notes
Prose Composition
Syllabus
Bibliography
Basic Concepts
Assignments
Darrow
Webster, Exordium
Webster, Narratio
Nixon
Periodicity
Pro Archia 3. 5
DIVINATIO IN QUINTUM CAECILIUM 1. 1
Travel
Guest Speaker Travel
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16822 | https://www.sciencedirect.com/topics/neuroscience/frameshift-mutation | Skip to Main content
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Frameshift Mutation
In subject area:Neuroscience
Frameshift mutation refers to the addition or deletion of nucleotides that alters the reading frame of the ribosome during translation, resulting in premature termination of the protein synthesis.
AI generated definition based on: Emery and Rimoin's Principles and Practice of Medical Genetics (Sixth Edition), 2013
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Chapters and Articles
You might find these chapters and articles relevant to this topic.
Chapter
Mutations and Repair
2013, Molecular Biology (Second Edition)David P. Clark, Nanette J. Pazdernik
2.6Frameshift Mutations Sometimes Produce Abnormal Proteins
Bases are read as codons, which are groups of three nucleotides, when translated into amino acids during protein synthesis (Ch. 13). The introduction or removal of one or two bases can have drastic effects since the alteration changes the reading frame of the afflicted gene. If a single base of a coding sequence is inserted or removed, the reading frame for all codons following the insertion or deletion will be changed (Fig. 23.09). The result will be a completely garbled protein sequence. Such frameshift mutations usually completely destroy the function of a protein, unless they occur extremely close to the far end. Insertion or deletion of two bases also changes the reading frame and alters protein function.
A mutation that alters the reading frame usually disrupts protein function completely.
However, insertion or deletion of three bases adds or removes a whole codon and the reading frame is retained. Apart from the single amino acid that is gained or lost, the rest of the protein is unchanged. If the deleted (or inserted) amino acid is in a relatively less vital region of the protein, a functional protein may be made. Adding or deleting more than three bases will give a similar result as long as the number is a multiple of three. In other words, a whole number of codons must be added or subtracted to avoid the consequences of changing the reading frame.
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Book2013, Molecular Biology (Second Edition)David P. Clark, Nanette J. Pazdernik
Chapter
Forward Mutations
2013, Brenner's Encyclopedia of Genetics (Second Edition)L.J. Reha-Krantz
Glossary
Base substitution mutation
: A mutation caused by the substitution of one base pair for another.
Frameshift mutation
: A mutation caused by deletion or insertion of one or a number of base pairs.
Mutagen
: An agent that damages or modifies DNA and can, depending on the ability of an organism to repair DNA, result in production of a mutation.
Mutation
: A change in the DNA sequence of an organism.
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Reference work2013, Brenner's Encyclopedia of Genetics (Second Edition)L.J. Reha-Krantz
Review article
Genome Editing Provides New Insights into Receptor-Controlled Signalling Pathways
2018, Trends in Pharmacological SciencesGraeme Milligan, Asuka Inoue
Effective targeting of a gene locus depends on functional domains/motifs of its encoded protein. For G protein α subunits, amino acid residues crucial for effective interactions with GPCRs are located within the extreme C-terminal region . Thus, a frameshift mutation at any position in the protein open-reading frame is expected to abolish G protein function owing to a lack of GPCR-induced activation. A frameshift mutation usually leads to the emergence of a premature stop codon. When this type of stop codon appears 5′ upstream of the last exon, the mRNA undergoes nonsense-mediated mRNA decay . This results in smaller amounts of a truncated protein than of the full-length native protein. Because NHEJ repair can introduce a random, in-frame mutation, it is preferable to design a sgRNA sequence to target a site encoding a residue crucial for protein function. For example, nucleotide-interacting residues and flexible loops (finger loop, middle loop, and lariat loop) are crucial for the function of G protein α subunits (GDP/GTP binding) and arrestins (GPCR binding) , respectively. For GPCRs, highly conserved sequence amino acid motifs (e.g., Asp-Arg-Tyr, DRY) at the cytoplasmic face of transmembrane domain 3 and (Asn-Pro-Xaa-Xaa-Tyr, NPxxY, in transmembrane domain 7) are key targets. An in-frame mutation near these residues is expected to ablate protein function. Another potentially effective target site is a splicing acceptor/donor sequence. Because RNA splicing occurs at conserved sequences (GG-GU and AG-G, for the donor and acceptor, respectively, where the hyphen denotes a cleavage site), disruption of these nucleotides can result in skipping of proper RNA splicing and loss of mature protein.
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Journal2018, Trends in Pharmacological SciencesGraeme Milligan, Asuka Inoue
Chapter
The Evolution of The Human Brain: Apes and Other Ancestors
2017, Evolution of Nervous Systems (Second Edition)E.J. Vallender
4.07.2.2.2.1Gene Loss
Although gene loss can occur through many mechanisms, including large-scale deletions, more often it is the result of nonsense mutations or frameshifts. The former causes a premature stop codon and is the result of a standard mutation from one nucleotide to another. Frameshift mutations are the result of insertions or deletions, not a multiple of three and usually small, within the coding region that change how the codons are translated into amino acids. These changes create completely different proteins from that point on and usually introduce stop codons. It is perhaps worth mentioning here that although the above described manners in which protein-coding genes are lost, similar effects can be achieved for the loss of noncoding genes or regulatory regions. Although we may not know or understand the exact mechanisms as well at this point, that it can and does occur is not disputed.
It is easy to see how either of these situations that result in the loss of a gene would have a functional effect. If a specific amino acid sequence is under strong selective constraint, eliminating it altogether must engender particularly strong consequences. And in fact, while we do see variation in any extant population that includes gene loss, it is rare that this is adaptive. Even when genes are lost between species, it would seem that often ecological or other circumstances have previously conspired to alleviate the negative constraint, making the loss an effect rather than a cause. This hypothesis, for instance, dominates our understanding of the large-scale pseudogenization that occurs in the olfactory subgenome of primates (Young et al., 2002). This in part explains why some olfactory receptors still appear to have undergone pseudogenization only after the human–chimpanzee divergence; it is not because some new selective pressure has arose differently between the species, but rather that they are simply decaying more slowly than speciation is occurring.
With that in mind, it is worth noting that one particular gene loss has been proposed to play a significant role in human brain evolution. A myosin heavy-chain protein that is expressed exclusively in the masticatory muscles of the head in nonhuman primates, MYH16, has been lost due to a frameshift mutation in humans (Stedman et al., 2004). This mutation was proposed to reduce mechanical stress on the skull that allowed for the coincident expansion of the human brain and head. But while it is clear, both from fossil evidence and from genetic evidence, that there were important dietary transitions that occurred roughly concurrently with the growth of the human brain, the relative cause and effect relationships remain unclear (Perry et al., 2005). Which came first, dietary change that enabled the loss of large chewing muscles or dietary change that caused the brain and skull to expand? Regardless, it is important and illustrative to consider these aspects of interrelatedness as species evolve. While it is possible to identify what changes have occurred, it is often difficult to identify which are causative.
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Reference work2017, Evolution of Nervous Systems (Second Edition)E.J. Vallender
Mini review
A gene for Crohn's disease is given the nod
2001, Trends in Pharmacological SciencesLuke O'Neill
The discovery by both groups that up to 15% of sufferers have mutant NOD2 is therefore of great interest. To date, three different mutations have been reported: the first is a frameshift mutation, reported by both groups, that leads to a truncation in the tenth leucine-rich repeat in NOD2; the other two mutations, reported by Hugot et al., give rise to non-conservative amino acid substitutions in either a leucine-rich repeat or an adjacent region. A particular strength of the work of Ogura and colleagues is that they tested the function of protein resulting from the frameshift mutation. This is often lacking in genetic association studies and is essential if the meaning of a mutation is to be understood. Somewhat surprisingly, the truncated NOD2 was unable to respond to lipopolysaccharide. One might predict a resultant constitutive or enhanced responsiveness, which would have been pro-inflammatory. However, the authors speculate that the deficit in sensing lipopolysaccharide might give rise to an exaggerated inflammatory response to bacteria by the adaptive immune system. A second possibility that Ogura et al. discuss is that the role of NOD2 might be to induce anti-inflammatory IL-10. Furthermore, the mutant NOD2 might become sensitive to other unknown pathogens.
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Journal2001, Trends in Pharmacological SciencesLuke O'Neill
Chapter
A contemporary view on the molecular basis of neurodevelopmental disorders
2020, Neuroprotection in Autism, Schizophrenia and Alzheimer's DiseaseIlse M. van der Werf, ... R. Frank Kooy
Mutational effects
In general, frameshift and nonsense mutations are considered as most severely hampering protein function as the part of the protein is not produced. To prevent these types of partial proteins to accumulate in the cell, the nonsense-mediated decay (NMD) process is started in case a stop codon is detected before the last exon-exon boundary in the processed mRNA fragment.85 NMD causes degradation of the erroneous mRNA fragment, thereby preventing the production of affected proteins, resulting in a loss-of-function mutation. At the protein level, this may be compared with the consequences of a microdeletion, which causes the heterozygous loss of the gene(s) encompassed by the deletion interval. Next to a reduced expression, loss-of-function mutations can occur when, for example, an active site of a protein is affected by the mutation or when a mutation is expected to cause a conformational change that generates a nonfunctional protein. Nonsense and frameshift variants that are located in the last exon of a gene can escape from NMD and still be translated. The resulting incorrect proteins may disturb cellular processes, for example, by being incorporated in a protein complex but not being active or by executing an alternative function due to the altered sequence. When a mutation induces a constitutively active protein or induces an abnormal function, it is designated as a gain-of-function mutation. Dominant negative mutations often result in an inactive protein that interferes with the function of the wildtype protein, such as an inactive channel subunit protein that incorporates in a channel complex and thereby inactivates the complete channel.86, 87
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Book2020, Neuroprotection in Autism, Schizophrenia and Alzheimer's DiseaseIlse M. van der Werf, ... R. Frank Kooy
Review article
The retinal pigmentation pathway in human albinism:
2022, Progress in Retinal and Eye ResearchReinier Bakker, ... Arthur A. Bergen
4.2.7.2Genetic variation and mutations
Only a few mutations in LRMDA causing albinism have been found. A study in 23 unrelated Iranian albinism patients reported a p.N89K missense mutation (Fig. 5) (Khordadpoor-Deilamani et al., 2016). The functional consequence of these mutations remains to be elucidated.
Next to the aforementioned missense mutation, both LRMDA stop and frameshift mutations have also been implicated in OCA7. The stop mutation is located very close to the C-terminus at residue 194 of 198 (p.R194) (Grønskov et al., 2013). The frameshift mutation is located at residue 23 of the protein (p.A23Rfs39) likely resulting in loss of functional protein (Grønskov et al., 2013).
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Journal2022, Progress in Retinal and Eye ResearchReinier Bakker, ... Arthur A. Bergen
Review article
Causative and susceptibility genes for Alzheimer’s disease: a review
2003, Brain Research BulletinA. Rocchi, ... L. Murri
In PSEN1 gene most pathogenic mutations are missense mutations, two are esanucleotide insertions and one is a trinucleotide deletion . These mutations are predominantly located in the highly conserved transmembrane domains. None of them is a frameshift mutation nor it causes a protein truncation, suggesting that structural alterations in the presenilins are not compatible with life. Only two splicing defects have been so far identified: a splice-site mutation in intron 8 of PS1, resulting in the in-frame deletion of exon 9 plus an amino acid substitution at codon 290 , and the splice donor site mutation in intron 4 of PS1, resulting in three different transcripts: one with the insertion of a threonine between codons 113 and 114, and two with partial or complete deletion of exon 4 [58,214,254]. The latter two mutations result in a frame-shift and premature stop codon, but probably the presence of the first type of transcript is sufficient for life.
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Journal2003, Brain Research BulletinA. Rocchi, ... L. Murri
Chapter
Duchenne Muscular Dystrophy
2013, Brenner's Encyclopedia of Genetics (Second Edition)J. Wicki, ... J.S. Chamberlain
See also
Female Carriers; Frameshift Mutation; Gene Expression; Gene Therapy, Human; Genetic Counseling; Genetic Diseases; Introns and Exons; Muscular Dystrophies; Mutation; Mutation, Nonsense; Mutation, Spontaneous; Recessive Inheritance and Recessiveness; Splicing.
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Reference work2013, Brenner's Encyclopedia of Genetics (Second Edition)J. Wicki, ... J.S. Chamberlain
Review article
SNARE Proteins: A Long Journey of Science in Brain Health and Disease
2019, NeuroscienceAnna Kádková, ... Jakob B. Sørensen
Involvement in disease
A homozygous frameshift mutation at position 220 (from ATG, the mutation is 220delG) in SNAP-29 was identified in two consangineous families, leading to a rare neurocutaneous syndrome, CEDNIK (for Cerebral Dysgenesis, Neuropathy, Ichtyosis, and palmoplantar Keratoderma) (Sprecher et al., 2005). The mutation led to a premature stop for translation, and a strong reduction in SNAP-29 transcript. The skin of patients was characterized by abnormal maturation of lamellar granules, and instead numerous small clear vesicles were present. By 8–15 months, psychomotor retardation was evident and brain MRI showed corpus callosum abnormality and cortical dysplasia (Sprecher et al., 2005). Two more siblings with another homozygous mutation (486insA) have been identified (Fuchs-Telem et al., 2011). The 22q11.2 deletion syndrome (di George syndrome) is caused by the heterozygous deletion of a part of chromosome 22 and is the most frequent microdeletion found in humans. It results in congenital heart problems, specific facial features, developmental delay and learning problems and it is associated with a higher risk of psychiatric disease, including schizophrenia and bipolar disorder (Saito et al., 2001). SNAP-29 is present on 22q11 and is frequently deleted in this syndrome. The resulting haploinsufficiency can uncover recessive mutations on the other allele, leading in some cases to CEDNIK syndrome (McDonald-McGinn et al., 2013). A polymorphism in the SNAP-29 promotor region is associated with schizophrenia within the 22q12 deletion population (Saito et al., 2001). Finally, in the general population copy number variations within SNAP-29 ranked 6th among all copy number variations associated with schizophrenia (Luo et al., 2014).
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Journal2019, NeuroscienceAnna Kádková, ... Jakob B. Sørensen
Related terms:
CRISPR
C-Terminus
Exon
Stop Codon
Amino Acid
Pervasive Developmental Disorder
Migraine
Missense Mutation
Nonsense Mutation
Codon
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1985 AIME Problems/Problem 11
Contents
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1 Problem
2 Solution 1
3 Solution 1 Simplified
4 Solution 2 (Calculus)
5 Video Solution
6 See also
Problem
An ellipse has foci at and in the -plane and is tangent to the -axis. What is the length of its major axis?
Solution 1
An ellipse is defined to be the locus of points such that the sum of the distances between and the two foci is constant. Let , and be the point of tangency of the ellipse with the -axis. Then must be the point on the axis such that the sum is minimal. (The last claim begs justification: Let be the reflection of across the -axis. Let be where the line through and intersects the ellipse. We will show that . Note that since is on the -axis. Also, since the entire ellipse is on or above the -axis and the line through and is perpendicular to the -axis, we must have with equality if and only if is on the -axis. Now, we have But the right most sum is the straight-line distance from to and the left is the distance of some path from to ., so this is only possible if we have equality and thus ). Finding the optimal location for is a classic problem: for any path from to and then back to , we can reflect (as above) the second leg of this path (from to ) across the -axis. Then our path connects to the reflection of via some point on the -axis, and this path will have shortest length exactly when our original path has shortest length. This occurs exactly when we have a straight-line path, and by the above argument, this path passes through the point of tangency of the ellipse with the axis.
The sum of the two distances and is therefore equal to the length of the segment , which by the distance formula is just .
Finally, let and be the two endpoints of the major axis of the ellipse. Then by symmetry so (because is on the ellipse), so the answer is .
Solution 1 Simplified
This assumes you know properties of ellipses. Like above, X is the point on the x-axis minimizing . By reflection, this equals ...But this just equals the length of the major axis. Solving like above, we get
Solution 2 (Calculus)
An ellipse is defined as the set of points where the sum of the distances from the foci to the point is fixed. The length of major axis is equal to the sum of these distances . Thus if we find the sum of the distances, we get the answer. Let k be this fixed sum; then we get, by the distance formula:
This is the equation of the ellipse expressed in terms of . The line tangent to the ellipse at the given point will thus have slope . Taking the derivative gives us the slope of this line. To simplify, let and . Then we get:
Next, we multiply by the conjugate to remove square roots. We next move the resulting form expression into form .
We know and . Simplifying yields:
To further simplify, let and . This means . Solving yields that . Substituting back and yields:
.
Solving for yields . Substituting back into our original distance formula, solving for yields .
Video Solution
1985 AIME #11
MathProblemSolvingSkills.com
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1985 AIME (Problems • Answer Key • Resources)
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Problem 10Followed by
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Inside out
There are 27 small cubes in a 3 x 3 x 3 cube, 54 faces being visible at any one time. Is it possible to reorganise these cubes so that by dipping the large cube into a pot of paint three times you can colour every face of all of the smaller cubes?
Age
14 to 16
Challenge level
Exploring and noticingWorking systematicallyConjecturing and generalisingVisualising and representingReasoning, convincing and proving
Being curiousBeing resourcefulBeing resilientBeing collaborative
Problem Student Solutions Teachers' Resources
Problem
Image
Imagine a (2x2x2) cube made from 8 smaller red cubes. Each small cube has 6 faces, making a total of 6 x 8 = 48 small faces.
On each side of the large cube you can see faces of 4 of the smaller cubes. By turning the large cube around you can see 24 small faces all together.
Image
You dip the cube into a pot of yellow paint and wait for it to dry. Now you separate all the smaller cubes and you can see that 24 of the 48 small faces are yellow and 24 are still red.
This suggests that you can rearrange the cubes so that only red faces are on the outside and yellow on the inside. Then, by dipping the cube into a pot of green paint it seems reasonable to assume you will end up with 8 small cubes whose 48 faces are now all painted yellow or green (24 of each), none that are still red and none that have been painted twice. Is this true?
Image
How about a 3 x 3 cube? This will have 27 smaller cubes, making a total of 162 faces, 6 x 9 = 54 being visible at any one time. This suggests you might need three pots of paint to colour all the faces of the smaller cubes without repetition and without having any red faces left.
Is this possible and how would you rearrange the cubes each time? Describe the system you are using.
How about a 4 x 4 x 4 cube and an n x n x n cube?
N.B. This problem is a 3D version of "On the Edge " which can be found on the site.
Student Solutions
Andre's (Tudor Vianu School) started of by giving his thoughts on the problem. Many thanks for this Andre - I think it is an excellent attempt to unpick what is happening.
For a 2 x 2 Cube
First, I analysed a 2x2x2 cube. After dipping it into the pot of yellow paint, each cube has three faces painted yellow and three faces painted red (the original colour). So, the other three faces of the cube remain red. Reversing each cube (so that each exterior vertex goes into the centre of the big cube) all the red faces remain at the exterior. After dipping the cube into the green pot, the rest of the face that remained red are now coloured green. So, it is sufficient to use two colours.
For a 3 x 3 Cube
Now, to analyse further, for increasing numbers of small cubes, I will create a table, with columns indicating: the side of the big cube, the total numbers of small cubes, the total number of faces, the number of faces at the exterior, and the number of colours that should be used (in principle, obtained as a division of the total number of faces by the number of exterior faces):
Side of cube Small cubes Faces (total)Faces (exterior)Colours
2 8 6 x 8 = 48 6 x 4 = 24 2
3 27 6x 27 = 162 6 x 9 = 54 3
4 64 6 x 64 = 384 6 x 16 = 96 4
n n x n x n 6 x n x n x n 6 x n x n n
Now, I analysed the situation with 3x3x3 cubes. I created 27 cubes from paper in order to analyse the situation experimentally. I see that in order to make possible to colour all the faces using 3 colours, each face must be coloured only once. After trying several times I observed that this is impossible, because the number of cubes with different numbers of coloured faces is very different:
After the first dipping:
there are 8 cubes with three coloured faces (vertices)
there are 12 cubes with 2 coloured faces (centres of edges)
there are 6 cubes with 1 coloured 1 face (centres of faces)
there is 1 cube with no coloured faces (centre of big cube)
I used in my solutions the following considerations:-
the cube from the centre must go into one vertex, to maximise its coloured faces in the following step
one cube from one vertex must go into the centre of the big cube
the other cubes from the vertices must go into the centres of the faces (6), and one in a centre of the edges
the cubes from the centres of the faces must go into the vertices
The third of these assumptions was incorrect. Someone else then completed the solution:
For the 3x3x3 cube, 2 of the corner slots and the middle slot can be used to swap 3 cubes around and colour then in the three goes. The rest of the 6 corner pieces are moved to 6 of the edge pieces, the 6 edges pieces are moved to the 6 centre slots and the 6 centre pieces are moved to the corners. Doing this after the second dipping as well will cover all 18 of these pieces in the three goes. Finally, the remaining 6 edge pieces can be left where they are, and simply rotated each time so that after the three dippings they are all covered. Thus a 3x3x3 cube can be totally coated.
For an n x n x n cube (if n is 6 or greater, or is 4), can be coloured as follows:
We start of with 8 corner pieces, 12(n-2) edge pieces, 6(n-2)^2 centres, and (n-3)^3 middles
4n blocks are looped around spending 2 dips in the corners, and (n-2) dips in the middles. This leaves 12(n-2) edges, 6(n-2)^2 centres and n(n-2)(n-4) middles.
6n(n-2) gaps are kept for blocks spending 2 times at the edge, 2 times at the centre, and (n-4) times in the middle
If n=4 then all the blocks are covered. If n is greater than or equal to 6, then there are still some edge and middle pieces we haven't describe. Blocks loop around these n(n-2)(n-4) slots, spending 6 times at the centres, and (n-6) times at the edges.
The 'missing' case is n=5. Here instead of the second point we would use 52 of the centre slots and 13 of the middle slots to move around a total of 65 cubes in to cover them in the five dips. We would then use 2 of the edge slots, two of the centre slots, and one middle to colour 5 more blocks in the 5 dips. Finally we would use the 21 edge slots and 14 middle slots to cover a total of 35 blocks using a total of 5 dips (so each block spends two 3 times at an edge and 2 times in the middle).
So an nnn block can be covered in n dips of paint as expected.
Teachers' Resources
This problem benefits from being able to handle real cubes using multilink or similar equipment.
Some sticky labels to identify which cubes have been painted and their locations may also prove beneficial.
Just trying to justify that in theory this should also be possible is a useful little proof.
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16825 | https://www.groundguitar.com/stevie-ray-vaughan-gear/ | Stevie Ray Vaughan
Born
: October 3, 1954
Died
: August 27, 1990 (aged 35)
Years Active
: 1971–1990
Genre(s)
: Blues Rock, Electric Blues, Texas Blues
Bands
: Stevie Ray Vaughan and Double Trouble
Main Guitar(s)
: Fender Stratocaster ("Number One")
Stevie Ray Vaughan's Gear Collection
Stevie Ray Vaughan was born in Dallas, Texas, in 1954. He started playing guitar young, copying older blues players like Albert King and Buddy Guy. By the late ’70s, he’d formed Double Trouble and was building a following in Austin. His first album, Texas Flood, came out in 1983 and put him on the map fast, huge tone, fast hands, but always controlled. He played a beat-up Strat through loud amps, no tricks, just raw blues with fire. Over the next few years, he put out more records, played with legends, and cleaned up from drugs and alcohol. In 1990, he died in a helicopter crash after a show in Wisconsin. He was 35.
Stevie Ray Vaughan's Electric Guitars
1965 Fender Stratocaster “Lenny”
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This guitar first caught Stevie’s attention around 1979, when he went with a couple of friends to a pawnshop in Austin, Texas. Unfortunately, it cost $350, and he couldn’t afford it at that time.
That however didn’t stop his wife, Lenora, who talked to a couple of Stevie’s friends to cash in $50 each, so they could give it to him as a birthday present. The plan worked out, and they presented the guitar to Stevie on October 3rd, 1980 at Steamboat Springs – a nightclub he often played at.
The guitar was originally a 3-tone sunburst maple-neck model with a rosewood fingerboard. It was refinished by the previous owner with a dark natural color, and it had an early 1900s mandolin-style pickguard inlayed behind the bridge, on top of the body.
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1962/63 Fender Stratocaster (Number One)
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Stevie acquired this guitar in 1974, at Ray Hennig’s Heart of Texas Music in Austin. The guitar was an old Stratocaster, characterized by the shop owner as one of the worst guitars that he ever had in his shop.
Nonetheless, it felt right for Stevie, and he asked Ray whether he could trade in his previous guitar for this one. After Ray agreed, they took apart the guitar, cleaned it up, and set it up according to Stevie’s preferences.
In 1974 he came by this particular day and was looking around as he’d usually do, going up and the rows of guitars taking the down, feeling them, when he discovered this old Stratocaster that I had hanging. He took it down, looked at this, felt of it, and fiddled around for a pretty good while.
I said – Stevie, that’s the worst guitar I guess I’ve ever traded for in the history of this business.
The guitar was an old Fender Stratocaster, a 59 model, that was traded to me maybe a week or two prior by Christopher Cross.
RAY HENNIG ON LEGENDARY GUITARIST Stevie Ray Vaughan
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Charley Custom Stratocaster
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This guitar was made in 1984 by Charley Wirz – the owner of Charley’s Guitar Shop in Dallas.
I saw it being built. Charley had made one for Jimmie, and he was like a proud father! They were both being built, and I think they were DiMarzio parts. Larry DiMarzio was the first guy to start making bodies and necks and pickups in that era. I think Charley told me they were mostly DiMarzio parts, but Van Zandt pickups. It was really like a Frankenstein or parts guitar.
Mark Pollock, Stevie Ray Vaughan Day By Day, Night After Night His Final Years, 1983-1990
Stevie’s Charley Stratocaster has a white Strat-style body with a rosewood neck, two controls (volume and tone), and three Danelectro lipstick pickups. On the back of the body is a hula girl sticker and the neckplate has the words “To Stevie Ray Vaughan, more in ’84” engraved on it.
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1961 Fender Stratocaster “Scotch”
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Stevie acquired this guitar sometime in 1985. According to the book Texas Flood: The Inside Story of Stevie Ray Vaughan, he bought it during an in-store appearance.
Most of the online sources claim that the guitar was originally intended to be a prize at one of Stevie’s shows. But, Stevie apparently liked it so much that he kept it for himself and instead offered another of his instruments as a prize.
If you happen to come across any information about this giveaway, or you happen to recall it happening, please be sure to post it in the comments.
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Tokai Springy Sound
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Stevie was photographed holding a maple-neck Tokai on the cover of his debut album Texas Flood, released in 1983. The cover was actually a drawing based on an existing photo, which is publically available online, and the drawing itself does not show the Tokai logo on the headstock. However, the original photo, as shown below, does.
What is presumably the same guitar, was also seen on a photo of Stevie performing live, taken likely sometime after the release of the album. If you happen to know the exact date and place, please leave it in the comments.
Aside from those two photos, and what can be concluded from the, not much is known about the guitar.
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1959 Fender Stratocaster “Yellow”
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This guitar was previously owned by Vince Martell, Vanilla Fudge’s lead guitarist, who sold it to Charley Wirz of Charley’s Guitar Shop in Dallas sometime in the early 80s.
Before we move further, please note that most of the info about this guitar comes from second-hand sources, as Charley and Stevie never really talked about it. So, it’s unclear where all this information that is available online originated from. So, we’d appreciate any help regarding finding sources and interviews where this guitar is mentioned either by those two, or someone close to them.
When Charley got the guitar, it had a hollowed-out body, because Vince apparently wanted to fit as many humbuckers in it as possible. Charley decided to ignore this approach, and instead make a new pickguard that would cover up the routed out holes, and instead placed just one single-coil pickup in the neck position. He also repainted the body in bright yellow paint.
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1962 Fender Stratocaster “Red”
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Stevie bought this guitar at Charley’s Guitar Shop sometime in 1983 or 1984 (sources vary on this). It was originally a sunburst Stratocaster, but it was repainted with a custom red finish, most likely by the previous owner (information about this guitar is very scarce).
The guitar remained in the stock condition up until around 1986 when Rene Martinez took the neck off to allegedly replace the neck on Stevie’s Number One. That neck couldn’t handle any more fret changes, so Rene suggested to Stevie that they should completely replace it.
Instead of the original neck, the red Stratocaster was fitted with a left-handed rosewood neck, which was apparently a knock-off (as suggested in the book Texas Flood: The Inside Story of Stevie Ray Vaughan). Around that same period, Stevie added the “SRV” sticker on the pickguard.
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Hamiltone Custom Stratocaster “Main”
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This guitar was made for Stevie by James Hamilton, a guitar builder from Buffalo, NY, who gave it to Stevie on April 29, 1984. But, the guitar was actually commissioned by Billy Gibbons, as a gift to Stevie.
“The Main” aka “the Couldn’t Stand the Weather guitar” features a neck-through-body design (unlike any other guitars Stevie played), and a two-piece maple body. The ebony fretboard is inlayed with abalone reading “Stevie Ray Vaughan”, designed by the artist Bill Narum.
Originally, the guitar was equipped with three EMG pickups, but ironically, during the recording of the Couldn’t Stand the Weather music video, the guitar itself couldn’t stand all the rain and water that it was exposed to, and the pickups were damaged. They were replaced with stock Fender single-coils.
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1963 Epiphone Riviera
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Stevie acquired this guitar in 1971, trading it for a 1951 Fender Broadcaster, known as “Jimbo”. He traded the guitar with Geoff Appold, who was a music teacher based in North Texas.
I left the “Jimbo” guitar for Stevie, and he took it to school and routed it out and put two P-90 pickups in it, or attempted to, which might explain why it had two volume knobs. He didn’t want me to see what he’d done, so he traded it for the Epiphone Riviera.
Jimmie Vaughan, Texas Flood: The Inside Story of Stevie Ray Vaughan
One interesting detail about Stevie’s Epiphone Riviera is the fact that it had a Gibson Varitone switch. This is something that was not fitted on Riviera models from the factory, so Stevie obviously did the mod himself.
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1951 Fender Broadcaster
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This was Stevie’s second electric guitar and one that he used in his early performances.
There are two different versions of the story of how Stevie came to acquire this guitar. Both agree that he inherited the guitar from his brother Jimmi, but they differ in some important details.
The first version of the story, tells that Jimmie left the Broadcaster in pieces at his parent’s house after he moved away. Stevie then found it, assembled it, and started using it.
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Tokai TST-50 (Rosewood, Black Pickguard)
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Stevie was photographed with this guitar on a promotional poster for Tokai guitars. The same guitar was auctioned in 2004, and labeled as “owned and used by Stevie Ray Vaughan on stage and in the studio”. Unfortunately, the auction house did not provide any specifics, at least not publically, so very little is actually known about this guitar. (STEVIE RAY VAUGHAN OWNED AND USED TOKAI GUITAR – Christie’s)
From what can be concluded from the photos, the guitar is a cherry sunburst model with a non-typical black colored pickguard – perhaps pointing towards the possibility that it was made specifically for SRV, and was based on his Number 1 Strat.
Recently, around 2016, Tokai released a model called TST-50 SRV, which was based on the guitar sold at auction in 2004. The model features a 2 piece alder body, maple neck with rosewood fretboard, and three Seymour Duncan STK-S7 pickups.
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Tokai AST-56
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Stevie was photographed playing this Tokai on March 24, 1985, in Worcester, Massachusetts. Apparently, it was one of the five Tokai’s that Stevie was provided after signing an endorsement deal with Tokai in 1985.
Very little is known about the guitar, aside from what was revealed in the 2020 auction (see Lot # 8I: Stevie Ray Vaughan’s Stage Used Tokai AST-56 Guitar). The only really interesting thing worth pointing out is the fact that the guitar didn’t have the “SRV” sticker when it was first photographed in 1985, but that changed by the time it was sold, in 1994.
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1957 Gibson ES-125T
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This was Stevie’s first electric guitar, which he inherited from his brother Jimmie. The guitar was a 3/4 size model with a single P90 pickup in the neck position.
JIMMIE VAUGHAN: About a year after I started, Daddy bought me a 3/4 Gibson with no cutaway, about as thick as a 335 with one pickup [a 1957 Gibson 125-T hollow body, serial number U142221], and a little brown Gibson amp.
(GARY WILEY, Vaughan cousin:) One time, Steve was listening intently to the radio and playing along on Jim’s guitar, which he had inherited [the Gibson 125-T]. My brother and I were bugging him to jump on our bikes, but he had to nail that part first.
Texas Flood: The Inside Story of Stevie Ray Vaughan
As far as when exactly Stevie used this guitar, most sources indicate that Jimmie started playing when he was 12, which was in 1963. If his father bought the guitar a year after he started playing, that would obviously be 1964, and probably at least another year passed before Stevie inherited it. So, 1965/66 seems to be the year in which Stevie acquired his first electric guitar – or when he was around 11/12 years old.
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Rickenbacker 360 Capri
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Stevie was seen playing this guitar in 1978 while gigging with Hubert Sumlin (see Stevie Ray Vaughan & Hubert Sumlin, Austin TX (1978)). There are also photos of Stevie with the Rickenbacker shown from the back with a sticker reading “Stingray”. This would indicate that he used the guitar somewhat extensively since he at least bothered to decorate it.
According to Stevie’s guitar tech Greg Sisk, during one of the gigs – which happened to be on Stevie’s birthday, Hubert Sumlin showed up with no guitar, at which point Stevie decided to give the Rickenbacker and an amp to Hubert (original source on this story needed).
The exact model of the guitar is at this point unknown. The most likely candidate would be the 360 Capri, which matches Stevie’s guitar in terms of finish, body binding, bridge style, and pickguard color. If you happen to be a vintage Rickenbacker expert, please do comment if you have any more knowledge. The photos of Stevie with the guitar can easily be found by simply searching “SRV Rickenbacker” on Google.
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Gibson Johnny Smith
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Stevie used this guitar to record the song Stang’s Swang, from his second album, Couldn’t Stand the Weather (source – Stevie Ray Vaughan – Day by Day, Night After Night: His Final Years). Aside from this, not much is known about the guitar.
Gibson Johhny Smith model was produced from 1981 to 1969. It featured a carved spruce top, carved maple back, ebony bridge and fretboard – with split block inlays. As far as pickups, the model came in two different configurations, either with a single mini-humbucker in the neck position or with an additional humbucker in the bridge. Stevie’s was the second version, with two humbuckers.
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Gibson ES-150 Charlie Christian
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According to the book Texas Flood: The Inside Story of Stevie Ray Vaughan, Stevie used this guitar to record the song Boot Hill, from the 1991 posthumous album The Sky Is Crying.
This is an electric-Spanish model from Gibson, made popular by Charlie Christian who bought his in 1936. It features a fully hollow body, with an arched solid spruce top and solid maple back and sides, and a mahogany neck with rosewood fretboard. It also has a single bar-style pickup in the neck position, which over the years became known as the “Charlie Christion pickup“.
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Danelectro Longhorn 4623
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Stevie was seen playing this guitar on a few occasions, including on September 27, 1985, at Hill Auditorium in Ann Arbor, Michigan, and at the Royal Oak Music Theatre in February 1986.
From the photos taken at Hill Auditorium, it is obvious that Stevie used the guitar only for a specific part of a song, at least on that occasion, because he had his number one Strat on his back at the same time while playing the Danelectro. The photos are available on Getty images, but we cannot show them here due to copyright restraints.
On other occasions, Stevie used the guitar to play the song “Tuff Enough”, as can be seen in a gig he played with The Fabulous Thunderbirds in 1986 (photo below).
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Stevie Ray Vaughan's Acoustic Guitars
Guild JF65-12
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Stevie used this guitar on MTV Unplugged filmed on January 30, 1990, to play Rude Mood and Pride and Joy.
This guitar wasn’t actually Stevie’s – he borrowed it from Timothy Duckworth, his friend and personal assistant. Allegedly, Timothy stated somewhere that Stevie’s hands were so strong that he accidentally cracked the neck.
The guitar is among Guild’s top-of-the-line models. It features a spruce top and sides, maple back, an abalone rosette, and ebony fretboard mother of pearl blocks with triangle-shaped abalone insets.
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1930s National Duolian
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Most famously, Stevie was seen holding this guitar on the cover of the In Step album. Stevie believed that the guitar previously belonged to Blind Boy Fuller, an American blues guitarist, and singer from the 1930s.
I’ve got a ’28 Dobro and I sometimes play some slide, but not very often. I go through phases where I feel comfortable about it. It’s funny, I’ll get into doing it again and get real confident with it, and then something happens…
Classic interview: Stevie Ray Vaughan, MusicRadar
Although Stevie refers to it as a ’28 Dobro, based on the serial number provided by the article seen here, the guitar is Duolian, which means that it couldn’ve been made earlier than 1930.
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Gibson L-1
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Stevie was seen using this guitar during an interview that he did in 1983 – watch Stevie Ray Vaughan, acoustic solo 1983, Dallas, Texas. The guitar was also seen sitting behind him during his performance at the MTV Unplugged in January 1990.
Based on the footage available, the guitar was a flat-top model, which means that it was made sometime between 1926 and 1937. It also doesn’t have a pickguard, nor any fingerboard binding, which would suggest that it was made after 1930. Lastly, it has a 12-fret neck (or the part of the neck extending past the body) which would mean that it was made no later than 1932.
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Roy Rogers Six-String Guitar
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Stevie’s first guitar ever was a Ray Rogers toy guitar, which he acquired around 1961.
I got my first guitar when I was seven. It was one of those Roy Rogers guitars; it had pictures of cowboys and cows on it, some rope. I had a blanket with the same sht on it, too.
Texas Flood: The Inside Story of Stevie Ray Vaughan
In case you’re unfamiliar with Roy Rogers, he was an American western singer and actor, commonly known as the “King of Cowboys”. During his active years, he had a lot of merchandise with his name on it, including shirts, lunch boxes, watches, and of course, guitars. Most commonly, these were toy guitars made by Jefferson Manufacturing Company of Philadelphia, but there were also some more “serious” models available.
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Stevie Ray Vaughan's Amps
Dumble Steel String Singer (Silverface)
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Stevie started using this amp sometime in early 1984, and it directly replaced the Dumble ODS that he used in late 1983.
Right now, I use a Dumble 150 watt. He calls it Steel String Singer, I call it King Tone Consoul, that’s s-o-u-l. It’s like an overgrown Fender tube amp. Some Dumbles, like the Overdrive Special, you’ve got to know what you’re doing with them, because they’ll get away from you and take you with ’em.
Stevie Ray Vaughan, Guitar Player magazine, 1984
Initially, he used a silverface Steel String Singer, occasionally paired with a couple of Fender Vibroverbs. In that scenario, SSS was presumably used for clean sound, while the Vibroverbs were used for overdrive.
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Fender Vibroverb Blackface
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Stevie used two of these amps from around early 80 to around 1985 initially for everything, but later on only for his dirty or distorted sound.
Stevie had a very simple setup, just two Vibroverb amps and no pedals. For the Hendrix stuff, Cutter would plug in a wah for the one tune.
Dan Opperman, Texas Flood: The Inside Story of Stevie Ray Vaughan
One of the two Vibroverbs was used to power a Fender Vibratone Leslie speaker.
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Marshall JMP 4140 Club and Country
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Stevie used this amp for his clear tone in the early days, circa the early 80s. The amp was a 1980 model 4140 Club and Country combo with two 12-inch speakers, 100 watts of output, and KT77 tubes. In some photos, the amp is seen with tape on the speaker grill, which was apparently done in an attempt to cut the higher frequencies.
This particular model was basically Marshall’s take on the Fender Twin Reverb, so it would sense that Stevie liked it for his clean sound. At this same time, he used two Fender Vibroverbs for his dirty or distorted sound.
The amp remained in Vaughan’s rig until the early to mid-80s when he started using a Dumble amps instead.
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Marshall Major (Model 1967) Amp
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Stevie used this amp on stage around 1987, paired with a Dumble Steel String Singer.
When I first started working with him, it was a Dumble Steel String Singer and a 200-watt Marshall head, and each one of those amps was playing through EV-loaded Dumble 4 x 12 cabinets – one was angled and one was flat.
Stevie Ray Vaughan’s guitar tech Rene Martinez
The goal behind this setup, according to Stevie, was to simplify things and get rid of all the various amps that he usually used.
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Fender Super Reverb
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Stevie usually used two Fender Super Reverbs on stage for his overdriven sound, from around the mid-80s to 1990. The amps were loaded with four ElecroVoice speakers,
According to Cesar Diaz, Stevie’s amp tech, he never really like the Super Reverbs, and thought they were not exactly for him. Also according to Cesar, the output transformers on the Super Reverbs were changed to those from a Twin Reverb.
Stevie used to have Super Reverbs, but they somehow never sounded quite right to him – too much power, you know? He also used to be real superstitious about the number 6. He’d always set the controls on his amps on 6 – the treble on 6, the bass on 6, and I’d back off the knobs with a screwdriver so that when it said 6 it was really on 10 (laughs). You have to do these things.
Cesar Diaz The Last Great Interview (SRV)
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Dumble Overdrive Special
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This is the first Dumble amp Stevie used live, starting from August 1983. The amp was an “Overdrive Special” model, played through a 4×12 Dumble speaker cabinet.
The amp however wasn’t used too long, and Stevie replaced it with a Dumble Steel String Singer by the beginning of 1984.
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Mother Dumble Amp
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According to Chris Layton, Stevie’s drummer, the amp that Stevie used to record the Texas Flood album was a Mother Dumble (300 watts). The amp belonged to Jackson Browne, who owned the studio where the album was recorded.
There was literally nothing between the guitar and the amp. It was just his Number One Strat plugged into a Dumble amp called Mother Dumble, which was owned by Jackson Browne. The real tone just came from Stevie, and that whole recording was so pure; the whole experience couldn’t have been more innocent or naïve.
Chris Layton, Texas Flood: The Inside Story of Stevie Ray Vaughan
As far as the exact model, it seems likely that this was a prototype of some sort, or perhaps a Dumbleland custom model. Therefore, it likely didn’t have a model name, and the “Mother Dumble” was just something that Jackson came up with (if you happen to know anything about this amp, leave it in the comments).
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Dumble Steel String Singer (Blackface)
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This is the second Dumble Steel String Singer that Stevie Ray Van used in his live rig, starting from around late 1986. Unfortunately, it’s unknown to which extent he used the amp, whether it was a backup, or whether it was used for a different purpose than his Silverface SSS (which was for clean sound).
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Fender Bassman Blackface
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This was one of Stevie’s early amps. According to Roddy Colonna, a drummer who played in a few of Stevie’s early bands, he used the amp while Roddy played with him.
If we look at the timeline posted over at the SRV Achieve website, the first band that had Roddy as the drummer, name Deryk Jones Party, was formed in the summer of 1971. So, based on this, Stevie used this amp sometime around that period.
Also according to Roddy, the amp had the word “Euphoria” stenciled on it, which led Stevie to assume that the amp was once owned by Eric Clapton.
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Stevie Ray Vaughan's Effects
Fender Vibratone Leslie Speaker
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Stevie used this effect for live performances of Cold Shot, Couldn’t Stand the Weather, and The Things (That) I Used to Do – usually powered with one of his Fender Vibroverb combo amps.
In case you’re unfamiliar with what the Fender Vibratone does – basically, it’s just a cabinet with a speaker inside of it that has the ability to rotate, which creates a “wobbly” sound. Originally, these speakers were intended to be used on Hammond organs, but were quickly adopted by musicians of all kinds, including of course guitarists.
Based on the photo and video material, he owned two different Vibratones, one earlier model, and one later one – with the large metal plate Fender Vibratone badge on the front.
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Vox Wah
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Stevie used a variety of different Vox wah pedals throughout his career. But, overall, he favored the original 60s models.
One of the wahs that he used, primarily in the studio, was a V846 model, that previously belonged to Jimi Hendrix. Stevie’s brother, Jimmie Vaughan, got this pedal from Hendrix after a gig in 1969, when Vaughan’s group opened for The Jimi Hendrix Experience in Fort Worth, Texas.
After the show, Jimi’s roadie asked Jimmie if he would swap his Vox wah pedal for Hendrix’s broken pedal plus some cash, a trade the young guitarist happily agreed to.
Texas Flood: The Inside Story of Stevie Ray Vaughan
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Roger Mayer Octavia
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This was Stevie’s most widely used Octavia pedal, which he “mained” until switching to a Tycobrae Octavia, sometime in 1990.
I was also the one to introduce Stevie to the Fuzzface. He didn’t know what one was until In Step. He had a Univibe, but he never used it, and I brought in a Tycobrae Octavia and the Fuzzfaces.
Cesar Diaz The Last Great Interview (SRV)
We used four pedals: a CryBaby wah-wah, an Ibanez Tube Screamer, an Octavia and a Dallas Arbitar Fuzz Face.
Interview: Stevie Ray Vaughan’s guitar tech Rene Martinez
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Dallas-Arbiter England Fuzz Face
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Stevie started using the English version of the Dallas Arbiter Fuzz Face sometime around the release of the In Step album, in 1989. Most notably, the effect can be heard during his performance of Voodoo Chile during his 1989 Austin City Limits performance.
Some sources also claim that he used it on the solo for the original release of Couldn’t Stand the Weather, but based on Cesar Diaz’s statements, that would be impossible.
I was also the one to introduce Stevie to the Fuzzface. He didn’t know what one was until In Step. […] I had two of the original ones marked NKT, which stands for New Market. They would tend to sound really good when they were nice and cool, but when they got hot, they’d go down. The fix for any old Fuzzface that has germaniums is to put it in the freezer. Take it out, play your song and replace it with another one that’s cold.
Cesar Diaz The Last Great Interview (SRV)
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Ibanez Tube Screamer
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The Ibanez Tube Screamer was one of the four key pedals used by Stevie. Allegedly, he used all three versions that were available in his lifetime, the original TS-808, the TS-9, and the “modern” TS-10. According to Don Opperman, he used the pedal as a boost, in conjecture with his wah – mostly just for the Hendrix-influenced songs.
For the Hendrix stuff, Cutter would plug in a wah for the one tune. One of the first things I did was to show Stevie this trick I had picked up from Joe Walsh, who used a Tube Screamer in conjunction with his wah as a boost to make it sound more expressive, switching the Tube Screamer on and off as he saw fit.
Dan Opperman – Texas Flood: The Inside Story of Stevie Ray Vaughan
Even though there are rumors that the TS-808 was his favorite, the TS-9 was the one that was seen most often on his pedalboard.
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Tycobrahe Octavia
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According to Stevie’s technician Cesar Diaz, Stevie started using a Tycobrae Octavia effect pedal sometime around the release of the In Step album in 1989.
I was also the one to introduce Stevie to the Fuzzface. He didn’t know what one was until In Step. He had a Univibe, but he never used it, and I brought in a Tycobrae Octavia and the Fuzzfaces.
Cesar Diaz The Last Great Interview (SRV)
Unfortunately, it’s unknown to which extent Stevie used this specific pedal in comparison to the Roger Mayer Octavia – which he also had. The latter essentially does the same job, and it was seen way more often on Stevie’s pedalboard.
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Shin-Ei Uni-Vibe
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This pedal was mentioned by Stevie’s technician, Cesar Diaz, who said that even though Stevie had one, he never used it. So, it’s likely that this was something Stevie bought when he was chasing the Hendrix tone, but he eventually decided that he preferred using the Fender Vibratone (which achieves a similar sound effect)
I was also the one to introduce Stevie to the Fuzzface. He didn’t know what one was until In Step. He had a Univibe, but he never used it, and I brought in a Tycobrae Octavia and the Fuzzfaces
Cesar Diaz The Last Great Interview (SRV)
More about this Effect
Stevie Ray Vaughan's Strings
GHS 1300 Guitar Strings
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Based on Stevie’s statements in interviews, in the early years, it sounds like he wasn’t really set on a particular set of strings, but used different gauges based on the condition of his fingers.
The gauges vary because it’s based on the shape my fingers are in. I go from an .011 to an .013 on the high E, which is the only one I lighten up on. As a rule, the others are .015, .019, .028, .038, .054 to .056 or even .058. The good thing about such heavy strings is that you can hit ’em hard and they don’t move—when you pop ’em, they stay there.
Texas Flood: The Inside Story of Stevie Ray Vaughan
Most likely, he used the GHS 1300 set, which measures .011, .015, .019, .028, .038, and .058, and occasionally swapped the high E string with a .013.
More about these Guitar Strings
Stevie Ray Vaughan's Accessories
Music Note Guitar Strap
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Stevie used these guitar straps for most of his iconic live performances, including Live at Montreux in 1982 and 1985, and at Rockpalast in 1984. Sometimes he used white straps with black notes, other times black straps with white notes, but they were essentially the exact same straps.
The original straps that Stevie used were made by Earth III, owned by Richard Oliveri who was based in Staten Island, New York. Shortly before Richard died in 1989, the company went out of business, and production stopped.
Just recently, Roy Jones, who worked at the original factory, restarted production. The result was the Earth III Music Note guitar strap, which is now available for pre-order from EarthThree.
More about this Guitar Accessory 2
Fender Medium Guitar Picks
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Most of the time, Stevie used Fender Medium 351 shape guitar picks. This was far from exclusive though, as he was seen using various different picks, with branding from different guitar shops, and made from different materials. But, they all did seem to be of medium thickness.
One important thing to point out is that he usually played with the back, or “fat” end of the pick.
More about this Guitar Accessory |
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Annotating is any action that deliberately interacts with a text to enhance the reader's understanding of, recall of, and reaction to the text. Sometimes called "close reading," annotating usually involves highlighting or underlining key pieces of text and making notes in the margins of the text. This page will introduce you to several effective strategies for annotating a text that will help you get the most out of your reading.
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By annotating a text, you will ensure that you understand what is happening in a text after you've read it. As you annotate, you should note the author's main points, shifts in the message or perspective of the text, key areas of focus, and your own thoughts as you read. However, annotating isn't just for people who feel challenged when reading academic texts. Even if you regularly understand and remember what you read, annotating will help you summarize a text, highlight important pieces of information, and ultimately prepare yourself for discussion and writing prompts that your instructor may give you. Annotating means you are doing the hard work while you read, allowing you to reference your previous work and have a clear jumping-off point for future work.
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Survey: This is your first time through the reading
•Look through the article/chapter/book.
•Ask if the article is a useful and trustworthy source. (Who wrote it? Who published it? Who is the audience?)
•Note the title--what does it tell you about the article’s topic/argument?
•Is there an Abstract (paragraph that summarizes topic, questions, research methods, findings)?
•Subheadings--what do they tell you?
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Skim: This is your second time through the reading
•Read the first few sentences of the first few paragraphs
•Identify the main thesis.
•Underline the thesis (the main argument or viewpoint, one or two sentences) and write it in your own words in the margin.
•Continue reading the first sentence or two of the body paragraphs.
•Highlight the point of each paragraph and summarize it in the margin in your own words.
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•Now that you have a good idea of the article’s thesis, read through the entire article and look for more details. Highlight supporting evidence.
•Write any questions you have in the margins.
•Circle any words you don’t recognize, look them up in a dictionary, and write their meanings in the margins.
Annotating Strategies
You can annotate by hand or by using document software. You can also annotate on post-its if you have a text you do not want to mark up. As you annotate, use these strategies to make the most of your efforts:
Include a key or legend on your paper that indicates what each marking is for, and use a different marking for each type of information. Example: Underline for key points, highlight for vocabulary, and circle for transition points.
If you use highlighters, consider using different colors for different types of reactions to the text. Example: Yellow for definitions, orange for questions, and blue for disagreement/confusion.
Dedicate different tasks to each margin: Use one margin to make an outline of the text (thesis statement, description, definition #1, counter argument, etc.) and summarize main ideas, and use the other margin to note your thoughts, questions, and reactions to the text.
Lastly, as you annotate, make sure you are including descriptions of the text as well as your own reactions to the text. This will allow you to skim your notations at a later date to locate key information and quotations, and to recall your thought processes more easily and quickly.
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16827 | https://www.maths.dur.ac.uk/users/alexander.stasinski/Student%20projects/2024-2025/ProjIII_Euclidean%20domains.html | Project III (MATH3382)
Project III (MATH3382) 2024-2025
Euclidean domains
Alexander Stasinski
Description
Euclidean domains are integral domains where we have a notion of division with remainder, and hence the Euclidean algorithm. As such, these rings play a fundamental role in number theory. In particular, it turns out that they all have unique factorisation into irreducible elements. The precise definition is:
Definition. Let R R be an integral domain. A Euclidean function (or norm) on R R is a function ϕ:R∖{0}→N∪{0}ϕ:R∖{0}→N∪{0}
such that
∀x,y∈R∖{0}∀x,y∈R∖{0}, we have ϕ(x)≤ϕ(x y)ϕ(x)≤ϕ(x y),
∀x∈R∀x∈R, y∈R∖{0}y∈R∖{0}, ∃q,r∈R∃q,r∈R such that x=q y+r x=q y+r with either r=0 r=0 or ϕ(r)<ϕ(y)ϕ(r)<ϕ(y).
If R R has a Euclidean function, it is called a Euclidean domain (ED).
Examples of EDs include the ring of integers Z Z with ϕ ϕ given by a↦|a|a↦|a|. (This is because we have division with remainder.) and the ring of polynomials F[x]F[x] for any field F F with ϕ ϕ given by f(x)↦deg f f(x)↦degf (division with remainder for polynomials over a field). It is also true that the Gaussian integers Z[i]Z[i], as well as the ring Z[√−2]Z[−2] are Euclidean domains.
There are also more exotic Euclidean domains and a host of variations and generalisations that we will study. For example, one can replace N∪{0}N∪{0} by any well-ordered set (where every non-empty subset has a least element).
It is not hard to prove that Euclidean domains are principal ideal domains (PIDs), and moreover, a more deeper theorem says that every PID is a unique factorisation domain (UFD).
One of the main questions is to determine when the ring of integers in a number field is a Euclidean domain and, more specifically, whether it is Euclidean with respect to the absolute value of the norm coming from the number field (it is then called norm-Euclidean). We will define these notions as we go along. A ring of integers need to be a PID, but if it is, it makes sense to ask whether it is Euclidean. This is not always the case. We will see that the ring Z[(1+√−19)/2]Z[(1+−19)/2] is not Euclidean, yet it is known to be a PID. On the other hand, we expect that as soon as the ring of integers has infinitely many units, then it is a PID iff it is Euclidean. This is because of the following theorem.
Theorem (Weinberger, Hooley).Assume the Generalised Riemann Hypothesis. If a ring of integers of a number field has infinitely many units and is a PID, then it is Euclidean.
It was only in 1994 that it was shown that a ring of integers can be Euclidean but not norm Euclidean: Clark showed that this holds for Z[(1+√69)/2]Z[(1+69)/2]. Even more recently, in 2004, the second example was given Z[√14]Z.
It is known that there are only finitely many norm-Euclidean quadratic rings (essentially rings given by adjoining a square root, as above). In contrast, it is believed that there are infinitely many Euclidean quadratic rings, but that remains an open problem.
Further directions for individual study include various generalisations of Euclidean domains (see Lemmermeyer article), studying the constructions of exotic Euclidean norms, like the one on Z[√14]Z, as well as Lenstra's notion of Euclidean ideal classes (certain elements that sometimes exist in the ideal class group).
Prerequisites
Algebra II
Co-requisites
Number Theory III
Resources
K. Conrad, Remarks about Euclidean domains, link.
H. W Lenstra, Jr, Lectures on Euclidean rings, link.
F. Lemmermeyer, The Euclidean algorithm in algebraic number fields, link.
email: Alexander Stasinski
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16828 | https://artofproblemsolving.com/wiki/index.php/Jensen%27s_Inequality?srsltid=AfmBOoqiasqATTKKPX-IpjYV65ysBew6LshWKC1k6MUcHhWXrA3sI7Ma | Art of Problem Solving
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Jensen's Inequality
Jensen's Inequality is an inequality discovered by Danish mathematician Johan Jensen in 1906.
Contents
[hide]
1 Inequality
2 Proof
3 Example
4 Problems
4.1 Introductory
4.1.1 Problem 1
4.1.2 Problem 2
4.2 Intermediate
4.3 Olympiad
Inequality
Let be a convex function of one real variable. Let and let satisfy . Then
If is a concave function, we have:
Proof
We only prove the case where is concave. The proof for the other case is similar.
Let . As is concave, its derivative is monotonically decreasing. We consider two cases.
If , then If , then By the fundamental theorem of calculus, we have Evaluating the integrals, each of the last two inequalities implies the same result: so this is true for all . Then we have which is exactly what we want! Hooray!😀😀😀
Example
One of the simplest examples of Jensen's inequality is the quadratic mean - arithmetic mean inequality. Taking , which is convex (because and ), and , we obtain
Similarly, arithmetic mean-geometric mean inequality (AM-GM) can be obtained from Jensen's inequality by considering .
In fact, the power mean inequality, a generalization of AM-GM, follows from Jensen's inequality.
Problems
Introductory
Problem 1
Prove AM-GM using Jensen's Inequality
Problem 2
Prove the weighted AM-GM inequality. (It states that when )
Intermediate
Prove that for any , we have .
Show that in any triangle we have
Olympiad
Let be positive real numbers. Prove that
(Source)
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16829 | https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/0d4d90053f2bb24f7551c78e2d1b21a2_MIT18_03SCF11_s30_5text.pdf | Green’s Formula, Laplace Transform of Convolution 1. Green’s Formula in Time and Frequency When we studied convolution we learned Green’s formula. This says, the IVP p(D)x = f (t), with rest IC (1) has solution x(t) = (w ∗ f )(t), where w(t) is the weight function. (2) (Remember, the weight function is the same as the unit impulse response.) The Laplace transform changes these equations to ones in the frequency variable s. p(s)X(s) = F(s) (3) 1 X(s) = F(s) = W(s)F(s), (4) p(s) where W(s) is the transfer function. Equation (2) is Green’s formula in time and (4) is Green’s formula in fre quency. In words, viewed from the t side, the solution to (1) is the convo lution of the weight function and the input. Viewed from the s side, the solution is the product of the transfer function and the input. 2. Convolution Comparing equations (2) and (4) we see that L(w ∗ f ) = W(s) F(s). (5) · It appears that Laplace transforms convolution into multiplication. Tech nically, equation (5) only applies when one of the functions is the weight function, but the formula holds in general. Theorem: For any two functions f (t) and g(t) with Laplace transforms F(s) and G(s) we have L( f ∗ g) = F(s) G(s). (6) · Remarks: 1. This theorem gives us another way to prove convolution is commutative. It is just the commutivity of regular multiplication on the s-side. L( f ∗ g) = F G = G F = L(g ∗ f ).
· · Green’s Formula, Laplace Transform of Convolution OCW 18.03SC 2. In fact, the theorem helps solidify our claim that convolution is a type of multiplication, because viewed from the frequency side it is multiplication. Proof: The proof is a nice exercise in switching the order of integration. We won’t use 0− and t+ in the integrals, since they would just clutter the expo sition. It is an amusing exercise to put them in and see that they transform correctly as we manipulate the integrals. We start by writing L( f ∗ g) as the convolution integral followed by the Laplace integral. Z ∞ L( f ∗ g) = ( f ∗ g)(t)e−st dt 0 Z ∞ Z t = f (t − u)g(u)e−st du dt. 0 0 Next, we change the order of integration (see the figure below). Z ∞ Z ∞ = f (t − u)g(u)e−st dt du. 0 u Finally, change variables in the inner integral: substitute v = t − u, dv = dt, (u a constant) Z ∞ Z ∞ = f (v)g(u)e−s(v+u) dv du 0 0 Z ∞ Z ∞ = f (v)e−sv dv g(u)e−su du 0 0 = F(s)G(s). u / t O t = u u / t O t = u Fig. 1. Changing the order from du dt to dt du. 3. Integration Rule If differentiation on the time side leads to multiplication by s on the frequency side then we should expect integration in time to lead to division by s. If f (t) is a function with Laplace transform F(s) then the integration 2 Green’s Formula, Laplace Transform of Convolution OCW 18.03SC rule states: Z t+ F(s) L 0− f (τ) dτ = s . Proof: One way to prove this is using the t-derivative rule. Let’s be clever and use convolution instead. The integral is exactly f (t) ∗ 1. Thus, Z t+ L 0− f (τ) dτ = L( f ∗ 1) = F(s) ∗L(1) = F( s s) . This is what we needed to show. 3 MIT OpenCourseWare 18.03SC Differential Equations Fall 2011 For information about citing these materials or our Terms of Use, visit: |
16830 | https://math.stackexchange.com/questions/3703064/im-stuck-trying-to-factor-x2-4-to-x-2x2 | factoring - I'm stuck trying to factor $x^2-4$ to $(x-2)(x+2)$ - Mathematics Stack Exchange
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I'm stuck trying to factor x 2−4 x 2−4 to (x−2)(x+2)(x−2)(x+2)
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I am trying to understand each step in order to get from x 2−4 x 2−4 to (x−2)(x+2)(x−2)(x+2)
I start from here and got this far...
x 2−4=x 2−4=
x∗x−4=x∗x−4=
x∗x+x−x−4=x∗x+x−x−4=
x∗x+x−2+2−x−4=x∗x+x−2+2−x−4=
x∗x+x−2+2−(x+4)=x∗x+x−2+2−(x+4)=
After this I try x(x−2)+2−(x+4)=x(x−2)+2−(x+4)=
and this clearly does not even equal the other factorings. I thought the x x could be factored out. I'm confused.
I know I can just insert a 2−b 2 a 2−b 2 into the difference of squares formula like so (a−b)(a+b)(a−b)(a+b) but I am practicing factoring. I'm just curious to see each and every step of the factoring.
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edited Jun 2, 2020 at 22:56
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2 Expand (x+2)(x−2)(x+2)(x−2) and do it in the other sens.hamam_Abdallah –hamam_Abdallah 2020-06-02 22:37:57 +00:00 Commented Jun 2, 2020 at 22:37
Add and subtract 2 x 2 x from x 2−4 x 2−4 and then carry on.user743391 –user743391 2020-06-02 22:39:04 +00:00 Commented Jun 2, 2020 at 22:39
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x 2−4=x 2−2 x+2 x−4=x(x−2)+2(x−2)=(x+2)(x−2)x 2−4=x 2−2 x+2 x−4=x(x−2)+2(x−2)=(x+2)(x−2)
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edited Jun 2, 2020 at 23:22
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Alternatively, recall the general formula
a 2−b 2=(a+b)(a−b).a 2−b 2=(a+b)(a−b).
This can be seen by expanding the RHS, or by writing
a 2−b 2=a 2−a b+a b−b 2=a(a−b)+b(a−b)=(a+b)(a−b).a 2−b 2=a 2−a b+a b−b 2=a(a−b)+b(a−b)=(a+b)(a−b).
Then, with a=x a=x and b=2 b=2, we have
x 2−4=x 2−2 2=(x+2)(x−2).x 2−4=x 2−2 2=(x+2)(x−2).
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answered Jun 2, 2020 at 22:44
YiFan TeyYiFan Tey
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I see, but I'm kind of confused why it's different when b=1 b=1. In that case you don't have to do −a b+a b−a b+a b you can do −a−b+a+b−a−b+a+b. As in a 2−a−b+a+b−b 2 a 2−a−b+a+b−b 2 when b=1 b=1.Ant –Ant 2020-06-02 22:47:16 +00:00 Commented Jun 2, 2020 at 22:47
@Renoldus I don't see how that's the case. I would factor x 2−1 x 2−1 via x 2−1=x 2−x+x−1=x(x−1)+(x−1)=(x+1)(x−1),x 2−1=x 2−x+x−1=x(x−1)+(x−1)=(x+1)(x−1), which you can check is the same as what I did above with general a,b a,b. But of course, there can be many different ways to factor the same thing.YiFan Tey –YiFan Tey 2020-06-02 22:48:35 +00:00 Commented Jun 2, 2020 at 22:48
I know it works your way with b=1 b=1 but this way only works when b=1 b=1 a 2−a−b+a+b−b 2 a 2−a−b+a+b−b 2 I think.Ant –Ant 2020-06-02 22:50:20 +00:00 Commented Jun 2, 2020 at 22:50
@Renoldus I'm not seeing how that way leads to a factorisation. (Perhaps it does, but I'm not seeing it.) Anyway, no need to obssess over slight differences in the specific method, so long as it works.YiFan Tey –YiFan Tey 2020-06-02 22:53:39 +00:00 Commented Jun 2, 2020 at 22:53
Just my observation. Because I spent time today doing this with b=1 and thought I had the factoring down because of it. It was the source of my blockage.Ant –Ant 2020-06-02 23:01:09 +00:00 Commented Jun 2, 2020 at 23:01
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x 2−4=x 2+2 x−2 x−4 x 2−4=x 2+2 x−2 x−4
=(x 2+2 x)−(2 x+4)=(x 2+2 x)−(2 x+4)
=x(x+2)−2(x+2)=x(x+2)−2(x+2)
(x+2)(x−2).(x+2)(x−2).
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answered Jun 2, 2020 at 22:41
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If x 2−4 x 2−4 can be factored then there exist a,b∈R a,b∈R such that
x 2−4=(x−a)(x−b)x 2−4=(x−a)(x−b)
Now multiplying out the rhs and collecting terms,
(x−a)(x−b)=x 2−(a+b)x+a b(x−a)(x−b)=x 2−(a+b)x+a b
So we need to solve the system of equations
a+b=0(1)(1)a+b=0 a b=−4(2)(2)a b=−4
From equation (1)(1) we have
a=−b a=−b
and plugging that into equation (2)(2),
−b 2=−4−b 2=−4
We have two solutions for b b.
If b=2 b=2 then a=−2 a=−2 and so
x 2−4=(x+2)(x−2)(ANS)(ANS)x 2−4=(x+2)(x−2)
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edited Jun 3, 2020 at 18:25
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16831 | https://www.youtube.com/watch?v=nzl2soZzvUI | Solving Systems of Equations with Distribution: Step-by-Step Exampl
Math Notes and Examples 2.0
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In this tutorial, learn how to solve a system of equations involving distribution with the example 5x - 4y = -10 and y = 2x - 5. I walk you through each step to help you understand how to solve when there is a number multiplying the varible you are trying to substitute. Perfect for students looking to strengthen their algebra skills! #SystemsOfEquations #Distribution #Algebra #MathTutorial #SolvingEquations #MathHelp #MiddleSchoolMath #HighSchoolMath
Transcript:
welcome back we're going to talk in this video about solving a system of equations by substitution a classic type of example and we're going to talk about a little bit why and go through the step by step to solve it first uh we're going to be solving by finding an x value and a y value that will satisfy the top equation will label it one and the bottom equation will label it to the first step here is to solve for a variable now this is why this is a class example we have a variable already solved for us we have y by itself we know that y = 2x - 5 and so that part is already done for us we can check that off the second step is to substitute the expression the variable equals in the other equation so I'm going to take 2x - 5 and put it into the other equation in substitution for y and let's write out what that looks like so I have 5x - 4 and then instead of Y in our first equation I put 2x - 5 cuz I know that's what y equals and then I continue it equals -10 so this is a version of number one but I substituted the expression y equals in for y and then if it's boxed up like this I can put parentheses around it four is being multiplied or -4 is being multiplied to this expression so I'll put parentheses around that and now I've substituted the third step is to solve the one variable equation so when I look at a an equation like this and it's one variable I first think distributive property I'm going to write out 5x and then I'm going to multiply -4 2x I get 8X and then multiply -4 -5 I get POS 20 and then write out the rest of it equals -10 now the next step is going to be to combine like terms we have 5x and -8x combining those together we're going to get -3x then we have our inverse operations I want to get X by itself so I'm going to subtract 20 from both sides I get -3x = -30 and then I'm going to divide both sides by -3 to get x = 10 so I've solved the one variable equation that means that the x value in this solution has to be equal to 10 now if xal 10 we're on to our last step is to substitute the value in place of the variable in one of the two original equations so I take my two original equations I know that if x = 10 10 I can put that in substitution for X so in the first equation and the second equation I can substitute 10 for x and this is what they look like and now it's really up to me which one I want to solve and I can look at these and say okay I have a equation one becomes a two-step equation and equation two down here just has 2 10 - 5 I know that 2 10 is 20 20 - 5 gives me 15 so I know the Y must be 15 in this bottom equation and I can always check to see if that's true I put it up here 5 10 is 50 then 4 15 is 60 so 50 - 60 does give me -10 so it's true for both equations and I can substitute Y in from the rest of the coordinate So my answer is 10 15 as my solution that satisfies both of these equations please feel free to leave a comment letting me know any questions you have as always be kind to yourself be kind to others I hope you have a great day |
16832 | https://en.wikipedia.org/wiki/Metal_aromaticity | Metal aromaticity - Wikipedia
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Metal aromaticity
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Concept of aromaticity extended to metals
Metal aromaticity or metalloaromaticity is the concept of aromaticity, found in many organic compounds, extended to metals and metal-containing compounds. The first experimental evidence for the existence of aromaticity in metals was found in aluminium cluster compounds of the type MAl−
4 where M stands for lithium, sodium or copper. These anions can be generated in a helium gas by laser vaporization of an aluminium / lithium carbonate composite or a copper or sodium / aluminium alloy, separated and selected by mass spectrometry and analyzed by photoelectron spectroscopy. The evidence for aromaticity in these compounds is based on several considerations. Computational chemistry shows that these aluminium clusters consist of a tetranuclear Al 2−
4 plane and a counterion at the apex of a square pyramid. The Al 2−
4 unit is perfectly planar and is not perturbed by the presence of the counterion or even the presence of two counterions in the neutral compound M
2 Al
In addition its HOMO is calculated to be a doubly occupied delocalized pi system making it obey Hückel's rule. Finally a match exists between the calculated values and the experimental photoelectron values for the energy required to remove the first 4 valence electrons. The first fully metal aromatic compound was a cyclogallane with a Ga 3 2- core discovered by Gregory Robinson in 1995.
D-orbital aromaticity is found in trinuclear tungstenW
3 O−
9 and molybdenumMo
3 O−
9metal clusters generated by laser vaporization of the pure metals in the presence of oxygen in a helium stream. In these clusters the three metal centers are bridged by oxygen and each metal has two terminal oxygen atoms. The first signal in the photoelectron spectrum corresponds to the removal of the valence electron with the lowest energy in the anion to the neutral M
3 O
9 compound. This energy turns out to be comparable to that of bulk tungsten trioxide and molybdenum trioxide. The photoelectric signal is also broad which suggests a large difference in conformation between the anion and the neutral species. Computational chemistry shows that the M
3 O−
9 anions and M
3 O 2−
9 dianions are ideal hexagons with identical metal-to-metal bond lengths. Tritantalum oxide clusters (Ta 3 O 3−) also are observed to exhibit possible D-orbital aromaticity.
The molecules discussed thus far only exist diluted in the gas phase. A study exploring the properties of a compound formed in water from sodium molybdate (Na
2 MoO
4·2H
2 O) and iminodiacetic acid also revealed evidence of aromaticity, but this compound has actually been isolated. X-ray crystallography showed that the sodium atoms are arranged in layers of hexagonal clusters akin to pentacenes. The sodium-to-sodium bond lengths are unusually short (327 pm versus 380 pm in elemental sodium) and, like benzene, the ring is planar. In this compound each sodium atom has a distorted octahedral molecular geometry with coordination to molybdenum atoms and water molecules. The experimental evidence is supported by computed NICS aromaticity values.
See also
[edit]
Metal cluster compound– Cluster of three or more metals
Catenation– Bonding of atoms of the same element into chains or rings
References
[edit]
^Feixas, Ferran; Matito, Eduard; Poater, Jordi; Solà, Miquel (13 September 2012). "Metalloaromaticity". Wiley Interdisciplinary Reviews: Computational Molecular Science. 3 (2): 105–122. doi:10.1002/wcms.1115. S2CID222199114.
^Observation of All-Metal Aromatic Molecules Xi Li, Aleksey E. Kuznetsov, Hai-Feng Zhang, Alexander I. Boldyrev, Lai-Sheng Wang Science Vol. 291. p. 859 2001doi:10.1126/science.291.5505.859
^ Jump up to: abKrämer, Katrina. "The search for the grand unification of aromaticity". Chemistry World.
^Observation of d-Orbital Aromaticity Xin Huang, Hua-Jin Zhai, Boggavarapu Kiran, Lai-Sheng Wang, Angewandte Chemie International Edition Volume 44, Issue 44, Pages 7251–54 2005doi:10.1002/anie.200502678
^Synthesis and structure of 1-D Na6 cluster chain with short Na–Na distance: Organic like aromaticity in inorganic metal cluster Snehadrinarayan Khatua, Debesh R. Roy, Pratim K. Chattaraj and Manish Bhattacharjee Chem. Commun., 2007, 135–37, doi:10.1039/b611693k
| hide v t e Chemical bonds |
| Intramolecular (strong) | | Covalent | Electron deficiency 3c–2e 4c–2e 8c–2e Hypervalence 3c–4e Agostic Bent Coordinate (dipolar) Pi backbond Metal–ligand multiple bond Charge-shift Hapticity Conjugation Hyperconjugation Aromaticity homo bicyclo | | Metallic | Metal aromaticity | | Ionic | | | |
| Intermolecular (weak) | | Van der Waals forces | London dispersion | | Hydrogen | Low-barrier Resonance-assisted Symmetric Dihydrogen bonds C–H···O interaction | | Noncovalent other | Mechanical Halogen Chalcogen Metallophilic (aurophilic) Intercalation Stacking Cation–pi Anion–pi Salt bridge | |
| Bond cleavage | Heterolysis Homolysis |
| Electron counting rules | Aromaticity Hückel's rule Baird's rule Möbius spherical Polyhedral skeletal electron pair theory Jemmis mno rules |
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16833 | https://math.stackexchange.com/questions/454510/about-monotonicity-of-power-functions | Stack Exchange Network
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About monotonicity of power functions
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$\begingroup$
Is the function: $\large{1\over(x-2)^3}+3$ monotonic? I understand it is strictly decreasing in its domain, but being discontinous makes it not monotonic. Is this right? Sorry if this question sounds trivial to you. Thanks in advance. Raul
algebra-precalculus
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edited Jul 29, 2013 at 2:44
Amzoti
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asked Jul 29, 2013 at 2:24
Raul RuedaRaul Rueda
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$\endgroup$
1
2
$\begingroup$ It's not necessarily the discontinuity that makes it non-monotonic, just the existence of two values $x \leq y$ with $f(x) \not\leq f(y)$. In this case 1 and 3 work. You can have discontinuous functions which jump around but are still monotonic. $\endgroup$
Andrew Poelstra
– Andrew Poelstra
2013-07-29 02:47:08 +00:00
Commented Jul 29, 2013 at 2:47
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1 Answer 1
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$\begingroup$
Overall, it's not monotonic.
Let's call the function $f$.
It's domain is $X = \mathbb R \setminus {2}$.
It's (strictly) monotonically decreasing on the interval $\left]-\infty, 2\right[$, and on the interval $\left]2, \infty \right[$.
But it's not monotone on its entire domain. You can see that $f(1) = 2$ and $f(3) = 4$, so $f(x)$ increases between $x=1$ and $x=3$, which destroys any hope of it being monotonically decreasing everywhere on its domain.
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answered Jul 29, 2013 at 3:07
bubbabubba
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$\endgroup$
2
$\begingroup$ Thank you so much for your replies! Now I dare to ask if we can accept in general that this type of functions: $f(x)={a\over(x-d)^n}+e$ are non-monotonic. I understand they can be decreasing or increasing depending on the sign of "a" and if "n" is even or odd. $\endgroup$
Raul Rueda
– Raul Rueda
2013-07-29 03:37:18 +00:00
Commented Jul 29, 2013 at 3:37
$\begingroup$ Let's assume that $a>0$. If $n$ is odd, the behaviour is more-or-less the same as in your specific example. If $n$ is even, then the function will be increasing on $\left]-\infty, d\right[$ and decreasing on $\left]d, \infty\right[$, so not even close to monotonic. The case $a<0$ can be analyzed in a similar way. $\endgroup$
bubba
– bubba
2013-07-29 04:00:36 +00:00
Commented Jul 29, 2013 at 4:00
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16834 | https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Enthalpy/Enthalpy_Change_of_Neutralization | Skip to main content
Enthalpy Change of Neutralization
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Bond Enthalpies
Enthalpy Change of Solution
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Jim Clark
Truro School in Cornwall
( \newcommand{\kernel}{\mathrm{null}\,})
The standard enthalpy change of neutralization is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water. Notice that enthalpy change of neutralization is always measured per mole of water formed. Enthalpy changes of neutralization are always negative - heat is released when an acid and and alkali react. For reactions involving strong acids and alkalis, the values are always very closely similar, with values between -57 and -58 kJ mol-1. That varies slightly depending on the acid-alkali combination (and also on what source you look it up in!).
Why do strong acids reacting with strong alkalis give closely similar values?
We make the assumption that strong acids and strong alkalis are fully ionized in solution, and that the ions behave independently of each other. For example, dilute hydrochloric acid contains hydrogen ions and chloride ions in solution. Sodium hydroxide solution consists of sodium ions and hydroxide ions in solution. The equation for any strong acid being neutralized by a strong alkali is essentially just a reaction between hydrogen ions and hydroxide ions to make water. The other ions present (sodium and chloride, for example) are just spectator ions, taking no part in the reaction.
The full equation for the reaction between hydrochloric acid and sodium hydroxide solution is:
NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l)(1)
but what is actually happening is:
OH−(aq)+H+(aq)→H2O(l)(2)
If the reaction is the same in each case of a strong acid and a strong alkali, it is not surprising that the enthalpy change is similar.
In a weak acid, such as acetic acid, at ordinary concentrations, something like 99% of the acid is not actually ionized. That means that the enthalpy change of neutralization will include other enthalpy terms involved in ionizing the acid as well as the reaction between the hydrogen ions and hydroxide ions. And in a weak alkali like ammonia solution, the ammonia is also present mainly as ammonia molecules in solution. Again, there will be other enthalpy changes involved apart from the simple formation of water from hydrogen ions and hydroxide ions. For reactions involving acetic acid or ammonia, the measured enthalpy change of neutralization is a few kJ less exothermic than with strong acids and bases.
For example, one source which gives the enthalpy change of neutralization of sodium hydroxide solution with HCl as -57.9 kJ mol-1:
NaOH(aq)+HCl(aq)→Na+(aq)+Cl−(aq)+H2O(3)
the enthalpy change of neutralization for sodium hydroxide solution being neutralized by acetic acid is -56.1 kJ mol-1 :
NaOH(aq)+CH3COOH(aq)→Na+(aq)+CH3COO−(aq)+H2O(4)
For very weak acids, like hydrogen cyanide solution, the enthalpy change of neutralization may be much less. A different source gives the value for hydrogen cyanide solution being neutralized by potassium hydroxide solution as -11.7 kJ mol-1, for example.
NaOH(aq)+HCN(aq)→Na+(aq)+CN−(aq)+H2O(5)
Neutralizing Strong acids vs. Weak Acids
The experimentally measured enthalpy change of neutralization is a few kJ less exothermic than with strong acids and bases.
Contributors and Attributions
Jim Clark (Chemguide.co.uk)
Bond Enthalpies
Enthalpy Change of Solution |
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16836 | https://math.stackexchange.com/questions/68293/what-is-the-difference-between-only-if-and-iff | Skip to main content
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I have read this question. I am now stuck with the difference between "if and only if" and "only if". Please help me out.
Thanks
logic
terminology
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edited Apr 13, 2017 at 12:20
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asked Sep 28, 2011 at 19:39
user2857user2857
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The moon is made of lemon meringue only if 1+1=2.
– Ilmari Karonen
Commented
Sep 28, 2011 at 20:08
1
Also try to understand in terms of plain translation. AiffB means A is true 'if' B is true & A is true 'only if' B is true.The 'only if' means that A is true in no other cases.'A if B' can be written as B => A.And 'A only if B' can be written as notB => notA. It is the property of => sign that c=>d is same as notd=>notc. Thus , you can replace notB=>notA by A=>B. Thus A iff B can be written as A=>B and B=>A . Of course what I am saying is same as what others have already said . I just wanted to emphasise how we can intuitively try to understand the logic from the meaning of 'if' and 'only if'.
– ameyask
Commented
Feb 11, 2014 at 11:18
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@Ilmari: So moon rocks are frozen lemon meringue?
– Asaf Karagila
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Jun 7, 2015 at 12:52
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The mathematician R.L. Moore used "only if" to mean "if and only if". This sounds weird to us now, because it goes against the accepted convention, but I can see what Moore was thinking. The statement "A only if B" sounds like the statement "A if B", except that you are also given an extra piece of information: not just A if B, but A only if B.
– littleO
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Jun 25, 2015 at 6:58
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Let's assume A and B are two statements. Then to say "A only if B" means that A can only ever be true when B is true. That is, B is necessary for A to be true. To say "A if and only if B" means that A is true if B is true, and B is true if A is true. That is, A is necessary and sufficient for B. Succinctly,
A only if B is the logic statement A⇒B.
A iff B is the statement (A⇒B)∧(B⇒A)
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edited Sep 29, 2011 at 2:27
answered Sep 28, 2011 at 19:48
michaelmichael
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@RossMillikan & Josh: Thanks for the answers. So I can conclude it as 'only if' is same as implies and 'iff' is same as equivalence?
– user2857
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Sep 28, 2011 at 20:05
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@Akito: that is correct
– Ross Millikan
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Sep 28, 2011 at 20:07
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Then shouldn't your first line read "A only if B is the logic statement B⇒A"?
– Michael Fulton
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Apr 11, 2017 at 19:06
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Nice explanation here criticalthinkeracademy.com/courses/2514/lectures/51574
– Devasish
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Jan 11, 2018 at 2:16
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@erised: Nope, the accepted answer is correct. What comes before the "only if" is the antecedent. For clarification, see "2. A only if B" in the link to which you (via @Devasish) refer.
– Nick The Swede
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Nov 22, 2018 at 14:13
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I will find a million dollars inside this locker only if I know the combination.
But that doesn't mean I will find a million dollars there if I know the combination. After all, there might be only a half million in there.
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answered Sep 28, 2011 at 20:25
Michael HardyMichael Hardy
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I don't know why the answer received so many votes when this is the correct one.
– Haelia101
Commented
Dec 3, 2016 at 6:42
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but the question is about comparing 'iff' and 'only if', not comparing 'if' and 'only if'
– hzh
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Dec 7, 2021 at 20:13
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A "only if B"
is the same as saying
"B is necessary" for A
which is the same as saying
A could not have happened without B
but that does mean that other things do not also need to happen for A to be true.
Therefore,
A→B
but it is not true that B→A because B being true does not guarantee A happened. There could also be other requirements for A to be true.
e.g.:
You are eligible to be president only if you are at least 35 years
old.
let p: "You are eligible to be president" and a: "You are at least 35 years old".
Here is is the case that p→a,
but it is not the case that a→p
In other words, a is necessary for p, but just because a is true does not mean that a is the one single requirement for p. Just because you're at least 35 years old does not mean that you are eligible to be president.
As far as the difference goes, (which I guess was the specific question), if and only if means just that.
p if and only if q means (p if q) AND (p only if q).
The bottom line is:
p if q
equates
if q, then p
which is the same as q→p
I just (hopefully well) explained that
p only if q
equates
p→q
Also,
q→p and p→q
is the same as saying p⟺q
So there you have. One statement is unidirectional, the other is bidirectional.
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edited Apr 10, 2019 at 15:09
AtilioA
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If A then B is true unless A is true and B is false and written A⟹B.
A only if B is true unless A is true and B is false, equivalent to if A then B.
A if B is true unless A is false and B is true, the converse of the above, and is written B⟹A
A iff B, also written A if and only if B, is true if A and B have the same truth value. It represents (A if B) and (A only if B) and is written A⟺B
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edited Sep 28, 2011 at 19:57
answered Sep 28, 2011 at 19:50
Ross MillikanRoss Millikan
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A real number is positive if and only if it is greater than zero.
A real number is an rational only if it has a finite decimal expansion. A real number, in general, however need not be rational.
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edited Jun 7, 2015 at 12:51
answered Sep 28, 2011 at 20:06
Asaf Karagila♦Asaf Karagila
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I'd be happy to improve this, if someone has some suggestions as to what's wrong.
– Asaf Karagila
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Jun 6, 2015 at 19:59
I'm not the downvoter, but it looks like you've explained the difference between 'if and only if' and 'if', while the question asked about 'if and only if' and 'only if'.
– user88319
Commented
Jun 6, 2015 at 21:38
You're right. There, now it's all better.
– Asaf Karagila
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Jun 7, 2015 at 12:51
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How I understood this:
A rectangle is a square only if it has all sides equal. In this case the sentence gives us information, how we should name rectangles fulfilling the condition. It does not say that only these rectangles are squares. There may be many other rectangles which do not have equal sides, but are squares as well, for example the definition of "a square" could be "a square is a rectangle with equal sides or a circle with radius larger than 2 cm". So from this sentence I only know how to name particular rectangle, but I don't know what a square is in general, I know only one case.
Each square is a rectangle with all sides equal (or using "if" a square exists only if it is a rectangle with its sides equal). In this sentence we know that it is necessary for squares to be rectangles with equal sides. However, it does not say, that each rectangle with equal sides is a square. The might be some rectangles with equal sides, that we could name "circles". So again, I know only one case.
A rectangle is a square if and only if it has equal sides means that 1. only each rectangle with equal sides can be called a square, but also 2. each square is a rectangle with equal sides. There are no other conditions for both. No other figure can be a square, and a rectangle with equal sides can be nothing but a square.
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answered Apr 10, 2017 at 10:31
VoitcusVoitcus
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The way logic is written can be confusing. Tip: whenever you see something like "A only if B", mentally insert "A [is true] only if B [is true]".
What does "A only if B" mean? It means "if A then B", or more precisely, "If A is true, then B is true".
"I cry only if I'm sad" = "If I cry, then I'm sad" = "If I'm crying, you know I'm sad"
What does "A if B" mean? It means "If B then A", or more precisely, "If B is true, then A is true", the reverse from before.
"I cry if I'm sad" = "If I'm sad, then I cry" = "If I'm sad, you know I'm crying"
What does "A if and only if B" mean? It means "A if B and A only if B". In other words, "If B is true, then A is true, and if A is true, then B is true". A and B are always both true or both false.
"I cry if and only if I'm sad" = "If I cry, then I'm sad, and if I'm sad, then I cry" = " You know I'm either crying and sad, or neither"
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edited Apr 4, 2020 at 2:14
answered Oct 8, 2018 at 22:21
JoseOrtiz3JoseOrtiz3
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A iff B is the statement
"if B then A" and "only if B then A"
(B⇒A)∧(notB⇒notA)
(B⇒A)∧(A⇒B)
A=B
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answered Feb 11, 2014 at 10:53
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If I say that an object is an apple only if it is a fruit (Fruit⇐Apple), then that means that something has to be a fruit in order for it to be an apple, but it does not have to be an apple. If something is a fruit, it can also be an orange or a banana.
However, if said that an object is an apple if and only if it is a fruit (Fruit⟺Apple), then that would once again mean that something has to be a fruit in order for it to be an apple, but here the main difference is that it would also have to be an apple and not an orange or a banana. If it is a fruit, then it's an apple, and if it is an apple, then it is a fruit.
In the second example, we have also added Fruit⇒Apple which, on it's own, means that if something is a fruit, then it has to be an apple. Another way of writing A⟺B is (A⇒B) & (A⇐B).
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answered Apr 11, 2017 at 17:08
John KJohn K
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I win the competition if I compete.
This sentence implies that if I compete, I necessarily win the competition, but not that I only win if I compete.
I win the competition only if I compete.
This sentence implies I only win if I compete, but not necessarily that if I compete, I win the competition.
I win the competition if and only if I compete.
This sentence implies that if I compete, I necessarily win the competition AND that I only win if I compete.
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answered Aug 13, 2023 at 2:27
Eduardo MEduardo M
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I am just putting an another comprehnsive answer which steals important points from other answers. Other answers don't break the IF AND ONLY IF correctly.
Let's try to prove the truth table of P IFF Q
```
Label P Q P IFF Q
tt T T T
tf T F F
ft F T F
ff F F T
```
Proof:
```
P IFF Q
<=> P IF AND ONLY IF Q
<=> (P IF Q) AND (P ONLY IF Q)
<=> (IF Q THEN P) AND (P THEN Q)
```
Let's put a truth table for our last expression now.
```
P | Q | (IF Q THEN P) | (P THEN Q) | R AND S <=> P IFF Q
| | (R) | (S) |
T T T T T
T F T F F
F T F T F
F F T T T
QED.
```
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answered Nov 14, 2019 at 19:01
Madhusoodan PMadhusoodan P
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16837 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5684575/ | Review of Routine Laboratory Monitoring for Patients with Rheumatoid Arthritis Receiving Biologic or Nonbiologic DMARDs - PMC
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. 2017 Oct 31;2017:9614241. doi: 10.1155/2017/9614241
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Review of Routine Laboratory Monitoring for Patients with Rheumatoid Arthritis Receiving Biologic or Nonbiologic DMARDs
William F C Rigby
William F C Rigby
1 Geisel School of Medicine, Dartmouth College, Lebanon, NH, USA
Find articles by William F C Rigby
1,✉, Kathy Lampl
Kathy Lampl
2 Ultragenyx Pharmaceutical, Inc., Novato, CA, USA
Find articles by Kathy Lampl
2, Jason M Low
Jason M Low
3 Genentech, Inc., South San Francisco, CA, USA
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3, Daniel E Furst
Daniel E Furst
4 Division of Rheumatology, Department of Medicine, David Geffen School of Medicine, University of California, Los Angeles, Los Angeles, CA, USA
5 University of Washington, Seattle, WA, USA
6 University of Florence, Florence, Italy
Find articles by Daniel E Furst
4,5,6
Author information
Article notes
Copyright and License information
1 Geisel School of Medicine, Dartmouth College, Lebanon, NH, USA
2 Ultragenyx Pharmaceutical, Inc., Novato, CA, USA
3 Genentech, Inc., South San Francisco, CA, USA
4 Division of Rheumatology, Department of Medicine, David Geffen School of Medicine, University of California, Los Angeles, Los Angeles, CA, USA
5 University of Washington, Seattle, WA, USA
6 University of Florence, Florence, Italy
Academic Editor: Ruben Burgos-Vargas
✉
Corresponding author.
Received 2017 Feb 16; Accepted 2017 Aug 15; Issue date 2017.
Copyright © 2017 William F. C. Rigby et al.
This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC5684575 PMID: 29225625
Abstract
Safety concerns associated with many drugs indicated for the treatment of rheumatoid arthritis (RA) can be attenuated by the early identification of toxicity through routine laboratory monitoring; however, a comprehensive review of the recommended monitoring guidelines for the different available RA therapies is currently unavailable. The aim of this review is to summarize the current guidelines for laboratory monitoring in patients with RA and to provide an overview of the laboratory abnormality profiles associated with each drug indicated for RA. Recommendations for the frequency of laboratory monitoring of serum lipids, liver transaminases, serum creatinine, neutrophil counts, and platelet counts in patients with RA were compiled from a literature search for published recommendations and guidelines as well as the prescribing information for each drug. Laboratory abnormality profiles for each drug were compiled from the prescribing information for each drug and a literature search including meta-analyses and primary clinical trials data.
1. Introduction
Rheumatoid arthritis (RA) is a chronic systemic inflammatory disease that, without treatment, leads to permanent joint damage and destruction. Patients with RA have an increased risk of comorbidities (most commonly cardiovascular [CV] disease and infection) and decreased survival (the standardized mortality ratio is ≈2, with no decrease observed over time) compared with the general population, and the majority of premature deaths are due to CV disease [1–5]. The higher risk of CV disease in patients with RA is generally thought to be due to the increased inflammatory burden, which causes accelerated atherosclerosis , as well as a greater prevalence of traditional risk factors (hypertension, dyslipidemia, and smoking) [1, 7, 8]. The elevated risk of infection in patients with RA may be due either to the immunomodulatory effects of RA itself or to the immunosuppressive effects of RA-related treatments . Patients with RA are also at an increased risk of renal impairment, which is often associated with CV risk factors .
Many of the drugs indicated for the treatment of RA can exacerbate comorbidity risks. In addition, the drugs themselves can cause adverse events. Close monitoring of patients receiving treatment for RA is therefore a critical part of patient care. The first line of treatment for RA is with conventional synthetic disease-modifying antirheumatic drugs (csDMARDs), usually methotrexate [11, 12]. However, approximately 50% to 70% of patients have an inadequate response to methotrexate alone [13–15]. In patients who have an inadequate response to csDMARDs, initiation of a biologic DMARD, as monotherapy or in addition to the csDMARD, may provide increased efficacy. Biologics indicated for RA include anti-tumor necrosis factor (aTNF) agents; the B-cell-targeting anti-CD20 monoclonal antibody rituximab; the T-cell costimulatory modulator abatacept; the interleukin 1 receptor antagonist anakinra; and the anti-interleukin 6 alpha receptor monoclonal antibody tocilizumab. Tofacitinib, a targeted small-molecule DMARD that inhibits JAK/STAT signaling, is also available in some countries. Lastly, glucocorticoids are frequently administered to control symptoms of RA.
Safety concerns associated with biologic and nonbiologic DMARDs include increased CV risk, liver and hematologic toxicity, renal impairment, infection, and bleeding. These concerns may be attenuated by identifying abnormal laboratory values early and adjusting medication use accordingly; however, specific recommendations for the frequency of laboratory monitoring in patients with RA are often unclear and left to the treating physician's discretion. There is a need to minimize phlebotomies and physician visits and the inconvenience of patient time spent on them. Because physicians may take into consideration the necessary laboratory monitoring rubrics when choosing between different DMARDs, the purpose of this article is to review the current guidelines for laboratory monitoring in patients with RA, in general, and during treatment. We also set out to provide an overview of the laboratory abnormality profile associated with each drug indicated for RA. Although laboratory testing can refer to much more, such as biomarkers and predictive markers for response, this review is limited to laboratory tests primarily concerned with drug toxicity and pharmacodynamics. Further, this report limits its review to monitoring guidelines for the most common laboratory tests for toxicity: serum lipids, liver aminotransferases, serum creatinine, absolute neutrophil counts (and potential for associated infections), and platelets. Although there is variation in physician views regarding monitoring requirements based on personal experience, this review covers the recommendations given in the prescribing information of each drug as well as by the American College of Rheumatology (ACR), the European League Against Rheumatism (EULAR), and the British Society for Rheumatology (BSR) [11, 12, 16–18].
2. Methods
For data on the effects of DMARDs on laboratory measures, PubMed was searched using the name of the agent in question and “rheumatoid arthritis” in combination with terms related to the laboratory measure, such as “lipid”, “cholesterol”, “cardiovascular”, “kidney”, “liver”, “neutrophil”, “neutropenia”, “platelet” and “thrombocytopenia”. Further relevant information was obtained from primary clinical trials, the prescribing information for the DMARDs, and the authors' own experiences.
3. Overview of Good Clinical Practice Recommendations for Laboratory Monitoring in Patients With RA
3.1. CV Risk and Serum Lipid Levels
The American Heart Association recommends that people aged 20 years or older and not diagnosed with CV disease have their cholesterol levels checked every 4 to 6 years as part of a cardiovascular risk assessment, and more often if the risk is elevated. Patients with RA have a 50% to 60% increased risk for CV-related death compared with the general population [1, 19]. Because of this increased risk, EULAR guidelines recommend that all patients with RA should undergo an annual CV risk assessment, as treatment and underlying inflammation may alter CV risk factors . However, individual risk profiles will vary; therefore, the EULAR guidelines suggest that the treatment and follow-up plan be determined on an individual basis. It has been noted that CV risk assessment can be easily incorporated into a routine RA visit by adding the determination of nonfasting lipids (total cholesterol [TC], low-density lipoprotein [LDL], and high-density lipoprotein [HDL]) to routine laboratory tests and that the ratio of TC to HDL is the most stable marker of lipid-associated CV risk in RA. Because the TC : HDL ratio does not require fasting, it is also the most convenient method of assessing lipid-associated CV risk . Intervention with statins in patients with RA is recommended at the same frequency as in the general population and in accordance with the national guidelines for the general population . Complicating CV risk assessment and subsequent treatment is the “lipid paradox,” an observed effect in which increased inflammatory burden is associated with decreased serum lipid levels [20, 21]. Because of this paradoxical effect between inflammation and serum lipid levels, a recent review suggested that traditional, lipid profile-based CV disease risk stratification should not be applied to patients with active RA; rather, lipid levels should be assessed after control of inflammation is achieved .
3.2. Kidney Function Monitoring
Kidney disease is relatively common (~8% to 15%) in patients with RA and may arise as a result of the treatments for RA or the presence of amyloidosis or vasculitis [10, 22]; the contributory role of inflammation in renal impairment remains unclear. Currently, specific guidelines on the recommended frequency of renal function monitoring in patients with RA are scarce, but a study from Couderc et al. on the prevalence of renal dysfunction in RA concluded that, regardless of the treatment regimen, at least annual creatinine measurements appear necessary for patients with RA . This approach is further strengthened by the frequent use of methotrexate and its clearance by the kidney; if abnormal renal function is present, more frequent monitoring is justifiable. Amyloidosis is one of the most severe renal complications of RA, with a reported incidence of 5% to 19% . However, it is likely that this incidence is quite low due to the increase in effective treatment options for RA . Given that the presenting feature of amyloidosis often is proteinuria, testing for proteinuria in patients with longer duration of and/or uncontrolled RA may be prudent .
3.3. Liver Function
Liver injury is generally not a manifestation of RA. However, medications for RA (usually long-term methotrexate or leflunomide therapy) are associated with abnormal liver function. For this reason, recommendations for monitoring liver enzymes (alanine aminotransferase [ALT] and aspartate aminotransferase [AST]) and performing liver function tests vary depending on individual therapeutic regimens.
3.4. Neutropenia
It has been shown that, after adjustment for several key factors, patients with RA are significantly more likely to develop serious infections compared with the general population (hazard ratio, 1.83 [95% CI, 1.52–2.21]) . Severe neutropenia is an uncommon feature of RA (Felty's syndrome), with most cases arising as a consequence of RA-related therapies . Neutropenia has been linked to an increased risk of infection; however, it is important to note that cases of therapy-associated severe neutropenia have been observed without incidence of infection . Because the risk of infection may increase with the severity and duration of neutropenia, routine monitoring of absolute neutrophil count (ANC) is necessary during RA treatment; recommendations for the frequency of monitoring ANCs are specific to individual medications.
3.5. Thrombocytopenia
Although thrombocytopenia is an uncommon feature of RA, its occurrence may be associated with RA-related therapies. Although cases of drug-induced thrombocytopenia have been associated with clinically important bleeding events , the association of bleeding events with RA therapy-induced thrombocytopenia is not known. Monitoring of platelet counts is important during RA treatment to assess the risk of internal bleeding, and, as with ANCs, the recommendations for the frequency of monitoring platelets in patients with RA are specific to the individual therapeutic regimen.
4. Monitoring Guidelines for Conventional Synthetic, Biologic, and Targeted Small-Molecule DMARDs in Patients with RA
Laboratory monitoring guidelines for serum lipids, liver enzymes, serum creatinine, neutrophils, and platelets in patients receiving DMARDs are described in the prescribing information for each drug, drug reports from the manufacturers, experts in the field, the ACR recommendations for laboratory monitoring in patients with RA during treatment with DMARDs, and the BSR guideline for DMARD therapy (Table 1) [16–18]. The EULAR guidelines for managing RA with biologic and conventional DMARDs do not make specific recommendations regarding the frequency of routine laboratory monitoring .
Table 1.
Laboratory monitoring guidelines for DMARDs in patients with RA.
| | Lipids | Liver enzymes (function, ALT, and AST) | Neutrophils and/or platelets | Serum creatinine | Ref(s) |
:---: :---: :---: |
| Conventional synthetic DMARDs |
| |
| Methotrexate | N/A | LFT every 1-2 months | CBC with differential and platelet counts at least monthly | N/A | PI |
| N/A | ALT and AST at baseline, every 2–4 weeks for the first 3 months, every 8–12 weeks the following 3–6 months, and every 12 weeks thereafter | CBC at baseline, every 2–4 weeks for the first 3 months, every 8–12 weeks for 3–6 months after initiation, and every 12 weeks thereafter | At baseline, every 2–4 weeks for the first 3 months, every 8–12 weeks the following 3–6 months, and every 12 weeks thereafter | ACR |
| N/A | ALT and/or AST every 2 weeks until stable dose for 6 weeks, then monthly for 3 months; at least every 12 weeks thereafter | Full blood count every 2 weeks until stable dose for 6 weeks, then monthly for 3 months; at least every 12 weeks thereafter | Creatinine/calculated GFR every 2 weeks until stable dose for 6 weeks, then monthly for 3 months; at least every 12 weeks thereafter | BSR [17, 18] |
| |
| Leflunomide | N/A | ALT levels ≥ monthly for 6 months after initiation; every 6–8 weeks thereafter | Platelet count, white blood cell count, and hemoglobin or hematocrit monitored at baseline and monthly for 6 months after initiation and every 6–8 weeks thereafter | N/A | PI |
| N/A | ALT and AST at baseline, every 2–4 weeks for the first 3 months, every 8–12 weeks for 3–6 months after initiation, and every 12 weeks thereafter | CBC at baseline, every 2–4 weeks for the first 3 months, every 8–12 weeks for 3–6 months after initiation, and every 12 weeks thereafter | At baseline, every 2–4 weeks for the first 3 months, every 8–12 weeks for 3–6 months after initiation, and every 12 weeks thereafter | ACR |
| N/A | ALT and/or AST every 2 weeks until stable dose for 6 weeks, then monthly for 3 months; at least every 12 weeks thereafter | Full blood count every 2 weeks until stable dose for 6 weeks, then monthly for 3 months; at least every 12 weeks thereafter | Creatinine/calculated GFR every 2 weeks until stable dose for 6 weeks, then monthly for 3 months; at least every 12 weeks thereafter | BSR [17, 18] |
| |
| HCQ/CQ | N/A | ALT and AST at baseline and none thereafter | CBC at baseline and none thereafter | At baseline | ACR [16, 73] |
| |
| Gold | N/A | ALT and/or AST every 2 weeks until stable dose for 6 weeks, then monthly for 3 months; at least every 12 weeks thereafter | Full blood count every 2 weeks until stable dose for 6 weeks, then monthly for 3 months; at least every 12 weeks thereafter | Creatinine/calculated GFR every 2 weeks until stable dose for 6 weeks, then monthly for 3 months; at least every 12 weeks thereafter | BSR [17, 18] |
| |
| Sulfasalazine | N/A | LFT at baseline and every 2 weeks during the first 3 months, monthly during the second 3 months, and every 3 months or as needed thereafter | CBC with differential at baseline and every 2 weeks during the first 3 months, monthly during the second 3 months, and every 3 months or as needed thereafter | N/A | PI |
| N/A | ALT and AST at baseline, every 2–4 weeks for the first 3 months, every 8–12 weeks for 3–6 months after initiation, and every 12 weeks thereafter | CBC at baseline, every 2–4 weeks for the first 3 months, every 8–12 weeks for 3–6 months after initiation, and every 12 weeks thereafter | At baseline, every 2–4 weeks for the first 3 months, every 8–12 weeks for 3–6 months after initiation, and every 12 weeks thereafter | ACR |
| N/A | ALT and/or AST every 2 weeks until stable dose for 6 weeks, then monthly for 3 months; at least every 12 weeks thereafter | Full blood count every 2 weeks until stable dose for 6 weeks, then monthly for 3 months; at least every 12 weeks thereafter | Creatinine/calculated GFR every 2 weeks until stable dose for 6 weeks, then monthly for 3 months; at least every 12 weeks thereafter | BSR [17, 18] |
| |
| Biologic DMARDs d |
| |
| Adalimumab a | N/A | N/A | N/A | N/A | |
| |
| Infliximab a,b | N/A | Same as for MTX b | Same as for MTX b | Same as for MTX b | |
| |
| Etanercept a | N/A | N/A | N/A | N/A | |
| |
| Golimumab a,b | N/A | Same as for MTX b | Same as for MTX b | Same as for MTX b | |
| |
| Certolizumab a | N/A | N/A | N/A | N/A | |
| |
| Tocilizumab | 4–8 weeks after initiation and every 24 weeks thereafter | ALT and AST levels 4–8 weeks after initiation and every 3 months thereafter | ANC 4–8 weeks after initiation and every 3 months thereafter | N/A | PI |
| |
| Rituximab b | N/A | Same as for MTX b | CBC and platelet counts at 2- and 4-month intervals during rituximab therapy | N/A | PI |
| |
| Abatacept a | None required | None required | N/A | N/A | PI |
| |
| Anakinra a | N/A | N/A | N/A | N/A | |
| |
| Targeted small-molecule DMARDs |
| |
| Tofacitinib | 4–8 weeks after initiation | Routine monitoring of all | At initiation, 4–8 weeks after initiation, and every 3 months thereafter | N/A | PI |
| |
| Glucocorticoids | At baseline, 1 month after initiation, and every 6–12 months thereafter | N/A c | N/A c | N/A c | Liu et al. |
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ACR, American College of Rheumatology; ALT, alanine aminotransferase; AST, aspartate aminotransferase; BSR, British Society for Rheumatology; CBC, complete blood count; DMARD, disease-modifying antirheumatic drug; HCQ/CQ, hydroxychloroquine/chloroquine; LFT, liver function test; MTX, methotrexate; N/A, not available; PI, prescribing information; RA, rheumatoid arthritis; ref, reference. a Adalimumab, infliximab, etanercept, golimumab, certolizumab, abatacept, and anakinra do not currently have a laboratory monitoring program; patients receiving these medications should follow the laboratory monitoring guidelines for any coadministered medications. b Infliximab, golimumab, and rituximab are indicated for RA only in combination with methotrexate. c Glucocorticoid-associated toxicity is dependent on lifetime cumulative dose and average daily dose. d At the time of this review, monitoring guidelines for baricitinib and sarilumab have not been established.
4.1. Conventional Synthetic DMARDs
There are no specific guidelines in place for monitoring lipid levels during csDMARD therapy (Table 1). With regard to liver toxicity, the ACR recommends that, for patients receiving methotrexate, leflunomide, or sulfasalazine, liver enzymes should be measured at baseline, every 2 to 4 weeks for the first 3 months, every 8 to 12 weeks for the 3 to 6 months after initiation, and every 12 weeks thereafter . This is slightly less stringent than the guidelines given in the prescribing information for methotrexate and leflunomide, which recommend liver enzyme and function tests every 1 to 2 months throughout methotrexate therapy and monitoring of ALT levels every 6 to 8 weeks throughout leflunomide therapy [29, 30]. The BSR recommends that the frequency of ALT and/or AST monitoring during methotrexate and leflunomide therapy falls between these ranges, with the specification that after 3 months of a stable dose of methotrexate or leflunomide, the monitoring be reduced to every 12 weeks, with more frequent monitoring in patients at higher risk of toxicity [17, 18].
Because methotrexate, leflunomide, and sulfasalazine all suppress cell proliferation, routine hematologic monitoring is necessary . For these drugs, the ACR recommends complete blood counts at baseline, every 2 to 4 weeks for the first 3 months, every 8 to 12 weeks for months 3 to 6, and every 12 weeks thereafter . These guidelines are also less stringent than those given in the prescribing information for methotrexate and leflunomide, which recommend a complete blood count with differential and platelet counts at least monthly throughout methotrexate therapy and platelet and white blood cell counts every 6 to 8 weeks throughout leflunomide therapy [29, 30]. The BSR recommendations again fall between these ranges, with the specification that a full blood count be performed every 2 weeks during methotrexate therapy until the dose is stable for 6 weeks, then monthly for 3 months, and at least every 12 weeks thereafter [17, 18].
If combinations of csDMARDs are considered, most guidelines suggest following the most stringent laboratory testing among the drugs being combined.
4.2. Biologic and Targeted Synthetic DMARDs: Liver and Lipid Monitoring
Among the biologic and targeted synthetic DMARDs, tocilizumab and tofacitinib are the only ones for which specific recommendations for monitoring serum lipids and liver function are given (Table 1). According to the prescribing information, lipids (TC, triglycerides, LDL-C, and/or HDL-C) should be measured 4 to 8 weeks after initiation for both tocilizumab and tofacitinib and every 24 weeks thereafter for tocilizumab [31, 32]. For tocilizumab, ALT and AST levels should be measured 4 to 8 weeks after initiation and every 3 months thereafter. The prescribing information for tofacitinib states that there should be routine monitoring of all liver enzymes .
4.3. Biologic and Targeted Synthetic DMARDs: Neutrophil and Platelet Count Monitoring
Biologic DMARDs are often associated with transient, sustained, or late-onset decreases in neutrophils and/or platelets. The prescribing information for both tocilizumab and tofacitinib suggests monitoring of ANCs 4 to 8 weeks after initiation and every 3 months thereafter [31, 32]. For rituximab, continued monitoring of complete blood counts, including ANCs, is recommended at 2- and 4-month intervals during rituximab therapy .
4.4. Biologic and csDMARD Combination Therapy
A recent, large observational study showed that, among patients with RA initiating biologic therapy, >70% of patients initiated the biologic in combination with a csDMARD [34, 35]. Concomitant methotrexate is often part of the therapeutic regimen for patients receiving aTNF agents, as it can reduce the development of anti-drug antibodies and increase the efficacy of aTNF therapy in the treatment of RA [36–38]. In particular, it has been shown that concomitant administration of methotrexate suppresses development of anti-adalimumab and anti-infliximab antibodies, maximizing efficacy and reducing the occurrence of certain adverse drug reactions [39, 40]. Laboratory monitoring for patients receiving combination therapy should follow the guidelines for both the biologic and csDMARD(s).
4.5. Glucocorticoids
Despite conflicting information on the relationship between glucocorticoids and dyslipidemia, Liu et al. recommend regular monitoring of lipids in patients receiving glucocorticoids (including low doses) for prolonged periods or at high doses: for patients scheduled for long-term systemic corticosteroid therapy, serum lipid levels should be assessed at baseline, 1 month after glucocorticoid initiation, and then every 6 to 12 months thereafter . No recommendations are in place for monitoring of liver function, neutrophils, or platelets for patients receiving glucocorticoids.
5. Profiles of Laboratory Abnormalities and Associated Clinical Sequelae in Patients with RA
5.1. Lipids
Despite an increased risk of CV events in patients with RA, growing evidence suggests that patients with active, untreated RA have lower serum TC and LDL-C levels than the general population . Serum lipid levels in patients with RA can respond to changes in the inflammatory burden as well as to DMARD therapy and/or glucocorticoid therapy (Table 2). Consistent with this paradox in lipid levels and disease activity, achievement of reduced inflammation with DMARD therapy often results in increased serum lipid levels [20, 21].
Table 2.
Effects of nonbiologic and biologic DMARDs on lipid levels in patients with RA.
| | Cholesterol | TGs | Ref(s) |
:---: :---: |
| TC | LDL | HDL |
| Conventional synthetic DMARDs |
| Methotrexate | ↑ or = | ↑ or = | ↑ | N/A | |
| Hydroxychloroquine/chloroquine | ↓ or = | ↓ or = | ↑ or = | ↓ | |
| Sulfasalazine | ↑ | = | ↑ | N/A | |
| |
| Biologic DMARDs a |
| Adalimumab | ↑ or = | = | ↑ | = | |
| Infliximab | ↑ | ↑ | ↑ | ↑ or = | |
| Etanercept | ↓ | ↑ or ↓ | ↑ or ↓ | ↓ | [77, 78] |
| Golimumab | ↑ | ↑ | ↑ | ↑ | |
| Tocilizumab | ↑ | ↑ | = | ↑ | [31, 77] |
| Rituximab b | N/A | N/A | N/A | N/A | |
| Abatacept | ↑ or = | = | ↑ | ↑ | |
| |
| Targeted small-molecule DMARD |
| Tofacitinib | ↑ | ↑ | ↑ | N/A | [32, 81] |
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=, no change; ↑, increase; ↓, decrease; DMARD, disease-modifying antirheumatic drug; HDL, high-density lipoprotein; LDL, low-density lipoprotein; N/A, not available; RA, rheumatoid arthritis; ref, reference; TC, total cholesterol; TG, triglyceride. a No data were available for leflunomide, anakinra, or certolizumab. b Lipid and cholesterol levels were not studied in the RA rituximab clinical trials.
Most csDMARDs, including methotrexate and sulfasalazine, are associated with increased TC, LDL-C, and HDL-C levels (Table 2) [42, 43]. Hydroxychloroquine/chloroquine, however, has been reported to decrease TC and LDL-C levels, which is consistent with previously reported associations between antimalarials and favorable lipid profiles .
The degree to which biologic therapy affects serum lipid levels is complex, with differing reports; the effect of a biologic or targeted nonbiologic therapy itself on lipid levels is also complicated by the frequent administration of concurrent csDMARDs, which limits the ability to compare the effects between therapies (Table 2) [34, 35]. Tocilizumab and tofacitinib have been shown to increase TC and LDL-C and, to a lesser extent, HDL-C in patients with RA [27, 45, 46]. In a long-term (up to 4.6 years) study of pooled tocilizumab clinical trials, TC and LDL-C levels increased by week 6 and remained relatively stable at subsequent time points . The MEASURE study demonstrated that tocilizumab induced quantitative changes in lipoprotein profiles, including elevations in LDL-C, but also altered HDL particles towards an anti-inflammatory composition, whereby proinflammatory HDL-associated serum amyloid A and secretory phospholipase A2 significantly decreased from baseline during 24 weeks of tocilizumab treatment . An analysis of pooled tofacitinib clinical trials in patients with RA demonstrated dose-dependent increases in serum TC, LDL-C, and HDL-C within the first 1 to 3 months of therapy, which remained stable thereafter . A recent meta-analysis demonstrated that patients with RA treated with aTNF agents had a significant increase in HDL-C in the first 2 to 6 weeks of therapy, after which HDL-C remained stable . Despite modest increases in TC and LDL-C levels with aTNF use at 6 months, there was no significant overall effect on the atherogenic index . Overall, the trend towards elevated TC and LDL-C levels in patients with RA appears consistent among the different biologic and targeted nonbiologic therapies; however, there is no apparent associated increase in the risk of atherosclerosis, and some studies have suggested that biologic therapy actually reduces CV risk [45, 49–52].
The relationship between lipid levels and CV risk in patients with RA receiving DMARDs is not well understood. In studies that measured lipid levels following conventional synthetic, biologic, or targeted small-molecule DMARD therapy, there was no evidence that the elevated lipid levels with any of the DMARDs were associated with increased risk of CV disease, and indeed the reverse may be true: a large, prospective study of 1240 patients with RA demonstrated that methotrexate use significantly reduced the risk of CV mortality . Additionally, results from the Corrona registry demonstrated that patients with RA who received aTNF agents had a reduced risk of CV events compared with patients who received csDMARDs; however, glucocorticoid use, which may be a surrogate marker of increased inflammation, was associated with a dose-dependent increase in risk of CV events .
Glucocorticoids have varying effects on lipid levels in patients with RA. In a prospective study of 42 patients with newly diagnosed RA being treated with csDMARDs, there was no significant difference in lipid levels at 12 months between corticosteroid users and nonusers . In contrast, in patients with RA from the COBRA trial treated with sulfasalazine monotherapy or sulfasalazine plus methotrexate with or without high but rapidly tapered prednisone, HDL-C levels increased by 50%; this elevation occurred much more quickly in steroid users than nonusers (16 and 40 weeks, resp.) .
5.2. Liver Enzymes
Increases in liver aminotransferases have been noted in patients with RA following administration of DMARDs (Table 3). Among the csDMARDs, methotrexate and leflunomide are most frequently associated with elevations in liver enzymes; this association correlates with longer duration of use. In a controlled trial comparing safety and efficacy of methotrexate versus that of leflunomide in patients with RA, 999 patients were randomized to leflunomide (n = 501) or methotrexate (n = 498) and followed for 2 years . Of these patients, 25% of patients who received methotrexate and 6.4% of patients who received leflunomide had elevated ALT and AST levels > 3 × upper limit of normal (ULN) during the first year (alcohol use was not accounted for in this study). Over the 2 years, 4.2% of patients discontinued methotrexate and 1.6% discontinued leflunomide due to elevated serum liver enzymes. This is comparable to the results of a systematic literature review that demonstrated that 3.7% of patients with RA who received methotrexate discontinued due to liver toxicity (mean treatment duration of 55.8 months and mean dose of 10.5 mg/week) .
Table 3.
Proportions of patients who experienced increases in liver enzymes during RA treatment a.
| % of patients b | ≤6 months | >6 months | Clinical sequelae |
:---: :---: |
| ALT or AST > 3 × ULN | ALT or AST > 3 × ULN | |
| Conventional synthetic DMARDs |
| |
| Methotrexate | 3.6% | >2 × ULN: 12.9% 25% | 3.7% of patients discontinued due to liver toxicity 4.2% discontinued by 2 years due to liver toxicity |
| |
| Leflunomide | >5 × ULN: 3.0% | 1.5%–6.4% [30, 56] | Due to grade 2 hepatotoxicity, 1 patient (1%) was withdrawn from LEF treatment and 1 patient (1%) continued LEF at a reduced dose . 1.6% of patients discontinued by 2 years |
| |
| LEF + MTX | 6.8% | 3.8% for LEF + MTX; (0.8% for MTX + placebo) 17% | Four patients (4.5%) were withdrawn from treatment due to persistent elevation of plasma liver enzyme level |
| |
| Sulfasalazine | 4% | N/A | N/A |
| |
| Biologic DMARDs c |
| |
| Adalimumab | 3.5%d <1% (w/MTX) | <1% (w/MTX) | N/A |
| |
| Infliximab | >5 × ULN: <1%e | ≥3 × ULN: 5% (w/MTX)c | N/A |
| |
| Etanercept | Grades 2–4: 0% | Grades 2–4: 0% | N/A |
| |
| Golimumab | Any ALT or AST elevation: 0% (monotherapy); 4.4%–6.3% (w/MTX) | N/A | N/A |
| |
| Tocilizumab | 3.4%–6.5% [31, 87] | 11% (TCZ + MTX); 2%-3% (MTX → TCZ) 10.8% | There were no serious liver function disorders during TCZ treatment . Seven patients (1.1%) prematurely discontinued due to elevated liver aminotransferases (with concomitant MTX) |
| |
| Anakinra | N/A | Patients with elevated liver enzymes: <1% | N/A |
| |
| Targeted small-molecule DMARD |
| |
| Tofacitinib | 1.1%–1.6% (w/MTX) ≤1.0% | <1% 3.2% (w/MTX) | N/A |
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ALT, alanine aminotransferase; AST, aspartate aminotransferase; DMARD, disease-modifying antirheumatic drug; LEF, leflunomide; MTX, methotrexate; N/A, not available; RA, rheumatoid arthritis; TCZ, tocilizumab; ULN, upper limit of normal. a For ALT and AST levels, grade 3 was >5 to 10 × ULN and grade 4 was >10 × ULN. b Data are the proportions of patients with ALT or AST ≥ 3 × ULN except where noted. c No data were available for rituximab, certolizumab, or abatacept. d Many of these patients were also taking methotrexate or nonsteroidal anti-inflammatory drugs. e Company medical letter (personal communication), data on file.
Most published reports of elevated liver enzymes in patients receiving biologic or targeted synthetic DMARDs have involved patients who received concomitant methotrexate or leflunomide. In studies of patients who received biologic DMARDs, higher percentages of patients had elevations in AST or ALT with coadministration of methotrexate than with administration of biologic monotherapy (Table 3). For example, among patients who received golimumab + methotrexate combination therapy in a phase 3 trial, 4% to 6% had elevations in ALT and AST, respectively, whereas, among patients who received golimumab monotherapy, none had any elevation in AST or ALT . In the tocilizumab ACT-RAY trial, 7.7% of patients who received tocilizumab + methotrexate experienced ALT > 3 × ULN by 6 months, whereas 1.2% of patients who received tocilizumab monotherapy experienced ALT > 3 × ULN . In the tofacitinib monotherapy trials, there was no difference in the proportion of patients experiencing increases in liver enzymes > 3 × ULN among those who received tofacitinib monotherapy compared with those who received placebo .
5.3. Neutrophils
Neutropenia is a common adverse event in patients receiving treatment for RA, with the main complication being infection due to bacteria and/or fungi . Mild to moderate decreases in ANCs are often associated with DMARD therapy. With the exception of rituximab, acquired neutropenia in patients with RA receiving biologic or csDMARD therapy is generally transient, with ANCs recovering within a few days of treatment and rarely leading to infection.
Reports of grade 3 or 4 neutropenia (based on the Common Terminology Criteria for Adverse Events) in patients who received csDMARDs are scarce. For clinical trials in which neutropenia was reported, the proportions of patients experiencing neutropenia are shown in Table 4. In a recent RA clinical trial with a methotrexate monotherapy arm, 9.8% of patients who received methotrexate monotherapy experienced grade 1 or 2 neutropenia (grades 1 and 2 are described as mild and moderate, resp.), but only 0.4% experienced grade 3 neutropenia (described as severe or medically significant but not immediately life-threatening) and 0% experienced grade 4 neutropenia (described as life-threatening or disabling) . In a study of patients with juvenile RA who received methotrexate or leflunomide, <1% experienced grade 3 or 4 decreases in neutrophil counts by week 16 .
Table 4.
Proportions of patients who experienced neutropenia during RA treatment in clinical trials a.
| % of patients b | ≤6 months | >6 months | Clinical sequelae |
:---: :---: |
| Grade 3 or 4 | Grade 3 or 4 |
| Conventional synthetic DMARDs c |
| |
| Methotrexate | 0.4% | N/A | N/A |
| |
| Leflunomide | Grade 4: 0% | N/A | N/A |
| |
| Sulfasalazine | N/A | 0% | N/A |
| |
| Biologic DMARDs c |
| |
| Adalimumab | Grade 2 or 3: <1.0% | ≤1.0% | N/A |
| |
| Infliximab | N/A | Grade 2, 3, or 4: 1.7%–8.8%d | N/A |
| |
| Etanercept | N/A | 0% | N/A |
| |
| Golimumab | ≥1 abnormal value: 2.3%e | N/A | N/A |
| |
| Tocilizumab | Grade 3:4.8% Grade 4:<1.0% | N/A | One patient experienced a serious infection of empyema temporally associated with grade 3 neutropenia. None of the patients with grade 4 neutropenia experienced serious infection within 30 days of observed neutropenia |
| |
| Rituximab | LON: 4.6%–6% [64, 65] | N/A | LON was defined as an ANC < 1.0 × 10 9/L occurring 4 weeks after the last rituximab infusion. Most cases resolved spontaneously or following administration of G-CSF or GM-CSF. Febrile neutropenia and neutropenia-related infections requiring hospitalization have been observed |
| |
| Abatacept | N/A | N/A | N/A |
| |
| Anakinra | 0.4% | 0.5% | While neutropenic, 1 patient developed cellulitis. In most cases, grade 3 or 4 toxic events were sporadic and were not associated with progressive decreases nor were indicative of drug-related impairment |
| |
| Targeted small-molecule DMARDs |
| |
| Tofacitinib | <1% Grade 2 or 3: 0.9%–3.1% [60, 91, 92] Grade 4: 0% | Grade 2 or 3: 1.3% | N/A |
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ANC, absolute neutrophil count; DMARD, disease-modifying antirheumatic drug; G-CSF, granulocyte colony-stimulating factor; GM-CSF, granulocyte-macrophage colony-stimulating factor; LON, late-onset neutropenia; N/A, not available; RA, rheumatoid arthritis; TCZ, tocilizumab. a Neutropenia grades were defined as follows: grade 2, ≥1000 to <1500 cells/mm 3; grade 3, ≥500 to <1000 cells/mm 3; grade 4, <500 cells/mm 3. b Data are the proportions of patients experiencing grade 3 or 4 ANCs except where noted. c Data were not available for hydroxychloroquine/chloroquine or certolizumab. d From a cohort of patients with polyarticular juvenile idiopathic arthritis who received infliximab plus methotrexate. e Company medical letter (personal communication), data on file.
Among studies reporting ANCs following administration of biologic DMARDs, <1% of patients who received adalimumab or etanercept in combination with methotrexate were reported to experience grade 3 or 4 decreases (Table 4) [62, 63]. In the pooled long-term (up to 4.6 years) safety analysis of the tocilizumab RA trials, 4.8% of patients experienced grade 3 decreases in ANCs, and <1% of patients (14 out of 4009 observed patients) experienced grade 4 decreases in ANCs; 1 patient experienced a serious infection of empyema temporally associated with grade 3 neutropenia, and no patients with grade 4 neutropenia experienced serious infection within 30 days of observed neutropenia . In a Phase III clinical trial, 1.1% to 1.6% of patients who received tofacitinib experienced grade 2 or 3 neutropenia at month 3 . Reports of neutropenia in patients with autoimmune disease treated with rituximab are infrequent; however, there have been reports of late-onset neutropenia in 1.3% to 5.8% of patients with RA who received rituximab, observed at medians of 21 to 23 weeks, with variable incidences of associated infections [64–66]. Thus, the relationship between ANC and infection risk may differ among various biologics.
5.4. Platelets
In addition to causing decreases in ANCs, DMARD treatment can lead to a transient decrease in platelet counts that generally recovers after 1 week . In clinical trials, <1% of patients have been reported to experience grade 3 or higher (severe/medically significant or worse) decreases in platelet counts with conventional synthetic, biologic, or targeted small-molecule DMARDs (Table 5).
Table 5.
Proportions of patients who experienced decreases in platelet counts during RA treatment in clinical trials a.
| % of patients b | ≤6 months | >6 months | Clinical sequelae |
:---: :---: |
| Grade 3 or 4 | Grade 3 or 4 |
| Conventional synthetic DMARDs c |
| |
| Methotrexate | 0%–1.3%[61, 67, 95] | N/A | N/A |
| |
| Leflunomide | 0% | 0%d | N/A |
| |
| Biologic DMARDs c |
| |
| Adalimumab | 0.2%e | N/A | N/A |
| |
| Infliximab | N/A | 0%d | N/A |
| |
| Golimumab | ≥1 abnormal value: 0.2%b | N/A | N/A |
| |
| Tocilizumab | 0.2% | <1.0% | One serious bleeding event of hemorrhagic stomatitis occurred in a patient with grade 4 thrombocytopenia |
| |
| Rituximab | N/A | N/A | N/A |
| |
| Abatacept | N/A | N/A | N/A |
| |
| Anakinra | 0% | 0% | N/A |
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DMARD, disease-modifying antirheumatic drug; N/A, not available; RA, rheumatoid arthritis; a Thrombocytopenia grades were defined as follows: grade 2, 50,000 to <75,000 cells/mm 3; grade 3, 25,000 to <50,000 cells/mm 3; grade 4, <25,000 cells/mm 3. b Data are the proportions of patients experiencing grade 3 or 4 platelet counts except where noted. c No data were available for sulfasalazine, hydroxychloroquine/chloroquine, certolizumab, etanercept, or tofacitinib. d Patients received leflunomide plus infliximab. e Company medical letter (personal communication), data on file.
5.5. Serum Creatinine
For monitoring and assessment of kidney function, serum creatinine may be measured as opposed to creatinine clearance. The prescribing information for cyclosporine and tacrolimus both notes elevated serum creatinine following administration and recommends close monitoring of renal function. Among the DMARDs discussed here, only tofacitinib has prescribing information that reports drug-associated increases in serum creatinine levels. The mean increase in serum creatinine in patients treated with tofacitinib in clinical trials was <0.1 mg/dL over 12 months of treatment. In the long-term extensions, however, up to 2% of patients discontinued tofacitinib due to an increase in creatinine > 50% above baseline. The clinical significance of the observed serum creatinine elevations is unknown .
6. Discussion
This review is among the first to comprehensively examine the differences in laboratory monitoring recommendations for each DMARD indicated for the treatment of RA. In particular, guidelines for monitoring serum lipids, liver aminotransferases, serum creatinine, and neutrophil and platelet counts were summarized. Information was gathered from the prescribing information for each drug and from ACR, BSR, and EULAR recommendations. Furthermore, an overview of available information on the laboratory abnormality profiles associated with each drug indicated for RA was given.
Regardless of the choice of treatment, good clinical practice dictates that routine laboratory monitoring for patients with RA is important. Due to the increased risk of CV disease, it is recommended that patients with RA have at least yearly monitoring of serum lipids . In addition, due to the high prevalence of renal impairment in patients with RA, once-yearly monitoring of kidney function is also advised .
RA therapies have the potential to exacerbate CV and renal comorbidities; in addition, many drugs indicated for RA have inherent risks associated with them. Although it is recommended that patients with RA be monitored for CV risk (including serum lipid levels) at least yearly, serum lipid levels should be assessed more frequently in those receiving tocilizumab or tofacitinib [31, 32]. When determining an appropriate monitoring regimen, consideration should be given to the observation that high-grade inflammation in patients with active RA is associated with reductions in the levels of TC, HDL-C, and LDL-C; this is seen in other inflammatory states (e.g., autoimmune and sepsis) and is at least partially reversible with anti-inflammatory treatment . Whereas levels of serum lipids tend to predict CV disease risk in the general population, an opposite relationship has been observed in patients with RA: a 2011 retrospective cohort study showed that, in patients with RA, lower levels of TC and LDL-C were associated with increased risk of CV disease . Inflammation itself is associated with lower lipid levels, and therefore baseline levels (prior to treatment) may be relatively low, and although lipid levels may increase with DMARD treatment, they often stay within the acceptable reference range. Because of this relationship, the increases in lipids associated with particular biologic DMARDs may actually be a reflection of their efficacy in reducing inflammation, and the use of statins to reduce the associated increase in cholesterol is recommended.
Liver toxicity can be a significant problem with long-term use of methotrexate or leflunomide, warranting frequent monitoring of liver function, particularly ALT and AST levels. Although it may appear that, due to elevations in lipid levels, the recommendations for patients receiving tocilizumab suggest more frequent laboratory monitoring than those for patients receiving aTNF therapy, this is not necessarily true. The aTNF agents are usually administered in combination with methotrexate to reduce immunogenicity and increase efficacy, and infliximab and golimumab are indicated for patients with RA only when administered in combination with methotrexate. In these patients, the monitoring frequency that is recommended for liver toxicity or decreases in ANC due to methotrexate is advised (Table 6).
Table 6.
Summary of recommended frequencies of laboratory monitoring for patients with RA receiving DMARDs a.
| | Lipids | AST and ALT | Neutrophils and platelets |
:---: :---: |
| MTX, LEF, SSZ | — | Initially: every 2–4 weeks After ~1–3 months: every 1–3 months After ~6–2 months: every 3 months or based on clinical judgment | Initially: every 2–4 weeks After ~1–3 months: every 1–3 months After ~6–12 months: every 3 months or based on clinical judgment |
| |
| GLU | Initially: 1 month after initiation After 1 month: every 6–12 months | — | — |
| |
| TCZ | Initially: 4–8 weeks after initiation After 1–2 months: every 6 months | Initially: 4–8 weeks after initiation After 1-2 months: every 3 months | Initially: 4–8 weeks after initiation After 1-2 months: every 3 months |
| |
| TOF | Initially: 4–8 weeks after initiation | Routine | Initially: 4–8 weeks after initiation After 1-2 months: every 3 months |
| |
| RTX | — | — | Every 2-3 months |
| |
| aTNF | — | aTNFs administered in combination with MTX should follow the MTX monitoring guidelines | aTNFs administered in combination with MTX should follow the MTX monitoring guidelines |
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ALT, alanine aminotransferase; AST, aspartate aminotransferase; aTNF, anti-tumor necrosis factor agent; DMARD, disease-modifying antirheumatic drug; GLU, glucocorticoid; LEF, leflunomide; MTX, methotrexate; RA, rheumatoid arthritis; RTX, rituximab; SSZ, sulfasalazine; TCZ, tocilizumab; TOF, tofacitinib. a Infliximab, golimumab, and rituximab are indicated for RA only when administered in combination with MTX. Monitoring frequency should follow that of the recommendations for MTX.
Although high-dose glucocorticoid treatment for short-term duration is generally considered to be safe, glucocorticoid-associated toxicity is related to both the average daily dose and the lifetime cumulative dose. In a chronic disease such as RA, the frequent use of even low-dose glucocorticoids over long periods commonly reaches cumulative dose thresholds that predispose patients to increased risk of adverse events and mortality [41, 71, 72]. The risks associated with approaching the cumulative glucocorticoid dose thresholds in patients who experience an increase in disease activity (“flare”) while receiving a biologic DMARD should be weighed when considering addition of chronic low-dose glucocorticoid treatment versus a short-term course of high-dose glucocorticoids (dose and duration). It may be prudent for clinicians to keep a record of their patients' cumulative glucocorticoid dose over time.
While this review provides a comprehensive overview of the laboratory monitoring recommendations for drugs indicated for RA, there are some important limitations. All potential guidelines were not reviewed (i.e., liver and renal subspecialty guidelines), which may add to the diversity of recommendations. Neither a systematic literature review nor a meta-analysis was done; this is a descriptive article, as we were interested in those recommendations most familiar to rheumatologists. Further, recommendations were made based on our personal opinions and our reviews to help readers, but each reader is encouraged to reach their own considered opinion. The overview and comparison of guidelines presented here may help to inform and assist physicians in the choices of treatment based on the necessity of laboratory monitoring. Ultimately, what is best for the patient regarding the recommendations and individual patient-risk factors must be considered in all cases.
In closing, the frequency of drug-related toxicities associated with RA therapy is low and therefore easily managed in the vast majority of patients. Although screening guidelines exist, clear evidence of the benefit of monitoring in preventing harm to the patient is lacking. Moreover, does the prior experience of the patient shape the need for laboratory monitoring? For instance, if a patient is stably controlled on methotrexate for 2 years with testing every 2 to 3 months, do they still require that frequency of testing? If not, what frequency is required? In the United States, for a patient with limited means and a high deductible, lab work alone is likely to constitute their major out-of-pocket expense. As a result, it is often difficult to get these patients to agree to be tested more than once or twice a year. Similarly, if someone is on etanercept monotherapy without signs of infection, what frequency of lab testing, if any, is required? While guidelines exist, it is not clear that their rigor has been amply demonstrated. These questions are ones that we await answers to in the coming years.
Acknowledgments
This study was funded by F. Hoffmann-La Roche/Genentech, Inc. Support for third-party writing assistance, furnished by Ellen Mercado, Ph.D., of Health Interactions, Inc., was provided by F. Hoffmann-La Roche/Genentech, Inc.
Conflicts of Interest
William F. C. Rigby has acted as a consultant for Roche/Genentech; Daniel E. Furst has received grant/research support from AbbVie, Actelion, Amgen, BMS, Novartis, Pfizer, and Roche/Genentech and has acted as a consultant for AbbVie, Actelion, Amgen, BMS, Cytori, Novartis, Pfizer, and Roche/Genentech. Jason M. Low is an employee of Genentech, Inc. Kathy Lampl was an employee of Genentech, Inc., during the drafting of this manuscript.
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Abstract
1. Introduction
2. Methods
3. Overview of Good Clinical Practice Recommendations for Laboratory Monitoring in Patients With RA
4. Monitoring Guidelines for Conventional Synthetic, Biologic, and Targeted Small-Molecule DMARDs in Patients with RA
5. Profiles of Laboratory Abnormalities and Associated Clinical Sequelae in Patients with RA
6. Discussion
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Chapters and Articles
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Pneumomediastinum
Presentation
•
: Most common: Chest pain, cough, shortness of breath
•
: ± crepitus in neck or chest (from subcutaneous emphysema)
○
: “Hamman sign”: Precordial systolic crepitation on auscultation; variably present
•
: Secondary PM likely to present with rib fracture, pneumothorax, hemothorax, intracranial injury, respiratory distress, tachycardia
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URL:
Chapter
Pneumomediastinum
2017, Diagnostic Imaging: Pediatrics (Third Edition)
CLINICAL ISSUES
Presentation
•
: Most common: Chest pain, cough, shortness of breath
•
: ± crepitus in neck or chest (from subcutaneous emphysema)
○
: “Hamman sign”: Precordial systolic crepitation on auscultation; variably present
•
: Secondary PM likely to present with rib fracture, pneumothorax, hemothorax, intracranial injury, respiratory distress, tachycardia
Demographics
•
: Spontaneous & secondary PM more common in males
Natural History & Prognosis
•
: Spontaneous PM: Self-limited, recurrence rare
○
: Morbidity related to underlying pulmonary condition
•
: Secondary PM: Mortality related to other injuries
•
: Complications uncommon, can include pseudotamponade, mediastinitis, pneumothorax
Treatment
•
: Isolated spontaneous PM in stable patient
○
: Supportive care, emergency department observation
○
: No evidence to support further studies
•
: Spontaneous PM with symptom progression
○
: Evaluate & treat underlying pulmonary condition
•
: Secondary PM
○
: CT or esophagram if underlying aerodigestive tract injury specifically suspected
○
: Treat underlying condition
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Read full chapter
URL:
Book2017, Diagnostic Imaging: Pediatrics (Third Edition)
Review article
On-the-Field Emergencies
2023, Clinics in Sports MedicineAlexander J. Tomesch MD, CAQ-SM, ... Olivia Keller-Baruch MD
Clinics care points
•
: PM is a rare diagnosis and even less common in athletics
•
: Even if occurring in a collision sport, the cause is likely secondary to a spontaneous PM
•
: Common symptoms include subcutaneous emphysema, substernal chest pain, dyspnea, odynophagia, and cough
•
: Hamman sign is most specific, however, only occurs 12% to 5% of the time
•
: EKG changes can occur secondary to PM
•
: Treatment is primarily supportive care
○
: Oxygen, antibiotics, and dietary changes have been suggested, but there are limited to no data to support this
•
: Return to sports and air travel is again limited to expert opinion, but athletes can likely start back into activities following the resolution of symptoms
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Journal2023, Clinics in Sports MedicineAlexander J. Tomesch MD, CAQ-SM, ... Olivia Keller-Baruch MD
Chapter
Boerhaave's Syndrome
2004, Encyclopedia of GastroenterologyMakau Lee
Glossary
: Spontaneous, transmural tear of the esophagus, with free perforation.
Hamman sign
: Mediastinal crunching sound with heartbeat.
: Infection involving the mediastinum.
: Pain on swallowing foods or liquids.
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URL:
Reference work2004, Encyclopedia of GastroenterologyMakau Lee
Chapter
Boerhaave's Syndrome
2004, Encyclopedia of GastroenterologyMakau Lee
Diagnosis
Clinically, patients with Boerhaave's syndrome may present with chest, neck, and abdominal pain and odynophagia, dysphagia, hoarseness, aphonia, vomiting, hematemesis, and respiratory distress. Physical examination may reveal subcutaneous crepitation, mediastinal crunching sound with the heartbeat (also called Hamman sign), fever, and shock. Of note, only one-third of patients with Boerhaave's syndrome present with the classic Mackler triad of chest pain, vomiting, and subcutaneous emphysema. Leukocytosis is common. Radiographic findings include pleural effusion, mediastinal widening, hydropneumothorax, and pneumomediastinum. Esophagrams performed with water-soluble contrast medium (such as gastrograffin) are generally diagnostic. The differential diagnosis should include Mallory–Weiss tear, esophageal intramural hematoma, peptic ulcer disease and its complications (such as bleeding and perforation), aortic dissection, myocardial infarction, pericarditis, pulmonary embolism, spontaneous pneumothorax, and pancreatitis.
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Reference work2004, Encyclopedia of GastroenterologyMakau Lee
Chapter
Esophageal Trauma
2009, Parkland Trauma Handbook (Third Edition)Richard Hershberger MD, Stacey Woodruff MD
II Evaluation
A
: Signs and symptoms
1
: Present in 60% to 80% of injuries
2
: Signs and symptoms depend on location of injury, size of the defect, extent of contamination, and length of time from injury.
3
: Odynophagia
4
: Dysphagia
5
: Hematemesis
6
: Hemoptysis
7
: Cervical crepitus
8
: Prevertebral air
9
: Widened mediastinum
10
: Hoarseness
11
: Dyspnea
12
: Audible mediastinal crunching (Hamman's sign)
B
: Radiologic assessment
1
: Plain films aid in discerning trajectory of missiles.
2
: Lateral radiograph of the neck
a
: Signs that aid in diagnosis include prevertebral air, subcutaneous emphysema, mediastinal air, and mediastinal widening.
b
: Prevertebral air occurs in 39% of esophageal injuries
C
: Esophagography
1
: Sensitivity of 93%
2
: Gastrografin is used first, unless a tracheoesophageal fistula is suspected.
3
: Negative study should prompt further evaluation if clinical suspicion for injury remains high.
D
: Esophagoscopy
1
: When combined with esophagography a sensitivity of almost 100% is reached.
2
: Can be flexible or rigid
a
: Air used for insufflation with flexible scopes can worsen pneumomediastinum or pneumothorax.
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Book2009, Parkland Trauma Handbook (Third Edition)Richard Hershberger MD, Stacey Woodruff MD
Chapter
Pneumothorax and Pneumomediastinum
2007, Comprehensive Pediatric Hospital MedicineJoshua Nagler
Clinical Presentation
Most patients with spontaneous pneumomediastinum are asymptomatic. In those with symptoms, chest pain is the most common complaint. The pain is usually pleuritic and retrosternal and often radiates to the back, shoulder, or arms. Some patients will note dyspnea, dysphagia, or neck pain. Occasionally, patients will notice skin changes attributable to subcutaneous emphysema. More commonly, the presenting symptoms are related to the underlying illness (e.g., asthma) and do not result from the air collection in the mediastinal cavity.
On physical examination, precordial crepitus associated with systole, known as the Hamman sign, is pathognomonic.15 Other possible findings include palpable subcutaneous emphysema, neck swelling or torticollis, and in more severe cases, respiratory distress.
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URL:
Book2007, Comprehensive Pediatric Hospital MedicineJoshua Nagler
Review article
Correlation between Murmurs and Echocardiographic Findings; From an Imaging Cardiologist Point of View
2023, Current Problems in CardiologySahrai Saeed, ... Stig Urheim
Methods
The electronic database PubMed was systematically searched for relevant original studies and review articles in English. The keywords “Auscultation”, “Heart valve disease”, “Murmur” and “Transthoracic echocardiography” were used.
Normal Heart Sounds and Murmurs
A normal beating heart makes 2 sounds commonly characterized as “lub-dupp”. “Lub” is the first sound produced by the closure of mitral and tricuspid valves and “Dupp” or “Dub” is the second sound caused by the closure of aortic and pulmonary valves. A systematic auscultation should include 4 standard points, and the intermediate positions between these points with the patient at 45°, and at the left sternal edge while patient sitting forward, and at the apex and axilla lying on the left side.5 Carotid arteries should also be gently auscultated on both sides and primary stenotic sounds due to carotid artery stenosis should be differentiated from the heart murmurs (primarily due to aortic valve disease) radiating towards carotid artery sites.
Murmurs are unusual cardiac sounds caused by turbulent blood flow, forced through a narrowed heart valve (stenosis), intraventricular or arterial obstruction, or leaking backward through the valve when it should be closed (regurgitation or insufficiency). Murmurs are either physiological/benign or pathological (Fig 1 and Table 1). Some heart murmurs are innocent, commonly found in children, and may not require treatment or strict follow-up. In most cases, these murmurs disappear with age. Innocent murmurs can also be present in adults and an echocardiogram may not always clarify their etiology. Benign murmurs in healthy hearts are common during pregnancy, fever and in some other non-cardiac conditions such anemia, polycythemia, liver cirrhosis and thyrotoxicosis. These murmurs are often changeable in character, intensity and over the course of the day/night. A careful TTE, if needed combined with change in posture and/or physical maneuvers (Valsalva, leg rising, etc) is required to reveal the cause of less severe murmurs. By contrast, pathological murmurs are caused by valvular stenosis or regurgitation, congenital heart disease (coarctation of aorta, patent ductus arteriosus, ventricular septal defect, atrial septal defect with subsequent increased pulmonary flow), rheumatic heart disease or infective endocarditis. Generally, systolic murmurs are caused by aortic and pulmonary valve stenosis, and mitral and tricuspid valve regurgitation, and diastolic murmurs include aortic and pulmonary regurgitation murmurs and mitral and tricuspid valve rumbles.
Table 1. Auscultatory characteristics/description of cardiac murmurs.
| Description | Criteria |
--- |
| Phase | • Early, mid or late systolic/diastolic, holosystolic • Continuous “machine-like” murmurs (PDA) • Mid-ejection click – late systolic murmur (Barlow) |
| Location | • Aortic valve: 2nd intercostal space at the RSB • Pulmonary valve: 2nd intercostal space at the LSB • Tricuspid valve: 4th intercostal space at the LSB • Mitral valve: 5th intercostal space at the left midclavicular line |
| Radiation | • Murmurs may radiate and can be audible at remote locations such as axilla, carotids, back (coarctation of the aorta) and epigastrium |
| Intensity | • Grade I: hardly audible (faint murmur) • Grade II: soft murmur • Grade III: easily audible but without a palpable thrill • Grade IV: easily audible with a palpable thrill • Grade V: loud murmur, audible with stethoscope lightly touching the chest • Grade VI: loudest murmur, audible with stethoscope not touching the chest |
| Pitch | • High or low frequency |
| Characteristics | • Blowing, harsh, musical, rumbling, squeaky • Loud ejection systolic murmur (aortic stenosis) • Normal splitting of the second heart sound (S2) • Second sound (S2): Preserved or silent (severe aortic stenosis) |
| Profile | • Crescendo: a murmur increasing in intensity • Decrescendo: a murmur decreasing in intensity • Crescendo-decrescendo: a murmur first increasing in intensity, reaches peak, and then declines in intensity • Plateau: stable intensity |
PDA, patent ductus arteriosus; RSB, right sternal border; LSB, left sternal border.
A newly detected murmur in a patient with acute myocardial infarction may be an alarming sign for serious complications, such as acute mitral regurgitation due to papillary muscle/chordae ruptur or ventricular septal rupture. Similarly, a new murmur in a patient with infective endocarditis may indicate valvular dysfunction or structural valve deterioration in patients with a prosthetic heart valve. Acute aortic syndromes, particularly aortic dissection type A and aortic aneurysm, may be associated with aortic regurgitation (AR) and diastolic murmur. In these life-threatening situations an urgent echocardiogram should be arranged to clarify the etiology of murmur and guide further treatment. Furthermore, a loud systolic murmur at the aortic area of the chest in elderly patients admitted with new syncope should immediately rise the suspicion of aortic stenosis (AS) and prompt a standard TTE.
In some young female patients with milder form of VHD or congenital heart disease, a relatively rapid increase in the intensity of murmur during pregnancy may suggest load-dependent worsening of the VHD, which is an indication for extra clinical and echocardiographic control to assess the hemodynamic significance of the associated VHD.
Murmurs in Aortic Valve Disease
After mitral regurgitation, AS is the most prevalent form of VHD caused by either progressive degeneration/calcification of a trileaflet aortic valve (TAV) or bicuspid aortic valve (BAV).10 A number of predictors of rapid hemodynamic progression including male sex, smoking, dyslipidemia, diabetes mellitus, hypertension, chronic kidney disease and coronary artery disease have been identified.11 In Western countries, degenerative AS is highly prevalent, while in South Asian and African countries, rheumatic heart disease is a common cause of VHD. The systolic murmur in AS has a crescendo-decrescendo characteristic, best heard along the left sternal border radiating to the upper right sternal border and carotid arteries, or in some cases to the left ventricle (LV) apex mistaken with mitral regurgitation murmur. During the progression of AS severity, the duration of the murmur increases, peaking at mid-to-late systole. However, the intensity of the murmur does not always correspond to the severity of AS. Despite a similar peak aortic jet velocity and severity of AS, the grade of murmur may vary from patient to patient (Fig 2–3). Heart rate and rhythm (particular rapid atrial fibrillation), stroke volume and systemic hypertension all may possibly affect the character and intensity of heart murmurs. Furthermore, LV systolic function can affect the intensity/severity of murmur in AS. For example in patients with severely reduced LV ejection fraction and advanced stage of low flow low gradient AS with reduced ejection fraction, the intensity of systolic murmur may be diminished. Finally, in patients who have smaller LV chambers, proximal septum hypertrophy and a paradoxical low flow, low gradient AS with preserved LV ejection fraction, a loud systolic murmur may be heard both at the aortic area and on the entire precordium (Fig 4). In these cases, subvalvular flow acceleration may augment the intensity and duration of systolic murmurs with late peaking.
Compared to systolic murmurs, diastolic murmurs in AR are more difficult to hear and require careful auscultatory skills. The severity of chronic AR is often difficult to assess by auscultation. Diastolic murmur starts with the second heart sound and is decrescendo, usually described as high-frequency blowing sounds. In sinus tachycardia or other tachyarrhythmias in which the duration of diastole is shortened, it may be even more difficult to hear an AR-related diastolic murmur, or it may be mistaken with AS murmurs because of shortening of the diastole or equalization of diastolic phase with systolic phase. A wide pulse pressure and typical bounding peripheral pulses may help confirm the diagnosis of aortic incompetence. Furthermore, in severe AR an Austin Flint rumble, a low-pitched mid-to-late diastolic rumble, may be best heard at the apex of the heart. The regurgitant jet in AR can cause fluttering/vibration of the anterior mitral valve leaflet and keeps it in partial closure state (functional mitral stenosis),12 leading to anterior mitral valve leaflet remodeling in the longer term.13 Furthermore, a diastolic mitral regurgitation due to elevated LV end-diastolic pressure (rapid filling by both regurgitant volume and antegrade mitral flow) may also indicate a severe AR – and often acute. Overall, the character and intensity of diastolic murmurs in patients with hemodynamically significant AR on TTE may vary. In a patient with severe AR, the diastolic murmur may be absent (Fig 5), while in other patients with less severe AR by echocardiography, the diastolic murmur may be easily heard on auscultation (Fig 6).
Other signs associated with AR which should be noted during clinical examination are “de Musset's sign” in which the head of the patient nods, together with “Traube's sign” or a “pistol shot” with the pulse auscultation over the femoral arteries.14 In severe AR, the nail beds may also show systolic capillary pulsations upon light compression, referred to as Quincke's sign.
Murmurs in Mitral and Tricuspid Valve Disease
The auscultatory findings are low- to medium-pitch diastolic rumbles (mid-diastolic murmur, often decrescendo) at the apex, heard best in left lateral decubitus position. The sound is caused by the turbulence of blood flow across the stenotic valve as the left atrium contracts during diastole (Fig 7). There is also often a loud first heart sound accompanied by an opening snap. The murmur in mitral regurgitation is often soft holosystolic heard best at the apex usually radiating to axilla or forward depending on the valve morphology. In rheumatic heart disease affecting the mitral valve, the valve leaflets are thickened with restricted motion both in systole and diastole, leading to both stenosis and regurgitation (Carpentier type 3A), and hence both systolic and diastolic murmurs may be audible. The anterior mitral valve leaflet may show a typically “Hockey stick” morphology in mitral stenosis. Other signs of pulmonary hypertension may also be present in patients with moderate to severe mitral stenosis.
A "mitral honk" murmur has been described in several conditions,15 including tricuspid and mitral valve prolapse, but it is more commonly described in the setting of mitral valve prolapse. The intensity of a prolapse murmur typically increases with change in position, inhalation, or Valsalva maneuver; that is manipulation of preload with resultant change in LV size. In Barlow's disease with mitral valve prolapse and regurgitation, a typical “late systolic murmur-mid ejection click” due to progressive superior shift of mitral valve coaptation line toward the left atrium during systole is described (Fig 8). Furthermore, a so-called “curling” movement which is an unusual systolic motion of the posterior mitral ring and the adjacent myocardium, may contribute to an abrupt systolic click.16 However, this phenomenon is still poorly understood and need further exploration.
Respiratory changes in venous return to the heart affect the character and intensity of mitral and tricuspid regurgitation rumbles. A decrease in venous return to the left side of the heart during inspiration may diminish the intensity of the mitral murmurs, whereas an increase in venous return during expiration can have the opposite effect. This makes it often difficult hear a mild/soft systolic murmur in younger patients with otherwise healthy heart (Fig 9). A septal bounce in constrictive pericarditis is another example of an exaggerated response to respiration-related inflow swing.17
Infective endocarditis, rheumatic heart disease, right atrial myxoma, carcinoid syndrome and Ebsteins’ anomaly are causes of tricuspid valve involvement and dysfunction leading to either regurgitation or stenosis. However, in our routine clinical practice we experience that systolic murmurs in tricuspid regurgitation are often difficult to hear (Fig 10).
Murmurs in Hypertrophic Cardiomyopathy
In patients with hypertrophic cardiomyopathy, either obstructive or non-obstructive phenotypes, a number of mitral leaflet abnormalities has been related to the mechanism of mitral systolic anterior motion (SAM), which causes both subaortic obstruction and mitral regurgitation. The obstruction typically causes a loud systolic ejection murmur at the left sternal edge, radiating to the right upper sternal edge and apex.18 The intensity of the murmur increases with reduction in preload (from the squatting to the standing position or during the strain phase of the Valsalva maneuver) or afterload. Most patients with obstruction also have a mitral regurgitation.18 Furthermore, a SAM-associated sound has often dynamic changes, and its assessment on auscultation is important to understand the distinct hemodynamic features of hypertrophic obstructive cardiomyopathy even in the era of advanced multimodality imaging. Of note, in patients with hypertrophic cardiomyopathy, a heart murmur or electrocardiographic abnormality (often T-wave inversion) prompt to echocardiographic or cardiac MR evaluation, and hence the correct diagnosis.
Finally, pericardial friction rub and pleural rub should be distinguished from cardiac murmurs. The former is caused by pericarditis and is often synchronous with heart rhythm, while the latter is caused by pleuritis and is synchronous with respiration. Noisy pneumothorax is another condition associated with sounds described in the literature as bubbling, scraping, clicking or crunching sounds.19 initially described by Louis Hamman in late 1930s (also called Hamman's sign) in a case of pneumomediastinum.20
Current Practice of Referring Individuals With Murmurs for Echocardiogram
Hospital doctors, particularly those who work in medical wards, are generally expected to better describe the grade, severity and possible etiology of murmurs, and may be therefore able to distinguish age-related mild expected murmurs (aortic sclerosis in the elderly) from more severe symptomatic murmurs suggesting significant VHD or complications in acute coronary artery syndromes. However, auscultation skills are dependent on the overall clinical experience. Some in-patients scheduled for surgery are referred for echocardiogram by anesthesiologists when a murmur is heard during preoperative clinical assessment. Knowing the cause of murmur and severity of underlying VHD is important regarding their implications for per-operative risk stratification and the course of post-operative recovery.
GPs may have a tendency to refer patients to a cardiologist for echocardiogram rather than provide a more in-depth description of the murmur. In a previous study, it was demonstrated that auscultation by GPs had a low sensitivity of only 44% and specificity of 69% for the detection of significant VHD on TTE.8 A possible explanation for this observation was that GPs might not have enough time to undertake a thorough auscultation due to their long patient lists.21-23 However, it is also known that some GPs may not routinely perform cardiac auscultation in their patients.20-21 In a 2015 European survey on 8860 people aged ≥60 years, more than half of the respondents (54.2%) reported that their GPs only occasionally or never used a stethoscope to check their heart sounds for a murmur, which in turn contributes to the underappreciation of VHD.21 A repeated VHD awareness survey in 2017 including 12820 people ≥60 years in 11 European countries aimed to re-evaluate the concerns and knowledge about VHD after 2 years.22 The survey showed that nearly 51% of the respondents still reported that they rarely or never underwent auscultation of their hearts by their GPs compared with 54% in 2015 (P<0.001). However, it is also important to highlight that some GPs tend to routinely perform serial assessment of the murmurs during consultations, and refer patients for TTE only when they notice progression in the intensity or change in the character of murmur in light of other clinically relevant information.
The Role of Murmur Clinics in Detecting VHD
It has been previously demonstrated that a specialist murmur clinic for patients referred for open access echocardiography by GPs, was feasible whether it was served by a consultant cardiologist or clinical scientists performing both auscultation and the point-of-care scan.5 Up to 70% of patients referred from the community with a murmur had structurally normal hearts and could be screened with a point-of-care scan. The sensitivity of auscultation by the scientist was 83% and by the cardiologist 91%. However, the point-of care scan had a sensitivity of 100% whether it was performed by the scientists or cardiologist. Although patients attend clinics for an already planned TTE, the scientist or the Cardiologist serving the clinic should always conduct a careful prior auscultation in order to interpret the echocardiographic findings in light of clinical examination. A number of uncommon abnormalities such as muscular ventricular septal defect or aortic coarctation would normally be associated with clear systolic murmurs but may not be detected by a point-of-care scan. On the other hand, diastolic murmurs as a result of mitral stenosis or pulmonary or aortic regurgitation may be difficult to hear on auscultation particularly in obese patients or those with chronic obstructive lung disease, but may be easily detected by a point-of-care echocardiogram, which further prompts a standard 2D or 3D TTE for a more in-depth assessment.
Larger prospective studies in the future should be performed to correlate the frequency of murmurs heard in adult life and clinical diagnosis with the subsequent echocardiographic findings. Murmurs related to congenital heart disease was briefly touched upon and was beyond the scope of this focused clinical review. It is a broad topic and should be discussed in more details in future works.
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Journal2023, Current Problems in CardiologySahrai Saeed, ... Stig Urheim
Chapter
Diving Medicine
2016, Murray and Nadel's Textbook of Respiratory Medicine (Sixth Edition)Alfred A. Bove MD, PhD, Tom S. Neuman MD
Pressure Effects: Boyle's Law
Relation of Gas Volume to Depth
Boyle's law states that, if the temperature of a fixed mass of an ideal gas is kept constant, volume and pressure are inversely related. Consequently, when the pressure is doubled, the volume is reduced to one half of the original volume. Because the gas volume is proportional to the absolute pressure, the volume change from the surface to 33 feet of seawater (from 1 to 2 ATA) is greater than the change from 33 to 66 feet (from 2 to 3 ATA).
Barotrauma
With increased pressure, volume in the lungs, middle ear, paranasal sinuses, and gastrointestinal tract are reduced. Displacement of tissues into the diminishing volume of these spaces may cause tissue injury and dysfunction of the organ involved. Barotrauma can affect a paranasal sinus with an occluded orifice, a residual air pocket left between a tooth filling and the base of the tooth, or the air space within a diving mask.
Pulmonary Barotrauma and Arterial Gas Embolism
The gas a diver breathes is pressurized to the ambient pressure so that pressure gradients from the breathing supply to the airways are not altered as the diver descends. Behnke8 and Polak and Adams9 described lung barotrauma in ascending divers due to inadequate exhalation during ascent and overexpansion of the lungs. Later studies provided further insight into mechanisms and prevention of pulmonary barotrauma.10 After breathing compressed air, persons who ascend to the surface from depths as shallow as 4 feet can experience pulmonary barotrauma.
Pathophysiology.
Under experimental conditions, transpulmonary pressures (i.e., the difference between intratracheal and intrapleural pressures) of 95 to 110 cm H2O are sufficient to disrupt the pulmonary parenchyma and force gas into the interstitium.10 Extra-alveolar gas will migrate through perivascular sheaths to cause mediastinal emphysema and pneumothorax.10 Gas can also dissect into the retroperitoneum and into the subcutaneous tissues of the neck. Extra-alveolar gas can pass into ruptured blood vessels, travel to the left side of the heart, and enter the arterial circulation as gaseous emboli. The dissemination of gas bubbles throughout the arterial circulation causes injury to other organ systems and to skeletal muscle, which is evident by a rise in serum creatine kinase level.11
Pulmonary barotrauma can be seen in divers who would not be considered at risk for lung overpressure. Occult lung disease may contribute to unexplained barotrauma and cerebral air embolism.12 Epidemiologic studies have not demonstrated a significant relationship between asthma and an increased risk for pulmonary barotrauma.13
Clinical Manifestations of Arterial Gas Embolism.
The brain is commonly involved. Within minutes of surfacing, the diver can experience loss of consciousness, hemiplegia, stupor, and confusion. Seizures, vertigo, visual disturbances, sensory changes, headache, and circulatory collapse are common. Most individuals fully recover if they are promptly recompressed.14
When they lose consciousness in the water, victims of arterial gas embolism frequently drown. Chest radiographs (Fig.78-3) may show a diffuse lung edema pattern. About 5% of patients immediately develop apnea, unconsciousness, and cardiac arrest. This catastrophic course results from filling of the heart and great vessels with air. Many of these individuals are unresponsive to cardiopulmonary resuscitation and advanced life support measures.15 A report of 31 patients with cerebral air embolism from diving included the following findings: 25% demonstrated pneumomediastinum; 10%, subcutaneous emphysema; 6%, pneumocardium; 3%, pneumoperitoneum; and 3%, pneumothorax. Fifty-two percent had pulmonary opacities indicating associated drowning.16
Mediastinal emphysema is generally associated with mild substernal pain that may be exacerbated by inspiration, coughing, or swallowing. Unless massive, this condition is not usually associated with circulatory compromise. On physical examination a crunching sound synchronous with cardiac action may be auscultated (the Hamman sign). The chest radiograph confirms the diagnosis. No treatment is usually necessary.
Subcutaneous emphysema causes swelling and crepitus in the base of the neck and supraclavicular fossa, sore throat, hoarseness, and dysphagia. Radiographs may be helpful in detecting subtle cases, but computed tomography (CT) scans are more sensitive and can be useful to confirm the diagnosis of barotrauma in doubtful cases. Extra-alveolar gas that ruptures into the pleural space will cause a pneumothorax. Laboratory evaluation may show an elevated hematocrit level and elevation of several serum enzyme levels.17 Treatment for arterial gas embolism requires recompression in a hyperbaric chamber (see later).
Middle Ear Barotrauma
Middle ear barotrauma is the most common diving-related disorder encountered in divers.18 The middle ear undergoes barotrauma when the eustachian tube is blocked during descent and the middle ear space cannot equilibrate with the increasing ambient pressure. The tympanic membrane is displaced inward and may rupture. The middle ear may fill with blood from engorged mucous membranes. Infection and hearing loss are complications. Symptoms during descent include pain in the affected ear that increases with depth. Relief of pain without proper equalization of the middle ear pressure usually indicates that the tympanic membrane has ruptured. Cold water entering the middle ear when the tympanic membrane ruptures may cause vertigo because of unilateral vestibular stimulation. Late complications include bacterial otitis media, serous otitis media, and chronic tympanic membrane perforation.18 In rare cases of middle ear barotrauma, the facial nerve is injured by the increased pressure and a temporary facial paralysis results.19 A modified Valsalva maneuver is commonly used to equilibrate middle ear pressure. Because middle ear barotrauma causes edema and hemorrhage in the middle ear, equalization is usually impossible to achieve until healing is complete. The presence of middle ear barotrauma usually prohibits diving until it is resolved.
Alternobaric Vertigo.
Vertigo may develop on ascent when the reduction of middle ear pressure is not uniform in both ears. The pressure imbalance causes differential stimulation of the labyrinths, resulting in what is called alternobaric vertigo. The sensation of vertigo may persist for 1 to 2 hours after diving and gradually disappears without therapy. Symptoms are similar to labyrinthitis and can include nausea, vomiting, and generalized malaise. Some subjects may be particularly susceptible to alternobaric vertigo if they have had previous injury or infection of the labyrinths. In susceptible individuals, use of moderate doses of antihistamines or decongestants may prevent symptoms. The disorder must be differentiated from vestibular DCS, which is usually associated with deeper, prolonged diving.
Inner Ear Barotrauma
Inner ear barotrauma may develop on descent in divers who perform a forceful Valsalva maneuver to equalize middle ear pressure. When the eustachian tube is blocked, middle ear pressure becomes progressively more negative relative to ambient pressure.18,20 When a Valsalva maneuver is then performed, intrathoracic pressure, central venous pressure, spinal fluid pressure, and inner ear pressure rise above ambient pressure, thereby increasing the gradient between the inner ear perilymph and the middle ear. The round or oval window can rupture, and perilymph can then leak from the inner ear to the middle ear.
Symptoms include vertigo, nausea, vomiting, tinnitus, and loss of hearing on the affected side. Severity may vary, and some divers complain of hearing loss, tinnitus, or vertigo only after diving.
Treatment varies from conservative therapy to surgical repair of the round or oval window. Vertigo and nausea can be treated with benzodiazepine medications. Tinnitus and reduced hearing may become chronic, particularly if no treatment is provided. Divers who exhibit clinical evidence of inner ear barotrauma with intact round and oval windows may have a pressure injury to the organ of Corti and the vestibular system.21
Inner ear DCS may be seen within 2 hours of surfacing and may include vertigo and hearing loss.22 The mechanism is poorly understood but may involve formation of bubbles in the inner ear or embolism of systemic bubbles. Divers with inner ear DCS have been reported to have a higher prevalence of patent foramen ovale (PFO), a condition that increases risk for right to left shunts of air bubbles and thus of decompression events; such an association suggests that the inner ear DCS may also result from air emboli.23,24 The proper therapy for inner ear DCS is hyperbaric recompression.
Sinus Barotrauma
When a sinus orifice is occluded, pressure within the sinus becomes negative with respect to ambient pressure, and mucosal blood vessels become engorged and eventually rupture. Pain over the affected sinus during descent and epistaxis on ascent are usual symptoms. Headache following a dive may indicate sphenoid sinus barotrauma. Treatment includes decongestants and maneuvers to drain the affected sinus. Persistence of blood in the sinus may result in bacterial sinusitis. Prevention is accomplished by avoiding diving with congestion of the nasopharynx and prudent use of decongestants. If a maxillary sinus orifice is occluded, the maxillary branch of the trigeminal nerve may be compressed during ascent and result in infraorbital paresthesias that usually resolve in 2 to 3 hours.25
Less Common Forms of Barotrauma
Facial barotrauma (mask squeeze) happens in the area of distribution of the diving mask. Facial edema, ecchymoses, and conjunctival hemorrhages can be noted after diving. Retro-orbital hematoma and diplopia have been described as complications.26,27 The disorder is self-limiting; no treatment is needed.
Tooth barotrauma leading to severe toothache results when air pockets under fillings or in areas of decay become compressed on descent. Careful dental work prevents this disorder.28
Gastrointestinal barotrauma results when air enters the stomach during diving due to faulty breathing apparatus or by air swallowing. On ascent the expanding air will distend the stomach or intestine. Gastric distention can occlude the esophageal-gastric junction and prevent eructation. Distention of the stomach may cause stomach rupture and pneumoperitoneum.29 The diver experiences abdominal pain, which increases during ascent. Treatment requires surgical repair of the ruptured viscus. Divers with previous gastric surgery may be prone to gastric air trapping.30
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Book2016, Murray and Nadel's Textbook of Respiratory Medicine (Sixth Edition)Alfred A. Bove MD, PhD, Tom S. Neuman MD
Chapter
MEDIASTINUM: ANATOMY, MASSES AND AIR
2008, The Chest X-Ray: A Survival GuideGerald de Lacey MA, MB, B Chir, FRCR, ... Laurence Berman MB, BS, FRCP, FRCR
PNEUMOMEDIASTINUM IN AN INFANT
See Chapter 15, pp. 208–210.
AN INTERESTING CONDITION—SPONTANEOUS PNEUMOMEDIASTINUM
Aetiology/pathology
Alveolar rupture. Thought to be linked to Valsalva's manoeuvre (e.g. athletic activity, defaecation, parturition, marijuana smoking, use of cocaine).
Clinical features
▪
: Chest pain and/or dyspnoea are common. Neck pain, dysphagia, back pain, shoulder pain, neck swelling also occur. Occasionally asymptomatic.
▪
: Physical signs may be absent. Hamman's sign (mediastinal systolic crunch heard over the precordium) occurs in less than 50% of cases.
The CXR
Air in the mediastinum. Air dissecting into the soft tissues of the neck is sometimes a particularly dominant feature.
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Book2008, The Chest X-Ray: A Survival GuideGerald de Lacey MA, MB, B Chir, FRCR, ... Laurence Berman MB, BS, FRCP, FRCR
Related terms:
Dyspnea
Pneumothorax
Plummer-Vinson Syndrome
Odynophagia
Pneumomediastinum
Respiratory Distress
Subcutaneous Emphysema
Auscultation
Thorax Pain
Asthma
View all Topics |
16839 | https://www.teacherspayteachers.com/browse/free?search=converse%20inverse%20contrapositive%20worksheets | Converse Inverse Contrapositive Worksheets | TPT
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Conditional Statements Quick Check
Created by
Rounding Out Math
This Conditional Statements Quick Check is a low-prep, easy-to-use worksheet that helps students identify converse, inverse, and contrapositive forms of conditional statements. It features clear, real-world and geometric examples that support critical thinking and logical reasoning. Students read each original conditional statement and its variation, then determine which logical form it represents by checking the appropriate box.
9 th - 10 th
Geometry
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Conditional Statements Sorting Activity
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Jessica Holman
This is a great way to get students to recognize the differences between converse, inverse and contrapositive. The students have to sort the different conditionals into piles. Students can work in pairs or alone. I allow students to use Expo to write the name of the pile on their desk. The letters on the ends make checking easy.
8 th - 12 th
For All Subjects, Geometry, Other (Math)
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Rated 4.9 out of 5, based on 10 reviews
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Geometry Worksheet: Converse, Inverse, Contrapositive
My Geometry World
$3.99 Original Price $3.99 Rated 5 out of 5, based on 4 reviews 5.0(4)
Conditional Statements Practice Chart: Inverse, Converse, and Contrapositive
Math Giraffe
$1.75 Original Price $1.75 Rated 4.8 out of 5, based on 48 reviews 4.8(48)
Logic in Geometry: Converse, Inverse, Contrapositive Worksheet
ScienceHelper
$1.25 Original Price $1.25
Geometry: Logic with Conditional Statements, Converse, Inverse, Contrapositives
CreativeMathLessons
$4.00 Original Price $4.00 Rated 4.93 out of 5, based on 47 reviews 4.9(47)
Conditional Statements Chart
Created by
Temte Teaching
A conditional statements chart to help students organize the vocabulary words (conditional statements, converse, inverse, contrapositive and biconditional) with the symbol, if-then statement and an example. I use this chart as the front page of my notes packet on the Conditional Statements chapter.
9 th - 10 th
Geometry, Math
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Conditional Statements Graphic Organizer for Logic and Proof Geometry Unit
Created by
Rise and Sine
Help your students master conditional statements with this free graphic organizer! Perfect for introducing or reviewing conditional statements as well as the inverse, converse, and contrapositive statements in geometry. It's the perfect no prep handout to support your geometry logic lessons. This organizer has a clear, student-friendly layout and works well for: class notesstudy guidehomework or test support'This a ready to use PDF for easy printing. Download today and easily provide your studen
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Geometry End of Year EOC Review
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AG Tutoring LLC
This Geometry End of Year EOC Review is ideal for high school geometry students preparing for end of year exams. Need more geometry practice? Check out my Geometry End of Year Review! Reviewed Skills Include:LogicConjunctionDisjunctionNegationConditionalConverseInverseContrapositiveBi-conditionalLaw of DetachmentLaw of SyllogismParallel and perpendicular linesMidpoint and distance formulaTrigonometric ratiosEquation of a circleAngle relationshipsProofsProof by definitionsProperties, postulates
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16840 | https://www.math.uzh.ch/gorodnik/nt/lecture3.pdf | LECTURE 3: CONGRUENCES 1. Basic properties of congruences We begin by introducing some definitions and elementary properties.
Definition 1.1. Suppose that a, b ∈Z and m ∈N. We say that a is congru-ent to b modulo m, and write a ≡b (mod m), when m | (a −b).
We say that a is not congruent to b modulo m, and write a ̸≡b (mod m), when m ∤(a −b).
Theorem 1.2. Let a, b, c, d be integers. Then (i) a ≡b (mod m) ⇐ ⇒b ≡a (mod m) ⇐ ⇒a −b ≡0 (mod m); (ii) a ≡b (mod m) and b ≡c (mod m) ⇒a ≡c (mod m); (iii) a ≡b (mod m) and c ≡d (mod m) ⇒a + c ≡b + d (mod m) and ac ≡ bd (mod m); (iv) If a ≡b (mod m) and d | m with d > 0, then a ≡b (mod d); (v) If a ≡b (mod m) and c > 0, then ac ≡bc (mod mc).
Proof. Verification of these properties is straightforward.
For instance, we prove (iii). Suppose that a ≡b (mod m) and c ≡d (mod m). Then a−b = um and c−d = vm for some integers u and v. Hence, (a+c)−(b+d) = (u+v)m, so that a + c ≡b + d (mod m). Also, ac −bd = (b + um)(d + vm) −bd = (ud + bv + uvm)m which implies that ac ≡bd (mod m).
□ Corollary 1.3. When p(t) is a polynomial with integral coefficients, it follows that whenever a ≡b (mod m), then p(a) ≡p(b) (mod m).
Proof. Use induction to establish that whenever a ≡b (mod m), then an ≡ bn (mod m) for each n ∈N.
□ The above corollary also extends to polynomials in several variables. In par-ticular, we see that if the polynomial equation p(x1, . . . xn) = 0 has an integral solution, then the congruence p(x1, . . . xn) ≡0 (mod m) is also solvable for all m ∈N. This provides a useful test for solvability of equations in integers.
The next theorem indicates how factors may be cancelled through congru-ences.
Theorem 1.4. Let a, x, y ∈Z and m ∈N. Then (i) ax ≡ay (mod m) ⇐ ⇒x ≡y (mod m/(a, m)).
In particular, if ax ≡ay (mod m) and (a, m) = 1, then x ≡y (mod m); (ii) x ≡y (mod mi) (1 ⩽i ⩽r) ⇐ ⇒x ≡y (mod [m1, . . . , mr]).
1 2 LECTURE 3 Proof. Observe first that when (a, m) = 1, then m | a(x−y) ⇐ ⇒m | (x−y).
Then the conclusion of whenever (a, m) = 1. When (a, m) > 1, on the other hand, one does at least have (a/(a, m), m/(a, m)) = 1, so that m | a(x −y) ⇐ ⇒ m (a, m) a (a, m)(x −y) ⇐ ⇒ m (a, m) (x −y).
This establishes the conclusion of part (i) of the theorem.
We now consider part (ii) of the theorem.
Observe first that whenever mi | (x −y) for (1 ⩽i ⩽r), then [m1, . . . , mr] | (x −y). On the other hand, if [m1, . . . , mr] | (x −y), then mi | (x −y) for (1 ⩽i ⩽r). The conclusion of part (ii) is now immediate.
□ We investigate existence of multiplicative inverse modulo m.
Theorem 1.5. Suppose that (a, m) = 1. Then there exists an integer x with the property that ax ≡1 (mod m). If x1 and x2 are any two such integers, then x1 ≡x2 (mod m). Conversely, if (a, m) > 1, then there is no integer x with ax ≡1 (mod m).
Proof. Suppose that (a, m) = 1. Then by the Euclidean Algorithm, there exist integers x and y such that ax+my = 1, whence ax ≡1 (mod m). Meanwhile, if ax1 ≡1 ≡ax2 (mod m), then a(x1 −x2) ≡0 (mod m). But (a, m) = 1, and thus x1 −x2 ≡0 (mod m). We have therefore established both existence and uniqueness of the multiplicative inverse for residues a with (a, m) = 1.
If (a, m) > 1, then (ax, m) > 1 for every integer x. But if one were to have ax ≡1 (mod m), then (ax, m) = (1, m) = 1, which yields a contradiction.
This establishes the last part of the theorem.
□ Now we examine the set of equivalence classes with respect to congruence modulo m.
Definition 1.6. (i) If x ≡y (mod m), then y is called a residue of x modulo m; (ii) We say that {x1, . . . , xm} is a complete residue system modulo m if for each y ∈Z, there exists a unique xi with y ≡xi (mod m); (iii) The set of integers x with x ≡a (mod m) is called the residue class, or congruence class, of a modulo m.
We also wish to consider residue classes containing integers coprime to the modulus, and this prompts the following observation.
Theorem 1.7. Whenever b ≡c (mod m), one has (b, m) = (c, m).
Proof. If b ≡c (mod m), then m | (b −c), whence there exists an integer x with b = c + mx. But then (b, m) = (c + mx, m) = (c, m), as desired.
□ Definition 1.8. A reduced residue system modulo m is a set of integers r1, . . . , rℓsatisfying (a) (ri, m) = 1 for 1 ⩽i ⩽ℓ, (b) ri ̸≡rj (mod m) for i ̸= j, LECTURE 3 3 (c) whenever (x, m) = 1, then x ≡ri (mod m) for some i with 1 ⩽i ⩽ℓ.
Theorem 1.9. The number of elements in a reduced residue system is equal to the number of integers n satisfying 1 ⩽n < m and (n, m) = 1.
Proof. We observe that every integer x can be written as x = qm + r with 0 ⩽r < m. Moreover, (x, m) = (r, m). Hence, we see that {n ∈N : 1 ⩽n < m, (n, m) = 1} is a reduced residue system modulo m.
Let r1, . . . , rℓand s1, . . . , sk be reduced residue systems modulo m. Then for every i = 1, . . . , ℓ, we have ri ≡sσ(i) (mod m) with some σ(i) = 1, . . . , k.
Similarly, for every j = 1, . . . , k, we have sj ≡rθ(j) (mod m) with some θ(j) = 1, . . . , ℓ. We deduce that ri ≡rθ(σ(i)) (mod m) and sj ≡rσ(θ(j)) (mod m). It follows from the properties of the reduced residue systems, that θ(σ(i)) = i and σ(θ(j)) = j. Hence, the maps σ and θ define a bijection between r1, . . . , rℓ and s1, . . . , sk. In particular, reduced residue systems have the same sizes.
□ The number of elements in a reduced residue system modulo m is denoted by φ(m) (Euler’s totient, or Euler’s φ-function).
2. Euler and Fermat theorems Theorem 2.1 (Euler, 1760). If (a, m) = 1 then aφ(m) ≡1 (mod m).
In the proof we use the following lemma Lemma 2.2. Suppose that (a, m) = 1. Then whenever {r1, . . . , rℓ} is a reduced residue system modulo m, the set {ar1, . . . , arℓ} is also a reduced residue system modulo m.
Proof. Since (a, m) = 1, it follows that whenever (ri, m) = 1 one has (ari, m) = 1.
If ari ≡arj (mod m), then it follows from Theorem 1.4(i) that ri ≡ rj (mod m). Hence we deduce that ari ̸≡arj (mod m) for i ̸= j.
It remains to verify property (c). Take x with (x, m) = 1. By Theorem 1.5, there exists an integer a′ such that aa′ ≡1 (mod m). Since {r1, . . . , rℓ} is a reduced residue system modulo m, a′x ≡ri (mod m) for some i. Then ari ≡(aa′)x ≡x (mod m). This shows that {ar1, . . . , arℓ} is a reduced residue system modulo m.
□ Proof of Theorem 2.1. Let {r1, r2, . . . , rφ(m)} be any reduced residue system modulo m, and suppose that (a, m) = 1. By Lemma 2.2, the system {ar1, . . . , arφ(m)} is also a reduced residue system modulo m. Then there is a permutation σ of {1, 2, . . . , φ(m)} with the property that ri ≡arσ(i) (mod m) for 1 ⩽i ⩽φ(m).
Consequently, one has φ(m) Y i=1 ri ≡ φ(m) Y i=1 (arσ(i)) = φ(m) Y j=1 (arj) = aφ(m) φ(m) Y j=1 rj (mod m).
4 LECTURE 3 But (r1 · · · rφ(m), m) = 1, and thus aφ(m) ≡1 (mod m).
□ Corollary 2.3 (Fermat’s Little Theorem, 1640). Let p be a prime number, and suppose that (a, p) = 1. Then one has ap−1 ≡1 (mod p).
Moreover, for all integers a one has ap ≡a (mod p).
Proof. Note that the set {1, 2, . . . , p−1} is a reduced residue system modulo p.
Thus φ(p) = p−1, and the first part of the theorem follows from Theorem 2.1.
When (a, p) = 1, the second part of the theorem is immediate from the first part. Meanwhile, if (a, p) > 1, one has p | a, so that ap ≡0 ≡a (mod p). This completes the proof of the theorem.
□ Fermat’s Little Theorem, and Euler’s Theorem, ensure that the computation of powers is very efficient modulo p (or modulo m).
Example 2.4. Compute 52016 (mod 41). Observe first that φ(41) = 40, and so it follows from Fermat’s Little Theorem that 540 ≡1 (mod 41), and hence 52016 = 540·50+16 = (540)50516 ≡516 (mod 41).
Note next that powers which are themselves powers of 2 are easy to compute by repeated squaring (the “divide and conquer” algorithm). Thus one finds that 52 = 25 ≡−16 (mod 41), 54 = (52)2 ≡(−16)2 = 256 ≡10 (mod 41), 58 = (54)2 ≡(10)2 = 100 ≡18 (mod 41), 516 = (58)2 ≡(18)2 = 324 ≡37 (mod 41).
Thus 52016 ≡37 (mod 41).
Theorem 2.5 (Wilson’s Theorem; Waring, Lagrange, 1771). For each prime number p, one has (p −1)! ≡−1 (mod p).
Proof. The proof for p = 2 and 3 is immediate, so suppose henceforth that p is a prime number with p ⩾5. Observe that when 1 ⩽a ⩽p −1, one has (a, p) = 1, so there exists an integer a unique modulo p with aa ≡1 (mod p).
Moreover, there is no loss in supposing that a satisfies 1 ⩽a ⩽p −1, and then a is a uniquely defined integer.
We may now pair offthe integers a with 1 ⩽a ⩽p −1 with their counterparts a with 1 ⩽a ⩽p −1, so that aa ≡1 (mod p) for each pair. Note that a ̸= a so long as a2 ̸≡1 (mod p).
But a2 ≡1 (mod p) if and only if (a −1)(a + 1) ≡0 (mod p), and the latter is possible only when a ≡±1 (mod p). Thus we find that p−2 Y a=2 a = Y a (aa) ≡1 (mod p), LECTURE 3 5 whence p−1 Y a=1 a ≡(p −1) ≡−1 (mod p).
□ The proof of Wilson’s Theorem motivates a proof of a criterion for the solubility of the congruence x2 ≡−1 (mod p).
Theorem 2.6. When p = 2, or when p is a prime number with p ≡1 (mod 4), the congruence x2 ≡−1 (mod p) is soluble.
When p ≡3 (mod 4), the latter congruence is not soluble.
Proof. When p = 2, x = 1 provides a solution.
Assume next that p ≡ 1 (mod 4), and write r = (p −1)/2, x = r!. Then since r is even, one has x2 = r! · (−1)rr! = (1 · 2 · · · r)((−1) · (−2) · · · (−r)) ≡(1 · 2 · · · r)((p −1) · (p −2) · · · (p −r)) = (p −1)! ≡−1 (mod p).
Thus, when p ≡1 (mod 4), the congruence x2 ≡−1 (mod p) is indeed soluble.
Suppose then that p ≡3 (mod 4). If it were possible that an integer x exists with x2 ≡−1 (mod p), then one finds that (x2)(p−1)/2 ≡(−1)(p−1)/2 ≡−1 (mod p), yet by Fermat’s Little Theorem, one has (x2)(p−1)/2 = xp−1 ≡1 (mod p) whenever (x, p) = 1. We therefore arrive at a contradiction, and this completes the proof of the theorem.
□ |
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How to conjugate Pensar in Spanish
To think, to believe Irregular Verb Top 100
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Table of Contents
Introduction
Indicative tenses of Pensar
Pensar in the Indicative Present
Pensar in the Indicative Preterite
Pensar in the Indicative Imperfect
Pensar in the Indicative Present Continuous
Pensar in the Indicative Informal Future
Pensar in the Indicative Future
Pensar in the Indicative Conditional
Pensar in the Indicative Present Perfect
Pensar in the Indicative Past Perfect
Pensar in the Indicative Future Perfect
Pensar in the Indicative Conditional Perfect
Subjunctive tenses of Pensar
Pensar in the Subjunctive Present
Pensar in the Subjunctive Imperfect
Pensar in the Subjunctive Future
Pensar in the Subjunctive Present Perfect
Pensar in the Subjunctive Past Perfect
Pensar in the Subjunctive Future Perfect
Imperative tenses of Pensar
Pensar in the Imperative Affirmative
Pensar in the Imperative Negative
Example sentences and usage
Downloadable cheat sheet (PDF)
Practice Pensar conjugations (free mobile app)
Introduction
Pensar is the Spanish verb meaning "to think". It is very simple to use, with a small number of use cases. For example: To reflect about something, to believe/ to think about something, to intend to do, or to plan something (estoy pensando empezar clases de español).
Similar verbs to pensar include: considerar (to consider).
| Item | Spanish | English |
---
| Infinitive | pensar | to think, to believe |
| Past participle | pensado | thought |
| Gerund | pensando | thinking |
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Indicative Tenses of Pensar
Pensar in the Indicative Present
The Indicative Present of pensar is used to talk about situations, events or thoughts that are happening now or in the near future. It is also used to talk about facts and truths. For example, "pienso", meaning "I think".
In Spanish, the Indicative Present is known as "El Presente".
| Pronoun | Spanish | English |
---
| Yo | pienso | I think |
| Tú | piensas | you think |
| Ella / Él / Usted | piensa | s/he thinks, you (formal) think |
| Nosotras / Nosotros | pensamos | we think |
| Vosotras / Vosotros | pensáis | you (plural) think |
| Ellas / Ellos / Ustedes | piensan | they think |
The red dot () above denotes an irregular conjugation.
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Pensar in the Indicative Preterite
The Indicative Preterite of pensar is used to talk about actions completed in the past, at a specific point in time. For example, "pensé", meaning "I thought".
In Spanish, the Indicative Preterite is known as "El Pretérito Indefinido".
| Pronoun | Spanish | English |
---
| Yo | pensé | I thought |
| Tú | pensaste | you thought |
| Ella / Él / Usted | pensó | s/he thought, you (formal) thought |
| Nosotras / Nosotros | pensamos | we thought |
| Vosotras / Vosotros | pensasteis | you (plural) thought |
| Ellas / Ellos / Ustedes | pensaron | they thought |
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Pensar in the Indicative Imperfect
The Indicative Imperfect of pensar is used to describe regular and repeated actions that happened in the past and descriptions of things you used to do. For example, "pensaba", meaning "I used to think".
In Spanish, the Indicative Imperfect is known as "El Pretérito Imperfecto".
| Pronoun | Spanish | English |
---
| Yo | pensaba | I used to think |
| Tú | pensabas | you used to think |
| Ella / Él / Usted | pensaba | s/he used to think, you (formal) used to think |
| Nosotras / Nosotros | pensábamos | we used to think |
| Vosotras / Vosotros | pensabais | you (plural) used to think |
| Ellas / Ellos / Ustedes | pensaban | they used to think |
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Pensar in the Indicative Present Continuous
The Indicative Present Continuous of pensar is used to talk about something that is happening continuously or right now. For example, "estoy pensando", meaning "I am thinking".
In Spanish, the Indicative Present Continuous is known as "El Presente Progresivo".
| Pronoun | Spanish | English |
---
| Yo | estoy pensando | I am thinking |
| Tú | estás pensando | you are thinking |
| Ella / Él / Usted | está pensando | s/he is thinking, you (formal) are thinking |
| Nosotras / Nosotros | estamos pensando | we are thinking |
| Vosotras / Vosotros | estáis pensando | you (plural) are thinking |
| Ellas / Ellos / Ustedes | están pensando | they are thinking |
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Pensar in the Indicative Informal Future
The Indicative Informal Future of pensar is used to talk about something that will happen in the future, especially in the near future. For example, "voy a pensar", meaning "I am going to think".
In Spanish, the Indicative Informal Future is known as "El Futuro Próximo".
| Pronoun | Spanish | English |
---
| Yo | voy a pensar | I am going to think |
| Tú | vas a pensar | you are going to think |
| Ella / Él / Usted | va a pensar | s/he is going to think, you (formal) are going to think |
| Nosotras / Nosotros | vamos a pensar | we are going to think |
| Vosotras / Vosotros | vais a pensar | you (plural) are going to think |
| Ellas / Ellos / Ustedes | van a pensar | they are going to think |
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Pensar in the Indicative Future
The Indicative Future of pensar is used to talk about something that will happen in the future. For example, "pensaré", meaning "I will think".
In Spanish, the Indicative Future is known as "El Futuro Simple".
| Pronoun | Spanish | English |
---
| Yo | pensaré | I will think |
| Tú | pensarás | you will think |
| Ella / Él / Usted | pensará | s/he will think, you (formal) will think |
| Nosotras / Nosotros | pensaremos | we will think |
| Vosotras / Vosotros | pensaréis | you (plural) will think |
| Ellas / Ellos / Ustedes | pensarán | they will think |
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Pensar in the Indicative Conditional
The Indicative Conditional of pensar is used to talk about something that may happen in the future, hypothesis and probabilities. For example, "pensaría", meaning "I would think".
In Spanish, the Indicative Conditional is known as "El Condicional Simple".
| Pronoun | Spanish | English |
---
| Yo | pensaría | I would think |
| Tú | pensarías | you would think |
| Ella / Él / Usted | pensaría | s/he would think, you (formal) would think |
| Nosotras / Nosotros | pensaríamos | we would think |
| Vosotras / Vosotros | pensaríais | you (plural) would think |
| Ellas / Ellos / Ustedes | pensarían | they would think |
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Pensar in the Indicative Present Perfect
The Indicative Present Perfect of pensar is used to describe actions that started recently (in the past) and are still happening now or things that have been done recently. For example, "he pensado", meaning "I have thought".
In Spanish, the Indicative Present Perfect is known as "El Pretérito Perfecto".
| Pronoun | Spanish | English |
---
| Yo | he pensado | I have thought |
| Tú | has pensado | you have thought |
| Ella / Él / Usted | ha pensado | s/he has thought, you (formal) have thought |
| Nosotras / Nosotros | hemos pensado | we have thought |
| Vosotras / Vosotros | habéis pensado | you (plural) have thought |
| Ellas / Ellos / Ustedes | han pensado | they have thought |
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Pensar in the Indicative Past Perfect
The Indicative Past Perfect of pensar is used to talk about actions that happened before another action in the past. For example, "había pensado", meaning "I had thought".
In Spanish, the Indicative Past Perfect is known as "El Pretérito Pluscuamperfecto".
| Pronoun | Spanish | English |
---
| Yo | había pensado | I had thought |
| Tú | habías pensado | you had thought |
| Ella / Él / Usted | había pensado | s/he had thought, you (formal) had thought |
| Nosotras / Nosotros | habíamos pensado | we had thought |
| Vosotras / Vosotros | habíais pensado | you (plural) had thought |
| Ellas / Ellos / Ustedes | habían pensado | they had thought |
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Pensar in the Indicative Future Perfect
The Indicative Future Perfect of pensar is used to talk about something that will have happened in the future after something else has already happened. For example, "habré pensado", meaning "I will have thought".
In Spanish, the Indicative Future Perfect is known as "El Futuro Perfecto".
| Pronoun | Spanish | English |
---
| Yo | habré pensado | I will have thought |
| Tú | habrás pensado | you will have thought |
| Ella / Él / Usted | habrá pensado | s/he will have thought, you (formal) will have thought |
| Nosotras / Nosotros | habremos pensado | we will have thought |
| Vosotras / Vosotros | habréis pensado | you (plural) will have thought |
| Ellas / Ellos / Ustedes | habrán pensado | they will have thought |
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Pensar in the Indicative Conditional Perfect
The Indicative Conditional Perfect of pensar is used to talk about something that would have happened in the past but didn’t due to another action. For example, "habría pensado", meaning "I would have thought".
In Spanish, the Indicative Conditional Perfect is known as "El Condicional Perfecto".
| Pronoun | Spanish | English |
---
| Yo | habría pensado | I would have thought |
| Tú | habrías pensado | you would have thought |
| Ella / Él / Usted | habría pensado | s/he would have thought, you (formal) would have thought |
| Nosotras / Nosotros | habríamos pensado | we would have thought |
| Vosotras / Vosotros | habríais pensado | you (plural) would have thought |
| Ellas / Ellos / Ustedes | habrían pensado | they would have thought |
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Subjunctive Tenses of Pensar
Pensar in the Subjunctive Present
The Subjunctive Present is used to talk about situations of uncertainty, or emotions such as wishes, desires and hopes. It differs from the indicative mood due to the uncertainty of the events which are being spoken about. For example, "piense", meaning "I think".
In Spanish, the Subjunctive Present is known as "El Presente de Subjuntivo".
| Pronoun | Spanish | English |
---
| Yo | piense | I think |
| Tú | pienses | you think |
| Ella / Él / Usted | piense | s/he thinks, you (formal) think |
| Nosotras / Nosotros | pensemos | we think |
| Vosotras / Vosotros | penséis | you (plural) think |
| Ellas / Ellos / Ustedes | piensen | they think |
The red dot () above denotes an irregular conjugation.
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Pensar in the Subjunctive Imperfect
The Subjunctive Imperfect is used to speak about unlikely or uncertain events in the past or to cast an opinion (emotional) about something that happened in the past. For example, "pensara", meaning "I thought".
In Spanish, the Subjunctive Imperfect is known as "El Imperfecto Subjuntivo".
| Pronoun | Spanish | English |
---
| Yo | pensara | I thought |
| Tú | pensaras | you thought |
| Ella / Él / Usted | pensara | s/he thought, you (formal) thought |
| Nosotras / Nosotros | pensáramos | we thought |
| Vosotras / Vosotros | pensarais | you (plural) thought |
| Ellas / Ellos / Ustedes | pensaran | they thought |
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Pensar in the Subjunctive Future
The Subjunctive Future is used to speak about hypothetical situations, and actions/events that may happen in the future. For example, "pensare", meaning "I will think".
In Spanish, the Subjunctive Future is known as "El Futuro de Subjuntivo".
| Pronoun | Spanish | English |
---
| Yo | pensare | I will think |
| Tú | pensares | you will think |
| Ella / Él / Usted | pensare | s/he will think, you (formal) will think |
| Nosotras / Nosotros | pensáremos | we will think |
| Vosotras / Vosotros | pensareis | you (plural) will think |
| Ellas / Ellos / Ustedes | pensaren | they will think |
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Pensar in the Subjunctive Present Perfect
The Subjunctive Present Perfect is used to describe past actions or events that are still connected to the present day and to speak about an action that will have happened by a certain time in the future. For example, "haya pensado", meaning "I have thought".
In Spanish, the Subjunctive Present Perfect is known as "El Pretérito Perfecto de Subjuntivo".
| Pronoun | Spanish | English |
---
| Yo | haya pensado | I have thought |
| Tú | hayas pensado | you have thought |
| Ella / Él / Usted | haya pensado | s/he has thought, you (formal) have thought |
| Nosotras / Nosotros | hayamos pensado | we have thought |
| Vosotras / Vosotros | hayáis pensado | you (plural) have thought |
| Ellas / Ellos / Ustedes | hayan pensado | they have thought |
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Pensar in the Subjunctive Past Perfect
The Subjunctive Past Perfect is used to speak about hypothetical situations, and actions/events that occurred before other actions/events in the past. For example, "hubiera pensado", meaning "I had thought".
In Spanish, the Subjunctive Past Perfect is known as "El Pretérito Pluscuamperfecto de Subjuntivo".
| Pronoun | Spanish | English |
---
| Yo | hubiera pensado | I had thought |
| Tú | hubieras pensado | you had thought |
| Ella / Él / Usted | hubiera pensado | s/he had thought, you (formal) had thought |
| Nosotras / Nosotros | hubiéramos pensado | we had thought |
| Vosotras / Vosotros | hubierais pensado | you (plural) had thought |
| Ellas / Ellos / Ustedes | hubieran pensado | they had thought |
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Pensar in the Subjunctive Future Perfect
The Subjunctive Future Perfect is used to speak about something that will have happened if a hypothetical situations occurs in the future. For example, "hubiere pensado", meaning "I will have thought".
In Spanish, the Subjunctive Future Perfect is known as "El Futuro Perfecto de Subjuntivo".
| Pronoun | Spanish | English |
---
| Yo | hubiere pensado | I will have thought |
| Tú | hubieres pensado | you will have thought |
| Ella / Él / Usted | hubiere pensado | s/he will have thought, you (formal) will have thought |
| Nosotras / Nosotros | hubiéremos pensado | we will have thought |
| Vosotras / Vosotros | hubiereis pensado | you (plural) will have thought |
| Ellas / Ellos / Ustedes | hubieren pensado | they will have thought |
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Imperative Tenses of Pensar
Pensar in the Imperative Affirmative
The Imperative Affirmative is used to give orders and commands, to tell someone to do something. For example, "piense", meaning "(to you formal) think!".
In Spanish, the Imperative Affirmative is known as "El Imperativo Afirmativo".
| Pronoun | Spanish | English |
---
| Yo - |
| Tú | piensa | (to you) think! |
| Ella / Él / Usted | piense | (to you formal) think! |
| Nosotras / Nosotros | pensemos | let's think! |
| Vosotras / Vosotros | pensad | (to you plural) think! |
| Ellas / Ellos / Ustedes | piensen | (to you plural formal) think! |
The red dot () above denotes an irregular conjugation.
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Pensar in the Imperative Negative
The Imperative Negative is used to give orders and commands, telling someone not to do something. For example, "no piense", meaning "(to you formal) don't think!".
In Spanish, the Imperative Negative is known as "El Imperativo Negativo".
| Pronoun | Spanish | English |
---
| Yo - |
| Tú | no pienses | (to you) don't think! |
| Ella / Él / Usted | no piense | (to you formal) don't think! |
| Nosotras / Nosotros | no pensemos | let's not think! |
| Vosotras / Vosotros | no penséis | (to you plural) don't think! |
| Ellas / Ellos / Ustedes | no piensen | (to you plural formal) don't think! |
The red dot () above denotes an irregular conjugation.
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Example sentences and usage
Pensemos en lo peor que podría pasar.Let's consider the worst that could happen.
¡No puedes esperar de mí que yo siempre piense en todo!You can't expect me to always think of everything!
¡Ni siquiera pienses en comerte mi chocolate!Don't you even think of eating my chocolate!
No pienso, luego no existo.I don't think, therefore I am not.
Estoy impaciente por escuchar qué piensas de este tema.I look forward to hearing your thoughts on this matter.
Yo pensé que era cierto.I thought it was true.
Yo pienso que mejor hubieras tomado un descanso.I think you had better take a rest.
A menudo pienso en el lugar donde te encontré.I often think about the place where I met you.
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16842 | https://stats.stackexchange.com/questions/14089/what-is-the-difference-between-mean-value-and-average | Skip to main content
What is the difference between "mean value" and "average"?
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Wikipedia explains:
For a data set, the mean is the sum of the values divided by the number of values.
This definition however corresponds to what I call "average" (at least that's what I remember learning). Yet Wikipedia once more quotes:
There are other statistical measures that use samples that some people confuse with averages - including 'median' and 'mode'.
Now that's confusing. Are "mean value" and "average" different from one another? If so how?
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edited Jul 25, 2020 at 11:53
gung - Reinstate Monica
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asked Aug 10, 2011 at 15:36
neydroydrecneydroydrec
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The mean you described (the arithmetic mean) is what people typically mean when they say mean and, yes, that is the same as average. The only ambiguity that can occur is when someone is using a different type of mean, such as the geometric mean or the harmonic mean, but I think it is implicit from your question that you were talking about the arithmetic mean.
– Macro
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Aug 10, 2011 at 15:43
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For more information about means, the different kinds that exist, & how they are related to each other, see this excellent CV question: Which "mean" to use and when?
– gung - Reinstate Monica
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Oct 30, 2012 at 18:59
Mean, or Expected Value - is a theoretical property of a certain probability. Average is the observed/measured outcome of a certain sample. If a measured average diverge too much from the expected mean, it's a sign that the underlying probability assumption, or one of its properties, is wrong. This is the main distinction between the terms that statisticians use.
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Mean versus average
The mean most commonly refers to the arithmetic mean, but may refer to some other form of mean, such as harmonic or geometric (see the Wikipedia article). Thus, when used without qualification, I think most people would assume that "mean" refers to the arithmetic mean.
Average has many meanings, some of which are much less mathematical than the term "mean". Even within the context of numerical summaries, "average" can refer to a broad range of measures of central tendency.
Thus, the arithmetic mean is one type of average.
Arguably, when used without qualification the average of a numeric variable often is meant to refer to the arithmetic mean.
Side point
It is interesting to observe that Excel uses the sloppier but more accessible name of AVERAGE() for its arithmetic mean function, where R uses mean().
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answered Aug 11, 2011 at 10:37
Jeromy AnglimJeromy Anglim
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You can end up with bizarre conversations like this: "So we look at the average return..." "Which average do you mean? A median? A weighted average?" "The mean return." "Oh, okay"... and it seems like everyone understood each other, ... except that the first person may actually be talking about the geometric mean of the returns. I've seen it happen.
– Glen_b
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Which of this denotes μ ?
– user208618
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@Isa The symbol μ is generally used to refer to the arithmetic mean of the population (x¯, on the other hand, is used to refer to the arithmetic mean of the sample).
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There are several "averages." Just think of this trick question: "What is the probability that the next person you meet has more than the average number of arms?"
The "mean" or "arithmetic mean" or "arithmetic average" is one average that you learned in the past. But the median (the value with half the observations greater and half less than it), the mode (the most common value), the geometric mean (multiply the values then take the nth root), the harmonic mean (the reciprocal of the mean of the reciprocals of the data), and others all fall under the general term "average."
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edited Aug 10, 2011 at 17:21
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Greg SnowGreg Snow
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"More than the average number of arms" sounds a bit strange to me (but I'm also not a native English speaker), but "more arms than the average person" (or "more arms than normal") sounds more natural to me, in which "average" of course also is used quite differently from any scientific definition of the word.
– HelloGoodbye
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Mean and average are generally used interchangeably (although I've seen them used as referring to population versus empirical).
They, like median and mode, are measures of central tendency, but in many cases, the other two are different.
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answered Aug 10, 2011 at 15:44
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The mean you described (the arithmetic mean) is what people typically intend when they say "mean" and, yes, that is the same as average. The only ambiguity that can occur is when someone is using a different type of mean, such as the geometric mean or the harmonic mean, but I think it is implicit from your question that you were talking about the arithmetic mean
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edited Aug 10, 2011 at 18:51
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I see "average" and "mean" used mostly as synonyms. One author who draws a clear distinction is Donald Wheeler, in "Advanced Topics in Statistical Quality Control." He declares that the "average" is a statistic determined by some arithmetic procedure, whereas "mean" is a parameter, specifying location of a distribution. By way of example, he writes that one could calculate an "average" telephone number, which would be meaningless (pun?). The average (a statistic) is an unbiased estimate of the mean (a parameter).
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answered Mar 5, 2013 at 17:35
StephenStephen
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This is needs to be a much higher rated answer. Although the average is a unbiased estimate of the mean only in Asymptopia. 'Mean' can be used to refer to the expected value of any population. 'Average' is statistic an empirical function of the population which is sometimes, but not always a good point estimate of the mean. For certain distributions (e.g. lognormal) there may be a better point estimate of the mean which is not the average.
– Dalton Hance
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A Complete Guide to Six Sigma Defects Per Million (2025)
Achieving consistent and predictable business process performance is crucial for delivering value to customers. However, variability, defects, and errors threaten this objective, eroding quality and driving up costs.
To systematically enhance process capability, organizations leverage Six Sigma – a rigorous methodology centered on statistical thinking and defect elimination. Professionals equipped with a Six Sigma certification are trained to apply these principles to streamline operations and reduce variability.
Transform Process Variability into Precision!
Get a Lean Six Sigma Green Belt Certification and master defect reduction techniques.
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DPMO offers a way to quantify how many defects or mistakes occur per million chances for them to happen.
By measuring defect frequency, DPMO establishes a baseline to compare process outcomes, pinpoint quality gaps, and steer improvements via evidence-based analysis. A lower DPMO score signifies a higher process sigma rating, lower variation from perfection, and superior reliability.
What is Six Sigma Defects per Million?
Six Sigma defects per million refer to a key metric and concept within the Six Sigma methodology for quality improvement and process enhancement. Specifically, it represents the number of defects per million opportunities that are likely to occur in a business process.
The Six Sigma methodology aims to improve process outputs by identifying and eliminating the root causes of defects and variations. A foundational premise is that all processes can be measured, analyzed, improved, and controlled.
Defects per million opportunities (DPMO) offers a standardized way of quantifying the presence of defects in a process. It measures how many defects could potentially happen for every million opportunities where an error or discrepancy might occur.
A lower DPMO value corresponds to better process performance and higher quality. As processes are improved using Six Sigma techniques, the DPMO rating drops, indicating that defects are being systematically reduced.
Role in Six Sigma Methodology
DPMO plays a pivotal role in Six Sigma initiatives and projects, a core focus of Six Sigma Green Belt certification programs. These programs teach practitioners to establish baselines, measure process capability, and set tangible goals for defect reduction.
Each Six Sigma process sigma level has a target DPMO value. For example, a Six Sigma process has a DPMO of only 3.4, meaning that statistically only 3.4 defects occur per million opportunities. This translates to near-perfect quality with 99.99966% accuracy.
Relation to process improvement
By providing a standardized defect rate, DPMO enables comparative analysis across different processes within an organization. It becomes easier to identify processes that need priority focus for improvement.
As DPMO ratings shift downwards through successive efforts, processes become more efficient and aligned with customer requirements. This drives increased satisfaction, potential for innovation, and cost competitiveness.
Six Sigma defects per million is a vital metric that catalyzes and Steers process improvement via the Six Sigma methodology. It quantifies quality issues and guides enhancement initiatives.
Deepen your understanding of Six Sigma methodologies with our comprehensive Green Belt training.
Master Six Sigma techniques →
Calculating Six Sigma Defects per Million
To quantify and monitor process performance using the Six Sigma methodology, defects per million opportunities (DPMO) need to be accurately calculated. This requires collecting and analyzing quality data to determine the current sigma rating and defect levels.
There are two formulas for arriving at DPMO based on the type of data available:
Population DPMO: When complete data for the entire population of products/services is accessible.
Sample DPMO: When only a sample data set is available due to the infeasibility of measuring the entire output.
Formulas for population vs sample data
Population DPMO formula:
DPMO = (Total number of defects / Total opportunities for defects) x 1,000,000
Here, the defects and opportunities pertain to the full population size.
Sample DPMO formula:
DPMO = (Number of defects / Number of units in sample x Opportunities for defect per unit) x 1,000,000
This estimates the defect rate based on a representative sample’s quality performance.
The steps for calculation involve:
Determining sample size
Identifying opportunities for defect per unit
Finding defect opportunities for sample
Counting total defects in the sample
Applying the formula
Accurately calculating DPMO is crucial for setting valid baselines for improvement initiatives. The population versus sample formula is applied appropriately based on data constraints.
Tools for calculation (Excel, Minitab, etc)
There are several software tools and platforms that can be leveraged to compute defects per million opportunities (DPMO) accurately and efficiently:
Excel Spreadsheets: Excel provides in-built statistical functions along with easy table/data manipulation features to calculate DPMO, a skill emphasized in foundational Six Sigma Yellow Belt certification for process analysts
Quality Management Software: Tools like Minitab, Quality Companion, SigmaXL, etc. specially meant for quality improvement techniques contain specific functions for DPMO calculation. They eliminate manual errors.
Online DPMO Calculators: Simple web-based calculators allow inputting the formula variables to automatically calculate DPMO. Useful for quick, one-time assessments.
Business Intelligence Software: Solutions like Tableau, Power BI, and Qlik can take in large-quality data sets and run DPMO computations seamlessly. Visualizations identify patterns.
Statistical Programming Software: Platforms such as SAS, R, and Python having statistical capabilities can customize DPMO analysis as per required assumptions/models. Handle complex scenarios.
Step-by-step manual calculation walkthrough
The steps to manually compute sample DPMO are:
Select sample size – For demonstration, let’s take the sample size as 500 units
Define defect opportunities per unit – For example, 5 defect opportunities per unit
Compute total opportunities for defects in sample = Sample size x Defect opportunities per unit Here, 500 units x 5 opportunities = 2500 opportunities
Count number of defects found across the sample = 135 defects
Apply the DPMO formula:
DPMO = (Number of defects / Number of units x Opportunities per unit) x 1,000,000
DPMO = (135/500 x 5) x 1,000,000 = 135,000
Therefore, the DPMO rating is 135,000 defects per million opportunities.
Following these step-by-step calculations manually or using appropriate tools facilitates accurate DPMO measurement for Six Sigma initiatives.
Learn how to efficiently calculate and analyze DPMO using Minitab with our Minitab Essentials course.
Calculate DPMO effectively →
Interpreting Six Sigma Defects per Million
Once the defects per million opportunities (DPMO) is accurately calculated for a process, the next step is interpreting its implication on quality and performance. The DPMO value signifies the short-term and long-term sigma rating which determines how well the process is functioning.
A lower DPMO denotes better quality and less variation from perfection. It translates to higher process sigma levels and indicates excellent consistency. On the other hand, a higher DPMO implies higher defects, lower capability, and a need for improvements.
Small vs large standard deviations
The DPMO specifically provides insight into the process variation by showing how widely dispersed the actual performance is from the ideal or desired performance.
A small standard deviation and DPMO indicate that most observations fall very close to the average while a large standard deviation and high DPMO suggest that the observations are widely spread out from average.
A DPMO value of zero represents zero defects and zero variation. As the DPMO score rises, it indicates larger deviations and frequency of defects. Appropriate data visualization techniques like statistical modeling, probability curves, histographs, etc. can facilitate interpretation.
By distinguishing between small and large standard deviations using DPMO, priority areas for reducing variability and enhancing process stability can be identified. This drives effective quality improvement initiatives.
Impact on process variability and quality
The defects per million opportunities (DPMO) have a direct correlation with process variability and quality. A lower DPMO indicates minimum variation and superior quality while a higher rating signals high variability and poor quality.
Every process has an inherent amount of natural variability that cannot be eliminated. However, special causes of variation can be addressed through process enhancements. As DPMO decreases, variability reduces, and quality improves.
For example, an automated manufacturing line with integrated inspection may have 30,000 DPMO whereas a manual assembly process may exhibit 90,000 DPMO for the same product. The automated line demonstrates better consistency and reliability.
Using standard deviation analysis to identify improvements
By using statistical analysis techniques, the element contributing most to higher standard deviation and DPMO can be identified. For instance, measurement system analysis can evaluate if inspection equipment variability is the root cause for wide quality fluctuations.
Other approaches like hypothesis testing, correlation analysis, process capability analysis, factorial experiments, etc. assess the influence of controllable factors like temperature, pressure, speed, etc. on output variability.
The factors showing maximum positive correlation and significance to variability become targets for improvement through control mechanisms, automation, etc. to reduce DPMO and enhance process capability. They help align overall performance to Six Sigma goal levels.
Six sigma defects per million in Practice
While defects per million opportunities (DPMO) provides a metric to evaluate the current state of a process, its true value lies in actually leveraging it to control and improve quality over time. This requires deploying the DPMO within a structured statistical process control (SPC) framework for sustaining gains.
Application in statistical process control (SPC)
SPC techniques like control charts plot the DPMO values over different periods to visually track fluctuations in process performance. Control limits are set at +/- 3 standard deviations from the centerline.
If the DPMO data falls outside control limits or shows concerning patterns, it signals the process is unstable and interventions are required. The out-of-control signals guide improvement priorities.
As improvements are implemented, the control charts reflect gains through lower DPMO rates and data points within limits. SPC solves process issues and maintains consistency.
Control charts and monitoring process performance
Control charts with DPMO data help identify special causes versus common causes of variation. By analyzing the patterns, specific assignable root causes of quality problems can be determined versus inherent process variation.
Stratification of DPMO by product type, machine, material, etc. further pinpoints trouble areas. Real-time monitoring through statistical warnings combined with drilling down to understand flair-ups delivers sustainable defect reduction.
Thus, leveraging the Six Sigma DPMO metric within statistical process control frameworks provides the right visibility to drive process excellence.
Capability analysis to meet specifications
Process capability analysis determines if a process can meet the quality specifications necessary for customer satisfaction. It uses statistical analysis to establish acceptance criteria and assess process potential.
The defects per million opportunities (DPMO) data feeds into capability ratio metrics like Cp, Cpk, Pp, and Ppk to determine process power. For example, Cpk above 1.33 signals the process is capable of a Six Sigma rating.
If DPMO is too high and capability too low, process reengineering may be required through technology, systems, competence enhancement, etc. Capability analysis identifies sustained potential.
Case studies and examples
Leading organizations globally leverage DPMO and statistical thinking for breakthrough improvements:
General Electric improved jet engine assembly defect rates from ~20,000 DPMO to under 10 DPMO using systematic problem-solving.
Tata Motors passenger vehicle plant in Pune, India achieved 30% cycle time reduction and 100% model changeover through DPMO-guided cell design.
The FDA’s Medical Device Innovation Consortium has DPMO-level quality targets for new product development pilots in healthcare.
Such success stories across manufacturing, healthcare, and services showcase the central role of defects per million opportunities metric within robust process control frameworks for managing and improving quality.
Discover advanced Six Sigma techniques and their real-world applications in our comprehensive Black Belt training.
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Common Mistakes and Challenges of Six Sigma Defects Per Million
While defects per million opportunities (DPMO) provide a valuable quality metric, some common pitfalls can undermine its accuracy and effectiveness for driving process improvements. Being aware of these mistakes can help teams derive maximum value.
Sample vs population data issues
Attempting to extrapolate a sample DPMO to the entire population can be misleading if the sample is not truly representative. Small samples may be inadequate to capture potential defects across product variants, materials, machines, operators, etc.
However, measuring the entire population may be impractical. Balancing sample size and homogeneity is key along with analyzing stability using statistical confidence intervals before generalizing DPMO.
Similarly, samples should not be drawn during abnormal process conditions. This inaccurately skews the DPMO versus normal functioning. Data interpretation must factor in variability over time.
Mitigating data issues enhances the integrity of defects per million opportunities for decision-making. Supplementary metrics also add perspective for holistic understanding.
Understanding variation origins
A thorough root cause analysis must supplement DPMO reviews to grasp why defects, rework, and variability occur in the first place. Without diagnosing the source, sustainable solutions are difficult to formulate.
Some defects and causes could be training gaps, inconsistent materials, faulty equipment, unclear instructions, etc. Distinguishing between special and common causes is also imperative before proceeding.
Predicted vs empirical standard deviation
The defect rate used to calculate DPMO may rely on predicted models or empirical observations. However, statistical assumptions in predictive models may not always hold.
For example, estimating the standard deviation based on experience may understate or overstate the real-world variability. Periodic empirical data gathering is advised to validate if the DPMO based on predictive assumptions mirrors ground reality.
Analytics helps determine the most appropriate statistical distribution and methods to accurately model defects and DPMO rather than implausible approximations.
Embracing analytics minimizes the fallacies of speculative defects per million opportunities. DPMO deprives decision-makers rather than misguides them.
Importance for Driving Process Improvements
The defects per million opportunities (DPMO) metric holds invaluable significance for systematically enhancing process capabilities and output quality over time. Some key ways in which DPMO catalyzes improvement are:
Sets Tangible Targets: The DPMO values at different sigma levels provide tangible goals for teams to aspire towards. Having quantifiable quality objectives fuels better planning and resourcing for achievements.
Enables Benchmarking: DPMO introduces standardization which permits benchmarking across product lines, manufacturing sites, suppliers, etc. to propagate best practices. Healthy competition on metrics is motivated.
Measures Progress: As improvement projects are rolled out, DPMO helps continually assess advancement made towards Six Sigma goals. Progress tracking prevents change initiatives from losing steam.
Sustains Momentum: Since DPMO can be monitored over time using statistical process control, new issues are rapidly detected before escalation. This sustains momentum on enhancing process stability and building resilience.
Identifies Priorities: DPMO trends can reveal pain points in the process that are leading to the most defects. Focusing projects on quick wins for the highest DPMO reduction gives maximum bang for the buck.
Six Sigma Defects per Million Opportunity plays an indispensable role as a change agent for steering quality improvement journeys and realizing process excellence.
Key Takeaways of Six Sigma Defects Per Million
Defects per million opportunities (DPMO) serve as a vital component of the Six Sigma methodology for quality enhancement across business processes and functions.
Some of the key lessons about effectively leveraging DPMO include:
Standardized Metric: DPMO introduces a consistent way of measuring process capability and improvement that permits comparisons across systems, products, etc.
Actionable Insights: By revealing shortfalls quantitatively, the defect per million metric guides quality improvement priorities through focused analytical problem-solving.
Requires Statistical Rigor: Practitioners must apply care and statistical proficiency in sampling, computation, analysis, and interpretation of DPMO data for reliable decision-making.
Continual Improvement: Driving processes up the sigma curve necessitates continual measurement, analysis, and improvement driven by DPMO data using methods like SPC.
Customer Centricity: A lower DPMO denoting higher quality and consistency results in enhanced customer value and satisfaction by reducing variability and defects.
In conclusion, organizations must embrace DPMO integrated with robust statistical techniques to accelerate process excellence in today’s competitive business landscape. Mastering Six Sigma defects per million analysis fuels the journey.
Best practices for leveraging Six Sigma defects per million
Organizations aiming to deeply entrench defects per million opportunities (DPMO) analysis for catalyzing process improvements must embrace several proven practices including:
Invest in Training: Statistically sound DPMO analysis requires competencies. Investments in orienting teams on techniques like measurement system analysis, correlation analysis, sampling, FMEA, etc. pay rich dividends.
Voice of Customer: Keep customer needs and perspectives central to DPMO goal-setting and improvement initiative prioritization. Visit the Gemba to observe pain points.
Rigor in Measurement: Adhere to the tightest standards possible in defective identification, counting, classification, and data capture to ensure the reliability of DPMO numbers.
Multilayer Drill-Down: Slice and dice DPMO data across various process dimensions such as product, batch, machine, material, etc. to reveal insights.
Cluster Analysis: Analyze patterns in DPMO data clustered by time, region, operator, etc. to diagnose if causes are sporadic or systematic.
Integrate with Lean Six Sigma: Blend DPMO analysis with complementary LSS tools covering value stream mapping, root cause analysis, mistake proofing, etc.
Following these best practices will lead to competitive differentiation through entrenching statistical thinking and robust process control regimes fueled by the Six Sigma defects per million metric.
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16845 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10742435/ | Acute Colonic Diverticulitis: CT Findings, Classifications, and a Proposal of a Structured Reporting Template - PMC
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Diagnostics (Basel)
. 2023 Dec 8;13(24):3628. doi: 10.3390/diagnostics13243628
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Acute Colonic Diverticulitis: CT Findings, Classifications, and a Proposal of a Structured Reporting Template
Francesco Tiralongo
Francesco Tiralongo
1 Radiology Unit 1, University Hospital Policlinico “G. Rodolico-San Marco”, 95123 Catania, Italy; davidegiuseppecastiglione@gmail.com (D.G.C.); corrado.ini@gmail.com (C.I.)
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1,, Stefano Di Pietro
Stefano Di Pietro
2 Department of Medical Surgical Sciences and Advanced Technologies “GF Ingrassia”, University Hospital Policlinico “G. Rodolico-San Marco”, 95123 Catania, Italy; s.dipietro@studium.unict.it (S.D.P.); dario.milazzo2015@libero.it (D.M.); sebastianogalioto1996@gmail.com (S.G.); pietrovalerio.foti@unict.it (P.V.F.); spalmucci@sirm.org (S.P.); basile.antonello73@gmail.com (A.B.)
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2, Dario Milazzo
Dario Milazzo
2 Department of Medical Surgical Sciences and Advanced Technologies “GF Ingrassia”, University Hospital Policlinico “G. Rodolico-San Marco”, 95123 Catania, Italy; s.dipietro@studium.unict.it (S.D.P.); dario.milazzo2015@libero.it (D.M.); sebastianogalioto1996@gmail.com (S.G.); pietrovalerio.foti@unict.it (P.V.F.); spalmucci@sirm.org (S.P.); basile.antonello73@gmail.com (A.B.)
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2, Sebastiano Galioto
Sebastiano Galioto
2 Department of Medical Surgical Sciences and Advanced Technologies “GF Ingrassia”, University Hospital Policlinico “G. Rodolico-San Marco”, 95123 Catania, Italy; s.dipietro@studium.unict.it (S.D.P.); dario.milazzo2015@libero.it (D.M.); sebastianogalioto1996@gmail.com (S.G.); pietrovalerio.foti@unict.it (P.V.F.); spalmucci@sirm.org (S.P.); basile.antonello73@gmail.com (A.B.)
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2, Davide Giuseppe Castiglione
Davide Giuseppe Castiglione
1 Radiology Unit 1, University Hospital Policlinico “G. Rodolico-San Marco”, 95123 Catania, Italy; davidegiuseppecastiglione@gmail.com (D.G.C.); corrado.ini@gmail.com (C.I.)
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1, Corrado Ini’
Corrado Ini’
1 Radiology Unit 1, University Hospital Policlinico “G. Rodolico-San Marco”, 95123 Catania, Italy; davidegiuseppecastiglione@gmail.com (D.G.C.); corrado.ini@gmail.com (C.I.)
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1, Pietro Valerio Foti
Pietro Valerio Foti
2 Department of Medical Surgical Sciences and Advanced Technologies “GF Ingrassia”, University Hospital Policlinico “G. Rodolico-San Marco”, 95123 Catania, Italy; s.dipietro@studium.unict.it (S.D.P.); dario.milazzo2015@libero.it (D.M.); sebastianogalioto1996@gmail.com (S.G.); pietrovalerio.foti@unict.it (P.V.F.); spalmucci@sirm.org (S.P.); basile.antonello73@gmail.com (A.B.)
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2, Cristina Mosconi
Cristina Mosconi
3 Department of Radiology, IRCCS Azienda Ospedaliero-Universitaria di Bologna, Sant’Orsola-Malpighi Hospital, 40138 Bologna, Italy; cristina.mosconi@aosp.no.it
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3, Francesco Giurazza
Francesco Giurazza
4 Interventional Radiology Department, Cardarelli Hospital of Naples, 80131 Naples, Italy; francescogiurazza@hotmail.it
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4, Massimo Venturini
Massimo Venturini
5 Department of Diagnostic and Interventional Radiology, Circolo Hospital, Insubria University, 21100 Varese, Italy; venturini.massimo@hsr.it
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5, Guido Nicola Zanghi’
Guido Nicola Zanghi’
6 Department of General Surgery, University of Catania, 95123 Catania, Italy; gzanghi@unict.it
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6, Stefano Palmucci
Stefano Palmucci
2 Department of Medical Surgical Sciences and Advanced Technologies “GF Ingrassia”, University Hospital Policlinico “G. Rodolico-San Marco”, 95123 Catania, Italy; s.dipietro@studium.unict.it (S.D.P.); dario.milazzo2015@libero.it (D.M.); sebastianogalioto1996@gmail.com (S.G.); pietrovalerio.foti@unict.it (P.V.F.); spalmucci@sirm.org (S.P.); basile.antonello73@gmail.com (A.B.)
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2, Antonio Basile
Antonio Basile
2 Department of Medical Surgical Sciences and Advanced Technologies “GF Ingrassia”, University Hospital Policlinico “G. Rodolico-San Marco”, 95123 Catania, Italy; s.dipietro@studium.unict.it (S.D.P.); dario.milazzo2015@libero.it (D.M.); sebastianogalioto1996@gmail.com (S.G.); pietrovalerio.foti@unict.it (P.V.F.); spalmucci@sirm.org (S.P.); basile.antonello73@gmail.com (A.B.)
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2
Editors: Andor WJM Glaudemans, Francesco Macrì
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1 Radiology Unit 1, University Hospital Policlinico “G. Rodolico-San Marco”, 95123 Catania, Italy; davidegiuseppecastiglione@gmail.com (D.G.C.); corrado.ini@gmail.com (C.I.)
2 Department of Medical Surgical Sciences and Advanced Technologies “GF Ingrassia”, University Hospital Policlinico “G. Rodolico-San Marco”, 95123 Catania, Italy; s.dipietro@studium.unict.it (S.D.P.); dario.milazzo2015@libero.it (D.M.); sebastianogalioto1996@gmail.com (S.G.); pietrovalerio.foti@unict.it (P.V.F.); spalmucci@sirm.org (S.P.); basile.antonello73@gmail.com (A.B.)
3 Department of Radiology, IRCCS Azienda Ospedaliero-Universitaria di Bologna, Sant’Orsola-Malpighi Hospital, 40138 Bologna, Italy; cristina.mosconi@aosp.no.it
4 Interventional Radiology Department, Cardarelli Hospital of Naples, 80131 Naples, Italy; francescogiurazza@hotmail.it
5 Department of Diagnostic and Interventional Radiology, Circolo Hospital, Insubria University, 21100 Varese, Italy; venturini.massimo@hsr.it
6 Department of General Surgery, University of Catania, 95123 Catania, Italy; gzanghi@unict.it
Correspondence: tiralongofrancesco91@hotmail.it
Roles
Andor WJM Glaudemans: Academic Editor
Francesco Macrì: Academic Editor
Received 2023 Oct 13; Revised 2023 Nov 25; Accepted 2023 Dec 6; Collection date 2023 Dec.
© 2023 by the authors.
Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
PMC Copyright notice
PMCID: PMC10742435 PMID: 38132212
Abstract
Acute colonic diverticulitis (ACD) is the most common complication of diverticular disease and represents an abdominal emergency. It includes a variety of conditions, extending from localized diverticular inflammation to fecal peritonitis, hence the importance of an accurate diagnosis. Contrast-enhanced computed tomography (CE-CT) plays a pivotal role in the diagnosis due to its high sensitivity, specificity, accuracy, and interobserver agreement. In fact, CE-CT allows alternative diagnoses to be excluded, the inflamed diverticulum to be localized, and complications to be identified. Imaging findings have been reviewed, dividing them into bowel and extra-intestinal wall findings. Moreover, CE-CT allows staging of the disease; the most used classifications of ACD severity are Hinchey’s modified and WSES classifications. Differential diagnoses include colon carcinoma, epiploic appendagitis, ischemic colitis, appendicitis, infectious enterocolitis, and inflammatory bowel disease. We propose a structured reporting template to standardize the terminology and improve communication between specialists involved in patient care.
Keywords: diverticulitis, colonic, computed tomography, structured report, abscess, fistula, perforation, peritonitis
1. Introduction
Diverticular disease is one of the most common gastrointestinal disorders, especially in industrialized countries; it includes diverticulosis, defined as the presence of one or more diverticula that can develop in either the colon or small bowel, and diverticulitis, characterized by inflammation of the same .
Pseudodiverticula of the colon or false diverticula are defined as herniation of the mucous and submucous layers through a wall defect of the muscular layer of an intestinal loop between the taenia coli and the mesentery at the point of penetration of the blood vessel , often present in the sigmoid colon; true diverticula contain all layers of the intestinal wall, are congenital in etiology, and are usually detected in the ascending colon .
Pseudodiverticula genesis is multifactorial; it is believed that the low-fiber diet, the altered gut microbiome, and genetic factors are implicated in the development of diverticular disease; in particular, the hardening of the feces produces an increase in intraluminal pressure, which leads to herniation of the internal parietal layers through loci minoris resistentiae [3,4].
Acute colonic diverticulitis (ACD) represents the most common complication of diverticular disease since it occurs in about 10–25% of patients. ACD represents an abdominal emergency and encompasses a variety of conditions, ranging from localized diverticular inflammation to fecal peritonitis [2,5].
The pathophysiological basis of diverticulitis is thought to lie in ostial obstruction of the diverticulum by feces or food debris, leading to mucosal damage and subsequent ischemia, micro-perforation, and infection; however, new studies suggest that the state of chronic inflammation and alterations in the gut microbiome are involved in the pathogenic process .
Contrast-enhanced computed tomography (CE-CT) represents the diagnostic tool of first choice in patients with suspected diverticulitis [7,8,9,10], showing high diagnostic sensitivity (98–99%), specificity (99–100%), accuracy (98−99%), and excellent interobserver agreement [10,11,12].
CE-CT allows, in the scenario of suspected diverticulitis, to 1. confirm the diagnosis and identify the inflamed diverticulum; 2. stage the disease, stratifying patients for operative versus nonoperative treatment; 3. identify complications, depicting the extracolonic disease extent; 4. provide a valid tool in preoperative surgical and radiological interventional planning; and 5. suggest alternative diagnoses presenting similar clinical symptoms and signs (such as neoplasm, inflammatory bowel disease, appendicitis, epiploic appendagitis, and colon ischemia) [12,13].
Treatment of ACD depends on whether it is uncomplicated or complicated and the degree of complications. In the case of uncomplicated diverticulitis, the use of antibiotics is not recommended, and management of the patient in an outpatient setting is recommended. In complicated diverticulitis, in addition to broad-spectrum intravenous antibiotics and intestinal rest, the lower stages are treated with non-operative management, in which interventional radiology intervenes through percutaneous drainage of the abscesses [8,14].
In cases of peritonitis or severe complications, surgical management is necessary, especially in an emergency setting; options include colostomy formation, colonic resection with the construction of an end-colostomy (Hartmann procedure), and colonic resection with primary anastomosis with or without diverting loop ileostomy .
This article aims to describe the main CT findings of ACD using the cases in our database, illustrate the main classifications, and propose a structured report template for use in the emergency setting.
CT Protocol
In our institution, in the clinical scenario of suspected ACD—with pain in the left lower quadrant, fever, and elevated inflammatory indexes—computed tomography (CT) examination is performed before and after intravenous administration of iodinated contrast medium. The image acquisition extends from the diaphragm to the pubic symphysis.
The technical parameters kV and mAs should be adjusted depending on the patient’s waist circumference to optimize the image quality and radiation dose.
The CT protocol consists of a non-contrast phase and a portal venous phase (acquired with a delay of 65–70 s). In the case of concomitant presence of clinical and laboratory signs of bleeding, an arterial phase (acquired with an automatic bolus tracking technique with the Region of Interest placed in the proximal abdominal aorta and a delay of 15 s) and a delayed phase were also acquired.
The CT examinations were performed using two different scanners: the Optima 660 (GE Healthcare, Chicago, IL, USA) and the Revolution EVO (GE Healthcare, Chicago, IL, USA).
A bolus of 80–120 mL of Iomeprol (Iomeron 350 or 400 mg/mL; Bracco Imaging, Milan, Italy), followed by a 30 mL saline flush, was administered using an automated contrast injection system at a flow rate of 3–4 mL/s. The CT images were reformatted in both the coronal and sagittal planes.
We believe that the use of oral contrast medium may not be indicated in the setting of ACD; the non-use does not compromise the effectiveness of the CT in the diagnosis , it is not recommended in clinical practice for long preparation times, and the large volume to be ingested and studies show that it does not provide any diagnostic benefit .
Rectal administration of contrast medium may be considered in cases of suspected sinus tract from the rectosigmoid colon to an adjacent pelvic organ or to evaluate for postoperative perforation or leakage .
2. CT Findings: What to Look for
Imaging findings in acute colonic diverticulitis can be divided into imaging findings related to the bowel wall and extra-intestinal wall signs.
2.1. Bowel Wall Imaging Findings
The primary finding of the intestinal wall in evaluating diverticulitis is the mural thickening (Figure 1). The intestinal wall, if distended, has a thickness ranging up to 3 mm ; if the segment of the colon is contracted, the wall thickness is considered normal up to 8 mm .
Figure 1.
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Axial contrast-enhanced CT images show acute left colonic diverticulitis and associated findings of perivisceritis (a,b). Once the longitudinal axis of the colon is identified (green straight line in (a)), bowel wall thickness is measured perpendicular to the centerline (9.6 mm, red segment in (a)), showing an increase in the maximum distance from the serosal-to-mucosal surface. Additional mesenteric findings (b), such as increased density of pericolic fat (arrow), thickening of latero-conal fascia (“comma sign”) (arrowhead), and abdominal free fluid (asterisk), are found.
For the correct mural measurement, it is necessary to identify the longitudinal centerline of the colon, and measurements were made perpendicular to the centerline, excluding abscess and lumen, to identify the maximum distance from the serosal-to-mucosal surface of the colon, including the folds and teniae coli (Figure 1). When a lumen could not be seen, the entire serosa-to-serosa distance was measured and divided in half . Dickerson et al. showed that maximum colonic wall thickness in patients with ACD predicts recurrent diverticulitis and may be helpful in stratifying patients according to the need for elective partial colectomy .
The length of the involved colon should also be measured and indicated based on the useful information provided with multiplanar reconstruction and 2D and 3D reformatting software . Often, the mural thickening is associated with an identifiable inflamed diverticulum; in these cases, it is identifiable as an “arrowhead sign” if a portion of fluid crosses the edematous neck of the diverticulum . A severity scale proposed by Dickerson et al. quantifies diverticula based on the distance from each other as few (more than 5 cm in between), mild (1–5 cm), moderate (<1 cm), and severe (no distance) .
2.2. Extra-Intestinal Wall Imaging Findings
2.2.1. Mesenteric Findings
An increased density of pericolic fat [fat stranding] and a small amount of pericolic fluid represent the main mesenteric findings, which, combined with the mural thickening, suggest a localized inflammatory process in uncomplicated diverticulitis (Figure 2) [10,12,13]. The degree of fat stranding can range from “dirty fat” to peridiverticular phlegmon. The phlegmon consists of an inflammatory mass, without walls, located near the inflamed colonic tract, round or oval in shape, and on CT, it presents with high attenuation compared to the mesenteric fat, without an enhancing wall (Figure 3) [3,5]. Multiplanar reconstruction allows for identifying minimum amounts of pericolic fat stranding in the case of horizontal colonic segments . Pereira et al. suggest that the presence of “disproportionate” fat stranding concerning mural thickening suggests the diagnosis of diverticulitis . A small amount of fluid on the root of the mesentery [comma sign] and thickening of the lateroconal fascia are additional signs of an inflammatory process (Figure 1) .
Figure 2.
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Axial contrast-enhanced CT images show wall thickening of the left colon (thick arrow) and the presence of perivisceritis, including the increased density of pericolic fat (fat stranding) (dotted arrow) and thickening of the left lateroconal fascia (so-called comma sign) (arrowhead).
Figure 3.
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Axial unenhanced (a), axial and coronal contrast-enhanced CT images (b,c) show wall thickening of the left colon and inflammatory mass, without walls, located near the inflamed colonic tract, oval in shape, without an enhancing wall, that represents a peridiverticular phlegmon (arrowheads).
2.2.2. Vascular Findings
The inflammatory process could cause engorgement of the mesenteric vessels at the involved colonic tract, which is appreciated on CT images as the “centipede” sign . ACD is one of the most common causes of pylephlebitis, ascending septic thrombophlebitis, characterized by inflammation and septic thrombosis of the mesenteric and portal venous systems [2,5]. Pylephlebitis represents the extension of the septic process to the efferent venous system from the inflamed bowel region . CE-CT imaging shows filling defects in the mesenteric or portal vein, gas or soft tissue density within the vein (representing purulent material), and circumferential stranding of the perivascular fat can be appreciated . Complications of thrombophlebitis include the development of liver abscesses, septic emboli, venous rupture, and pulmonary thromboembolism . Representing the diverticulum as an outpouching from a parietal defect from where the blood vessels penetrate, diverticular bleeding can be an occurrence, more appreciable in chronic diverticulitis [3,24]. On unenhanced CT, hyperdense intraluminal contents can suggest bleeding; after contrast medium, the presence of active contrast extravasation with enlarging contrast volume on the portal venous phase represents the finding of active bleeding (Figure 4).
Figure 4.
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Axial contrast-enhanced arterial (a), portal (b), and delayed (c) CT phases show the presence of arterial active contrast extravasation within a diverticulum (arrowheads), more appreciable in the later phases of the study (b,c); this is a typical finding of active bleeding. Following DSA (d,e) reveals active extravasation of iodinated contrast medium (arrow), allowing identification of the source of the bleeding, which was then treated with coil embolization (dotted arrow in (e)).
2.2.3. Findings of Complications
CT findings of acute complicated diverticulitis include extra-luminal free-air, abscess, intra-abdominal free-fluid, fistula, and bleeding .
The severe inflammatory state of the colonic wall could lead to mural necrosis and perforation of the bowel . Free air can be well-contained and self-limiting, consisting of localized gas adjacent to the inflamed colon (air bubbles or pockets), or it can spread into the intraperitoneal or retroperitoneal cavity [5,8]. Pericolic free air is defined as the presence of air bubbles or air collection within 5 cm of the inflamed bowel segment without distant air (Figure 5), or distant free air collections in the abdominal or retroperitoneal cavity with a distance >5 cm from the inflamed bowel segment (Figure 6) . CT can detect signs of perforation, including bowel wall discontinuity and extraluminal air, especially using lung window settings (Figure 6) .
Figure 5.
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Coronal contrast-enhanced CT images show wall thickening of the left colon with signs of perivisceritis (asterisk) and localized air bubbles adjacent to the inflamed colon (arrow).
Figure 6.
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Axial and coronal contrast-enhanced CT images (a,b) show an abscess, a fluid-containing mass with air and an enhancing wall (white arrow), near the inflamed colonic tract (dotted arrow) and signs of perforation, with distant air bubbles below the left hemidiaphragm (arrowhead in (a)); lung window axial CT image (c) shows distant free air anteriorly to the liver (arrowhead).
Diverticulitis can cause abscess formation. An abscess is defined as a fluid-containing mass with or without air and an avidly enhancing wall that can be intramural, pericolic, or distant from the inflamed colon (Figure 6 and Figure 7). It is necessary to report the inflamed colonic tract’s localization, the abscess’s maximum diameter (<4 cm or ≥4 cm), and the anatomical relationships with the surrounding structures, especially intestinal loops or organs, with which fistulas can develop. Regarding the dimensions, a diameter greater than or equal to 4 cm indicates percutaneous drainage (Figure 7) [8,10].
Figure 7.
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Axial (a) and coronal (b) contrast-enhanced CT images show an abscess near the sigmoid colon (arrowhead), located close to the uterine fundus (arrow). An axial unenhanced CT image (c) shows a 10 F pigtail catheter in the abscess collection.
The most common sites of distant abscesses are the liver, adnexa, lungs, brain, and spine [5,27]. The tubo-ovarian abscess can be caused by the close contiguity of the sigmoid colon with the adnexa; instead, colonic mucosal defects can cause the hematogenous spread of bacteria and induce the formation of abscesses in distant sites. The presence of an abscess in direct contact with another anatomical structure, violating its parietal integrity, leads to the formation of a fistula . Fistulas occur in about 2% of cases of acute diverticulitis , and the most common types, in descending order, are colovesical, colocutaneous, colovaginal, coloenteric, and colouterine . Colovesical fistulas are more frequent in men due to the protective effect of the uterus in women . It manifests clinically as dysuria, pneumaturia, or fecaluria, localized above the left posterior side of the bladder due to its proximity to the sigmoid colon . The CT shows the lack of a fatty cleavage plane between the colon and the bladder, the thickening of the organ wall, and the presence of free intravesical air (Figure 8) . Administration of contrast medium into the bladder or rectum and fluid-sensitive sequences in magnetic resonance imaging may help locate the sinus tract [3,5].
Figure 8.
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A sagittal contrast-enhanced CT image shows a fistulous tract (arrowhead) between the thickened wall of the sigmoid colon and the posterior bladder wall that appears thickened (dotted arrow) and the presence of free intravesical air (arrow). These findings are suggestive of a colovescical fistula.
The presence of air bubbles in the uterine cavity could be a sign of a myometrial abscess due to the formation of a colouterine fistula. In contrast, in hysterectomized patients, the colovaginal fistula should be excluded .
Peritonitis is a life-threatening complication with visceral perforation as its etiological cause. Peritonitis can be purulent when the perforation concerns a pericolic abscess or feculent [stercoraceous] when the perforation concerns a non-inflamed diverticulum [3,31]. On CT, the signs of peritonitis are represented by free fluid in the peritoneal cavity and thickening and enhancement of the peritoneal layers (Figure 9); Sartelli et al. defined diffuse fluid as the presence of abdominal effusion in at least two distant quadrants .
Figure 9.
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Axial contrast-enhanced CT images (a,b) show the presence of pericolic air bubbles and air collections (arrows) and diffuse peritoneal free fluid (asterisks); lung window axial CT (c) shows distant free air (arrowheads) below the abdominal wall.
3. Staging and Management
In 1978, Hinchey et al. classified the severity of ACD into four stages, based on intraoperative findings, to guide surgical management .
An exclusively peri-colonic phlegmonous change or abscess characterizes stage I, whereas a distant pelvic, retroperitoneal, or intra-abdominal abscess categorizes stage II. Stage III includes purulent peritonitis, while stage IV occurs when a large perforation of the loop causes fecal or stercoraceous peritonitis.
With the advent of CT, the Hinchey classification has changed to integrate the radiological findings; in 2005, Kaiser et al. modified the Hinchey classification based on CT findings. The modified Hinchey classification included stage 0 (mild clinical diverticulitis), in which the presence of a diverticulum is associated or not with colonic wall thickening; stage Ia (confined pericolonic inflammation/phlegmon), in which colonic wall thickening is associated with pericolic soft tissue changes; and stage Ib (pericolonic/mesocolic abscess), in which the pericolonic/mesocolic abscess appears .
To overcome the lack of categorization on chronic changes, Schreyer et al. developed a classification (Table 1) in which they include chronic diverticular disease, recurrent or chronic symptomatic diverticular disease (type 3), and diverticular bleeding (type 4).
Table 1.
Classifications in use for ACD.
| | Hinchey Classification | Modified Hinchey Classification by Kaiser et al. | CDD by Schreyer et al. | AAST Grade | WSES Classification by Sartelli et al. |
:--- :--- :--- |
| 0 | | Mild clinical diverticulitis | Asymptomatic diverticulosis | | Uncomplicated diverticulitis |
| 1 | Pericolic abscess or phlegmon | | Acute uncomplicated diverticular disease/diverticulitis | Colonic inflammation | |
| 1a | | Confined pericolic inflammation/phlegmon | Diverticulitis/diverticular disease without reaction of surrounding tissue | | Pericolic air bubbles or little pericolic fluid without abscess |
| 1b | | Pericolonic/mesocolic abscess | diverticulitis with phlegmonous reaction of surrounding tissue | | Abscess ≤ 4 cm |
| 2 | Pelvic, distant intraabdominal, or retroperitoneal abscess | Pelvic, distant intraabdominal, or retroperitoneal abscess | Acute complicated diverticulitis as in 1b, additionally | Colon microperforation or pericolic phlegmon without abscess | |
| 2a | | | Microabscess | | Abscess > 4 cm |
| 2b | | | Macroabscess | | Distant air (>5 cm from inflamed bowel segment) |
| 2c | | | Free perforation (2c1 purulent peritonitis; 2c2 fecal peritonitis) | | |
| 3 | Generalized purulent peritonitis | Generalized purulent peritonitis | Chronic diverticular disease recurrent or chronic symptomatic diverticular disease | Localized pericolic abscess | Diffuse fluid without distant free air (no hole in colon) |
| 3a | | | Symptomatic uncomplicated diverticular disease (SUDD) | | |
| 3b | | | Recurrent diverticulitis without complications | | |
| 3c | | | Recurrent diverticulitis with complications | | |
| 4 | Generalized fecal peritonitis | Generalized fecal peritonitis | Diverticular bleeding | Distant and/or multiple abscesses | Diffuse fluid with distant free air (persistent hole in the colon) |
| 5 | | | | Free colonic perforation with generalized peritonitis | |
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CDD—classification of diverticular disease; AAST—American Association for the Surgery of Trauma.
The American Association for the Surgery of Trauma (AAST) developed a grading scale from I (mild disease) to V (severe disease) for ACD that addresses clinical, radiologic, operative, and pathologic grades of disease .
Some pilot studies showed that, compared with the modified Hinchey classification, the AAST grade better predicted the decision to operate and was equivalent in predicting procedural intervention and complications .
In 2015, a new classification based on radiological findings was proposed and included in the WSES guidelines for managing ACD [8,10]. This classification divides ACD into two groups: complicated and uncomplicated. Uncomplicated ACD is characterized only by thickening of the wall of the diverticula, with increased density of the peri-colic fat. Complicated diverticulitis is divided into the Ia stage with pericolic air bubbles or little pericolic fluid without abscess; Ib with abscess ≤4 cm; IIa with abscess >4 cm; IIb with distant air; III with diffuse fluid without distant free air; and IV with diffuse fluid with distant free air.
At each stage, more appropriate management is suggested, from conservative treatment with broad-spectrum antibiotic therapy (stages Ib and Ia) to percutaneous drainage (stage IIa), up to surgical treatment, from resection with anastomosis (stage IIb or III in stable patients) to Hartmann’s resection (stage IIb or III in unstable patients or stage IV) .
A recent multicenter study comparing the three most used classifications showed that the AAST, WSES, and modified Hinchey classifications were similar in predicting complications, reintervention, and mortality rates . The area under the curve (AUC) for the need for surgery and the occurrence of major complications was higher for AAST and modified Hinchey scores .
4. Differential Diagnosis
The main differential diagnoses of acute colonic diverticulitis include colon carcinoma, epiploic appendagitis, ischemic colitis, appendicitis, infectious enterocolitis, and inflammatory bowel disease [2,12]. Colon adenocarcinoma must be excluded from the differential diagnoses, as it presents as an eccentric or circumferential mural thickening . Compared to ACD, the colonic segment is <5–10 cm involved, there are no diverticula, and there is an abrupt transition between the thickened segment and the normal loop (shoulder sign) (Figure 10) [2,12]. Local lymphadenopathy suggests colorectal cancer .
Figure 10.
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Axial (a), coronal (b), and sagittal (c) contrast-enhanced CT show stenosing annular colon adenocarcinoma (thick arrow in (a)), with its typical “apple core” appearance and associated malignant lymphadenopathy (thin arrow in (b)). There is also the presence of a colocolic fistula (arrowheads).
The epiploic appendages are protrusions of fatty tissue located on the antimesenteric side of the colon. In case of torsion or thrombosis of their vascular pedicle, they can manifest ischemia and inflammation and aim for acute diverticulitis with phlegmon . CE-CT shows an oval mass with adipose tissue density located on the antimesenteric side and with peripheral hyperattenuating (hyperattenuating ring sign), expression of inflamed fat, and a central focal area of hyperattenuating (central dot sign), which corresponds to the thrombosed vein within the inflamed appendage (Figure 11) . Compared with acute diverticulitis, the associated diverticular disease is not present in appendagitis without colonic wall thickening.
Figure 11.
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Axial contrast-enhanced CT shows an oval mass with adipose tissue density located on the antimesenteric side (circle), which presents fat inflammation, demonstrated by peripheral hyperdensity (arrowhead in detail) and central thrombosed vein (arrow in detail). These findings are consistent with epiploic appendagitis.
5. Proposed Reporting Template
Despite different classifications for diagnosing and managing ADC, to our knowledge, no reporting templates have been proposed for the structured reporting of acute diverticulitis and its complications.
Some studies demonstrated how greater standardization in reporting leads to improved communication and more complete reports and reduces variability and error rates [40,41].
Structured reporting facilitates the radiology resident’s learning, providing a systematic approach to recognizing the specific pathology’s key characteristics to be reported in the radiology report .
Structured reporting has been introduced in other abdominal pathologies, particularly in oncology abdominal imaging, such as rectal cancer, where it has become necessary in staging and restaging and facilitates clinical decision making .
The proposed reporting template concerns acute diverticulitis and is designed to be used with contrast-enhanced CT imaging, as shown in Figure 12.
Figure 12.
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Structured reporting template.
This template incorporates the classification scheme and terminology of the WSES classification while maintaining the flexibility to add free text for the qualitative aspects of the report. The model includes colonic wall findings and complication findings, divided into signs of perforation, abscess, free fluid, and vascular complications.
Our aim is to standardize the language, avoid the nuances expected by radiologists in their reports, improve communication between specialists involved in patient care, and standardize the multidisciplinary approach.
6. Conclusions
CE-CT is the diagnostic modality of choice if ACD is suspected. In fact, it plays an essential role in evaluating the extent, severity, and complications of this pathology and, therefore, allows the correct management and surgical strategy to be established.
In this paper, we have reviewed the most used classifications of ACD, proposing a new structured reporting template that, through a system of appropriate terminology, can reduce discrepancies between radiological reports and standardize the language among specialists in managing this condition.
Further studies are needed to validate improved outcomes in ACD patients using a standardized reporting template.
Author Contributions
Conceptualization, F.T. and A.B.; methodology, F.T. and A.B.; software, F.T.; validation, F.T., A.B., S.P. and P.V.F.; formal analysis, F.T., S.D.P. and D.M.; investigation, F.T. and D.M.; resources, A.B.; data curation, F.T., S.D.P. and D.M.; writing—original draft preparation, F.T.; writing—review and editing, F.T. and S.D.P.; visualization, S.G., C.M., F.G., M.V., G.N.Z., S.P., P.V.F., D.G.C. and C.I.; supervision, A.B.; project administration, A.B.; funding acquisition, A.B. All authors have read and agreed to the published version of the manuscript.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Written informed consent has been obtained from the patients to publish this paper.
Data Availability Statement
Not applicable.
Conflicts of Interest
The authors declare no conflict of interest.
Funding Statement
This research received no external funding.
Footnotes
Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.
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Abstract
1. Introduction
2. CT Findings: What to Look for
3. Staging and Management
4. Differential Diagnosis
5. Proposed Reporting Template
6. Conclusions
Author Contributions
Institutional Review Board Statement
Informed Consent Statement
Data Availability Statement
Conflicts of Interest
Funding Statement
Footnotes
References
Associated Data
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Alveolar Pressure
In subject area:Veterinary Science and Veterinary Medicine
Alveolar pressure is defined as the pressure within the lung alveoli that fluctuates in relation to atmospheric pressure during the respiratory cycle, specifically becoming more negative during inspiration and returning to baseline at peak inspiration and expiration.
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: Explain why surfactant (a) lowers surface tension, (b) reduces the work of breathing, (c) helps stabilize alveoli size, and (d) prevents low interstitial fluid pressures.
4.
: The relaxation volume–pressure curve shows that large lung volumes must be supported by large positive pressure (solid line in Figure 6.1.8). In dynamic breathing, alveolar pressure during inspiration is never positive! How do the lungs get to that position?
5.
: At rest, the pulmonary ventilation is about 6 L min−1. During exercise, it can increase markedly, depending on the size of the person and their physical fitness, to highs near 120 L min−1. What happens to alveolar pressure, PA, during inspiration and expiration during exercise, with respect to its value at rest?
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Book2017, Quantitative Human Physiology (Second Edition)Joseph Feher
Chapter
Pulmonary gas exchange
2014, Veterinary Anaesthesia (Eleventh Edition)
Spontaneous respiration
In a spontaneously breathing animal, active contraction of the inspiratory muscles lowers the normally subatmospheric intrapleural pressure still further by enlarging the relatively rigid thoracic cavity. The decrease in intrapleural pressure lowers the alveolar pressure (Fig. 9.1) so that a pressure gradient or driving force is set up between the exterior and the alveoli. This overcomes the airway resistance and air flows into the alveoli until, at the end of inspiration, the alveolar pressure becomes equal to the atmospheric pressure. During expiration the pressure gradient is reversed and air flows out of the alveoli.
The transpulmonary pressure is a measure of the elastic forces which tend to collapse the lungs (Fig. 9.1). There is no one single intrapleural pressure; in the ventral parts of the chest it is just sufficient to keep the lungs expanded but because of the influence of gravity acting on the lungs, in the dorsal parts of the chest the intrapleural pressure should be much more below atmospheric. However, it is not at all certain how uniform the pressure on the pleural surface of the lung really is. The hilar forces, the buoyancy of the lung in the pleural cavity and the different shapes of the lung and chest wall are all possible sources of local pressure differences. Thus, it is customary to measure the intraoesophageal pressure as being representative of the mean intrapleural pressure (Fig. 9.2).
The alveolar pressure changes generate airflow into and out of the lungs against a resistance in a way analagous to that stated by Ohm's Law for electricity, where:
So that:
Airway resistance is largely influenced by the lung volume because the elastic recoil of lung parenchyma exerts traction on the pleural surfaces and walls of airways (holding them patent) when the lungs are inflated above residual volume. As the lungs are further inflated, elastic recoil pressure increases, thus further dilating the airways and decreasing resistance to air flow. This relationship between airway resistance and lung volume is hyperbolic in nature, as shown in Figure 9. 3. Airway resistance also depends on the nature of airflow through the airway. With a clear airway and a low gas flow rate, intrapulmonary flow is largely laminar (streamlined) and airway resistance is also low, but obstruction or a high flow velocity will give rise to turbulence and a greatly increased resistance. Measurement of airway resistance must be made when gas is flowing. During IPPV, when the chest wall is intact, resistance to expansion of the lungs is also offered by the chest wall which then contributes to the total respiratory resistance.
Total respiratory resistance (Rrs) may be estimated by the application of an oscillating airflow to the airways with measurement of the resultant pressure and airflow changes. A technique was developed by Lehane et al. (1980) to measure airway resistance as a function of lung volume during a vital capacity manoeuvre and so to derive specific lower airways conductance, s.Glaw (conductance being the reciprocal of resistance) and the expiratory reserve volume (ERV). The method was modified by Watney et al. (1987, 1988) for use in anaesthetized and paralysed horses and dogs and it was demonstrated that, in ponies, xylazine, acepromazine, halothane and enflurane produce bronchodilation and a decrease in ERV while isoflurane appears to increase ERV. In dogs, it was concluded that both bronchoconstriction and changes in lung volume may be responsible for changes in airway resistance seen during hypoxia. During spontaneous breathing, changes in resistance may necessitate a great increase in the work of breathing. The effect of inhalation anaesthetics on total respiratory resistance in conscious horses was studied by Hall & Young (1992) who showed that halothane appeared to have no effect while enflurane and isoflurane seemed to increase it. In contrast in humans, neither isoflurane nor sevoflurane altered résistance, although desflurane at higher concentrations did cause an increase (Nyktari et al, 2011).
Resistance is not the only factor opposing movement of air in and out of the chest; a full analysis includes the effects of compliance and inertance. Adding the compliance and inertance forms the reactance and this can be combined with the resistance in one complex term called the ‘impedance’. If the impedance of the respiratory system is known then the resistance and reactance can be determined. A non-invasive method (Michaelson et al., 1975) that does not require patient cooperation has been adapted for use in conscious animals as described by Young and Hall (1989) for horses but it is difficult to use in anaesthetized, intubated animals because the impedance of the tube alone is much greater than that of a non-intubated animal. Small airways contribute little to the total lung resistance; although each one has a large individual resistance, there are large numbers in parallel so that the overall effect is small. This is important because small airway disease (which increases local resistance) is not detected by measurement of total airway resistance until the condition is well advanced.
Anaesthetic apparatus may afford resistance that is considerably higher than that offered by the animal's respiratory tract. However, it is unlikely that moderate expiratory resistance will cause serious problems in spontaneously breathing animals and, indeed, positive end expiratory pressure (PEEP) (see PEEP and CPAP below) has many positive advantages. However, resistance increases the work of breathing, and common sense suggests that apparatus resistance should be kept to a minimum. Purchase (1965a,b) studied the resistance afforded by four closed breathing systems used in horses and cattle and in three, all of which had internal bores of 5 cm, found it to be of the order of 1 cmH2O (0.1 kPa) per 100L/min at flow rates of 600 L/min. The resistance of endotracheal tube connectors was relatively high in comparison with that of the remainder of the apparatus.
During a breathing cycle, mean intrathoracic pressure may be above or below atmospheric pressure as a result of apparatus resistance. For example, if the expiratory flow through a piece of apparatus with a high resistance is great enough to induce turbulence, while the inspiratory rate is low (as it often is in horses) so that during inspiration the flow is laminar, the mean intrathoracic pressure will be above atmospheric. Conversely, if the inspiratory flow rate is greater, there may be a subatmospheric mean intrathoracic pressure. Mean intrathoracic pressures above atmospheric reduce the effect of the thoracoabdominal pump for venous return, with subsequent cardiovascular effects. Large subatmospheric mean intrathoracic pressures may be equally dangerous, perhaps by producing pulmonary oedema, but probably more importantly by reducing lung volume. Trapping of gas in the lungs occurs more readily at low lung volumes and gas trapping produces widespread airway obstruction with serious impairment of respiratory function.
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Book2014, Veterinary Anaesthesia (Eleventh Edition)
Chapter
Pulmonary gas exchange: artifical ventilation of the lungs
2001, Veterinary Anaesthesia (Tenth Edition)L.W. Hall MA, BSc, PhD, Dr(Hons Causa)Utrecht, DVA, DEVC, Hon DACVA, FRCVS, ... C.M. Trim BVSc, MRCVS, DVA, DACVA, DECVA
SPONTANEOUS RESPIRATION
In a spontaneously breathing animal active contraction of the inspiratory muscles lowers the normally subatmospheric intrapleural pressure still further by enlarging the relatively rigid thoracic cavity. The decrease in intrapleural pressure lowers the alveolar pressure (Fig. 8.1) so that a pressure gradient or driving force is set up between the exterior and the alveoli. This overcomes the airway resistance and air flows into the alveoli until at the end of inspiration the alveolar pressure becomes equal to the atmospheric pressure. During expiration the pressure gradient is reversed and air flows out of the alveoli.
The transpulmonary pressure is a measure of the elastic forces which tend to collapse the lungs and there is no one intrapleural pressure. In the ventral parts of the chest it is just sufficient to keep the lungs expanded. Because of the influence of gravity acting on the lungs, the intrapleural pressure in the dorsal parts of the chest should be much more below atmospheric, but it is not at all certain how uniform the pressure on the pleural surface ot the lung really is. The hilar forces, the buoyancy of the lung in the pleural cavity and the different shapes of the lung and chest wall are all possible sources of local pressure differences. Thus, it is customary to measure the intra-oesophageal pressure as being representative of the mean intrapleural pressure (Fig. 8.2)
The alveolar pressure changes generate airflow into and out of the lungs against a resistance in a way analagous to that stated by Ohm's Law for electricity, where:
So that:
Airway resistance is largely influenced by the lung volume because the elastic recoil of lung parenchyma exerts traction on the pleural surfaces and walls of airways (holding them patent) when the lungs are inflated above residual volume. As the lungs are further inflated, elastic recoil pressure increases, thus further dilating the airways and decreasing resistance to air flow. This relationship between airway resistance and lung volume is hyperbolic in nature, as shown in Fig. 8.3. Airway resistance also depends on the nature of airflow through the airway. With a clear airway and a low gas flow rate, intrapulmonary flow is largely laminar (streamlined) and airway resistance is also low, but obstruction or a high flow velocity will give rise to turbulence and a greatly increased resistance. Measurement of airway resistance must be made when gas is flowing. During IPPV when the chest wall is intact, resistance to expansion of the lungs is also offered by the chest wall which then contributes to the total respiratory resistance.
Total respiratory resistance (Rrs) may be estimated by the application of an oscillating airflow to the airways with measurement of the resultant pressure and airflow changes. A technique was developed by Lehane et al. (1980) to measure airway resistance as a function of lung volume during a vital capacity manoeuvre and so to derive specific lower airways conductance, s.Glaw, (conductance being the reciprocal of resistance) and the expiratory reserve volume (ERV). The method was modified by Watney for use in anaesthetized and paralysed horses and dogs (Watney et al., 1987; 1988) and it was demonstrated that in ponies xylazine, acepromazine, halothane and enflurane produce bronchodilatation and a decrease in ERV while isoflurane appears to increase ERV. In dogs, it was concluded that both bronchoconstriction and changes in lung volume may be responsible for changes in airway resistance seen during hypoxia. During spontaneous breathing changes in resistance may necessitate a great increase in the work of breathing. The effect of inhalation anaesthetics on total respiratory resistance in conscious horses was studied by Hall and Young (1992) who showed that halothane appeared to have no effect while enflurane and isoflurane seemed to increase it.
Resistance is not the only factor opposing movement of air in and out of the chest; a full analysis includes the effects of compliance and inertance. Adding the compliance and inertance forms the reactance and this can be combined with the resistance in one complex term called the ‘impedance’. If the impedance of the respiratory system is known then the resistance and reactance can be determined. A completely non-invasive method suggested by Michaelson et al. (1975) has, since the introduction of computers, been used quite extensively to determine the frequency dependence of resistance in man. Because it does not require patient cooperation it is relatively simple to use in conscious animals as described by Young and Hall (1989) for horses but it is dificult to use in anaesthetized, intubated animals because the impedance of the tube alone is much greater than that of a non-intubated animal. Commercially available apparatus is expensive but, nevertheless, it can provide a useful diagnostic tool for the identification of horses suffering from chronic obstructive pulmonary disease (COPD) which can pose problems during anaesthesia with IPPV because of a sudden increase airway resistance at the end of expiration (Gillespie et al., 1966). This increase does not usually present problems during anaesthesia with spontaneous breathing but may necessitate a prolonged expiratory period during IPPV. Unfortunately, small airways contribute little to the total lung resistance; although each one has a large individual resistance, there are large numbers in parallel so that the overall effect is small (Fig. 8.4). This is important because small airway disease (which increases local resistance) is not detected by measurement of total airway resistance until the condition is well advanced.
Anaesthetic apparatus may afford resistance that is considerably higher than that offered by the animal's respiratory tract. It is difficult to say at what value this apparatus resistance becomes intolerable because a healthy anaesthetized animal seems able to compensate for increases in resistance to airflow. Anaesthetized human subjects breathe at 80 % of control tidal volume against an inspiratory load of 10 cm H2O (Nunn & Ezi-Ashi, 1961). It is unlikely that moderate expiratory resistance will cause serious problems in spontaneously breathing animals provided the P aCO2 remains within acceptable limits. In halothane anaesthetized horses, Hall and Trim (1975) found that 10 and 20 cm H2O of expiratory resistance did not affect the P aO2 but was associated with increases in P aCO2 of about 1.2 mmHg (0.16 kPa) per min. However, common sense would seem to suggest that apparatus resistance should be kept to a minimum. Purchase (1965; 1965a) studied the resistance afforded by four closed breathing systems used in horses and cattle and in three, all of which had internal bores of 5 cm, found it to be of the order of 1 cm H2O (0.1 kPa) per 100l/min at flow rates of 600 l/min, which he judged to be quite acceptable. He also found that the resistance of endotracheal tube connectors was relatively high in comparison with that of the remainder of the apparatus.
During a breathing cycle mean intrathoracic pressure may be above or below atmospheric pressure as a result of apparatus resistance. For example, if the expiratory flow through a piece of apparatus with a high resistance is great enough to induce turbulence, whilst the inspiratory rate is lower (as it often is in horses) so that during inspiration the flow is laminar, the mean intrathoracic pressure will be above atmospheric. Conversely, if the inspiratory flow rate is greater there may be a subatmospheric mean intrathoracic pressure. Mean intrathoracic pressures above atmospheric may cause cardiovascular failure in hypovolaemic states by reducing the effect of the thoraco-abdominal pump for venous return. Large subatmospheric mean intrathoracic pressures may be equally dangerous, perhaps by producing pulmonary oedema, but probably more importantly by reducing lung volume. Trapping of gas in the lungs occurs more readily at low lung volumes and gas trapping produces widespread airway obstruction with serious impairment of respiratory function.
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Book2001, Veterinary Anaesthesia (Tenth Edition)L.W. Hall MA, BSc, PhD, Dr(Hons Causa)Utrecht, DVA, DEVC, Hon DACVA, FRCVS, ... C.M. Trim BVSc, MRCVS, DVA, DACVA, DECVA
Chapter
The Mechanics of Breathing
2017, Quantitative Human Physiology (Second Edition)Joseph Feher
The Pressure Drop Across the Lung and Chest Wall Can Be Estimated Using Ppl
The contribution of the chest wall and the lung to the total pressure, Prs, can be determined by recognizing that the intrapleural space divides the total pressures into two components, as shown in Figure 6.1.7. The difference between the alveolar pressure and the barometric pressure is Prs.
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Book2017, Quantitative Human Physiology (Second Edition)Joseph Feher
Chapter
The Respiratory Cycle
2015, Back to Basics in PhysiologyJuan Pablo Arroyo, Adam J. Schweickert
Composition of Alveolar Air
Alveolar air is a product of both alveolar ventilation, as we studied in the previous section, and the rate at which blood extracts O2 from and releases CO2 into the alveoli. However, keep in mind that in spite of alveolar ventilation being approximately 4,200 mL/min under steady state conditions, the turnover rate of alveolar air is surprisingly small. What this means is that the complete renewal of the air that is present in the alveoli does not happen with each breath. The main reason for this is the FRC. The FRC is 2300 mL, and with each breath approximately 350 mL of air gets inspired and expired. This means that only 15% of the FRC is exchanged with each breath. Think of this as a failsafe of sorts. A small turnover rate allows for a gradual gas exchange, which means that even though we are breathing in and out all the time, there are no sudden changes to blood gas concentrations.
So, what exactly are the partial pressures of O2 and CO2 in the alveoli? Table 4.1 shows the progressive changes in PO2 and PCO2 as air moves down the respiratory system. In a nutshell PO2 decreases and PCO2 increases the closer we get to the alveoli. This should be intuitive as O2 is being consumed and CO2 is being produced. The initial decrease in O2 from atmospheric to moist tracheal air is due to the increase in PH2O from 0 to 47 mmHg. Then PO2 continues to decrease and CO2 continues to increase until the air reaches the alveoli. However, in order for exchange to take place appropriately the alveolar pressure of O2 or PAO2 should be approximately 100 mmHg at sea level, and the alveolar pressure of CO2 or PACO2 should be approximately 40 mmHg. Commit these numbers to memory.
Table 4.1. Approximate Standard Partial Pressures of Gases at Sea Level on an Average Day
| Partial Pressures of Gases in mmHg |
| Empty Cell | Atmospheric Air | Moist Tracheal Air | Alveolar Air | Expired Air |
| PO2 | 160 | 150 | 104 | 120 |
| PCO2 | 0 | 0 | 40 | 27 |
| PH2O | 0 | 47 | 47 | 47 |
| PN2 | 600 | 563 | 569 | 566 |
| Ptotal | 760 | 760 | 760 | 760 |
Key
At sea level the alveolar pressure of O2 (PAO2) is approximately 100 mmHg, and the alveolar pressure of CO2 (PACO2) is approximately 40 mmHg.
Clinical Correlate
FiO2: Fraction of inspired oxygen
As we said, 21% of the atmospheric air is O2. Therefore the fraction of air we inspire that is O2 or (FiO2) is usually 21%. However, when a patient is sick from a respiratory illness (or even from anemia or shock), giving O2 is an imperative part of stabilizing the patient. A simple way to get more O2 to a patient is by increasing the amount of O2 inhaled with each breath. Thus the blood will pick up more O2 and will end up delivering more O2 to cells in the body. Therefore, increasing the partial pressure of O2 by providing supplemental O2, and thus altering the FiO2, can be critical in unstable patients.
The PAO2 and the PACO2
Remember, the goal is to get O2 into the body! To do this, we need to increase the amount of O2 in the alveoli. So let’s take a look at two factors that regulate the alveolar pressure of O2 (PAO2):
•
: The rate at which O2 is brought in by the ventilation system. In broad terms this means that the higher the alveolar ventilation (VA) the more the PAO2 is going to approximate the pressure of O2 in the atmospheric air.
•
: The rate of O2 extraction by the pulmonary capillaries. In a nondiseased state the amount of O2 that the pulmonary capillaries extract from the alveoli is directly related to the amount of O2 being consumed by the body. This means that the body’s metabolic rate is directly related to the pulmonary extraction of O2.
The relationship between PAO2, alveolar ventilation, and O2 consumption is represented in Figure 4.3A. Alveolar ventilation in L/min is on the X-axis, while pressure of O2 in mmHg is on the Y-axis. The top curve represents a standard O2 consumption of 250 mL/min. Maintaining a PAO2 of approximately 100 mmHg requires a VA of approximately 5 L (point 1). If we were to maintain O2 consumption stable at 250 mL/min and we changed the VA, the PAO2 would follow VA; that is, as VA increases the PAO2 would increase and if VA decreases the PAO2 would decrease. However in the setting of increased metabolic requirements (e.g., aerobic exercise), there is an increase in O2 consumption. Imagine if we run to catch the bus and this tripled O2 consumption of O2 from 250 mL/min to 750 mL/min (bottom curve). If VA stayed at 5 L/min, the PAO2 would be around 50 mmHg!! So in order to return the PAO2 to 100 mmHg, VA must also triple and increase from around 5 L/min to 15 L/min (point 2).
Key
PA denotes alveolar pressures whereas Pa denotes arterial pressures.
In contrast to O2 the goal with CO2 is to get it out of the body, and curiously enough there are also two factor that regulate the alveolar pressure of CO2 (PACO2):
•
: The rate at which the pulmonary capillary exchanges CO2 with the alveoli. Essentially in a non-diseased state this amounts to the amount of CO2 that is being produced by the body. This means that the more CO2 the body produces, the larger the gradient for exchange with the alveolus. This is a different side of the same energy production coin. Remember Chapter 1? As you consume O2 through aerobic respiration to produce ATP, you produce CO2. Therefore the more O2 you consume the more CO2 you produce.
•
: The rate at which the ventilation system clears the alveoli of CO2. Alveolar ventilation (VA), is also in charge of clearing the alveolar air of CO2. The more air is moved in and out of the alveoli, the more the PACO2 will decrease as it approximates the PCO2 in the atmospheric air, which is close to zero. This means that the higher the rate of alveolar ventilation (VA), the lower the pressure of CO2 at the alveoli level is going to be. The relationship between alveolar ventilation and arterial pressure of CO2 or PaCO2 is summarized in the alveolar CO2 equation or PACO2 equation, which states that:
where:
•
: PaCO2 is the pressure of CO2 in the arterial blood.
•
: VCO2 is the amount of CO2 produced by the body and delivered to the lungs.
•
: VA is alveolar ventilation.
This relationship is based on the idea that the pressure of CO2 in the alveolus (PACO2) and the pressure of CO2 in arterial blood (PaCO2) is essentially equivalent (remember CO2 diffuses extremely fast). The take-home message from this equation is that if VA decreases the PaCO2 will increase and if VA increases the PaCO2 will decrease.
Key
If VA decreases the PaCO2 will increase. If VA increases the PaCO2 will decrease.
This relationship is graphed in Figure 4.3B. The axis are the same as those in Figure 4.3A, but this time the curves represent PACO2. The bottom-most curve represents a normal production of 200 mL of CO2 per minute. At this rate of CO2 production VA needs to be around 5 L/min to maintain a PACO2 of 40 mmHg. As defined by the PaCO2 equation, if the production of CO2 remains stable at 200 mL/min, increases in VA will decrease the PACO2, while a decrease in VA will increase CO2. If we were to increase the production of CO2 from 200 mL/min to 600 mL/min because we ran to catch the bus (top curve), we would have to increase VA from 5 L/min to 15 L/min in order to maintain a PACO2 of 40 mmHg.
The body, being the amazing feat of biological engineering that it is, tries to make efficient use of all its biological processes. Therefore the same alveolar ventilation that brings in O2 is in charge of clearing CO2! This is represented in Figure 4.3C, where we have superimposed Figures 4.3A and 4.3B, and you can see how PAO2, PACO2, and VA are interrelated. Increases in VA will result in both an increase in the PAO2 and a decrease in PACO2, while a decrease in VA will result in a decrease in PAO2 and an increase in PACO2. Think about it like this: If you were to hold your breath right now, you would decrease VA, and by doing so the PAO2 would decrease and the PACO2 would increase. (If you’re hypoventilating, you’re still consuming O2 without bringing any new O2 in, and you’re still producing CO2 without dumping any of it out into the atmosphere!) Conversely if you were to increase VA by hyperventilating, the PAO2 would increase and the PACO2 would decrease. (If you’re hyperventilating, you’re bringing in more O2 than is being consumed, and dumping out more CO2 than is being produced.)
However, the increase in PAO2 is not of the same magnitude as the decrease in PACO2. Why? Well, for starters take a look at the amount of CO2 that is produced (200 mL) and the amount of O2 that is consumed (250 mL). Not exactly a 1:1 ratio is it? The relationship between O2 that is consumed to the CO2 that is produced is called the Respiratory Quotient (RQ). The RQ will depend on the molar ratios of O2 consumption to CO2 production, from the substrate that is being used as fuel by the body. When we use glucose the ratio is 1:1; that is, you consume one mole O2 for every mole CO2 you produce. However the body also uses fat and protein as fuel, which aren’t as efficient. The RQ in the body approximates 0.8 (200 mL CO2/250 mL O2). In other words, for every mole of O2 consumed, the body produces approximately 0.8 moles of CO2. Therefore the RQ helps us estimate how much O2 we consumed in order to produce a given amount of CO2. Why is this important? Well, in clinical practice it’s extremely difficult to directly measure the PAO2, so we calculate it using something called the Alveolar Gas equation. To do so we need to understand the RQ.
The Alveolar Gas Equation, or, How Much O2 is in There!
In the previous section we stated that the two factors that regulated the PAO2 were the renewal of O2 through ventilation and the rate of consumption. Since it is relatively difficult to measure exactly how much O2 is in the alveoli, wouldn’t it be nice if we could put this in a formula to calculate the PAO2? Well, there is a formula and it’s called the alveolar gas equation; it looks something like this:
However, before we delve into the specifics of each variable let’s take a step back. Consider this: If we’re trying to calculate the PAO2 we need to know two things:
•
: How much O2 is being inspired
•
: How much O2 is being consumed
Therefore, here’s a simplified version of the alveolar gas equation:
Since most of the time we are not breathing pure O2, we’ll need to calculate the partial pressure of O2. We can calculate how much O2 is being inspired from partial pressure of O2 in the air that is being breathed in (in this case we’ll use atmospheric air) and the PH2O in the airways with the following formula:
where:
: PATM=Atmospheric pressure (at sea level it would be 760 mmHg).
: FiO2 (Fraction of Inspired Oxygen)=The percentage partial pressure of O2 in the air that the patient is breathing in. Since O2 makes up 21% of the air we breathe under normal conditions the FiO2 would be 0.21 if no extra oxygen is added. (If O2 is added to the mix by providing supplemental oxygen to the patient the FiO2 will increase.)
: PH2O=Partial pressure of H2O in the system. At normal body temperature this is equal to 47 mmHg.
If we plug the numbers into our formula we come up with the following:
Then:
So:
This number is the same number we found in Table 4.1 for moistened tracheal air. So far, so good! Now our simplified alveolar gas equation looks like this:
How do we calculate O2 consumption? Remember what we mentioned about RQ in the previous section? RQ is the ratio of CO2 produced to O2 consumed. So if we know the PACO2 we can estimate how much O2 is being consumed. Lucky for us the CO2 in arterial blood or PaCO2 is almost identical to the PACO2. CO2 diffuses extremely rapidly, and is generally not affected by issues that can alter O2 diffusion (keep in mind that CO2 diffuses about 20 times as fast as O2). Therefore if PaCO2≈PACO2 together with the RQ, we can then calculate how much O2 is being consumed:
where:
: PaCO2=The alveolar pressure of CO2, which can be inferred from an arterial blood sample. Assuming that diffusion is occurring unimpinged, the amount of CO2 in the blood, the PaCO2 is going to be very similar to the PACO2, and is therefore used as a surrogate.
: RQ=The molar ratio of CO2 produced to O2 consumed depending on the fuel being consumed (carbohydrates vs fats vs proteins). In the body it approximates 0.8 and is product of the combined metabolism of carbohydrates, fats, and proteins.
Key
The RQ is the amount of O2 in mmHg that has to be consumed to account for the amount of CO2 in mmHg that is being produced.
Now that we understand the components let’s look at the entire alveolar gas equation:
or
So, let’s plug in the numbers under standard conditions and see what we come up with.
As we saw previously, if we solve the first half first we come up with 150 mmHg.
Dividing by 40 by 0.8 yields 50 mmHg. Let’s go over this again. This number is telling us that for every 40 mmHg of CO2 that we produce we are consuming 50 mmHg of O2. So if your PACO2 is 40, it’s because we are consuming 50 mmHg of O2. Therefore, we subtract the amount of O2 that is being consumed (50 mmHg) from the amount that is being brought into the lungs (150 mmHg). So:
Then:
Success! The PAO2 we calculated is right on the money! Taking small variations into account, the normal PAO2 is approximately 100 mHg.
Key
The alveolar gas equation is a way to estimate the PAO2, and is calculated with the following formula: PAO2=FiO2 (PATM – PH2O) – PaCO2/RQ.
Knowing the pressures of O2 and CO2 at the level of the alveolus, however, is only half the battle! We now need to understand what happens in the blood that allows for the diffusion of these gases from the air and into the blood, and how hemoglobin allows this phenomenon to take place in quantities that are compatible with life.
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Book2015, Back to Basics in PhysiologyJuan Pablo Arroyo, Adam J. Schweickert
Chapter
The respiratory system: Anatomy, physiology, and adaptations to exercise and training
2014, The Athletic Horse (Second Edition)PIERRE LEKEUX, ... DAVID R. HODGSON
Gravitational factors
It has been shown that there is a vertical gradient of pulmonary blood flow, with the ventral regions receiving more perfusion per unit lung volume than the dorsal regions (Amis et al., 1984; Franklin et al., 2012). According to the relative magnitudes of pulmonary arterial, venous, and alveolar pressures, blood flow in the lung can be divided into four zones (Fishman, 1985).
In zone 1, at the top of the lung, there is no blood flow, because the mean pulmonary arterial pressure is too low to overcome the hydrostatic pressure imposed by the column of blood connecting the pulmonary artery to the apical blood vessels. Therefore, alveolar pressure exceeds both pulmonary arterial and venous pressures, and the collapsible capillaries remain closed. However, because the mean pulmonary arterial pressure is about 15 to 18 mm Hg (i.e., 20 to 25 cmH2O), it may be sufficient to perfuse the vertical height of the lung above the heart. This zone is probably small in most horses. Because this lung region is unperfused, it does not participate in gas exchange and represents the “alveolar dead space.” During exercise, the increased pulmonary arterial pressure probably improves the recruitment of the vessels in zone 1 and therefore makes the distribution of perfusion more homogeneous. The alveolar dead space is estimated to be 0.8 L to 1 L in the resting horse or about 3.5% of the functional residual capacity and is very likely to disappear in the exercising horse (Franklin et al., 2012; Lekeux et al., 1992).
In zone 2, pulmonary arterial pressure is greater than the alveolar pressure, the latter being, in turn, greater than venous pressure. Therefore, the capillary is open for a part of its length, until the point where alveolar pressure exceeds intravascular pressure. Consequently, blood flow in zone 2 is determined by the respective values of alveolar and arterial pulmonary pressures (and is independent of venous pressure); it, therefore, increases down this zone of lung, according to the progressive increase of pulmonary arterial pressure as a result of the hydrostatic gradient.
In zone 3, both pulmonary arterial and venous pressures exceed alveolar pressure; capillaries are perfused throughout their length and are increasingly distended down this zone.
A zone 4 is sometimes described in which the pulmonary blood flow decreases as a result of a compressing interstitial pressure on the vessel. Although it is well established that pulmonary blood flow is distributed in a vertical direction with respect to gravity, Hakim et al. (1987) suggested that in humans there also could be a gradient from the center to the periphery. Because local or peripheral vascular resistance rises in proportion to distance from the lung hilum, the center of each lobe will be better perfused than its periphery.
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Book2014, The Athletic Horse (Second Edition)PIERRE LEKEUX, ... DAVID R. HODGSON
Review article
Respiratory Physiology, Diagnostics, and Disease
2007, Veterinary Clinics of North America: Small Animal PracticeAndrew M. Hoffman DVM, DVSc
Air flow arises from voluntary and involuntary signals that drive muscular effort to expand the chest by (1) rib cage expansion and (2) caudal displacement of the diaphragm (Fig. 1). During inspiration, expansion of the chest shifts pleural pressure to a more negative value. Pleural pressure is often cited as the driving force to overcome elastic recoil and resistance (ie, respiratory system impedance). When pleural pressure descends to a more negative value (eg, from −2 to −7 cm H2O), alveolar gas within the lung expands. Air flows into the lung down its pressure gradient but ceases to flow once alveolar pressure returns to zero (atmospheric pressure) at peak inspiration. Flow then reverses during expiration because of transient positive alveolar pressure created by elastic recoil, a process that does not require muscular effort at rest. At end-expiration, alveolar pressure returns to baseline and flow comes to a halt again. In sum, flow is directly attributable to fluctuations in alveolar pressure relative to atmospheric pressure that cycle from negative to positive during each breath. Hence, the appearance of alveolar pressure and flow waveforms (on a strip-chart) are superimposable, and inspiratory flow is often depicted in the negative direction to correspond with negative alveolar pressure. The relevance of alveolar pressure to flow is most important in discussion of the measurement of Raw.
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Journal2007, Veterinary Clinics of North America: Small Animal PracticeAndrew M. Hoffman DVM, DVSc
Chapter
Gas Exchange at Rest and during Exercise in Mammals
2015, Comparative Biology of the Normal Lung (Second Edition)Peter D. Wagner, ... Kim E. Longworth
12Partial Pressures of Alveolar Gas
Fick's second law of diffusion (Eqn (7)) indicates that the flux of a gas will be driven by the difference in partial pressures of the gas across a barrier, e.g., the alveolar-capillary membrane, and Eqn (10) shows that the rates at which respiratory gases are transferred across the lung will be functions of both the mean alveolar pressures of the gases and their mean pressures in the pulmonary capillary blood. The values of these gas pressures are affected by different factors for different gases. The effects of hemoglobin's concentration and allosteric binding properties on the partial pressures of O2 and CO2 in the pulmonary capillary blood have been discussed previously.
The partial pressure of CO2 in alveolar gas (, torr) is a function of the rates at which CO2 is being delivered to the alveolus for removal (, ml [STPD] CO2/(min kg)) and the rate at which it is being removed from the alveolus, or alveolar ventilation (, ml [BTPS]/(min kg)), as stated in the classic alveolar ventilation equation (Fenn et al., 1946):
(26)
where K is a factor that converts gas fractions to partial pressures and BTPS volumes to STPD. Minute ventilation and standard vary in mammals according to allometric equations with indistinguishable mass exponents (Stahl, 1967; Calder, 1984; Schmidt–Nielsen, 1984). If the respiratory exchange ratio were similar and dead space were approximately a constant fraction of tidal volume in all mammals (no data conflict with these assumptions), then it would appear that in all mammals at rest must be nearly identical, approximately 40 torr.
Lahiri (1975) suggested that very small mammals might have lower blood and alveolar than larger mammals, however, measurements of from tissue gas pockets found no difference between values in shrews, bats, and pocket mice, and those in larger mammals (Tenney and Morrison, 1967). Alveolar is generally regarded as being approximately 40 torr and independent of body size in all mammals, except unusually adapted species, e.g., fossorial rodents (Dejours, 1981; Schmidt–Nielsen, 1984). Because of the high diffusivity of CO2, (Table 5), is expected to usually be nearly identical to , with little diffusion limitation.
Whereas is determined by the ratio of to , the calculation of a value for , is complicated by the fact that it is affected by differences in the rates at which O2 is being removed from the lung and the rate at which CO2 is entering it , expressed as the respiratory exchange ratio . Ideal alveolar can be calculated using modifications of the concepts derived by Fenn et al. (1946):
(27)
where is inspired and is the inspired fraction of O2. The is a function, therefore, of (determined by and ) and (determines R). The assessment of – indicates whether the lung is functioning well as a gas exchanger, or whether oxygenation of the arterial blood is hampered by diffusion limitation, right to left vascular shunts, or mismatching of alveolar ventilation and perfusion. If resting and R are similar for different-sized mammals, there is no reason to expect or to vary systematically with body size.
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Book2015, Comparative Biology of the Normal Lung (Second Edition)Peter D. Wagner, ... Kim E. Longworth
Chapter
The Alveolar–Capillary Unit and V/Q Matching
2015, Back to Basics in PhysiologyJuan Pablo Arroyo, Adam J. Schweickert
The Alveolar–Arterial Difference: How Good is the Lung at Exchanging O2 and CO2?
In Chapters 4 and 5Chapter 4Chapter 5 we learned the specific pressures of O2 and CO2 in the alveolus, the arterial blood, and the venous blood. But what good does it do us to know all these different PA, Pa, and Pv’s? (Remember: A=alveolar, a=arterial, v=venous.) In order to answer this, let’s take a look at the functional unit of the lung: the alveolar capillary unit (Figure 6.1). As we’ve seen before, it is composed of the alveoli and the pulmonary capillaries that abut the alveolar wall. This is where the exchange of O2 and CO2 takes place. Look at Figure 6.1A to get a brief overview of the exchange process that’s taking place at the level of the membrane. Blood that is coming in from the right side of the heart (blood from the venous side of the circulation) has a very low PVO2 40 mmHg (partial pressure of O2 in the veins) and a high PvCO2 45 mmHg (partial pressure of CO2 in the veins). This means that compared to the alveolar pressures, O2 has a gradient of 60 mmHg from alveolus to the blood and CO2 has a gradient of 5 mmHg from the blood to the alveolus. If our membrane works perfectly and the diffusion of O2 and CO2 happens without a problem, then the pressures of O2 and CO2 in the arterial blood will be identical to those in the alveoli. (Remember, diffusion happens until equilibrium has been reached; once the equilibrium has been reached—in this case the pressures of O2 and CO2 are equal on both sides of the membrane—diffusion will cease.) The membrane however, is not as simple as we would initially presume. In fact, it is made up of seven, yes, seven different layers, all of which are “sandwiched” together with an average approximate thickness of 1 μM (Figure 6.1B). What does this mean for exchange? Well, as we said earlier, if the membrane is working properly, there’s no problem. However, anything that increases the thickness of any of the layers of the membrane will increase the distance that O2 and CO2 need to travel, thereby decreasing diffusion.
Key
Increasing the thickness of the membrane or decreasing the surface area available for exchange decreases diffusion.
As you can see, maintaining the PAO2 and the PACO2 close to their target levels is extremely important to maintain a constant and adequate exchange of O2 and CO2 with the pulmonary capillaries. As we saw in Chapter 4, the alveolar gas equation
gives us a pretty good estimation of what was going on in the alveoli, but as we just saw, respiratory membrane function is also important. So, do you think there’s something else that we can calculate in order to evaluate the function of our respiratory membrane? Yes there is! It’s called the Alveolar–Arterial O2 Difference, or Δ(A–a), and it’s commonly referred to as the A–a gradient. Basically, in the simplest terms, this number is telling us the difference between the alveolar O2 (PAO2) and the arterial O2 (PaO2), and that’s exactly how it’s calculated:
Key
The alveolar–arterial difference (Δ(A–a)) is the difference between the PAO2 and the PaO2. It is a measure at how efficient the lungs are at exchanging O2.
The normal value for the Δ(A–a) is actually a moving target, since it varies by patient, and there are many different reference ranges, especially in the elderly in whom the Δ(A–a) can be larger and be within normal limits. However, for the sake of simplicity, we’ll establish that anything greater than a 10 mmHg difference warrants further workup as it could be potentially altered.
Now, reader beware! Despite it being referred to commonly as the A–a gradient, as we’ll see in the following section the Δ(A–a) formula does not represent a gradient, it represents the difference in O2 between all the alveolar air and all the blood going through the lungs. This is a crucial distinction. If you look at Figure 6.1A, why the Δ(A–a) is not a gradient is not really obvious. So what the heck is going on? Well, it’s all a matter of perspective. The lungs are composed of millions upon millions of alveolar–capillary units, millions of little Figure 6.1 As put together, and the PaO2 represents the average pressure of O2 in all the arterial blood coming from all the pulmonary capillaries, not just a single unit. This means that the number we’re actually measuring when we measure the PaO2 is the average of all the alveolar capillary units put together, some that are super efficient and some that aren’t as efficient. So in reality the Δ(A–a) has the O2 of some alveoli that are awesome at their job (e.g., PaO2=100 mmHg), and other alveoli that pretty much stink at it (e.g., PaO2=60 mmHg). The blood coming from all these different alveoli gets mixed and the resultant is, for example, a PaO2 of 80 mmHg. This doesn’t mean that all the alveoli in the lungs have a gradient of 20 mmHg however (100 mmHg of PAO2 – 80 mmHg PaO2); it means that we have some alveoli that are working and some that are not. It provides us the whole picture of how lungs are working.
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Book2015, Back to Basics in PhysiologyJuan Pablo Arroyo, Adam J. Schweickert
Chapter
The Respiratory Cycle
2015, Back to Basics in PhysiologyJuan Pablo Arroyo, Adam J. Schweickert
The Alveolar Gas Equation, or, How Much O2 is in There!
In the previous section we stated that the two factors that regulated the PAO2 were the renewal of O2 through ventilation and the rate of consumption. Since it is relatively difficult to measure exactly how much O2 is in the alveoli, wouldn’t it be nice if we could put this in a formula to calculate the PAO2? Well, there is a formula and it’s called the alveolar gas equation; it looks something like this:
However, before we delve into the specifics of each variable let’s take a step back. Consider this: If we’re trying to calculate the PAO2 we need to know two things:
•
: How much O2 is being inspired
•
: How much O2 is being consumed
Therefore, here’s a simplified version of the alveolar gas equation:
Since most of the time we are not breathing pure O2, we’ll need to calculate the partial pressure of O2. We can calculate how much O2 is being inspired from partial pressure of O2 in the air that is being breathed in (in this case we’ll use atmospheric air) and the PH2O in the airways with the following formula:
where:
: PATM=Atmospheric pressure (at sea level it would be 760 mmHg).
: FiO2 (Fraction of Inspired Oxygen)=The percentage partial pressure of O2 in the air that the patient is breathing in. Since O2 makes up 21% of the air we breathe under normal conditions the FiO2 would be 0.21 if no extra oxygen is added. (If O2 is added to the mix by providing supplemental oxygen to the patient the FiO2 will increase.)
: PH2O=Partial pressure of H2O in the system. At normal body temperature this is equal to 47 mmHg.
If we plug the numbers into our formula we come up with the following:
Then:
So:
This number is the same number we found in Table 4.1 for moistened tracheal air. So far, so good! Now our simplified alveolar gas equation looks like this:
How do we calculate O2 consumption? Remember what we mentioned about RQ in the previous section? RQ is the ratio of CO2 produced to O2 consumed. So if we know the PACO2 we can estimate how much O2 is being consumed. Lucky for us the CO2 in arterial blood or PaCO2 is almost identical to the PACO2. CO2 diffuses extremely rapidly, and is generally not affected by issues that can alter O2 diffusion (keep in mind that CO2 diffuses about 20 times as fast as O2). Therefore if PaCO2≈PACO2 together with the RQ, we can then calculate how much O2 is being consumed:
where:
: PaCO2=The alveolar pressure of CO2, which can be inferred from an arterial blood sample. Assuming that diffusion is occurring unimpinged, the amount of CO2 in the blood, the PaCO2 is going to be very similar to the PACO2, and is therefore used as a surrogate.
: RQ=The molar ratio of CO2 produced to O2 consumed depending on the fuel being consumed (carbohydrates vs fats vs proteins). In the body it approximates 0.8 and is product of the combined metabolism of carbohydrates, fats, and proteins.
Key
The RQ is the amount of O2 in mmHg that has to be consumed to account for the amount of CO2 in mmHg that is being produced.
Now that we understand the components let’s look at the entire alveolar gas equation:
or
So, let’s plug in the numbers under standard conditions and see what we come up with.
As we saw previously, if we solve the first half first we come up with 150 mmHg.
Dividing by 40 by 0.8 yields 50 mmHg. Let’s go over this again. This number is telling us that for every 40 mmHg of CO2 that we produce we are consuming 50 mmHg of O2. So if your PACO2 is 40, it’s because we are consuming 50 mmHg of O2. Therefore, we subtract the amount of O2 that is being consumed (50 mmHg) from the amount that is being brought into the lungs (150 mmHg). So:
Then:
Success! The PAO2 we calculated is right on the money! Taking small variations into account, the normal PAO2 is approximately 100 mHg.
Key
The alveolar gas equation is a way to estimate the PAO2, and is calculated with the following formula: PAO2=FiO2 (PATM – PH2O) – PaCO2/RQ.
Knowing the pressures of O2 and CO2 at the level of the alveolus, however, is only half the battle! We now need to understand what happens in the blood that allows for the diffusion of these gases from the air and into the blood, and how hemoglobin allows this phenomenon to take place in quantities that are compatible with life.
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16847 | https://brainly.com/question/45554380 | [FREE] Why do 1-propanol and 2-propanol have different ΔT values? A) Molecular weight B) Intermolecular forces C) - brainly.com
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Why do 1-propanol and 2-propanol have different ΔT values?
A) Molecular weight
B) Intermolecular forces
C) Boiling points
D) Solubility
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1-propanol and 2-propanol have different ΔT values primarily due to their differing intermolecular forces, specifically hydrogen bonding capabilities, which affect their boiling points.
Explanation
1-propanol and 2-propanol have different ΔT values mainly due to the differences in their intermolecular forces, which in turn affect their boiling points. Both 1-propanol and 2-propanol have similar molecular weights and molar masses, so this is not the distinguishing factor. The key difference lies in the hydrogen bonding potential of 1-propanol, which has a hydroxyl group at the end of the carbon chain, allowing for stronger hydrogen bonds compared to 2-propanol, where the hydroxyl group is in the middle of the chain. Therefore, 1-propanol exhibits stronger intermolecular forces, leading to a higher boiling point compared to 2-propanol.
The difference in the position of the hydroxyl group causes a variation in the molecule's ability to engage in hydrogen bonding. Hydrogen bonds are significantly stronger than other types of dipole-dipole interactions, contributing to the higher ΔT and boiling points observed for alcohols. In addition, the concept of polarizability also suggests that the molecular shape and the distribution of charge affect the intermolecular forces and thus impact the boiling points of substances.
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The different ΔT values for 1-propanol and 2-propanol are primarily due to their distinct intermolecular forces, particularly the differences in their hydrogen bonding capabilities. 1-propanol has stronger hydrogen bonds because of its linear structure compared to the more branched structure of 2-propanol, leading to a higher boiling point. Consequently, the correct answer is B) Intermolecular forces.
Explanation
1-propanol and 2-propanol differ in their ΔT values primarily because of their intermolecular forces.
Intermolecular forces are the forces that hold molecules together, and they can significantly impact the physical properties of substances, such as boiling points and freezing points.
1-propanol (CH₃CH₂CH₂OH) has an -OH (hydroxyl) group at the end of its carbon chain. This structure allows it to engage in stronger hydrogen bonding because the hydroxyl group can form hydrogen bonds with other 1-propanol molecules easily.
In contrast, 2-propanol (CH₃CHOHCH₃) has the hydroxyl group attached to the middle carbon of its chain. While it can also form hydrogen bonds, the branching of the carbon chain reduces the overall strength of hydrogen bonding compared to 1-propanol. Therefore, 1-propanol exhibits stronger intermolecular forces than 2-propanol.
Because of these differences in hydrogen bonding, the boiling point of 1-propanol is higher than that of 2-propanol, leading to different ΔT values when subjected to the same heating conditions.
In summary:
1-propanol has stronger hydrogen bonds due to its molecular structure.
2-propanol, having a branched structure, experiences relatively weaker hydrogen bonding.
The stronger intermolecular forces in 1-propanol result in a higher boiling point and therefore a higher ΔT value than in 2-propanol.
Examples & Evidence
For instance, if you were to heat equal amounts of 1-propanol and 2-propanol, you would observe that 1-propanol requires more energy (higher temperature) to reach its boiling point compared to 2-propanol, due to the stronger hydrogen bonds present in 1-propanol.
The differences in boiling points and intermolecular forces can be observed in standard chemical data tables, where 1-propanol has a boiling point of 97.4 °C while 2-propanol has a boiling point of 82.4 °C, reflecting the stronger hydrogen bonding in 1-propanol.
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### Free Chemistry solutions and answers
Community Answer Two of the liquids, acetone and 1-propanol, have nearly the same molecular masses but different ΔT values. Why? Explain using intermolecular forces.
Community Answer I have a lab about Boiling Points and Intermolecular Forces. using these alcohols 1-butanol, 1-propanol, isopropanol boiling point of each was 117.7 c 97 c 82 c -Based on your results, compare polarity and molecular mass with boiling point of a liquid. Be sure to discuss all relevant intermolecular forces in your answer. Is this what was expected? - Was there a difference in boiling points seen between 1-propanol and isopropanol? Should a difference be seen? Explain with reference to intermolecular forces.
Community Answer Classify each of the characteristics as a property of butane or 1-propanol. Butane, CH,CH,CH.CH 1-Propanol, CH,CH,CH, OH Answer Bank more polar lower boiling point weaker intermolecular forces more soluble in H,O
Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain?
Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized
Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __
Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta
Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction.
Community Answer 4.8 52 Which statements are TRUE about energy and matter in stars? Select the three correct answers. Al energy is converted into matter in stars Only matter is conserved within stars. Only energy is conserved within stars. Some matter is converted into energy within stars. Energy and matter are both conserved in stars Energy in stars causes the fusion of light elements
Community Answer 4.5 153 The pH of a solution is 2.0. Which statement is correct? Useful formulas include StartBracket upper H subscript 3 upper O superscript plus EndBracket equals 10 superscript negative p H., StartBracket upper O upper H superscript minus EndBracket equals 10 superscript negative p O H., p H plus P O H equals 14., and StartBracket upper H subscript 3 upper O superscript plus EndBracket StartBracket upper O upper H superscript minus EndBracket equals 10 to the negative 14 power.
New questions in Chemistry
Place the following transitions of the hydrogen atom from smallest to largest frequency of light absorbed: (a) n=3 to n=6 (b) n=4 to n=9 (c) n=2 to n=3 (d) n=1 to n=2
How many molecules are in 0.400 moles of N O 3 ?
What type of chemical bond holds C a 2+ and O 2− together in CaO? A. metallic bond B. covalent bond C. ionic bond D. hydrogen bond
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16848 | https://library.snls.org.sz/archive/doc/wikipedia/wikipedia-terodump-0.1/tero-dump/wikipedia/gr/Graph_of_a_function.html | Graph of a function - Wikipedia
<<UpContents
Graph of a function
Mathematically, the graph of a function is the collection of all pairs (x, f(x)) of the function.
The graph of the function
f(x)=\left{\begin{matrix} a, & \mbox{if }x=1 \ d, & \mbox{if }x=2 \ c, & \mbox{if }x=3. \end{matrix}\right.is {(1,a), (2,d), (3,c)}.
The graph of the cubic polynomial on the real line
f(x)=x^3-9x
is {(x,x 3-9 x) : x is a real number}. If the set is plotted on a Cartesian plane, the result is
Therefore the graph of a function on real numbers is identical to the graphic representation of the function. For general functions, the graphic representation cannot be applied and the formal definition of the graph of a function suits the need of mathematical statements, e.g., the closed graph theorem[?] in functional analysis.
The concept of the graph of a function is generalised to the graph of a relation. Note that althrough a function is always identified with its graph, they are not the same because it will happen two functions with different codomain could have the same graph. For example, the cubic polynomial mentioned above is a surjection if its codomain is the real numbers but it is not if its codomain is the complex field.
wikipedia.org dumped 2003-03-17 with terodump |
16849 | https://www.acep.org/sonoguide/basic/cardiac | American College of Emergency Physicians
'Sonoguide Home Home
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August 18, 2020
Cardiac
Michael I. Prats, MD, FACEP and David P. Bahner, MD, FACEP
I. Introduction and Indications
Cardiac ultrasound can be used to quickly diagnose many serious and potentially life-threatening pathologies.
It is reasonable to perform a focused cardiac ultrasound on any critically ill patient to help with diagnosis and guide management.
Specific indications include:1,2
Trauma
Hypotension
Concern for right heart strain in suspected pulmonary embolism
Concern for acute heart failure
Syncope
Cardiac arrest
Chest pain or dyspnea
Assessment of volume responsiveness
Emergency ultrasound potentially improves survival in penetrating cardiac injury,3 can accurately calculate systolic function,4 identify pericardial effusions,5 diagnosis acute heart failure,6 diagnose right heart strain,7 predict fluid responsiveness,8 and predict short-term outcomes in cardiac arrest.9 Competency in many of these areas does not take advanced training.10
II. Anatomy
Normally the heart sits in the left thorax with the apex pointing inferiorly and approximately 60 degrees to the left.
The point of maximal impulse may be palpated on the chest to estimate the patient’s cardiac apex.
Each patient may have different orientation of their cardiac structure and transducer position may need to be adjusted accordingly.
This is especially common in patients with pathologies such as cardiomegaly or chronic lung disease.
Echocardiography is a dynamic assessment and the structures must be examined through the entire cardiac cycle.
The valves can be used to determine cardiac cycle in a patient in normal sinus rhythm.
Atrio-ventricular valves (mitral and tricuspid) are open during diastole and closed during systole.
The aortic and pulmonic valves are open during systole and closed during diastole.
The thinner walled right and left atria sit superior to the thicker walled right and left ventricles.
Blood enters the heart through the inferior and superior vena cava in the right atrium. It then passes through the tricuspid valve into the right ventricle, through the pulmonic valve into the pulmonary arteries. After flowing through the pulmonary circulation, blood returns to the left atrium via four pulmonary veins, then passes through the mitral valve into the left ventricle, and exits the heart through the aortic valve into the ascending aorta.
There are distinctions between the normal left and right ventricles:
The left ventricle is larger with thicker walls and a rounded apex. Contraction is torsional, towards the apex.
The right ventricle is more tapered toward the apex. Contraction is more linear, towards the apex.
The cardiac anatomy appears differently depending on the plane of imaging.
Illustration 1. Transducer location for the four focused cardiac views
Subxiphoid or Subcostal View: The liver is used as an acoustic window and is seen at the left of the screen and near field. The right ventricle being the most anterior is adjacent to the liver. The apex will be to the right of the screen. By fanning inferiorly, the inferior vena cava (IVC) is seen in short axis. Often the hepatic veins can be seen in the liver draining into the IVC.
Figure 1. Subxiphoid View. RA = Right atrium. RV = Right ventricle. IVS = interventricular septum. LV = left ventricle. MV = Mitral valve. LA = left atrium. L = Liver Video 1. Normal Subxiphoid View
Parasternal Long Axis View: This view captures the flow of blood through the left side of the heart. The apex is to the left of the screen. The mitral valve leaflets are seen and often chordae tendinae connecting them to papillary muscles. Two cusps of the aortic valve (usually the non-coronary and right coronary cusp) are seen. The left ventricular outflow tract (LVOT) is the term for the aortic root and proximal ascending aorta. A portion of the right ventricle is seen in the near field.
Figure 2. Parasternal long axis view. RV = Right ventricle. LV = left ventricle. MV = mitral valve. LVOT = left ventricular outflow tract. LA = left atrium Video 2. Normal parasternal long axis view
Parasternal Short Axis View: This view is a cross sectional view of the left and right sides of the heart. These can be “sliced” at various levels between the base and the apex. By fanning the probe towards the right shoulder, one can visualize the aortic valve in cross section. The “Mercedes Benz” sign shows all three of the leaflets. The right ventricular outflow tract and pulmonary valve can be seen in this view. At the base, the mitral valve is seen inside the left ventricle. At the mid ventricular level, the papillary muscles are seen within the left ventricle. This is the most commonly used parasternal short axis view in point-of-care ultrasound in the acute care setting. At the apex, the left ventricle has tapered and no structures are seen within.
Figure 3. Parasternal short axis view at the mid ventricular level. RV = right ventricle. IV SEPT = interventricular septum
Video 3. Normal Parasternal short axis view at the mid ventricular level
Apical Four Chamber View: This view shows all four chambers of the heart. The apex is toward the near field with the interventricular septum pointing to approximately the 12 o’clock position. By fanning superiorly, the apical five chamber view can be obtained, with the “fifth chamber” being the LVOT with the aortic valve.
Figure 4. Apical Four Chamber view. RV = Right ventricle. LV = Left ventricle. TV = Tricuspid valve. RA = Right atrium. LA = Left atrium. MV = Mitral valve Video 4. Normal Apical Four Chamber View
III. Scanning Technique, Normal Findings and Common Variants
Scanning Technique
Following the I-AIM model,11 once the indication has been recognized, acquisition of images can take place. This involves optimizing the probe, patient, and picture.
To visualize the heart between ribs, transducers with smaller footprints and low frequencies (2-4 MHz) are ideal. The phased or microconvex arrays are commonly used.
The curvilinear transducer can also be used and may be preferred for the subxiphoid view; however, rib shadows may limit adequate views on the chest.
Figure 5. Phased array probe
There are two accepted conventions for probe position indicator which are 180 degrees of each other.12 Emergency physicians tend to favor the indicator on the left side of the screen as this is the convention for other forms of ultrasonography. Cardiology and other specialties will often have the indicator on the right side of the screen. Importantly, both techniques result in identical images. The instructions below are assuming screen indicator is on the left side of the screen.
The patient can be supine or semi-upright. Left lateral decubitus can assist in obtaining parasternal and apical views. The supine position may allow better subcostal visualization.
In all cases, video clips are preferable to still images. Attempt to obtain all four focused cardiac views or corroborate findings.
Obtain subxiphoid view. Palpate xiphoid process, place probe inferior with the indicator to the patient's right. Initially orient slightly to the right shoulder to visualize liver as acoustic window. Rock toward left shoulder until heart is visualized. Deep inspiration of the patient may improve visualization
Figure 6. Probe position for acquisition of subxiphoid view
Fan inferiorly to visualize the IVC in short axis. Rotate the transducer 180 degrees clockwise to visualize IVC in long axis. Note overall diameter and collapsibility with respirations approximately 2-3 cm distal from the right atria or just distal to the insertion of the hepatic vein.
Figure 7. Subcostal Inferior Vena Cava in Long Axis. IVC = inferior vena cava. Caudate = caudate lobe of the liver. RA = right atrium
Obtain parasternal long axis view. The indicator should point to the patient's left hip initially, but then the position can be modified to best visualize the aortic valve, mitral valve, and cardiac apex. The transducer is moved in large circular motions on the left anterior chest to determine the patient’s optimal window. This is generally in the third or fourth intercostal space.
Figure 8. Probe position for acquisition of parasternal long axis view
Obtain parasternal short axis. From the parasternal long axis, rotate the transducer 90 degrees clockwise such that the indicator now points to the right hip. Fan through the aortic valve, left ventricle base, mid, and apex.
Figure 9. Probe position for acquisition of parasternal short axis view
Obtain apical four chamber view. Palpate the point of maximal impulse. Place the transducer, with indicator pointed to patient’s right side, over this point, which is generally at the inframammary fold or under the left breast in women. Slide and rock the probe to ensure the interventricular septum is at approximately the 12 o’clock position. You can then fan superiorly to visualize the LVOT in the apical 5 chamber view.
Figure 10. Probe position for acquisition of apical four chamber view
E-point Septal Separation
This is a technique that is commonly employed to provide an objective measurement of systolic function of the left ventricle. Using M-mode, capture the distance between the anterior leaflet of the mitral valve and the interventricular septum. A measure >7mm correlates with reduced cardiac function.13 This measurement may not be accurate in cases of aortic regurgitation, mitral stenosis, hypertrophic cardiomyopathy, and non-sinus rhythm.
Figure 11. Normal E-point septal separation
IVC Measurements
The IVC can be used as a measure of volume responsiveness, although this is more accurate in mechanically ventilated patients than in those spontaneously breathing.8,14,15 This is likely most helpful in the extremes. A patient with a small IVC that is highly collapsible would likely benefit from fluids. M-mode can be used to measure inspiratory and expiratory diameters although caution must be taken to prevent measuring the IVC at inconsistent points due to overall movement from breathing. A patient with a large IVC that has no change in diameter with respirations, is less likely to benefit. Keep in mind that a large plethoric IVC will be present in signs of obstructive shock and any cause of elevated right sided cardiac pressures.
Figure 12. Inferior Vena Cava Respiratory variation on M-mode
Video 5. Normal Inferior Vena Cava with mild respiratory collapse
Video 6. Inferior Vena Cava with >50% respiratory collapse
Video 7. Large inferior vena cava with minimal respiratory collapse
TAPSE
The tricuspid annular plane systolic excursion (TAPSE) is a measurement of the right-sided systolic function. In the emergency department, this is most commonly employed to assess for signs of right heart strain in a patient with suspected or confirmed pulmonary embolism. This is a good prognostic test, although insufficiently sensitive for the diagnosis of any pulmonary embolism.16 On the apical four chamber view, place the m-mode cursor over the lateral annulus of the tricuspid valve. This provides a waveform that can be measured from peak to trough. Measurements less than 16 mm are considered indicative of poor right ventricular function and correlate with increased mortality.17
Video 8. Using M-mode to obtain TAPSE
Figure 13. Normal TAPSE as measured on M-mode
Mitral Valve Inflow
The mitral valve inflow is obtained by placing the pulse wave doppler gate at the tip of mitral leaflets in the apical four chamber view. In cases of suspected cardiac tamponade, a respiratory variability of greater than 25% is concerning for an exaggerated variation due to tamponade physiology.18 Although more advanced, this measurement can also be used for diastology.
Figure 14. Normal Mitral valve inflow on pulse wave doppler
Pathology
Effusion, Trauma, and Tamponade
Pericardial effusions are common and range from benign to life threatening.
In contrast to an effusion, epicardial fat pads are isolate to the anterior of the heart and often have increased echogenicity.
In the setting of trauma, significant effusion may signify cardiac injury.
The parasternal long axis view can be helpful in distinguishing between pericardial (anterior to descending thoracic aorta) and pleural (posterior to the DTA) effusions.
Findings of tamponade on echocardiography include right ventricular diastolic collapse, right atrial systolic collapse, plethoric IVC, and increased inflow variation.19
Pericardial tamponade is potentially lethal without intervention. Intuitively POCUS can decrease time to intervention although to date this is supported only by weak evidence.20 Video 9. Subxiphoid view of pericardial effusion with cardiac tamponade physiology. Note circumferential pericardial effusion, right ventricular diastolic collapse
Heart Failure
There are many patients who present to the emergency department with concern for acute heart failure or fluid overload. Ultrasound can decrease the time to diagnosis in these patients.21,22
Findings include reduced systolic function or diastolic dysfunction. Poor systolic function can be visually estimated which has been shown to be equivalent to measurements. Alternately, you can measure EPSS, fractional shortening, or many other measurements.
Isolated diastolic dysfunction can be present without systolic dysfunction. This is a complex determination often requiring many advanced measurements; however, measuring the mitral valve inflow E and A wave as well as tissue doppler of the mitral annulus to obtain the e’ and a’ waves can be helpful in this diagnosis.23
In addition to focused echocardiography, evaluating the lung fields for B-lines will increase the accuracy for the diagnosis.24
Video 10. Parasternal long axis view of hyperdynamic systolic function Video 11. Parasternal long axis view of severely reduced systolic function
Right Heart Strain
Pulmonary embolism represents a broad spectrum of disease. Patients with evidence of right heart strain have worse prognosis and may benefit from more advanced therapies.
Focused ultrasound has been shown to accurately detect right heart strain more accurately than CT or serum markers with sensitivity of 100% and specificity of 99% in one study.7
Right heart strain alone is not sensitive for the diagnosis of pulmonary embolism.
Findings suggestive of right heart strain are:
Increased RV diameter. Usually 60% of the LV's size, when the RV size approaches the same size of the LV it is pathologically enlarged.
Septal bowing. When the right ventricle bows towards the left this indicates increased right-sided pressures. This appears as a “D-sign” on a parasternal short axis.
McConnell’s sign. This is an akinetic right ventricle with hyperkinetic or preserved function of the apex.Video 12. Apical four chamber view of right ventricular dilation and septal bowing Video 13. Parasternal short axis view with D sign Video 14. McConnell’s Sign
A TAPSE can be measured to evaluate the right heart systolic function. A TAPSE <16mm signifies increased mortality.17
It can be difficult to distinguish acute from chronic right heart strain as all of the above findings can occur in chronic pathology. Right ventricular hypertrophy as measured by the anterior wall thickness >5mm indicates possible chronic elevated pressures. The 60/60 rule, although complex, can also be helpful in this scenario.25
Cardiac Arrest
Ultrasound has proven useful in cardiac arrest at prognosticating survival, although the best studies have not provided information on neurologic outcomes.9
Ultrasound in cardiac arrest is assessing both the presence of cardiac activity and evidence of etiology of the cardiac arrest. Patients with cardiac activity have odds ratio of 3.6 for survival to admission.26
Isolated valve fluttering or myocardial twitching likely does not indicate a perfusing rhythm. Patients without cardiac activity have low likelihood of survival.
Patients with severely reduced cardiac activity may present pulseless but benefit from more tailored therapy compared to standard ACLS treatment algorithms.27
Care must be taken to avoid delaying chest compressions while performing ultrasound during arrest.28
Transesophageal echocardiography can be performed in the emergency department during cardiac arrest and can offer more cardiac windows and ongoing cardiac assessment during compressions.29
Video 15. Cardiac standstill Video 16. Cardiac Arrest with initial cardiac activity, cardiac standstill, and compressions
V. Pearls and pitfalls
Scan in a systematic fashion. Moving the probe in a coordinated fashion allows access to cardiac windows as the operator manipulates the transducer. Often larger circular motions are needed to find the best sonographic window. Then smaller refining movements can improve the image.
Find sonographic landmarks that can be used as points of reference to identify surrounding anatomy. Pattern recognition of key cardiac shapes (especially the location of the apex, valves, and papillary muscles) help the novice sonographer anchor their scanning and adjust the angle of the probe to recreate the intended standard images.
Epicardial fat pads are usually isolated to the anterior heart and have internal echoes. Pericardial fluid usually collects in the dependent posterior pericardial space and can be seen surrounding the myocardium anterior to the descending aorta. A left sided pleural effusion will be located posterior to the descending aorta.
Figure 15. Parasternal long axis showing pericardial effusion anterior to descending thoracic aorta
Clotted blood in the pericardial space may have more hypoechoic appearance.
Cardiac tamponade is a clinical diagnosis. Place the echocardiographic findings in the context of the patient’s clinical status.
Not all acute heart failure exacerbations have poor systolic function. This is why knowledge of diastolic measurements can be helpful.
When comparing ventricular size, ensure that you are visualizing the correct ventricle. These findings can be helpful: the descending thoracic aorta lies behind the left atrium, the left ventricular outflow tract will come from the left ventricle, and the moderator band is present in the right ventricle.
Right heart strain alone is not sensitive for the diagnosis of acute pulmonary embolism. Additionally, when present, right heart strain can be from acute or chronic causes.
The absence of cardiac activity during cardiac arrest is associated with non-survival, but it does not rule out survival. There is little evidence that the presence of cardiac activity on ultrasound is associated with survival with a good neurologic outcome.Be cautious of “scope creep.” There are many other cardiac findings that experienced sonologists will be able to assess such as regional wall motion abnormalities, valvular abnormalities, cardiac tumors, vegetations, cardiomyopathies, and advanced doppler measurements. Recognize that the majority of these are not supported by national guidelines, have variable accuracies when performed at the point-of-care, and should only be utilized by users with firm understanding of the strengths and limitations of these examinations.
VI. References
Ultrasound guidelines: Emergency, point-of-care and clinical ultrasound guidelines in medicine. [policy statement]. Ann Emerg Med. 2017;69(5):e27-e54.
Labovitz AJ, Noble VE, Bierig M. Focused cardiac ultrasound in the emergent setting: a consensus statement of the American Society of Echocardiography and American College of Emergency Physicians. J Am Soc Echocardiogr. 2010;23(12):1225-30.
Plummer D, Brunette D, Asinger R, Ruiz E. Emergency department echocardiography improves outcome in penetrating cardiac injury. Ann Emerg Med. 1992;21(6):709-12.
Moore CL, Rose GA, Tayal VS, et al. Determination of left ventricular function by emergency physician echocardiography of hypotensive patients. Acad Emerg Med. 2002;9(3):186-93.
Mandavia DP, Hoffner RJ, Mahaney K, et al. Bedside echocardiography by emergency physicians. Ann Emerg Med. 2001;38(4):377-82.
Russell FM, Ehrman RR, Cosby K. Diagnosing acute heart failure in patients with undifferentiated dyspnea: a lung and cardiac ultrasound (LuCUS) protocol. Acad Emerg Med. 2015;22(2):182-91.
Weekes AJ, Thacker G, Troha D. Diagnostic accuracy of right ventricular dysfunction markers in normotensive emergency department patients with acute pulmonary embolism. Ann Emerg Med. 2016; 68(3):277-91.
Barbier C, Loubières Y, Schmit C. Respiratory changes in inferior vena cava diameter are helpful in predicting fluid responsiveness in ventilated septic patients. Intensive Care Med. 2004;30(9):1740-6.
Tsou PY, Kurbedin J, Chen YS. Accuracy of point-of-care focused echocardiography in predicting outcome of resuscitation in cardiac arrest patients: A systematic review and meta-analysis. Resuscitation. 2017;114:92-9.
Bustam A, Noor Azhar M, Singh Veriah R, et al. Performance of emergency physicians in point-of-care echocardiography following limited training. Emerg Med J. 2014;31(5):369-73.
Bahner DP, Hughes D, Royall NA. I-AIM: a novel model for teaching and performing focused sonography. J Ultrasound Med. 2012;31(2):295-300.
Moore C. Current issues with emergency cardiac ultrasound probe and image conventions. Acad Emerg Med. 2008;15(3):278-84.
McKaigney CJ, Krantz MJ, La Rocque CL, et al. E-point septal separation: a bedside tool for emergency physician assessment of left ventricular ejection fraction. Am J Emerg Med. 2014;32(6):493-7.
Nagdev AD, Merchant RC, Tirado-Gonzalez A, et al. Emergency department bedside ultrasonographic measurement of the caval index for noninvasive determination of low central venous pressure. Ann Emerg Med. 2010;55(3):290-5.
Corl KA, George NR, Romanoff J. Inferior vena cava collapsibility detects fluid responsiveness among spontaneously breathing critically-ill patients. J Crit Care. 2017;41:130-7.
Daley J, Grotberg J, Pare J. Emergency physician performed tricuspid annular plane systolic excursion in the evaluation of suspected pulmonary embolism. Am J Emerg Med. 2017;35(1):106-11.
Lobo JL, Holley A, Tapson V. Prognostic significance of tricuspid annular displacement in normotensive patients with acute symptomatic pulmonary embolism. J Thromb Haemost. 2014;12(7):1020-7.
Shyy W, Knight RS, Kornblith A, et al. Point-of-care diagnosis of cardiac tamponade identified by the flow velocity paradoxus. J Ultrasound Med. 2017;36:2197–2201.
Goodman A, Perera P, Mailhot T, et al. The role of bedside ultrasound in the diagnosis of pericardial effusion and cardiac tamponade. J Emerg Trauma Shock. 2012;5(1):72-5.
Alpert EA, Amit U, Guranda L, et al. Emergency department point-of-care ultrasonography improves time to pericardiocentesis for clinically significant effusions. Clin Exp Emerg Med. 2017;4(3):128-32.
Zanobetti M, Scorpiniti M, Gigli C. Point-of-care ultrasonography for evaluation of acute dyspnea in the ED. Chest. 2017;151(6):1295-1301.
Russell FM, Ehrman RR. A modified lung and cardiac ultrasound protocol saves time and rules in the diagnosis of acute heart failure. J Emerg Med. 2017;52(6):839-845.
Ferre RM, Chioncel O, Pang PS, et al. Acute heart failure: the role of focused emergency cardiopulmonary ultrasound in identification and early management. Eur J Heart Fail. 2015;17(12):1223-7.
Al Deeb M, Barbic S, Featherstone R, et al. Point-of-care ultrasonography for the diagnosis of acute cardiogenic pulmonary edema in patients presenting with acute dyspnea: a systematic review and meta-analysis. Acad Emerg Med. 2014;21(8):843-52.
Fields JM, Davis J, Girson L. Transthoracic echocardiography for diagnosing pulmonary embolism: A systematic review and meta-analysis. J Am Soc Echocardiogr. 2017;30(7):714-723.e4.
Gaspari R, Weekes A, Adhikari S. A retrospective study of pulseless electrical activity, bedside ultrasound identifies interventions during resuscitation associated with improved survival to hospital admission. A REASON Study. Resuscitation. 2017;120:103-7.
Gaspari R, Weekes A, Adhikari S. Emergency department point-of-care ultrasound in out-of-hospital and in-ED cardiac arrest. 2016;109:33-9.
Huis In 't Veld MA, Allison MG, Bostick DS. Ultrasound use during cardiopulmonary resuscitation is associated with delays in chest compressions. Resuscitation. 2017;119:95-8.
Fair J, Mallin M, Mallemat H. Transesophageal echocardiography: Guidelines for point-of-care applications in cardiac arrest resuscitation. Ann Emerg Med. 2018;71(2):201-7.
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Basic Cardiovascular Imaging Page Sonoguide Ultrasound
Related Articles
Femoral Nerve Block
E-FAST (Extended Focused Assessment with Sonography in Trauma)
Physics and Technical Facts for the Beginner
General Nerve Block Contraindications and Precautions
Allergy to local anesthetic agents
Active infection at injection site
Risk of compartment syndrome or need to monitor neurologic function
Uncooperative patient
Pre-existing neurologic deficit that could prevent the patient from communicating paresthesias or pain during the nerve block
Obesity obscuring nerve visualization
Coagulopathy (relative)
General Procedure Setup
Minimize ambient light to improve image quality. Placing a spotlight on the block site and darkening the rest of the room can have a dramatic effect on image quality.
Place machine in direct field of vision of practitioner. Avoid placement that requires twisting or head turning.
For large volume blocks place the patient on a cardiac monitor, establish IV access and have emergency airway equipment available.
Provide analgesics and light sedation as needed.
Obtain consent and document a pre-block neurologic exam.
Injection Precautions
Remove all air from the needle prior to beginning.
Ensure the needle is clearly visualized prior to advancing.
Never inject if the patient complains of intense pain or pressure as this could signify intraneural injection.
Anesthetic should inject easily with one finger on the plunger. High pressure can be due to intraneural placement.
Avoid intravascular injection. Always aspirate for blood prior to injecting anesthetic. If injection fails to produce a resulting bolus on the ultrasound image, stop injecting and reposition. If intravascular injection of local anesthetic occurs, then the physician must recognize the potentially life-threatening complication of local anesthetic systemic toxicity (LAST). LAST can lead to cardiovascular collapse. If this occurs, the provider should follow ACLS protocols and administer intralipid emulsion therapy.
In-plane (long-axis) approach
Video 1. In-plane approach
The needle enters the skin at the short side of the transducer and the probe is slid toward the needle until seen
Course of the needle is parallel to and transects the ultrasound beam
Bevel faces the transducer
The transducer is readjusted frequently
Across the axis of the needle to maintain visualization
Rotated to ensure the ultrasound beam remains parallel to the needle
Advantages
Needle shaft and tip are well visualized
Great depth and trajectory information
Generally easier to master
Disadvantages
May require the operator to use the needle in their non-dominant hand
Visualizing a thin needle with a thin ultrasound beam can be challenging
Important Application Information
Providers should choose the appropriate needle gauge, length and type
Larger gauge needles are easier to locate with ultrasound and should be considered for beginners
Several needles are available to perform nerve blocks:
Cutting point needles
the most common found in the ED
easily pass through tissue and theoretically could more easily penetrate a nerve if accidentally contacted
Spinal needle (quincke tip)
non-cutting, long bevel needle
more difficult to penetrate through tissue
readily available in a variety of lengths and gauges in most EDs
used with success in several ED based US-guided nerve block studies
Short-beveled or blunt-point needles
generally used by anesthesiologists and can be special ordered
more difficult to pass through tissue and therefore provide feedback to the operator as they pass through tissue planes
theoretically less likely to cause damage if brought into contact with a nerve. However, no studies have confirmed decreased neural injury with short or blunt point needles
Needle Orientation: In-Plane vs Out-of-Plane Approach
Figure 1. Needle Orientations2 needle orientations for ultrasound guided procedures, the in-plane and out-of-plane approach
In-plane (long-axis) approach
Video 1. In-plane approach
The needle enters the skin at the short side of the transducer and the probe is slid toward the needle until seen
Course of the needle is parallel to and transects the ultrasound beam
Bevel faces the transducer
The transducer is readjusted frequently
Across the axis of the needle to maintain visualization
Rotated to ensure the ultrasound beam remains parallel to the needle
Advantages
Needle shaft and tip are well visualized
Great depth and trajectory information
Generally easier to master
Disadvantages
May require the operator to use the needle in their non-dominant hand
Visualizing a thin needle with a thin ultrasound beam can be challenging
Out-of-plane (short-axis) approach
Video 2. Out-of-plane approach
The needle enters the skin at the long side of the transducer and the probe is slid towards the needle until seen
Course is perpendicular to the ultrasound beam
Bevel faces the transducer
The needle tip and shaft can look the same on ultrasound
The procedure involves allowing the needle to catch up to the ultrasound beam but never pass it:
Identify the needle tip after piercing the skin
Slide the probe away from the needle until the tip is no longer visualized
Advance the needle only until it is seen again
Slide the probe away from the needle again until it is no longer seen and then advance the needle as above
This process is repeated down to the target
This catch-up motion prevents the needle tip from being lost or passing beyond the beam to damage deeper structures
Out-of-plane (short-axis) Approach
Video 1. Out-of-plane approach
The needle enters the skin at the long side of the transducer and the probe is slid towards the needle until seen
Course is perpendicular to the ultrasound beam
Bevel faces the transducer
The needle tip and shaft can look the same on ultrasound
The procedure involves allowing the needle to catch up to the ultrasound beam but never pass it:
Identify the needle tip after piercing the skin
Slide the probe away from the needle until the tip is no longer visualized
Advance the needle only until it is seen again
Slide the probe away from the needle again until it is no longer seen and then advance the needle as above
This process is repeated down to the target
This catch-up motion prevents the needle tip from being lost or passing beyond the beam to damage deeper structures
Advantages
Easier visualization of surrounding structures (e.g. artery and nerve)
Needle path can be short
Operators who perform vascular access in short axis may find this approach more comfortable
Disadvantages
Tip of the needle and shaft of the needle can look the same
Needle tracking can be difficult for novices
Visualizing the Needle
During the block, keeping the tip in view is of critical importance. Doing so will result in the highest quality block and avoid the most devastating complications.
A few factors play into how well the tip displays:
Approach
The tip is generally easier to visualize with the in-plane approach.
Needle orientation
Choosing a shallow angle will also help, as it presents more of the needle to the ultrasound beam for reflection.
The needle tip should always be oriented such that the bevel is facing toward the probe; this way the broadest, most echogenic portion of the tip is available for reflection.
Needle gauge
The larger needles are easier to visualize.
Never advance the needle if you are unsure of the tip position.
If you lose the tip, slowly rock the probe on its face to scan through the needle. You may need to rotate the probe to align it with the axis of the needle.
If you are still unsure, withdraw the needle slightly and watch for tissue movement.
If the tip is not clear, inject 0.5 cc of local anesthetic. This will deform the tissue surrounding the tip, and the anechoic fluid bolus should highlight the tip.
If these attempts fail to localize the tip, withdraw the needle and choose a different angle or puncture site.
Ensure to inject all air from needles as air in the tissue will prevent visualization of deeper structures.
Anisotropy
Video 1. Anisotropy
Nerves can exhibit strong anisotropy
Anisotropy refers to the changes in echogenicity of a structure as the angle of insonation is changed.
Care must be taken to adjust the angle of insonation to ensure optimal nerve visualization.
Tendons, ligaments and muscle also exhibit anisotropy.
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16850 | https://www.youtube.com/watch?v=14QVGcG1_z8 | 4.123 Equal Fractions with Denominators of 10, 100, 1000
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Equal Fractions with Denominators of 10, 100, 1000
Compare two equivalent fractions with denominators of 10, 100, 1000 to find the missing number.
Try this skill yourself at
Transcript:
today we're going to learn how to make fractions with different denominators equal to each other let's look at some examples all of these examples will involve denominators of 10 100 at 1 000. that's going to make things a little bit easier but it will still be a good way to understand these concepts the other thing to keep in mind for all of these questions is our goal is going to be the same find the missing number we'll have three numbers in the problem and one will be missing and we need to find that missing number let's get into it in order to go from a denominator of 10 to a denominator of 100 we have to multiply by 10 but if the denominator got multiplied by 10 in order to keep the fraction equal you have to do the same thing to the numerator so they're both multiplied by 10 which means the numerator on the right hand side must be 10 times the numerator on the left hand side 10 times 8 which is going to be 80. so we can put 80 in our box and check our answer looks good here we go again we're going from 10 to 100. that means we have to multiply by 10 which means the numerator gets multiplied by 10 as well and 9 times 10 will give us our answer which is 90. one more we're going from 10 to 100. how do we do that we multiply by ten so we also multiply the numerator by ten four times ten is forty now we're going back the other direction rather than going from 10 to 100 we're going from 100 down to 10. well if before we were multiplying by 10 to go the other way we have to divide by 10 but we have to do the same thing to the numerator and the denominator so we divide both by 10. 100 divided by 10 gives us the ten ninety divided by ten gives us a numerator of nine going from left to right again to go from ten to a hundred multiply by ten so do the same thing to the numerator and the denominator giving us thirty one more example to go from 10 to 100 multiply by 10 then do the same thing in the numerator 10 times 8 is going to be 80. so we keep seeing the same pattern figure out what you need to do to the denominator to make the two fractions equal then do the same thing with the numerator as long as you keep that in mind you can't get into too much trouble here's another example in order to go from 10 to 100 we have to multiply by 10 again so you do the same thing in the numerator 10 times 10 gives us 100 10 times 9 gives us 90. now we're going back the other way in order to go from 100 to 10 you divide by 10. we have to do the same thing to the numerator so we also divide 70 by 10 which will give us 7. back to the right to go from 10 to 100 multiply by 10 but do the same thing for the numerator and the denominator ten times ten becomes a hundred ten times six becomes sixty one last example to go from a hundred to ten simply divide by 10 but you have to do the same thing to the numerator so 60 divided by 10 will be 6. and that's it |
16851 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_10?srsltid=AfmBOoqNQroGt93a3VQvf4dGgUqRrhDii8V4CLfJ1b6oy73es9Ni03fY | Art of Problem Solving
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2012 AMC 8 Problems/Problem 10
Contents
[hide]
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3 (a simpler version of Solution 2)
5 Solution 4 (Complementary Counting)
6 Video Solution
7 See Also
Problem
How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
Solution 1
For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of , since all of the valid 4-digit number will always be greater than . The best way to solve this problem is by using casework.
There can be only two leading digits, namely or .
When the leading digit is , you can make such numbers.
When the leading digit is , you can make such numbers.
Summing the amounts of numbers, we find that there are such numbers.
Solution 2
Notice that the first digit cannot be , as the number is greater than . Therefore, there are three digits that can be in the thousands.
The rest three digits of the number have no restrictions, and therefore there are for each leading digit.
Since the two 's are indistinguishable, there are such numbers .
Note by algebramaster2: 1 and 2 are the only digits that can be in the thousands place as 0 in the thousands place will not make the number 4 digit anymore.
Solution 3 (a simpler version of Solution 2)
We can list out the four digits in the number. The first digit of the number can’t be 0 since the number is greater than . This leaves us with three integers for the digit in the thousands place. We have no restrictions in the hundreds place except for the fact that it can’t be the integer we chose for the thousands place. Continuing this pattern, there are choices for the thousands place, for the hundreds, for the tens, and for the ones. Adding this up we get which is choice .
—-jason.ca
Solution 4 (Complementary Counting)
There are ways to arrange the numbers in . The number of ways to arrange 2012 with 0 as the first digit is 3 because there are three places to put the 1 and the 2s are the same. which is choice .
~ Edited by GeometryMystery
Video Solution
~savannahsolver
See Also
2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 9Followed by
Problem 11
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AJHSME/AMC 8 Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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16852 | https://hypertextbook.com/facts/2005/MayaBarsky.shtml | The PhysicsFactbook
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Index of Refraction of Air
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| Bibliographic Entry | Result(w/surrounding text) | StandardizedResult |
---
| Serway, Raymond., Faughn, Jerry S. "The Law of Refraction." College Physics. Sixth edition, Pacific Grove, CA: Brooks/Cole-Thomson Learning, 2003: 692. | "Indices of Refraction for Various Substances, Measure with Light of Vacuum Wavelength 589nm… Air 1.000293" | 1.000293 |
| Encarta. Indexes of Refraction. Microsoft 2005. | "Substance Refractive Index… Air 1.0003" | 1.0003 |
| Lide, David R. "Index of Refraction of Air." Handbook of Chemistry and Physics. 75th edition. Boca Raton, FL, CRC Press Inc., 1994: 10-302. | | | | Index of Refraction of Air | | wavelength vac: | (n-1) × 108 | | 200 nm | 32408 | | 900 nm | 27419 | | 1.00027419-1.00032408 |
| Selloy, Samuel M. "Index of Refraction of Air." Handbook of Chemistry and Physics. 48th edition, Cleveland OH, The Chemical Rubber Co., 1967: E-160. | [table] | 1.0002000-1.0002739 |
| Weisstein Eric. Index of Refraction. Wolfram Research. 2005. | "Index of Refraction… Air 1.00029" | 1.00029 |
For thousands of years, people have noted that a straight stick placed in water appears to be broken at an angle where it enters the water. This is the origin of the term "refraction,'' which means "broken back.''
Every material that light can travel through has an index of refraction, denoted by the letter n. The velocity of light in a vacuum is 3.0 × 108 m/s. The index of refraction equals the ratio of the velocities of light in vacuum (c) to that in the medium (v), that is n = c/v. Light slows down when traveling through a medium, thus the index of refraction of any medium will be greater than one. The index of refraction of air is 1.0003, which is very similar to the index of refraction in a vacuum (1.0000), therefore, in most problems, these indices are used interchangeably. The index of refraction is also based on the wavelength of the incident light, where light of a longer wavelength refracts less than light of a shorter wavelength.
The actual law of refraction was discovered in the early 1600s by a Dutch mathematician and geodesist, Willebrord Snell van Royen, which is now termed Snell's law. When light is refracted at a surface, the angles of the incident and refracted rays are related by Snell's law, n1sinθ1 = n2sinθ2. n represents the refractive indices of material 1 and material 2 and are the angles of light traveling through these materials with respect to the normal. The index of refraction of air, along with other materials is essential to many optics and laser applications.
Maya Barsky -- 2005 |
16853 | https://www.chegg.com/homework-help/questions-and-answers/use-built-matlab-function-fzero-find-root-f-x-x2-sinx-4-0-initial-guess-x0-05-b-use-false--q14751349 | Solved a) Use built-in Matlab function fzero to find the | Chegg.com
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a) Use built-in Matlab function fzero to find the root of f(x) = x2|sinx|-4=0, initial guess x0=0.5. b) Use false-position method to find the root of f(x)= x2|sinx|-4=0, initial guess [a, b]=[0, 4]. Accept your result when |f(xi)|<510-7. Print out all iterations of xi and f(xi).
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Question: a) Use built-in Matlab function fzero to find the root of f(x) = x2|sinx|-4=0, initial guess x0=0.5. b) Use false-position method to find the root of f(x)= x2|sinx|-4=0, initial guess [a, b]=[0, 4]. Accept your result when |f(xi)|<510-7. Print out all iterations of xi and f(xi).
a) Use built-in Matlab function fzero to find the root of f(x) = x 2|sinx|-4=0, initial guess x0=0.5.
b) Use false-position method to find the root of f(x)= x 2|sinx|-4=0, initial guess [a, b]=[0, 4].
Accept your result when |f(x i)|<510-7. Print out all iterations of x i and f(x i).
Here’s the best way to solve it.Solution 100%(1 rating) Share Share Share done loading Copy link Here’s how to approach this question This AI-generated tip is based on Chegg's full solution. Sign up to see more!
Write a Matlab function f(x) that returns y equal to (x.2).×(|sin(x)|)−4. Code for file "f.m" function y = f( x ) y=(x.^2).(abs(sin(x)))-4; end save the above file by the name f.m a) code for the file "fzro.m" fun=@f; x0=0.5; z=fzero(fun,x…
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16854 | https://ckrao.wordpress.com/2010/12/07/why-circles-map-to-ellipses-under-linear-maps/ | Why circles map to ellipses under linear maps | Chaitanya's Random Pages
Chaitanya's Random Pages
December 7, 2010
Why circles map to ellipses under linear maps
Filed under: mathematics — ckrao @ 11:16 am
Here is an interesting fact worth pondering. Take a circle and stretch it along some direction. It becomes an ellipse. Now take the ellipse and stretch it again, this time in a different direction. Continue stretching or shrinking in other directions if you like. No matter what directions are used, the final image is an ellipse, having two perpendicular axes (ignoring the degenerate case of a line segment). The direction of these axes depend on the direction and amount of stretching, but the existence of perpendicular axes always remains. This did not seem immediately obvious to me – surely one can be clever about the way one stretches and remove the perpendicularity of the axes!
A linear transformation of a circle is a sequence of stretches along axes, and such a sequence can always be reduced to at most two perpendicular stretches. That is, circles map to ellipses under linear maps.
One way of seeing why this is the case is that the equation of an ellipse is a quadratic form, and applying linear transformations does not change its degree (i.e. a conic remains a conic). Since the only bounded conic sections are ellipses, the result follows. However I would be more satisfied with a coordinate-free proof. Here is one argument that I read recently and outline below. (See more via Trefethen and Bau, Planetmath, and also the nice explanation due to Aubrey Poor here)
Let T be a linear transformation (map) from n-dimensional to m-dimensional space. T is usually represented by an m by n matrix, but we will not need that here. We will assume complex vector spaces for generality, but a 2-dimensional real vector space is easier to visualise. We wish to show that a unit sphere maps to an ellipsoid, by which we mean there exist perpendicular axes of the sphere that each map to perpendicular axes in the m dimensional space.
Consider the image of the unit sphere under T. There exists a vector such that is maximal (by the extreme value theorem). Let where and are unit vectors and (we ignore the degenerate case ). We wish to show that any vector perpendicular to maps to a vector perpendicular to . Suppose where (such an orthogonal decomposition is possible). We identify a vector that maps to something as least as long as . Consider the unit vector
Then
The length of this vector is at least
where equality holds when . In other words, to prevent from having length greater than , we require . This means as we wished to show.
By an inductive argument we may then show that there exist orthonormal vectors in the n-dimensional space and orthonormal vectors in the m-dimensional space such that
In other words there exist orthonormal bases in the row space and column space of T which map to each other. This is a restatement of the singular value decomposition (SVD), that spheres map to ellipsoids (generalised to any dimension). The are known as singular values and are interpreted as the semi-axes lengths of the ellipsoidal image of the unit sphere. Stacking the v’s and u’s into matrices (and adding orthonormal vectors from nullspaces if necessary) gives us the alternate form
where is the matrix representation of the linear transformation T, and has the singular values down its diagonal and 0 entries elsewhere.
The SVD (which applies to any matrix) is the generalisation of the spectral theorem (which applies to normal matrices) and has wide applications, from solving least squares problems (via the pseudo-inverse) to finding low-rank approximations to matrices (enabling compression).
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16855 | https://www.oxfordreference.com/abstract/10.1093/acref/9780198821489.001.0001/acref-9780198821489-e-1629?rskey=VTSVc0&result=1 | Family - Oxford Reference
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A Dictionary of Biology (8 ed.)
Edited by: Robert Hine
Publisher:Oxford University Press Print Publication Date:2019 Print ISBN-13:9780198821489 Published online:2019 Current Online Version:2019 eISBN:9780191860812
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date: 29 September 2025
family
Source:A Dictionary of Biology Author(s):Robert HineRobert Hine
1. (in taxonomy) A category used in the classification of organisms that consists of one or several similar or closely related genera. Similar families are grouped into an order. Family names end in -... ...
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Preface
Credits
The List of Entries by Subject
SI units
Simplified phylogenetic tree of the animal kingdom
Simplified phylogenetic tree for plants
Geological time scale
Navigating the body
Model organisms and their genomes
Major mass extinction of species
Nobel prizewinning contributions to biology
Evolution
Useful websites
Oxford University Press
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16856 | https://www.chemspider.com/Chemical-Structure.171.html | Accessed: Sun, 21 Sep 2025 09:36:11 GMT
Acetic acid
Acetic acid
| | |
--- |
| Molecular formula: | C2H4O2 |
| Average mass: | 60.052 |
| Monoisotopic mass: | 60.021129 |
| ChemSpider ID: | 171 |
Wikipedia articleWikipedia article
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Verified
200-580-7
[EINECS]
64-19-7
[RN]
Acetic acid
[Wiki] [IUPAC name – generated by ACD/Name] [IUPAC index name – generated by ACD/Name]
Acid, Acetic
Acide acétique
[French]
[IUPAC name – generated by ACD/Name]
Acido acetico
[Italian]
AcOH
[Formula]
ättiksyra
[Swedish]
azido azetikoa
[Basque]
azijnzuur
[Dutch]
Essigsäure
[German]
[IUPAC name – generated by ACD/Name]
Ethanoic acid
etikkahappo
[Finnish]
Glacial acetic acid
HOAc
[Formula]
kwas octowy
[Polish]
Kyselina octova
[Czech]
MFCD00036152
[MDL number]
MFCD00198163
[MDL number]
Unverified
(2H3)Acetic (2H)acid
10.Methanecarboxylic acid
109945-04-2
[RN]
1112-02-3
[RN]
120416-14-0
[RN]
147416-04-4
[RN]
149748-09-4
[RN]
1563-80-0
[RN]
159037-04-4
[RN]
16651-47-1
[RN]
17217-83-3
[RN]
1794892-02-6
[RN]
2-Mercapto-5-chlor-benzoxazol-7-sulfonsure, Kaliumsalz
3913-68-6
[RN]
42204-14-8
[RN]
498-63-5
[RN]
55511-07-4
[RN]
57745-60-5
[RN]
63459-47-2
[RN]
88-32-4
[RN]
99%
AA
Acetic acid (JP17/NF)
Acetic acid 1 mol/L
Acetic Acid, GlenDry, anhydrous
Acetic acid-1-13C
Acetic acid-1-13C,d4
Acetic acid-12C2
Acetic acid-13C2
Acetic acid-17O2
Acetic acid-18O2
Acetic acid-2,2,2-d3
Acetic acid-2-13C
Acetic acid-C,C,C-d3
Acetic Acid-d4
Acetic-2,2,2-d3 Acid
acetol
C2:0
Essigsaeure
Ethanoate
Ethylic acid
Glacial Acetic
Glacial acetic acid (JP17)
[Formula]
[Formula]
Methanecarboxylic acid
Methanecarboxylic Acid, Acetic Acid
methyl carboxylic acid
[MDL number]
Pyroacetic acid
Database IDs |
16857 | https://stackoverflow.com/questions/47085624/mean-of-means-how-to-aggregate-means-of-different-sample-sizes | Stack Overflow
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Mean of Means - How to aggregate means of different sample sizes
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Modified 7 years, 11 months ago
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How do I think about taking the average of several different averages?
Here is some data:
library(dplyr) month <- c("January", "January","January", "February", "March", "April", "April", "May", "June", "July") year <- c(2014, 2014, 2014, 2014, 2014, 2014, 2014, 2014, 2014, 2014) v1 <- c(0, 1, 0, 1, 0, 0, 1, 0, 1, 1) df <- data.frame(month, year, v1)
As you can see, I have different sample sizes for different months. January's sample size is 3, April's sample size is 2, etc.
I can take the average of each of them, gaining a mean for every month:
df %>% group_by(year, month) %>% summarize_all(mean)
However, how do I get a correct mean for the year 2014, given that I have the mean of several months, where each monthly mean had varying sample sizes?
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asked Nov 2, 2017 at 22:29
John StudJohn Stud
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Is this a statistics question? You just group_by(year) assuming each row is a sample as it appears in your data. Alternatively, you can use weighted.mean()
ssp3nc3r
– ssp3nc3r
2017-11-02 22:44:16 +00:00
Commented Nov 2, 2017 at 22:44
Yes, I suppose it is a statistics question. I am not certain if it requires any sort of weighting given that the monthly averages were created with different sample sizes.
John Stud
– John Stud
2017-11-02 22:47:43 +00:00
Commented Nov 2, 2017 at 22:47
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ssp3nc3r
– ssp3nc3r
2017-11-02 22:57:35 +00:00
Commented Nov 2, 2017 at 22:57
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Both methods of averaging give you the same value:
Here's simply taking the yearly average:
df %>% group_by(year) %>% summarise(year_avg = mean(v1))
Compare with taking the average of monthly averages where each month has a different sample:
df %>% group_by(year, month) %>% summarise(month_avg = mean(v1), samples = n()) %>% summarise(year_avg = weighted.mean(month_avg, samples))
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answered Nov 2, 2017 at 22:55
ssp3nc3rssp3nc3r
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Don't have the rep to comment. Your question is unclear, what is it you want to calculate? Do you want the mean monthly average?
Given the lack of data in some months I wonder whether it's appropriate to calculate a mean in each month. If you just want the straightforward mean for 2014 then there is no need to group into months, you can just calculate the sample mean.
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answered Nov 2, 2017 at 22:46
ASeatonASeaton
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John Stud
John Stud
I have a varying amount of daily binary observations that I am aggregating into monthly data by taking the mean. However, I am also interested in not just looking at the mean for each month, I am also curious as to the mean for the year. I see your point now that I can just take the mean from the un-aggregated data instead of trying to take the mean of several means that were calculated for the monthly level.
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16858 | https://www.youtube.com/watch?v=ID5K1Yt7Yho | AutoIonization of Water, Ion Product Constant - Kw, Calculating H3O+, OH-, and pH Using Ice Tables
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Posted: 27 Nov 2017
This acids and bases chemistry video tutorial provides a basic introduction into the auto-ionization of water. It explains how to calculate the hydroxide ion [OH-] concentration given the hydronium ion [H3O+] concentration using the ion product constant of water - Kw. it explains how to determine if a solution is acidic, basic, or neutral based on the concentration of H3O+ at 25C. It explains how to calculate the pH of a neutral solution when the temperature increases and how to tell if the auto ionization of water is an endothermic or an exothermic process. Finally, it shows you how to calculate the pH of a very dilute basic solution of 3 x 10^-7 M NaOH. This tutorial contains plenty of examples and practice problems with the formulas and equations as well.
Acids-Bases - Free Formula Sheet:
Chapter 14 - Video Lessons:
Acids and Bases - Introduction:
The 7 Strong Acids to Memorize:
Conjugate Acid-Base Pairs:
pH and pOH Calculations:
Estimate The pH Without a Calculator:
Autoionization of Water - Kw:
Which Acid Is Stronger?
Acidic, Basic, & Neutral Salts:
pH of Weak Acids:
Buffer Solutions:
Polyprotic Acid Base Equilibria:
Acid Base Titration Curves:
Acids and Bases - Practice Test:
Ksp - Molar Solubility & Ice Tables:
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Entropy - 2nd Law of Thermodynamics:
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Transcript:
in this tutorial we're going to talk about how to solve problems associated with the autoionization of water so let's start with this one we're given the concentration of h3o plus also known as the hydronium ion and our goal is to calculate the concentration of hydroxide at 25 degrees celsius so what formula do we need to use now let's talk about the autoionization of water water can react with itself to form two ions the hydronium ion h2o plus and hydroxyl ion so this water molecule acts as the acid it donates the proton and this one act as a base it accepts the proton and so we get the conjugate acid h3o plus and the conjugate base hydroxide because water can act as an acid and as a base it's known as an amphoteric substance now the equilibrium expression for this reaction is going to be h3o plus times o h minus now this is dissolved in water so it's an aqueous phase now water is in the liquid phase so we cannot include it in the equilibrium expression so you shouldn't write h2o squared on the bottom so this is the ion product constant kw and it's equal to the product of h2o plus and oh minus now at 25 degrees celsius kw is one times 10 to the negative 14. now this value changes with temperature so keep that in mind it's temperature dependent so now we can calculate the concentration of hydroxide it's simply going to be kw which is 1 times 10 to the minus 14 divided by the hydronium concentration of h3o plus so we have to take that number divided by 2 times 10 to the minus 6. and so the hydroxide concentration in this example is 5 times 10 to the negative 9 moles per liter or molarity so that's it for this problem number two the concentration of h3o plus is three times ten to the negative five moles per liter is the solution acidic basic or neutral now if the concentration of h3o plus at 25 degrees celsius if it's greater than 1 times 10 to negative 7 the solution will be acidic now if the temperature is not specified assume that it's 25 degrees celsius now if the concentration of h2o plus equals 1 times 10 to the minus 7 moles per liter at 25 degrees celsius then the solution is going to be neutral if it's less than its number then we have a basic solution now ten to the negative five is greater than ten to negative seven so therefore this particular solution is acidic number three the kw constant for water at 50 degrees celsius is 5.5 times 10 to the negative 14. calculate the h2o plus concentration so let's make an ice table so liquid water is going to ionize into two ions the hydronium ion and the hydroxide ion so we have pure water so initially this is going to be zero and as a result the reaction has no choice but to shift to the right so this is going to increase by x and that's going to increase by x so kw which is h3o plus times o h minus that's going to be x times x so we can say that 5 times or 5.5 times 10 to negative 14 is x times x or x squared and x is the concentration of h3o plus and hydroxide for this problem so to calculate x we need to take the square root of both sides so x which equals the concentration of h3o plus that's going to be 2.345 times 10 to the negative seven moles per liter so this is the answer to part a now let's move on to part b now let's move on to part b what is the ph of the solution the ph of the solution is simply the negative log of the hydronium ion concentration which we already have so it's going to be negative log 2.345 times 10 to the minus 7. so the ph is 6.63 by the way is the solution acidic basic or neutral now notice the temperature is different it's not 50. i mean it's not 25 it's 50. so we can't just compare h2o plus to 1 times 10 to the minus 7. that only applies at 25 degrees celsius so we need something else to describe if it's going to be acidic basic or neutral if the h2o plus concentration if it's greater than the hydroxide concentration then overall it's acidic if it's less than the hydroxide concentration then the solution is going to be basic but if they're equal it's neutral we still have pure water at a different temperature and notice that x is equal to both h2o plus and oh minus which means that those two are equal to each other so what we have is a neutral solution because h3o plus equals the hydroxide concentration so just keep that in mind now let's talk about part c what is the relationship between kw and temperature well we know that kw is 1 times 10 to the negative 14 at a temperature of 25 degrees celsius now in this example kw is higher it's 5.5 times 10 to the negative 14 at a temperature of 50 degrees celsius so we can see that as the temperature increases the autoionization of water increases so we have a higher ion product constant now is the autoionization of water is it exothermic or endothermic so let's write the reaction so the change in enthalpy for this reaction do you think it's going to be positive which is associated with an endothelic reaction or a negative which is associated with an exothermic reaction now let's review some basics in le chatelier's principle if we have an endothermic reaction if you increase the temperature the reaction is going to shift to the right now if the reaction is exothermic if you increase the temperature it's going to shift to left so make sure you understand that now in our example we increased the temperature and kw increased now kw is proportional to the concentrations of h3o plus and oh minus so if kw increased that means that the concentration of oh minus had to increase and h3o plus had to increase so the value of the products went up now if the reaction were to go to the left that is towards the reactants the value of the reactants will go up and the products will go down however if the reaction shifts to the right then the reactants will decrease in value and the products will increase in value so because kw went up which means that the products had to go up therefore it had to shift to the right which means that we have an endothermic reaction now let's make a short summary of what we just talked about now if the temperature goes up and if the equilibrium constant goes up as well then this means that the reactions shift to the right and so what we have is an endothermic reaction now keep in mind k is the ratio of the products to the reactants if you increase the numerator of a fraction the value of the fraction go up so as the reaction shifts to the right the value of the products go up and the value of the reactants go down which causes k to increase if it's due to a change in temperature if the temperature doesn't change then k will stay the same now let's say if we increase the temperature and if k decreased instead of increased so what this means is that the reaction shifts to left and so therefore this would be an exothermic reaction whereas this one is an endothermic reaction now if it's exo and if it shifts to the left that means that the reactants increase in value and the products decrease in value if you increase the denominator the value of the fraction goes down and if you decrease the numerator the value of the fraction also goes down so by shifting to the left this will cause a decrease in k so make sure you understand that so if an increase in temperature leads to an increase in k value it's an endothermic reaction if it leads to a decrease in k value it's exo number four calculate the ph of a three times ten to the minus seven moles per liter sodium hydroxide solution now you might think that this is an easy problem but it's actually a challenge problem go ahead and try it if you want to now here's a question for you what is the hydroxide concentration is it three times ten to minus seven the answer is it's not three times ten to minus seven so there's something we need to understand right now we have sodium hydroxide and it's going to dissolve into the sodium cation and hydroxide and so this concentration is the concentration for both of these ions due to the dissolution of naoh now keep in mind water is also present so water can dissociate into h3o plus and oh minus and this is going to be 1 times 10 to the minus 7. now if this value is large we could ignore this number however this value is small and so this value is not too small relative to this number so we need to take both into account so initially once we dissolve this amount of naoh the initial concentration of hydroxide is the sum of these two so let's make an ice table so we have initial change in equilibrium so initially this is going to be four times ten to the minus seven that's due to this plus the ionization of water and this is going to be 1 times 10 to the negative 7 due to the autoionization of water alone now i didn't specify the temperature so always assume it's 25 degrees celsius unless specified otherwise which means kw is one times ten to negative fourteen now in what direction will the reaction shift will it shift to the right or will it shift to the left in order to determine this we need to calculate the reaction quotient so kw is equal to the product of h3o plus and oh minus and the reaction quotient q is also going to be products over reactants so it's going to be these two multiplied to each other and keep in mind this is a liquid so we can't include that in the equilibrium expression now the difference between q and k is q is associated with the equilibrium values k i mean k is associated with the equilibrium values i want to make that correction q is associated with the initial values so if i confuse you just remember q is associated with the initial values k is associated with the equilibrium values so these are initial values so i'm going to calculate q using those values so h3o plus is 1 times ten to the minus seven and q i mean hydroxide rather i'm just making a lot of mistakes today hydroxide is four times ten to the minus seven so if we multiply those two numbers the q value is going to be 4 times 10 to the negative 14. now why is this important you need to know that if q is greater than k the reaction is going to shift to left if q is less than k it's going to shift to the right and if q equals k it's at equilibrium so this number is greater than kw which is 1 times 10 to the minus 14. so since q is greater than k the reaction is going to shift to the left now let's say if it shift to the right this would be plus x then plus x however because it shifts to the left because it's going this way this is going to be minus x and minus x and that's why it's important to determine in what direction the reaction is going to shift because that's going to affect your answer it's going to be the difference between a right answer and a wrong answer if you get a wrong answer on the test what good is that now let's go ahead and add the first two rows so for the last row this is going to be 1 times 10 to the minus 7 minus x and then 4 times 10 to the minus 7 minus x so kw is going to be the product of h3o plus and oh minus so at 25 degrees celsius kw is 1 times 10 to the minus 14. at equilibrium h3o plus is 1 times 10 to the minus 7 minus x as it shifts to the left this is going to decrease and the hydroxide concentration is 4 times 10 to the minus 7 minus x now we no longer need the ice table but keep in mind that h3o plus is 1 times 10 to the minus 7 minus x because we're going to get that value and then use that to calculate the ph of the solution so let's get rid of a few things we're going to have to use the quadratic equation for this problem so let's foil 1 times 10 to the minus 7 multiplied by four times ten to the minus seven one times four is four and negative seven plus negative seven is negative fourteen so we get this number and then we have this times negative x so that's 1 times 10 to the minus 7 multiplied by x and then if we multiply these two that's going to be negative 4 times 10 to the negative 7 times x and then negative x times negative x is positive x squared and all of that is equal to one times ten to the negative fourteen now we need to write our expression in standard form for a quadratic equation so we want to write it like this ax squared plus bx plus c so this is the x squared term let's write that first and here we have two like terms so negative one minus four is negative five so we got negative five times ten to the minus seven multiplied by x now i'm gonna move this to the other side so basically i'm going to subtract this number by that number so 4 minus 1 is 3. so on the right it's going to be positive 3 times 10 to the negative 14 and that's equal to 0. so now we could use the quadratic formula x is equal to negative b plus or minus the square root of b squared minus 4ac divided by 2a so b is the number in front of x which is this number so x is going to be negative times negative 5 times 10 to the minus 7 plus or minus the square root of negative five times ten to negative seven almost wrote 17 there squared minus four the number in front of x squared is a one so a is one and this is the c value so that's three times ten to the minus fourteen divided by two a or two times one which is two so we can cancel these two negative signs and so that's going to be positive 5 times 10 to the negative 7 and then negative 5 squared that's positive 25. 10 to the negative 7 squared that's going to be negative 7 times 2 so that's 10 to the minus 14. and negative 4 times 3 that's going to be negative 12 times 10 to negative 14. so here we have 25 minus 12 and so that's 13. so we have 5 times 10 to the negative 7 plus or minus the square root of thirteen times ten to negative fourteen over two so now let's go ahead and take the square root of that number and let's just clear away this now so what we have is five times ten to the minus seven plus or minus three point six zero six times ten to the minus seven divided by two now we have two possible answers plus or minus now if you recall the h2 plus concentration was one times ten to the minus seven minus x so if we use a plus sign five plus three point six that's going to be eight point six times ten to the minus seven and if you take one times ten to the minus seven minus eight point six times ten to the minus seven that's going to give you a negative number which is not possible so there's no point in using the positive sign we have to use the negative sign so 5 times 10 to the minus 7 minus 3.606 times 10 to the minus seven and then divided by two so that's going to give us an x value of 6.97 times 10 to the minus 8 which is less than this number so if we take this number minus x we're going to get a positive result which is a good sign so now let's calculate the h2o plus concentration and to check our work we're going to calculate the hydroxide ion concentration so let's start with h3o plus so we know it's uh this number minus x so it's one times ten to the minus seven minus six point nine seven times ten to the minus eight and so that's going to come out to be three point zero three times ten to negative eight now let's calculate the hydroxide ion concentration so at equilibrium it was 4 times 10 to the minus 7 minus x or minus six point nine seven times ten to the minus eight and so that is equal to 3.303 times 10 to the minus 7. now let's make sure that our answers are correct the product of h2o plus and hydroxide should equal kw so let's see if we can get an answer that's close to it so h3o plus is three point zero three times ten to the minus eight and hydroxide is three point three zero three times ten to the minus seven so let's multiply those two numbers and so this gives us 1.0008 times 10 to the minus 14 which is good enough so that means that these values are indeed correct now the last thing that we need to do is calculate the ph of the solution which is negative log of the h2o plus concentration so let's plug in that number negative log of 3.03 times 10 to the minus 8. and so the ph of the solution is 7.52 so this is at 25 degrees celsius so if the ph is above 7 that means that it's basic and we did have a small excess of hydroxide whenever the hydroxide ion concentration is greater than the h2o plus concentration as we can see it here it's going to be a basic solution and the ph is slightly above seven so that's an agreement with the numbers that we see here and so that's it for this video thanks for watching and have a good day you |
16859 | https://jillianstarrteaching.com/open-number-lines-addition-subtraction/ | Published Time: 2024-04-14T22:25:31-04:00
Open Number Lines for Addition and Subtraction - Overcoming the 3 Most Common Obstacles!
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Open Number Lines for Addition and Subtraction – Overcoming the 3 Most Common Obstacles!
Open number lines are the Swiss Army knife of number lines: useful, practical, and multi-purpose!
One way to use open number lines is to use them as a model for addition and subtraction. Unlike regular number lines, open number lines are created by the mathematician. “Open” in the term open number line means that it does not have a set start and end point. Instead, students decide where to start and how to decompose a number to add or subtract.
There are multiple ways to solve problems using open number lines, which allows students to approach problems in a way that makes sense to them. If you’re reading this and thinking, “A visual support that can be differentiated to the needs of each learner and also acts as a perfect tool to help students share their thinking?!” The answer is, yes, open number lines do it all.
But today I’m here to share the other side of this conversation. While open number lines are an incredibly versatile tool, I have found that students often stumble on a few common obstacles, which can make things frustrating for students and their teachers. Let’s face these obstacles head-on so that we can take full advantage of the greatness that is open number lines!
Obstacle #1 – Where to begin with open number lines?
Starting with open number lines from scratch can leave students scratching their heads! Because open number lines do not exist until students create them, the first obstacle kids encounter is how to know where to even start their number lines. Fear not, every problem drops a hint!
When solving addition equations, students should already have an understanding of the commutative property (even if they don’t know what it’s called!) so that they know that, for example, 45 + 76 and 76 + 45 will have the same sum. This property of addition is something students should be familiar with from their work with addition in first grade.
When creating their open number lines to solve 45 + 76, students can choose to start at 45 or 76 because of our old pal, the commutative property. When students first begin doing this work, it really will not matter which addend they choose to start at. However, as they progress, students will likely begin to realize that, like when they are counting on to solve problems like 4 + 7, it can be much more efficient to start at the larger addend. When using open number lines, starting at the larger addend like 76 will mean fewer jumps to get to the sum.
Here are two examples of what an open number line might look like for 45 + 76:
Of course, not all problems will be addition problems with two addends. Instead, students will have to solve problems like 82 – 34 and 35 + __ = 59. Subtraction equations can be approached using the idea of counting back to subtract. Students can start at 82 and jump back 34 on the number line to see where they land.
Similarly, missing addend equations Like 35 + __ = 59 can be solved by starting at the sum, in this case, 59, and jumping back 35 to see where they land.
Those aren’t the only ways to solve those problems though! We want our students to understand subtraction flexibly. Subtraction is not only taking away, but also the distance between two numbers. So 82 – 34 can also be solved by starting at 82 and jumping back to 34. This time, the answer is not where we landed on the number line, but the jump it took to get there. Don’t worry – more on this later!
35 + __ = 59 can be solved that way too. Students can start at 59 and jump back until they land at 35. OR! They can start at 35 and jump forward until they land at 59. In either case, their answer is the distance between the two numbers.
Now, I realize that I showed you, the teacher, how to model different problems on open number lines, but how do we get kids to think deeply about approaching these problems? Here are some prompts I use with my students when they are feeling stuck:
What kind of problem are you solving?
What information are you trying to find? (The sum, the addend, the difference, etc.)
What if this was a smaller problem like 5 + __ = 9 instead of 35 + __ = 59, how would you solve it then?
Tell me how you are thinking about solving this problem. How can we model that on the number line?
Obstacle #2 – Friendly numbers only on open number lines
The second obstacle I see students stumble upon is not using friendly numbers when moving on an open number line. “Friendly numbers” are what I call numbers that make computation easier. Not using friendly numbers can cause students to make calculation mistakes because they are attempting to add or subtract numbers mentally that they are not yet able to solve accurately, defeating the purpose of using an open number line.
Instead, we want to think flexibly about decomposing numbers to add or subtract them efficiently. Let me show you what I mean. Let’s say we are solving 57 + 28. Students who are not using friendly numbers might use the open number line to start at 57 and then jump forward 5 as their first step when adding 28. While 5 is often considered a friendly number, in this case, students need to add 57 and 5 which may be tricky for some students.
One way to use friendly numbers to solve that problem would be to decompose the 28 when adding to reach a multiple of 10. What does this look like? If we start at 57, we might first add 3 first to land at 60. Then I still need to add 25 but I can efficiently add a multiple of ten to a two-digit number so adding 60 and 25 is easy peasy.
Using friendly numbers is also helpful when subtracting. Suppose a student is solving 72 – 26. As mentioned earlier, they could start at 72 and jump back 26, start at 72 and jump back to land at 26, or start at 26 and jump forward to 72. Regardless of the approach, decomposing the numbers to reach multiples of tens makes the calculations much more efficient!
Obstacle #3 – So what’s my answer?
Here is where it all comes together. Students have done all of the work using open number lines to solve the problem but then they’re faced with the million-dollar question: where is my answer? The final obstacle students face when using open number lines is figuring out what the heck the answer is! Is the answer the jumps, the number they landed at, where they started, or something else?
Some problems and approaches are easier for students. Problems like 45 + 76 tend to be straightforward when determining the answer. Students are often able to start at one number, jump on the number line, and find their answer at the spot on which they land.
Some problems like 62 – 26 or 35 + __ = 59 are tougher nuts to crack. If students do not have a solid understanding of the properties of addition and subtraction equations those kinds of problems can create confusion. For example, a student who does not understand the meaning of the equal sign as “is the same as” rather than “the answer” may say the answer to 35 + __ = 59 is 59.
Another student who does not have a solid understanding of the relationship between addition and subtraction may attempt to solve 62 – 26 by starting at 26 and jumping to 62. In this case, the answer would be the total number of jumps it took to get to 62. In other words 26 + __ = 62 with the blank representing the jumps. However, without that important understanding, students may look at their open number line and say their answer is 62 because that is where they landed.
Smaller Problems to Larger Problems on an Open Number Line
To support students who are facing this obstacle, it can be very helpful to have them solve similar problems using smaller numbers. Instead of solving 72 – 26, I might ask a student to show me how they can use open number lines to solve 22 – 6 using the approach they were attempting to use with the larger problem.
For example, I might help them to start at 6 and jump to 22 using friendly numbers. Then we’d discuss what the answer is to the smaller problem, 22 – 6. Once students are comfortable using their approach with these smaller numbers, I’ll have them work back up to the larger problem.
Some students often just need a simple reminder to look again at the problem they are trying to solve! For example, when solving 35 + __ = 59, students may just need the reminder to name the “mystery number” rather than one of the numbers already in the equation. I’ll ask, “Which of the numbers you see is the one that was a mystery when you first started solving?” Everyone loves a good mystery and this phrasing has resonated with so many of my students!
While second graders likely have experience with number paths and even number lines, the transition to open number lines can be tricky. Getting ahead of the common obstacles young mathematicians face will certainly help them to be successful along the number path to number line progression in no time!
Addition & Subtraction Resources
Addition & Subtraction Centers Pack
Make a Ten to Add Centers
2-Digit Addition & Subtraction Centers
Single Digit Addition Logic Puzzles
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Ways to Make Ten – Why Kids NEED to be Fluent with Making 10Why I Don’t Use Base-Ten Blocks to Introduce Addition and Subtraction (and what I use instead)7 Favorite Subtraction with Regrouping Games and Activities
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16860 | https://www.youtube.com/watch?v=Y7cwsliV94E | How to calculate f(x)=1/1-x as a Taylor Series Polynomial at 0
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Posted: 9 May 2020
How to calculate a Taylor Polynomial of degree 5 at 0 for 1/(1-x)
A useful Taylor Polynomial which can be used in many scenarios.
Take the derivative of 1/(1-x) at 0 for 5 terms.
Then take the values of each of these derivatives at 0 and input into the general formula for a Taylor Polynomial , this Taylor Series could also be known as a Maclaurin series.
This is also known as 1+x+x^2+x^3+x^4.....
taylorseries
taylorserrano
polynomials
polynomial
maclaurinseries
maclaurin
derivatives
triginometry
calc1
calculus
The graph plotter used in this video is at the link below
The LaTex used to make this video is below
\documentclass{article}
\usepackage{fancyhdr}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{xfrac}
\usepackage{gensymb}
\usepackage{polynom}
\usepackage{scalerel}
\usepackage{stackengine}
\usepackage{xcolor}
\usepackage{bigints}
\usepackage{physics}
\usepackage{lastpage}
\usepackage{mathtools}
\usepackage{wasysym}
\newcommand{\mathsym}{{}}
\newcommand{\unicode}{{}}
\newcounter{mathematicspage}
\addtolength{\oddsidemargin}{-.875in}
\addtolength{\evensidemargin}{-.875in}
\addtolength{\textwidth}{1.75in}
\addtolength{\topmargin}{-.875in}
\addtolength{\textheight}{1.75in}
\begin{document}
\large
Calculate the Taylor Polynomial for 5 terms at point $0$ for the function $f(x)=\cfrac{1}{1-x}$.\
\
\
\
\
The formula for a Taylor series of degree n at a point a for any function is as follows.\
\
\
\
\
[
T_n(x)=f(a)+\cfrac{f^{'}(a)}{1!}(x-a)^1+\cfrac{f^2(a)}{2!}(x-a)^2+\cfrac{f^3(a)}{3!}(x-a)^3+\cfrac{f^4(a)}{4!}(x-a)^4+\cfrac{f^5(a)}{5!}(x-a)^5......\cfrac{f^n(a)}{n!}(x-a)^n
]
\
\
\
Differentiating the function $\cfrac{1}{1-x}\ $ 5 times and the values of each derivative at $0$.\
\
\begin{align}
& f(x)=\cfrac{1}{1-x} & f(0)=1 \
& f^{'}(x)=\cfrac{1}{(1-x)^2} & f^{'}(0)=1 \
& f^2(x)=\cfrac{2}{(1-x)^3} & f^2(0)=2 \
& f^3(x)=\cfrac{6}{(1-x)^4} & f^3(0)=6 \
& f^4(x)=\cfrac{24}{(1-x)^5} & f^4(0)=24 \
& f^5(x)=\cfrac{120}{(1-x)^6} & f^5(0)=120 \
\end{align}
[
T_5(x)=f(0)+\cfrac{f^{'}(0)}{1!}(x-0)+\cfrac{f^2(0)}{2!}(x)^2+\cfrac{f^3(0)}{3!}(x)^3+\cfrac{f^4(0)}{4!}(x)^4+\cfrac{f^5(0)}{5!}(x)^5
]
\
So the Taylor Polynomial of degree 5 at point $0$ for $\cfrac{1}{1-x}$ is\
[
T_5(x)=1+x+\cfrac{2(x^2)}{2!}+\cfrac{6(x^3)}{3!}+\cfrac{24(x^4)}{4!}+\cfrac{120(x^5)}{5!}
]
So this simplifies to
[
T_5(x)=1+x+x^2+x^3+x^4+x^5
]
\clearpage
\end{document}
1 comments
Transcript:
okay so in this video we're going to calculate Taylor polynomial for five terms at the point zero for the function 1 over 1 minus X and as always always remember to find the three important things to remember when you try to calculate a Taylor polynomial so the degree n which is five terms of n is five so that's the first thing needs to be important to remember the point a in this case point zero so it's at the zero which were calculating it for and a function of course which is one over one minus X and then you've got your generic formula for Taylor polynomial which is here and basically what you would do is you will substitute the ends for fives and the AES for zeroes so in this function here this TN of X which is the generic name for Taylor polynomial five degrees so T five of X and then the function at the value of zero then the value of the first derivative of 0 divided by 1 factorial then X minus a well that's in this case it's just going to be X because X 1 is zero to power of 1 just X to power one then we've got the second derivative divided by 2 factorial and that's calculated at 0 as well and the X minus a also would be x squared and so on and so on but also remember the pattern in Taylor polynomials you've got a 2 2 2 3 3 & 3 4 4 & 4 & 5 5 & 5 but that's the end values basically as we've got here at the end which is the generic formula for the stages of the Taylor polynomial so next thing to do we need to differentiate the function so differentiating 1 differentiate 1 divided by 1 minus X so we're going to differentiate it 5 times and calculate the derivative or the value of 0 so we start off draw a nice little table here I'm going to show you a nice easy shortcut way of differentiating 1 over 1 minus X and then to calculate the function at 0 is pretty straightforward just plug in 0 and calculate it so the value that the function is 1 over 1 minus X and the value at 0 is just 1 over 1 1 minus 0 is 1 so that's 1 so that's straightforward to calculate the first line then the first derivative that we could imagine is here this 1 minus X is to the power of minus 1 but when it's minus 1 you've got it's in the denominator so therefore when you increase the power you increase the power because you're going up and negative if this was a positive you would decrease it by 1 so you imagine that's a minus 1 so 1 over 1 minus X to the minus 1 and then take the derivative of the denominator which is minus 1 1 minus X derivative of minus X is minus 1 derivative of 1 is 0 so 0 0 minus 1 so 1 over 1 minus x squared is what we get so you basically got a minus 1 here and a 1 so that's minus 1 1 times minus 1 is minus 1 multiply it by the derivative of the bottom which is minus 1 minus 1 times 1 so 1 is 1 therefore you get the one on the top and then you 1 minus X is then in brackets and then squared but you're going to increase the power because we're going in negative territory so effectively this is 1 divided by 1 minus x squared is actually 1 minus X to the power of minus 2 so we get 2 there so again using that formula 1 times minus 2 because that's a always going to be a minus so that's minus 2 times y minus 1 minus 2 times minus 1 is plus 2 then increase the power we're going from minus 2 to minus 3 so that's minus 3 then your next derivative 2 times 3 is 6 but it's a minus because the 3 is in the denominator and then minus 6 times the derivative of minus 8 which is minus 1 is plus 6 and then increase your power to the 4 it's got six times minus 4 is minus 24 times minus one is plus 24 so I mean let me get the 24 24 times minus five is 120 120 which is opposed to being minus 120 because I said minus five said minus 120 times that minus one is plus 120 so then to calculate the values of all these derivatives basically in the denominator we've got 1 minus 0 squared 1 minus 0 cubed 1 minus 0 to the fourth 1 minus 0 to the 5 so all these denominators are all basically 1 whatever 1 is the power of anything is always 1 so 1 divided by 1 is 1 so the first derivative is 1 2 divided by 1 is 2 plus 2 6 divided by 1 is 6 and so on 24 and the 5th 120 so then what we do now is we've got the values of all our derivatives we're just going to plug those values in to the Taylor polynomial formula and then we end up the first stage obviously we're going to put this substitute all the A's in so the F of 0 first derivative at 0 X minus 0 I've left the first one in there so you can see why we've left it there as x squared X cubed because you don't want to keep writing 0 and then now we're going to substitute all these values in to our formula so the fifth derivative is 120 so let's substitute your 120 in there the fourth derivative is 24 so we substitute for 24 in there third derivative is 6 substitute was 6 in now the second review is to substitute with 2 in there and the value is 0 and the derivative is both 1 so basically substitute 1 there and a 1 there so that's a lot what we've got so we've got T 5 of X is 1 plus X plus 2 x squared over 2 factorial and then these are so on and so on so to see how he got to that the value at 0 is 1 so that's why we put a 1 there first derivative is 1 1 over 1 factorial times X is X the second derivative was 2 so 2 divided by 2 factorial by Z cancels each other out and then we just got the x squared same with the third derivative which was 6 so 6 divided by 3 factorial or 3 factorial is 6 so again these cancel each other out and you see there's a pattern for B now fourth derivative is 24 4 factorial is also 24 so these will also cancel out and the same with the 5th one 125 factorial is 120 because 1 times 2 times 3 times 4 times 5 is 120 so again these cancel out so this makes a nice little mean formula so basically we probably imagine going all the way along here plus X to the 6 plus X to the 7 plus X to the 8th again improving the accuracy of our Taylor polynomial each time so just to simplify that up we can just write it as this t 5 of x equals 1 plus X plus x squared plus X cubed plus X to the 4 plus X to the 5 so now what we want to do is just to check our Taylor polynomial against the graph of the original function 1 divided by 1 minus X so here it is so this orange line is our Taylor polynomial and we was asked to calculate it at zero so this is the point we're interested in here but this dotted line is the graph of 1 divided by 1 minus X and also you see goes here because they would go off with an asymptote at 1 because you can't have 1 as the value of x he would be undefined so 1 divided by 1 minus 1 which we've won at the 0 so that's why the line never touches this plus 1 sign and where we discussed to calculate at 0 it's very accurate so here it touches it and goes all the way up here as it approaches 1 it starts to peel away a little bit but as it goes down to negative values they tend to go off in different directions but 1 over 1 minus X tends to converge to 0 and our Taylor polynomial tend to diverge down to negative infinity so but at the point we was interested in it is very accurate so we conclude that I'll Taylor polynomial is very good it's accurate and that's your answer all right thanks for watching and any comments leave below but very much |
16861 | https://www.merriam-webster.com/thesaurus/primitive | Est. 1828
Synonyms of primitive
adjective
as in rudimentary
as in ancient
as in naive
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primitive 1 of 2
Definition of primitive
as in rudimentary
belonging to or characteristic of an early level of skill or development primitive wooden tools were used before the Iron Age
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old-fashioned
antediluvian
homespun
old-time
out-of-date
outworn
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aged
dated
musty
fusty
unmodernized
passé
oldfangled
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advanced
evolved
high
complex
late
developed
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mature
modern
civilized
grown
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perfected
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current
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mod
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2
as in ancient
relating to or occurring near the beginning of a process, series, or time period a primitive period in church history when members still lived in communes
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ancient
prehistoric
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early
primeval
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antediluvian
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antique
hoary
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aged
age-old
passé
dateless
Antonyms & Near Antonyms
late
complex
higher
advanced
high
evolved
developed
full-blown
full-fledged
full-scale
3
as in naive
lacking in worldly wisdom or informed judgment her taste in art was still primitive, but she knew schlock when she saw it
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naive
simple
innocent
immature
unsophisticated
inexperienced
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ingenuous
uncritical
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green
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primitive
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noun
as in barbarian
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barbarian
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heathen
noble savage
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It can be used to achieve a lot of other cryptographic primitives. —Quanta Magazine, 1 Aug. 2024 Meanwhile, humans have regressed into being primitives that the apes hunt for sport, or feed out of pity from beside their campfire. —EW.com, 2 Nov. 2023
Noun
There was barely any surviving footage of the show, except for a primitive audio recording made by Short. —Brent Lang, Variety, 6 Sep. 2025 The body shell, fenders, and low-slung doors were hung on a ladder-type frame, primitive compared to the unibody MGB roadster that debuted in 1962. —Robert Ross, Robb Report, 29 Aug. 2025 See All Example Sentences for primitive
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The common factor in both these scenes, however, is the lack of any sort of even rudimentary security in place.
— Erik Kain, Forbes.com, 17 Sep. 2025
People have monitored those temperatures since the 1870s, when whalers threw buckets into the sea to take rudimentary measurements.
— Gregory Barber, Quanta Magazine, 15 Sep. 2025
Definition of rudimentary
Adjective
The effect was Romantic, reminiscent of an ancient church on the cover of a Penguin edition of Coleridge.
— Dana Goodyear, New Yorker, 22 Sep. 2025
Combining ancient techniques The RMIT team drew inspiration from designs like Shigeru Bans iconic Cardboard Cathedral.
— Mrigakshi Dixit, Interesting Engineering, 22 Sep. 2025
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Adjective
Abby is portrayed as a naive and innocent receptionist, but is edgy.
— Lisa Stardust, PEOPLE, 19 Sep. 2025
The clash of pristine tailoring with naïve florals, seed beading and distressed leather.
— Blue Carreon, Forbes.com, 17 Sep. 2025
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Noun
Bautista will have the plum role of the immortal barbarian known as The Kurgen, who has been killing other immortals across the centuries in order to absorb their essence.
— Borys Kit, HollywoodReporter, 7 Aug. 2025
This barbarian conceptualized this atrocity and brought it to reality.
— Sun Sentinel Editorial Board, Sun Sentinel, 21 July 2025
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Adjective
On either side of the cool, free-flowing river, hollows flanking the water conceal outtakes from another timefrom prehistoric sites dating back thousands of years to untamed wilderness and waterfalls left wholly untouched.
— Katie Strasberg Rousso, Southern Living, 19 Sep. 2025
Much of this prehistoric area is buried underwater today, but not all of it.
— Andrew Paul, Popular Science, 18 Sep. 2025
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Adjective
Mahle views its range-extenders as far more than simple independent add-on, looking at them as one component in a holistic driving architecture in which smaller batteries can replace larger ones while still alleviating, rather than intensifying, range anxiety.
— Prabhat Ranjan Mishra, Interesting Engineering, 20 Sep. 2025
For a similarly simple, comfortable option thats a little more sophisticated, try a stylish ivory sweater (yes, white after Labor Day is definitely acceptable in France), particularly in a rollneck fashion that will suit silhouettes of all types.
— Lane Nieset, Travel + Leisure, 20 Sep. 2025
Definition of simple
Noun
The 'Kill Tony' host, Hinchcliffe, famously appeared at last year's massive Netflix roast for Tom Brady and generated buzz for his savage lines aimed at the NFL legend.
— Matthew Couden, MSNBC Newsweek, 15 Sep. 2025
Texas is gearing up for war as a savage, flesh-eating fly appears poised for a US invasion and is expanding its range of victims.
— Beth Mole, ArsTechnica, 8 Aug. 2025
Definition of savage
Adjective
This is where Birminghams industrial heritage meets the primal spectacle of live fire cooking.
— Rai Mincey, Forbes.com, 18 Sep. 2025
Alior Richard Durham channeling Aliwould write just months after the fight that some part of him had always rebelled against the primal allure of his sport.
— Vann R. Newkirk II, The Atlantic, 16 Sep. 2025
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16862 | https://www.studocu.com/en-ca/messages/question/13201540/show-that-x2-y2-z2-xy-yz-zx-for-all-x-y-z-r | [Solved] Show that x 2 y 2 z 2 xy yz zx for all x y z R - Mathematical Reasoning (MATH 1P66) - Studocu
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6. Show that x 2 + y 2 + z 2 ≥ xy + yz + zx, for all x, y, z ∈ R.Mathematical Reasoning (MATH 1P66)
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Brock University
Mathematical Reasoning
Question
Show that x 2 y 2 z 2 xy yz zx for all x y z R
Brock University
Mathematical Reasoning
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Anonymous Student
3 months ago
Show that x 2 + y 2 + z 2 ≥ xy + yz + zx, for all x, y, z ∈ R.
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Answer Created with AI
3 months ago
To prove that (x^2 + y^2 + z^2 \geq xy + yz + zx) for all (x, y, z \in \mathbb{R}), we can use the algebraic identity:
[ (x - y)^2 + (y - z)^2 + (z - x)^2 \geq 0 ]
Expanding and rearranging the terms, we get:
[ x^2 + y^2 + z^2 - 2xy - 2yz - 2zx \geq 0 ]
Adding (2xy + 2yz + 2zx) to both sides, we obtain:
[ x^2 + y^2 + z^2 \geq xy + yz + zx ]
This proves that (x^2 + y^2 + z^2 \geq xy + yz + zx) for all (x, y, z \in \mathbb{R}).
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16863 | https://pmc.ncbi.nlm.nih.gov/articles/PMC12408600/ | False positive results in the Widal test in adults immunized with the Commirnaty (Pfizer-BioNtech) and BBIBP-CorV (Sinopharm) vaccines against COVID-19 - PMC
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False positive results in the Widal test in adults immunized with the Commirnaty (Pfizer-BioNtech) and BBIBP-CorV (Sinopharm) vaccines against COVID-19
Jeel Moya-Salazar
Jeel Moya-Salazar
1 Faculties of Medicine, Universidad Señor de Sipán, Chiclayo, Peru
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1,, Lizbeth Y Ciprian
Lizbeth Y Ciprian
2 Clinical Laboratory Area, Clínica Celestial de Ayacucho, Ayacucho, Peru
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2, Víctor Rojas-Zumaran
Víctor Rojas-Zumaran
3 Department of Pathology, Hospital Nacional Docente Madre Niño San Barotlomé, Lima, Peru
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3, Belén Moya-Salazar
Belén Moya-Salazar
4 School of Medicine, Universidad Norbert Wiener, Lima, Peru
5 Infectious Unit, Nesh Hubbs, Lima, Peru
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4,5, Eliane A Goicochea-Palomino
Eliane A Goicochea-Palomino
6 Faculties of Health Science, Universidad Tecnológica del Perú, Lima, Peru
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6, Hans Contreras-Pulache
Hans Contreras-Pulache
4 School of Medicine, Universidad Norbert Wiener, Lima, Peru
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4,
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1 Faculties of Medicine, Universidad Señor de Sipán, Chiclayo, Peru
2 Clinical Laboratory Area, Clínica Celestial de Ayacucho, Ayacucho, Peru
3 Department of Pathology, Hospital Nacional Docente Madre Niño San Barotlomé, Lima, Peru
4 School of Medicine, Universidad Norbert Wiener, Lima, Peru
5 Infectious Unit, Nesh Hubbs, Lima, Peru
6 Faculties of Health Science, Universidad Tecnológica del Perú, Lima, Peru
Edited by: Mohammed Noushad, Dar Al Uloom University, Saudi Arabia
Reviewed by: Jeff Bolles, Francis Marion University, United States
Abu Taiub Mohammed Mohiuddin Chowdhury, Ministry of Health and Family Welfare, Bangladesh
✉
Correspondence: Jeel Moya-Salazar, jeelmoya@gmail.com; moyasalazarjeel@uss.edu.pe; Hans Contreras-Pulache, hans.contreras@uwiener.edu.pe
Received 2025 Mar 14; Accepted 2025 Jul 29; Collection date 2025.
Copyright © 2025 Moya-Salazar, Ciprian, Rojas-Zumaran, Moya-Salazar, Goicochea-Palomino and Contreras-Pulache.
This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
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PMCID: PMC12408600 PMID: 40917845
Abstract
Introduction
Vaccination against COVID-19 has generated a dramatic reduction in deaths and infections worldwide. However, there may be cross-reactivity with numerous biochemical and immunological markers. The Widal test for the detection of typhoid fever is an antigen–antibody test that can be affected by vaccination, causing errors in the results, so we determined the frequency of false positive results of the Widal test in adults vaccinated with Commirnaty (Pfizer -BioNtech) and BBIBP-CorV (Sinopharm) vaccines.
Methods
We conducted a cross-sectional study analyzing the titers of the serum agglutinins (S. typhi O and H antigen, and S. paratyphi A and B antigen) in 50 adults.
Results
The proportion of false positives for O, H, A, and B antigens was 60, 44, 8, and 40%, respectively (total false positives = 38%). The Wildal tests’ results of patients with Commirnaty vaccine had higher false-positive dilution titers for O antigen [titer 1/400 (7.1%) and 1/800 (9.5%)] and H antigen [titer 1/400 (4.8%) and 1/800 (17.6%)]. We found differences in the paratyphic B by age and gender (p< 0.05).
Conclusion
Our results suggest misleading results of the Widal test in patients vaccinated against COVID-19, mainly in those vaccinated with Commirnaty. It is important to monitor and evaluate the results of routine immunological tests to ensure the quality of medical care.
Keywords: COVID-19, false positive, vaccine, Salmonella, Widal test, SARS-CoV-2, in vitro, Peru
1. Introduction
Almost a year after the World Health Organization (WHO) declared COVID-19 infection to be a world health emergency, the production of vaccines against SARS-CoV-2 escalated, which allowed the unprecedented start of secondary prevention programs by the end of 2020 (1). Since then, all the countries have developed their immunization plans and have been promoting the COVID-19 vaccination, which have various available commercial types and booster doses (2).
The Pfizer-BioNTech COVID-19 Vaccine (CONMIRNATY) has been one of the most accepted, safe, and available in most of the countries in the midst of 2021, which made it be one of the most used (3). However, in Peru, Sinopharm vaccines (BBIBP-CorV) have been initially distributed among the vulnerable populations such as the armed forces, health personnel, and older people (4). In the midst of corruption scandals and political problems (5, 6), Peruvian vaccination programs have used BBIBP-CorV vaccines, achieving the reduction of severe cases and deaths due to COVID-19, even when there are differences in the composition of each vaccine (7).
With the use vaccines, cross-reactivity has been reported in COVID-19 patients with serological markers, among them, the serological test for the detection of typhoid fever (8, 9). It is possible that, due to the production of antibodies because of the use of vaccines against COVID-19, interferences with other diagnostic tests based on antigen–antibody reaction might have been generated, which could have caused false-positive results. Cross-reactivity in vaccinated patients with more than one dose, which progressively produces a higher level of antibodies against SARS-CoV-2, could also interfere with tests based on antigen–antibody reactions such as the Widal test (typhoid fever) for salmonellosis (10). In Peru, research that quantifies the frequency of false positive results in patients vaccinated against COVID-19 has not been carried out. Hence, the typhoid fever test could produce uncertainty with regard to their results in the context of a mass vaccination of the population. In addition, the Peruvian population has a high risk for typhoid fever (11) and, therefore, understanding the frequency of false positive results, what types of altered serological markers there are, and in what proportion these cross-reactions appear is unavoidable.
We aimed to determine the frequency of false positive results of the Widal test for adults vaccinated with Commirnaty and BBIBP-CorV vaccines against COVID-19. During the pandemic, Peru was among the first countries to deploy BBIBP-CorV vaccines to its population. Despite the clinical trials being intentionally modified, the Peruvian government still approved and used the vaccine, overlooking certain details about its efficacy, side effects, and physiological responses. The hypotheses of this study were the following: (i) there exists a high frequency of false positive results in patients vaccinated with BBIBP-CorV and Commirnaty; (ii) there exist differences between the types of COVID-19 vaccines used; and (iii) patients with post-immunization adverse events have presented with higher titers of typhoid fever. Taking into account the need to achieve a quality assurance of laboratory tests, this study is certainly important in order to achieve the continuous improvement of laboratory processes during the COVID-19 pandemic.
2. Materials and methods
2.1. Study design, population and inclusion criteria
This cross-sectional study was developed in the Clinical Laboratory Area of the Celestial Clinic, in the Municipality of Ayacucho (2,746 meters above the sea level) in Peru. All the patients attended the occupational health consultation for their monthly work evaluation; hence, none of them presented with fever or symptoms related with salmonellosis in the initial medical check-up. Vital functions (i.e., blood pressure, temperature) were assessed at the clinic triage and and then underwent a comprehensive clinical evaluation in general medicine offices. The inclusion criteria were the following: healthy patients over 18 years of age of both genders, admitted for their routine control of COVID-19, vaccinated against COVID-19 (Commirnaty and BBIBP-CorV) during 2021, and with Widal test’s result for typhoid fever. We excluded foreign patients, those with a recent diagnosis of COVID-19, pregnant females, patients with hypertensive or tuberculosis treatment, with dyslipidemias or diabetes. As this study evaluated all patients treated in 2021, the sample was non-probabilistic and convenience sampling to cover the total number of patients treated.
2.2. Vaccination and Widal slide agglutination test
To identify the immunized patients, they were asked to show their vaccination cards in physical format and the doses were verified in the vaccination system of the Ministry of Health.1 The immunized patients were included in the study by filling out the written informed consent. The Widal test was administered following the immunoserology recommendations guide of the Ministry of Health (12), and the data were included in a data collection card designed by the authors.
2.3. Variables, data gathering, and analysis
Demographic variables were considered (age, gender, residence, occupation), variables of immunization (vaccine type, dose, vaccination date, adverse events, and use of post-vaccination analgesics), and results of the Widal test (for example, titer of serum agglutinins -antibodies against Salmonella antigens: O-somatic and flagellar H-). An early-morning sample of blood (3–4 mL) was taken from the patients following the guidelines of CLS H18-A4 between September and November, 2021 (13).
For agglutination on slides of the Widal test, we used QCA reagents (Barcelona, Spain). We used two drops of centrifuged blood serum with one drop of O and H antigens and parathypic A and B. Then it was shaken for 2 mins in an orbital shaker, and the formation of the antigen–antibody reaction was determined by visible agglutination on the card (positive result). We considered for each positive sample, serial dilutions of the serum of 1:20, 1:40, 1:80, 1:160, and on. The positive titer value is determined by the formation of visible clumps at 1:80 or higher dilutions from the mixture of each antigen with the patient’s serum. To ensure the quality of the results, positive and negative controls were performed using commercially available reagents (14). All results were confirmed using another antibody test from a different manufacturer (IgG/IgM Montest Rapid Test, Mont Group, Lima, Peru). The method used to identify false positive results was based on the difference between the antibody estimate and the Widal agglutination test.
We used the IBM SPSS v24.0 (Armonk, United States) for data analysis. Initially, we used descriptive statistics for the estimation of averages and standard deviation for the continuous variables and measures of frequency and central tendency for categorical variables. The titers of the serum agglutinins (S. typhi O and H antigen, and S. para-typhi A and B antigen) were determined following the recommendations of the manufacturer. The Kolgomorov-Smirnov test was used to determine data normality; similarly, the unpaired T test was used to see differences among the false positive results between the vaccines; and, finally, Pearson correlation coefficient test was used to determine the correlation between the characteristics of the administered vaccines and cross-reactivity with serum agglutinins of the Widal test. We used binary logistic regression to determine the predictor variables for these false-positive results and used diagnostic tests to find the specificity, negative predictive value and proportion of false positives for each Widal agglutinins. A threshold of significance of p< 0.05 and a confidence interval at 95% were considered for all the tests.
2.4. Ethical aspects
Written informed consent has been distributed and administered to study participants. The study has followed the guidelines of the declaration of Helsinki (15) and maintains the safeguarding of the results in accordance with Law 25,717 (Peru’s Data Protection Law) (16). In addition, this study has been approved by the Board of Directors of the Clinic (Oficio N° 101–12–2021-21) and by the Ethics Committee of the Universidad Norbert Wiener (VRI-N-089-2022).
3. Results
3.1. Demographic data
The average age of the patients was 41.88 ± 9.29 (95% CI: 39.30–44.46) and 46 (92%) were male. A total of 42 (84%) of the patients were vaccinated with the their Commirnaty vaccine as their last dose; 49 (98%) patients received the two doses and the average of weeks between vaccination and the Widal test was 28.54 ± 26.75 days (95% CI: 21.13–35.95). Although 38 (76%) patients did not have post-vaccination adverse events, 45 (90%) had consumed anti-inflammatory drugs (Table 1).
Table 1.
Baseline characteristics of population enrolled in the study (N = 50).
| Variable | Categories | N | % | p-value |
:---: :---:
| Age group (years) | <30 | 6 | 12 | 0.075 |
| 31–40 | 14 | 28 |
| 41–50 | 22 | 44 |
| >50 | 8 | 16 |
| Gender | Male | 46 | 92 | 0.001 |
| Female | 4 | 8 |
| Vaccine | Commirnaty | 42 | 84 | 0.001 |
| BBIBP-CorV | 8 | 16 |
| Doses | 1 | 1 | 2 | 0.001 |
| 2 | 49 | 98 |
| Adverse event | None | 38 | 76 | 0.078 |
| Fever | 3 | 6 |
| Pain | 9 | 18 |
| Treartment of symptoms | None | 5 | 10 | 0.255 |
| NSAIDs | 45 | 90 |
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Includes fever and body ache/headache. Includes arm, body, and headache. NSAIDs, nonsteroidal anti-inflammatory drugs.
3.2. False positive results
In total we found 20 (40%) and 28 (56%) patients with negative results for typhoid O and H, respectively, while 46 (92%) patients had negative results for paratyphoid A and 30 (60%) for paratyphoid B. We found no significant differences between Widal’s test results of typhi and paratyphic (p> 0.05; Figure 1). The proportion of false positives for O, H, A and B antigens was 60, 44, 8, and 40%, respectively (total false positives = 38%). Likewise, O antigen specificity was 40%; H antigen was 56%; An anti-gen was 92%; and B paratyphic, 60%.
Figure 1.
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Antibody titer of Salmonella serum agglutinins (Test de Widal) in patients vaccinated against COVID-19. (A) Global frequency of dilutions of Salmonella antigens. (B) Frequency of dilutions of Salmonella antigens according gender. p> 0.05.
The Wildal tests’ results of patients with the Commirnaty vaccine had higher false positive dilution titers for O antigen [titer 1/400 = 3 (7.1%) and 1/800 4 (9.5%)] compared with BBIBP-CorV vaccine results. For the H antigen, a high frequency of titers was evidenced in 1/400 (two patients, 4.8%) and in 1/800 (six patients, 17.65%) patients who had been vaccinated with the Commirnaty vaccine (Table 2). The BBIBP-CorV vaccine had a low frequency of false-positive results with high titers for paratyphi A (1/800) and B (1/400), both positive in one patient (12.5%). The samples of patients with Commirnaty vaccines showed higher titers against paratyphi B [titer 1/400 = 10 (23.8%) and 1/800 = 3 (7.1%)].
Table 2.
False positive results and dilution titer of Widal antigen agglutinins according to the type of vaccine administered.
| Result/Titer | O-antigen | H-antigen | A-antigen | B-antigen |
:---: :---:
| BBIBP-CorV | Commirnaty | BBIBP-CorV | Commirnaty | BBIBP-CorV | Commirnaty | BBIBP-CorV | Commirnaty |
| Negativo | 4 (50) | 16 (38.1) | 6 (75) | 22 (52.4) | 7 (87.5) | 39 (92.9) | 7 (87.5) | 23 (54.8) |
| 1/160 | 1 (12.5) | 8 (19) | 1 (12.5) | 7 (16.7) | 0 (0) | 1 (2.4) | 0 (0) | 4 (9.5) |
| 1/320 | 3 (37.5) | 11 (26.2) | 1 (12.5) | 5 (11.9) | 0 (0) | 2 (4.8) | 0 (0) | 2 (4.8) |
| 1/400 | 0 (0) | 3 (7.1) | 0 (0) | 2 (4.8) | 0 (0) | 0 (0) | 1 (12.5) | 10 (23.8) |
| 1/800 | 0 (0) | 4 (9.5) | 0 (0) | 6 (17.6) | 1 (12.5) | 0 (0) | 0 (0) | 3 (7.1 |
| p-value | 0.418 | 0.256 | 0.352 | 0.804 |
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3.3. Prediction analysis
The analysis of the characteristics of the patients vaccinated against COVID-19 only demonstrated the differences of the paratyphi B with age (p = 0.005) and gender (p = 0.037). The bivariate analysis showed that the variables included (such as age, vac-cine type, vaccination date) were not predictors of the false-positive results of the Widal test (Table 3).
Table 3.
Binary regression of the predictive variables of false positive results of the serum agglutinins of the Widal test.
| Variables | O-antigen | H-antigen | A-antigen | B-antigen |
:---: :---:
| B | SE | p- value | 95% CI | B | SE | p-value | 95% CI | B | SE | p-value | 95% CI | B | SE | p-value | 95% CI |
| Age group (years) | 1.208 | 1.029 | 0.247 | −0.865 to 3.281 | −0.707 | 0.933 | 0.452 | −2.585 to 1.171 | 1.916 | 1.364 | 0.167 | −0.831 to −4.663 | −1.747 | 0.909 | 0.061 | −3.578 to 0.085 |
| Gender | 0.001 | 0.031 | 0.974 | −0.062 to 0.064 | 0.012 | 0.028 | 0.681 | −0.045 to 0.069 | −0.011 | 0.041 | 0.791 | −0.094 to 0.072 | −0.055 | 0.028 | 0.051 | −0.111 to 0.001 |
| Vaccine | −0.016 | 0.044 | 0.711 | −0.105 to 0.072 | −0.071 | 0.040 | 0.083 | −0.151 to 0.010 | 0.023 | 0.058 | 0.693 | −0.094 to 0.140 | −0.052 | 0.039 | 0.187 | −0.130 to 0.026 |
| Dosis of vaccine | 0.016 | 0.016 | 0.349 | −0.018 to 0.049 | 0.012 | 0.015 | 0.443 | −0.019 to 0.042 | 0.012 | 0.022 | 0.599 | −0.032 to 0.056 | 0.012 | 0.015 | 0.398 | −0.017 to 0.042 |
| Adverse events | 0.040 | 0.050 | 0.425 | −0.060 to 0.140 | −0.052 | 0.045 | 0.257 | −0.142 to 0.038 | −0.047 | 0.066 | 0.476 | −0.180 to 0.085 | −0.051 | 0.044 | 0.254 | −0.139 to 0.038 |
| Treartment of symptoms | −0.004 | 0.036 | 0.904 | −0.077 to 0.068 | 0.019 | 0.033 | 0.554 | −0.046 to 0.085 | −0.010 | 0.048 | 0.832 | −0.106 to 0.086 | −0.010 | 0.032 | 0.832 | −0.740 to 0.054 |
| Days post vaccination | 4.120 | 3.033 | 0.181 | −1.989 to 10.228 | −2.382 | 2.748 | 0.391 | −7.917 to 3.154 | 5.555 | 4.019 | 0.174 | −2.540 to 13.651 | 2.595 | 2.679 | 0.338 | −2.801 to 7.992 |
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p-value <0.05 (significant).
No patient with a false-positive result for the serum agglutinins was reported, but we did found positivity to the O, H and A antigens and four (8%) patients with positive results for A, H, and B antigens (Figure 2).
Figure 2.
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Positive Widal test results in samples from patients vaccinated against COVID-19. In the right the result for Salmonella antigens O-somatic (titer 1/400) is shown, and in the left the positive result for para-typhus A (titer 1/320) is shown.
4. Discussion
In this study, we determined a high frequency of false-positive results of the Widal test in patients vaccinated against COVID-19. O and H typhus presented the most false positive results and with higher dilution titers in comparison with paratyphi (A and B). In addition, although we detected that the highest frequency of false positive results in patients immunized with Commirnaty (Pfizer-BioNtech) vaccines, there were not significant differences between the results of those patients vaccinated with BBIBP-CorV vaccines (Sinopharm). Moreover, the demographic, clinical, and immunization variables were not predictors in regard to false positives.
This study had different strengths. First, to the best of our knowledge, this is the first study that evaluated false-positive seropositivity of the Widal test in patients vaccinated against COVID-19. Although there are numerous studies that have demonstrated that other infectious and chronic diseases can lead to positive results of antigen–antibody tests of serological detection tests of SARS-CoV-2 (8–11, 17) so far, none of them has characterized the effects of immunization in other routine tests such as the Widal test. Second, this study has studied the effects of the Widal test results, by comparing two vaccines extensively used in the vaccination plans against COVID-19. Therefore, the results of the frequency of cross-reactivity of antibody tests with each vaccine can indicate us the subsequent effects and how the laboratories should monitor their processes to ensure reliable results.
The decrease of the rate of infection and deaths due to the vaccines has caused a reduction of the pandemic effects in the world, which allowed the improvement of the restrictions established during 2020–2021 (7). However, as it has been noticed in patients misdiagnosed with a recent SARS-CoV-2 infection (18), there might also exist immune response antibodies which can cause cross-reactivity with other tests based on serology, such as the Widal test or the dengue test (8, 11, 19). This can lead to false-positive results which might affect the quality assurance of laboratory results.
In the beginning of 2020, false-positive dengue results were reported in patients with COVID-19 serology. Yan et al. (9), through their study in Singapore, described a number of cases with serological rapid tests with positive results for dengue’s IgM and IgG antibodies that turned out to be misleading results after administering SARS-CoV-2 tests. In Indonesia, Luhulima et al. (17) showed that a patient aged 43 who suffered from dengue haemorrhagic fever had a positive COVID-19 result after being administered a serological ELISA IgG and IgM test. Subsequent determination indicated the presence of IgG antibody due to dengue which caused the cross-reaction antibody (17). Also, the study by Boukli et al. (8) reported a high incidence (10%) of false-positive results, which led to discontinuation of the use of the commercial chemiluminescent microparticle immunoassay Liaison SARS-CoV-2 S1/S2 IgG at the Saint-Antoine Hospital (Paris, France). These cross-reactivity antibody results were consistently observed in patients suffering from acute infectious diseases, especially from Epstein–Barr virus or hepatitis B virus infection (7). Consistent with these results on rapid tests and automated equipment, more than 35% of our patients obtained false positives after the administration of the Widal test with the Química Clínica Analítica (QCA) reagent for any of the serum agglutinins tested.
It is clear that there continues to be an overlap in the diagnosis of diseases with the same clinical threshold as COVID-19 at the symptoms level. For Salmonellosis, some cases of co-infection have been reported in low- and middle-income countries (10, 20, 21), but false-positive results have also been reported. Babu and Srees showed that six patients in India had false-positive Widal’s test results for typhoid fever when they developed COVID-19 (8). Our results are partially consistent with the study previously mentioned as they identified false-positive seropositivity in the Widal test post-immunization against SARS-CoV-2. However, our false-positive results are independent of the clinical, demographic, and vaccination characteristics of the patients, and are possibly due to the concentration of antibodies caused by the vaccine and to non-specific antibody binding and activation (22, 23).
The frequency of false positives in patients immunized with Commirnaty (Pfizer-BioNtech) or BBIBP-CorV (Sinopharm) vaccines did not show significant differences, although the dilution titers were higher with the former. The results of the Widal test of patients with the Commirnaty vaccine showed higher false positive dilution titers for 3/4 serum agglutinins (O, H and B), which would indicate a greater humoral response with the Commirnaty vaccine (Pfizer-BioNtech) (24), a greater nonspecific binding of these antibodies during the Widal test, or both. Therefore, it is important to define the performance characteristics and false-positive results of typhoid diagnostic tests during the COVID-19 pandemic in order to achieve quality assurance of clinical testing laboratory.
This study had limitations that must be recognized. First, the study was unicentric, with a population from the Peruvian Andes. Therefore, there might be differences with the results in regard to other populations. Second, the Widal test was used in this study; however, the performance and the proportion of false-positive results may vary between commercial kits. Third, we compared cross-reactivity results for the Widal test post immunization with Commirnaty (Pfizer-BioNtech) or BBIBP-CorV (Sinopharm) vaccines; however, there are other available vaccines that could change the false-positive rate and performance. In addition, booster doses of vaccines continue to increase during the pandemic, which could impact the production of antibodies that cause the interference of results of the Widal and other tests.
5. Conclusion
In conclusion, this study demonstrated, for the first time, the misleading results of the Widal test in patients vaccinated against COVID-19. The proportion of false-positive results was higher in those vaccinated with Commirnaty (Pfizer-BioNtech) but showed no notable differences and it seems that clinical, demographic, and vaccination characteristics are not predictors of these results. Thus, serological tests for the detection of typhoid fever may be affected, causing false-positive results and compromising the diagnostic processes of health centers, which might generate delays in the detection of this pathogen, the use of other diagnostic tests, and diagnostic uncertainty in patients and physicians.
Acknowledgments
We are grateful with Gonzalo Moscoso for their scientific support and for their help reviewing this work. In addition, we thank the Nesh Hubbs technical team for the advice on data analysis and the Clínica Celestial de Ayacucho for access to patients.
Funding Statement
The author(s) declare that no financial support was received for the research and/or publication of this article.
Footnotes
1
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 authors.
Ethics statement
The studies involving humans were approved by Board of Directors of the Clinic (Oficio N° 101–12–2021-21) and by the Ethics Committe of the Universidad Norbert Wiener (VRI-N-089-2022). The studies were conducted in accordance with the local legislation and institutional requirements. The human samples used in this study were acquired from a by- product of routine care or industry. Written informed consent for participation was not required from the participants or the participants’ legal guardians/next of kin in accordance with the national legislation and institutional requirements.
Author contributions
JM-S: Conceptualization, Data curation, Formal analysis, Investigation, Methodology, Resources, Software, Validation, Visualization, Writing – original draft, Writing – review & editing. LC: Conceptualization, Data curation, Formal analysis, Resources, Writing – original draft. VR-Z: Conceptualization, Project administration, Software, Validation, Writing – review & editing. BM-S: Data curation, Methodology, Visualization, Writing – original draft. EG-P: Formal analysis, Investigation, Software, Visualization, Writing – review & editing. HC-P: Investigation, Methodology, Resources, Supervision, Validation, Writing – review & editing.
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.
Generative AI statement
The author(s) declare that no Gen AI was used in the creation of this manuscript.
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.
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16864 | https://learn.microsoft.com/en-us/windows/win32/procthread/scheduling-priorities | Read in English
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Scheduling Priorities
07/14/2025
Threads are scheduled to run based on their scheduling priority. Each thread is assigned a scheduling priority. The priority levels range from zero (lowest priority) to 31 (highest priority). Only the zero-page thread can have a priority of zero. (The zero-page thread is a system thread responsible for zeroing any free pages when there are no other threads that need to run.)
The system treats all threads with the same priority as equal. The system assigns time slices in a round-robin fashion to all threads with the highest priority. If none of these threads are ready to run, the system assigns time slices in a round-robin fashion to all threads with the next highest priority. If a higher-priority thread becomes available to run, the system ceases to execute the lower-priority thread (without allowing it to finish using its time slice) and assigns a full time slice to the higher-priority thread. For more information, see Context Switches.
The priority of each thread is determined by the following criteria:
The priority class of its process
The priority level of the thread within the priority class of its process
The priority class and priority level are combined to form the base priority of a thread. For information on the dynamic priority of a thread, see Priority Boosts.
Priority Class
Each process belongs to one of the following priority classes:
IDLE_PRIORITY_CLASS
BELOW_NORMAL_PRIORITY_CLASS
NORMAL_PRIORITY_CLASS
ABOVE_NORMAL_PRIORITY_CLASS
HIGH_PRIORITY_CLASS
REALTIME_PRIORITY_CLASS
By default, the priority class of a process is NORMAL_PRIORITY_CLASS. Use the CreateProcess function to specify the priority class of a child process when you create it. If the calling process is IDLE_PRIORITY_CLASS or BELOW_NORMAL_PRIORITY_CLASS, the new process will inherit this class. Use the GetPriorityClass function to determine the current priority class of a process and the SetPriorityClass function to change the priority class of a process.
Processes that monitor the system, such as screen savers or applications that periodically update a display, should use IDLE_PRIORITY_CLASS. This prevents the threads of this process, which do not have high priority, from interfering with higher priority threads.
Use HIGH_PRIORITY_CLASS with care. If a thread runs at the highest priority level for extended periods, other threads in the system will not get processor time. If several threads are set at high priority at the same time, the threads lose their effectiveness. The high-priority class should be reserved for threads that must respond to time-critical events. If your application performs one task that requires the high-priority class while the rest of its tasks are normal priority, use SetPriorityClass to raise the priority class of the application temporarily; then reduce it after the time-critical task has been completed. Another strategy is to create a high-priority process that has all of its threads blocked most of the time, awakening threads only when critical tasks are needed. The important point is that a high-priority thread should execute for a brief time, and only when it has time-critical work to perform.
You should almost never use REALTIME_PRIORITY_CLASS, because this interrupts system threads that manage mouse input, keyboard input, and background disk flushing. This class can be appropriate for applications that "talk" directly to hardware or that perform brief tasks that should have limited interruptions.
Priority Level
The following are priority levels within each priority class:
THREAD_PRIORITY_IDLE
THREAD_PRIORITY_LOWEST
THREAD_PRIORITY_BELOW_NORMAL
THREAD_PRIORITY_NORMAL
THREAD_PRIORITY_ABOVE_NORMAL
THREAD_PRIORITY_HIGHEST
THREAD_PRIORITY_TIME_CRITICAL
All threads are created using THREAD_PRIORITY_NORMAL. This means that the thread priority is the same as the process priority class. After you create a thread, use the SetThreadPriority function to adjust its priority relative to other threads in the process.
A typical strategy is to use THREAD_PRIORITY_ABOVE_NORMAL or THREAD_PRIORITY_HIGHEST for the process's input thread, to ensure that the application is responsive to the user. Background threads, particularly those that are processor intensive, can be set to THREAD_PRIORITY_BELOW_NORMAL or THREAD_PRIORITY_LOWEST, to ensure that they can be preempted when necessary. However, if you have a thread waiting for another thread with a lower priority to complete some task, be sure to block the execution of the waiting high-priority thread. To do this, use a wait function, critical section, or the Sleep function, SleepEx, or SwitchToThread function. This is preferable to having the thread execute a loop. Otherwise, the process may become deadlocked, because the thread with lower priority is never scheduled.
To determine the current priority level of a thread, use the GetThreadPriority function.
Base Priority
The process priority class and thread priority level are combined to form the base priority of each thread.
The following table shows the base priority for combinations of process priority class and thread priority value.
| Process priority class | | Thread priority level | Base priority |
--- --- |
| IDLE_PRIORITY_CLASS | | THREAD_PRIORITY_IDLE | 1 |
| THREAD_PRIORITY_LOWEST | 2 |
| THREAD_PRIORITY_BELOW_NORMAL | 3 |
| THREAD_PRIORITY_NORMAL | 4 |
| THREAD_PRIORITY_ABOVE_NORMAL | 5 |
| THREAD_PRIORITY_HIGHEST | 6 |
| THREAD_PRIORITY_TIME_CRITICAL | 15 |
| BELOW_NORMAL_PRIORITY_CLASS | | THREAD_PRIORITY_IDLE | 1 |
| THREAD_PRIORITY_LOWEST | 4 |
| THREAD_PRIORITY_BELOW_NORMAL | 5 |
| THREAD_PRIORITY_NORMAL | 6 |
| THREAD_PRIORITY_ABOVE_NORMAL | 7 |
| THREAD_PRIORITY_HIGHEST | 8 |
| THREAD_PRIORITY_TIME_CRITICAL | 15 |
| NORMAL_PRIORITY_CLASS | | THREAD_PRIORITY_IDLE | 1 |
| THREAD_PRIORITY_LOWEST | 6 |
| THREAD_PRIORITY_BELOW_NORMAL | 7 |
| THREAD_PRIORITY_NORMAL | 8 |
| THREAD_PRIORITY_ABOVE_NORMAL | 9 |
| THREAD_PRIORITY_HIGHEST | 10 |
| THREAD_PRIORITY_TIME_CRITICAL | 15 |
| ABOVE_NORMAL_PRIORITY_CLASS | | THREAD_PRIORITY_IDLE | 1 |
| THREAD_PRIORITY_LOWEST | 8 |
| THREAD_PRIORITY_BELOW_NORMAL | 9 |
| THREAD_PRIORITY_NORMAL | 10 |
| THREAD_PRIORITY_ABOVE_NORMAL | 11 |
| THREAD_PRIORITY_HIGHEST | 12 |
| THREAD_PRIORITY_TIME_CRITICAL | 15 |
| HIGH_PRIORITY_CLASS | | THREAD_PRIORITY_IDLE | 1 |
| THREAD_PRIORITY_LOWEST | 11 |
| THREAD_PRIORITY_BELOW_NORMAL | 12 |
| THREAD_PRIORITY_NORMAL | 13 |
| THREAD_PRIORITY_ABOVE_NORMAL | 14 |
| THREAD_PRIORITY_HIGHEST | 15 |
| THREAD_PRIORITY_TIME_CRITICAL | 15 |
| REALTIME_PRIORITY_CLASS | | THREAD_PRIORITY_IDLE | 16 |
| THREAD_PRIORITY_LOWEST | 22 |
| THREAD_PRIORITY_BELOW_NORMAL | 23 |
| THREAD_PRIORITY_NORMAL | 24 |
| THREAD_PRIORITY_ABOVE_NORMAL | 25 |
| THREAD_PRIORITY_HIGHEST | 26 |
| THREAD_PRIORITY_TIME_CRITICAL | 31 |
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Additional resources
Documentation
SetPriorityClass function (processthreadsapi.h) - Win32 apps
Sets the priority class for the specified process. This value together with the priority value of each thread of the process determines each thread's base priority level.
Context Switches - Win32 apps
The scheduler maintains separate queues of executable threads for each priority level.
GetThreadPriority function (processthreadsapi.h) - Win32 apps
Retrieves the priority value for the specified thread. This value, together with the priority class of the thread's process, determines the thread's base-priority level.
Thread Ordering Service - Win32 apps
The thread ordering service controls the execution of one or more client threads. It ensures that each client thread runs once during the specified period and in relative order.
SetThreadPriority function (processthreadsapi.h) - Win32 apps
Sets the priority value for the specified thread. This value, together with the priority class of the thread's process, determines the thread's base priority level.
Priority Boosts - Win32 apps
Each thread has a dynamic priority.
Priority Inversion - Win32 apps
Priority inversion occurs when two or more threads with different priorities are in contention to be scheduled.
Scheduling - Win32 apps
The system scheduler controls multitasking by determining which of the competing threads receives the next processor time slice. The scheduler determines which thread runs next using scheduling priorities.
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16865 | https://www.youtube.com/watch?v=iO7vb3HLHqU | Rectangular Prism - Volume, Surface Area and Diagonal Length, Rectangles, Geometry
The Organic Chemistry Tutor
9670000 subscribers
3461 likes
Description
372247 views
Posted: 29 Jul 2017
This geometry video tutorial explains how to find the volume of a rectangular prism as well as how to find the surface area of a rectangular prism and its diagonal length. This video contains plenty of examples and practice problems as well as how to conceptually generate the formulas needed to find the volume, surface area and diagonal length of a rectangle.
3D Shapes - Faces, Edges, & Vertices:
Rectangular Prism - Volume:
Rectangular Prism - Surface Area:
Rectangular Prism - Diagonal Length:
Triangular Prism - Volume:
Triangular Prism - Surface Area:
Triangular Prism - Lateral Area:
Hexagonal Prism - Surface Area:
Cube - Volume:
Cube - Surface Area:
Cube - Diagonal Length:
Pyramid - Volume:
Pyramid - Surface Area:
Square Pyramids - Volume:
Cylinder - Surface Area:
Sphere - Volume:
Sphere - Surface Area:
Cone - Volume:
Cone - Surface Area:
Final Exams and Video Playlists:
Transcript:
in this video we're going to talk about how to calculate the volume the surface area and the diagonal length of a rectangle so let's try this problem what is the volume of a rectangle that has a length width and height of 12 inches 6 inches and 5 inches respectively so let's draw a picture so here is a rectangular prism and it has a length of 12 inches a width of 6 inches and a height of 5. so how can we find the volume of this prism the volume of any prism is basically the area of the base multiplied by the height the height of the prism is 5. the length is 12 and the width is 6. so we already have the height in this equation we don't have to worry about it but what is the area of the base in this particular picture this is the base of the rectangular solid based on the way strong and the area of the base the area of that blue portion is basically the left times the width so we can replace b with l times w so the volume of a rectangular prism is simply the length times the width times the height the length is 12 inches the width is six inches and the height is five inches twelve times five is sixty and sixty times six well six times six is thirty six sixty times 6 is 360. you just have to add the extra zero now what about the units so what are the units of volume in this problem inches times inches times inches that's equal to inches cube or cubic inches so that's the volume in this example it's 360 cubic inches number two what is the surface area of a rectangular prism with the dimensions 15 centimeters by eight centimeters by nine centimeters so let's say that this is the length this is the width and this is the height well let's begin by drawing a picture just like we did before doesn't have to be perfect just a simple rough sketch my prism looks bigger in the back but you can work with it so let's say this is the length width and the height how can we come up with an equation that was that's going to help us to to find a surface area of this rectangular prism the surface area is basically the area of all six faces of this prism or this rectangle so the area of the front is the same as the area of the back face or the just the back side of the rectangle so that's highlighted in blue notice that it's the width times the height same thing on this side this is the width and this portion is the height so the area of the front and the back is just w times h and because there's two of them it's going to be 2 wh now the six sides we need to cover so far we've covered the front and back so just two out of six now the next thing is you have the area of the bottom surface and the area of the top surface so the bottom surface is l units long and it's w units y same thing for the top so the area is l times w but because we have the top and the bottom we got to multiply by two so it's going to be two l times w now the last thing we need to worry about is the right side and the left side the right side has a length of l but a height of h so the area is lh but we've got to multiply by 2. so this is the formula that you need to calculate the surface area of a rectangular prism now let's go ahead and plug in everything that we have so the width is eight centimeters and the height is nine centimeters now the length is 15 centimeters and the width is eight centimeters and then two times lh that's gonna be two times fifteen times nine so now we just gotta do some math eight times nine that's 72 and two times 72 is 144. now 2 times 15 is 30 and what's 30 times 8 well 3 times 8 is 24 if you add the zero that's going to be 240. and here we have 2 times 15 again that's 30 30 times 9 is 270. so now we just got to add so what's 240 plus 270 well 200 plus 200 is 400 40 plus 70 is 110 so 400 plus 110 that's going to be the 510 now 500 plus 144 that's 644. if you add 10 to it that's going to be 654. so that's the surface area and whenever you're dealing with area the units is going to be unit squared so it's going to be square centimeters or centimeters squared so that is the surface area of this figure 654 square centimeters so now you know how to find it number three what is the length of the diagonal of a rectangular prism with the dimensions 14 centimeters by nine by seven so once again this is the left the width and the height so let's start with a picture so we need to find the distance between one edge of the rectangular prism to the other edge so basically we're looking for the length of the yellow line so how can we go ahead and find that how can we find the length of that line well before we come up before we write the equation let's talk about how to get it so you could understand why the equation is the way it is so notice that you could form a right triangle so i'm going to draw another diagonal that remains on the bottom surface and notice the triangle that forms this is a right angle so i'm going to call the length of the diagonal which we're trying to find the yellow line let's call it d we want to find the distance between these two points and this side is the height of the prism that's h and the left for the red diagonal that is on the bottom surface let's call this z so we have this triangle the hypotenuse is d which is what we're trying to find h is the height and z is the red diagonal that's on the bottom face of the rectangular prism now for any right triangle we have the pythagorean theorem c squared is equal to a squared plus b squared so in this case d squared is equal to z squared plus h squared now notice that there's another diagonal that we could focus on i'm going to highlight it in green so this is z as we mentioned before but notice that we could form another right triangle where the hypotenuse is right here so this is l and this is w so for that second triangle which looks like this we have a hypotenuse of z we have the width and the length so therefore using the pythagorean theorem z squared is equal to l squared plus w squared now if we combine these two equations if we replace z squared with what it's equal to l squared plus w squared we're going to have this equation which is what we want d squared is equal to l squared plus a w squared plus h squared and now let's take the square root of both sides so here's the final equation d is equal to the square root of l squared plus w squared plus h squared so if you have the left the width and the height you can find the diagonal length using that excuse me using that equation which comes from the pythagorean theorem so now let's go ahead and find the answer so it was 14 by nine by seven so those are the dimensions of this particular rectangular prism and this is the length with and the height so let's go ahead and use this formula to calculate the length of the diagonal so l is fourteen w is nine h is seven fourteen squared is one ninety six nine squared is eighty-one and seven squared is 49. so if we add those three numbers we should get a 326 if i typed in everything correctly and the square root of 326 as a decimal is 18.055 and the unit is going to be centimeters so make sure you get an answer that is greater than each of these numbers individually the length of the diagonal has to be longer than the longest side of the rectangle it's just the way the math works so this is the answer so now you know how to find the volume the surface area and the length of the diagonal using these equations so that's all you need to know for rectangular prisms so thanks for watching this video and have a great day you |
16866 | https://arxiv.org/pdf/2008.12396 | Exact sharp-fronted travelling wave solutions of the Fisher-KPP equation
Scott W. McCue 1, Maud El-Hachem 1, and Matthew J. Simpson ∗11School of Mathematical Sciences, Queensland University of Technology, Brisbane, Queensland 4001, Australia September 3, 2020
Abstract
A family of travelling wave solutions to the Fisher-KPP equation with speeds c = ±5/√6can be expressed exactly using Weierstraß elliptic functions. The well-known solution for
c = 5 /√6, which decays to zero in the far-field, is exceptional in the sense that it can be written simply in terms of an exponential function. This solution has the property that the phase-plane trajectory is a heteroclinic orbit beginning at a saddle point and ends at the origin. For c = −5/√6, there is also a trajectory that begins at the saddle point, but this solution is normally disregarded as being unphysical as it blows up for finite z. We reinter-pret this special trajectory as an exact sharp-fronted travelling solution to a Fisher-Stefan
type moving boundary problem, where the population is receding from, instead of advanc-ing into, an empty space. By simulating the full moving boundary problem numerically, we demonstrate how time-dependent solutions evolve to this exact travelling solution for large time. The relevance of such receding travelling waves to mathematical models for cell migration and cell proliferation is also discussed.
Keywords: Fisher-Kolmogorov; Painlev´ e property; Weierstraß elliptic functions; Moving bound-ary problem
∗
To whom correspondence should be addressed. E-mail: matthew.simpson@qut.edu.au
1
arXiv:2008.12396v2 [nlin.PS] 2 Sep 2020
1 Introduction
For various applications in ecology and cell biology, the Fisher-KPP equation [1–3]
∂u ∂t = ∂2u∂x 2 + u(1 − u), (1) provides a very well studied model for the growth and spread of a population of species or cell types [4–6]. One key mathematical result is that, with the associated initial and boundary conditions
u(x, 0) = F (x), 0 < x < ∞, (2)
∂u ∂x = 0 on x = 0 , u → 0 as x → ∞ , (3) the time-dependent solution evolves towards a travelling wave profile U (z), where z = x − ct ,as t → ∞ [1–3]. A combination of phase-plane analysis and simple asymptotics demonstrates that the travelling wave speed c satisfies c ≥ 2 and is selected by the far-field behaviour of the initial condition F (x) in (2) [4–6]. In this work, we are motivated by our recent studies [7–9] where we have restricted the Fisher-KPP equation (1) to hold on the moving domain 0 < x < s (t), together with a Stefan-type moving boundary condition, to give the so-called Fisher-Stefan model
∂u ∂t = ∂2u∂x 2 + u(1 − u), 0 < x < s (t), (4)
∂u ∂x = 0 on x = 0 , (5)
u = 0 , ds
dt = −κ ∂u ∂x on x = s(t). (6) Here κ is a parameter that relates the leakage of the population at the boundary to the speed of the boundary. In this context, we and others have provided new interpretations for travelling wave solutions (1) for c < 2, including slowly moving fronts that advance with speed 0 <c < 2 [7, 10–13], stationary profiles for c = 0 and receding fronts with speed c < 0 [8, 9]. Travelling wave solutions for c < 2 are interesting because they are normally disregarded as being unphysical (because they do not satisfy the boundary conditions and/or are not restricted to 0 < U < 1 for all z ∈ R) [1–6]. In this letter we focus on travelling wave solutions to (1) for the special values c = ±5/√6. 2For c = 5 /√6 there is a well known exact solution [4, 14]
U =
(
1 + ( √5 − 1) e z/ √6)−2
, (7) as shown in Figure 1(a)-(b). Other exact travelling wave solutions to (1) for c = ±5/√6, which can be written in terms of Weierstraß elliptic functions , are normally disregarded as being unphysical in the usual way . However, in the context of the Fisher-Stefan model (4)-(6), we provide a new physical interpretation of one of these solutions. In particular, we claim that one of the profiles for c = −5/√6 corresponds to a receding travelling wave to (4)-(5) with a special value of κ = −0.906 . . . . In this way, we illustrate a second physically realistic exact travelling wave solution to (1) for c = ±5/√6. In section 2 we review the exact travelling solutions to the Fisher-KPP equation for c =
±5/√6, taken from Ablowitz & Zeppetella . This derivation involves Weierstraß elliptic functions . By using phase-plane analysis, we demonstrate the qualitative behaviour of this family of solutions and highlight the two special trajectories that evolve to ( U, V ) = (1 , 0) as
z → −∞ . In section 3 we link this special trajectory to solutions of the Fisher-Stefan model and demonstrate numerically that for κ = 0 .906 . . . , initial conditions evolve to this travelling wave solution for large time. Finally, we providing concluding remarks in section 4.
2 Travelling wave solutions for c = ±5/√6
To study travelling wave solutions of (1) we write u = U (z), where z = x − ct , to give d2U
dz2 + c dU
dz + U (1 − U ) = 0 . (8) We discuss the domain of interest and the boundary conditions below, but for the moment we highlight the physically relevant boundary condition
U → 1−, dU
dz → 0−, as z → −∞ , (9) which applies for two special cases considered below. We rewrite (8) in the usual way as dU
dz = V, (10) dV
dz = −cV − U (1 − U ). (11) 3(a) (c) (b) (d) Figure 1: Exact solutions to the Fisher-KPP model . (a)-(b) shows the exact solutions for c = 5 /√6 in the physical plane and the phase plane, respectively. Similarly, (c)-(d) shows the exact solutions for c = −5/√6 in the physical plane and the phase plane, respectively. (a), (c) show the travelling wave profiles U (z) that satisfy (9) (thick black), horizontal arrows are superimposed to emphasize the difference in direction. (b),(d) show various phase plane trajectories. Equilibrium points in the phase planes are shown with black discs, and trajectories corresponding to the special travelling wave solutions are plotted (thick black). Various other trajectories (thin coloured lines) are superimposed, and the corresponding curves are given in (a),(c) using the same colours as in (b),(d). Some trajectories are numbered to emphasise the point that the numbered trajectories in (a)-(d) are identical. 4One point to note here is that this system is reversible under the substitution z → − z, V → − V ,
c → − c, which means that the phase-plane for (10)-(11) for c < 0 is simply a reflection about the U -axis of the phase-plane for c > 0. The concern here is with special values c = ±5/√6. For these values, we may solve (8) exactly, as explained by Ablowitz & Zeppetella . A summary of the working is as follows. We start by letting U = f (z)w(z) and substituting into (8). Then, by forcing f ′′ + cf ′ + f = 0, we find f w ′′ + (2 f ′ + cf )w′ = f 2g2. We choose the linearly independent solution f = e λz , where
λ = ( −c + √c2 − 4) /2, so that w′′ + √c2 − 4w′ = e λz g2. The equations simplify by setting
w = w(s), s = h(z). The left-hand side of this differential equation reduces to a single term if
h′′ + √c2 − 4h′ = 0, which suggests we choose h = e −√c2−4z . Finally, with c2 = 25 /6 we end up with the second-order differential equation for w to be d 2w/ ds2 = 6 w which, upon multiplying both sides by d w/ ds, integrates directly to
( dw
ds
)2
= 4 w3 − ω, (12) where ω is a constant. The first-order ode (12) is separable and is solved exactly in terms of the Weierstraß p-function ℘(z; 0; ω) . Rewriting the solution in terms of U and V gives
U = e −2z/ √6℘
(
e−z/ √6 − k; 0; ω
)
, (13)
V = −√63 e−2z/ √6 (
℘
(
e−z/ √6 − k; 0; ω
)
2e −z/ √6℘′ (
e−z/ √6 − k; 0; ω
))
, (14) where ℘′ is the derivative of ℘ . It is noteworthy that this exact solution is possible because (8) has the Painlev´ e property for c = ±5/√6. In other words, for these special values of c, the movable singularities of solutions to (8) are poles . It is worth plotting the exact solution(s) (13)-(14) both in the form U = U (z) and in the phase-plane. As the travelling wave solutions are invariant to translations in z, we fix each solution in the z-direction by setting U = U0 at z = 0. For a given point in the phase plane, (U, V ) = ( U0, V 0), we determine the two constants k and ω by solving the nonlinear algebraic system U0 = ℘(1 − k; 0; ω), V0 = −(√6/3) ( ℘(1 − k; 0; ω) + 2 ℘′(1 − k; 0; ω)) numerically, for example with Newton’s method. The system can be sensitive and may require a close initial guess. Fixing c = 5 /√6 for the moment, travelling wave profiles U (z) are shown in Figure 1(a), while corresponding trajectories in the phase-plane are shown in Figure 1(b). Each of these 5curves could be drawn using the exact solution (13)-(14) or just as easily be generated using numerical solutions to (10)-(11). As is well known, the phase-plane in Figure 1(b) is charac-terised by two fixed points, namely the saddle point at (1 , 0) and the stable node at (0 , 0). All of the trajectories in Figure 1(b) enter the stable node (0 , 0); a linearisation about (0 , 0) demonstrates that U ∼ const e −2z/ √6 as z → ∞ . As z decreases, we see in both Figures 1(a) and (b) that, with one exception, each solution (thin coloured curves) blows up (with U → ±∞ )at a finite value of z. This is because ℘(ζ; 0 , ω ) has a double pole at ζ = 0. The exception is the heteroclinic orbit (thick black curve) that joins the two fixed points; this trajectory cor-responds to the well-known exact solution (7), which notably satisfies the physically realistic boundary condition (9). The simplification from (13)-(14) to (7) in this case arises because this special case corresponds to taking the limit k → 0, which is in effect pushing the singularity to
z = −∞ . Numerical solutions to (1)-(3) with appropriate initial conditions evolve to (7), as we demonstrate in section 3. Now turning to c = −5/√6, we show results in Figure 1(c)-(d). Here it is convenient to reflect our phase-plane about the U -axis, so the heteroclinic orbit just mentioned is in the upper-half plane. The trajectories in the phase-plane are still given by (13)-(14), except that we must make the changes V → − V , z → − z. Five of the solutions shown in Figure 1(c)-(d) are the same as in Figure 1(a)-(b) (to enable a straightforward comparison across the figures, we have labelled these solutions 1○– 5○); these are trajectories that start at the stable node (0 , 0), except now we see that U ∼ const e 2z/ √6 as z → −∞ . As we follow these trajectories for increasing z, we again note that they blow up at a finite value of z (with U → −∞ ). Other trajectories also blow up for finite z (this time with U → ∞ ), except for the separatrix (solid black curve) which eventually enters the saddle point (1 , 0). We now interpret the separatrix in Figure 1(d) in terms of an exact sharp-fronted travelling wave solution of (1). As explained in Ablowitz & Zeppetella , in order to extract this special case from (13)-(14), we must choose the constants k and ω such that ℘(ζ; 0 , ω ) has one of its double poles at ζ = −k. This is a numerical constraint which can be enforced by solving the system
U0 = ℘(1 − k; 0; ω), ℘′(k; 0; ω)/℘ (k; 0; ω)2 = 0 , (15) where 0 < U 0 < 1. In Figure 1(c) we have chosen U0 = 0 .2, but of course this value is arbitrary and, in effect, corresponds to a translation in z. The key observation of this special solution is that it satisfies the physically realistic boundary condition (9) as z → −∞ . As we see in Figure 1(c), as z increases, this solution for U is very flat until it decreases sharply to U = 0 6(a) (b) Figure 2: Time-dependent PDE solutions . (a) Numerical solution of (1)-(3) showing the initial condition (green), intermediate-time solutions at t = 3, 6 and 9 (blue). The solution at
t = 9 is superimposed with (7) (dashed orange). (b) Numerical solution of (4)-(6) showing an initial condition (green), intermediate-time solutions at t = 3, 6 and 9 (blue). The solution at
t = 9 is superimposed with (13)-(14) subject to (15) (dashed orange). at some finite value of z = z∗ where V = V ∗ = −2.251 . . . , and then continues to decrease as
z increases further. While this solution satisfies (9), it is normally disregarded as it is negative for all z > z c. In the following section we show how this profile is a receding travelling wave solution for the Fisher-Stefan moving boundary problem (4)-(6).
3 Time-dependent solutions to the Fisher-Stefan model
Figure 2(a) shows time-dependent solutions of (1)-(3). Here, we see that a carefully-chosen initial condition with the appropriate decay at infinity evolves to a travelling wave solution that is visually indistinguishable from (7). To make this point we show the initial condition in green, together some intermediate-time solutions in blue. The latest solution is superimposed with (7) in dashed orange, confirming that the time-dependent solutions converge to the exact solution reasonably rapidly. Similar results in Figure 2(b) show time-dependent solutions of (4)-(6). Here we have used a simple step function for an initial condition, which is shown in green, and have carefully chosen our parameter κ to be κ = c/V ∗ = −0.906 . . . . Again we show three intermediate-time solutions in blue. The superimposed exact solution (13)-(14) with (15) for c = −5/√6 in orange compares extremely well with the numerical solution at t = 9. 74 Conclusion
We present a new interpretation of an exact travelling wave solution of the Fisher-KPP model. The Fisher-KPP model is one of the most well-studied reaction-diffusion equations with ap-plications including wound healing [18–21] and ecological invasion [22, 23]. For cell biology applications, the Fisher-KPP model is often used because cells are thought to move randomly, by diffusion, as well as proliferating logistically [19, 24]. Experimental observations of mov-ing cell fronts can be described by travelling wave solutions of the Fisher-KPP model , or generalisations of the Fisher-KPP model . For the dimensional Fisher-KPP model with diffusivity D, proliferation rate λ and carrying capacity density K, the speed of the travelling wave solution is c ≥ 2√λD . In reality, fronts of cells may move at a slower speed or even retreat . One way to deal with this is to write down a more complicated model with more than one species or with a different source term, like an Allee effect . Even with a single species model that retains a logistic growth term, we can introduce nonlinear degenerate diffusion, leading to the Porous-Fisher model , which gives rise to travelling wave solutions with c ≥ √λD/ 2 [29, 30]. An alternative to all of these modifications is to simply retain Fisher (1) but include a moving boundary as in (4)-(6) . This has appealing features, namely: a sharp moving front, which seems biologically reasonable; retains the classical logistic growth term with an easy to measure λ; allows for solutions of all speeds c < 2√λD , including negative speeds . Despite the apparent simplicity of the Fisher-KPP model, exact solutions are relatively elu-sive but of high interest since they provide important mathematical insight and can be used as benchmarks for testing numerical methods . It is well-known that the travelling wave solution (7) can be written down exactly for a special wave speed c = 5 /√6 [14,15]. This special travelling wave is consistent with the usual view that travelling wave solutions of the nondi-mensional Fisher-KPP have positive speed c > 2, whereas solutions with c < 2 are normally disregarded on the grounds of being unphysical . In our work we take a different point of view and re-formulate the Fisher-KPP model with a moving boundary, often called the Fisher-Stefan model . The Fisher-Stefan model has several attractive features: (i) travelling wave solutions of the Fisher-Stefan model have a well-defined front without needing to introduce the compli-cation degenerate nonlinear diffusion; (ii) the Fisher-Stefan model gives rise to travelling wave solutions with ∞ < c < 2, which is more flexible that the usual Fisher-KPP and Porous-Fisher models since it can be used to model both invasion and retreat; and (iii) the Fisher-Stefan model provides a simple physical interpretation for an exact solution with c = −5/√6. This 8overlooked solution can be expressed exactly using Weierstraß elliptic functions and, as we show numerically, this solution is the long-time limit of our moving boundary problem (4)-(6). 95 Appendix
5.1 Numerical method for the phase plane
To construct the phase planes in the main document we solve the first order dynamical system dU
dz = V, (16) dV
dz = −cV − U (1 − U ), (17) numerically, using a two-stage Runge-Kutta method, also called Heun’s method , with a constant step size, d z = 1 × 10 −3. This choice of discretisation leads to results that are visually independent of the numerical discretisation. Our main interest is in examining phase plane trajectories that either enter or leave the saddle (1 , 0) along the stable or unstable manifold, re-spectively. Therefore, it is important that the initial condition we chose when solving Equations (16)–(17) are on the appropriate stable or unstable manifold, and sufficiently close to (1 , 0). To choose this point we use the MATLAB eig function to calculate the eigenvalues and eigen-vectors when c = ±5/√6. The vector field associated with Equations (16)–(17) are plotted on the phase planes using the MATLAB quiver function . MATLAB implementations of these algorithms are available on GitHub.
5.2 Phase plane visualisation
Results in Figure 1 (main document) show regions of the phase plane that are deliberately scaled so that we can easily see the details of the trajectories associated with c = ±5/√6. Since each phase plane in Figure 1 (main document) is shown on a different scale, it is easy to forget that these phase planes are closely related, with the only difference being the sign of c. To make this point we show here, in Figure 3, the phase plane shown over a wider domain with
c = ±5/√6. We show, in thick red lines, the trajectories associated with the travelling waves with c = ±5/√6. For completeness, we also show many other phase plane trajectories, in thin black lines. These additional trajectories are not associated with travelling waves. Note that the thick red trajectory in the third and fourth quadrant are associated with the invading travelling wave with c = 5 /√6, whereas the thick red trajectory in the first quadrant is associated with the receding travelling wave with c = −5/√6. To plot these trajectories on the same phase plane, the trajectory associated with the receding travelling wave c = −5/√6 is obtained by 10 integrating the dynamical system in the other direction, effectively by setting d z = −1 × 10 −3.
Figure 3: Phase plane visualisation . Phase plane showing the two equilibria (black discs). The two trajectories corresponding to the travelling wave solutions (thick red lines) and a range of other trajectories (thin black lines) are shown.
5.3 Numerical method for the Fisher–KPP model
We solve the Fisher–KPP equation
∂u ∂t = ∂2u∂x 2 + u(1 − u), (18) on 0 ≤ x ≤ L, with L chosen to be sufficiently large so that the time–dependent solutions of the partial differential equation (PDE) model have sufficient time to approach a travelling wave. We discretise the domain with a uniform finite difference mesh, with spacing ∆ x, and we approximate the spatial derivatives in Equation (18) using a central finite difference approxi-mation. The resulting system of coupled ordinary differential equations are integrated through 11 time using an implicit Euler approximation, giving rise to
uj+1
i
− uji
∆t =
( uj+1
i−1
− 2uj+1
i
uj+1
i+1
∆x2
)
uj+1
i
(1 − uj+1
i
), (19) for i = 2 , . . . , m − 1, where m = L/ ∆x + 1 is the total number of spatial nodes, and j indexes time so that we approximate u(x, t ) by uji , where x = ( i − 1)∆ x and t = j∆t.For all numerical solutions of Equation (18) we enforce no–flux boundary conditions at x = 0 and x = L, giving
uj+1 2 − uj+1 1 = 0 , (20)
uj+1
m
− uj+1
m−1
= 0 . (21) In this work we wish to study the solution of Equation (18) with a travelling wave speed greater than the minimum speed, c > 2. Therefore, care is taken to choose the initial condition to achieve this. For the initial condition we set
u(x, 0) =
12 0 ≤ x < 10 ,
e
−
2
√6 (x−10)
2 10 ≤ x < L,
(22) where the exponential decay rate of the initial condition is carefully chosen to be u(x, 0) ∼
e−2x/ √6, because this choice leads to travelling waves with c = 5 /√6 . To advance the discrete system from time t to t + ∆ t we solve the system of nonlinear alge-braic system, Equations (19)–(21), using Newton–Raphson iteration with convergence tolerance
ε. The resulting systems of linear equations are solved efficiently using the Thomas algorithm. All numerical solutions of Equation (18) are obtained by setting ∆ x = 1 × 10 −4, ∆ t = 1 × 10 −2,
L = 60 and ε = 1 × 10 −8. We find that these choices lead to results that are independent of the numerical discretisation. Using the numerically–generated time–dependent solutions we also estimate the travelling wave speed. To estimate c∗ we specify a contour value, u(x, t ) = u∗, and at the end of each time step, we use linear interpolation to find x∗ such that u(x∗, t ) = u∗, and we then calculate c∗ = [ x∗(t + ∆ t) − x∗(t)] /∆t. MATLAB implementations of these algorithms are available on GitHub. 12 5.4 Numerical method for the Fisher–Stefan model
We solve
∂u ∂t = ∂2u∂x 2 + u(1 − u), (23) for 0 < x < s (t) numerically by using a boundary fixing transformation ξ = x/s (t) so that we have
∂u ∂t = 1
s2(t)
∂2u∂ξ 2 + ξs(t)ds(t)dt∂u ∂ξ + u(1 − u), (24) on the fixed domain, 0 < ξ < 1. Here s(t) is the length of the domain that we will discuss later. To close the problem we also transform the appropriate boundary conditions
∂u ∂ξ = 0 at ξ = 0 , (25)
u = 0 at ξ = 1 . (26) Spatially discretising Equations (24)–(26) with a uniform finite difference mesh, with spac-ing ∆ ξ, approximating the spatial derivatives using central differences and an implicit Euler approximation for the temporal derivative gives,
uj+1
i
− uji
∆t = 1(sj )2
( uj+1
i−1
− 2uj+1
i
uj+1
i+1
∆ξ2
)
ξsj
( sj+1 − sj
∆t
) ( uj+1
i+1
− uj+1
i−1
2∆ ξ
)
uj+1
i
(1 − uj+1
i
),
(27) for i = 2 , . . . , m − 1, where m = 1 /∆ξ + 1 is the total number of spatial nodes on the finite difference mesh, and the index j represents the time index so that we approximate u(ξ, t ) by
uji , where ξ = (i − 1) ∆ ξ and t = j∆t.Discretising the boundary conditions, Equations (25)–(26), gives
uj+1 2 − uj+1 1 = 0 , (28)
uj+1
m
= 0 . (29) The initial condition for the Fisher–Stefan problem is u(x, 0) = 1 on 0 ≤ x ≤ s(0). To advance the discrete system from time t to t + ∆ t we solve the system of nonlinear algebraic equations, Equations (27)–(29), using Newton–Raphson iteration with convergence tolerance ε.13 During each iteration of the Newton–Raphson algorithm we update s(t) using the discretised Stefan condition
sj+1 = sj − κ∆tsj ∆ξ
( uj+1
m−2
2 − 2uj+1
m−1
3uj+1
m
2
)
, (30) where we set uj+1
m
= 0 to be consistent with (29). All results in this work are obtained by setting κ = −0.906 . . . , ∆ ξ = 1 ×10 −5, ∆ t = 1 ×10 −2,
ε = 1 ×10 −8 and s(0) = 100. Again, we find these choices lead to results that are independent of the numerical discretisation, and we also use the time–dependent PDE solution to estimate c∗ by tracking the time evolution of the leading edge, c∗ = ( sj+1 −sj )/∆t. MATLAB implementations of these algorithms are available on GitHub.
Acknowledgements: This work is supported by the Australian Research Council (DP200100177). 14 References
RA Fisher. The wave of advance of advantageous genes. Annals of Eugenics. 7 (1937) 355-369. AN Kolmogorov, PG Petrovskii, NS Piskunov. A study of the diffusion equation with increase in the amount of substance, and its application to a biological problem. Moscow University Mathematics Bulletin. 1 (1937) 1-26. J Canosa. On a nonlinear diffusion equation describing population growth. IBM Journal of Research and Development. 17 (1973) 307-313. JD Murray. Mathematical Biology I: An Introduction. Third edition, Springer, New York, (2002). L Edelstein-Keshet. Mathematical Models in Biology. SIAM, Philadelphia, (2005). M Kot. Elements of Mathematical Ecology. Camridge University Press, Cambridge, (2003). M El–Hachem, SW McCue, W Jin, Y Du, MJ Simpson. Revisiting the Fisher-Kolmogorov-Petrovsky-Piskunov equation to interpret the spreading–extinction dichotomy. Proceed-ings of the Royal Society A: Mathematical, Physical and Engineering Sciences. 475 (2019) 20190378. M El–Hachem, SW McCue, MJ Simpson. A sharp-front moving boundary model for ma-lignant invasion. Physica D: Nonlinear Phenomena. 412, 132639. M El–Hachem, SW McCue, MJ Simpson. Invading and receding sharp-fronted travelling waves. arXiv:2008.00662 Y Du, Z Lin. Spreading–vanishing dichotomy in the diffusive logistic model with a free boundary. SIAM Journal on Mathematical Analysis. 42 (2010) 377-405. Y Du, H Matano, K Wang. Regularity and asymptotic behavior of nonlinear Stefan prob-lems. Archive for Rational Mechanics and Analysis 212 (2014) 957-1010. Y Du, H Matsuzawa, M Zhou. Sharp estimate of the spreading speed determined by non-linear free boundary problems. SIAM Journal on Mathematical Analysis 46 (2014) 375-396. Y Du, B Lou. Spreading and vanishing in nonlinear diffusion problems with free boundaries. Journal of the European Mathematical Society 17 (2015) 2673-2724. 15 P Kaliappan. An exact solution for travelling waves of ut = Du xx + u − uk. Physica D: Nonlinear Phenomena. 11 (1983) 368-374. M Ablowitz, A Zeppetella. Explicit solutions of Fisher’s equation for a special wave speed. Bulletin of Mathematical Biology. 41 (1979) 835-840. M Abramowitz, IA Stegun (Eds). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Dover Publications, New York (1965). PA Clarkson. Painlev´ e Equations — Nonlinear Special Functions. In: Marcell´ an F., Van Assche W. (eds) Orthogonal Polynomials and Special Functions. Lecture Notes in Mathe-matics, vol 1883. Springer, Berlin, Heidelberg (2006) BG Sengers, CP Please, ROC Oreffo. Experimental characterization and computational modelling of two–dimensional cell spreading for skeletal regeneration. Journal of the Royal Society Interface. 4 (2007) 1107-1117. JA Sherratt, JD Murray. Models of epidermal wound healing. Proceedings of the Royal Society of London: Series B. 241 (1990) 29-36. ST Johnston, JV Ross, BJ Binder, DLS McElwain, P Haridas, MJ Simpson. Quantifying the effect of experimental design choices for in vitro scratch assays. Journal of Theoretical Biology. 400 (2016) 19-31. ST Vittadello, SW McCue, G Gunasingh, NK Haass, MJ Simpson. Mathematical models for cell migration with real–time cell cycle dynamics. Biophysical Journal. 114 (2018) 1241-1253. JG Skellam. Random dispersal in theoretical populations. Biometrika. 38 (1951) 196-218. J Steele, J Adams, T Sluckin. Modelling paleoindian dispersals. World Archaeology. 30 (1998) 286–305. PK Maini, DLS McElwain, D Leavesley. Traveling waves in a wound healing assay. Applied Mathematics Letters. 17 (2004) 575-580. DJ Warne, RE Baker, MJ Simpson. Using experimental data and information criteria to guide model selection for reaction–diffusion problems in mathematical biology. Bulletin of Mathematical Biology. 81 (2019) 1760-1804. KJ Painter, JA Sherratt. Modelling the movement of interacting cell populations. Journal of Theoretical Biology. 225 (2003) 327-339. 16 NT Fadai, MJ Simpson. Population dynamics with threshold effects gives rise to a diverse family of Allee effects. Bulletin of Mathematical Biology. 82 (2020) 74. TP Witelski. Merging traveling waves for the porous–Fisher’s equation. Applied Mathe-matics Letters. 8 (1995) 57-62. F S´ anchez Garduno, PK Maini. An approximation to a sharp type solution of a density– dependent reaction–diffusion equation. Applied Mathematics Letters. 7 (1994) 47-51. JA Sherratt, BP Marchant. Nonsharp travelling wave fronts in the Fisher equation with degenerate nonlinear diffusion. Applied Mathematics Letters. 9 (1996) 33-38. MJ Simpson, KA Landman. Characterizing and minimizing the operator spit error for Fisher’s equation. Applied Mathematics Letters. 19 (2006) 604-612. MathWorks eig . Retrieved August 2020 from MathWorks quiver . Retrieved August 2020 from M.J. Simpson, K.A. Landman, T.P. Clement. Assessment of a non-traditional operator split algorithm for simulation of reactive transport. Mathematics and Computers in Simulation. 70 (2005) 44-60. 17 |
16867 | https://www.quora.com/What-is-the-truth-value-of-p-or-q-when-p-is-true-and-q-is-false | Something went wrong. Wait a moment and try again.
Truth Value
Laws of Boolean Algebra
Logic (philosophy)
Logical Operators
Propositional Calculus
Boolean Functions
Mathematical Logic
5
What is the truth value of "p or q" when p is true and q is false?
·
Sep 7
Truth value: true.
Reasoning (concise):
The logical disjunction "p or q" (inclusive OR) is true if at least one of p or q is true.
Given p = true and q = false, at least one operand (p) is true, so p ∨ q = true.
Truth-table reference:
T ∨ T = T
T ∨ F = T
F ∨ T = T
F ∨ F = F
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In the conditional “if p, then q,” given that p is false and q is true, what is the truth value of the conditional?
Let p = true; q = false; r = true, and s= false. What is the truth value of either r and s or either p or q?
Assuming that p and r are false and that q and s are true, what is the truth value for (p→q) ⋀(q→r)?
David Seed
MSc numerical maths
·
Author has 2.5K answers and 1.5M answer views
·
1y
this is a simple case.
the proposition p or q is true, if either or both of the propositions {p,q} is true.
eg so if p means “Smith is old” and q means “Smith is male”
when Smith is old and Smith is not male,
then the composite “Smith is old or male” ie (p OR q) is a true statement.
As with many such puzzles, the real problems arise when trying to define the statements unambiguously. eg what exactly is male and what is old.
eg suppose i name my Hermaphrodite clown fish “smith”.
Kele Douglas Keli'imakekauonu'uanuokona Perkins
MA/Philosophy (CSULB); 2022 US Chess Local TD of the Year
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Author has 174 answers and 239.5K answer views
·
2y
Q. What is the truth value of "p or q" when p is true and q is false?
A. The truth-value of the disjunction ‘p v q’ is ‘true’ (also written as ‘T’ or sometimes as ‘1’) whenever at least one of its disjuncts (in this case, p) is true.
P.S. The answer remains the same in two other cases: (1) when p is false and q is true, and (2) when both p and q are true.
Sun P
PhD from Columbia University
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Author has 369 answers and 276.3K answer views
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4y
Related
Why do we consider if p, then q as true when p is false and q is false?
Question: Why do we consider if p, then q as true when p is false and q is false?
”If P then Q” is logically equivalent to “Q OR not P.” What that means is either Q must be true OR P must be false for the statement to be true.
In other words, the truth table looks like this (T=true, F=False):
Ok, so what the heck is going on here? Well, let’s parse it out in terms of something more familiar. Suppose Tiny Timmy is taking a tough test (applause for alliteration skills), and the totalitarian teacher (dying applause) makes a rule: “If you are taking a test, then cell phones must be turned off!”
Ok
Question: Why do we consider if p, then q as true when p is false and q is false?
”If P then Q” is logically equivalent to “Q OR not P.” What that means is either Q must be true OR P must be false for the statement to be true.
In other words, the truth table looks like this (T=true, F=False):
Ok, so what the heck is going on here? Well, let’s parse it out in terms of something more familiar. Suppose Tiny Timmy is taking a tough test (applause for alliteration skills), and the totalitarian teacher (dying applause) makes a rule: “If you are taking a test, then cell phones must be turned off!”
Ok, fair enough; let’s use the above table to check if Timmy is obeying the rule. The first column tells us if he is taking the test (we write “T” if taking, “F” if not taking). The second column tells us if his phone is off (“T” if off, “F” if on). The third column tells us if he is a good boy and obeys the rule (“T” if obeys, “F” if he is badass).
Day 1: Timmy takes a test and his cell phone is off. Has he obeyed the rule? Sure has. That’s the first row.
Day 2: Timmy takes a test and his cell phone is on. Has he obeyed the rule? Nope-he was a bad boy. That’s the second row.
Day 3: Timmy doesn’t have a test today. Yippie! Now, does it matter if his cell phone is off or on? Nope—either way, he was a good boy and obeyed the rule. That’s the third and fourth rows.
Here is a more schematic way to look at it with Venn diagrams.
”If P then Q” (a statement about logic) is schematically equivalent to saying “P is a subset of Q” (a statement about sets), because being in state P ensures we’re in state Q.
The first box is a world where “If P then Q” is true. The next two boxes are worlds where “If P then Q” is violated. We use a light green shade to indicate a state compatible with “If P then Q.” We use a bold reddish peach shade (excuse me if I am color blind, lol) to indicate a state that is incompatible with “If P then Q.”
In the first world, everything is shaded light green because, well, the entire world is the “If P then Q” world by definition. Notice that we don’t have to be in the circle P to be in the world where P implies Q.
Therefore, any point not in circle P in the next two worlds is shaded green because, as we just found, such a point is compatible with “If P then Q.”
However, the first world never contains a state where we are in circle P and not in circle Q—that is forbidden by “If P then Q.” But the next two worlds contain those states. Thus, we shade those regions peach—they reveal unmistakably that we are not in the “If P then Q” world.
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Jason Chu
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4y
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Why do we consider if p, then q as true when p is false and q is false?
Think of an implication as a promise. And the implication is false if the promise is not kept, and true otherwise.
Say your parents promise to give you $100 if you get an A in Logic, we write
get-an-A-in-Logic get-100-dollars
The only situation that you can accuse your parents of breaking their promise is when you actually get an A and they do not give you the $100 . If you don’t get an A, and they don’t give you $100 (), you can not blame them for breaking the promise. Actually, even if you don’t get an A, and they give you
Think of an implication as a promise. And the implication is false if the promise is not kept, and true otherwise.
Say your parents promise to give you $100 if you get an A in Logic, we write
get-an-A-in-Logic get-100-dollars
The only situation that you can accuse your parents of breaking their promise is when you actually get an A and they do not give you the $100 . If you don’t get an A, and they don’t give you $100 (), you can not blame them for breaking the promise. Actually, even if you don’t get an A, and they give you $100 anyway (), you can’t really say they break their promise.
Dan Christensen
Creator of proof-checking freeware DC Proof 2.0
·
Author has 2.8K answers and 834.3K answer views
·
4y
Related
If p⟶q is false, what is the truth value of ((¬p) ∧q) ⟷(p∨q)? Explain your answer?
Here is the truth table :
Notice that is true in only one case: when is true and is false (on line 2).
Here is the truth table for :
Notice that in the only case when is true, is false (see line 2 of both truth tables). I will leave the conclusion to you.
Here is the truth table :
Notice that is true in only one case: when is true and is false (on line 2).
Here is the truth table for :
Notice that in the only case when is true, is false (see line 2 of both truth tables). I will leave the conclusion to you.
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Justin Wang
Competitive Highschool Mathematics
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8y
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In the conditional “if p, then q,” given that p is false and q is true, what is the truth value of the conditional?
Currently taking philosophy,
Here’s what we know
From Modus Tollens we also know
Now if Q is true yet P is false, it gives little to no information on the statement’s soundness. Assuming that is a logical fallacy known as Assuming the Consequent
For example,
If it rains the sidewalk will be wet.
By Modus Ponens
It rained, therefore the sidewalk is wet
By Modus Tollens
The sidewalk is not wet, therefore it did not rain
Assuming the Consequent
The sidewalk is wet, therefore it rained
The problem w
Currently taking philosophy,
Here’s what we know
From Modus Tollens we also know
Now if Q is true yet P is false, it gives little to no information on the statement’s soundness. Assuming that is a logical fallacy known as Assuming the Consequent
For example,
If it rains the sidewalk will be wet.
By Modus Ponens
It rained, therefore the sidewalk is wet
By Modus Tollens
The sidewalk is not wet, therefore it did not rain
Assuming the Consequent
The sidewalk is wet, therefore it rained
The problem with this is that rain may not be the sole cause of the wetness of the sidewalk. If it rains then the sidewalk must be wet, but if the sidewalk is wet, you can inductively judge if it rained, however guaranteeing it is a logical fallacy
To conclude, the given information tells little to no information on the statement because it's valid for the Consequent to occur without its Antecedent so it does not provide information to support nor deny the claim that
Quite nice.
Alex Eustis
Ph.D. in Mathematics, University of California, San Diego (Graduated 2013)
·
Upvoted by
David Joyce
, Professor of Mathematics at Clark University · Author has 4.6K answers and 23.8M answer views
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2y
Related
How do you prove without a truth table that p or q is logically equivalent to, if not p then q?
I like this question, because I can’t stand truth tables. Overzealous math teachers talk about truth tables a lot, but no actual mathematician thinks in terms of truth tables when doing deductive reasoning. It’s simply not how our minds work or how deductive reasoning works.
Instead, I will borrow Martin Jansche’s idea and go with a Fitch-style natural deduction system. If you’re learning proof-based mathematics, it would be very instructive to learn how to construct these Fitch-style proofs yourself. (There are a bunch of exercises you can do at your own pace.)
To the uninitiated this might loo
I like this question, because I can’t stand truth tables. Overzealous math teachers talk about truth tables a lot, but no actual mathematician thinks in terms of truth tables when doing deductive reasoning. It’s simply not how our minds work or how deductive reasoning works.
Instead, I will borrow Martin Jansche’s idea and go with a Fitch-style natural deduction system. If you’re learning proof-based mathematics, it would be very instructive to learn how to construct these Fitch-style proofs yourself. (There are a bunch of exercises you can do at your own pace.)
To the uninitiated this might look like useless nonsense, but in my opinion it’s much closer to how a mathematician actually uses deductive reasoning to construct proofs. In particular, this style of proof exemplifies how assumptions actually work in mathematics, which is a far more important concept than any “truth table” ever was.
I’ll explain this line-by-line, but first, a general overview of what’s going on here. The proof consists of a sequence of statements a.k.a. sentences. The first one is the hypothesis and the last one is the conclusion. But in addition to being linear, the proof has a nested structure. Lines 2–6 comprise a sub-proof where we assume as a (sub)-hypothesis and deduce as a (sub)-conclusion. This allows us to conclude the implication
at one level “up”, i.e. outside the sub-proof. There is even a third nested proof level, where we assume and arrive at a contradiction, allowing the conclusion at level 2.
As for the right column, those are the justifications. The syntax is rather funny-looking, but it just means that we are using a particular rule of inference identified as “" but better known as modus ponens, with lines 1 and 3. We have and , therefore . When using any rule of inference, you are only allowed to invoke previous lines within that sub-proof or any of its parents (but not its siblings!). This makes sense because each sub-proof starts out with an assumption. A rule of inference can use only the assumptions that are currently in effect, at that level of sub-proof and its direct ancestors.
Some of the rules determine what you are allowed to conclude outside the sub-proof, at one level up. Others allow you to make use of the sentences you’ve already proved (or assumed) to deduce other sentences at the same level.
As for the E and I, for the most part, each logical symbol (like ) has an “I” rule and an “E” rule. The “I” rule tells you how to infer that symbol. For instance the rule tells you that if you assume in a sub-proof and deduce , then you get to infer outside the sub-proof. The “E” rule tells you what you can extract from that symbol, under what conditions. For example the rule is modus ponens as mentioned above; it tells you that you can extract from and .
So let’s just go through the whole thing line by line.
L1: , the hypothesis.
L2: Begin a new subproof by assuming . (Making an assumption requires no justification.)
L3: Begin a new subproof by assuming .
L4: Use modus ponens with L1 and L3 to conclude . (Discussed above)
L5: The symbol means “contradiction”. A contradiction follows from any statement and its logical negation, in this case (L2) and (L4).
L6: From a contradiction, we get to conclude the negation of whatever assumption led to that contradiction; in this case . That is the rule, better known as the law of non-contradiction.
L7: Because we assumed and were able to deduce from that (on the same level), we now know that . This is the rule. I don’t even know what the “official” name of this rule of inference is. I’d simply call it “definition of implies”. That follows from is the very meaning of the notation .
Here is the same proof, but written out in natural language:
It is given that P implies Q. Now suppose Q is false. Then P cannot be true, for if it were, Q would follow. Therefore, if Q is false, then P is false.
You might notice that the natural language proof is a lot shorter, which is why it is preferred among mathematicians. Perhaps not as “rigorous” or whatever, but once you get used to writing proofs like this, no one really uses the Fitch style anymore. But it’s a useful learning tool.
My point is that even a natural-language “paragraph proof”, in essence, follows this same structure. Sub-proofs, a.k.a. lemmas. What follows from what. What leads to a contradiction. Et cetera. For instance if you go to Groupprops, a handy group theory reference website, it's all “this implies that”. Finite non-nilpotent and every proper subgroup is nilpotent implies not simple. You get the idea. Groupprops usually tries to be very clear about “facts used” in what order, and occasionally even follows a two-column proof style.
In general the nested structure of the assumptions is not always explicitly written out (although it is when writing out the statement of a theorem or lemma). Most of the time, it must be inferred from context which assumptions are in effect, where, and at what point the sub-proof “drops down” a level and the assumption goes away. That takes some getting used to.
Anyway, I know this is probably way more information that the question really asked for, but I am trying to get at the question behind the question, if that makes sense.
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Keith Ramsay
Ph.D. in mathematics
·
Author has 5.4K answers and 8.2M answer views
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4y
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Why do we consider if p, then q as true when p is false and q is false?
There are different conceptions of “if-then”. Some do not automatically do this.
Here is a reason why at least sometimes we want for “if then ” to be considered true in spite of and both being false. Suppose I said in some context that and are integers, and that if and are even, then is even. Now suppose at some later point, it turns out and . Do we want now to go back and say that my “if-then” statement was invalid? I don’t think so. Depending on the values of and , we get all three cases: and both true, false and true, and and both false.
If one has wha
There are different conceptions of “if-then”. Some do not automatically do this.
Here is a reason why at least sometimes we want for “if then ” to be considered true in spite of and both being false. Suppose I said in some context that and are integers, and that if and are even, then is even. Now suppose at some later point, it turns out and . Do we want now to go back and say that my “if-then” statement was invalid? I don’t think so. Depending on the values of and , we get all three cases: and both true, false and true, and and both false.
If one has what is called a “truth-functional” concept of “if-then”, this tends to force you to consider “if then ” to be true all of the time when and are false. (We have to be careful about what we mean by a “truth-value” however.) In mathematics, one ordinarily uses the law of excluded middle, thinking also that each statement is in fact true or false even if we don’t know which it is. This tends to result in mathematicians using a truth-functional form of “if-then” that is equivalent to “either is false or is true”.
Intuitionistic logic, which doesn’t include the law of excluded middle as a deductive rule, does not make “if then ” equivalent to “either is false or is true”. The connective “if-then” cannot be reduced to any combination of “and”, “or”, and negation. Intuitionistic logic still does make “if then ” hold when is false, because it obeys the rule of “ex falso quodlibet” (from the false, anything). So to some extent the two issues are independent of each other.
One reason why avoiding ex falso quodlibet is not more popular than it might be is that we tend to want to keep three other properties of logic. One is that from p one can deduce “ or ”. Another is that from “ or ” and “ is false” one can deduce . And a third is that if we can deduce from , then we can deduce “if then ”. So one tends to want “if is true and is false, then ” to hold, as well as “if then ” in cases where is false.
One logic which does not have ex falso quodlibet is called minimal logic. It is essentially intuitionistic logic with ex falso quodlibet removed.
As far as I’m aware, the typical alternative view on “if-then” considers it necessary for to be relevant in some way to for “if then ” to hold. If this seems interesting to you, you could try looking up “relevance logic”.
Grice’s implicatures seem to me to help explain why people sometimes object to a remark of the form “if then ” in cases where is false. To put it a little simplistically, our problem is sometimes that if the speaker knows already that is false, why are they even bringing into it? My example above where it could seem reasonable for me to say an “if-then” when the “if” part is false is one where I hypothetically don’t know yet that it is. My sense is however that it is worth separating out this kind of objection from the claim that the statement being made is incorrect. There are a lot of contexts in which such a statement would be pointless, but should probably be counted as correct anyway.
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Working with three-dimensional coordinates Working with 3D coordinates
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16869 | https://web.williams.edu/Mathematics/sjmiller/public_html/math/papers/erdos_small2021discrete10final.pdf | DISTINCT ANGLE PROBLEMS AND VARIANTS HENRY L. FLEISCHMANN, HONGYI B. HU, FAYE JACKSON, STEVEN J. MILLER, EYVINDUR A. PALSSON, ETHAN PESIKOFF, AND CHARLES WOLF ABSTRACT. The Erd˝ os distinct distance problem is a ubiquitous problem in discrete geometry. Less well known is Erd˝ os’ distinct angle problem, the problem of finding the minimum number of distinct angles between n non-collinear points in the plane. The standard problem is already well understood. However, it admits many of the same variants as the distinct distance problem, many of which are unstudied.
We provide upper and lower bounds on a broad class of distinct angle problems. We show that the number of distinct angles formed by n points in general position is O(nlog2(7)), providing the first non-trivial bound for this quantity. We introduce a new class of asymptotically optimal point configurations with no four cocircular points. Then, we analyze the sensitivity of asymptotically optimal point sets to perturbation, yielding a much broader class of asymptotically optimal configurations. In higher dimensions we show that a variant of Lenz’s construction admits fewer distinct angles than the optimal configurations in two dimensions.
We also show that the minimum size of a maximal subset of n points in general position admitting only unique angles is Ω(n1/5) and O(nlog2(7)/3). We also provide bounds on the partite variants of the standard distinct angle problem.
CONTENTS 1.
Introduction 2 1.1.
Background 2 1.2.
Summary of Results and Methods 2 2.
Erd˝ os Angle Problems 4 2.1.
Unrestricted Point Sets 4 2.2.
No Three Collinear Points 5 2.3.
Restricting Cocircularity 5 2.4.
General Position: No three points on a line nor four on a circle 7 3.
The Robustness of Efficient Point Configurations 9 4.
The Pinned Angle Problem 11 5.
Partite Sets 12 6.
Maximal Subsets of Points with Distinct Angles 14 6.1.
Definitions and Upper Bounds 14 6.2.
A Probabilistic Lower Bound on Maximal Distinct Angle Subsets in general position 14 7.
Higher Dimensions: Lenz’s Construction 17 8.
Future Work 20 References 21 Date: August 24, 2021.
2020 Mathematics Subject Classification. 52C10 (primary), 52C35 (secondary), 52C30, 52B15, 52B11.
Key words and phrases.
Erd˝ os Problems, Discrete Geometry, Angles, Restricted Point Configurations, Maximal Subsets, This work was supported by NSF grant 1947438 and Williams College. E. A. Palsson was supported in part by Simons Foun-dation grant #360560.
1 1. INTRODUCTION 1.1. Background. In 1946, Erd˝ os published a paper titled “On sets of distances of n point", introducing the problem of finding asymptotic bounds on the minimum number of distinct distances among sets of n points in the plane . This simply stated problem proved to be surprisingly challenging and is now known as the Erd˝ os distance problem. Indeed, the original question was only finally resolved by Guth and Katz in 2015 .
Over time, many variations of the problem were introduced: restricting the point sets, studying subsets with no repeated distances, and many other quantities. We study variations of a related problem, introduced by Erd˝ os and Purdy . What is A(n), the minimum number of distinct angles formed by n not all collinear points on the plane? Unlike in the distance setting, an extra restriction of non-collinearity is required to prevent the degenerate case of at most two angles. When this problem was proposed, the regular n-gon was conjectured to be optimal (yielding n −2 angles), and a lower bound of (n −2)/2 angles was proven for point sets without three collinear points. Since then, the problem and all other analogues of distinct distance problems with angles have gone untouched. We study this problem of distinct angles in many of the settings originally considered for distinct distances, providing exact or asymptotic bounds, depending on the problem. We summarize our results below.
1.2. Summary of Results and Methods. Note that throughout, unlike Erd˝ os, we do not count angles of 0 or π to avoid some degenerate behaviors. This is consistent with the current literature on related repeated angle questions (see, for example, ).
1.2.1. Erd˝ os Angle Problems. We begin with the most natural extension of the Erd˝ os distance problems to angles: what is the least number of distinct angles determined by n not all collinear points in the plane?
Given that the known low angle constructions contain obvious structures, such as many points on a line or on a circle, it is natural to also consider the problem over restricted point sets. Our main results in this section are summarized in the following.
• We provide a construction of a polygon projected onto a line, yielding a point configuration with no four points on a circle admitting n −2 distinct angles, the same as the conjectured optimal regular n-gon.
• We also provide a point configuration in general position admitting less than cnlog2(7) for some constant c, the first nontrivial bound for this problem. The configuration relies on enumerating classes of triangle equivalent up to edge translation and then projecting onto a generic plane. While and also use a projection onto a generic plane for the distance problem analogue, the properties of orthogonal projections are much more convenient for distances. Attempting to directly apply their results fails in a dramatic fashion due to additional complexity added by angles.
In addition, for completeness, we provide a known upper bound on A(n) of n −2 from the regular n-gon, conjectured by Erd˝ os and Purdy to be the optimal configuration, and a lower bound of n/6 by partial progress towards the Weak Dirac Conjecture of Erd˝ os and Dirac. We also consider similar problems on restricted point sets, as in the distance setting. Under a restriction forbidding three points on a line, we provide bounds on the restricted quantity Ano3l(n) (these were known by Erd˝ os).
1.2.2. The Robustness of Efficient Point Configurations. Having identified efficient point configurations in the polygon and the projected polygon, we ask how resilient they are to perturbation. Erd˝ os investigated a similar question for distances in . We prove combinatorially the surprising result that both the polygon defines1 O(nk) angles with k points perturbed, so long as they all remain on the circle, and we prove an analogous result for the projected polygon. Consequently, if any constant number of points in a regular polygon are moved to random positions on the circle, the construction still defines O(n) angles (an optimal 1We write f = O(g) if there exists constant C such that f(n) ≤Cg(n) for sufficient large n. Conversely, we write f = Ω(g) if g = O(f). Lastly, we write f = Θ(g) if f = O(g) and f = Ω(g).
2 number asymptotically), even though moving even one point off the circle experimentally gives a super-linear number of distinct angles. In that vein, we provide conjectures about the number of angles in several perturbed optimal configurations in which points may no longer lie exactly on a circle or line.
1.2.3. The Pinned Angle Problem. We subsequently examine the angle equivalent of a prominent Erd˝ os distance problem: namely, given n points, what is the minimum number of distances determined between one “pinned” point and the rest, in the worst case? This problem remains open in the distance setting for convex configurations of points, and is conjectured to be ⌊n/2⌋by Erdos in . An upper bound of ⌊n/2⌋is obtained by considering the regular n-gon, and the current best lower bound of (13/36 + 1/22701) n+O(1) is obtained by Nivasch, Pach, Pinchasi, and Zerbib .
Denoting the analogous angular quantity allowing any configuration of points as ˆ A(n) (with the pinned point as the center-point of the angles), we bound it between n/6 and n −2 using related A(n) proofs. This in turn also provides an upper bound on ˆ AΣ(n), the sum of the number of distinct angles determined by each point.
1.2.4. Partite Sets. Given a partite set, the question of distances determined between the two sets has been studied by Elekes in the unrestricted setting, but remains unsolved in general. We ask the analogous question in the angular setting: how many angles are defined by a k-partite set, where each point is in a distinct set? We provide low angle configurations in the unrestricted case, establish linear lower and upper bounds on partite sets without three collinear points, and completely solve the problem in a particular case.
1.2.5. Maximal Subsets of Points with Distinct Angles. One prominent variation of the Erd˝ os distance prob-lem asks: what is minimum maximal subset of n points such that no distance is repeated? None of the numerous variants in the distance setting have been fully resolved, although a number of upper and lower bounds have been proven by a variety of authors. For a complete picture of these problems in the distance setting see [1, 3, 18, 12, 15].
We ask the analogous question for angles. We upper bound this configuration in general, showing R(n) ≤ A(n)1/3. Then, we employ a probabilistic method similar to that in to show a lower bound of Ω(n1/5).
1.2.6. Higher Dimensions: Lenz’s Construction. The construction consists of multiple unit circles, each in a disjoint pair of dimensions. We show that, just as it has been for repeated angle problems in higher dimen-sions and the unit distance problem, Lenz’s construction also provides a good upper bound of 2⌈2n/d⌉−2 on Ad(n), the least number of distinct angles defined by n points in d dimensions (see page 499 of ). This construction demonstrates that, for a fixed number of points, increasing the dimension decreases the upper bound on the number of distinct angles dramatically. This behavior aligns with the behavior in the distance setting with the integer lattice. We also provide a higher dimensional upper bound for the maximal subset question.
Variant Lower Bound Upper Bound A(n) n/6 n −2 Ano3l(n) (n −2)/2 n −2 Ano4c(n) n/6 n −2 Agen(n) Ω(n) O(nlog2(7)) ˆ A(n) n/6 n −2 ˆ AΣ(n) n/6 + n −1 3n −6 Rgen(n) Ω(n1/5) O(nlog2(7)/3) Ad(n) 2 2 ⌈2n/d⌉−2 TABLE 1. Summary of Results.
3 We provide a tabular summary of our results in this section for convenience. Each parameter is described informally in Table 1 but formally defined in its respective section.
We note several other miscellaneous bounds over the course of the paper, but do not include them here because we only provide an upper or lower bound or they do not fit nicely into the structure of the table.
These include: variants of angle sum bounds in Section 4, partite set bounds in Section 5, and maximal subset bounds in non-general position in Section 6, to name a few.
2. ERD ˝ OS ANGLE PROBLEMS 2.1. Unrestricted Point Sets. We begin by considering the most broad, non-trivial version of the distinct angles problem. This is the version of the problem originally posed by Erd˝ os .
Definition 2.1. For a point set P ⊂R2, let A(P) denote the number of distinct angles in (0, π) determined by points in P. Then, let A(n) := min |P|=n A(P), where P is not all collinear.
We begin by showing A(n) = Θ(n) with explicit upper and lower bounds. First, we give an upper bound using the regular polygon: Lemma 2.2. A(n) ≤n −2.
Proof. Consider the point configuration given by the vertices of an n-sided regular polygon. Upon inscribing the polygon in a circle, notice that distinct angles are in bijection to the arclengths on the circle. We may fix a point as the central point of our angles by the symmetry of the polygon. There are then exactly n −2 possible arc lengths subtending angles with this central point.
□ Remark 2.3. When n is odd, we may alternatively use an (n −1)-gon with an extra point in the center.
Adding the center point to an even regular polygon does not increase the number of nonzero angles defined, and so, if n = 2m + 1, we achieve a slightly better bound: A(2m + 1) ≤2m −2.
We now use progress on the Weak Dirac Conjecture to provide a lower bound on A(n). In 1961, based on a stronger conjecture of Dirac’s, Erd˝ os conjectured in the following.
Conjecture 2.4 (Weak Dirac Conjecture). Every set P of n non-collinear points in the plane contains a point incident to at least ⌈n/2⌉lines of L(P), where L(P) is the set of lines formed by points.
While the Weak Dirac Conjecture is open, significant progress has been made. Let ℓ(n) be the largest proven lower bound proven for the Weak Dirac Conjecture, i.e., every set P of n points not on a line in the plane contains a point incident to at least ℓ(n) lines of L(P). Then, we have the following.
Theorem 2.5. For n > 3, A(n) ≥ℓ(n)−1 2 ≥n 6 .
Proof. Fix a set P of n non-collinear points in the plane. Let p ∈P be incident to at least ℓ(n) lines of L(P). Fix another point q. Note that for any fixed nonzero angle θ < π, there are exactly two possible lines where r must lie on for ∠qpr = θ. Since p is incident to ℓ(n) −1 lines without q, p is the center angle of at least (ℓ(n) −1)/2 distinct angles. Therefore A(n) ≥ℓ(n) −1 2 .
We have ℓ(n) ≥⌈n/3⌉+ 1 from . As such, we have A(n) ≥n/6, as desired.
□ Notably, this argument is known (see Conjecture 10 in 6.2 of ), but is included for completeness.
4 2.2. No Three Collinear Points. Given that any collinear point set defines at most two angles, it is intu-itively clear why restricting the number of collinear points might result in interesting behavior. We briefly consider such point sets in this section.
Definition 2.6. Let Ano3l(n) = min|P|=n A(P), where P contains no collinear triples.
Note that the regular n-gon contains no collinear triples, and so as with A(n), we have an upper bound of Ano3l(n) ≤n −2. The usual stipulation on this bound holds. See Remark 2.3. Our restrictions on the point set allow for a stronger lower bound. This bound was known by Erd˝ os but is included for completeness.
Lemma 2.7. For n > 3, Ano3l(n) ≥n−2 2 .
Proof. Fix a point p ∈P. As no three points are on a line, p determines n −1 distinct lines with each of the other points. The result follows by fixing another point q and repeating the argument for Theorem 2.5.
□ We can easily generalize this restriction to no k points on a line. However, the lower bound given by repeating this argument with k ≥4 points on a line is always weaker than that in Theorem 2.5. Moreover, in those cases the regular polygon remains an upper bound.
2.3. Restricting Cocircularity. Since the regular polygon construction requires many points on a circle, it is natural to wonder how the bounds change when we require that no four points lie on a circle. This setting is not specifically studied in the context of distances but merits special attention for angles given the seeming optimality of the regular n-gon. We provide the following definition.
Definition 2.8. Let Ano4c(n) = min|P|=n A(P), where P contains no co-circular quadruples.
We then have the following lemma.
Lemma 2.9. For n > 3, Ano4c(n) ≤n −2.
Proof. Consider the vertices of a regular n-gon. Fix a vertex p. Then, if n is even, there is a vertex q directly opposite p. In that case, let ℓbe the line perpendicular to pq. If n is odd, there are instead two vertices of minimal distance from p. In that case, let ℓinstead be the line those two vertices. Then, for each vertex r other than p in the regular n-gon, project r onto ℓat the intersection of pr and ℓ. This is the stereographic projection of the points onto ℓvia p. Let the n −1 projected points on ℓand p define the projected polygon configuration, P.
Note that P contains no four cocircular points (Figure 1).
We can now count the number of angles in this configuration. Let α = π/n, the angle subtended by an arc between consecutive points in a regular n-gon.
Note that the angles formed in the case of p being the center of the angle are exactly the n −2 angles of a regular n-gon, iα = iπ/n for 1 ≤i ≤n −2.
Next, we count angles of the form ∠pqiqj where qi and qj lie on line ℓ. We do not consider when all three points are on ℓas that forms degenerate angles. We may assume that qi and qj are both on the same half of the line by the reflectional symmetry of the configuration. We consider two cases: i > j or i < j.
Suppose first that n is even. Then there will be a point q0 at the center of the line. First, we count angles with i > j. Notice that ∠pqiqj = ∠pqiq0. The other two angles in △pqiq0 are iα and π/2, so ∠pqiqj = π 2 −iπ n = (n/2 −i)α, for 1 ≤i ≤(n−2)/2. Since n is even, these are integer multiples of α. Moreover, since 1 ≤n/2−i ≤n−2, angles of this form are already accounted for in the angles with center p. Next, we examine the case of i < j.
Then 0 ≤i ≤n/2 −1. Except for the angle with center at q0, which has value π/2 (accounted for already 5 FIGURE 1. Projecting Regular Polygons onto a Line.
in case 1), all these angles are the supplements of angles i > j. That is, ∠pqiqj = π −∠pqiq0. Thus, we achieve angles of π −(n/2 −i)π n = π 2 + πi n , where 1 ≤i ≤(n −2)/2. All these angles are also accounted for in the case of angles with p as the center, so, when n is even, we have n −2 distinct angles.
Now suppose n is odd. In this case, there will not be point q0 opposite p and on ℓ, but we introduce one that is not in the configuration for later convenience. As before, we first count angles with i > j. There is a special angle that we will add to this count, namely ∠pq1q−1. Then, in effect, we are considering angles ∠pqiq0 for i > 0 as in the odd case. Each of these angles is in the triangle △pq0qi, whose other angles are π/2 and iα −α/2, for 1 ≤i ≤(n −1)/2. Then ∠pqiqj = π 2 −iπ n + π 2n = (⌈n/2⌉−i)π n .
This is an integer multiple of π/n , and, further, 1 ≤⌈n/2⌉−i ≤n −2. Thus, all these angles are already accounted for in case 1. Next, we examine the case of i < j. Then 1 ≤i ≤(n −3)/2. All these angles are the supplements of angles i > j. That is, ∠apipj = π −∠apip0. Thus, we achieve angles of π −(⌈n/2⌉−i)π n = π 2 + πi n where 1 ≤i ≤(n −1)/2. All these angles are also accounted for in case 1, and so when n is odd, we have n −2 distinct angles.
Therefore, this configuration determines exactly n −2 angles.
□ Remark 2.10. The projected polygon construction and the regular polygon both give n −2 angles for n points. The former contains collinear points but no four points on a circle, while the latter contains the opposite.
6 Additionally, there are infinitely many such “one point off the line" configurations yielding ≤cn angles, for some c. Fix α < π/(n−1). Fix some p and some line ℓ. Space the remaining n−1 points on ℓsuch that ∠rps = α for consecutive r and s on ℓ. This configuration forms at most 3n angles in general. We revisit this in Section 3.
Note that Theorem 2.5 provides a lower bound of n/6 distinct angles here as well.
2.4. General Position: No three points on a line nor four on a circle. Now that we have illustrated con-structions determining O(n) angles that forbid either three points on a line or four points points on a circle, we consider configurations that forbid both. Erd˝ os and others have investigated this problem extensively in the distance setting. While the best known lower bound in the distance setting is trivially Ω(n), the best known upper bound is n2O(√log n) from . In this section we provide a nontrivial upper bound on this quantity.
Definition 2.11. Let Agen(n) := min |P|=n A(P), where P is in general position.
We use a construction inspired by the projective construction in Theorem 1 of to provide an upper bound. We take higher dimensional hypercubes and project their points down to a generic plane. Unlike with distances, we have very little control over the triangles in the projection. As such, we proceed by very careful combinatorics.
Let Qd be the d-dimensional hypercube with vertices of the form p = (x1, x2, . . . , xd) and xi = 0 or 1 for each i. For any 2-dimensional plane Π in Rd, let T be the orthogonal projection of the points in Qd onto Π. It is possible to choose Π satisfying the followingl conditions.
(1) T(p1) = T(p2) if and only if p1 = p2, and (2) P := T(Qd) ⊂Π is in general position.
In addition, since orthogonal projections are self-adjoint and idempotent, we have p1 −p2 = p3 −p4 = ⇒d(T(p1), T(p2)) = d(T(p3), T(p4)) (2.1) by the following computation.
⟨T(p1 −p2), T(p1 −p2)⟩= ⟨p1 −p2, T(T(p1 −p2))⟩ = ⟨p3 −p4, T(p1 −p2)⟩ = ⟨T(p3 −p4), p1 −p2⟩ = ⟨T(T(p3 −p4)), p3 −p4⟩ = ⟨T(p3 −p4), T(p3 −p4)⟩.
Unfortunately, T does not preserve the distance between points in Qd. That is, d(p1, p2) = d(p3, p4) ̸ = ⇒d(T(p1), T(p2)) = d(T(p3), T(p4)).
This means two congruent triangles in Qd need not be congruent after projection. However, by (2.1) and SSS-congruence, two congruent triangles with equal difference vector edges remain congruent after projec-tion. This inspires the following definition.
Definition 2.12. Given a triangle ∆with vertices in Qdd, define the equivalence class [∆]Qd as the set of all triangles congruent to ∆whose vertices lie in Qd and edges correspond to (individually) translated copies of the edges of ∆.
It suffices to characterize the equivalence classes of translated congruent triangles in Qd to bound the number of angles in P. We do so in the following lemma.
7 Lemma 2.13. The number of equivalence classes of triangles in Qd is 7d −3d+1 + 2 12 .
Proof. We begin by counting the number of unordered triples of distinct binary k-tuples such that no coor-dinate of the triple is fixed for all three. By “fixed," we mean equal among all three k-tuples. Let ak denote the number of such triples.
At each coordinate in the triple of k-tuples, the possible values for each of the triples are 0 or 1. Since no coordinate of the triples is fixed, each coordinate must either have exactly one 1 or one 0 among the three k-tuples. There are then (3 · 2)k ways to choose whether there will be one 1 or one 0 and the choice of tuple for that singleton for each of the k coordinates. This imposes an ordering which we divide out at the end. Now, note that although it is impossible to repeat a k-tuple thrice since no coordinate is fixed, we overcount instances with a repeated k-tuple. A k-tuple can be chosen to repeat twice in 2k ways and the choice of repeated k-tuple forces the choice of the third k-tuple. Such triples can be ordered in 3 ways. After subtracting off such pairs, the remaining triples are all distinct and thus can be ordered in 3! ways. We have ak = 6k −3 · 2k 6 .
Now we use ak to count tk, the number of triangles in Qd with exactly k unfixed coordinates. As there are d k ways to choose the unfixed coordinates, ak ways to choose the values in those unfixed coordinates, and 2d−k ways to choose the values of the fixed coordinates, we get tk = 6k −3 · 2k 6 d k 2d−k = d k 6k2d−k −3 · 2d 6 .
Finally, we prove that all triangles in the same equivalence class have the same number of fixed coor-dinates. We also prove that the size of the equivalence class of triangles with a given number of fixed coordinates is constant.
First observe that, given a triangle in Qd, we may get an equivalent triangle by flipping any combination of its fixed coordinates (from 0 to 1 or vice versa), thereby translating each vertex of the triangle by the same amount. Moreover, if a coordinate is not fixed, then any translation to a equivalent triangle cannot be nonzero in that coordinate as it would then lead to some point having a coordinate that is both nonzero and not one. The only other way to achieve an equivalent triangle is to translate the edges individually (keeping the difference vectors corresponding to them identical, but altering their relative orientations). This can happen in at most one way.
Moreover, flipping each coordinate of each of the three points yields a congruent triangle composed of the same difference vectors (by flipping, we mean changing from 0 to 1 and vice versa). This follows from the observation that, if a coordinate in a difference vector is 0, the subtracted coordinates must be equal and remain equal under swapping each coordinate. If a coordinate is 1 or −1, it will swap sign but still be the same vector.
Thus, we have that the size of the equivalence class of each triangle with k unfixed points in a Qd is exactly 2d−k+1. Moreover, all triangles in the equivalence class have exactly k unfixed points.
8 Putting this all together, we have that the number of equivalence classes of triangles in Qd is d X k=1 d k (6k2d−k −3 · 2d) 6 · 2d−k+1 = d X k=1 d k (6k/12 −2k−2) = 1 12 " d X k=1 d k 6k −3 d X k=1 d k 2k # = 7d 12 −1 12 −3d 4 + 1 4 = 7d −3d+1 + 2 12 .
The above follows from standard binomial formula identities.
□ Since triangles in the same equivalence class are congruent under the projection from Qd to a specially chosen generic plane, Lemma 2.13 provides an upper bound of O(7d) distinct angles on point configurations in general position with 2d points. To establish this result for n not a power of two, pick the least d such that n < 2d and apply the upper bound to a subset of n vertices in Qd. This proves the following theorem.
Theorem 2.14. Agen(n) = O(nlog2(7)).
3. THE ROBUSTNESS OF EFFICIENT POINT CONFIGURATIONS Before we consider several variants of this problem, we discuss a very natural question: how far can point sets stray from our best constructions (regular polygons and projected polygons) while still being “near-optimal?” In the distance setting, a point set is called near-optimal if it admits O(n/√log n) angles, like the √n × √n integer lattice. Erd˝ os asked if such sets have lattice-like structure, containing Ω(n1/2) points on a line. The question has gone unsolved even after replacing 1/2 by any ϵ > 0. For some partial results related to this problem, see , , and , in which the authors bound the number of points on various algebraic curves in near-optimal point sets.
Formally, we have the following definition.
Definition 3.1. Let P1, P2, . . . ⊂R2 be a sequence of point configurations with |Pn| = n. Then Pn is near-optimal if A(Pn) = O(n).
For example, the sequence of point configurations of regular n-gons is near-optimal. Perhaps surprisingly, these configurations are reasonably robust to point perturbation. We begin by studying points on a circle in a way reminiscent of a regular n-gon.
Proposition 3.2. For a fixed k ≥0, let Sk n be the collection of n points on a circle with n−k points forming a regular (n −k)-gon and the remaining k placed arbitrarily. Then max P∈Sk n A(P) = Θ(nk) Proof. Fix a configuration P ∈Sk n. Since points in P lie on a circle, all its angles are incident angles. Thus the number of distinct angles is bounded by the number of distinct arc lengths. We divide the set of arcs into three cases. Suppose an arc is the minor one formed between p, q ∈P. We then have the following cases.
(1) Both p, q are on the polygon. There are at most n −k −2 distinct arc lengths of this form.
(2) Neither of p, q are on the polygon. Then, they are among the k arbitrarily placed points. There are at most 2 k 2 = k2 −k distinct arc lengths of this form.
(3) We have p is on the polygon and q is not. There are at most 2(n −k)k = 2nk −2k2 distinct arc lengths of this form.
9 Then, in total, P determines at most 2nk −k2 + n −2k −2 angles. Since k ≤n, this quantity is O(nk).
We show that this bound is tight. Choose the k points to be placed at multiples of 2π between n−k−1 n−k · 2π and 2π such that all arcs formed with the added point are iteratively not admitted among the points in the configuration. Then there are n−k−2 arcs between the point at n−k−1 n−k ·2π and the other non-zero polygonal points, moving clockwise about the circle. Notably, they can be extended by 1 to k arcs. By the choice of these arcs, for any length rational arc chosen, all the extensions by 1 to k arcs are distinct. Thus, at least (n −k −2)k = Ω(nk) distinct angles are formed (as the ends of the arc can be used as the end points of an angle with 0 as the center).
□ From this, it follows that such configurations are near-optimal for any k constant in n.
We now discuss a related, more challenging problem. Let T k n be the collection of n point configurations having n −k points on a circle and no circle containing more. Denote by T(n) the maximum quantity k ≤n/2 satisfying min P∈T k n A(P) = Θ(n).
What can be said about the value of T(n)? We restrict k to n−k = Ω(n) in order to avoid cases that reduce to configurations with a negligible number of points on a circle.
Consider a new point at the center of a regular n-gon. When n is even, the new point generates no new angles (see Remark 2.3). When n is odd, it generates exactly ⌈n/2⌉additional angles. To see this, say n = 2m + 1. One new angle of 4mπ 2m+1 is from the angle whose center is at the new point and the ends at the original n-gon. Another n more angles of iπ n −π 2n for 1 ≤i ≤n are from those with the new point as an end point; they are equivalent to the original arcs on the n-gon with half of the 2π/n arc cut out.
Thus, T(n) ≥1. We conjecture this bound to be tight and this point in the center is the only way to achieve it.
Conjecture 3.3. T(n) = 1.
The aforementioned optimal point configuration, the projected polygon, has n −1 points be on a line. To what extent can we perturb this configuration while remaining near-optimal?
Proposition 3.4. For a fixed k ≥0, let Lk n be the collection of planar n-point configurations with with n−k (including the off-line point) in the projected polygon construction from Lemma 2.9 and the remaining k placed on the line arbitrarily. Then max P∈Lk n = A(P) = Θ(nk).
Proof. We may divide the possible angles into four cases. Observe an angle ∠pqr in the following cases.
(1) Say q is the point off the line, and p, r are both projected points. Then the number of such angles is bounded by the number of angles in the projected polygon construction n −k −2.
(2) Say q is the point off the line, and neither of p, r are projected points. Then the number of angles in this case is bounded by k 2 = (k2 −k)/2.
(3) Say q is the point off the line, p is a projected point, and r is not (without loss of generality). Then the number of such angles is this bounded by (n −k −1)k.
(4) Say q is on the line. Then all nontrivial ∠pqr have one leg along the line and the other at the point off the line. Thus, for each q on the line, there are at most two possible angles, and they are supplementary amongst themselves. Then the number of these angles is bounded by 2(n −1).
Then in total we O(nk) distinct angles formed.
Now we show this bound is tight. Choose the k points to be placed following the rightmost point on the line such that the angles they form with the point off the line as the center are all unique (we can do this 10 iteratively by the infinitude of the real line). Then, by construction, there are at least k X i=1 (n −k + i −1) = k(n −k) + k 2 = Ω(nk) distinct angles in the configuration, counting only angles centered at the point off the line.
□ From this, if follows that such configurations are near-optimal for any k constant in n.
Now, we consider the quantity analogous to T(n). Let Mk n be the collection of all n point configurations having n −k points on a line and no lines containing more. Denote by M(n) the maximum k satisfying min P∈Mk n A(P) = Θ(n).
What can be said about the value of M(n)?
It is immediate that M(n) ≥1 by the projected polygon. In addition, from Remark 2.10, there are many other near-optimal configurations in Mk n. Consider a new point that is the original point off the line reflected over the line. It generates O(n) additional angles.
Thus, M(n) ≥2. We conjecture this bound to be tight and the only pair of points to achieve this are those symmetric about the line.
Conjecture 3.5. M(n) = 2.
4. THE PINNED ANGLE PROBLEM We now pivot to a variant originally considered in the context of distinct distance problems. Among n points in the plane, in the worst case, what is the maximum number of distinct angles centered at some “pinned” point? For example, see Corollary 6 in for the original problem for distances.
Definition 4.1. For a point set P and a point p in it, let Ap(P) denote the number of distinct angles in (0, π) formed in P with p as the center. Then, let ˆ A(n) := min |P|=n max p∈P Ap(P), where P is a not all collinear planar point set.
Theorem 4.2. We have ˆ A(n) = Θ(n).
Proof. Regular polygons give an upper bound of n −2. Let ℓ(n) be the current best lower bound on the Weak Dirac Conjecture (see Conjecture 2.4). Using the same logic as Theorem 2.5, this gives a lower bound of (ℓ(n) −1)/2. This is at least n/6, completing the proof.
□ This use of the Weak Dirac Conjecture is known (see Conjecture 10, Section 6.2 in ). A related classical question for distinct distances is determining the average number of distinct distances admitted by a pinned point. We present its analogy for angles here.
Definition 4.3. Let ˆ AΣ(n) := min |P|=n X p∈P Ap(P), where P is a not all collinear planar point set.
Theorem 4.4. We have ˆ AΣ(n) ≤3n −6.
Proof. The construction from Lemma 2.9 has one point off the line as the center of n−2 distinct angles. All the points on the line, apart from the endpoints, are the center of exactly two non-degenerate angles. These contribute another 2n −4 to the sum. Hence, ˆ AΣ(n) ≤3n −6.
□ We also have the following for ˆ A′ Σ(n), this quantity forbidding three collinear points.
11 Theorem 4.5. We have ˆ A′ Σ(n) = Θ(n2).
Proof. Since no three points are on a line, we can repeatedly remove points and apply the bound from Theorem 2.5. This gives n X i=c i 6 = Ω(n2).
To get the upper bound, use the fact that every point of a regular n-gon is the center of exactly n −2 distinct angles.
□ 5. PARTITE SETS Another well known variant of the distinct distances problem is that of distances between points in partite sets. See , for example. We introduce a similar problem for angles. We make heavy use of the best lower bound of the Weak Dirac Conjecture on n points, ℓ(n), in this section. As a reminder, from , we have ℓ(n) ≥⌈n/3⌉.
Definition 5.1. Given bipartite P, Q ⊂R2, denote by A(P, Q) the number of distinct angles in (0, π) whose central vertex lies in a different set than its two end points. Where P ∪Q is not all collinear, let A(m, n) := min |P|=m,|Q|=n A(P, Q).
For simplicity, assume m ≤n. We begin by providing upper bounds on both unrestricted and restricted point sets.
Lemma 5.2. We have A(m, n) ≤m.
Proof. We utilize the projected polygon construction from Lemma 2.9. Assign the point off the line, q, to be in Q and the m leftmost points on the line in P. Notably, the points in P do not cross the center point of the points on the line. Now, the angles of the form ∠p1qp2, the angles with center in Q, form the angles iπ n+m for 1 ≤i ≤m −1. Let r be the rightmost point on the line. From Lemma 2.9, we have two cases for the the angles of the form ∠qpir. If n + m is even, they form the angles π 2 − jπ n + m for n−m 2 ≤j ≤n+m−2 2 (or, for n = m, j ≥1 and these angles form an angle of π/2 for pi the orthogonal projection of q onto the line). In the case n + m odd, the angles ∠qpir are π 2 − jπ n + m − π 2(n + m) for n−m+1 2 ≤j ≤n+m−1 2 . Namely, these angles are computed by completing the angles of the right triangle containing q, pi, and the orthogonal projection of q. Substituting the ranges of the angles for both, we find that the only angles in the configuration are lπ n+m for 1 ≤l ≤m, implying the result.
□ Note that this result implies that no unrestricted lower bound in terms of n + m can exist.
Next, we provide an upper bound on the case of no three collinear points in P or Q.
Lemma 5.3. We have Ano3l(m, n) ≤n −2.
Proof. Let P form a subset of the vertices of a regular n-gon of size m and Q the vertices of a regular n-gon, both inscribed in the same circle. The bipartite angles formed in this configuration are all incident angles subtended by arcs of the size subtending angles in a regular n-gon. Thus, the distinct angles formed in this configuration are a subset of the angles of a regular n-gon, implying the result from Lemma 2.2.
□ Now, we provide a lower bound in the restricted case of no three collinear points within sets P or Q.
12 Lemma 5.4. We have Ano3l(m, n) ≥⌊(n −1)/2⌋.
Proof. Fix a point p ∈P to be a center point and a q ∈Q to be a non-center point. By the argument in Lemma 2.7, since no three points are collinear, at most 2 of each of the remaining n −1 points in Q can form the same angle with pq with center p. However, one other point in the Q can be collinear to p and q, not contributing any angle, yielding ⌈(n −2)/2⌉= ⌊(n −1)/2⌋.
□ Corollary 5.5. We have Ano3l(m, n) ≥ ( m+n 2 −1)/2 .
This allows us to completely solve this problem in the case of m = 1.
Lemma 5.6. We have Ano3l(1, n) = ⌊(n −1)/2⌋.
Proof. Let Q be the vertices of a regular n-gon. Inscribe these vertices in a circle. Let the singular point p in the other set be the center of the circle. Then the number of angles of the form ∠q1pq2 for q1, q2 ∈Q is n/2 −1 for n even and (n −1)/2 for n odd by counting subtending arclengths. This yields A(1, n) ≥ ⌊(n −1)/2⌋. Combining with Lemma 5.4, we achieve the desired result.
□ We now consider k-partite sets.
Definition 5.7. Let n = Pk i=1 ri. Let A(r1, r2, . . . , rk) denote the minimum number of distinct angles determined by point sets in R2 of respective sizes r1, . . . , rk with each of the following stipulations: (1) each angle is formed by three points in distinct sets, (2) r1 ≥r2 ≥· · · ≥rk and k ≥3, and (3) not all points are collinear.
We begin with upper bounds on unrestricted and restricted point sets.
Lemma 5.8. We have A(r1, r2, . . . , rk) ≤2(n −r1 −1).
Proof. Let S = n −r1. We follow a similar proof to Lemma 5.2. Let the n points be in an n point projected regular polygon configuration. Let the point off the line, p, be in the r1 set. Let the leftmost rk points be in the rk set, the next leftmost in the rk−1 set, and so on, with the remaining rightmost points on the line in the r1 set. Note that all angles must include p. The angles iπ/n for 1 ≤i ≤S −1 are exactly those formed by angles with p as the center. We assume without loss of generality that S is at most n/2 as, from Lemma 2.9, there are n −2 total angles in the configuration. From the proof of Lemma 5.2, the angles centered at p and the acute angles centered at some point in an ri set for i > 1 overlap completely, yielding S −1 angles (note that the rightmost center point has no endpoint for an angle opening to the right, thus the minus one).
Moreover, since we may assume S ≤n/2, all of S −1 supplemental angles are obtuse. Thus, they do not overlap at all, yielding the desired bound of 2S −2 angles.
□ As before, we are unable to provide a lower bound only in terms of n, as, from the above, there are configurations with large r1 which exhibit very few distinct angles.
We next provide an upper bound on A(r1, r2, . . . , rk) in the restricted case of no three points on a line.
Lemma 5.9. We have Ano3l(r1, r2, . . . , rk) ≤n −max(2, rk + 1).
Proof. Place the n points as the vertices of a regular n-gon. Assign the first r1 to the first partite set, the next r2 to the second, and so on, continuing about the circle circumscribing the polygon clockwise. Then, since all points are on a circle, the angles in the configuration each correspond exactly to the arc subtending them.
But, by construction, the arcs subtending angles may contain at most n −max(3, rk + 2) points . As such, the distinct angles in this configuration are exactly iπ/n for 1 ≤i ≤n −max(2, rk + 1). This implies the desired bound.
□ 13 If we forbid three points on a line within each of constituent sets, we have a stronger bound than Lemma 5.4 for rk−1 and rk small.
Lemma 5.10. Ano3l(r1, r2, . . . rk) ≥⌈n−rk−1−rk 2 ⌉.
Proof. Fix a point in the rk-set and a point in rk−1-set. Now, since there are no three collinear points, the points in the other sets make each angle with these points at most twice. Thus, Ano3l(r1, r2, . . . rk) ≥ ⌈n−rk−1−rk 2 ⌉.
□ 6. MAXIMAL SUBSETS OF POINTS WITH DISTINCT ANGLES 6.1. Definitions and Upper Bounds. Another variant of the Erd˝ os distinct distance problem is the follow-ing: given n points in a plane, how many points must we remove so that the remaining points determine no repeated distances? This problem has been studied extensively in the context of distances, with varying restrictions on the points set [3, 16, 18, 12]. We study an analogous problem for angles, proving the first nontrivial lower and upper bounds.
Definition 6.1. For a point set P, let R(P) be the maximum size of Q ⊆P such that Q determines no repeated angle. Then, let R(n) := min |P|=n R(P), where the minimum is taken over non-collinear point sets P of n points.
In general, configurations with a low number of angles provide a reasonable upper bound for R(n).
Lemma 6.2. Let P ⊆R2 be a planar point configuration of n points with no three collinear points. Then R(P) ≤(2A(P))1/3.
Proof. Fix P ⊆R2, a planar point configuration of n points with no three collinear points. Then, the subsets of the point configuration determine at most A(P) distinct angles. Moreover, as there are no three collinear points in P, any subset S of P admits 3 |S| 3 not necessarily distinct angles. Thus, if a subset S has no repeated angles, it must be that 3 |S| 3 ≤A(P).
This implies |S| ≤(2A(P))1/3.
□ Using Lemma 2.2 and Theorem 2.14, we get the following bounds.
Corollary 6.3. We have R(n), Rno3l(n) = O(n1/3).
Rno4c(n), Rgen(n) = O(nlog2(7)/3).
Remark 6.4. Notably, Lemma 6.2 does not provide an especially strong bound for Rno4c(n), since the construction from Lemma 2.9 has n −1 points on a line and removing the point off the line thus yields a subset with all (trivially) distinct angles.
6.2. A Probabilistic Lower Bound on Maximal Distinct Angle Subsets in general position. Now we provide a lower bound on Rgen(n). The proof is in many ways reminiscent of Charalambides’ proof of this for distances (see Proposition 2.1 in ).
To proceed, we define and bound several quantities.
14 Definition 6.5. For a point set P, let Q3(P) := {(p, q, r) ∈P3 : p, q, r distinct, ∠pqr = ∠qrp}, Q4(P) := {(p, q, r, s) ∈P4 : p, q, r, s distinct, ∠pqr = ∠pqs, ∠pqr = ∠rsp, ∠pqr = ∠qrs, or ∠pqs = ∠qrs}, Q5(P) := {(p, q, r, s, t) ∈P5 : p, q, r, s, t distinct, ∠pqr = ∠sqt, ∠pqr = ∠qst, ∠pqr = ∠rst}, Q6(P) := {(p, q, r, s, t, u) : p, q, r, s, t, u distinct ∠pqr = ∠stu}.
Remark 6.6. Q3(P) is the collection of pairs of equal angles overlapping at all three points. It is also the collection of isosceles triangles, over counting up to a factor of 3.
Q4(P) is the collection of pairs of equal angles overlapping at two points. The first case is when the angles share a central point and one endpoint. The second case is when they share both endpoints. The third case is when their center points are endpoints for the other angle (and since that gives two points overlapping, the other two endpoints do not overlap). The fourth and final case is when the endpoints of one angle are the center and an endpoint of the other. See Figure 2.
or FIGURE 2. Four Point Repeated Angle Configurations.
Q5(P) is the collection of pairs of equal angles overlapping at one point. The first case is when the angles share a central point. The second case is when the center of one angle is an endpoint of the other.
The third case is when an endpoint of one angle is also an endpoint of the other. See Figure 3.
Q6(P) is the collection of pairs of equal angles without overlaps.
These cases are all encompassing as, for each number of repeated points, it involves some matching of angle endpoints/centers to angle endpoints/centers. These cases are an exhaustive list of such matchings.
For that which follows, we assume the point configuration is in general position in the plane. That is, no three points in the configuration are on a line and no four are on a circle. The argument fails without restricting to sets with no four points on a circle, as remarked afterward.
Definition 6.7. For each 3 ≤i ≤6, let qi(n) := max |P|=n |Qi(P)|, where P is a planar point set in general position.
15 FIGURE 3. Five Point Repeated Angle Configurations.
Lemma 6.8. We have q3(n) = O(n7/3).
Pach and Sharir show the number of isosceles triangles in the plane is O(n7/3) . We count such triangles twice in Q3(n) if the triple forms an isosceles triangle and thrice if it forms an equilateral triangle.
This makes no difference asymptotically.
Lemma 6.9. We have q4(n) = O(n3).
Proof. Let P be n points in general position. We show that there are at most cn3 quadruples in each case of Q4(P), for some constant c, implying the desired bound. In each of the cases below, we use the same naming convention as in Definition 6.5. Our case numbering also correspond to those in Figure 2.
Fix p, q and r, which may be done in less than n3 ways.
Case 1: We count the number of ways to choose s such that ∠pqs = α = ∠pqr. Since α is determined by p, q and r, then s must be on a ray from q forming angle α with qp. There are at most two such rays, with one being − → qr. Since the points are in general position, s cannot be on − → qr. On the other ray, there are at most one such point s. So, there are at most n3 quadruples in this case.
Case 2: Let C be the circle determined by the three points and C′ be its reflection across the segment ← → pr. Let ℓbe the perpendicular bisector of segment ← → pr. We show that any point s forming ∠psr = α has to lie on the outer perimeter of the figure C ∪C′.
First, assume point s lies in the same half space as q with respect to the line ← → pr and the same half space as r with respect to ℓ. Let s′ be the point projected onto C projected from s along ← → ps. By comparing the triangles △psr and △ps′r, we have that ∠ps′r = ∠psr iff ∠prs = ∠prs′. Therefore, it must be that s = s′.
The previous argument can be performed when s is on the other side of ℓ, by projecting along ← → rs instead. However, in either of these cases, s ∈C, which violates the restriction of no four points on a circle. That is, the second diagram in Case 2 of Figure 2 is impossible given our restrictions.
This argument can be repeated when s is on the other side of ← → pr, but instead on circle C′. Since there are already two points on this circle, there is most one choice for s in this case. Hence, there are at most n3 quadruples in this case.
Case 3: We choose s such that ∠qrs = α. Since α is determined by p, q and r, s must be on a ray from r forming angle α with qr. There are at most two such rays. Each such ray contains at most two point, yielding at most two options for s. So, there are at most 2n3 quadruples in this case.
16 Case 4: This case is extremely similar to Cases 1 and 3. We choose s such that ∠sqr = α = ∠pqr.
As before, there are at most two lines which intersect qr at q at an angle of α. Each line contains at most one additional point in P, yielding an upper bound of 2n3 quadruplets in this case.
Since each four cases can occur in at most cn3 ways, for c = 2, then q4(n) = O(n3), as desired.
□ Remark 6.10. If four points on a circle is allowed, then we instead get q4(n) = Θ(n4), crippling the proof.
This can be achieved with a regular n-gon. Fix two of its vertices as the overlapping endpoints. They partition the vertices of the polygon into major and minor arcs, with the major one having least (n −2)/2 vertices. Then, any choice of two vertices from the major arc will yield a configuration as in Case 2. Thus there are Ω(n4) such configurations in this case. As such, forbidding four points on a circle is necessary.
Lemma 6.11. We have q5(n) = O(n4).
Proof. Let P be n points in general position. As in Lemma 6.9, we show that each case of Q5(P) occurs at most cn4 times, for constant c, implying the desired bound.
Case 1: Fix p, q, r, and s, done in less than n4 ways. As in Lemma 6.9, there are at most two choices for t, yielding at most 2n4 options for this case.
Case 2: This follows exactly as Case 1, yielding a bound of at most 2n4 options for this case.
Case 3: This follows exactly as Cases 1 and 2, yielding a bound of at most 2n4 options for this case.
The above casework implies the result.
□ Lemma 6.12. We have q6(n) = O(n5).
Proof. Fix n points in general configuration and from it points p, q, r, s, and t, in less than n5 ways. There are then exactly two ways to choose u so that ∠pqr = ∠stu. Then u must lie on one of two lines containing t. Since there are no three collinear points, there are at most two choices of u. The result follows.
□ Theorem 6.13. We have Rgen(n) = Ω(n1/5).
Proof. Let P ⊂R2 be a point set of size n and let Q ⊂P be a set in which each element of Q is chosen independently and uniformly from P with probability p. The probability p will be specified below.
Each occurrence of some configuration from S6 i=3 Qi in Q generates a repeated angle. Let Q′ ⊂Q be the points remaining after one point from each configuration is removed. Indeed, Q′ is free of repeated angles and |Q′| ≤Rgen(P).
Taking expectations we obtain E[|Q′|] ≥E[|Q|] − 6 X i=3 E[|Qi|] = pn − 6 X i=3 piqi(n).
Using Lemmas 6.8–6.12, there exist some constant c > 1 such that for all n > N for some N, we get E[|Q′|] ≥np −c(p3n7/3 −p4n3 −p5n4 −p6n5).
Setting p = c−1n−4/5, for n > N we have E[|Q′|] ≥c−1n1/5 −c−2n−1/15 −c−3n−1/5 −c−4 −c−5n1/5 = Ω(n1/5).
By the first moment method, there exists a subset of size Ω(n1/5) without repeated angles.
□ 7. HIGHER DIMENSIONS: LENZ’S CONSTRUCTION In general, the minimum number of distinct angles among n points in Rd should decrease as lower dimensional spaces can be embedded in higher dimensional ones. In this section, we provide a construction that demonstrates that this is indeed the case.
17 Definition 7.1. Let Ad(n) be the minimum number of distinct angles on three points determined by n non-collinear points in d-dimensional space.
In dimension d for d ≥4, Lenz gives a construction for a low upper bound on Ad(2n), as described in .
Construct a unit regular n-gon centered at the origin in the x1x2-plane and another unit regular n-gon centered at the origin in the x3x4-plane. This is Lenz’s construction. Now, we upper bound the number of distinct angles in this configuration. From Lemma 2.7, there are n −2 distinct angles between points in the same n-gon. There are then two other cases to consider. We may assume without loss of generality the points lie in four dimensions, as the extra dimensions make no difference in the computation.
Let the three points be (1) x = (cos(θ), sin(θ), 0, 0) (2) y = (cos(ψ), sin(ψ), 0, 0) (3) z = (0, 0, sin(φ), cos(φ)), where θ, ψ, φ ∈{2πi/n : 0 ≤i ≤n −1} and θ ̸= ψ.
Case 1: The endpoints are in the same polygon.
In this case, z is the center of the angle. We compute α = ∠xzy using α = arccos ⟨x −z, y −z⟩ ∥x −z∥∥y −z∥ .
(7.1) For the computations that follow, here is a useful trigonometric identity: cos(θ1 −θ2) = cos(θ1) cos(θ2) + sin(θ1) sin(θ2).
(7.2) Now, for convenience, x −z = (cos(θ), sin(θ), −sin(φ), −cos(φ)) y −z = (cos(ψ), sin(ψ), −sin(φ), −cos(φ)).
Substituting, we have ⟨x −z, y −z⟩= cos(θ) cos(ψ) + sin(θ) sin(ψ) + sin2(φ) + cos2(φ) = 1 + cos(θ −ψ), where the second step follows from applying Equation 7.2. Similarly, we have ∥x −z∥= ∥y −z∥= √ 2.
(7.3) Substituting into (7.1), we have α = arccos 1 + cos(θ −ψ) 2 .
We know θ −ψ = 2πk/n for nonzero −n+1 ≤k ≤n−1. Since cosine is an even and periodic with 2π, the image of cos(θ −ψ) are exactly cos(2πk/n) for 1 ≤k ≤⌈n−1 2 ⌉. Because arccosine is injective on [−1, 1], it yields exactly ⌈n−1 2 ⌉values of α.
Case 2: The endpoints are in different polygons.
Here, we may assume x is the center of the angle. We compute α = ∠yxz using α = arccos ⟨y −x, z −x⟩ ∥y −x∥∥z −x∥ .
(7.4) Again, for convenience, y −x = (cos(ψ) −cos(θ), sin(ψ) −sin(θ), 0, 0) z −x = (−cos(θ), −sin(θ), sin(φ), cos(φ)).
18 Substituting, ⟨y −x, z −x⟩= −cos(ψ) cos(θ) + cos2(θ) −sin(ψ) sin(θ) + sin2(θ) = 1 −cos(θ −ψ), where the last step is another application of Equation (7.2). As before, ∥z −x∥= √ 2. However, we also have ∥y −x∥= q cos2(ψ) + cos2(θ) −2 cos(ψ) cos(θ) + sin2(ψ) + sin2(θ) −2 sin(ψ) cos(θ) = p 2 −2 cos(θ −ψ) = √ 2 p 1 −cos(θ −ψ).
Now, substituting into Equation (7.4), we have α = arccos p 1 −cos(θ −ψ) 2 !
.
As in Case 1, we have n−1 2 possible values for cos(θ −ψ).
Now, since in both cases arccosine is injective on the domain, these α are duplicate angles if and only if 1 + cos(θ −ψ) = p 1 −cos(θ −ψ).
Since both sides are non-negative we may square both sides. Thus, we have a duplicate angle if and only if we have a solution to cos(θ −ψ)(cos(θ −ψ) + 3) = 0.
So, cos(θ −ψ) must equal zero. Since we are only considering θ −ψ = 2πk/n for 1 ≤k ≤ n−1 2 , this occurs only for k = n/4. Hence, we overcount between these two cases exactly once if and only if 4 | n.
As a result of these computations, we have the following lemma.
Lemma 7.2. The number of distinct angles in Lenz’s construction with 2n points is at most 2n −4 if 4 | n and at most 2n −3 otherwise.
We can extend Lenz’s construction to get even better bounds on the minimum number of distinct angles in higher dimensions. In dimension d ≥6, we may now have three unit regular n-gons in disjoint pairs of coordinates. Crucially, adding the third n-gon adds at most one angle, formed by points on three different polygons. As the distance between points is √ 2, three points always yield an equilateral triangle and an angle of π/3.
If the dimension allows, you may then add even more n-gons in disjoint coordinates. After the third, the additional ones do not add any additional distinct angles. Since subsets of the vertices of n-gons have a subset of the angles, you can make point sets of any size using Lenz-like constructions.
Theorem 7.3. Fix d ≥2. For n > d + 1, we have that 2 ≤Ad(n) ≤ n −2, d = 2, 3 (7.5) 2 ln 2 m −3, d = 4, 5 (7.6) 2 n ⌊d/2⌋ −2, d ≥6.
(7.7) For 3 ≤n ≤d + 1, we have Ad(n) = 1.
(7.8) 19 Proof. The lower bound of 2 for (7.5), (7.6), (7.7) follows from the fact that the d-dimensional simplex has exactly d + 1 vertices and it is the largest point configuration in d-dimensional space with all points equidistant. As such, in each of these cases, there are at least two distances between the points and thus more than one distinct angle.
Now note that l n ⌊d/2⌋ m ≥3 if n > d + 1. (7.6) and 7.7 then follow from Lemma 7.2 and the above discussion of generalized Lenz’s constructions. In fact, in special cases for d ≥6 we get a slightly lower bound. For n ⌊d/2⌋a multiple of 3 or 4, we may reduce the bound by 1 (and 2 if a multiple of 12).
For (7.6) we also may reduce the bound by 1 if n 2 is a multiple of 4.
For (7.5), for d = 2 or 3, this follows from Lemma 2.7 by using a regular n-gon.
For (7.8), note that for n = d + 1, we can arrange the points to form the vertices of a d-dimensional regular simplex, yielding all points equidistant from one another. This means all angles are π/3 as they are the angle of an equilateral triangle. We can take any subset of the vertices of such a simplex to get the same result.
□ Crucially, this construction implies that no uniform lower bound greater than 4 for a fixed n and varying d can exist.
From this, we can also give an upper bound on the higher dimensional version of the quantity R(n) from Definition 6.1.
Definition 7.4. Let Rd(P) be the maximum size of any Q ⊆P ⊆Rd such that Q defines no angle twice.
Over the set of all n non-collinear points, define Rd(n) = min |P|=n Rd(P).
We again make use of a variation of Lenz’s construction to provide an upper bound.
Proposition 7.5. We have Rd(n) ≤ 2 n ⌊d/2⌋ −4 1 3 .
Proof. We use the variation of Lenz’s construction from Lemma 7.2. Distribute the points as evenly as possible amongst the largest possible regular polygons in disjoint pairs of dimensions as normal. Note that there cannot be points on three different circles as they form an equilateral triangle. Also note that there cannot be two points on one circle and one on another as that forms an isosceles triangle. As such, we may apply Lemma 6.2 to the largest polygon to achieve our desired bound.
□ Remark 7.6. For n ≤ d+1 2 , there is a two distance set of that many points (see Lemma 3.1 of ). Thus, in such sets all but two points must be removed, yielding Rd(n) = 2 for n ≤ d+1 2 .
8. FUTURE WORK Future research may take distinct angles problems in a number of new directions: (1) We have shown that n/6 ≤A(n) ≤n −2. Further, we have identified two non-collinear point configurations which define exactly n −2 angles, the regular n-gon and its projection onto the line.
Whether these are in fact the optimal configurations is open (though they are conjectured to be so), and even if they are, there may be others which also define n−2 angles. Note that we have observed that, excluding angles of 0 and π, one may add a point to the center of an even sided regular polygon without adding any angles. See Remark 2.3. This does not contradict Erd˝ os’ initial conjecture in , as he included 0 angles.
(2) Prove Conjecture 3.3 and Conjecture 3.5 regarding optimal point configurations.
(3) One may similarly improve our bounds on Ano3ℓ(n), Ano4c, and Agen. In general position, an optimal construction has yet to be conjectured.
20 (4) The question of distinct angles in higher dimensional space has yet to be explored deeply, and one may generalize any of our bounded quantities to the general setting. Further research may also investigate higher analogues of angles like three-dimensional solid angles.
(5) We bounded Rgen(n), the size of the largest distinct-angle subset of an n point configuration. Al-ternatively, by viewing the point configuration as a complete graph on n vertices, we may define Rgen(n) as the number of vertices in the largest complete distinct-angle sub-graph. Instead of re-moving vertices, one might ask about removing edges until all angles left are distinct.
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O. E. Raz, O. Roche-Newton, and M. Sharir, Sets with few distinct distances do not have heavy lines, Discrete Math. 338 (2015), 1484-1492.
Sheffer, J. Zahl, and F. de Zeeuw, Few distinct distances implies no heavy lines or circles, Combinatorica 36 (2016), 349-364.
Email address: henryfl@school.edu DEPARTMENT OF MATHEMATICS, UNIVERSITY OF MICHIGAN, ANN ARBOR, 48109 Email address: hongyih@andrew.cmu.edu DEPARTMENT OF MATHEMATICAL SCIENCES, CARNEGIE MELLON UNIVERSITY, PITTSBURGH, 15213 Email address: alephnil@umich.edu DEPARTMENT OF MATHEMATICS, UNIVERSITY OF MICHIGAN, ANN ARBOR, 48109 Email address: sjm1@williams.edu, Steven.Miller.MC.96@aya.yale.edu DEPARTMENT OF MATHEMATICS AND STATISTICS, WILLIAMS COLLEGE, WILLIAMSTOWN, MA 01267 Email address: palsson@vt.edu 21 DEPARTMENT OF MATHEMATICS, VIRGINIA TECH, BLACKSBURG, VA 24061 Email address: ethan.pesikoff@yale.edu DEPARTMENT OF MATHEMATICS, YALE UNIVERSITY, NEW HAVEN, CT 06511 Email address: charles.wolf@rochester.edu DEPARTMENT OF MATHEMATICS, ROCHESTER, NY, 14627 22 |
16870 | http://101qs-media.s3.amazonaws.com/others/_669_squarepyramidformula.pdf | Square pyramidal number 1 Square pyramidal number Geometric representation of the square pyramidal number 1 + 4 + 9 + 16 = 30.
In mathematics, a pyramid number, or square pyramidal number, is a figurate number that represents the number of stacked spheres in a pyramid with a square base. Square pyramidal numbers also solve the problem of counting the number of squares in an n × n grid.
Formula The first few square pyramidal numbers are: 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819 (sequence A000330 in OEIS).
These numbers can be expressed in a formula as This is a special case of Faulhaber's formula, and may be proved by a straightforward mathematical induction. An equivalent formula is given in Fibonacci's Liber Abaci (1202, ch. II.12).
In modern mathematics, figurate numbers are formalized by the Ehrhart polynomials. The Ehrhart polynomial L(P,t) of a polyhedron P is a polynomial that counts the number of integer points in a copy of P that is expanded by multiplying all its coordinates by the number t. The Ehrhart polynomial of a pyramid whose base is a unit square with integer coordinates, and whose apex is an integer point at height one above the base plane, is (t + 1)(t + 2)(2t + 3)/6 = Pt + 1. Relations to other figurate numbers The square pyramidal numbers can also be expressed as sums of binomial coefficients: The binomial coefficients occurring in this representation are tetrahedral numbers, and this formula expresses a square pyramidal number as the sum of two tetrahedral numbers in the same way as square numbers are the sums of two consecutive triangular numbers. In this sum, one of the two tetrahedral numbers counts the number of balls in a stacked pyramid that are directly above or to one side of a diagonal of the base square, and the other tetrahedral number in the sum counts the number of balls that are to the other side of the diagonal. Square pyramidal numbers are also related to tetrahedral numbers in a different way: The sum of two consecutive square pyramidal numbers is an octahedral number.
Augmenting a pyramid whose base edge has n balls by adding to one of its triangular faces a tetrahedron whose base edge has n − 1 balls produces a triangular prism. Equivalently, a pyramid can be expressed as the result of subtracting a tetrahedron from a prism. This geometric dissection leads to another relation: Square pyramidal number 2 Besides 1, there is only one other number that is both a square and a pyramid number: 4900, which is both the 70th square number and the 24th square pyramidal number. This fact was proven by G. N. Watson in 1918.
Another relationship involves the Pascal Triangle: Whereas the classical Pascal Triangle with sides (1,1) has diagonals with the natural numbers, triangular numbers, and tetrahedral numbers, generating the Fibonacci numbers as sums of samplings across diagonals, the sister Pascal with sides (2,1) has equivalent diagonals with odd numbers, square numbers, and square pyramidal numbers, respectively, and generates (by the same procedure) the Lucas numbers rather than Fibonacci.
In the same way that the square pyramidal numbers can be defined as a sum of consecutive squares, the squared triangular numbers can be defined as a sum of consecutive cubes.
Squares in a square A 5 by 5 square grid, with three of its 55 squares highlighted.
A common mathematical puzzle involves finding the number of squares in a large n by n square grid. This number can be derived as follows: • The number of 1×1 boxes found in the grid is .
• The number of 2×2 boxes found in the grid is . These can be counted by counting all of the possible upper-left corners of 2×2 boxes.
• The number of k×k boxes (1 ≤ k ≤ n) found in the grid is .
These can be counted by counting all of the possible upper-left corners of k×k boxes.
It follows that the number of squares in an n by n square grid is: That is, the solution to the puzzle is given by the square pyramidal numbers.
The number of rectangles in a square grid is given by the squared triangular numbers.
Notes Beck, M.; De Loera, J. A.; Develin, M.; Pfeifle, J.; Stanley, R. P. (2005), "Coefficients and roots of Ehrhart polynomials", Integer points in polyhedra—geometry, number theory, algebra, optimization, Contemp. Math., 374, Providence, RI: Amer. Math. Soc., pp. 15–36, MR2134759.
References • Abramowitz, M.; Stegun, I. A. (Eds.) (1964). Handbook of Mathematical Functions. National Bureau of Standards, Applied Math. Series 55. pp. 813. ISBN 0486612724.
• Beiler, A. H. (1964). Recreations in the Theory of Numbers. Dover. pp. 194. ISBN 0486210960.
• Goldoni, G. (2002). "A visual proof for the sum of the first n squares and for the sum of the first n factorials of order two". The Mathematical Intelligencer 24 (4): 67–69.
• Sigler, Laurence E. (trans.) (2002). Fibonacci's Liber Abaci. Springer-Verlag. pp. 260–261. ISBN 0-387-95419-8.
Square pyramidal number 3 External links • Weisstein, Eric W., " Square Pyramidal Number (http:/ / mathworld. wolfram. com/ SquarePyramidalNumber.
html)" from MathWorld.
Article Sources and Contributors 4 Article Sources and Contributors Square pyramidal number Source: Contributors: 4pq1injbok, Akashsoham, Anton Mravcek, Arminius, Balaji Krishnakumar, CRGreathouse, David Eppstein, Essap, Favonian, Fieldday-sunday, Genjix, GerundandGerundive, Giftlite, Graham87, Gwalla, H-J, JamesBWatson, Joeldudesx, Jyril, Linas, Materialscientist, MathHisSci, MatrixFrog, Michael Hardy, Nick, Nishantsah, Nolanus, Oleg Alexandrov, Ozob, PedroFonini, Poli, PrimeFan, Quentar, Richard777, Rjwilmsi, Robert Illes, Robertd, Romancio, Santiperez, Tentacles, TexMurphy, Timotheus Canens, XJamRastafire, 53 anonymous edits Image Sources, Licenses and Contributors Image:Square pyramidal number.svg Source: License: Public Domain Contributors: Original uploader was David Eppstein at en.wikipedia File:Squares in a square grid.svg Source: License: Public Domain Contributors: David Eppstein License Creative Commons Attribution-Share Alike 3.0 Unported //creativecommons.org/licenses/by-sa/3.0/ |
16871 | https://theintactone.com/2019/07/10/bs-u4-topic-4-price-quantity-and-value-indices/ | the intact one
Read MBA, BBA, B.COM Notes
Price, Quantity and Value Indices
Price index (PI)
A price index (PI) is a measure of how prices change over a period of time, or in other words it is a way to measure inflation. There are multiple methods on how to calculate the inflation (or deflation), in this guide we will take a look at a couple of methods on how to do so. Inflation is one of the core metrics monitored by the FED in order to set interest rates.
The general formula for the price index is the following:
PI1,2 = f(P1,P2,X)
Where:
Quantity index numbers
Quantity index numbers measure the change in the quantity or volume of goods sold, consumed or produced during a given time period. Hence it is a measure of relative changes over a period of time in the quantities of a particular set of goods.
Just like price index numbers and value index numbers, there are also two types of quantity index numbers, namely
Let us take a look at the various methods, formulas, and examples of both these types of quantity index numbers.
Value Index Number
The value index number compares the value of a commodity in the current year, with its value in the base year. What is the value of a commodity? It is nothing but the product of the price of the commodity and the quantity. So the value index number is the sum of the value of the commodity of the current year divided by the sum of its value in the chosen base year. The formula is as follows,
v01 = (∑p1q1/∑p0q0)× 100
or alternatively, v01 = (∑V1/∑V0)× 100
In this case of a value index number, we do not apply any weights. Since these are considered to be inherent in the value of a commodity. Thus we can say that a value index number is an aggregate of values.
The value index number is not a very popular statistical tool. Price and quantity index numbers give a clearer picture of the economy for study and analysis. They even help in the formulation and implementation of economic policies.
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16872 | https://www.reddit.com/r/learnmath/comments/1cmfo6r/question_about_the_sum_of_fractions_with_odd/ | Question about the sum of fractions with odd denominators : r/learnmath
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Can the sum of some fractions with odd denominators and fractions with their greatest common denominator be equal to 1? If my question is not clear enough, here is an example: For 1/2+1/3+1/6 we have that the sum is 1 2. I am asking about natural numbers 3. I am asking about strictly odd denominators 3 more notes if my question still isn't well articulated. 3. If we have 1/3 and 1/7 and 1/5 for example, in the sum we must also include 1/105 1/35 1/21 1/15 (but not 1/1) 4. We can have as much numbers as we want 5. I am asking for a sum that is strictly equal to 1
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16873 | https://radiopaedia.org/articles/trigeminal-nerve?lang=us | Trigeminal nerve | Radiology Reference Article | Radiopaedia.org
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Trigeminal nerve
Last revised by Rohit Sharma on 1 May 2025
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Citation, DOI, disclosures and article data
Citation:
Gaillard F, Gajera J, Southi J, et al. Trigeminal nerve. Reference article, Radiopaedia.org (Accessed on 01 Sep 2025)
DOI:
Permalink:
rID:
2207
Article created:
2 May 2008, Frank Gaillard
Disclosures:
At the time the article was created Frank Gaillard had no recorded disclosures.
View Frank Gaillard's current disclosures
Last revised:
1 May 2025, Rohit Sharma
Disclosures:
At the time the article was last revised Rohit Sharma had no financial relationships to ineligible companies to disclose.
View Rohit Sharma's current disclosures
Revisions:
93 times, by 33 contributors - see full revision history and disclosures
Systems:
Central Nervous System, Head & Neck
Sections:
Anatomy
Synonyms:
Trigeminal nerve (V)
Fifth cranial nerve
Nervus trigeminus
Nervus cranialis V
Trigeminal nerve (CN V)
The trigeminal nerveis the fifth(CN V)cranial nerve and its primary role is relaying sensory information from the face and head, although it does provide motor control to the muscles of masticationvia the mandibular division (TA: nervus trigeminus or nervus cranialis V).
It is both large and complicated and has multiple brainstem nuclei (sensory and motor) as well as many interconnections with other cranial nerves. It swaps parasympathetic fibers and taste fibers somewhat haphazardly and divides into numerous terminal branches.
On this page:
Article:
Gross anatomy
Variant anatomy
Related pathology
Related articles
References
Images:
Cases and figures
Gross anatomy
Nuclei
There are four cranial nerve nuclei:three sensory and one motor.
The sensory nuclei are arranged in a column which spans from the midbrain through the pons and medulla and into the upper cervical cord.
1. mesencephalic nucleus:proprioceptive fibers for muscles of the face, orbit, mastication, and tongue; extends through the whole length of the midbrain, lateral to the cerebral aqueduct
2. main sensory nucleus:located in the upper pons, lateral to the motor nucleus, is responsible for touch sensation for all three trigeminal divisions
3. spinal nucleus:lower pons to upper cervical cord (as far as the third cervical segment), is responsible for pain and temperature; additionally it receives afferent fibers from the facial nerve, glossopharyngeal nerve and vagus nerve
The motor nucleus is located in the upper pons and gives off the smaller motor root (see below) which bypasses the trigeminal ganglionjoining the mandibular division and innervates the muscles of mastication as well as the mylohyoid, anterior belly of digastric, tensor tympani and tensor palatini muscles.
Motor and sensory roots
Although the trigeminal nerve is usually described singularly, it actually emerges/enters from the brainstem as multiple nerve roots 8.
The sensory root is large and single and enters the anterolateral aspect of the pons.
Usually superomedial to, and within 4 mm of the sensory root,one or more motor roots emerge. In ~50% of cases, this is a single nerve root, with two or three roots being less common 8. They are most often located superomedial to the sensory root but can be directly superior, medial or even superolateral 8,9.
These nerve roots can be identified on routine MRI imaging 8.
The motor nerve does not join the trigeminal ganglion, but rather joins the mandibular division as it exits via foramen ovale 9.
Nerve root entry zone and transition zone
The site where the nerve roots exit the brainstem is known as the nerve root entry zone. At, or near, this location oligodendrocytes, that supply insulating myelin to the nerve fibers, give way to Schwann cells. The transition zone between these two areas can measure approximately 2 mm in length and is located within 3 to 4 mm of the nerve root zone 7. This is relevant when assessing for neurovascular compression (causing trigeminal neuralgia) as the transition zone is far more vulnerable than other areas 7.
Intracranial extra-axial course
Having exited the mid pons anteriorly,the cisternal portion of the trigeminal nerve courses antero-superiorly through the prepontine cistern (cisternal portion), and crosses the porus trigeminusto enter a prolongation of dura at the apex of the petrous temporal bone known as the Meckel cave (cavernous portion) where its sensory fibers form the trigeminal ganglion (also known as the Gasserian or semilunar ganglion).It then divides into three main branches known as divisions: ophthalmic, maxillary and mandibular.
Ophthalmic division/nerve (V1 or Va)
Courses anteriorly in the lateral wall of the cavernous sinus inferior to the trochlear nerveand is crossed medially by the oculomotor nerve. Just before entering the orbit, the tentorial nerve arises and ascends to supply a large portion of the falx and supratentorial dura. The ophthalmic division then divides into 3 terminal branches before each passes through the superior orbital fissure separately:
frontal nerve (superior orbital fissure outside the tendinous ring)
supraorbital nerve (supraorbital notch)
supratrochlear nerve (supratrochlear notch)
lacrimal nerve (superior orbital fissure outside the tendinous ring)
nasociliary nerve (superior orbital fissure through the tendinous ring)
small communicating branch to the ciliary ganglion
short ciliary nerves
long ciliary nerves
infratrochlear nerve
posterior ethmoidal nerve (posterior ethmoidal foramen)
anterior ethmoidal nerve (anterior ethmoidal foramen then the cribriform plate)
Maxillary division/nerve (V2 or Vb)
Courses anteriorly low in the lateral wall of the cavernous sinus inferior to the ophthalmic division.Just before exiting the skull it runs along the floor of the middle cranial fossaand gives off the middle meningeal nerve which ascends to supply the anterior dura of the middle cranial fossa. It then passes through the foramen rotundum in the greater wing of the sphenoid bone to exit the skull and enter the superior aspect of the pterygopalatine fossa. It gives two branches (pterygopalatine nerves) to the pterygopalatine ganglion, but also receives parasympathetic nerves from the ganglion via the greater petrosal nerve. It then divides into the:
zygomatic nerve(inferior orbital fissure)
zygomaticotemporal nerve(zygomaticotemporal foramen)
zygomaticofacial nerve(zygomaticofacial foramen)
posterior superior alveolar nerve(pterygomaxillary fissure)
infraorbital nerve(inferior orbital fissure then infraorbital foramen)
middle superior alveolar nerve
anterior superior alveolar nerve
nasopalatine nerve(sphenopalatine foramen)
posterior superior nasal nerves(sphenopalatine foramen)
greater palatine nerve(greater palatine foramen)
lateral posterior inferior nasal nerve(unnamed foramen)
lesser palatine nerve(lesser palatine foramen)
pharyngeal nerve(palatovaginal canal)
Mandibular division/nerve (V3 or Vc)
Courses inferiorly through the foramen ovaleto enter the infratemporal fossa,hence it does not pass through the cavernous sinus. It consists of a sensory root and a smaller motor root, the latter which bypasses the trigeminal ganglion inferiorly. These roots pass through the foramen ovale separately and then unite just below the foramen.It immediately gives off nervus spinosus and nerve to medial pterygoidfrom the main trunk. It then descends in the infratemporal fossa passing between the tensor veli palatini and lateral pterygoid muscles before dividing into anterior and posterior divisions:
anterior division (4 branches,all motor except one)
deep temporal nerves
lateral pterygoid nerves
masseteric nerve
buccal nerve (sensory only)
posterior division (3 branches, all sensory except one)
auriculotemporal nerve
lingual nerve
inferior alveolar nerve
nerve to mylohyoid muscle (motor only)
incisive nerve
mental nerve
Variant anatomy
Variations can arise in any of the trigeminal nerve's major branches and its subsequent divisions.
ophthalmic division/nerve (V1)
the nasociliary nerve can distribute sensory branches to the muscles of the eye, i.e. superior rectus andmedial rectus11
maxillary division/nerve (V2)
the maxillary nerve can be bifid 12
mandibular division/nerve (V3)
unilateral and bilateral accessory foramen ovaleby which branches of the mandibular nerve travel directly to the deep muscles of theinfra temporal fossa (unilateral foramina being more common than bilateral foramina) 13
Related pathology
trigeminal neuralgia
trigeminal schwannoma
Quiz questions
Question 3408
Report problem with question
Which of the following does NOT pass through the cavernous sinus?
abducens nerve
facial nerve
internal carotid artery
oculomotor nerve
trochlear nerve
SubmitSkip question
References
1. Chummy S. Sinnatamby. Last's Anatomy. (2011) ISBN: 9780702033957 - Google Books
2. Carmine D. Clemente. Anatomy. (2011) ISBN: 9781582558899 - Google Books
3. Kyung Won Chung, Harold M. Chung. Gross Anatomy. (2012) ISBN: 9781605477459 - Google Books
4. Sheth S, Branstetter B, Escott E. Appearance of Normal Cranial Nerves on Steady-State Free Precession MR Images. Radiographics. 2009;29(4):1045-55. doi:10.1148/rg.294085743 - Pubmed
5. Robert H. Whitaker, Neil R. Borley. Instant Anatomy. (2000) ISBN: 9780632054039 - Google Books
6. Keith L. Moore, Arthur F. Dalley, A. M. R. Agur. Clinically Oriented Anatomy. (2013) ISBN: 9781451119459 - Google Books
7. Haller S, Etienne L, Kövari E, Varoquaux A, Urbach H, Becker M. Imaging of Neurovascular Compression Syndromes: Trigeminal Neuralgia, Hemifacial Spasm, Vestibular Paroxysmia, and Glossopharyngeal Neuralgia. AJNR Am J Neuroradiol. 2016;37(8):1384-92. doi:10.3174/ajnr.A4683 - Pubmed
8. Yousry I, Moriggl B, Holtmannspoetter M, Schmid U, Naidich T, Yousry T. Detailed Anatomy of the Motor and Sensory Roots of the Trigeminal Nerve and Their Neurovascular Relationships: A Magnetic Resonance Imaging Study. J Neurosurg. 2004;101(3):427-34. doi:10.3171/jns.2004.101.3.0427 - Pubmed
9. Joo W, Yoshioka F, Funaki T, Mizokami K, Rhoton A. Microsurgical Anatomy of the Trigeminal Nerve. Clin Anat. 2014;27(1):61-88. doi:10.1002/ca.22330 - Pubmed
10. FIPAT. Terminologia Anatomica. 2nd Ed. FIPAT.library.dal.ca. Federative International Programme for Anatomical Terminology, 2019.
11. Blodi F. Eugene Wolff's Anatomy of the Eye and Orbit. Arch Ophthalmol. 1977;95(7):1284. doi:10.1001/archopht.1977.04450070182024
12. Somayaji K & Rao M. Anatomy and Clinical Applications of the Maxillary Nerve in Dentistry: A Literature Review. Dent Update. 2012;39(10):727-30, 733-5. doi:10.12968/denu.2012.39.10.727 - Pubmed
13. Krmpotić-Nemanić J, Vinter I, Jalšovec D. Accessory Oval Foramen. Annals of Anatomy - Anatomischer Anzeiger. 2001;183(3):293-5. doi:10.1016/s0940-9602(01)80237-5 - Pubmed
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Inferolateral trunk
Posterior superior alveolar nerve
Dura mater
Temporalis muscle
Cavernous sinus haemangioma
Otic ganglion
Nervus spinosus
Superior salivary nucleus
Infraorbital foramen
Transition zone (nerve)
Gasperini syndrome
Digastric muscle
Palatovaginal canal
Muscles of the soft palate
Foramen ovale contents (mnemonic)
Superior orbital fissure
Inferior alveolar nerve
Cavernous sinus contents (mnemonic)
Pharyngeal nerve
Petrous part of temporal bone
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Cases:
Trigeminal schwannoma
Neurovascular compression syndrome - trigeminal nerve
Trigeminal neuralgia
Neurovascular compression of trigeminal nerve
Schwannoma of the maxillary nerve
Trigeminal amyloidoma
Trigeminal neuropathy due to ADEM
Trigeminal neuralgia due to vertebrobasilar dolichoectasia
Cranial nerves and brainstem nuclei (illustrations)
Bilateral stenosis of the foramina of Monro
Trigeminal schwannoma
Bilateral plexiform neurofibromas of the trigeminal and facial nerves - NF1
Epidermoid cyst of the cerebellopontine cistern
Teflon microvascular decompression
Trigeminal nerve cutaneous distribution (Gray's anatomy)
Mandibular division of the trigeminal nerve and submandibular and otic ganglia (Gray's illustration)
Mandibular division of the trigeminal nerve (Gray's illustration)
Pterygopalatine ganglion (Gray's illustration)
Maxillary division of the trigeminal nerve (Gray's illustration)
Maxillary and mandibular divisions of the trigeminal nerve (Gray's illustration)
Load more cases
Multiple choice questions:
Question 3408
Question 3299
Question 3182
Question 2679
Question 2621
Question 2136
Question 1993
Question 1874
Question 1574
Question 1455
Question 1454
Question 52
Related articles: Anatomy: Brain
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neostriatum
nucleus accumbens
canal of Gratiolet
pituitary gland
posterior pituitary and stalk(part of diencephalon)
ectopic posterior pituitary
anterior pituitary
inferior hypophyseal arterial circle
diencephalon
thalamencephalon
thalamus
interthalamic adhesion
lateral geniculate nucleus
medial geniculate nucleus
metathalamus
epithalamus
habenula
stria medullaris
pineal gland
subthalamus
subthalamic nuclei
hypothalamus
supraoptic nucleus
paraventricular nucleus
mammillary bodies
tuber cinereum
brainstem[+][+]
midbrain(mesencephalon)
tectal plate
tegmentum
cerebral peduncles
corpora quadrigemina
posterior perforated substance
periaqueductal grey matter
pons(part of metencephalon)
facial colliculus
superior olivary nucleus
medulla oblongata(myelencephalon)
olive
inferior olivary nucleus
nucleus of the tractus solitarius
nucleus ambiguus
dorsal vagal motor nucleus
white matter
grey matter
non-cranial nerve
substantia nigra
red nucleus
superior colliculi
inferior colliculi
cranial nerve nuclei
oculomotor nucleus
Edinger-Westphal nucleus
trochlear nucleus
motor nucleus of CN V
mesencephalic nucleus of CN V
main sensory nucleus of CN V
spinal nucleus of CN V
abducent nucleus
facial nucleus
superior salivatory nucleus
cochlear nuclei
vestibular nuclei
inferior salivatory nucleus
solitary tract nucleus
ambiguus nucleus
dorsal vagal motor nucleus
hypoglossal nucleus
cerebellum(part of metencephalon)[+][+]
vermis
cerebellar hemisphere
cerebellar tonsil
dentate nucleus
cerebellar peduncles
superior cerebellar peduncle
middle cerebellar peduncle
inferior cerebellar peduncle
cranial meninges(meninx primitiva)[+][+]
dura mater(pachymeninx)
falx cerebri
tentorium cerebelli
falx cerebelli
petroclinoid ligaments
diaphragma sellae
carotid cave
leptomeninges
arachnoid mater
Liliequest membrane
arachnoid granulations
aberrant arachnoid granulations
subarachnoid lymphatic-like membrane
pia mater
velum interpositum
cavum velum interpositum
extradural space
subdural space
subarachnoid space
blood supply of the meninges
innervation of the meninges
CSF spaces[+][+]
ventricular system
lateral ventricles
septum pellucidum
cavum septum pellucidum
cavum vergae
interventricular foramen (of Monro)
choroidal fissure
third ventricle
lamina terminalis
supraoptic recess
infundibular recess
pineal recess
suprapineal recess
aqueduct of Sylvius
fourth ventricle
superior medullary velum
inferior medullary velum
obex
area postrema
rhomboid fossa
calcar avis
foramen of Magendie
foramina of Lushka
tela choroidea
cerebrospinal fluid
choroid plexus
Bochdalek's flower basket
subarachnoid cisterns
suprasellar cistern
empty sella sign
interpeduncular cistern
ambient cistern
transverse fissure
quadrigeminal cistern
pericallosal cistern
prepontine cistern
cerebellopontine cistern
premedullary cistern
Sylvian cistern
cisterna magna
mega cisterna magna
Meckel cave
cranial nerves (mnemonic)
olfactory nerve (CN I)
optic nerve (CN II)[+][+]
optic chiasm
optic tract
oculomotor nerve (CN III)
trochlear nerve (CN IV)
trigeminal nerve (CN V) (mnemonic)
trigeminal ganglion
abducens nerve (CN VI)
facial nerve (CN VII) (segments mnemonic | branches mnemonic)[+][+]
geniculate ganglion
greater (superficial) petrosal nerve
Vidian nerve
external petrosal nerve
nerve to stapedius
nervus intermedius
vestibulocochlear nerve (CN VIII)[+][+]
vestibular ganglion (Scarpa's ganglion)
glossopharyngeal nerve (CN IX)[+][+]
Jacobson nerve
lesser petrosal nerve
vagus nerve (CN X)[+][+]
Arnold nerve
spinal accessory nerve (CN XI)
hypoglossal nerve (CN XII)
functional neuroanatomy[+][+]
Brodmann areas
cortical motor system
extrapyramidal system
cortical sensory system
pain and temperature sensation
vibration and proprioception sensation
auditory/speech system
Broca's area (Brodmann area 44)
Wernicke's area (Brodmann area 22)
visual system
olfactory system
default mode network
CNS development[+][+]
brain development
neural plate
neural tube
prosencephalon
telencephalon
paraphysis elements
Rathke pouch
diencephalon
mesencephalon
rhombencephalon
metencephalon
myelencephalon
cerebral vascular supply[+][+]
arteries
vascular territories
anterior circulation
posterior circulation
circle of Willis
internal carotid artery (ICA)(segments)
inferolateral trunk
meningohypophyseal trunk
inferior hypophyseal artery
capsular arteries (of McConnell) (variable)
superior hypophyseal artery
anterior cerebral artery (ACA)
anterior communicating artery (ACOM)
medial lenticulostriate arteries
recurrent artery of Heubner
medial frontobasal artery
frontopolar artery
orbitofrontal artery
pericallosal artery
callosomarginal artery
pericallosal moustache
middle cerebral artery (MCA)
M1 branches
lenticulostriate arteries
medial lenticulostriate arteries
lateral lenticulostriate arteries
anterior temporal artery
temporopolar artery
M2 branches
posterior communicating artery (PCom)
ophthalmic artery
lacrimal artery
supraorbital artery
posterior ethmoidal artery
anterior ethmoidal artery
internal palpebral artery
supratrochlear artery(frontal artery)
dorsal nasal artery
posterior ciliary arteries
short posterior ciliary arteries
circle of Zinn
long posterior ciliary arteries
anterior choroidal artery
vertebral artery
posterior inferior cerebellar artery (PICA)
basilar artery
anterior inferior cerebellar artery (AICA)
labyrinthine artery
pontine arteries
superior cerebellar artery (SCA)
posterior cerebral artery (PCA)
calcarine artery
splenial artery
posterior choroidal artery
medial posterior choroidal artery
lateral posterior choroidal artery
normal variants
intracranial arterial fenestration
internal carotid artery (ICA)
absent ICA
aberrant ICA
anterior cerebral artery (ACA)
azygos ACA
middle cerebral artery (MCA)
accessory MCA
duplicated MCA
MCA fenestration
posterior cerebral artery (PCA)
fetal origin of PCA / fetal PCOM
artery of Percheron
basilar artery
basilar artery fenestration
basilar artery hypoplasia
persistent carotid-vertebrobasilar artery anastomoses(mnemonic)
persistent primitive trigeminal artery
persistent otic artery
persistent hypoglossal artery
persistent proatlantal artery
persistent proatlantal intersegmental artery
vertebral artery
vertebral artery hypoplasia
ophthalmic artery
meningo-ophthalmic artery
cerebral venous system
dural venous sinuses
basilar venous plexus
cavernous sinus(mnemonic)
clival diploic veins
inferior petro-occipital vein
inferior petrosal sinus
inferior sagittal sinus
intercavernous sinus
internal carotid artery venous plexus of Rektorzik
jugular bulb
marginal sinus
occipital sinus
sigmoid sinus
sphenoparietal sinus
straight sinus
superior petrosal sinus
superior petrosal vein
superior sagittal sinus
torcula herophili
transverse sinus
cerebral veins
superficial veins of the brain
superior cerebral veins (superficial cerebral veins)
inferior cerebral veins
superficial middle cerebral vein
superior anastomotic vein(of Trolard)
inferior anastomotic vein (of Labbe)
deep veins of the brain
great cerebral vein (of Galen)
basal vein of Rosenthal
lateral mesencephalic vein
peduncular vein
anterior pontomesencephalic vein
transverse pontine veins
anterior medullary vein
internal cerebral vein
thalamostriate vein
septal cerebral vein
superior cerebellar vein
precentral cerebellar vein
superior vermian vein
venous circle of Trolard
normal variants
persistent falcine sinus
glymphatic pathway
Related articles: Anatomy: Head and neck
Anatomy: Head and neck
skeleton of the head and neck
cranial vault
scalp (mnemonic)
galea aponeurotica
fontanelle
anterior fontanelle
posterior fontanelle
anterolateral (sphenoidal) fontanelle
posterolateral (mastoid)fontanelle
sutures
calvarial
coronal suture
sagittal suture
lambdoid suture
accessory occipital bone sutures
metopic suture
squamosal suture
sphenosquamosal suture
squamomastoid suture
facial
frontozygomatic suture
frontomaxillary suture
frontolacrimal suture
frontonasal suture
temporozygomatic suture
zygomaticomaxillary suture
parietotemporal suture(parietomastoid suture)
occipitotemporal suture (occipitomastoid suture)
sphenofrontal suture
sphenozygomatic suture
spheno-occipital suture(not a true suture)
lacrimomaxillary suture
nasomaxillary suture
internasal suture
basal/internal
frontoethmoidal suture
petrosquamous suture
petroclival suture
sphenoethmoidal suture
sphenopetrosal suture
skull landmarks
nasion
glabella
bregma
vertex
lambda
inion
pterion
asterion
basion
opisthion
obelion
frontal bone
supratrochlear foramen
supraorbital foramen
temporal bone
squamous part
MacEwen triangle
mandibular fossa
sigmoid plate
petrous part
petrous apex
Fallopian canal
jugular fossa
inferior tympanic canaliculus
petrous ridge
Dorello canal
petromastoid canal
mastoid part
mastoid antrum
mastoid air cells
Koerner septum
mastoid canaliculus
mastoid foramen
tympanic part
tympanosquamous fissure
styloid process
stylomastoid foramen
styloid apparatus
stylohyoid ligament
parietal bone
parietal foramen
occipital bone
clivus
inferior median clival canal
foveolapharyngicarecess
occipital condyle
bathrocephaly
skull base (foramina)
anterior cranial fossa
anterior ethmoidal foramen
posterior ethmoidal foramen
foramen cecum
cribriform plate
middle cranial fossa (mnemonic)
foramen rotundum
foramen ovale (mnemonic)
foramen spinosum
foramen Vesalii
foramen lacerum
carotid canal
pterygoid canal
posterior cranial fossa
condylar canal
jugular foramen
jugular spine
hypoglossal canal
foramen magnum
facial bones
midline single bones
sphenoid bone
body
pituitary fossa
sella turcica
bridging of the sella turcica
dorsum sellae
pneumatized dorsum sellae
tuberculum sellae
optic strut
persistent hypophyseal canal
vomerovaginal canal
jugum sphenoideum
lesser wing
greater wing
pterygoid processes
palatovaginal canal
ethmoid bone
cribriform plate
crista galli
olfactory fossa
Keros classification
labyrinth of ethmoid
lamina papyracea
vomer
mandible
temporomandibular joint
articular disc
retrodiscal zone
pterygoid fovea
lingula
sphenomandibular ligament
stylomandibular ligament
mandibular foramen
mandibular canal
mental foramen
paired bilateral bones
maxilla
incisive canal
incisive foramen
palatine bone
sphenopalatine foramen
greater palatine foramen (canal)
lesser palatine foramina (canal)
nasal bone
lacrimal bone
zygoma (zygomatic bone)
zygomaticofacial foramen
zygomaticotemporal foramen
zygomatic arch
cervical spine
hyoid bone
laryngeal cartilages
arytenoid cartilage
corniculate cartilage
cuneiform cartilage
cricoid cartilage
thyroid cartilage
muscles of the head and neck
muscles of the tongue(mnemonic)
extrinsic muscles of the tongue
genioglossus muscle
hyoglossus muscle
styloglossus muscle
palatoglossus muscle
intrinsic muscles of the tongue
superior longitudinal muscle of the tongue
inferior longitudinal muscle of the tongue
transverse muscle of the tongue
vertical muscle of the tongue
muscles of mastication
temporalis muscle
masseter muscle
medial pterygoid muscle
lateral pterygoid muscle
facial muscles
epicranius muscle
occipitofrontalis muscle
frontalis muscle
occipitalis muscle
temporoparietalis muscle
circumorbital and palpebral muscles
orbicularis oculi muscle
corrugator supercilii muscle
levator palpebrae superioris muscle
nasal muscles
procerus muscle
nasalis muscle
compressor naris muscle
dilator naris muscle
myrtiformis muscle
depressor septi nasalis muscle
levator labii superioris alaeque nasalis muscle
buccolabial muscles
elevators, retractors and evertors of the upper lip
levator labii superioris alaeque nasalis muscle
levator labii superioris muscle
zygomaticus major muscle
zygomaticus minor muscle
levator anguli oris muscle
malaris muscle
risorius muscle
depressors, retractors and evertors of the lower lip
depressor labii inferioris muscle
depressor anguli oris muscle
mentalis muscle
compound sphincter
orbicularis oris muscle
incisivus labii superioris muscle
incisivus labii inferioris muscle
muscle of mastication
buccinator muscle
modiolus
muscles of the middle ear
stapedius muscle
tensor tympani muscle
orbital muscles
extraocular muscles
superior rectus muscle
inferior rectus muscle
lateral rectus muscle
medial rectus muscle
superior oblique muscle
inferior oblique muscle
levator palpebrae superioris muscle
superior tarsal muscle
muscles of the soft palate
tensor veli palatini muscle
levator veli palatini muscle
palatopharyngeusmuscle
palatoglossus muscle
muscle of the uvula
pharyngeal muscles
superior pharyngeal constrictor muscle
Passavant cushion
middle pharyngeal constrictor muscle
inferior pharyngeal constrictor muscle
Killian dehiscence
stylopharyngeus muscle
salpingopharyngeus muscle
suprahyoid muscles
digastric muscle
geniohyoid muscle
mylohyoid muscle
mylohyoid boutonniere
stylohyoid muscle
infrahyoid muscles
sternohyoid muscle
sternothyroid muscle
thyrohyoid muscle
omohyoid muscle
intrinsic muscles of the larynx
muscles of the neck
platysma muscle
longus colli muscle
longus capitis muscle
scalenus anterior muscle
colliscalene triangle
scalenus medius muscle
scalenus posterior muscle
scalenus pleuralis muscle
sternocleidomastoid muscle
suboccipital muscles
rectus capitis posterior major muscle
rectus capitis posterior minor muscle
obliquus capitis superior muscle
obliquus capitis inferior muscle
accessory muscles of the neck
levator glandulae thyroideae muscle
deep cervical fascia
superficial layer of the deep cervical fascia
middle layer of the deep cervical fascia
pharyngobasilar fascia
sinus of Morgagni
deep layer of the deep cervical fascia
deep spaces of the neck
anterior cervical space
buccal space
carotid space
carotid sheath
danger space
deep cervical fascia
alar fascia
infratemporal fossa
masticator space
parapharyngeal space
stylomandibular tunnel
parotid space
pharyngeal (superficial) mucosal space
perivertebral space
posterior cervical space
pterygopalatine fossa
pterygomaxillary fissure
retropharyngeal space
suprasternal space (of Burns)
visceral space
surgical triangles of the neck
anterior triangle
digastric triangle
carotid triangle
muscular triangle
submental triangle
posterior triangle
occipital triangle
suboccipital triangle
supraclavicular triangle
orbit
bony orbit
anterior ethmoidal notch
orbital apex
optic canal
superior orbital fissure (mnemonic)
inferior orbital fissure
infraorbital foramen
tendinous ring
supraorbital ridge
orbital spaces
extraconal
myofascial cone
intraconal
ocular globe
conjunctiva
cornea
sclera
uvea
iris
ciliary body
choroid
ora serrata
vitreous body
Cloquet's canal
lens
optic nerve-sheath complex
fovea
orbital septum
naso-orbitoethmoid region
eye movements
ocular adductors
ocular abductors
ocular elevators
ocular depressors
ocular internal rotators
ocular external rotators
orbital blood supply
ophthalmic artery
superior ophthalmic vein
inferior ophthalmic vein
orbital nerve supply
lacrimal apparatus
lacrimal gland
nasolacrimal drainage apparatus
lacrimal canaliculi
lacrimal sac
nasolacrimal duct
ear
inner ear
internal acoustic meatus(mnemonic)
Bill bar
falciform crescent
porus acusticus internus
labyrinth
osseous labyrinth
cochlea
modiolus
spiral lamina
helicotrema
cochlear aqueduct
semicircular canals
vestibule
macula cribrosa
vestibular aqueduct
perilymph
membranous labyrinth
cochlear duct (scala media)
organ of Corti (Spiral organ)
scala vestibuli
scala tympani
perilymphatic duct
semicircular ducts
saccule
utricle
endolymphatic duct
endolymph
middle ear
hypotympanum
Eustachian tube
mesotympanum
cochlear promontory
fissula ante fenestram
facial recess
pyramidal eminence
sinus tympani
oval window
round window
ossicular chain
malleus
incus
stapes
incudomalleolar joint
incudostapedial joint
stapediovestibular joint
epitympanum
anterior epitympanic recess
Prussak space
external ear
external auditory meatus
foramen tympanicum
porus acusticus externus
external auditory canal
scutum
tympanic membrane
pars tensa
pars flaccida
tympanic annulus
paranasal sinuses
frontal sinuses
maxillary sinuses
maxillary ostium
accessory maxillary ostium
ethmoid air cells (sinuses)
ethmoidal infundibulum
ethmoid bulla
bulla lamella
suprabullar recess
retrobullar recess
agger nasi air cells
infraorbital ethmoidal air cells
sphenoethmoidal air cells
supraorbital air cells
sphenoid sinuses
sphenoethmoidal recess
extramural air cell
ostiomeatal complex
ostiomeatal narrowing due to variant anatomy
upper respiratory tract
nose
external nose
nasal cartilages
nasal vestibule
anterior naris
nasal ala
nasal sill
columella
nasal septum
pyriform aperture
nasal cavity
inferior meatus
Woodruff plexus
nasal concha
inferior
middle
concha bullosa
paradoxical middle turbinate
hiatus semilunaris
uncinate process
superior
supreme
Kiesselbach plexus
Schneiderian epithelium
oral cavity
palate
hard palate
soft palate
uvula
floor of mouth
retromolar trigone
sublingual space
submandibular space
submental space
tongue
root of tongue
teeth(dental terminology)
supernumerary teeth
mesiodens
hypodontia
pharynx
nasopharynx
Rosenmüller fossa
torus tubarius
oropharynx
vallecula
hypopharynx
pyriform sinus (fossa)
larynx
supraglottic space
aryepiglottic folds
epiglottis
omega epiglottis
false vocal cords
laryngeal ventricle
laryngeal saccule
laryngeal vestibule
glottis
subglottic space
true vocal cords
anterior commissure of the larynx
viscera of the neck
Waldeyer ring
nasopharyngeal tonsils (adenoids)
palatine tonsils
lingual tonsils
thyroid gland
pyramidal lobe of thyroid
ectopic thyroid
Zuckerkandl tubercle
thyroglossal duct
parathyroid gland
salivary glands and ducts
major salivary glands
parotid gland
parotid duct
accessory parotid gland
submandibular gland
submandibular (Wharton) duct
sublingual gland
duct of Rivinus
minor salivary glands
esophagus
trachea
blood supply of the head and neck
arterial supply
common carotid artery
carotid body
carotid bifurcation
internal carotid artery (segments)
artery of Bernasconi and Cassinari
caroticotympanic artery
persistent stapedial artery
ophthalmic artery
supraorbital artery
lacrimal artery
central artery of the retina
supratrochlear artery
dorsal nasal artery
external carotid artery (mnemonic)
superior thyroid artery
superior laryngeal artery
ascending pharyngeal artery
lingual artery
facial artery
inferior labial artery
superior labial artery
angular artery
occipital artery
posterior auricular artery
(internal) maxillary artery (mnemonic)
deep auricular artery
middle meningeal artery
greater (descending) palatine artery
inferior alveolar artery
mental artery
sphenopalatine artery
anterior ethmoidal artery
posterior ethmoidal artery
infraorbital artery
masseteric artery
buccinator artery
deep temporal arteries
Vidian artery
superficial temporal artery
subclavian artery
vertebral artery
internal thoracic artery
thyrocervical trunk
inferior thyroid artery
suprascapular artery
ascending cervical artery
transverse cervical artery
dorsal scapular artery
costocervical trunk
variants
thyroidea ima artery
venous drainage
internal jugular vein (mnemonic)
jugular bulb
asymmetrically large jugular bulb
pharyngeal vein
common facial vein
angular vein
supraorbital vein
supratrochlear vein
lingual vein
superior thyroid vein
middle thyroid vein
external jugular vein (mnemonic)
posterior auricular vein
retromandibular vein
superficial temporal vein
(deep/internal) maxillary vein
pterygoid venous plexus
facial vein
anterior jugular vein
posterior external jugular vein
suprascapular vein
transverse cervical vein
cavernous sinus
facial-cavernous anastomoses
anterior condylar confluence
innervation of the head and neck
cranial nerves
olfactory nerve (CN I)
optic nerve (CN II)
oculomotor nerve (CN III)
trochlear nerve (CN IV)
trigeminal nerve (CN V)(mnemonic)
trigeminal ganglion
ophthalmic division
tentorial nerve
frontal nerve
supra-orbital nerve
supratrochlear nerve
lacrimal nerve
nasociliary nerve
small communicating branch to the ciliary ganglion
short ciliary nerves
long ciliary nerves
infratrochlear nerve
posterior ethmoidal nerve
anterior ethmoidal nerve
maxillary division
middle meningeal nerve
zygomatic nerve
zygomaticotemporal nerve
zygomaticofacial nerve
infra-orbital nerve
posterior superior alveolar nerve
middle superior alveolar nerve
anterior superior alveolar nerve
nasopalatine nerve
posterior superior nasal nerves
greater palatine nerve
lateral posterior inferior nasal nerve
lesser palatine nerves
pharyngeal nerve
mandibular division
nervus spinosus
nerve to medial pterygoid
anterior division
deep temporal nerves
lateral pterygoid nerves
masseteric nerve
buccal nerve
posterior division
auriculotemporal nerve
lingual nerve
inferior alveolar nerve
nerve to mylohyoid
incisive nerve
mental nerve
abducens nerve (CN VI)
facial nerve (CN VII)
chorda tympani
nerve to stapedius
vestibulocochlear nerve (CN VIII)
vestibular ganglion(Scarpa's ganglion)
glossopharyngeal nerve (CN IX)
Jacobson nerve
vagus nerve (CN X)
superior laryngeal nerve
external laryngeal nerve
internal laryngeal nerve
recurrent laryngeal nerve(inferior laryngeal nerve)
non-recurrent laryngeal nerve
(spinal) accessory nerve (CN XI)
hypoglossal nerve (CN XII)
parasympathetic gangliaof the head and neck
ciliary ganglion
pterygopalatine ganglion
otic ganglion
submandibular ganglion
cervical sympathetic ganglia
superior cervical ganglion
middle cervical ganglion
inferior cervical ganglion
stellate ganglion
greater occipital nerve
third occipital nerve
cervical plexus
muscular branches
longus capitis
longus colli
scalenes
scalenus anterior
scalenus medius
scalenus posterior
geniohyoid
thyrohyoid
ansa cervicalis
omohyoid (superior and inferior bellies separately)
sternothyroid
sternohyoid
phrenic nerve
accessory phrenic nerve
contribution to the accessory nerve (CN XI)
cutaneous branches
less occipital nerve
greater auricular nerve
transverse cervical nerve
supraclavicular nerve
brachial plexus
pharyngeal plexus
lymphatic drainage of the head and neck
cervical lymph node groups
Delphian node
intraparotid lymph nodes
jugular trunk
cervical lymph node levels
supraclavicular lymph nodes
embryological development of the head and neck
branchial apparatus
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Cases and figures
Figure 1: ophthalmic division
Figure 2: maxillary division
Figure 3: mandibular division
Figure 4: sensation
Figure 5: motor nucleus
Figure 6: cranial nerve origins
Figure 7: upper pons - CN V (diagram)
Figure 8: mid pons - CN V (diagram)
Figure 9: lower pons - CN V (diagram)
Figure 10: connections
Figure 11: cavernous sinus
Figure 12: trigeminal nerve
Figure 13: cranial nerves
Figure 14: ophthalmic division
Figure 15: nerves of the orbit
Figure 16: maxillary and mandibular divs
Figure 17: maxillary division
Figure 18: pterygopalatine ganglion
Figure 19: mandibular division
Figure 20: mandibular division
Figure 21: nerves of face, scalp, neck
Figure 22: cranial nerves
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16874 | https://brainly.com/question/474424 | [FREE] True or false: The equation \sec^2 x - 1 = \tan^2 x is an identity. - brainly.com
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True or false: The equation sec 2 x−1=tan 2 x is an identity.
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Use the trigonometric identities:
sec x=cos x 1tan x=cos x sin xsin 2 x+cos 2 x=1
sec 2 x−1=tan 2 x(cos x 1)2−1=(cos x sin x)2 cos 2 x 1−1=cos 2 x sin 2 x∣×cos 2 x 1−cos 2 x=sin 2 x∣+cos 2 x sin 2 x+cos 2 x=1 true
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Physics for K-12 - Sunil Singh
Introductory Physics - Building Models to Describe Our World - Ryan D. Martin, Emma Neary, Joshua Rinaldo, Olivia Woodman
University Physics Volume 1 - William Moebs, Samuel J. Ling, Jeff Sanny
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The equation sec 2 x−1=tan 2 x is a true identity. This is verified through the use of trigonometric identities and the Pythagorean identity. Thus, it holds for all values of x where these functions are defined.
Explanation
To determine if the equation sec 2 x−1=tan 2 x is an identity, we can utilize trigonometric identities. Recall that: sec x=c o s x 1 and tan x=c o s x s i n x.
Starting with the left side of the equation:
sec 2 x−1=(cos x 1)2−1
=cos 2 x 1−1
=cos 2 x 1−cos 2 x
We know from the Pythagorean identity that sin 2 x+cos 2 x=1. Thus, we can express 1−cos 2 x as sin 2 x:
=cos 2 x sin 2 x
=tan 2 x
Now we can write:
sec 2 x−1=tan 2 x
This shows that the left side equals the right side for all values of x where these functions are defined. Thus, we conclude that sec 2 x−1=tan 2 x is indeed a true identity.
Examples & Evidence
For example, if you plug in x=0, both sides of the equation yield 0 and confirm their equality. Similarly, at other values like x=4 π, both sides also yield 1, further supporting the identity.
The use of fundamental trigonometric identities, specifically the Pythagorean identity, supports that the relationship between secant and tangent is accurately represented in the given equation.
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16875 | https://artsandculture.google.com/usergallery/OwKykw3me9WaIw | The four seasons: a journey through time — Google Arts & Culture
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The four seasons: a journey through time
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This user gallery has been created by an independent third party and may not represent the views of the institutions whose collections include the featured works or of Google Arts & Culture.
Spring, summer, autumn, and winter. In times before we used the seasons to enumerate the time we had lived. In years long gone the seasons controlled whether we lived or died. This exhibit “The Four Seasons: A Journey Through Time”, encapsulates the feeling of each quarter of the year and carries the viewer across various times and cultures while we all take a journey through a cosmic year. The works selected are representations of the seasons by various cultures from around the world. For example, take the 145.2cm by 94.7cm oil on silk painting Mountain Village in Spring by Kawai Gyokudo. The warm yellow and natural landscape exemplifies the feeling of renewal and rejuvenation that is closely associated with spring. Very much akin to Kawai Gyokudo’s work, there is the oil painting Blow Blow Thou Winter Wind by John Everett Millais. Although the subject matter is a different season, Millais’ work masterfully represents the feelings of isolation and scarcity that are linked to winter. The feeling of impending death or the end is also portrayed by Millais’ work. Both of these artists skillfully express the many emotions that are felt with the respective seasons through their works.
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Spring Evening, Arnold Böcklin, 1879, From the collection of: Museum of Fine Arts, Budapest
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Mountain Village in Spring, KAWAI Gyokudo, c.1913, From the collection of: The Museum of Modern Art, Saitama
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Summer Day, Gloucester, Massachusetts, Willard LeRoy Metcalf, 1895, From the collection of: Huntington Museum of Art
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Summer, Giuseppe Arcimboldo, 1563, From the collection of: Kunsthistorisches Museum Wien
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Golden autumn. Slobodka, Isaac Levitan, 1889, From the collection of: The State Russian Museum
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Autumn on the Hudson, Gifford Beal, 1940, From the collection of: Hudson River Museum
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Hunters in the Snow (Winter), Pieter Bruegel the Elder, 1565, From the collection of: Kunsthistorisches Museum Wien
Blow Blow Thou Winter Wind, John Everett Millais, 1892, From the collection of: Auckland Art Gallery Toi o Tāmaki
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This user gallery has been created by an independent third party and may not represent the views of the institutions whose collections include the featured works or of Google Arts & Culture.
Huntington Museum of Art Museum of Fine Arts, Budapest Kunsthistorisches Museum Wien The Museum of Modern Art, Saitama Auckland Art Gallery Toi o Tāmaki Hudson River Museum The State Russian Museum
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16876 | https://www.youtube.com/watch?v=s1V_Tfzmm3o | 2.1 Displacement, Velocity, and Acceleration | General Physics
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Chad provides an introduction to Kinematics with the topics of displacement, velocity, and acceleration. Displacement is defined as the change in position or the straight line distance between two points and the associated direction and has SI units of meters (m). Displacement is a vector quantity and is contrasted with distance which is a scalar. Velocity is defined as the displacement divided by the change in time and has SI units of meters per second (m/s). Velocity is a vector quantity, having both magnitude and direction, and is contrasted with speed which is a scalar. The concepts of average velocity and instantaneous velocity are also introduced, and it is shown that the magnitude of the instantaneous velocity is equal to the slope of a line tangent to the curve of a plot of position vs time. Acceleration is defined as the change in the velocity divided by the change in time and has SI units of meters per second squared (m/s^2). Acceleration is a vector quantity, having both magnitude and direction.
Chad also introduces motion diagrams to compare and contrast zero velocity, constant velocity, and varying velocity. He presents similar motion diagrams for acceleration to compare and contrast zero acceleration (i.e. constant velocity), uniform acceleration (i.e. constant acceleration), and varying acceleration.
00:00 Lesson Introduction
00:50 Displacement and Displacement vs Distance
03:59 Velocity and Velocity vs Speed
13:03 Acceleration
21:08 Introduction to Kinematics Calculations
Check out Chad's General Physics Master Course:
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Transcript:
Lesson Introduction displacement velocity and acceleration going to be the topics of this lesson we're getting into chapter two of my new General Physics playlist and this is a chapter on Motion in one dimension now some might call that kinematics in one dimension either way so the key is we're going to focus on one dimension we're going to actually start doing some real physics so chapter one was all about equipping us with some of the tools we need to do physics like units and vectors but now we're going to start doing some actual physics my name is Chad and welcome toad ad prep where my goal is to take the stress out of learning science now if you're new to the channel we've got comprehensive playlists for General chemistry organic chemistry General Physics and high school chemistry and on Chads prep.com you'll find premium Master courses for the same that include study guides and a ton of practice you'll also find comprehensive prep courses for the DAT the MCAT and the oat all right so we're Displacement and Displacement vs Distance going to start with displacement so and first thing you need to know is that displacement is a vector not a scaler and you might remember that that means it has both magnitude and Direction so whereas scalers have only magnitude so it has both magnitude and Direction and we can contrast this with distance so distance has only magnitude it's just a scalar it doesn't have a direction associated with it so you might think of displacement as the change in position which is why it's often symbolized as Delta X now in Motion in one dimension we're only going to consider motion happening in one dimension largely so and and X is kind of the easiest first Dimension to introduce but technically displacement could be defined as say Delta y if it's in the y direction and things of this sort so however it's customary to introduce it simply as Delta X for displacement here and technically again it's displacement in the X Direction but in this particular chapter Motion in one dimension we're only going to be consuming uh uh we're only going to be considering one dimension and it'll be the X Dimension typically all right so displacement symbolize Delta X and the SI unit is meter it has units of length here and again want to contrast it with distance here so we've got two different uh paths here they both start at the origin and they both end up at the point 3 0 and so as a result the displacement is exactly the same in both cases you might think of displacement again as the straight line distance between your initial starting point and your final destination so and whether you actually take the straight line path doesn't matter displacement doesn't care it only matters where you start and where you end up where you start and where you end up and if it's the same in both cases then the displacement is going to be the same as well all right now the distance is definitely going to be different here notice the straight line path would be the shortest possible distance and its magnitude in this case would be the same as the magnitude of the displacement so however in this case here we can see that this distance traveled is actually going to be longer on that pathway even though the displacement is exactly the same so want to make sure you got that distinction again displ is a vector has both uh magnitude and Direction so distance is just a scalar with no Direction associated with it all right so if we took a look and said what is the displacement here so well in this case in going from the origin to 30 that's a displacement of three and assuming this was in units of meters we could say meters but we don't the units aren't given so we don't really know but it's three and we might say in the positive X Direction so notice it's not just magnitude we got to give a direction here and here it was in the positive X Direction it wasn't in the negative X Direction it was in the positive X Direction and notice I put this on a kind of classic cartisian here so but for the purpose of this problem we might just get rid of the y- axis and stuff like this although I really want to set this up for dealing with both the X and Y directions in the next chapter so uh why it's introduced this way cool same thing here if I said calculate the displacement it'd be exactly the same it' still be three in the positive X Direction so second we're going to move on to Velocity here and just like displacement velocity is a vector it has Velocity and Velocity vs Speed both magnitude and Direction now in this case I might tell you that I'm going down the freeway doing 65 miles hour so and that sounds like a velocity in a customary everyday vernacular we might actually call that a velocity however physics is really specific because I didn't include a Direction with that that is actually not a velocity it's not a vector it is simply a scaler and the scaler equivalent we call speed so speed is the scaler I can say I'm doing 65 mph yep yes you are that's uh magnitude no Direction but if I say I'm doing 65 miles hour due north that now is a vector quantity has both magnitude and Direction and that would be considered a velocity all right so uh we like to use miles per hour uh in driving and things of this sort but the SI unit is me/ second you might recall that that's a derived unit unit of length over time and that's actually how velocity is defined so the technical definition of velocity is an equation it's the change in position over the change in time you might recall that change in position that's displacement so it's really the displacement over the change in time now it works out quite often that you know your given time period you say from Time Zero to time 10 seconds well then delta T is going to be 10 seconds and so sometimes people leave this Delta off but it technically is a change in time because what if you didn't start at Time Zero what if you had some big journey and you decided Well from this point in the Journey to the End or you know something like that so it is you know proper to put the Delta there although if you see it left off on occasion no it just means they're starting from time zero so but technically this would be position final minus position initial over time final minus time initial the change in anything is always final minus initial and again if you've got time final minus time initial where the initial time is zero that's why it just turns into the time Final in such cases while you'll see it written sometimes improperly uh just like that you might even see me do it on occasion all right so there's your definition of velocity and again it has both magnitude and Direction so but there a couple different situations we might deal with when it comes to Velocity we might deal with constant velocity so and constant velocity means your velocity is not changing so and keep in mind this means you're still moving but your speed is not going up or down down if you will and I said speed I just want to make sure you realize the magnitude of your velocity is not changing and so it's just like you're cruising down the road doing 65 mph down the freeway and that speed is not changing that would be constant velocity so but it also means that your direction is not changing so it turns out that if your magnitude is not changing but your direction is like you're you're taking a turn at a constant speed that actually is not no longer considered constant velocity we have a a whole section on like rotation motion and things of this sort where we'll deal with such situations and so uh sometimes you can have your speed not changing and your velocity changing as in such cases where you're you're on a curve all right so but we might still in this section we're going to be dealing with really largely Motion in one dimension and so we don't have to worry about going around a curve and stuff just yet uh and so in this case when we deal with constant velocity the other option is going to be a varying velocity where the velocity is changing so what we're going to find is the only the only situation we're really going to consider is one the the velocity is varying uniformly it's either increasing or decreasing at a constant rate we'll find out that rate is what we call acceleration but we'll get there in a little bit so but that's the only other situation we're going to deal with and so we'll find out in such cases that it's advantageous to try and calculate the average velocity so and just like if you want to calculate the average of two things well you take like say you want to take uh the average of my weight today and my average of my weight tomorrow well you take my weight today and my weight tomorrow and add them together and divide by two and that's a essentially what we're going to do here as well in dealing with say a varying velocity is you're just going to take the initial velocity and the final velocity and add them together and divide by two or add them together and multiply by a half same diff you'll see it written both ways so another equation that'll be customary when dealing with a varying velocity on occasion as well all right so we've introduced a couple equations one was the definition of velocity and one is just convenient to use when we're dealing with a varying velocity so and now we're going to introduce some really simplistic motion diagrams uh the last lesson in this particular chapter is going to be all about motion diagrams we'll give a little more comprehensive treatment to it so but here we're definitely going to talk about some really introductory motion diagrams where we're plotting position on the y- axis versus time and I know this is a little confusing CU you're like you're plotting the X position on the Y AIS yes yes I am and it is customary sorry I didn't invent the rules here so but we're plotting the position in the only Dimension we're concerned with the X dimension on the y- axis and then time on the x-axis well we can see in this first graph we've just got a horizontal line notice that means that the position never changes as time progresses so well if you got no change in position that means you've got no displacement and if you got a big zero for your displacement well then your velocity is zero as well you might recall that slope is defined as rise over run or the change in the Y over the change in the X well in this case the change in the Y and the change in the x is going to correspond to the change in position over the change in time well the change in position over the change in time that is your definition of velocity so the slope on this position versus time graph is velocity and so if you've got a horizontal line that means your slope is zero indicating that your velocity is zero another way to recognize that all right the next one here so we've got a nonzero slope now so it is a positive slope it's uphill and in this case that's going to correspond to a positive velocity so also because it's a straight line it's a constant slope and again on a position versus time graph your rise over your run is your change in position over change of time it is your velocity and so if you've got a constant slope you've got a constant velocity now you might being tempted to be like well sh the velocity was constant back here well yes constantly zero but I definitely want to distinguish between a velocity that is Zero versus an actual constant velocity and you should see that you know in this case with a positive slope it means your position's going up as time progresses you're moving forward it's often customary in dealing with Motion in one dimension to Define your direction as just the positive X Direction versus the negative X Direction and so the the sign often is your direction so to speak so in this case we have a positive slope that is a positive velocity which in this case would mean we're moving forward in the X Direction in the positive X Direction whereas had we had a downhill slope a negative slope that would be a negative velocity which again in Motion in one dimension would mean we're moving backwards in the negative X Direction so to speak all right finally we've got a situation where we've got a slope that is not constant in this case and because your slope is not constant again the slope is equal to your velocity if the slope's not constant your velocity is not constant we'd call this a varying velocity now one thing to note you can tell here that anywhere along this graph here your slope is positive the entire time and you can see just in terms of X versus T you say well what's actually happening here um well your position is going up as time progresses that's true so this thing is moving forward but we can also see a little more than that and so in such cases where your velocity varies it's often customary to talk about what's known as the instantaneous velocity so let's say I want to know what's the instantaneous velocity right at that point along the way and it turns out what you do is you draw a line tangent to your curve right at that point so and the slope of that tangent so and again slope on your position versus time graph equals velocity the slope of that tangent line is the instantaneous slope at that point and is equal to the instantaneous velocity at that particular Point well we can see on this graph all along the way the slope is getting more and more and more and more and more positive which means the velocity is getting more and more and more positive and if the velocity is getting more and more positive that means this object whatever it is is moving faster and faster and faster in in the positive X Direction cool that's velocity so the third term we're going to Define here is acceleration so and just like displacement and velocity Acceleration acceleration is a vector not a scalar and so it has both magnitude and Direction now in the case of displacement it was the vector and we kind of said something like distance could be the scalar version uh with velocity we said velocity is Vector and then speed was the scalar we're going to have no scalar equivalent for acceleration acceleration is a vector it has both magnitude and Direction now if you look at the definition of acceleration so well first let start with the unit it's a derived SI unit of met per second squared so we can kind of see where that comes from based on the definition so now acceleration again being a vector I probably should have been using this a little more liberally I'm I'm really bad about remembering those arrows sorry I forgot them on both displacement uh as well as on uh velocity so but acceleration is equal to the change in velocity over the change in time that is the textbook definition for acceleration all right so to have an acceleration you have to have a velocity that is changing and students often struggle with the difference between velocity so and acceleration now if you have a velocity it means you are moving that's that's it if you have an acceleration it means not only are you moving but most of the time it means that your speed is changing now technically mean that your direction is changing but we'll deal with that in a whole other unit for now it's going to mean not only that you're moving but that how fast you're moving is changing so a lot of people think of of velocity and acceleration in similar fashion they think like oh I'm moving really fast you know that must be I'm accelerating or you know something like this again it's not about how fast you are moving or how slow you're are moving if you're moving you have a velocity period it's is your speed changing so think about it this way so a lot of people think of like you know if I was going down the road on a motorcycle at 3 300 mil hour that'd be crazy right so the wind blowing in my face and all sorts of craziness going on and they would think it you know because it's such a a rush and and and such a kind of forceful experience they might associate a little more to it than they should but think about the same thing if you were doing 300 miles an hour sitting on an airplane so taking a nap so and now you don't have the wind in your face and stuff like that and so I want you to think of that situation for here to start because that's the one I want to want to go with I I don't want the the adrenaline of of riding a motorcycle and the wind hitting you in the face and all that stuff as part of this discussion I just want a nice calm airplane ride now here's the deal if you're flying through the air at a cruising velocity of 300 m/ second in some particular direction and I said me second 300 miles hour in some particular direction so as long as it's just constantly 300 miles per hour you have a constant velocity you do not have an acceleration so and it should feel very calm in your your ride however think about taking off in that airplane so as you're taking off that plane is not only moving but it needs to get to a velocity that's fast enough to generate enough lift on the wings to get that plane into the air and so it's not only just moving it has velocity but it has a velocity that is speeding up faster and faster and faster as it goes down the runway and that's what throws you back in your seat so and the key is you have a velocity that is changing in this case it's increasing and so you have an acceleration and so often times you can distinguish between when you have an acceleration and when you don't is if you're being thrown back in your seat so to speak that's key you have an acceleration whereas if as all as you're moving is constant velocity you're not really being thrown back in your seat and the reason I want to you know not use the motorcycle analogy is a lot of people feel like oh I'm riding the motorcycle at 300 miles hour I'm being thrown back well you're being thrown back because of air resistance not because you're accelerating and that's why I didn't want to go with that example why the airplane's a much better calmer example as as long as you're at cruising velocity and it's constant there's no acceleration and it's a very calm ride but right as you take off and you're thrown back in your seat that's when you have acceleration so hopefully you see that difference all right so we want to deal with three situations analogous to what we saw with velocity but now instead of plotting position versus time we're going to plot velocity on the y-axis time on the x-axis and so in this case in this graph here notice this is no longer about position and so my question for on this first one is are you moving are you moving and that's tricky don't answer so quickly because if we were doing position versus time if this was position well then this would mean your position's constantly the same value and not changing and so you know you're not moving but that's not the case here here it's your velocity that is not changing but it's not zero if your velocity was zero down here that means you're not moving but here you've got a positive velocity it's somewhere up the y- AIS you're moving but your slope is zero and your velocity is not changing so we might call this constant velocity so keep in mind here that we've got velocity on the y-axis time on the x- axis and again slope is equal to rise over run and rise over run in this case is equal to the change in velocity over the change in time the change in velocity over the change in time the slope on this graph is now equal to acceleration whereas again that slope on the position versus time graph was equal to Velocity this is where students start to get confused again we're going to have a whole lesson on these motion diagrams to make sure you really understand the difference so but in this case the slope on the velocity versus time that is the definition of acceleration and so if your slope is zero that means your acceleration equals zero as well okay let's go on to the next one here so here we've got again velocity versus time you've still got a constant slope here but now you can see your velocity is not the same the whole time your velocity is not constant your velocity on the y- axis is going up and up and up and up and up as time progresses and so now you've got a velocity that is going up over time you now are not at constant velocity you actually have an acceleration and we might call this uniform acceleration or constant acceleration because again on a velocity versus time graph the slope is equal to the acceleration well because this is a straight line it means you have a constant slope constant slope means constant acceleration I.E which means more more commonly is going to be called uniform acceleration you can call it constant acceleration there's nothing wrong with that but you're going to hear this word uniform used pretty commonly and then finally in this last one here now it's not a straight line it's a curve and so in this case your slope is changing all the time and if your slope is equal to your acceleration then your acceleration is changing every time and here we're going to have a varying acceleration now here's the deal you are going to deal with problems in kinematics over the next couple of chapters that deal with this situation where there's no acceleration I.E constant velocity you're going to deal with mathematical problems calculations where you have a uniform acceleration a nonzero but constant acceleration but mathematically you're probably never going to deal with any kind of calculations dealing with a variant acceleration you might deal with it graphically and have to identify it graphically or something like this but you're probably not going to do any calculations with it throughout the entirety of this course FYI it would be a horrendous pain in the butt to treat okay so that is acceleration so now we've got displacement velocity acceleration and let's do our first actual physics calculations so now we're going to do our first actual physics calculations and if you're taking this Introduction to Kinematics Calculations as part of my master course uh then you've got the study guide and the questions are right on the study guide with plenty of room for you to work out the answers and take notes and all that stuff so if you don't I will make sure the the the questions actually end up on the screen here but the first question here says a man makes one complete revolution around a 400 met circumference circular track what is the magnitude of his displacement so we're going to go one full time and that circumference is 400 m the question is what is the displacement well again displacement only cares about where you start and where you finish and you kind of think of it as the straight line distance between those two and so in this case the displacement would be zilch now the distance traveled would not be zero the distance traveled would be 400 m of circumference here right so but the displacement is indeed zero so definitely just showing this example to make sure you realize the distinction between displacement and distance all right next question related to the first one it says a man makes 1/ half of a revolution around a 400.0 meter circumference circular track what is the magnitude of his displacement we'll find out that 400.0 part was all about uh sigfigs uh so it would make this kind of a little bit trivial uh all right a little bit different than the first question here so in this case we're starting at a certain point and we're only going to go halfway around the track now again we know the circumference of the whole track is 400 m and the question is what's his displacement now well it's not zero and it's not zero because the point where he started and the port where he ended up are not the same point so but again we don't want the distance and again if you think well a full circle is 400 MERS so a half circle would mean he's traveled 200 M well that's true the distance distance he's traveled is 200 M but again that's not the question the question is what is the displacement and in this case that is the straight line distance between the point you start at and the point you finish at which is definitely going to be less than the path we took less than 200 met but how do we figure out what that is well it's going to require you to remember just a little bit of geometry from back in the day you might recall that the circumference of a circle is equal to 2 pi r and so we can use the the circumference given of 400 m to figure out the radius then use the radius to to figure out the diameter because this would be the diameter of that Circle the other thing you might realize is that 2 the radius is the diameter and so sometimes you'll see the circum circumference simply written as Pi times diameter and we can use that so but again whether you use the first one and solve for the radius and then double it or just solve for the diameter directly take your pick I'm going to solve for it directly it looks a little bit easier and so in this case we're going to have 400.0 M = < the diameter and we'll just divide through by pi and the diameter is going to equal 400.0 m all over Pi now here you're going to see why I made it 400.0 MERS not just 400 truth be told that was a recent edit uh cuz I didn't want only one Sig fig so now the it ends in a zero right at the decimal that zero is significant that four is definitely significant and so these zeros are now in between significant figures and their significant so now we've got four significant figures if it was just 400 without the 0 at the end that would only be one significant figure and I definitely didn't want only one sign all right but we're definitely going to let our calculator do the work for us here when dividing by pi and so we're going to do 400 divided by pi we're going to get7 and some change and the change here I want to go at least five five digits if I want to end up with four sig figs uh and so it's going to be 3 2 and again my four sig figs here there's your first there's your second there's your third there's your fourth and the reason I included the fifth digit is just so I know if I need to leave that and round down or to change it and round up well in this case with a two following we're just going to round down and leave it alone so we'll just get rid of that two and there is your diameter of that Circle which in this case again is the magnet itude of the displacement and no said just asked for the magnitude because it'd be hard to ask for the direction you know based on what was given here and stuff like that obviously the way I drew it you could say well straight down chat across the track but technically if you were at a track you wouldn't be traveling straight down you'd be traveling across the track and that would be in certain direction but that's why I only asked for the magnitude and not the associated Direction all right the next question here says that a car travels 300 miles due north in 5 hours what is its average velocity during this journey all right so in this case just the definition of velocity and it's displacement over time here all right in this case both of those are given so it travels a displacement of 300 miles du North in a time of I believe that was 5 hours yeah 5 hours and so in this case 300 divid 5 we can see this out to 60 MPH so it's not SI units but the question ask us to give SI units but we're not done so because the question says what is the velocity not simply what is the magnitude of the Velocity so but again the velocity here is just going to point in the same direction as uh uh uh the displacement in this case and so it's going to be North cool next oh by the way sigfigs we had one SigFig provided in both our numbers and so our answer has to show up with one Sig SigFig well 60 fortunately has one SigFig all right next question a car accelerates uniformly from 0 to 60 mph in 6 seconds what is the magnitude of its acceleration during this time all right we look at that definition of acceleration we've been provided with and acceleration is just equal to the change in velocity over the change in time and again this didn't actually ask for SI units and I'm not expecting you to convert this to SI units but I want you to especially for those who in the US to use units you're familiar with and come comtable with before we go full on metric system here so in this case that's going to be final velocity minus initial velocity and one thing to note a lot of people and a lot of textbooks write initial velocity as V not you probably see me using these interchangeably I will try to be consistent but I probably will fail on occasion so but it's like V at Time Zero is what that means so whether it's V initial with a little I or V not either way but often V final is used in either case all right so change in velocity final minus initial all over the the change in time and so in this case we're going from 0 to 60 M hour so that's going to be 60 miles hour minus 0 all over the course and I believe it was 6 seconds so this going to have a little bit of funky units here so but 60 - 0 is 60 divid 6 is 10 and it's going to be 10 miles per hour per second what this means is that its velocity is speeding up 10 miles hour every second and so initially it's going zero 1 second later it' be have a velocity now or a magnitude of velocity of 10 mil per hour another second later it'd be up to 20 mil per hour another second later it'd be up to 30 miles hour it's getting faster and faster and faster and the rate at which that's happening is 10 miles hour faster every second okay now this is not your normal unit notice if we wrote this out a little bit differently we could write 10 miles per hour per second and this actually looks worse not better so and again these are not the normal units we're going to see and definitely not SI units but again this is going to be why I want you to think about uh the SI unit of meters per second squared as me/ second per second just like we have miles per hour per second you can see like oh 10 miles per hour per second means it's getting faster 10 miles per hour every second same thing in the next problem we're going to see an acceleration of 10 m/s squared but I want you to think of that again as 10 meters per second per second but we're going to use the exact same set of numbers here but go metric system instead with different units so now a car accelerates uniformly from 0 to 60 m/s in 6 seconds what is the magnitude of its acceleration during this time the process is exactly the same and so change in velocity over change in time which is again is V final minus V initial all over that change in time which now is 60 m/ second- 0 all over 6 seconds and now again my preference would be for you to look at this as being 60 m/ second I'm sorry 10 m/ second per second but the way you're going to formally write that is 10 m per second squared that's the unfortunate thing because the moment we make it 10 meters per second squared students don't think of it the same way they don't think oh when I see that that means it's getting faster and faster and faster 10 m/ second every second and that's why it's my strong preference that you look at the units for acceleration and divide it up as velocity over time meters per second per second and when you see this well it started at zero so 1 second later its velocity or the or its magnitude it's velocity at speed would be 10 meters per second faster so Time Zero it was at zero velocity at Time 1 second it's now got a velocity of 10 m/ second at time equals 2 seconds it's now got a velocity of 20 m/s at time equals 3 seconds it's now got a velocity of 30 m/ second it's getting faster and faster and faster now one thing to note especially with Motion in one dimension again direction is often defined when we're dealing with motion and direction is either forward or backward which means either positive or negative which usually refers to positive X Direction Negative X Direction so one thing to note in dealing with accelerations so in this case we saw that the acceleration came out positive because the final velocity was higher than the initial velocity but said this had been the opposite let's say instead this thing had been cruising down the road at 60 m/ second and hit the brakes and slowed down to zero over the course of 6 seconds and now of a sudden you would have had 0 - 60 m/ second over 6 seconds and you would have got -10 m/ second per second and it turns out the positive versus the negative that's actually considered a direction in this case like the positive X Direction the negative X Direction it's part of the direction when we're dealing with Motion in one dimension and so now of a sudden what does that actually mean well in this case what a positive acceleration versus a negative acceleration means is is it in the same direction as your velocity or opposite direction as your velocity so in the first case with a positive acceleration a positive acceleration is in you know is in the same direction as your velocity and your velocity is getting faster and faster and faster so but in the second case so it turns out your acceleration actually is in the opposite direction so our velocities were positive the whole time from 0 to 60 or in this case from 60 down to zero but they were in the positive the forward Direction the whole time so but the acceleration is in the negative Direction and so here's the deal when your acceleration and your velocity point in the same direction it means you're speeding up faster and faster and faster but when your acceleration and velocity point in opposite directions like when you put your brakes on on your car so now you're getting slower and slower and slower you're slowing down so big distinction to understand there between uh the signs on acceleration and velocity so that'll become important we start doing other kinds of problems and we'll find out that uh there are certain things we can do when your displacement your velocity your acceleration all point in the same direction so and turns out your sign is not going to be the most important thing in your calculations but if your displacement your velocity acceleration don't all point in the same direction you know keeping track of your signs becomes super important as we'll see in the next lesson now if you've liked this lesson then like the video Happy study |
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16878 | https://www.youtube.com/watch?v=brNolnM75R0 | Thermodynamics: Steady Flow Energy Balance (1st Law), Turbine
Raili Taylor
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Posted: 18 Mar 2019
Solution to the following problem (Thermodynamics: An Engineering Approach, CBK, 8th Edition, 5-46)
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500 C, and 80 m/s, and the exit conditions are 30 kPa, 92 percent quality, and 50 m/s. The mass flow rate of the steam is 12 kg/s. Determine (a) the change in kinetic energy, (b) the power output, and (c) the turbine inlet area.
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Transcript:
Introduction in this video I'm planning on going over this problem this is out of thermodynamics an engineering approach cbk the eighth edition and this problem is most likely in other editions as well although it'll have a different number so this is for a turbine and it says that steam flows steadily through an adiabatic turbine and it gives us the inlet conditions there are some of the inlet conditions some of the exit conditions the mass flow rate and it wants us to calculate the change in kinetic energy power output and the turbine Inlet area so what I'm going to do this problem already has a drawing but I always do it another drawing anyway just because as I I feel like as I as I draw a diagram of the problem and write down the information it helps me to kind of understand what I'm solving what I already know and what I'm looking for so I'm gonna call this the problem set up and you want to approach all of your problems like this like you want to go through these steps to set up your problem so we're gonna have a problem set up then we're going to make assumptions you need to make assumptions otherwise a lot of these problems aren't solvable without using a quite a bit more complicated math and probably a computer a numerical solution to solve your problem so that's why you need to make a sumption a lot of these equations also have assumptions built into them so if you're not making a certain assumption you can't use all of the equations that you might want to use and so you need to be really clear what your assumptions are how you're setting up the problem and then after we do those two things we're going to go through what equations we're going to use to solve this and then very last we're going to plug in numbers and calculate the answers so a lot of people jump right into just trying to write down equations and plug in numbers and that's not how you want to set up these problems you might get the right answer sometimes doing it that way but a lot of times you won't and you really want to learn how to set up your problems not just for this class but for all of your future engineering classes as well so let's go ahead and start setting this up Drawing a turbine so I'm gonna draw a turbine even though there's already a drawing of one I'm just gonna draw one so that I'm pretty clear what's going on so the turbine house flow in and flow out and it says that we have steam and since it's a turbine there's going to be some work out and this this work is this is likely rotating a shaft and a generator and producing electrical work so that there's gonna be some work out so the steam is we're basically extracting work from the steam as it condenses all right now let's write down some of our information so it says the steam flows steadily the steadily is a huge clue that means that we can assume that this is steady flow so this this is going to go under our assumptions but I'll just write it down now since it's right there in the problem sometimes your problem won't say that you're that your system is or whatever it is that you're analyzing is steady or it flows steadily or a steady flow but whenever you have these problems where you have mass flowing in and mass flowing out you and it's not in a start up or shutdown phase you pretty much are going to assume that it's steady flow it wouldn't make sense otherwise like if you think about if you think about this turbine we have steam flowing into the turbine and steam flowing out steady flow means that the the mass flow-rate doesn't change with time so in other words your your time derivatives are zero so if we have steam so if you think about this okay let's say that we have this turbine we just turned it on and we have steam flowing into the turbine well initially during that startup your mass inside the turbine is going to be changing with time so your mass is changing with time inside the turbine during the startup and once its operating steadily so here's so steadily again or and steady flow it is that means that the mass inside the turbine isn't changing with time and that just intuitively you know that that means that the mass flow in has to be equal to the mass flow out so what we can say by the steady flow is that m dot 2 is equal to m dot 1 is equal to m dot and it actually tells us what the mass flow rate is so the mass flow rate is 12 kilograms per second you'll see different terminology for this steady flow so steady flow in this problem they're just saying flow flow steadily so you'll see this different terminology you'll also see steady state so all of these mean that you're that basically you can assume that the mass inside your system in this case our system is the turbine isn't changing with time so I was gonna draw the boundary so this is what we're considering to be our system all right so let's write down some more stuff it's adiabatic that means that this turbine is likely well insulated so there's no heat in or out so right away we know that Q dot is zero so there's no heat in or out and we can also just add this to our assumption all right now it says that so now it's giving us Inlet conditions this is our inlet this is our outlet so the inlet it says that the pressure is 4 mega Pascal's the temperature is 500 degrees Celsius and it tells us that the velocity of the fluid is 80 meters per second and then it gives us the exit condition so at the exit the pressure is 30 kilo Pascal's the and then it gives us a quality 0.92 so since it gives us a quality that means that we know the outlet the phase of the outlet is a saturated mixture so it's a mixture of liquid and vapor going in since it says it's steam we're gonna we're gonna look up this pressure and temperature and verify but this is most likely a superheated vapor flowing in and then we have a saturated mixture flowing out and then it tells us the velocity out is 50 meters per second and then we want to find the change in kinetic energy so we're looking for the change in kinetic energy and the power output so this work isn't given but we want to find it and the turbine Inlet area so we're looking for the area of the inlet alright so we have our problem set up and I know that there was a drawing here that has exactly what we wrote down but I think it's good to just ignore this and rewrite the information because like I said as you're writing it down you're you're going to get ideas of assumptions you can make and really understand the problem and the problem set up so we have our basically our problem set up we have the drawing known we have the knowns and unknowns and we have some assumptions so let's make some more assumption one assumption I want to make so the steady flow and adiabatic are already here but there's not going to be it doesn't say anything about this turbine being really tall so basically we can assume that there's not a large elevation difference between the inlet and the outlet so we're going to assume that the change in potential energy is Setting up equations zero all right now let's write down some equations and I want to point out that setting these problems up writing down your assumptions and the equations you're going to solve is probably the most important part of these problems just plugging in numbers into an equation and possibly getting the right answer or not getting the right answer doesn't mean that you'd like you need to know how to set up these problems and so I would consider this part and the part I'm about to do probably the most important part of these problems so what you don't want to do is just skip all of this write down an equation and try and plug in numbers and see if you get the right answer because you might get the right answer but you might not and if you do get the right answer you might have just happened to plug numbers into your equation right but you want to know exactly where your values are coming from what the assumptions in the equations you're using are and what exactly is going on in your system and why you're using the equations you're using okay so let's look at what we have before we write down the equations so basically we have a turbine we have what is most likely superheated steam going into the turbine and we have a saturated mixture coming out and so if there's a large pressure drop basically we're extracting energy from the steam and getting work we're getting work out that's likely being used to generate electricity and it doesn't tell us t2 I don't know if we're going to need t2 but I'll just write that down anyway just so we know that we don't have it since we know the pressure and the quality we have we have the information we need to look up data on the tables for the outlet we have the pressure and temperature at the inlet so we we have everything we need to look up data on the tables and we're looking for Energy balance the work out the change in kinetic energy and the area of the inlet so right away we we know that we're going to need to do an energy balance and so I'm going to write down the energy balance for a single stream steady flow because that's what this is so it's if we determine it was steady flow at a single stream we only have one stream in and one stream out so the first law or energy balance for the single stream steady flow is Q minus W is equal to m dot H 2 minus H 1 plus B 2 squared minus B 1 squared over 2 plus G Z 2 minus Z 1 and I like writing down all of the terms even though we know that some of these terms are zero and I did derive a general energy balance for a turbine in a previous video and I think in that energy balance we assume that the potential energy was zero and I believe we also assumed that Q or the the heat input her output was 0 and so theoretically you could just take that equation and write it down and it wouldn't have the potential energy term or the heat term but the problem is what if your turbine what if you do need to consider the potential energy I don't think you'll I don't think you'll need to for a turbine unless this giant but the turbines are often well insulated but if they're not then you could potentially have heat into our out of the system and so anyway but we know that the the change in potential energy is zero q is zero in this case and it's giving us some velocity so we need to keep our kinetic energy term as well so then we have the work is equal to m dot H 2 minus H 1 plus B 2 squared minus B 1 squared over 2 and then since we're this term is the change in kinetic energy and it's on a per mass basis since we're multiplying it by the mass flow rate here so this term is the change in kinetic energy which is one of the things who are looking for so I'm going to actually just change that term in this equation to the change in kinetic energy and then and then we have the negative work is equal to m dot H 2 minus H 1 plus Delta ke and then I'm just going to write down the equation for Delta ke so this is actually what we're going to be solving so this is equal to V 2 squared minus V 1 squared over 2 normally I would just solve that directly in the energy balance so I would actually solve this equation but since it's asking specifically for the change in kinetic energy I'm just going to calculate it separately and then plug it plug what I calculate into this equation so basically then I'm not having to solve for kinetic energy term twice and then we have so and the other thing we're looking for is the work we know the mass flow rate because it was given we can calculate the change in kinetic energy because we're given the velocities and we need the enthalpies so we're gonna get the enthalpies from the tables we have all of the information we need to look those up the last thing that it asked us to calculate was the area of the inlet so we can calculate that because we know that the mass flow rate so m dot is equal to the area multiplied by the velocity divided by the specific volume and we can look this up on the table so I'm just going to put table I think these are all the equations we need I'm going to solve for the kinetic energy first and then I'm gonna plug that into the energy balance and solve for the work and then very last I'm gonna solve for the for the area so now Data analysis let's look up data and at this point in this analysis this would either be looking up data on the tables getting your like if you're using specific heats to calculate your change in enthalpy you would look up your specific heat at this point so this is really just getting all of the data that you need to solve the problem just depending on how you're solving it so at the inlet the pressure is four mega Pascal's and t1 is 500 degrees Celsius and this is 4000 kiloPascals so what I want to do first I mean I'm fairly positive this is a superheated vapor but I just want to verify that to make sure that I'm assuming the right phase because if it's not a superheated vapor we can't use the superheated vapor table to look at the enthalpy and the specific volume so we wanted to determine the phase so what we're going to do is look up the the temperature of saturated liquid at 4,000 kiloPascals and compare that to the temperature of our system so T sat at 4,000 kiloPascals is equal to 250 point 3 3 degrees Celsius and so our temperature is greater than T sat so this is in fact a superheated vapor so then we can use the superheated vapor table to look up our enthalpy and specific volume so we're going to look up p1 and t1 on the soup on the superheated vapor table and this is for water so v1 is equal to zero point zero eight six four four meters cubed per kilogram and the enthalpy is equal to three four four six point zero kilojoules per kilogram all right so we have our information at the inlet now let's look up our information at the outlet so we already know that this is a saturated mixture because it gave us the quality if you have a quality that means that you have a saturated mixture because the quality doesn't make sense for say a compressed liquid or superheated vapor because your quality is just telling you what percentage of vapor you have so p2 is equal to 30 kiloPascals and the quality is equal to 0.92 and so we know that our enthalpy is equal to the enthalpy of the saturated liquid plus X multiplied by the enthalpy of a prize and remember this is just the enthalpy of the set of the saturated vapor minus the enthalpy of the liquid but we can look we can look this up directly so that's what we're going to do so the enthalpy of the saturated liquid is equal to two eighty nine point two seven kilojoules per kilogram and this is equal to two three three five point three kilojoules per kilogram so then we can calculate the enthalpy so h2 remember I'm just looking these up on the saturated table for for water and I'm looking at the pressure table because I know the pressure so the enthalpy at the outlet is equal to two eighty nine point two seven kilojoules per kilogram plus the quality which is 0.92 x - 3 three 5.3 kilojoules per kilogram and this is equal to two so h2 the enthalpy at the outlet is equal to two four three seven point seven five kilojoules per kilogram so this is the data we need for the outlet so now we should have everything we need to do our calculations so now last we're going to do our calculations all right so I'm going to calculate the kinetic energy first so if we go back up and look at the equations for solving so first I'm going to calculate the kinetic energy then I'm going to plug what I get for that into the energy balance and solve the the energy balance for the work and we know that enthalpy at the inlet and outlet because we just looked them up and then last I'm going to calculate the area from this equation for the mass flow right so I'm going to go down here and I'll rewrite these equations so we know we're solving so the change in kinetic energy is equal to v2 squared minus v1 squared over two and this is equal to 50 squared meter squared second squared minus 80 squared meter squared second squared over two and so this is where it can get a little bit confusing we have units here of kilojoules per kilogram but it looks like over here we have units of meter squared second squared so we need to use this conversion that one kilojoule per kilogram is equal to 1,000 meters squared second squared so we're gonna put this conversion in so we can get the kilojoules per kilogram so we have one kilojoule per kilogram over 1,000 meters squared second squared the meter squared second squared cancels and we're left with kilojoules per kilogram which are the units we want so the change in kinetic energy is equal to negative one point nine five kilojoules per kilogram so that's our change in kinetic energy now we want to solve our energy balance so that was this equation so down here so and make sure you keep your signs for the for the work and the heat we cancel out the heat but don't change any signs at this point and then this is plus the change in kinetic energy which we just calculated alright so now I'm going to multiply this by a negative one just so they can get the work and then I'm going to have negative M and then still H two minus H one plus the change in kinetic energy now we can just plug in all of our numbers so we have the work is equal to twelve kilograms per second multiplied by and then we have the enthalpies that we looked up so this enthalpy at the outlet is 24 3/7 0.75 kilojoules per kilogram minus the enthalpy at the inlet was 344 6.0 kilojoules per kilogram and then we have plus our kinetic energy which we calculated here so this is going to be negative one point nine five kilojoules per kilogram now let's just make sure our units work out so we have kilogram here that's gonna cancel out all of these kilograms and we're left with kilojoules per second which is what we want because we're looking for the we're looking for the we're looking so this work is this dot means that this is on a per time so this is the units for this or like kilojoules per second or kilowatt so this is a unit of power but basically we want kilojoules per second so that means that means there are units look correct so this is going to work out to the work is equal to twelve thousand one hundred and twenty two kilowatts and then we can convert that into twelve point one megawatt and the work is positive because the work is being done by the system all right now last we want to calculate the area of the end so we know that MDOT is equal to the area multiplied by the velocity over the specific volume and just keep track of these they're both represented by V's so a little bit confusing but this is specific volume the cursive B is there's velocity all right so what I'm gonna do is solve this equation for the area and then I have everything I need and I know the mass flow rate I know the velocity and I know the specific volume at the inlet because I looked it up so if I solve this for the area all I need to do is plug in values so the area is equal to m dot multiplied by the specific volume over the velocity so this is equal to 12 kilograms per second multiplied by zero point zero eight six four four meters cubed per kilogram and remember this is what we looked up and then we have eighty meters per second for the inlet velocity just make sure our units work out so the kilograms cancel seconds cancel and this cancels we're left with meters squared which is what we want for an area so this works out to 0.01 three meters squared |
16879 | https://www.ninds.nih.gov/sites/default/files/2025-05/multiple-sclerosis-hope-through-research.pdf | Multiple Sclerosis Hope Through Research National Institute of Neurological Disorders and Stroke National Institutes of Health NINDS health-related material is provided for information purposes only and does not necessarily represent endorsement by or an official position of the National Institute of Neurological Disorders and Stroke or any other Federal agency. Advice on the treatment or care of an individual patient should be obtained through consultation with a physician who has examined that patient or is familiar with that patient’s medical history. All NINDS-prepared information is in the public domain and may be freely copied. Credit to the NINDS or the NIH is appreciated. i Table of Contents What is Multiple Sclerosis?......................................... 1 Myelin and the immune system............................ 1 What are the signs and symptoms of MS?................... 3 MS exacerbation ................................................. 6 Conditions associated with MS............................. 6 What causes MS? ....................................................... 7 Genetic susceptibility........................................... 7 Infectious factors and viruses............................... 9 Environmental factors .......................................... 9 How is MS diagnosed?.............................................. 10 How is MS treated?................................................... 11 Treatments for attacks........................................ 11 Disease-modifying treatments............................ 11 Managing MS symptoms.................................... 15 Complementary and alternative therapies .......... 18 What research is being done?................................... 19 Intramural research programs on MS.................. 22 Translational research........................................ 23 Focus on progressive MS therapies .................... 24 Focus on biomarkers.......................................... 24 Where can I get more information? ........................... 25 1 What is Multiple Sclerosis? Multiple Sclerosis (MS) is the most common disabling neurological disease of young adults with symptom onset generally occurring between the ages of 20 to 40 years. In MS, the immune system cells that normally protect us from viruses, bacteria, and unhealthy cells mistakenly attack myelin in the central nervous system (brain, optic nerves, and spinal cord), a substance that makes up the protective sheath (called the myelin sheath) that coats nerve fibers (axons). MS is a chronic disease that affects people differently. A small number of those with MS will have a mild course with little to no disability, whereas others will have a steadily worsening disease that leads to increased disability over time. Most people with MS, however, will have short periods of symptoms followed by long stretches of relative quiescence (inactivity or dormancy), with partial or full recovery. Women are affected more frequently with MS compared to men. The disease is rarely fatal and most people with MS have a normal life expectancy. New treatments can reduce long-term disability for many people with MS. Currently there are still no cures and no clear ways to prevent the disease from developing. Myelin and the immune system MS attacks axons in the central nervous system protected by myelin, which are commonly called white matter. MS also damages the nerve cell bodies, which are found in the brain’s gray matter, as well as the axons themselves in the brain, spinal cord, and optic nerves that transmit visual information from the eye to the brain. As the disease progresses, the outermost layer of the brain, called the cerebral cortex, shrinks (what is known as cortical atrophy). 2 The term multiple sclerosis refers to the distinctive areas of scar tissue (sclerosis—also called plaques or lesions) that result from the attack on myelin by the immune system. These plaques are visible using magnetic resonance imaging in the white and/or gray matter of people who have MS. Plaques can be as small as a pinhead or as large as a golf ball. During an MS exacerbation, most of the myelin, and to a lesser extent the axons within the affected area, is damaged or destroyed by different types of immune cells (also known as inflammation). The symptoms of MS depend on the severity of the inflammatory reaction as well as the location and extent of the plaques, which primarily appear in the brain stem, cerebellum (involved with balance and coordination of movement, among other functions), spinal cord, optic nerves, and the white matter around the brain ventricles (fluid-filled spaces). Myelin is a substance that makes up the protective sheath that insulates nerve fibers. In MS, myelin is damaged. The illustration above shows a normal, healthy myelin sheath, and one damaged by MS.
3 What are the signs and symptoms of MS? The natural course of MS is different for each person, which makes it difficult to predict. The onset and duration of MS symptoms usually depends on the specific type but may begin over a few days and go away quickly or develop more slowly and gradually over many years. There are four main types of MS, named according to the progression of symptoms over time: • Relapsing-remitting MS. Symptoms in this type come in attacks and, in-between attacks, people recover or return to their usual level of disability. The occurrence of symptoms in this form of MS is called an attack, or in medical terms, a relapse or exacerbation. The periods of disease inactivity or quiescence between MS attacks is referred to as remission. Weeks, months, or even years may pass before another attack occurs, followed again by a period of inactivity. Most people with MS (approximately 80%) are initially diagnosed with this form of the disease. • Secondary-progressive MS. People with this form of MS usually have had a previous history of MS attacks, but then start to develop gradual and steady symptoms and deterioration in their function over time. Most individuals with severe relapsing-remitting MS may go on to develop secondary progressive MS if they are untreated. • Primary-progressive MS. This type of MS is less common and is characterized by progressively worsening symptoms from the beginning with no noticeable relapses or exacerbations of the disease, although there may be temporary or minor relief from symptoms. 4 • Progressive-relapsing MS. This rarest form of MS is characterized by a steady worsening of symptoms from the beginning, with acute relapses that can occur over time during the disease course. There are some rare and unusual variants of MS. One of these is Marburg variant MS (also called malignant MS), which causes swift and relentless symptoms and decline in function, which can result in significant disability or even death shortly after disease onset. Balo’s concentric sclerosis, which causes concentric rings of myelin destruction that can be seen on an MRI, is another variant type of MS that can progress rapidly. Early MS symptoms often include: • Vision problems such as blurred or double vision, or optic neuritis, which causes pain with eye movement and a rapid loss of vision • Muscle weakness, often in the hands and legs, and muscle stiffness accompanied by painful muscle spasms • Tingling, numbness, or pain in the arms, legs, trunk, or face • Clumsiness, particularly difficulty staying balanced when walking • Bladder control problems • Intermittent or more constant dizziness MS may also cause later symptoms such as: • Mental or physical fatigue which accompanies the early symptoms during an attack • Mood changes such as depression or difficulty with emotional expression or control • Cognitive dysfunction—problems concentrating, multitasking, thinking, learning, or difficulties with memory or judgment 5 Vision problems, such as blurred or double vision or optic neuritis, are among the early symptoms of MS.
Muscle weakness, stiffness, and spasms may be severe enough to affect walking or standing. In some cases, MS leads to partial or complete paralysis and the use of a wheelchair is not uncommon, particularly in individuals who are untreated or have advanced disease. Many people with MS find that weakness and fatigue are worse when they have a fever or when they are exposed to heat. MS exacerbations may occur following common infections. Pain is rarely the first sign of MS, but pain often occurs with optic neuritis and trigeminal neuralgia, a disorder that affects one of the nerves that provides sensation to different parts of the face (see Conditions associated with MS section below). Painful limb spasms and sharp pain shooting down the legs or around the abdomen can also be symptoms of MS. Many individuals with MS may experience difficulties with coordination and balance. Some may have a continuous trembling of the head, limbs, and body, especially during movement. 6 MS exacerbation An exacerbation—which is also called a relapse, flare-up, or attack—is a sudden worsening of MS symptoms, or the appearance of new symptoms that lasts for at least 24 hours. MS relapses are thought to be associated with the development of new areas of damage in the brain. Exacerbations are characteristic of relapsing-remitting MS. An exacerbation may be mild, or severe enough to significantly interfere with life’s daily activities. Most exacerbations last from several days to several weeks, although some have lasted for as long as a few months. When the symptoms of the attack subside, an individual with MS is said to be in remission, characterized by disease quiescence. Conditions associated with MS Transverse myelitis (an inflammation of the spinal cord) may develop in those with MS. Transverse myelitis can affect spinal cord function over several hours to several weeks before partial or complete recovery. It usually begins as a sudden onset of lower back pain, muscle weakness, abnormal sensations in the toes and feet, or difficulties with bladder control or bowel movements. This can rapidly progress to more severe symptoms, including arm and/or leg paralysis. In most cases, people recover at least some function within the first 12 weeks after an attack begins. Neuromyelitis optica is a disorder associated with transverse myelitis as well as optic nerve inflammation (also known as optic neuritis). People with this disorder usually have abnormal antibodies (proteins that normally target viruses and bacteria) against a specific channel in optic nerves, the brain stem, or spinal cord, called the aquaporin-4 channel. These individuals 7 respond to certain treatments, which are different than those commonly used to treat MS.
Trigeminal neuralgia is a chronic pain condition that causes sporadic, sudden burning or shock-like facial pain. The condition is more common in young adults with MS and is caused by lesions in the brain stem, the part of the brain that controls facial sensation.
What causes MS?
Researchers are looking at several possible explanations for why the immune system attacks central nervous system myelin, including: • Fighting an infectious agent (for example, a virus) that has components that mimic components of the brain (called molecular mimicry) • Destroying brain cells because they are unhealthy • Mistakenly identifying normal brain cells as foreign There is also something known as the blood-brain barrier, which separates the brain and spinal cord from the immune system. If there is a break in this barrier, it exposes the brain to the immune system. When this happens, the immune system may misinterpret structures in the brain, such as myelin, as “foreign.” Research shows that genetic vulnerabilities combined with environmental factors may cause MS.
Genetic susceptibility MS itself is not inherited, but susceptibility to MS may be inherited. Studies show that some individuals with MS have one or more family member or relative who also have MS. 8 Current research suggests that dozens of genes and possibly hundreds of variations in the genetic code (called gene variants) combine to create vulnerability to MS. Some of these genes have been identified, and most are associated with functions of the immune system. Many of the known genes are similar to those that have been identified in people with other autoimmune diseases as type 1 diabetes, rheumatoid arthritis, or lupus. Trigeminal neuralgia is caused by lesions in the brain stem which controls facial sensation.
9 Infectious factors and viruses Several viruses have been found in people with MS, but the virus most consistently linked to the development of MS is the Epstein-Barr virus (EBV) which causes infectious mononucleosis. Only about 5 percent of the population has not been infected by EBV. These individuals are at a lower risk for developing MS than those who have been infected. People who were infected with EBV in adolescence or adulthood and who therefore develop an exaggerated immune response to EBV are at a significantly higher risk for developing MS than those who were infected in early childhood. This suggests that it may be the type of immune response to EBV that may lead to MS, rather than EBV infection itself. However, there is still no proof that EBV causes MS and the mechanisms that underlie this process are poorly understood. Environmental factors Several studies indicate that people who spend more time in the sun and those with relatively higher levels of vitamin D are less likely to develop MS or have a less severe course of disease and fewer relapses. Bright sunlight helps human skin produce vitamin D. Researchers believe that vitamin D may help regulate the immune system in ways that reduce the risk of MS or autoimmunity in general. People from regions near the equator, where there is a great deal of bright sunlight, generally have a much lower risk of MS than people from temperate areas such as the United States and Canada. Studies have found that people who smoke are more likely to develop MS and have a more aggressive disease course. People who smoke tend to have more brain lesions and brain shrinkage than non-smokers. The reasons for this are currently unclear. 10 How is MS diagnosed? There is no single test used to diagnose MS. The disease is confirmed when symptoms and signs develop and are related to different parts of the nervous system at more than one interval in time and after other alternative diagnoses have been excluded. Doctors use different tests to rule out or confirm the diagnosis. In addition to a complete medical history, physical examination, and a detailed neurological examination, a doctor may recommend: • MRI scans of the brain and spinal cord to look for the characteristic lesions of MS. A special dye or contrast agent may be injected into a vein to enhance brain images of the active MS lesions.
• Lumbar puncture (sometimes called a spinal tap) to obtain a sample of cerebrospinal fluid and examine it for proteins and inflammatory cells associated with the disease. Spinal tap analysis also can rule out diseases that may look like MS. • Evoked potential tests, which use electrodes placed on the skin and painless electric signals to measure how quickly and accurately the nervous system responds to stimulation. Research suggests that people who spend more time in the sun—which helps the skin to produce vitamin D—are less likely to develop MS.
11 How is MS treated? There is no cure for MS, but there are treatments that can reduce the number and severity of relapses and delay the long-term disability progression of the disease. Treatments for attacks • Corticosteroids, such as intravenous (infused into a vein) methylprednisolone, are prescribed over the course of 3 to 5 days. Intravenous steroids quickly and potently suppress the immune system and reduce inflammation. They may be followed by a tapered dose of oral corticosteroids. Clinical trials have shown that these drugs hasten recovery from MS attacks, but do not alter the long-term outcome of the disease. • Plasma exchange (plasmapheresis) can treat severe flare-ups in people with relapsing forms of MS who do not have a good response to methylprednisolone. Plasma exchange involves taking blood out of the body and removing components in the blood’s plasma that are thought to be harmful. The rest of the blood, plus replacement plasma, is then transfused back into the body. This treatment has not been shown to be effective for secondary progressive or chronic progressive MS. Disease-modifying treatments Current therapies approved by the U.S. Food and Drug Administration (FDA) for MS are designed to modulate or suppress the inflammatory reactions of the disease. They are most effective for relapsing-remitting MS at early stages of the disease.
12 Injectable medications include: • Beta interferon drugs are among the most common medications to treat MS. Interferons are signaling molecules that regulate immune cells. Potential side effects of these drugs include flu-like symptoms (which usually fade with continued therapy), depression, or elevation of liver enzymes. Some individuals will notice a decrease in the effectiveness of the drugs after 18 to 24 months of treatment. If flare-ups occur or symptoms worsen, doctors may switch treatment to alternative drugs.
• Glatiramer acetate changes the balance of immune cells in the body, but how it works is not entirely clear. Side effects are usually mild and consist of local injection site reactions or swelling. Brand Name Chemical Name Avonex Rebif Interferon beta-1a Betaseron Extavia Interferon beta-1b Plegridy Peginterferon beta-1a Copaxone Glatopa Glatiramer acetate Infusion treatments include: • Natalizumab is administered intravenously once a month. It works by preventing cells of the immune system from entering the brain and spinal cord. It is very effective but is associated with an increased risk of a serious and potentially fatal viral infection of the brain called progressive multifocal leukoencephalopathy (PML). Natalizumab is generally recommended only for individuals who have not responded well to or who are unable to tolerate other first-line therapies. 13 • Ocrelizumab is administered intravenously ever y six months and treats adults with relapsing or primary progressive forms of MS. It is the o nly FDA-approved disease-modifying therapy for primary-progressive MS. The drug target s the circulating immune cells that produce antib odies, which also play a role in the formatio n of MS lesions. Side effects include infusion -related reactions and increased risk of in fections. Ocrelizumab may increase the risk of cancer as well. • Alemtuzumab is administered for 5 consecutiv e days followed by 3 days of infusions one ye ar later. It targets proteins on the surface of imm une cells. Because this drug increases the ris k of autoimmune disorders it is recommende d for those who have had inadequate respon ses to two or more MS therapies. • Mitoxantrone, which is administered intravenously four times a year, has been approved for especially severe forms of relapsing-remitting and secondary progressive MS. Side effects include the development of certain types of blood cancers in up to one percent of those with MS, as well as with heart damage. This drug should be considered as a last resort to treat people with a form of MS that leads to rapid loss of function and for whom other treatments did not work. e Brand Name Chemical Name Novantrone Mitoxantrone Tysabri Natalizumab Ocrevus Ocrelizumab Lemtrada Alemtuzumab 14 Oral treatments include: • Fingolimod is a once-daily medication that reduces the MS relapse rate in adults and children. It is the first FDA-approved drug to treat MS in adolescents and children ages 10 years and older. The drug prevents white blood cells called lymphocytes from leaving the lymph nodes and entering the blood, brain, and spinal cord. Fingolimod may result in a slow heart rate and eye problems when first taken. Fingolimod can also increase the risk of infections, such as herpes virus infections, or in rare cases be associated with PML. • Dimethyl fumarate is a twice-daily medication used to treat relapsing forms of MS. Its exact mechanism of action is not currently known. Side effects of dimethyl fumarate are flushing, diarrhea, nausea, and lowered white blood cell count. • Teriflunomide is a once-daily medication that reduces the rate of proliferation of activated immune cells. Teriflunomide side effects can include nausea, diarrhea, liver damage, and hair loss. • Cladribine is administered as two courses of tablets about one year apart. Cladribine targets certain types of white blood cells that drive immune attacks in MS. The drug may increase the risk of developing cancer and should be considered for individuals who have not responded well to other MS treatments. • Diroximel fumarate is a twice-daily drug similar to dimethyl fumarate (brand name Tecfidera) but with fewer gastrointestinal side effects. Scientists suspect these drugs, which have been approved to treat secondary progressive MS, reduce damage to the brain and spinal cord by making the immune response less inflammatory, although their exact mechanism of action is poorly understood. 15 • Siponimod tablets (Mayzent) is taken orally and has a similar mechanism of action to fingolimod. Siponimod has been approved by the FDA to treat secondary-progressive MS. Clinical trials have shown that cladribine, diroximel fumarate, and dimethyl fumarate decrease the number of relapses, delay the progress of physical disability, and slow the development of brain lesions. Brand Name Chemical Name Gilenya Fingolimod Tecfidera Dimethyl fumarate Aubagio Teriflunomide Mayzent Siponimod Mavenclad Cladribine Vumerity Diroximel fumarate Managing MS symptoms MS causes a variety of symptoms that can interfere with daily activities but can usually be treated or managed. Many of these issues are best treated by neurologists who have advanced training in the treatment of MS and who can prescribe specific medications to treat these problems. Eye and vision problems are common in people with MS but rarely result in permanent blindness. Inflammation of the optic nerve (optic neuritis) or damage to the myelin that covers the nerve fibers in the visual system can cause blurred or grayed vision, temporary blindness in one eye, loss of normal color vision, depth perception, or loss of vision in parts of the visual field. Uncontrolled horizontal or vertical eye movements (nystagmus), “jumping vision” (opsoclonus), and double vision (diplopia) are common in people with MS. Intravenous steroid medications, special eyeglasses, and periodically resting the eyes may be helpful. 16 Muscle weakness and spasticity is common in MS. Mild spasticity can be managed by stretching and exercising muscles using water therapy, yoga, or physical therapy. Medications such as gabapentin or baclofen can reduce spasticity. It is very important that people with MS stay physically active because physical inactivity can contribute to worsening stiffness, weakness, pain, fatigue, and other symptoms.
Tremor, or uncontrollable shaking, develops in some people with MS. Assistive devices and weights attached to utensils or even limbs are sometimes helpful for people with tremor. Deep brain stimulation and drugs, such as clonazepam, also may be useful.
Problems with walking and balance occur in many people with MS. The most common walking problem is ataxia—unsteady, uncoordinated movements—due to damage to the areas of the brain that coordinate muscle balance. People with severe ataxia generally benefit from the use of a cane, walker, or other assistive device. Physical therapy also can reduce walking problems. The FDA has approved the drug dalfampridine to improve walking speed in people with MS. Physical therapy may help improve balance and walking problems.
17 Fatigue is a common symptom of MS and may be both physical (for example, tiredness in the arms or legs) and cognitive (slowed processing speed or mental exhaustion). Daily physical activity programs of mild to moderate intensity can significantly reduce fatigue, although people should avoid excessive physical activity and minimize exposure to hot weather conditions or ambient temperature. Other drugs that may reduce fatigue include amantadine, methylphenidate, and modafinil. Occupational therapy can help people learn how to walk using an assistive device or in a way that saves physical energy. Stress management programs, relaxation training, membership in an MS support group, or individual psychotherapy may help some people. Pain from MS can be felt in different parts of the body. Trigeminal neuralgia (facial pain) is treated with anticonvulsant or antispasmodic drugs, or less commonly painkillers. Central pain, a syndrome caused by damage to the brain and/or spinal cord, can be treated with gabapentin and nortriptyline. Treatments for chronic back or other musculoskeletal pain may include heat, massage, ultrasound, and physical therapy.
Problems with bladder control and constipation may include urinary frequency, urgency, or the loss of bladder control. A small number of individuals retain large amounts of urine. Medical treatments are available for bladder-related problems. Constipation is also common and can be treated with a high-fiber diet, laxatives, and stool softeners. Sexual dysfunction can result from damage to nerves running through the spinal cord. Sexual problems may also stem from MS symptoms such as fatigue, cramped or spastic muscles, and psychological factors. Some of these problems can be corrected with medications. Psychological counseling also may be helpful. 18 Clinical depression is frequent among people with MS. MS may cause depression as part of the disease process and chemical imbalance in the brain. Depression can intensify symptoms of fatigue, pain, and sexual dysfunction. It is most often treated with cognitive behavioral therapy, and selective serotonin reuptake inhibitor (SSRI) antidepressant medications, which are less likely than other antidepressant medications to cause fatigue. Inappropriate and involuntary expressions of laughter, crying, or anger—symptoms of a condition called pseudobulbar affect—sometimes are associated with MS. These expressions are often incongruent with mood; for example, people with MS may cry when they are actually happy or laugh when they are not especially happy. The combination treatment of the drugs dextromethorphan and quinidine can treat pseudobulbar affect, as can other drugs such as amitriptyline or citalopram. Cognitive impairment—a decline in the ability to think quickly and clearly and to remember easily— affects up to three-quarters of people with MS. These cognitive changes may appear at the same time as the physical symptoms or they may develop gradually over time. Drugs such as donepezil may be helpful in some cases. Complementary and alternative therapies Many people with MS benefit from complementary or alternative approaches such as acupuncture, aromatherapy, ayurvedic medicine, touch and energy therapies, physical movement disciplines such as yoga and tai chi, herbal supplements, and biofeedback. Because of the risk of interactions between alternative and conventional therapies, people with MS should discuss all the therapies they are using with their 19 doctor, especially herbal supplements. Herbal supplements have biologically active ingredients that could have harmful effects on their own or interact harmfully with other medications. What research is being done?
The National Institute of Neurological Disorders and Stroke (NINDS), a component of the National Institutes of Health (NIH), is the leading federal funder of research on the brain and nervous system, including research on MS. In addition to NINDS, other NIH Institutes—including the National Institute of Allergy and Infectious Diseases (NIAID)—fund research on multiple sclerosis. More information on NIH efforts on multiple sclerosis research and on other disorders can be found using NIH RePORTER ( a searchable database of current and past research projects supported by NIH and other federal agencies. RePORTER also includes links to publications and patents citing support from these projects. Some people with MS use complementary or alternative therapies like yoga and tai chi. 20 Although researchers have not been able to identify the cause of MS with any certainty, there has been excellent progress in other areas of MS research— especially in the development of new treatments to prevent exacerbations of the disease. New discoveries are constantly changing MS treatment options and helping to reduce MS-related disability. Research projects being conducted by NINDS scientists or through NIH grants to universities and other sites throughout the United States cover a wide range of topics such as comorbidities, mechanisms of cognitive impairment, blood-brain barrier breakdown in MS, the role of sleep and circadian rhythms, rehabilitation strategies, and telehealth. Other topics include: • Biomarkers to accurately diagnose MS and monitor disease progression, including blood and imaging tests (such as MRI) • Genetic and environmental risk factors for MS such as low Vitamin D or the Epstein-Barr virus • The role of the gut microbiome and diet in MS • Mechanisms that underlie gender differences in the incidence and presentation of MS • MS risk factors and disease course in African American and Hispanic populations and disparities in care • The role of the immune system in MS, including its function in the central nervous system (CNS) • The role and crosstalk of various cell types in the CNS with relation to MS • Basic functions of myelination, demyelination, and axonal degeneration, and strategies to overcome axonal and myelin loss 21 Scientists sponsored by NIH’s NIAID are testing an experimental stem cell treatment (called autologous hematopoietic stem cell transplantation, or AHSCT) against the best available biologic therapies for severe forms of relapsing MS. Investigators in the BEAT-MS (BEst Available Therapy versus autologous hematopoietic stem cell transplant for Multiple Sclerosis) trial are removing some of the person’s immune cells and then infusing some of the person’s own blood-forming stem cells to reset the immune system so it no longer attacks the central nervous system. For more information about BEAT-MS and how to apply to participate in this study or other clinical studies, visit www.clinicaltrials.gov. Genetic research funded by NINDS is exploring the roles of “susceptibility genes”—genes that are associated with an increased risk for MS. Several candidate genes have been identified and researchers are studying their function in the nervous system to discover how they may lead to the development of MS. Other studies aim to develop better neuroimaging tools, such as more powerful MRI methods, to diagnose MS, track disease progression, and assess treatments. NINDS scientists are collecting magnetic resonance imaging of the brain and spinal cord and scans of the retina, along with other clinical and biological data, from more than 100 individuals with MS and 50 individuals without the disease over a period of years to observe changes in the course of MS over time. Investigators also are using MRI to study the natural history of MS and to help define the mechanism of action and cause of side effects of disease modifying therapies. 22 Intramural research programs on MS NINDS and other NIH Institutes have a very active MS intramural research program (scientists working at NIH). NINDS Intramural scientists have: • Established and continue to develop magnetic resonance imaging as a critical tool for examining the natural course of the disease in humans, monitoring disease progression, assessing effects of treatments in clinical trials, and understanding MS biology • Played an important role in understanding why some patients develop a rare and potentially fatal brain infection (progressive multifocal leukoencephalopathy) when taking potent MS drugs, and they are developing new treatments for this infection. • Unraveled mechanisms by which viruses, especially the Epstein-Barr virus, contribute to the development of MS • Conducted next-generation treatment trials targeting specific mechanisms of disease progression, using advanced MRI and fluid biomarkers as outcome measures • Developed the first MRI method to visualize the lymph vessels surrounding the brain, which play a critical role in neuro-immune communication 23 Translational research NIH supports translational studies to develop therapies that will stop or reverse the course of the disease, focusing on pathways that modify immune system function, repair damaged myelin, or protect neurons from damage. Researchers are also developing animal models of MS to more accurately predict drug response in human disease. Currently available animal models share some of the disease mechanisms and symptoms of MS but do not fully mimic the disease, especially in its clinically progressive phase. Scientists are developing better animal models that closely resemble MS in humans. Testing potential therapies in more accurate models may lead to successful treatments in humans with the disease.
24 Focus on progressive MS therapies Scientists continue to study the biology and mechanisms of relapsing-remitting MS while increasing efforts to stop or prevent the steady decline in function that occurs in progressive MS. In the MS-SPRINT trial, the NINDS NeuroNEXT clinical trials network tested the drug ibudilast as a potential neuroprotective drug for progressive MS and showed that the drug slowed the rate of brain shrinkage as compared to a placebo. NINDS Intramural scientists are conducting proof-of-concept clinical trials to address a key driver of clinical progression called the “slowly expanding lesion.” Focus on biomarkers As part of a larger effort to develop and validate effective biomarkers (signs that may indicate risk of a disease or be used to monitor its progression) for neurological disease, NINDS is supporting two definitive multicenter MS studies: • The Central Vein Sign in MS (CAVS-MS) study, which is testing whether a rapid MRI approach designed by NINDS Intramural scientists, can use the detection of a central vein passing through brain plaques to differentiate MS from other common neurological disorders that can mimic MS. The goal is to develop a reliable imaging test for MS in order to achieve a rapid but accurate diagnosis and reduce misdiagnosis, which may affect up to 20 percent of people currently diagnosed with MS. • A study to test whether a simple new blood test, which measures small amounts of neuron-derived proteins (called neurofilaments), can be used to predict the severity of disease and help determine whether MS drugs are working to protect the brain tissues. 25 Where can I get more information?
For more information on neurological disorders or research programs funded by the National Institute of Neurological Disorders and Stroke, contact the Institute’s Brain Resources and Information Network (BRAIN) at: BRAIN P.O. Box 5801 Bethesda, MD 20824 800-352-9424 www.ninds.nih.gov Information also is available from the following organizations: Multiple Sclerosis Association of America 856-488-4500 800-532-7667 Multiple Sclerosis Foundation 954-776-6805 888 673-6287 National Multiple Sclerosis Society 800-344-4867 Accelerated Cure Project for Multiple Sclerosis 781-487-0008 American Autoimmune Related Diseases Association 586-776-3900 800-598-4668 26 Myelin Repair Foundation 408-871-2410 National Ataxia Foundation 763-553-0020 National Organization for Rare Disorders (NORD) 203-744-0100 National Rehabilitation Information Center 301-459-5900 800-346-2742 301-459-5984 (TTY) Paralyzed Veterans of America 202-872-1300 800-555-9140 27 Prepared by: Office of Neuroscience Communications and Engagement National Institute of Neurological Disorders and Stroke National Institutes of Health Bethesda, Maryland 20892 NIH Publication No. 20-NS-75 August 2020 NIH . . . Turning Discovery Into Health |
16880 | https://www.onemathematicalcat.org/algebra_book/online_problems/calc_percent_inc_dec.htm | Calculating Percent Increase and Decrease
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Calculating Percent Increase and Decrease
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Changing Decimals to Percents
Changing Percents to Decimals
Writing Expressions Involving Percent Increase and Decrease
Problems Involving Percent Increase and Decrease
More Problems Involving Percent Increase and Decrease
When a quantity grows (gets bigger), then we can compute its percent increase.
When a quantity shrinks (gets smaller), then we can compute its percent decrease.
These concepts are thoroughly explored on this page.
Percent Increase
When a quantity grows (gets bigger), then we can compute its percent increase:
PERCENT INCREASE=(new amount−original amount)original amount PERCENT INCREASE=(new amount−original amount)original amount
Some people write this formula with 100%100% at the end, to emphasize that since it is percent increase, it should be reported as a percent.
So, here's an alternative way to give the formula:
PERCENT INCREASE=(new amount−original amount)original amount⋅100%PERCENT INCREASE=(new amount−original amount)original amount⋅100%
Recall that 100%=100⋅1 100=1.100%=100⋅1 100=1. So, 100%100% is just the number 1.1. Multiplying by 1 1 doesn't change anything except the name of the number! (See examples below.)
By the way, there's a very optimistic percent T-shirt here. Wear it and watch people smile!
Visualizing Percent Increase
0,0
original + 75% of original = 75% increase
Percent to increase by:
Type a nonnegative number in the box above, and then:
Note:
If percent increase=75%,percent increase=75%,
then the formula
percent increase=(new−original)original percent increase=(new−original)original
becomes
75%=(new−original)original 75%=(new−original)original
and solving for ‘new’ gives:
new=original+75%(original)new=original+75%(original)
Percent Decrease
When a quantity shrinks (gets smaller), then we can compute its percent decrease:
PERCENT DECREASE=(original amount−new amount)original amount PERCENT DECREASE=(original amount−new amount)original amount
or
PERCENT DECREASE=(original amount−new amount)original amount⋅100%PERCENT DECREASE=(original amount−new amount)original amount⋅100%
Both formulas have the following pattern:
Percent Inc/Dec=change in amount original amount Percent Inc/Dec=change in amount original amount
or
Percent Inc/Dec=change in amount original amount⋅100%Percent Inc/Dec=change in amount original amount⋅100%
Note that when you compute percent increase or decrease, you always compare how much a quantity has changed to the original amount.
Note also that the numerator in these formulas is always a positive number (or zero, if the quantity doesn't change at all).
Visualizing Percent Decrease
0,0
original - 25% of original = 25% decrease
Percent to decrease by:
Type a number between 0 0 and 100 100 in the box above, and then:
Note:
If percent decrease=25%,percent decrease=25%,
then the formula
percent decrease=(original−new)original percent decrease=(original−new)original
becomes
25%=(original−new)original 25%=(original−new)original
and solving for ‘new’ gives:
new=original−25%(original)new=original−25%(original)
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Examples
Question:A price rose from $5$5 to $7.$7. What percent increase is this?
Solution:Which is the original price?
Answer: $5$5
This will be the denominator.
% increase=(7−5)5=2 5=0.40=40%% increase=(7−5)5=2 5=0.40=40%
or
% increase=(7−5)5⋅100%=2 5⋅100%=2⋅100 5%=2⋅20%=40%% increase=(7−5)5⋅100%=2 5⋅100%=2⋅100 5%=2⋅20%=40%
Notes:
No matter which version of the formula you choose to use, be sure to give your answer as a percent.
The units have been suppressed (left out) in the calculations above. This is common practice when it is known that units will cancel, since it makes things look simpler.
Here is the same result, with the units in place:
% increase=$7−$5$5=$2$5=units have cancelled2 5=0.40=40%% increase=$7−$5$5=$2$5=2 5⏞units have cancelled=0.40=40%
In a correct use of the formulas for percent increase and decrease, the units of the numerator and denominator will always be the same, so the units will always cancel.
Question:A quantity decreased from 90 90 to 75.75.What percent decrease is this?
Solution:Which is the original quantity?
Answer: 90 90
This will be the denominator.
% decrease=(90−75)90=15 90≈0.1667=16.67%% decrease=(90−75)90=15 90≈0.1667=16.67%
Note: In the exercises below, if an answer does not come out exact, then it is rounded to two decimal places.
Question:An item went on sale for $13$13 from $16.$16.What percent decrease is this?
Solution:Which is the original price?
Answer: $16$16
This will be the denominator.
% decrease=(16−13)16=0.1875=18.75%% decrease=(16−13)16=0.1875=18.75%
Master the ideas from this section by practicing below:
When you're done practicing, move on to:
Problems Involving Percent Increase and Decrease
Concept Practice
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Click ‘Check your answer’ to check!
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16881 | https://math.stackexchange.com/questions/4929136/how-to-find-closest-average-for-a-set-of-variables-from-multiple-sets-of-numbers | optimization - How to Find Closest Average for a set of variables from Multiple Sets of Numbers - Mathematics Stack Exchange
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How to Find Closest Average for a set of variables from Multiple Sets of Numbers
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Asked 1 year, 3 months ago
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I'm working on a problem where I have several sets of numbers and I need to find the closest average for a set of target values by combining numbers from these sets.
For example, let's say I want to achieve the following target averages:
x = 5
y = 10
I have two sets of numbers:
a = {2, 8, 4, 15, 3, 6, 5}
b = {5, 2, 1, 7, 5, 3, 2}
To illustrate, if I combine the first and second index from set aa, I get an average for xx as:
x = (2 + 8) / 2 = 5
However, if I use these indices, the corresponding values from set b give an average for y as:
y = (5 + 2) / 2 = 3.5
The objective is to find combinations of up to 256 different indices (absolutely no more) from these sets to get the closest possible average to the target values for all variables. My actual data sets are much larger than the example provided.
I am looking for a method or algorithm to efficiently find the best combination of indices to get the closest approximation to my target averages.
Any guidance or suggestions on how to approach this problem would be greatly appreciated!
Thank you!
optimization
programming
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asked Jun 7, 2024 at 20:25
4DBug4DBug
21 1 1 bronze badge
4
indices can be used twice 4DBug –4DBug 2024-06-07 20:27:06 +00:00 Commented Jun 7, 2024 at 20:27
Suppose one set of indices gets x¯x¯ somewhat close and y¯y¯ very close, and another set of indices gets x¯x¯ very close and y¯y¯ somewhat close. How do you decide which set of indices is better?MJD –MJD 2024-06-07 20:30:56 +00:00 Commented Jun 7, 2024 at 20:30
This is my understanding of the question. OP has N N finite sequences of numbers, each the same length K K, say s i={s i 1,s i 2,…s i K}s i={s i 1,s i 2,…s i K}. Each sequence s i s i also has an associated target value t i t i. OP wants to choose a single vector of weights w 1…w K w 1…w K, with W=∑w i W=∑w i no more than 256 256, such that for the i i th sequence, the weighted mean s i¯=1 W∑j=1 K w i s i j s i¯=1 W∑j=1 K w i s i j is as close to t i t i as possible. It's not clear yet how one measures the overall closeness of the set of s i¯s i¯ to the targets t i t i since there is more than one.MJD –MJD 2024-06-07 20:44:59 +00:00 Commented Jun 7, 2024 at 20:44
I would just take the sum of all the errors and use the smallest result to find the 'closest'4DBug –4DBug 2024-06-07 20:52:13 +00:00 Commented Jun 7, 2024 at 20:52
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Here is an integer optimization approach, I hope for a faster approach.
Suppose you have r r such sets and the targets are t j t j.
Let the list be A j=[A j 1,A j 2,…,A j k]A j=[A j 1,A j 2,…,A j k].
Let x i x i be a nonnegative integer indicating how many times index i i is chosen.
We want to impose condition that 1≤∑x i≤256 1≤∑x i≤256 and we want ∑k i=1 A j i x i∑k i=1 x i∑i=1 k A j i x i∑i=1 k x i to be close to t j t j.
One possible formulation is we want |∑k i=1(A j i−t j)x i||∑i=1 k(A j i−t j)x i| to be small for each j j.
Let say we want to minimize ∑r j=1|∑k i=1(A j i−t j)x i|∑j=1 r|∑i=1 k(A j i−t j)x i|.
One possible formulation is
min∑j=1 r y j min∑j=1 r y j
subject to
y j≥∑i=1 k(A j i−t j)x i,∀j∈{1,…,r}y j≥∑i=1 k(A j i−t j)x i,∀j∈{1,…,r}
y j≥∑i=1 k−(A j i−t j)x i,∀j∈{1,…,r}y j≥∑i=1 k−(A j i−t j)x i,∀j∈{1,…,r}
1≤∑x i≤256 1≤∑x i≤256
x i≥0,x i∈Z x i≥0,x i∈Z
Of course, one doesn't have to use the sum of absolute value as the measure. Depending on the optimization criteria, we can come out with different formulation.
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answered Jun 7, 2024 at 20:47
Siong Thye GohSiong Thye Goh
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16882 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10659817/ | Single-Center Outcomes of WATCHMAN™ Implantation with Comparison to Oral Anticoagulant and Dual Antiplatelet Therapy - PMC
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. 2023 Oct 21;15(10):e47444. doi: 10.7759/cureus.47444
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Single-Center Outcomes of WATCHMAN™ Implantation with Comparison to Oral Anticoagulant and Dual Antiplatelet Therapy
Francis Demiraj
Francis Demiraj
1 Department of Neurology, Florida Atlantic University Charles E. Schmidt College of Medicine, Marcus Neuroscience Institute, Boca Raton, USA
Find articles by Francis Demiraj
1,✉, Michael S Benrubi
Michael S Benrubi
2 Department of Neurology, Nova Southeastern University Dr. Kiran C. Patel College of Osteopathic Medicine, Boca Raton, USA
Find articles by Michael S Benrubi
2, Denis Babici
Denis Babici
3 Department of Neurology, Florida Atlantic University Charles E. Schmidt College of Medicine, Boca Raton, USA
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3, Eti Muharremi
Eti Muharremi
4 Department of Neurology, Columbia University Irving Medical Center, New York, USA
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4, Ronald Pachon
Ronald Pachon
5 Department of Cardiology, University of Miami, Miami, USA
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5, Ahmed Osman
Ahmed Osman
6 Department of Cardiology, Broward General Medical Center, Fort Lauderdale, USA
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6
Editors: Alexander Muacevic, John R Adler
Author information
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Copyright and License information
1 Department of Neurology, Florida Atlantic University Charles E. Schmidt College of Medicine, Marcus Neuroscience Institute, Boca Raton, USA
2 Department of Neurology, Nova Southeastern University Dr. Kiran C. Patel College of Osteopathic Medicine, Boca Raton, USA
3 Department of Neurology, Florida Atlantic University Charles E. Schmidt College of Medicine, Boca Raton, USA
4 Department of Neurology, Columbia University Irving Medical Center, New York, USA
5 Department of Cardiology, University of Miami, Miami, USA
6 Department of Cardiology, Broward General Medical Center, Fort Lauderdale, USA
✉
Francis Demiraj fd341@mynsu.nova.edu
✉
Corresponding author.
Received 2023 Aug 30; Accepted 2023 Oct 21; Collection date 2023 Oct.
Copyright © 2023, Demiraj et al.
This is an open access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
PMC Copyright notice
PMCID: PMC10659817 PMID: 38021921
Abstract
Background
The WATCHMAN™ device is a Food and Drug Administration(FDA)-approved device that reduces the risk of stroke from atrial fibrillation (AF) in those who have a contraindication to taking oral anticoagulation. A key aspect of this device implantation is the choice of medical therapy in the months after device implantation with Vitamin K antagonist oral anticoagulants (OAC) being the mainstay of therapy but dual antiplatelet therapy (DAPT) poses as a potential alternative to patients who have a contraindication to OAC use.
Methods
Our single-center study retroactively followed 150 patients post-WATCHMAN™ implantation and evaluated outcomes at 12 months post-implantation in two cohorts, those treated with OAC or DAPT. Our results were obtained via chart review of a single-center electronic medical records system.
Results
In our study, 67.33% of study patients were males and 49.33% were on OAC compared to 36.00% that were on DAPT. Ten patients were not able to undergo device implantation. With this analysis, we found similarly low rates of complications such as stroke and device-associated thrombosis (DAT) in both groups. Our DAPT cohort did have a higher number of gastrointestinal (GI) bleeding but this was not significant in our analysis.
Discussion
Our study compares to larger trials that show similar outcomes between OAC and DAPT post-implantation of the WATCHMAN™ device. The increased number of GI bleeding in our DAPT cohort could be the result of the underlying advanced age and comorbidity of that patient cohort.
Conclusion
Our results suggest that DAPT is a safe alternative to OAC for patients undergoing WATCHMAN™ implantation.
Keywords: atrial fibrillation, dual antiplatelet therapy (dapt), watchman device, prevention of ischemic stroke, direct oral anticoagulant therapy
Introduction
Atrial fibrillation (AF) predisposes affected patients to an increased propensity to form clots via several mechanisms. In AF, the left atrium incompletely contracts, which causes a stasis of coagulation factors in the left atrial appendage. The WATCHMAN™ device addresses this pathology by serving to seal off the left atrial appendage and eliminate the opportunity for coagulation in this location of the heart. The WATCHMAN™ device is an FDA-approved device to reduce the risk of stroke from AF in those who have a contraindication to taking anticoagulation .
Based on the post-implantation treatment protocols from the PROTECT AF and PREVAIL trials, the vast majority of WATCHMAN™ implantations described in the literature were accompanied by warfarin anticoagulation for 45 days, followed by dual antiplatelet therapy (DAPT) for six months post-procedure and aspirin thereafter [2,3].For patients who cannot tolerate oral anticoagulation (OAC) for 45 days post-procedure, DAPT therapy has been used for six months post-implantation as an alternative.
Currently, the optimal DAPT regimen and its duration after WATCHMAN™ implantation are still under debate and might be patient-specific. OAC remains the standard therapy in patients with low bleeding risk, whereas the use of antiplatelet agents may be indicated in some clinical settings when the risk of thromboembolism is balanced by the risk of bleeding . The role of non-vitamin K antagonist oral anticoagulants (NOACs), which have seen increases in usage in recent years, is unknown.
The goal of our work was to retrospectively evaluate post-WATCHMAN™ implantation medication therapy and examine complication rates at 12 months post-implantation for our cohort. The DAPT regimen was evaluated in comparison to OAC use and special attention was given to the rate of device-associated thrombus (DAT), given the DAT rates shown in the PROTECT-AF, PREVAIL, CAP, and CAP2 trials .
Materials and methods
This retrospective observational study evaluated 150 patients who underwent WATCHMAN™ implantation from November 2015 to May 2019. All procedures involving human participants were in accordance with the ethical standards of the institutional research committee, the 1964 Helsinki Declaration, and its later amendments or comparable ethical standards. The Broward Health IRB approved the conduct of this human study (approval number BRH00025709).Implantation drug regimens and outcomes were obtained from a retroactive chart review of a single health system's electronic medical record. The inclusion criteria for this study were patients who were implanted with the WATCHMAN™ device at the same single center and were followed in the same health system. All WATCHMAN™ procedures conducted in our study were performed by the same team of two cardiac electrophysiologists who performed device implantation via the standard femoral vein approach. All patients included in the study underwent transesophageal echocardiography (TEE) at 45 days and six months post-procedure. The choice of medical therapy after device implantation was based solely on the patient’s ability to tolerate OAC therapy. Patients who were not able to tolerate OAC were placed on DAPT.Common reasons for OAC intolerance were history of bleeding, previous adverse reactions to OAC, documented medication nonadherence, and increased occupational risk of being on an OAC. Patients in our OAC cohort received OAC for 45 days followed by DAPT until their six-month TEE. Patients in our DAPT cohort received DAPT from the time of implantation until their six-month TEE. All patients in the DAPT and OAC groups had successful implantation of the WATCHMAN™ device as visualized by their six-month TEE. No patient in these two groups required prolongation of their medical therapy. Both groups were placed on aspirin indefinitely after successful TEE at the six-month mark.Patients were excluded if their 12-month period monitoring was incomplete due to lack of follow or transfer to another health system. These patients were not part of the 150 included in the study and comprised a small portion (<10 percent) of those analyzed for inclusion. Evaluation of clinical course outcomes was assessed by using one-year post-device implantation complication rates such as stroke, gastrointestinal (GI) bleeds, DAT, and death.
To assess the difference in the rate of complications between the two groups (DAPT and OAC) and the association with gender, we conducted a two-sample Z-test of proportions using Prism statistical software. This test compares the proportion of patients who experienced complications in each group, and tests whether this proportion is influenced by gender. The test statistic is computed as follows:z=(p1−p2) ⁄√p∗(1−p)∗[ (1n⁄1)+(1n⁄2)].
P1 and p2 are the sample proportions of patients who experienced complications in the DAPT and OAC groups, respectively; n1 and n2 are the sample sizes of the two groups; and p is the pooled proportion of patients who experienced complications in both groups. The p-value is obtained by comparing the test statistic to the standard normal distribution. The P-value <0.05 indicates that the null hypothesis of no difference between the proportions can be rejected, meaning that there is a significant difference in the rate of complications between the two groups.
Results
Figure 1 depicts the timeline and composition of treatment after WATCHMAN™ implantation used in our study.
Figure 1. Timeline of antithrombotic treatment after left atrial appendage occlusion with the WATCHMAN™ device based on bleeding risk as recommended by the EHRA/EAPCI consensus statement.
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EHRA, European Heart Rhythm Association; EAPCI, European Association of Percutaneous Coronary Interventions
Our results from our 150-patient cohort revealed similarities and some differences between treatment regimens. As shown in Table 1, our patient makeup consisted of 101 (67.33%) males and 49 (32.60%) females.74 (49.33%) patients were on OAC therapy compared to 54 (36.00%) on DAPT therapy. In the comparison of OAC therapy between genders, there was a greater proportion of males in this cohort, which was statistically significant (p=<0.01). No statistically significant difference was seen between genders on DAPT (p=0.22). There was also no significant difference in gender in the cohort that was not able to undergo implantation (p=0.08).Our results did reveal some concerns with WATCHMAN™ device implantation as 12 patients (8.00%) were not on any DAPT or OAC therapy during implant. Ten (6.66%) patients could not be implanted with the device, which is a higher-than-expected finding.
Table 1. Composition of participant gender and medication regimen.
OAC, oral anticoagulation; DAPT, dual antiplatelet therapy
Gender of patients Frequency n (%)
Males 101 (67.33)
Females 49 (32.60)
Drug regimen at implant Male gender frequency (%)Female gender frequency n (%)P-value
OAC 74 (49.33)54 (53.46)20 (40.82)<0.01
DAPT 54 (36.00)31 (30.69)23 (46.94)0.22
None 12 (8.00)9 (8.91)3 (6.12)0.08
Unsuccessful implant 10 (6.66)7 (6.93)3 (6.12)0.19
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In a comparison between complications, our results were encouraging for DAPT in place of OAC use. As shown in Table 2, there was no statistically significant difference in rates of ischemic stroke or DAT. Even though there was a higher number of GI bleeding in the DAPT cohort, this difference was not statistically significant (p=1.91). Overall, the rate of complications between the DAPT and OAC cohort were shown to be similar in our analysis.The 12 (8.00%) patients who were not placed on any medical therapy immediately post-device implantation did not suffer any ischemic cerebrovascular events in the 12-month period post-implant.
Table 2. Complications at 12 months post-WATCHMAN™ implantation.
OAC, oral anticoagulation; DAPT, dual antiplatelet therapy
Complications at 12 months OAC cohort n (%)DAPT cohort n (%)P-value
Device-associated thrombus 2 (2.70)0 (0)0.22
Stroke 2 (2.70)1 (1.85)0.75
Gastrointestinal bleed 2 (2.70)5 (9.26)1.91
Complications 6 (8.11)6 (11.11)1.44
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Discussion
A key aspect of WATCHMAN™ device implantation is the choice of subsequent medical therapy consisting of either OAC or DAPT to reduce stroke risk and other complications . Based on the 2019 American Heart Association/American College of Cardiology/Heart Rhythm Society (AHA/ACC/HRS) Focused Update of the 2014 AHA/ACC/HRS Guideline for the Management of Patients with Atrial Fibrillation, it is recommended to give aspirin (81-325 mg) with warfarin for 45 days post-WATCHMAN™ implantation. Warfarin is switched to clopidogrel (75 mg) after an absence of DAT and significant peri-device leak (jet width≤5 mm) on control TEE conducted around the 45th-day mark. Clopidogrel and aspirin are to be continued for up to six months post-procedure. Aspirin is continued indefinitely . DAT is a key marker of any study regarding left atrial appendage closure with the WATCHMAN™ device. The exact mechanism of device-related thrombosis is unclear, but it likely involves incomplete endothelization of tissue surrounding the device, leading to thrombus formation. In a review of the PROTECT AF trial, transesophageal echocardiography revealed a DAT rate of 5.7% in patients who had undergone WATCHMAN™ implantation and were placed on OAC therapy for 45 days . Compared to our study, this represents a higher rate of DAT in comparison to our OAC cohort (2.7%) and the DAPT cohort (0%). The PROTECT AF trial did have a significantly larger patient cohort of 707 patients and was conducted at 59 hospitals. Notably, it occurred from 2005 to 2008 and could be a consequence of the novelty of device implantation. A more recent review from 2008 to 2015 revealed a DAT rate of 3.9% in patients who had undergone WATCHMAN™ implantation and OAC medical therapy with 95% resolution of affected patients . A meta-analysis of several large studies including the PROTECT AF trial examined the incidences of DAT between OAC and DAPT therapy, with results showing a higher incidence in DAPT therapy but treatment therapies revealed similar safety and efficacy endpoints . DAT can increase the risk of neurological complications, but it is considered an unlikely complication of WATCHMAN™ implantation, which was seen in our study as well. Further research is needed to better understand the specific pathophysiology of DAT and identify any potential patient groups at increased risk of this complication .
An encouraging finding in our study was the limited incidence of stroke in both the OAC and DAPT patient cohorts. Given that the prevention of ischemic cerebral events is the treatment goal for WATCHMAN™ implantation, the success of this device in either treatment cohort in preventing these occurrences is paramount to the device’s continued use. The ASA Plavix feasibility study found comparable results to our DAPT cohort with all stroke episodes occurring in a small number of patients and made the case for DAPT being an effective alternative to OAC in the setting of WATCHMAN™ implantation . Given that patients can have absolute contraindications to OAC use, it is critical to have well-studied alternatives to ensure adequate ischemic outcomes.
A concerning finding in our study was the rate of GI bleeding in the DAPT treatment group with five reported cases of GI bleeds even though this difference was not significant. Fortunately, in our cohort these five bleeding events were non-fatal. Given that a bleeding history could be the reason for many of the patients were placed on DAPT and not OAC, it is difficult to ascertain that DAPT was the cause for the increased bleeding risk. In the ASA Plavix feasibility study, the history of bleeding tendencies (93%) was the main contraindication for OAC use. The EWOLUTION study revealed the advanced age and comorbidities in the DAPT cohort showed a major bleeding rate of 2.1% excluding periprocedural events, which was significantly less than the expected Vitamin K antagonists HAS-BLED percentage of 5.1 . Given the increased comorbidity and bleeding risk tendency of DAPT cohorts, DAPT appears to be a safe alternative in patients who cannot be placed on OAC therapy.
The strengths of our study included the 150 patient enrollment with a significant portion of DAPT patients. Given that all patients were followed at a single center and performed by the same team of two cardiac electrophysiologists, uniformity could be expected in the procedural aspects of WATCHMAN™ implantation as well as patient follow-up. A weakness in our study was the higher proportion of males included in the study, which was significant in the OAC cohort. Given that female gender is an independent risk factor for ischemic stroke in the setting of AF, our cohort results could have undervalued the potential risk of ischemic cerebral vascular events with OAC and DAPT treatments . Since our study was only observational, our patients were not matched according to their CHA2DS2-VASc scores and HAS-BLED scores, two scoring systems used for stratifying clotting and bleeding risk, respectively. Patients with higher HAS-BLED scores carry higher bleeding risks, creating potential confounding variables as seen by other research in regard to DAPT use among higher-risk patients, especially in their GI bleeding propensity. Further research can also analyze the outcomes of NOAC use post-WATCHMAN™ implantation, given that these agents have decreased the risk of bleeding .
Conclusions
Based on our single-center experience, our DAPT and OAC patient cohorts post-WATCHMAN™ device implantation fared similarly in a 12-month outcome analysis in regard to ischemic stroke and DAT. Although our DAPT cohort experienced a higher number of GI bleeding, it was not statistically significant in our analysis. GI bleeding is a common indication for patients not being placed on OAC and creates a possible confounding variable. Our work is in line with other larger studies that suggest DAPT is a safe and effective alternative to OAC therapy.
Acknowledgments
The authors would like to acknowledge Florida Heart Rhythm Associates for data collection.
The authors have declared that no competing interests exist.
Human Ethics
Consent was obtained or waived by all participants in this study. Broward Health Institutional Review Board issued approval BRH00025709. All procedures involving human participants were in accordance with the ethical standards of the institutional research committee and the 1964 Helsinki Declaration and its later amendments or comparable ethical standards
Animal Ethics
Animal subjects: All authors have confirmed that this study did not involve animal subjects or tissue.
References
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16883 | https://economics.stackexchange.com/questions/54811/different-elasticities-of-substitution | microeconomics - Different elasticities of substitution - Economics Stack Exchange
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Different elasticities of substitution
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I have been reading into generalizations of the concept of elasticity of substitution for more goods/inputs and three main possibilities emerged:
Hicksian EOS
Allen-Uzawa EOS
Morishima EOS
HICKS
As I understand it, the Hicksian EOS is just an application of EOS on pairs of goods while holding others constant:
σ H i,j=∂x j/x i x j/x i∂M R S i,j M R S i,j σ i,j H=∂x j/x i x j/x i∂M R S i,j M R S i,j
where M R S i,j=M U i/M U j M R S i,j=M U i/M U j. This can be computed for any i i and j j.
The Hicksian EOS should therefore measure the EOS between two goods/inputs while others are considered constant.
However, further than this, things start to get weird. For example, I have stumbled upon this:
σ??i,j=∂U∂x i∂U∂x j U⋅∂2 U∂x i∂x j σ i,j??=∂U∂x i∂U∂x j U⋅∂2 U∂x i∂x j
which has different implication than the previous elasticity, however it could still belong to Hicks?
Allen-Uzawa
What concerns Allen-Uzawa EOS, it is strange how many different computations are offered there:
There is this:
σ A U i,j=x 1∂U∂x 1+…x n∂U∂x n∂U∂x i∂U∂x j F i,j F σ i,j A U=x 1∂U∂x 1+…x n∂U∂x n∂U∂x i∂U∂x j F i,j F
where F F should be determinant and F i,j F i,j should be co-factor of ∂2 U∂x i∂x j∂2 U∂x i∂x j in this determinant.
However, there is also this:
σ i,j=X P D i,j S j σ i,j=X P D i,j S j
where X P D i,j X P D i,j is cross-price elasticity of demand and S j S j is a share of j j input in total cost.
Morishima
The last one is the weirdest:
σ M i,j=X P D j,i−P D j,j σ i,j M=X P D j,i−P D j,j
Where X P D X P D is a crossprice elasticity of j j input according to i i price and P D P D is own-price elasticity of demand.
The question:
Right now, this seems to me to be a little too chaotic. Almost all of these almost seem to be completely unrelated. Then, there is of course the fact that in later definition the prices were brought in, while the original definition defines the elasticity of substitution as a propriety of utility function (concretely, the curvature of the indifference curve). So I am a little bit confused right now and I would like to ask the following:
What do these have in common?
How do they differ?
How do those latter ones really generalize the former one?
Do these serve as a measure of how much goods are substitutes or complements?
Why the later ones operate with prices if the original is solely the propriety of a function?
What would be the connection between crossprice and EOS?
microeconomics
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elasticity
elasticity-of-substitution
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edited Mar 23, 2023 at 9:27
AthaeneusAthaeneus
asked Mar 21, 2023 at 20:32
AthaeneusAthaeneus
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This note only answers the last of your question. All these elasticities tend to disappear from the empirical literature since the publication of the influencial paper by
Blackorby, C. and R. R. Russel, 1989, Will the real elasticity of substitution please stand up? (A comparison ofthe Allen/Uzawa and Morishima elasticities). The American Economic Review, 79, 882-888.
The (own- or) cross-price elasticity of x∗j x j∗ wrt p k p k is defined by
∂x∗j∂p k(p,w)p k x∗j(p,w)∂x j∗∂p k(p,w)p k x j∗(p,w)
which has an unambiguous interpretation, and supplanted these weird alternative concepts.
The Morishima elasticity is directly obtained from these elementarity elasticities and also has a clear interpretation as it measures the sensibility of the optimal quantity ratio, x∗j/x∗k x j∗/x k∗ wrt a percentage change in the price ratio p j/p k p j/p k.
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answered Apr 1, 2024 at 12:41
BertrandBertrand
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My biggest problem with cross-price elasticity was that it defines substitutes and complements with respect to price, which did not seem that sound to me for a primary concept. After all, substitutes and complements must ensue from tastes (how you like something, whether you need products together or you can exchange one for another). Therefore, I have expected that the primary definitional concept should be derived from preferences only and that is why I searched for elasticity of substitution (for which, the Robinson's satisfied this realism).Athaeneus –Athaeneus 2024-04-02 05:05:34 +00:00 Commented Apr 2, 2024 at 5:05
@Athaeneus: complementarity or subsitutability relationships are not universal properties but conditional to price and budget. Two goods may be complements for low budget, and become substitutes when the budget increases (or when relative prices change). In other words, if you want to consider only the utility function, you may have to evaluate the cross derivatives of U over an infinity of virtual consumption bundles, without being sure which ones will effectively be chosen when a budget constraint will restrict the choice field. Is this convenient? It depends on your objective.Bertrand –Bertrand 2024-04-02 19:29:54 +00:00 Commented Apr 2, 2024 at 19:29
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To everyone interested in the same problem, I have found a perfect article that goes into great length and detail explaining what are the interconnections and attributes/characteristics of each of these elasticities. It is a very good starting point and I can recommend it by all means.
It is by João Eustáquio de Lima (2000) ALTERNATIVE DEFINITIONS OF ELASTICITY OF SUBSTITUTION: REVIEW AND APPLICATION
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answered Mar 30, 2024 at 19:36
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16884 | https://cooljugator.com/etymology/en/comply | comply
English word comply comes from Latin ad ((direction) toward, to, on, up to, for.), Latin pleo (To fill, to fulfill.), Latin -esco (Forms verbs from adjectives meaning "become (adjective)".), Latin con-, Vulgar Latin complesco
ad (Latin)
(direction) toward, to, on, up to, for.
pleo (Latin)
To fill, to fulfill.
-esco (Latin)
Forms verbs from adjectives meaning "become (adjective)".
con- (Latin)
Used in compounds to indicate a being or bringing together of several objects. Used in compounds to indicate the completeness, perfecting of any act, and thus gives intensity to the signification of the simple word.
complesco (Vulgar Latin)
complere (Latin)
complesco (Latin)
(Vulgar Latin) I fulfill, accomplish.
complir (Old French)
To accomplish; to complete; to do.
compli (Old French)
comply (English)
(archaic) To be ceremoniously courteous; to make one's compliments.. (archaic) To enfold; to embrace.. (archaic) To fulfill; to accomplish.. To yield assent; to accord; agree, or acquiesce; to adapt oneself; to consent or conform.
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16885 | https://mathworld.wolfram.com/Erdos-HeilbronnConjecture.html | Erdős-Heilbronn Conjecture -- from Wolfram MathWorld
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Erdős-Heilbronn Conjecture
Erdős and Heilbronn (Erdős and Graham 1980) posed the problem of estimating from below the number of sums where and range over given sets of residues modulo a prime , so that . Dias da Silva and Hamidoune (1994) gave a solution, and Alon et al.(1995) developed a polynomial method that allows one to handle restrictions of the type , where is a polynomial in two variables over .
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{{6, -7, 10}, {0, 3, -1}, {0, 5, -7}}
References
Alon, N.; Nathanson, M.B.; and Ruzsa, I.Z. "Adding Distinct Congruence Classes Modulo a Prime." Amer. Math. Monthly102, 250-255, 1995.Dias da Silva, J.A. and Hamidoune, Y.O. "Cyclic Spaces for Grassmann Derivatives and Additive Theory." Bull. London Math. Soc.26, 140-146, 1994.Erdős, P. and Graham, R.L. Old and New Problems and Results in Combinatorial Number Theory. Geneva, Switzerland: L'Enseignement Mathématique Université de Genève, Vol.28, 1980.Lev, V.F. "Restricted Set Addition in Groups, II. A Generalization of the Erdős-Heilbronn Conjecture." Electronic J. Combinatorics7, No.1, R4, 1-10, 2000.
Referenced on Wolfram|Alpha
Erdős-Heilbronn Conjecture
Cite this as:
Weisstein, Eric W. "Erdős-Heilbronn Conjecture." From MathWorld--A Wolfram Resource.
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16886 | https://pubchem.ncbi.nlm.nih.gov/compound/Methanediol | Methanediol | CH4O2 | CID 79015 - PubChem
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compound Summary
Methanediol
PubChem CID
79015
Structure
Molecular Formula
CH 4 O 2
Synonyms
Methanediol
Methylene glycol
463-57-0
dihydroxymethane
CHEBI:48397
View More...
Molecular Weight
48.041 g/mol
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Dates
Create: 2005-03-26
Modify: 2025-09-27
Description
Methanediol is the simplest member of the class of methanediols that is methane in which two of the hydrogens have been substituted by hydroxy groups. It is a member of methanediols, an aldehyde hydrate and a one-carbon compound.
ChEBI
The CIR Expert Panel concluded that formaldehyde and methylene glycol are safe for use in cosmetics when formulated to ensure use at the minimal effective concentration, but in no case should the formalin concentration exceed 0.2% (w/w), which would be 0.074% (w/w) calculated as formaldehyde or 0.118% (w/w) calculated as methylene glycol. Additionally, formaldehyde and methylene glycol are safe in the present practices of use and concentration in nail hardening products. However, formaldehyde and methylene glycol are unsafe in the present practices of use and concentration in hair smoothing products (a.k.a. hair straightening products). Formalin is an aqueous solution wherein formaldehyde (gas) has been added to water to a saturation point, which is typically 37% formaldehyde (w/w). Because of the equilibrium between formaldehyde and methylene glycol in aqueous solution, formalin is composed of both formaldehyde and methylene glycol.
International Journal of Toxicology 32(Suppl. 4):5-32, 2013
Cosmetic Ingredient Review (CIR)
Formaldehyde Solution is a saturated solution of formaldehyde, water, and typically another agent, most commonly methanol. In its typical form, formalin is 37% formaldehyde by weight (40% by volume), 6-13% methanol, and the rest water.
NCI Thesaurus (NCIt)
1 Structures
1.1 2D Structure
Structure Search
Get Image
Download Coordinates
Chemical Structure Depiction
Full screen Zoom in Zoom out
PubChem
1.2 3D Conformer
PubChem
2 Names and Identifiers
2.1 Computed Descriptors
2.1.1 IUPAC Name
methanediol
Computed by Lexichem TK 2.7.0 (PubChem release 2025.04.14)
PubChem
2.1.2 InChI
InChI=1S/CH4O2/c2-1-3/h2-3H,1H2
Computed by InChI 1.07.2 (PubChem release 2025.04.14)
PubChem
2.1.3 InChIKey
CKFGINPQOCXMAZ-UHFFFAOYSA-N
Computed by InChI 1.07.2 (PubChem release 2025.04.14)
PubChem
2.1.4 SMILES
C(O)O
Computed by OEChem 2.3.0 (PubChem release 2025.04.14)
PubChem
2.2 Molecular Formula
CH 4 O 2
Computed by PubChem 2.2 (PubChem release 2025.04.14)
PubChem
2.3 Other Identifiers
2.3.1 CAS
463-57-0
CAS Common Chemistry; ChemIDplus; EPA DSSTox; European Chemicals Agency (ECHA)
2.3.2 Related CAS
9015-98-9
Compound: Polymethylene glycol
CAS Common Chemistry
2.3.3 European Community (EC) Number
207-339-5
European Chemicals Agency (ECHA)
2.3.4 ChEBI ID
CHEBI:48397
ChEBI
2.3.5 DSSTox Substance ID
DTXSID50196801
EPA DSSTox
2.3.6 HMDB ID
HMDB0254518
Human Metabolome Database (HMDB)
2.3.7 Metabolomics Workbench ID
56967
Metabolomics Workbench
2.3.8 NCI Thesaurus Code
C62229
NCI Thesaurus (NCIt)
2.3.9 Nikkaji Number
J196.183G
Japan Chemical Substance Dictionary (Nikkaji)
2.3.10 RXCUI
1797646
NLM RxNorm Terminology
2.3.11 Wikidata
Q2365760
Wikidata
2.3.12 Wikipedia
Methanediol
Wikipedia
Dihydroxymethylidene
Wikipedia
2.4 Synonyms
2.4.1 MeSH Entry Terms
methylene glycol
Medical Subject Headings (MeSH)
2.4.2 Depositor-Supplied Synonyms
Methanediol
Methylene glycol
463-57-0
dihydroxymethane
CHEBI:48397
DTXSID50196801
6Z20YM9257
RefChem:156911
DTXCID50119292
207-339-5
Methandiol
formaldehyde hydrate
Methanediol-
UNII-6Z20YM9257
Dihydroxymethylidene
EINECS 207-339-5
Paraformaldehyde, powder
SCHEMBL4597
UNII-T0H3L6C7I5
SCHEMBL15938
SCHEMBL28460
SCHEMBL125283
SCHEMBL444487
SCHEMBL2473800
SCHEMBL7597667
T0H3L6C7I5
SCHEMBL30814232
SCHEMBL31478808
NS00080778
Aldacide; Flo-Mor; Granuform; Paraform; TransFix
F543822
Q332879
Q2365760
PubChem
3 Chemical and Physical Properties
3.1 Computed Properties
Property Name
Property Value
Reference
Property Name
Molecular Weight
Property Value
48.041 g/mol
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
XLogP3-AA
Property Value
-1
Reference
Computed by XLogP3 3.0 (PubChem release 2025.04.14)
Property Name
Hydrogen Bond Donor Count
Property Value
2
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Hydrogen Bond Acceptor Count
Property Value
2
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Rotatable Bond Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Exact Mass
Property Value
48.021129366 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
Monoisotopic Mass
Property Value
48.021129366 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
Topological Polar Surface Area
Property Value
40.5 Ų
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Heavy Atom Count
Property Value
3
Reference
Computed by PubChem
Property Name
Formal Charge
Property Value
0
Reference
Computed by PubChem
Property Name
Complexity
Property Value
2.8
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Isotope Atom Count
Property Value
0
Reference
Computed by PubChem
Property Name
Defined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Defined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Covalently-Bonded Unit Count
Property Value
1
Reference
Computed by PubChem
Property Name
Compound Is Canonicalized
Property Value
Yes
Reference
Computed by PubChem (release 2025.04.14)
PubChem
3.2 SpringerMaterials Properties
Chemical shift
Lineshape
SpringerMaterials
3.3 Chemical Classes
3.3.1 Cosmetics
Cosmetic ingredients (Methylene Glycol) -> CIR (Cosmetic Ingredient Review)
International Journal of Toxicology 32(Suppl. 4):5-32, 2013
Cosmetic Ingredient Review (CIR)
4 Spectral Information
4.1 1D NMR Spectra
4.1.1 13C NMR Spectra
Source of Sample
J. R. Ebdon, P. E. Heaton Polymer 18, 971 (1977)
Copyright
Copyright © 1980, 1981-2025 John Wiley & Sons, Inc. All Rights Reserved.
Thumbnail
SpectraBase
4.1.2 17O NMR Spectra
Copyright
Copyright © 2016-2025 W. Robien, Inst. of Org. Chem., Univ. of Vienna. All Rights Reserved.
Thumbnail
SpectraBase
4.2 IR Spectra
4.2.1 FTIR Spectra
Source of Sample
Bayer (by P.Mueller)
Copyright
Copyright © 1989, 1990-2025 Wiley-VCH GmbH. All Rights Reserved.
Thumbnail
SpectraBase
5 Related Records
5.1 Related Compounds with Annotation
Follow these links to do a live 2D search or do a live 3D search for this compound, sorted by annotation score. This section is deprecated (see the neighbor discontinuation help page for details), but these live search links provide equivalent functionality to the table that was previously shown here.
PubChem
5.2 Related Compounds
Same Connectivity Count
8
Same Parent, Connectivity Count
83
Same Parent, Exact Count
76
Mixtures, Components, and Neutralized Forms Count
1265
Similar Compounds (2D)
View in PubChem Search
Similar Conformers (3D)
View in PubChem Search
PubChem
5.3 Substances
5.3.1 PubChem Reference Collection SID
500800329
PubChem
5.3.2 Related Substances
All Count
1532
Same Count
72
Mixture Count
1460
PubChem
5.3.3 Substances by Category
PubChem
5.4 Entrez Crosslinks
PubMed Count
1480
Protein Structures Count
2
PubChem
6 Chemical Vendors
PubChem
7 Use and Manufacturing
7.1 Uses
Cosmetic Ingredient Review Link
CIR ingredient: Methylene Glycol
International Journal of Toxicology 32(Suppl. 4):5-32, 2013
Cosmetic Ingredient Review (CIR)
EPA CPDat Chemical and Product Categories
The Chemical and Products Database, a resource for exposure-relevant data on chemicals in consumer products, Scientific Data, volume 5, Article number: 180125 (2018), DOI:10.1038/sdata.2018.125
EPA Chemical and Products Database (CPDat)
7.1.1 Household Products
Household & Commercial/Institutional Products
Information on 2 consumer products that contain Methylene glycol in the following categories is provided:
• Personal Care
Consumer Product Information Database (CPID)
8 Toxicity
8.1 Toxicological Information
8.1.1 Toxicity Summary
Cosmetic Ingredient Review Conclusion
The CIR Expert Panel concluded that formaldehyde and methylene glycol are safe for use in cosmetics when formulated to ensure use at the minimal effective concentration, but in no case should the formalin concentration exceed 0.2% (w/w), which would be 0.074% (w/w) calculated as formaldehyde or 0.118% (w/w) calculated as methylene glycol. Additionally, formaldehyde and methylene glycol are safe in the present practices of use and concentration in nail hardening products. However, formaldehyde and methylene glycol are unsafe in the present practices of use and concentration in hair smoothing products (a.k.a. hair straightening products). Formalin is an aqueous solution wherein formaldehyde (gas) has been added to water to a saturation point, which is typically 37% formaldehyde (w/w). Because of the equilibrium between formaldehyde and methylene glycol in aqueous solution, formalin is composed of both formaldehyde and methylene glycol.
International Journal of Toxicology 32(Suppl. 4):5-32, 2013
Cosmetic Ingredient Review (CIR)
Cosmetic Ingredient Review Finding(s)
Safe in the present practices of use and concentration. Ingredient, concentration, and use information are available in documents discoverable at
Safe for use in cosmetics, with qualifications
The ingredient is unsafe for use in cosmetics
International Journal of Toxicology 32(Suppl. 4):5-32, 2013
Cosmetic Ingredient Review (CIR)
9 Literature
9.1 Consolidated References
PubChem
9.2 NLM Curated PubMed Citations
Medical Subject Headings (MeSH)
9.3 Springer Nature References
Springer Nature
9.4 Thieme References
Thieme Chemistry
9.5 Wiley References
Wiley
9.6 Nature Journal References
Ritson et al. Prebiotic synthesis of simple sugars by photoredox systems chemistry. Nature Chemistry, doi: 10.1038/nchem.1467, published online 30 September 2012
Nature Chemistry
Liu et al. Prebiotic photoredox synthesis from carbon dioxide and sulfite. Nature Chemistry, DOI: 10.1038/s41557-021-00789-w, published online 11 October 2021
Nature Chemistry
9.7 Chemical Co-Occurrences in Literature
PubChem
9.8 Chemical-Gene Co-Occurrences in Literature
PubChem
9.9 Chemical-Disease Co-Occurrences in Literature
PubChem
9.10 Chemical-Organism Co-Occurrences in Literature
PubChem
10 Patents
10.1 Depositor-Supplied Patent Identifiers
PubChem
Link to all deposited patent identifiers
PubChem
10.2 WIPO PATENTSCOPE
Patents are available for this chemical structure:
PATENTSCOPE (WIPO)
10.3 Chemical Co-Occurrences in Patents
PubChem
10.4 Chemical-Disease Co-Occurrences in Patents
PubChem
10.5 Chemical-Gene Co-Occurrences in Patents
PubChem
10.6 Chemical-Organism Co-Occurrences in Patents
PubChem
11 Interactions and Pathways
11.1 Protein Bound 3D Structures
View 2 proteins in NCBI Structure
PubChem
12 Classification
12.1 MeSH Tree
Medical Subject Headings (MeSH)
12.2 NCI Thesaurus Tree
NCI Thesaurus (NCIt)
12.3 ChEBI Ontology
ChEBI
12.4 ChemIDplus
ChemIDplus
12.5 Household Products Database Tree
Consumer Product Information Database (CPID)
12.6 EPA CPDat Classification
EPA Chemical and Products Database (CPDat)
12.7 NORMAN Suspect List Exchange Classification
NORMAN Suspect List Exchange
12.8 EPA DSSTox Classification
EPA DSSTox
12.9 Consumer Product Information Database Classification
Consumer Product Information Database (CPID)
12.10 MolGenie Organic Chemistry Ontology
MolGenie
13 Information Sources
Filter by Source
CAS Common ChemistryLICENSE The data from CAS Common Chemistry is provided under a CC-BY-NC 4.0 license, unless otherwise stated. Polymethylene glycol Methanediol
ChemIDplusLICENSE Methylene glycol ChemIDplus Chemical Information Classification
EPA DSSToxLICENSE Methylene glycol CompTox Chemicals Dashboard Chemical Lists
European Chemicals Agency (ECHA)LICENSE Use of the information, documents and data from the ECHA website is subject to the terms and conditions of this Legal Notice, and subject to other binding limitations provided for under applicable law, the information, documents and data made available on the ECHA website may be reproduced, distributed and/or used, totally or in part, for non-commercial purposes provided that ECHA is acknowledged as the source: "Source: European Chemicals Agency, Such acknowledgement must be included in each copy of the material. ECHA permits and encourages organisations and individuals to create links to the ECHA website under the following cumulative conditions: Links can only be made to webpages that provide a link to the Legal Notice page. Methanediol
ChEBIMethanediol ChEBI Ontology
Cosmetic Ingredient Review (CIR)LICENSE Methylene Glycol
NCI Thesaurus (NCIt)LICENSE Unless otherwise indicated, all text within NCI products is free of copyright and may be reused without our permission. Credit the National Cancer Institute as the source. NCI Thesaurus
Consumer Product Information Database (CPID)LICENSE Copyright (c) 2024 DeLima Associates. All rights reserved. Unless otherwise indicated, all materials from CPID are copyrighted by DeLima Associates. No part of these materials, either text or image may be used for any purpose other than for personal use. Therefore, reproduction, modification, storage in a retrieval system or retransmission, in any form or by any means, electronic, mechanical or otherwise, for reasons other than personal use, is strictly prohibited without prior written permission. Methylene glycol Household Products Classification Consumer Products Category Classification
EPA Chemical and Products Database (CPDat)LICENSE EPA CPDat Classification
Human Metabolome Database (HMDB)LICENSE HMDB is offered to the public as a freely available resource. Use and re-distribution of the data, in whole or in part, for commercial purposes requires explicit permission of the authors and explicit acknowledgment of the source material (HMDB) and the original publication (see the HMDB citing page). We ask that users who download significant portions of the database cite the HMDB paper in any resulting publications. Methanediol
Japan Chemical Substance Dictionary (Nikkaji)
Metabolomics WorkbenchMethanediol
Nature Chemistry
NLM RxNorm TerminologyLICENSE The RxNorm Terminology is created by the National Library of Medicine (NLM) and is in the public domain and may be republished, reprinted and otherwise used freely by anyone without the need to obtain permission from NLM. Credit to the U.S. National Library of Medicine as the source is appreciated but not required. The full RxNorm dataset requires a free license. methylene glycol
SpectraBaseMETHANEDIOL METHANEDIOL Mixture of polyols by reduction of ''formose''; the latter is obtained by polyaddition of formaldehyde
Springer Nature
SpringerMaterialsMethanediol
Thieme ChemistryLICENSE The Thieme Chemistry contribution within PubChem is provided under a CC-BY-NC-ND 4.0 license, unless otherwise stated.
WikidataLICENSE CCZero methanediol
Wikipediamethanediol Dihydroxymethylidene
Wiley
PubChem
Medical Subject Headings (MeSH)LICENSE Works produced by the U.S. government are not subject to copyright protection in the United States. Any such works found on National Library of Medicine (NLM) Web sites may be freely used or reproduced without permission in the U.S. methylene glycol MeSH Tree
NORMAN Suspect List ExchangeLICENSE Data: CC-BY 4.0; Code (hosted by ECI, LCSB): Artistic-2.0 NORMAN Suspect List Exchange Classification
MolGenieLICENSE CC-BY 4.0 MolGenie Organic Chemistry Ontology
PATENTSCOPE (WIPO)SID 403385081
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1 Structures Expand this menu
2 Names and Identifiers Expand this menu
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16887 | https://math.stackexchange.com/questions/2083201/which-integration-formula-for-frac1a2-x2-is-correct | Stack Exchange Network
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Which integration formula for $\frac{1}{a^2-x^2}$ is correct?
In my book integration formula for $\frac{1}{x^2-a^2}$ is given as $\frac{1}{2a}\ln(\frac{x-a}{x+a})$.
From the above formula we can write the formula for integration of $\frac{1}{a^2-x^2}$ as $-\frac{1}{2a}\ln(\frac{x-a}{x+a}) = \frac{1}{2a}\ln(\frac{x+a}{x-a}) $ [as we know $\ln(\frac{1}{x})=-\ln(x)$]
But, in my book the integration of $\frac{1}{a^2-x^2}$ is written as $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $.
We know that the integration in the second case cannot be $\frac{1}{2a}\ln(\frac{x+a}{x-a}) $ and $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $ at the same time. Which formula is the correct one and why?
4 Answers 4
Neither is correct.
It's not restrictive to assume $a>0$ (for $a=0$ the antiderivative has a different form; for $a<0$ change the intervals below accordingly).
The function $\frac{1}{x^2-a^2}$ is defined for $x\ne\pm a$ and is negative over $(-a,a)$, positive over $(-\infty,-a)$ and $(a,\infty)$.
The sign of $\frac{x-a}{x+a}$ is the same as the sign of $x^2-a^2$ (for $x\ne\pm a$, of course), so it's clear that
$$
\frac{1}{2a}\ln\frac{x-a}{x+a}
$$
cannot be an antiderivative over $(-a,a)$, because it is undefined there.
There's a simple way out. An antiderivative of $1/x$ is $\ln|x|$ (up to an arbitrary constant over $(-\infty,0)$ and an arbitrary constant on $(0,\infty)$, with no connection between each other).
The partial fraction decomposition is
$$
\frac{1}{x^2-a^2}=\frac{1}{2a}\frac{1}{x-a}-\frac{1}{2a}\frac{1}{x+a}
$$
so we can write
$$
\int\frac{1}{x^2-a^2}\,dx=
\frac{1}{2a}(\ln|x-a|-\ln|x+a|)=
\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|
$$
up to arbitrary constants in the intervals $(-\infty,-a)$, $(-a,a)$ and $(a,\infty)$.
Since $1/(a^2-x^2)=-1/(x^2-a^2)$, we have
$$
\int\frac{1}{a^2-x^2}\,dx=
-\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|=
\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|^{-1}=
\frac{1}{2a}\ln\left|\frac{x+a}{x-a}\right|
$$
again up to constants as before.
We do have that $$I =\int \frac {1}{a^2-x^2} dx =\frac {1}{2a}\ln \frac {x+a}{x-a} $$ But however we apply the absolute value function to arguments of logarithm functions in order to extend the antiderivative's domain giving us: $$I =\int \frac {1}{a^2-x^2} dx =\frac {1}{2a} \ln |\frac {x+a}{x-a}|$$Hope it helps.
WLOG, $a>0$. As there are singularities at $x=-a$ and $x=a$, the expressions can differ in the intervals $(-\infty,-a)$, $(-a,a)$ and $(a,\infty)$, depending on the signs of $x+a$ and $x-a$.
Correct expressions are
$$\int\frac{dx}{x^2-a^2}=\frac1{2a}\log\left|\frac{x+a}{x-a}\right|=\frac1{2a}\log\left|\frac{a+x}{a-x}\right|.$$
(But you are not allowed to evaluate the definite integrals across the singularities.)
Then the effect of a change of sign is
$$\int\frac{dx}{a^2-x^2}=-\frac1{2a}\log\left|\frac{x+a}{x-a}\right|=\frac1{2a}\log\left|\frac{x-a}{x+a}\right|=\frac1{2a}\log\left|\frac{a-x}{a+x}\right|.$$
Let $f(x)=\frac{1}{x^2-a^2}$.
$f$ is continuous at $(-\infty,-|a|),(-|a|,|a|)$ and $(|a|,+\infty)$.
in each interval
$$2af(x)=\frac{1}{x-a}-\frac{1}{x+a}$$
and $$\int f(x)dx=\frac{1}{2a}\ln(\frac{|x-a|}{|x+a|}).$$
the final expression depends on which interval $J$ we want the antiderivative.
for example, if $a>0$ and $J=(-a,a)$, we have
$$\int f(x)dx=\frac{1}{2a}\ln( \frac{a-x}{x+a} )+C$$
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㈠线线平行
①两直线共面且无公共点(线线平行的定义)
②同位角相等，两直线平行。
③内错角相等，两直线平行。
④同旁内角互补，两直线平行。
⑤利用相似证平行
⑥垂直于同一条直线的两直线平行(只适用于平面内)
⑦根据平行四边形性质推平行(对边平行且相等)
⑧一条直线与一个平面平行，则过这条直线的任一平面与此平面的交线，与该直线平行。
(a∥α，a⊂β，α∩β=b ⇒ a∥b)
⑨如果两个平行平面同时和第三个平面相交，那么它们的交线平行。
(α∥β，α∩γ=a. β∩γ=b⇒ a∥b)
⑩垂直于同一平面的两直线相互平行。
(a⊥α，b⊥α ⇒ a∥b)
㈡线面平行;
①平面外一条直线与此平面内的一条直线平行则该直线与此平面平行。
(a⊄α，b⊂α 且a∥b，a∥α)
②两个平面平行，其中一个平面内任意一条直线都平行于另一个平面。
(α∥β，a⊂α ⇒a∥β)
③平面外的两条平行线，若其中一条与这个平面平行，则另一条也与此平面平行。
(a⊄α ，b⊄α，a∥b，a(b)∥α⇒ b(a)∥α)
④两个平面互相垂直，如果其中一个平面外的一条直线垂直于另一个平面，那么这条直线和这个平面平行。
(α⊥β，l⊄α，l⊥β⇒l∥α)
⑤两平行平面外的一条直线，平行于两平行平面中的一个，那么它也平行于另外一个。
(l⊄α，l⊄β，l∥α(β)，α∥β⇒l∥α)
⑥平面外的一条直线，垂直于这个平面的一条垂线，那么这条直线与这个平面平行。
(l⊄α，a⊥α，l⊥a⇒l∥α)
㈢面面平行
①一个平面内的两条相交直线与另一个平面平行，则这两个平面平行。
(l⊂α，a⊂α，l∩a=B，l∥β，a∥β⇒α∥β)
②一个平面内的两条相交直线分别平行于另一个平面内的两条相交直线，那么这两个平面平行。
(l⊂α，a⊂α，l∩a=B，b⊂β，c⊂β，b∩c=D，l∥b，a∥c⇒α∥β)
③平行于同一平面的两平面平行。
(α∥β，γ∥β⇒α∥γ)
④垂直于同一条直线的两个平面互相平行。
(l⊥α，l⊥β⇒α∥β)
⑤如果两个平面的两条垂线互相平行，则这两个平面互相平行。
(l⊥α，a⊥β，l∥a⇒α∥β)
㈣线线垂直
①线面垂直则这条直线垂直于平面内的任意一条直线。
(l⊥α，b⊂α⇒l⊥b)
②一条直线垂直于两条平行线中的一条，则它也垂直于另一条。
(a∥b，c⊥a(b)⇒c⊥b(a))
③两条平行线中的一条直线，垂直于某个平面内的一条直线，则另一条直线也垂直于这条直线。
(a∥b，a(b)⊥c⇒b(a)⊥c)
④【三垂线定理】平面内的一条直线，如果与穿过这个平面的一条斜线在这个平面上的射影垂直，那么它也和这条斜线垂直。(可逆)
有图
⑤三角形相似。
⑥勾股定理。
⑦菱形◇对角线垂直。
⑧直径所对圆周角是直角。
㈤线面垂直
①一条直线与一个平面内的两条相交直线都垂直，则该直线与此平面垂直。
(l⊥m，l⊥n，m∩n=A，m⊂α，n⊂α⇒l⊥α)
②两个平面垂直，则其中任一平面内垂直于交线的直线都与另一平面垂直。
(α⊥β，α∩β=A，b⊂β，b⊥a⇒b⊥α)
③如果两个相交平面都垂直于第三个平面，那么它们的交线也垂直于第三个平面。
(α∩β=l，α⊥γ，β⊥γ⇒l⊥γ)
④一条直线垂直于两个平行平面中的一个，那么它也垂直于另一个。
(α∥β，l⊥α(β)⇒l⊥β(α))
⑤两条平行线中的一条垂直于某个平面，则另一条也垂直于这个平面。
(a∥b，a(b)⊥α⇒b(a)⊥α)
㈥面面垂直(二面角为90º)
①一个平面过另一个平面的垂线，则这两个平面垂直。
(a⊥α，a⊂β⇒α⊥β)
②一条直线和一个平面平行，若这条直线和另一个平面垂直，则这两个平面垂直。
(a∥α，a⊥β⇒α⊥β)
③如下图↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
④利用空间向量证明面面垂直。
⑤如果两个平面相交，如果它们所成的二面角为90°，则证明这两个平面互相垂直。
【补充:】【空间几何中常用概念补充:】
㈠等角定理:空间中如果两个角的两边分别对应平行，那么这两个角相等或互补。
㈡三角形的五心
1.重心:三角形中，三条中线的交点。
【三角形重心的性质:】
①角与重心的连线把三角形分成面积相等的两半。
②角与重心的连线把对边分成相等的两半。
③三条中线把三角形分成六个面积相等的三角形。
④重心把中线分成2/3与1/3两段。
2.垂心:三角形中，三条高的交点。
【三角形垂心的性质:】
①锐角三角形的垂心在三角形内；直角三角形的垂心在直角顶点上；钝角三角形的垂心在三角形外.
②三角形的垂心是它垂足三角形的内心；或者说,三角形的内心是它旁心三角形的垂心；
③垂心H关于三边的对称点,均在△ABC的外接圆上.
④ △ABC中,有六组四点共圆,有三组(每组四个)相似的直角三角形,且AH·HD=BH·HE=CH·HF.
⑤ H、A、B、C四点中任一点是其余三点为顶点的三角形的垂心(并称这样的四点为一—垂心组).
⑥ △ABC,△ABH,△BCH,△ACH的外接圆是等圆.
⑦ 在非直角三角形中,过H的直线交AB、AC所在直线分别于P、Q,则 AB/AP·tanB+AC/AQ·tanC=tanA+tanB+tanC.
⑧ 设O,H分别为△ABC的外心和垂心,则∠BAO=∠HAC,∠ABH=∠OBC,∠BCO=∠HCA.
⑨ 锐角三角形的垂心到三顶点的距离之和等于其内切圆与外接圆半径之和的2倍.
⑩ 锐角三角形的垂心是垂足三角形的内心；锐角三角形的内接三角形(顶点在原三角形的边上)中,以垂足三角形的周长最短.
⑪设锐角△ABC内有一点P,那么P是垂心的充分必要条件(PBPCBC+PBPAAB+PAPCAC=ABBCCA)
⑫设H为非直角三角形的垂心,且D、E、F分别为H在BC,CA,AB上的射影,H1,H2,H3分别为△AEF,△BDF,△CDE的垂心,则△DEF≌△H1H2H3.
⑬三角形垂心H的垂足三角形的三边,分别平行于原三角形外接圆在各顶点的切线.
3.内心:三角形中，三条角平分线的交点 (即三角形内切圆的圆心)
4.外心:三角形中，三条垂直平分线的交点。(即三角形外接圆的圆心)
5.旁心:旁心是三角形两条外角平分线和一条内角平分线的交点，它到三边的距离相等。
㈢直三棱柱:直三棱柱是各个侧面的高相等，底面是三角形，上表面和下表面平行且全等，所有的侧棱相等且相互平行且垂直于两底面的棱柱。上下表面三角形可以是任意三角形。
㈣正三棱柱:正三棱柱是直三棱柱的特殊情况，即上下面是正三角形。
㈤正三棱锥:正三棱锥是锥体中底面是正三角形，三个侧面是全等的等腰三角形的三棱锥。
㈥正四面体:正四面体必须每个面都是全等的等边三角形的四面体。
编辑于 2023-02-14 14:48・山西
高中数学
高考立体几何
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21 条评论
默认
最新
云水边静沐暖阳
面面平行判定的第五个符号语言有错误，请及时更正
2020-01-14
回复7
小鱼
修远
更改了吗
2020-07-07
回复2
修远
作者
好的，多谢
2020-02-24
回复1
神马浮云
大哥，我想问哈第四个的第四个，三垂线定理，如果平面内的直线和斜线垂直，又和投影垂直，那不是两个直角了么，
2020-02-20
回复2
神马浮云
修远
感谢大佬
2020-02-25
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修远
作者
这条定理说的是只要平面内的那条直线垂直于穿过平面的那条直线的射影，说明它和那条直线也垂直。
可以反过来用，你可以搜一下
2020-02-24
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历史学圣剑
谢谢您勒。
2020-02-17
回复2
许安
学几何我学得我都快没了
2020-05-13
回复1
冽暮
看成设了，绷不住了
2024-02-26
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a哦sss
麻烦问一下下，等角定理可以用来证明线线平行吗？如果可以的话，怎么证明欸谢谢啦
2023-10-02
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玛卡巴卡
可以直接用吗
2023-06-09
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修远
作者
可以
2023-06-14
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殊荣
麻烦说一下垂心那块的字母都是哪条边上的，谢谢~~
2020-05-20
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许安
太感谢了
2020-05-13
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毋宁
恶！
2020-03-21
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自律
谢谢你啦
2020-03-11
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高三数学冲刺复习立体几何的证明方法总结 =================== 一、平行的证明 1、线线平行： 1）平行的传递性 2）平面几何中中位线的性质、平行线定理 3）线面平行的性质 4）面与面平行的性质 5）直线与平面垂直的性质 2、线面平行的证明：1）线面平行… 教育界大神 发表于高中数学轻... 高中数学：立体几何中添加辅助线的策略 ================== 阿巴酱分享一道不错的平面几何趣题 ============= 如下图，在△ABC中，D,E分别在直线BA,CA上，满足BD=CE.直线BE交直线CD于F. 求证：点F的轨迹为一条直线. 这题看起来似乎不好下手，那我们就从特殊情况入手. 观察点F的运动我们可以得到： F∈… 铃声末响春日暖 初中几何基本知识汇总 ========== XIAOying
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16889 | https://www.mdedge.com/cutis/article/67069/atopic-dermatitis/kaposis-varicelliform-eruption-case-report-and-review/pdf | Kaposi's Varicelliform Eruption: A Case Report and Review of the Literature | MDedge
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Kaposi's Varicelliform Eruption: A Case Report and Review of the Literature
Cutis. 2004 February;73(2):115-122
February 1, 2004|Cutis
Kramer SC;Thomas CJ;Tyler WB;Elston DM
Author and Disclosure Information
✕
Drs. Kramer, Thomas, Tyler, and Elston report no conflict of interest. The authors report discussion of off-label use for intravenous vaccinia immune globulin under an investigational new drug protocol. Drs. Kramer and Thomas are residents from the Department of Dermatology, Dr. Tyler is an Associate in Pathology, and Dr. Elston is an Associate in Dermatology and Laboratory Medicine from the Departments of Dermatology and Pathology, Geisinger Medical Center, Danville, Pennsylvania.
Sasha C. Kramer, MD; Chadwick J. Thomas, MD; William B. Tyler, MD; Dirk M. Elston, MD
Accepted for publication December 23, 2003. Drs. Kramer and Thomas are residents from the Department of Dermatology, Dr. Tyler is an Associate in Pathology, and Dr. Elston is an Associate in Dermatology and Laboratory Medicine from the Departments of Dermatology and Pathology, Geisinger Medical Center, Danville, Pennsylvania.
Disseminated herpes or vaccinia in the setting of underlying skin diseases is known as Kaposi's varicelliform eruption (KVE). Patients typically present with disseminated vesicopustules in the areas of the most severe involvement of their underlying skin disease. We report a case of eczema herpeticum in a woman with a long-standing history of atopic dermatitis (AD). This report also reviews the literature on eczema herpeticum and eczema vaccinatum (EV), summarizes clinical and histopathologic characteristics and treatment, and discusses the recommendations of the Centers for Disease Control and Prevention for smallpox vaccination.
Disseminated herpes or vaccinia in the setting of underlying skin diseases is known as Kaposi’s varicelliform eruption (KVE). Patients typically present with disseminated vesicopustules in the areas of the most severe involvement of their underlying skin disease. We report a case of eczema herpeticum in a woman with a long-standing history of atopic dermatitis (AD). This report also reviews the literature on eczema herpeticum and eczema vaccinatum (EV), summarizes clinical and histopathologic characteristics and treatment, and discusses the recommendations of the Centers for Disease Control and Prevention for smallpox vaccination.
Patients with chronic inflammatory skin diseases, particularly atopic dermatitis (AD), are at risk for dissemination of cutaneous viral infections. Infection is most commonly caused by herpes simplex virus (HSV); however, it also may occur with coxsackievirus or vaccinia. The term Kaposi's varicelliform eruption (KVE) is used synonymously with eczema herpeticum when HSV infects eczematous skin. When KVE occurs in a patient who has received or has come in close contact with someone who has received the smallpox vaccination, it also is referred to as eczema vaccinatum (EV). The pathogenesis of KVE may be related to impaired immune surveillance or simply may represent a mechanical phenomenon secondary to decreased epithelial barrier function. As the threat of bioterrorism with smallpox increases, physicians must address the question of safety when vaccination is considered in individuals with a history of atopy.
Case Report
A 40-year-old woman with long-standing AD presented with a 5-day history of painful vesicles that had started on her right arm and gradually spread to involve the rest of her body. She had been evaluated by a physician and had been placed on prednisone, cephalexin, and triamcinolone without improvement. The patient did not have any preceding history of oral ulcerations or erosions but did report a history of intermittent "cold sores." On examination, her face, chest, arms, abdomen, back, and upper thighs were packed with confluent vesicopapules; some areas were eroded and weeping a yellow serous fluid (Figures 1 through 3). Direct fluorescent antibody (DFA) test yielded positive results of HSV-1 and HSV-2. A diagnosis of eczema herpeticum was made, and treatment with valacyclovir and cephalexin was initiated. Results of a bacterial culture yielded Staphylococcus and Streptococcus species. Biopsy results confirmed cytopathic changes diagnostic of herpesvirus infection with focal keratinocyte necrosis and acantholysis (Figures 4 and 5).
Comment
KVE was first described in 1887 by Moritz Kaposi who was Professor and Chairman of Dermatology at the University of Vienna School of Medicine.1 Kaposi initially thought the condition was secondary to a fungal infection, but the discovery of inclusion bodies histologically suggested a viral etiology.2 The term KVE is now used to describe disseminated herpes simplex, vaccinia, or coxsackievirus in the setting of certain underlying skin diseases.1
Eczema herpeticum is a term often used synonymously with HSV-associated KVE because eczema is the most common underlying skin condition seen in KVE.1 KVE also has been reported to occur in the setting of Darier disease,3,4 cutaneous T-cell lymphoma,5 pityriasis rubra pilaris,6 familial benign chronic pemphigus,7 congenital ichthyosiform erythroderma, seborrheic dermatitis, Wiskott-Aldrich syndrome,8 psoriasis, and lupus erythematosus.9 Additionally, KVE has been reported in patients who have disruption of the epidermal barrier either as a result of irritant contact dermatitis caused by vigorous scrubbing of the face with a facial cleanser,10 following a skin graft,11 in the setting of second-degree burns,12 or after dermabrasion.2 It also has been reported to occur in the setting of multiple myeloma.13
The literature presents conflicting data regarding immunologic defects in response to herpesvirus infection in patients with AD. Although it has been suggested that patients with AD have depressed cell-mediated immunity to HSV, studies have failed to confirm this.14,15 Some authors have postulated that decreased numbers of circulating natural killer cells and a decrease of IL-2 receptors cause patients with atopic eczema to be more susceptible to herpetic infection.15 It may be that the spread of infection is related purely to mechanical factors rather than to immune surveillance.
KVE can present in a primary form or a recurrent form.16 The primary form presents with clusters of umbilicated vesicles and vesicopustules that usually occur in areas where skin has been most affected by the underlying skin disease.17 The lesions gradually spread and are accompanied by systemic symptoms such as fever, malaise, and lymphadenopathy.1 Milder cases may have lesions limited to the head and neck.16 Over time, the vesicles may become hemorrhagic and later develop into erosions that can become secondarily infected.17 More severe cases can result in scarring. Recurrent cases usually are more limited with fewer systemic symptoms.16
Herpetic keratitis is a serious ocular sequela. Fortunately, despite the frequent involvement of vesicopustules on the face, ocular herpetic infection is rare in the setting of KVE.18 One study reported 3 patients with KVE with positive HSV conjunctival culture results but no visible ocular disease.8 KVE can be associated with viremia and involvement of the lungs, liver, brain, and gastrointestinal tract.1 Prior to the availability of antiviral therapy, deaths occurred secondary to rhabdomyolysis and renal failure.18 Bacterial infection of the eroded skin can progress to bacterial sepsis.
Differential Diagnosis and Diagnosis
References
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16890 | https://tasks.illustrativemathematics.org/content-standards/tasks/344 | Illustrative Mathematics
Typesetting math: 100%
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Two Interpretations of Division
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Alignments to Content Standards:3.OA.A.3
Student View
Task
Maria cuts 12 feet of ribbon into 3 equal pieces so she can share it with her two sisters. How long is each piece?
Maria has 12 feet of ribbon and wants to wrap some gifts. Each gift needs 3 feet of ribbon. How many gifts can she wrap using the ribbon?
IM Commentary
Both of the questions are solved by the division problem 12÷3 but what happens to the ribbon is different in each case. In the first case, the number of pieces of ribbon is fixed at 3 and the question is asking "how long is each piece?" (12 feet divided into 3 pieces gives 4 feet per piece.) In the second question, the size of each piece is fixed and the question is "how many pieces does one get?" (12 feet divided by 3 feet per gift gives 4 gifts.) The problem can be solved with a drawing of a tape diagram or number line. For problem 1, the line must be divided into 3 equal parts. The second problem can be solved by successive subtraction of 3 feet to see how many times it fits in 12.
In this case it is particularly helpful for the teacher to require students to justify their answers with a diagram. The way in which a student represents the problem can reveal whether or not he or she really understands the distinction between the two types of division problems shown here.
Solutions
Solution: Tape Diagram
This question asks, "how long is each piece?" so is a "How many in each group?" division problem:
3×?=12
12÷3=4, so each child gets a piece of ribbon that is 4 feet long.
This question asks, "how many pieces does one get?" so is a "How many groups?" division problem:
?×3=12
12÷3=4, so Maria can wrap 4 gifts.
Solution: Number Line
This question asks, "how long is each piece?" so is a "How many in each group?" division problem: 3×?=12
12÷3=4, so each child gets a piece of ribbon that is 4 feet long.
This question asks, "how many pieces does one get?" so is a "How many groups?" division problem:
?×3=12
12÷3=4, so Maria can wrap 4 gifts.
Two Interpretations of Division
Maria cuts 12 feet of ribbon into 3 equal pieces so she can share it with her two sisters. How long is each piece?
Maria has 12 feet of ribbon and wants to wrap some gifts. Each gift needs 3 feet of ribbon. How many gifts can she wrap using the ribbon?
Print Task
Typeset May 4, 2016 at 18:58:52. Licensed by Illustrative Mathematics under a
Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. |
16891 | https://www.sage.com/en-us/blog/what-is-markup-percentage/ | Skip to Content
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If the mention of markup makes you feel like you’re presenting on Shark Tank and you try to avoid answering – you’ve come to the right place.
Whether you’re a seasoned entrepreneur or just starting out, having a solid grasp on markup is critical to ensure you get your selling prices right.
In this article, we will cover the essentials of markup, the differences between markup and margin, and how they both impact your pricing strategy.
We also have a markup calculator that can quickly give you an answer. Or, you can use this to sense check your own calculations, so you can learn to confidently set your own prices knowing you have covered all your costs.
By the end of this read, you’ll have a clear understanding of what markup is and how to calculate it so you can make informed pricing strategy choices.
Most importantly, you will never get your markup and margin confused again.
Here’s what we’ll cover
Markup calculator
What is markup?
Markup percentage or multiplier
How to calculate selling price using markup percentage
How to calculate cost price from selling price and markup
What is the difference between margin and markup?
Markup vs. Margin calculation example
What do I need to consider when I calculate markup?
Should I include overhead costs in the markup calculation?
Markup calculator
The markup calculator can be used in a variety of ways for your own products.
It can also be used for market research on competitor products.
Use our markup calculator to find out:
What a product should be sold for, based on the cost and the desired markup.
How much a product should cost, based on the selling price and a markup percentage.
What is the markup percentage of a product, based on the selling price and the cost price.
Simply input either your selling price, cost price or markup percentage into the required fields and press calculate.
Toggle through the buttons at the top, to calculate either selling price, cost price or markup percentage.
What is markup?
Markup is the percentage increase over the cost price of an item. You add the percentage to the cost price of a product to determine its selling price.
It’s the amount you’re “marking up” the price from what you paid for it.
Markup is calculated by dividing the profit (selling price minus cost) by the cost price and then multiplying by 100.
Markup formula
Markup = ((Selling price – Cost price) / Cost price) x 100
Example, if you sell a product for $100 that costs you $60 to produce, your markup would be:
Markup = ((100 − 60/60) × 100) = 66.67%
This means you’re selling the product for 66.67% more than it cost to produce.
In simple terms
Markup uses the cost price as the base and margin uses the selling price as the base.
Because of this, markup percentages will always be higher than margin percentages for the same item.
Markup percentage or multiplier
Markup is often expressed as a multiplier. For example, the fashion industry standard markup is 2.5.
This is equivalent to 150% markup.
If a dress costs $100 to manufacture, this would be sold for $250.
Many businesses use accounting software to apply consistent markup rules, track cost of goods sold, and ensure accurate pricing across products. These tools can also generate reports that help monitor profit margins and pricing strategies over time.
Common multiple to markup percentages include:
| | | |
---
| MULTIPLE | MARKUP | MARGIN |
| 1.25 | 25% | 20% |
| 1.5 | 50% | 33.33% |
| 2 | 100% | 50% |
| 2.5 | 150% | 60% |
| 4 | 300% | 75% |
| 5 | 400% | 80% |
| 10 | 900% | 90% |
How to calculate selling price using markup percentage
Determining the selling price of your products using a markup percentage is a straightforward process.
By adding a specified percentage to the cost of your product, you can ensure that your selling price covers your costs and provides the desired profit.
Here’s a step-by-step guide on how to calculate the selling price using markup percentage:
Determine the cost price
The cost price is the total cost incurred to produce or purchase the product.
This includes manufacturing costs, shipping fees, and any other expenses directly associated with getting the product ready for sale (read below).
Decide on the markup percentage
The markup percentage is the percentage by which you want to increase the cost price to arrive at the selling price.
This percentage should account for your desired profit margin and other indirect costs.
Apply the markup percentage
Use the markup formula to calculate the selling price:
Selling Price = Cost Price + (Cost Price × Markup Percentage)
Alternatively, this can be simplified to:
Selling Price = Cost Price × (100% + Markup Percentage)
How to calculate cost price from selling price and markup
Starting with a selling price and reversing the calculation to determine the cost price is useful if you are thinking of launching a new product.
You can also estimate what your competitors’ cost price might be on comparable products.
If you know the selling price and the markup percentage applied, you can easily reverse-calculate to find the original cost price.
Formula for calculating cost price
Cost Price = Selling Price / (100% + Markup Percentage)
Knowing how to calculate the cost price from the selling price and markup allows you to understand the base cost of your products, enabling you to adjust prices strategically without compromising on profitability.
Accurate cost-price calculations also help to value inventory, budget for future purchases, and manage cash flow effectively.
What is the difference between margin and markup?
Let’s clarify the distinction between margin and markup.
These two terms are so often confused and if you get it wrong, you could be selling goods at a loss.
Markup and margin are both business terms used to refer to profitability, but they calculate profit in slightly different ways.
Understanding both markup and margin is crucial for businesses to set effective pricing strategies and analyze profitability.
Have a look at the differences so that you can ensure you make the right calculations.
| | |
--- |
| MARGIN | MARKUP |
| The percentage of the selling price that is profit | The percentage added to the cost price to arrive at the selling price |
| Margin = ((Selling Price − Cost Price) / Selling Price) × 100 | Markup = ((Selling Price − Cost Price) / Cost Price) × 100 |
| Calculated based on the selling price | Calculated based on the cost price |
| If a product costs $60 and sells for $100, the margin is 40% | If a product costs $60 and sells for $100, the markup is 66.67% |
| Helps understand the profitability of sales | Helps determine the selling price needed to achieve desired profits |
Why the difference matters
Understanding the difference between margin and markup is important because it affects your pricing strategy and profitability.
Pricing accuracy: Getting confused between markup and margin can lead to product underpricing, and a reduction in profit.
Competitive analysis: Markup is useful for product comparison in different industries, to ensure that you are competitively pricing your products.
Financial analysis: Margin is important for financial reporting, as it directly impacts your profit and loss statements. It also helps you understand the profitability of individual products and overall business performance.
Pricing strategy: Both metrics are important for strategic planning. Markup is useful for setting initial selling prices, and margin helps to evaluate ongoing profitability.
Markup vs. Margin calculation example
Consider you own a food truck, and you want to set the selling price for a burger.
You know your cost to make the burger is $5.00 and a 50% markup would give you a competitive advantage.
Markup = $5.00 + ($5.00 × 50%) = $7.50
Another way to calculate this is:
Your cost price = $5.00
You want a markup of 50% = $2.50
Add your markup to the cost price = $7.50
If you want a 50% margin, work backward from the desired margin.
Margin = $5.00 / 50% = $10.00
A 50% margin on a burger costing $5.00 would need a selling price of $10.00
A 50% margin results in a higher selling price ($10.00) compared to a 50% markup on the same burger ($7.50).
This example shows why you need to understand the difference between margin and markup and make sure you get your calculations right.
What do I need to consider when I calculate markup?
When setting your markup, there are several factors to consider to make sure you set a profitable price that covers your costs and is competitive.
1. Cost of goods sold (COGS)
Overhead costs are essential because they relate to all the indirect expenses required to operate your business, failing to account for these costs can result in underpricing (read below).
2. Profit margin
Determine the profit margin you want to achieve.
This involves understanding your business goals to ensure that the markup not only covers costs but also provides a satisfactory profit.
3. Market conditions
Analyze the pricing strategies of your competitors. Competitive pricing can help you position your product effectively in the market and attract customers.
High-demand products can typically sustain higher markups, whereas low-demand items might require lower markups to boost sales.
4. Customer perception
Ensure that the markup reflects the perceived value of the product to the customer.
Luxury items can command higher markups due to their perceived value and exclusivity.
Also, consider the target market’s willingness and ability to pay.
Markups should balance profitability with customer affordability to maintain sales volume.
5. Industry standards
Research average markups within your industry to ensure your pricing aligns within your vertical.
This helps in setting competitive prices while maintaining profitability.
6. Product life cycle
For new or innovative products, initial markups might be higher to capitalize on early adopters.
Over time, markups may be adjusted as the product moves through its life cycle.
Adjust markups based on seasonal demand.
Higher markups can be applied during peak seasons, while lower markups might be necessary during off-peak times to stimulate sales.
7. Sales strategy
Factor in planned discounts and promotional activities.
Ensure that even after discounts, the selling price remains profitable.
Consider offering lower markups on bulk purchases to encourage higher sales volumes, which can lead to economies of scale and increased overall profitability.
8. Economic factors
Monitor inflation rates, currency fluctuations, and other economic conditions that might affect the cost of goods and adjust your markup accordingly to maintain profitability.
9. Legal and regulatory requirements
Ensure that your pricing complies with legal and regulatory requirements, avoiding practices that could be considered unfair or deceptive.
Should I include overhead costs in the markup calculation?
Finally, don’t forget to include all your overhead costs when considering your markup.
Your costs of product are not limited to the direct cost of the goods, but also the indirect costs required for the running of the business:
Direct costs: Include all direct costs associated with producing or purchasing the product, such as materials, labor, and manufacturing expenses.
Indirect costs: Consider overhead costs such as rent, utilities, salaries, and administrative expenses that contribute to the overall cost structure.
To include overhead costs in your markup calculation, follow these steps:
Calculate total overhead costs
Identify all indirect costs associated with running your business.
This includes rent, utilities, salaries, office supplies, and other administrative expenses.
Determine overhead rate
Allocate overhead costs to individual products or services.
This can be done by determining an overhead rate, which is typically a percentage of direct costs or a fixed amount allocated based on the number of units produced or sold.
Add overhead to direct costs
Combine overhead costs with direct costs to get the total cost of producing or acquiring a product.
If you don’t consider your overhead costs then you could underprice your products with detrimental impact on your business.
Failing to consider your overheads and indirect costs will mean your cash flow will gradually decline and your income might not be enough to ensure you have sufficient cash flow to continue trading.
Once you’ve set your markup, monitoring your revenue and profit is crucial to ensure that the chosen markup is working for you.
As you grow, you can use cash flow management software to gain a real-time view of your revenue and profit and adjust your pricing as needed.
That knowledge and ability to act quickly is essential for retail operations.
Cloud financial solutions such as Sage Intacct include all these tools and enable that awareness and ability to act across multiple stores, locations, and organizations.
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16892 | https://www.webqc.org/balanced-equation-CaC2+H20+CO2=Ca(OH)2+CH2CHCO2H | CaC2 + H20 + CO2 = Ca(OH)2 + CH2CHCO2H - Balanced chemical equation, limiting reagent and stoichiometry
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Balance Chemical Equation - Online Balancer
Enter a chemical equation to balance:
Balanced equation:
6 CaC 2 + 16 H 2 O + 3 CO 2 = 6 Ca(OH)2 + 5 CH 2 CHCO 2 H
| Reaction stoichiometry | Limiting reagent |
--- |
| Compound | Coefficient | Molar Mass | Moles | Weight |
| Reagents |
| CaC 2 | 6 | 64.10 | | |
| H 2 O | 16 | 18.02 | | |
| CO 2 | 3 | 44.01 | | |
| Products |
| Ca(OH)2 | 6 | 74.09 | | |
| CH 2 CHCO 2 H | 5 | 72.06 | | |
Units: molar mass - g/mol, weight - g.
| Full ionic equation |
| 6 CaC 2 + 16 H 2 O + 3 CO 2 = 6 Ca{+2} + 12 OH{-} + 5 CH 2 CHCO 2 H |
| Balancing step by step using the algebraic method |
| Let's balance this equation using the algebraic method. First, we set all coefficients to variables a, b, c, d, ... a CaC 2 + b H 2 O + c CO 2 = d Ca(OH)2 + e CH 2 CHCO 2 H Now we write down algebraic equations to balance of each atom: Ca: a 1 = d 1 C: a 2 + c 1 = e 3 H: b 2 = d 2 + e 4 O: b 1 + c 2 = d 2 + e 2 Now we assign a=1 and solve the system of linear algebra equations: a = d a 2 + c = e 3 b 2 = d 2 + e 4 b + c 2 = d 2 + e 2 a = 1 Solving this linear algebra system we arrive at: a = 1 b = 2.6666666666667 c = 0.5 d = 1 e = 0.83333333333333 To get to integer coefficients we multiply all variable by 6 a = 6 b = 16 c = 3 d = 6 e = 5 Now we substitute the variables in the original equations with the values obtained by solving the linear algebra system and arrive at the fully balanced equation: 6 CaC 2 + 16 H 2 O + 3 CO 2 = 6 Ca(OH)2 + 5 CH 2 CHCO 2 H |
Direct link to this balanced equation:
Please tell about this free chemistry software to your friends!
Instructions on balancing chemical equations:
Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below
Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide
To enter an electron into a chemical equation use {-} or e
To enter an ion, specify charge after the compound in curly brackets: {+3} or {3+} or {3}.
Example: Fe{3+} + I{-} = Fe{2+} + I2
Substitute immutable groups in chemical compounds to avoid ambiguity.
For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,
but PhC2H5 + O2 = PhOH + CO2 + H2O will
Compound states [like (s) (aq) or (g)] are not required.
If you do not know what products are, enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.
Reaction stoichiometry could be computed for a balanced equation. Enter either the number of moles or weight for one of the compounds to compute the rest.
Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reagents. The limiting reagent row will be highlighted in pink.
Examples of complete chemical equations to balance:
Fe + Cl 2 = FeCl 3
KMnO 4 + HCl = KCl + MnCl 2 + H 2 O + Cl 2
K 4 Fe(CN)6 + H 2 SO 4 + H 2 O = K 2 SO 4 + FeSO 4 + (NH 4)2 SO 4 + CO
C 6 H 5 COOH + O 2 = CO 2 + H 2 O
K 4 Fe(CN)6 + KMnO 4 + H 2 SO 4 = KHSO 4 + Fe 2(SO 4)3 + MnSO 4 + HNO 3 + CO 2 + H 2 O
Cr 2 O 7{-2} + H{+} + {-} = Cr{+3} + H 2 O
S{-2} + I 2 = I{-} + S
PhCH 3 + KMnO 4 + H 2 SO 4 = PhCOOH + K 2 SO 4 + MnSO 4 + H 2 O
CuSO 45H 2 O = CuSO 4 + H 2 O
calcium hydroxide + carbon dioxide = calcium carbonate + water
sulfur + ozone = sulfur dioxide
Examples of the chemical equations reagents (a complete equation will be suggested):
H 2 SO 4 + K 4 Fe(CN)6 + KMnO 4
Ca(OH)2 + H 3 PO 4
Na 2 S 2 O 3 + I 2
C 8 H 18 + O 2
hydrogen + oxygen
propane + oxygen
Understanding chemical equations
A chemical equation represents a chemical reaction. It shows the reactants (substances that start a reaction) and products (substances formed by the reaction). For example, in the reaction of hydrogen (H₂) with oxygen (O₂) to form water (H₂O), the chemical equation is:
H 2 + O 2 = H 2 O
However, this equation isn't balanced because the number of atoms for each element is not the same on both sides of the equation. A balanced equation obeys the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction.
Balancing with inspection or trial and error method
This is the most straightforward method. It involves looking at the equation and adjusting the coefficients to get the same number of each type of atom on both sides of the equation.
Best for: Simple equations with a small number of atoms.
Process: Start with the most complex molecule or the one with the most elements, and adjust the coefficients of the reactants and products until the equation is balanced.
Example:H 2 + O 2 = H 2 O
1. Count the number of H and O atoms on both sides. There are 2 H atoms on the left and 2 H atom on the right. There are 2 O atoms on the left and 1 O atom on the right.
2. Balance the oxygen atoms by placing a coefficient of 2 in front of H 2 O: H 2 + O 2 = 2H 2 O
3. Now, there are 4 H atoms on the right side, so we adjust the left side to match: 2H 2 + O 2 = 2H 2 O
4. Check the balance. Now, both sides have 4 H atoms and 2 O atoms. The equation is balanced.
Balancing with algebraic method
This method uses algebraic equations to find the correct coefficients. Each molecule's coefficient is represented by a variable (like x, y, z), and a series of equations are set up based on the number of each type of atom.
Best for: Equations that are more complex and not easily balanced by inspection.
Process: Assign variables to each coefficient, write equations for each element, and then solve the system of equations to find the values of the variables.
Example: C 2 H 6 + O 2 = CO 2 + H 2 O
1. Assign variables to coefficients:a C 2 H 6 + b O 2 = c CO 2 + d H 2 O
2. Write down equations based on atom conservation:
2 a = c
6 a = 2 d
2 b = 2c + d
Assign one of the coefficients to 1 and solve the system.
a = 1
c = 2 a = 2
d = 6 a / 2 = 3
b = (2 c + d) / 2 = (2 2 + 3) / 2 = 3.5
Adjust coefficient to make sure all of them are integers. b = 3.5 so we need to multiply all coefficient by 2 to arrive at the balanced equation with integer coefficients: 2 C 2 H 6 + 7 O 2 = 4 CO 2 + 6 H 2 O
Balancing with oxidation number method
Useful for redox reactions, this method involves balancing the equation based on the change in oxidation numbers.
Best For: Redox reactions where electron transfer occurs.
Process: identify the oxidation numbers, determine the changes in oxidation state, balance the atoms that change their oxidation state, and then balance the remaining atoms and charges.
Example: Ca + P = Ca 3 P 2
1. Assign oxidation numbers:
Calcium (Ca) has an oxidation number of 0 in its elemental form.
Phosphorus (P) also has an oxidation number of 0 in its elemental form.
In Ca 3 P 2, calcium has an oxidation number of +2, and phosphorus has an oxidation number of -3.
Identify the changes in oxidation numbers:
Calcium goes from 0 to +2, losing 2 electrons (oxidation).
Phosphorus goes from 0 to -3, gaining 3 electrons (reduction).
Balance the changes using electrons: Multiply the number of calcium atoms by 3 and the number of phosphorus atoms by 2.
Write the balanced Equation:3 Ca + 2 P = Ca 3 P 2
Balancing with ion-electron half-reaction method
This method separates the reaction into two half-reactions – one for oxidation and one for reduction. Each half-reaction is balanced separately and then combined.
Best for: complex redox reactions, especially in acidic or basic solutions.
Process: split the reaction into two half-reactions, balance the atoms and charges in each half-reaction, and then combine the half-reactions, ensuring that electrons are balanced.
Example: Cu + HNO 3 = Cu(NO 3)2 + NO 2 + H 2 O
1. Write down and balance half reactions:Cu = Cu{2+} + 2{e} H{+} + HNO 3 + {e} = NO 2 + H 2 O
2. Combine half reactions to balance electrons. To accomplish that we multiple the second half reaction by 2 and add it to the first one:Cu + 2H{+} + 2HNO 3 + 2{e} = Cu{2+} + 2NO 2 + 2H 2 O + 2{e}
3. Cancel out electrons on both sides and add NO 3{-} ions. H{+} with NO 3{-} makes HNO 3 and Cu{2+} with NO 3{-} makes Cu(NO 3)3:Cu + 4HNO 3 = Cu(NO 3)2 + 2NO 2 + 2H 2 O
Learn to balance chemical equations:
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16893 | https://discrete.openmathbooks.org/more/mdm/sec_counting-binom.html | Pascal's Triangle and Binomial Coefficients
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More Discrete Mathematics via Graph Theory
Richard Grassl, Oscar Levin
Contents
IndexPrevUpNext
Annotations
ContentsPrevUpNext
Front Matter
Colophon
Acknowledgements
Preface
1 Graph Theory
Definitions
Walking Around Graphs
Planar Graphs and Euler's Formula
Applications of Euler's Formula
Coloring
Matching in Bipartite Graphs
2 Basic Combinatorics
Pascal's Triangle and Binomial Coefficients
Proofs in Combinatorics
Counting with Equivalence
Counting with Recursion
The Catalan Numbers
3 Advanced Combinatorics
Counting Partitions
The Principle of Inclusion and Exclusion: the Size of a Union
Generating Functions
Stirling Numbers of the Second Kind
Bell Numbers
Integer Partitions
Back Matter
Group Projects
Background Material
Hints to Selected Activities
GNU Free Documentation License
List of Symbols
Index
Colophon
Authored in PreTeXt
Section 2.1 Pascal's Triangle and Binomial Coefficients
¶
Let's start our investigation of combinatorics by examining Pascal's Triangle. If you are already familiar with this mathematical object, try to see it with fresh eyes. Look at the triangle as if you have no idea where it comes from. What do you notice?
Activity 63
Treating Pascal's Triangle as nothing more than a triangular array of numbers, what do you notice about it? Are there any patterns to the numbers? Try to find some patterns and then see if you can answer (or have already answered) the following:
(a)
How are the entries in the triangle found? What would the next row down be?
(b)
What is the sum of the entries in the n n th row?
(c)
What is 1−5+10−10+5−1?1−5+10−10+5−1? What pattern is this an example of, and does it always work?
(d)
What is 1+3+6+10+15?1+3+6+10+15? What pattern is this an example of, and does it always work?
(e)
What is 1+7+15+10+1?1+7+15+10+1? What pattern is this an example of, and does it always work?
(f)
Are there any groups of numbers in the triangle that are particularly recognizable?
We would like to understand why the patterns you discovered above exist, and to prove that they really do exist everywhere in the triangle. There are at least two ways we can proceed (and we will try to proceed in both ways to maximize our understanding). First, we can use notation to describe the entries in the triangle, provide a definition for what each entry is, and prove our results entirely based on these definitions. Second, we can ascribe some meaning to the entries in the triangle, saying what the numbers represent, and then make arguments about those representations.
The advantage to pursuing both these approaches is that doing so creates a feedback loop. A fact we establish using pure symbolic manipulation allows us to conclude something about the real world and what these numbers represent. That helps us understand the applications better, which in turn might help us make arguments about those applications, which can then establish a purely symbolic fact about the triangle.
Let's pause and just for fun consider a few discrete math questions that probably have nothing to do with Pascal's Triangle or each other. Probably.
Activity 64
The integer lattice is the set of all points in the Cartesian plane for which both the x x and y y coordinates are integers.
A lattice path is one of the shortest possible paths connecting two points on the lattice, moving only horizontally and vertically. For example, here are three possible lattice paths from the points (0,0)(0,0) to (3,2):(3,2):
(a)
How many lattice paths are there between (0,0)(0,0) and (3,1)?(3,1)? Draw or list them all (you might want to invent some notation for describing a path without drawing it).
(b)
How many lattice paths are there between (0,0)(0,0) and (2,2)?(2,2)? Draw or list them all to be sure.
(c)
How many lattice paths are there between (0,0)(0,0) and (3,2)?(3,2)? How can you be sure?
Activity 65
Recall that a subset A A of a set B B is a set all of whose elements are also elements of B;B; we write A⊆B.A⊆B. For example, some of the subsets of B={1,2,3,4,5}B={1,2,3,4,5} are {2,4,5},{2,4,5},{1,2,3,4,5}{1,2,3,4,5} and ∅.∅. Recall also that the cardinality of a set is simply the number of elements in it.
Since we will often consider sets of the form {1,2,3,…,n},{1,2,3,…,n}, let's adopt the notation [n][n] for this set.
(a)
Write out all subsets of ={1,2,3}.={1,2,3}. Then group the subsets by cardinality. How many subsets have cardinality 0? How many have cardinality 1? 2? 3?
(b)
Write out all subsets of ,, and determine how many have each possible cardinality.
(c)
How many subsets of are there of each cardinality? Try to answer this questions without writing out all 32 subsets.
Activity 66
Some definitions: A bit is either 0 or 1 (bit is short for “binary digit”). Thus a bit string is a string of bits. The length of a bit string is the number of bits in the string; the weight of a bit string is the number of 1's in the string (or equivalently, the sum of the bits). A n n-bit string means a bit string of length n.n.
We will write B n k B k n to mean the set of all n n-bit strings of weight k.k. So for example, some of the elements in B 5 3 B 3 5 are,
11100 10101 01101.11100 10101 01101.
(a)
Write out all 3 3-bit strings. Group these by weight. How many strings are there of each weight?
(b)
Write out all 4 4-bit strings. You might want to use the list you had in the previous part as a starting point (how would you do this?). Again, group these strings by weight and see how many there are of each type.
(c)
How many 5 5-bit strings of weight 3 are there? List all of these, and say how they relate to some of the 4 4-bit strings you found above.
We are not quite ready to consider the algebraic approach yet, but here is a fun algebra exercise.
Activity 67
Multiply out (and collect like terms): (x+y)3.(x+y)3. Repeat for (x+y)4(x+y)4 and (x+y)5.(x+y)5.
(a)
What is the coefficient of x 3 y 2 x 3 y 2 in the expansion of (x+y)5?(x+y)5?
(b)
What are the coefficients of x 2 y 2 x 2 y 2 and x 3 y x 3 y in the expansion of (x+y)4?(x+y)4? How does this relate to the previous question? Why does this make sense?
(c)
If you believe the connection between coefficients and Pascal's triangle you have discovered above, find the coefficient of x 7 y 4 x 7 y 4 in the expansion of (x+y)11(x+y)11 using Pascal's triangle.
Hopefully by now you have some questions that are begging to be answered. Here are two that we will focus on specifically:
Why are the answers to all the counting questions above the same as each other?
Why are the answers to all the counting questions above numbers in Pascal's triangle?
We could answer the first question by answering the second: if we can say why each answer is a particular entry in Pascal's triangle, then we know two answers will be equal because they are in the same place on the triangle. But we can do better. In fact, we should be able to explain why these counting questions correspond to each other without knowing any of the answers.
Activity 68
(a)
Explain why the number of lattice paths from (0,0)(0,0) to (k,n−k)(k,n−k) is the same as number of n n-bit strings of weight k.k. You do not need to give a formal proof here, just explain the idea.
(b)
Explain why the number of n n-bit strings of weight k k is the same as the number of k k-element subsets of [n]={1,2,3,…,n}.[n]={1,2,3,…,n}. Again, an informal argument is fine.
(c)
Explain why the number of k k-element subsets of [n][n] is the same as the coefficient of x k y n−k x k y n−k in the expansion of (x+y)n.(x+y)n.
Hint
This is a little harder. It might help to notice that there is really nothing special about ([n]) except that it contains (n) elements. In fact, what can you say about the number of (k)-element subsets of any (n)-element set? Then, what (n)-element set would we look at when expanding ((x+y)^n\text{?})
Are you satisfied with the explanations you gave above? Do you think they would count as a formal mathematical proof that the counting questions have the same answer? One way to make this sort of an argument more precise is to involve precisely defined mathematical objects.
Activity 69
Consider functions f:X→Y f:X→Y where X X and Y Y are finite sets. In fact, let's say X=.X=. If you need a refresher on basic ideas about functions, check out Section B.3.
(a)
Can you say anything at all about Y Y if you know there is some function f:X→Y?f:X→Y? Give some examples of sets Y Y and functions f.f.
(b)
What if f:X→Y f:X→Y is injective (in other words, one-to-one)? Which of the examples you found above no longer work? What can you conclude about Y?Y?
(c)
What if f:X→Y f:X→Y is surjective (i.e., onto)? Now what can you say about Y?Y?
(d)
What if f:X→Y f:X→Y is bijective (so both injective and surjective)? State the general principle.
The previous activity was meant to help you discover what is usually called the bijection principle: Two sets X X and Y Y have the same cardinality if and only if there is a bijection f:X→Y f:X→Y.
If you want to prove that two sets have the same number of elements, all that is requires is to define a function from one set to the other, and prove that the function is a bijection. For us, those two sets will be the set of outcomes we are counting. So for example, we can show that the set of n n-bit strings of weight k k has the same cardinality as the set of k k-element subsets of [n].[n]. This will prove that the counting questions, “how many n n-bit strings of weight k k are there?” and, “how many k k-element subsets of [n][n] are there?” have the same answer.
Activity 70
(a)
Define a bijection f:X→Y f:X→Y where X X is the set of lattice paths from (0,0)(0,0) to (k,n−k)(k,n−k) and Y Y is the set of n n-bit strings of weight k.k. Prove that f f really is a bijection.
(b)
Define a bijection f:X→Y f:X→Y where X X is the set of n n-bit strings of weight k k and Y Y is set of k k-element subsets of [n]={1,2,3,…,n}.[n]={1,2,3,…,n}. Again, prove that f f is a bijection.
(c)
Can you now conclude that the number of lattice paths from (0,0)(0,0) to (k,n−k)(k,n−k) is the same as the number of k k-element subsets of [n]?[n]? Can you easily give the bijection (and be sure it is one)?
So all these counting questions correspond to each other. Why do the answers always fall in Pascal's triangle? To answer this, we need to start with a definition for the numbers in Pascal's triangle.
First, some notation: let's write (n k)(n k) for the k k th entry in row n.n. For both n n and k,k, start counting at 0. So for example (5 1)=5,(5 1)=5, and (n 0)=(n n)=1(n 0)=(n n)=1 for all n n (or at least the ones we can see). We will read (n k)(n k) as “n n choose k k” for reasons that will become clear soon.
Now we need to decide what (n k)(n k) should be defined as. It is not enough to simply define it as the specific entry in Pascal's triangle, since we haven't yet defined Pascal's triangle (we will want to define Pascal's triangle as the triangle of the numbers (n k)(n k)). Here are some choices:
Possible Definitions for (n k)(n k)
For each integer n≥0 n≥0 and integer k k with 0≤k≤n 0≤k≤n the number
(n k),(n k),
read “n n choose k k” is:
the number of n n-bit strings of weight k,k, that is, (n k)=|B n k|.(n k)=|B k n|.
(n k)(n k) is the number of subsets of a set of size n n each with cardinality k.k.
(n k)(n k) is the number of lattice paths of length n n containing k k steps to the right.
(n k)(n k) is the coefficient of x k y n−k x k y n−k in the expansion of (x+y)n.(x+y)n.
(n k)(n k) is the number of ways to select k k objects from a total of n n objects.
(n 0)=(n n)=1(n 0)=(n n)=1 and for all n≥1 n≥1 and 1<k<n 1<k<n we have (n k)=(n−1 k−1)+(n−1 k).(n k)=(n−1 k−1)+(n−1 k).
You have already prove that choices 1, 2 and 3 are equivalent, and we have good reason to believe these are also equivalent to choice 4.
What about choice 5? This is new, although not really, as it captures all of the previous four: Each of our counting problems above can be viewed in this way:
How many bit strings have length 5 and weight 3? We must choose 3 3 of the 5 bits to be 1's. There are (5 3)(5 3) ways to do this, so there are (5 3)(5 3) such bit strings.
How many subsets of {1,2,3,4,5}{1,2,3,4,5} contain exactly 3 elements? We must choose 3 3 of the 5 elements to be in our subset. There are (5 3)(5 3) ways to do this, so there are (5 3)(5 3) such subsets.
How many lattice paths are there from (0,0) to (3,2)? We must choose 3 of the 5 steps to be towards the right. There are (5 3)(5 3) ways to do this, so there are (5 3)(5 3) such lattice paths.
What is the coefficient of x 3 y 2 x 3 y 2 in the expansion of (x+y)5?(x+y)5? We must choose 3 of the 5 copies of the binomial to contribute an x.x. There are (5 3)(5 3) ways to do this, so the coefficient is (5 3).(5 3).
In fact, choice 5 is usually taken as the definition of (n k),(n k), which is why we say n n choose k.k. Although interestingly enough, the other common name for (n k)(n k) is a binomial coefficient, which refers to the coefficients of the binomial (x+y)n.(x+y)n.
What does all this have to do with Pascal's triangle though? If you are like me, you think of the entries in the triangle as the result of adding the two entries above it. That is, Pascal's triangle contains those numbers described by choice 6 above.
Recurrence relation for (n k)(n k)
For any n≥1 n≥1 and 0<k<n:0<k<n:
(n k)=(n−1 k−1)+(n−1 k).(n k)=(n−1 k−1)+(n−1 k).
The last piece of the puzzle is to connect this definition to all the others. Of course given the work done above, it would be enough to prove that any one of the models for binomial coefficients match the recurrence for Pascal's triangle, but it is instructive to check all four.
Activity 71
(a)
Prove that bit strings satisfy the same recurrence as Pascal's triangle. That is prove that |B n k|=|B n−1 k−1|+|B n−1 k|.|B k n|=|B k−1 n−1|+|B k n−1|.
(b)
Prove that subsets satisfy the same recurrence as Pascal's triangle. Make sure you clearly state what you are proving.
(c)
Prove that lattice paths satisfy the same recurrence as Pascal's triangle. What exactly do you need to prove here?
(d)
Prove that binomial coefficients (the actual coefficients of the expansion of the binomial (x+y)n(x+y)n) satisfy the same recurrence as Pascal's triangle.
At last we can rest easy that can use Pascal's triangle to calculate binomial coefficients and as such find numeric values for the answers to counting questions. How many pizza's containing three distinct toppings can you make if the pizza place offers 10 toppings? Well the answer should be (10 3),(10 3), and Pascal's triangle says that is 120. |
16894 | https://www.cs.cmu.edu/~odonnell/papers/csp-sos-lower-bounds.pdf | Sum of squares lower bounds for refuting any CSP Pravesh K. Kothari∗ Ryuhei Mori† Ryan O’Donnell‡ David Witmer‡ January 16, 2017 Abstract Let P : {0, 1}k →{0, 1} be a nontrivial k-ary predicate. Consider a random instance of the constraint satisfaction problem CSP(P) on n variables with ∆n constraints, each being P applied to k randomly chosen literals. Provided the constraint density satisfies ∆≫1, such an instance is unsatisfiable with high probability. The refutation problem is to efficiently find a proof of unsatisfiability.
We show that whenever the predicate P supports a t-wise uniform probability distribution on its satisfying assignments, the sum of squares (SOS) algorithm of degree d = Θ( n ∆2/(t−1) log ∆) (which runs in time nO(d)) cannot refute a random instance of CSP(P).
In particular, the polynomial-time SOS algorithm requires e Ω(n(t+1)/2) constraints to refute random instances of CSP(P) when P supports a t-wise uniform distribution on its satisfying assignments. Together with recent work of Lee et al. [LRS15], our result also implies that any polynomial-size semidef-inite programming relaxation for refutation requires at least e Ω(n(t+1)/2) constraints.
More generally, we consider the δ-refutation problem, in which the goal is to certify that at most a (1 −δ)-fraction of constraints can be simultaneously satisfied. We show that if P is δ-close to supporting a t-wise uniform distribution on satisfying assignments, then the degree-Θ( n ∆2/(t−1) log ∆) SOS algorithm cannot (δ + o(1))-refute a random instance of CSP(P). This is the first result to show a distinction between the degree SOS needs to solve the refutation problem and the degree it needs to solve the harder δ-refutation problem.
Our results (which also extend with no change to CSPs over larger alphabets) subsume all previously known lower bounds for semialgebraic refutation of random CSPs. For every con-straint predicate P, they give a three-way hardness tradeoffbetween the density of constraints, the SOS degree (hence running time), and the strength of the refutation. By recent algorithmic results of Allen et al. [AOW15] and Raghavendra et al. [RRS16], this full three-way tradeoffis tight, up to lower-order factors.
∗Princeton University and IAS. kothari@cs.princeton.edu †Department of Mathematical and Computing Sciences, Tokyo Institute of Technology. mori@is.titech.ac.jp ‡Computer Science Department, Carnegie Mellon University.
Supported by NSF grant CCF-1618679.
{odonnell,dwitmer}@cs.cmu.edu 1 Introduction Where are the hard problems?
In computational complexity, we have a comprehensive theory of worst-case hardness, assuming P ̸= NP. The theory is particular rich in the context of constraint satisfaction problems (CSPs) — optimization tasks that are both simple to state and powerfully expressive. (See, e.g., [BJK05, Rag08].) But despite our many successes in the theory of NP-completeness and NP-hardness-of-approximation, we know relatively little about the nature of hard instances. For example, 3-SAT is conjecturally hard to solve — or even approximate to factor 7 8 + ϵ — in 2o(n) time. But what do hard(-seeming) instances look like? How can we generate one? These sorts of questions are a key part of understanding what makes various algorithmic problems truly hard. They are particularly important for CSPs, as these are nearly always the starting point for hardness reductions; the ability to find hard instances for CSPs yields the ability to find hard instances for many other algorithmic problems.
In some sense, a single instance can never be “hard” because its solution can always be hard-coded into an algorithm. Thus it is natural to turn to random instances, and the theory of average-case hardness. Uniformly random instances of CSPs are a particularly simple and natural source of hard(-seeming) instances. Furthermore, they arise as the fundamental object of study in many disparate areas of research, including cryptography [ABW10], proof complexity [BSB02], hardness of approximation [Fei02], learning theory [DLSS14], SAT-solving [SAT], statistical physics [CLP02], and combinatorics.
1.1 Random CSPs Let Ωbe a finite alphabet and let P be a collection of nontrivial predicates Ωk →{0, 1}. An input I to the problem CSP(P) consists of n variables x1, . . . , xn, along with a list E of m constraints (P, S), where P is a predicate from P, and S ∈[n]k is a scope of k distinct variables. We often think of the associated “factor graph”: that is, the bipartite graph with n “variable-vertices”, m “constraint-vertices” of degree k, and edges defined by the scopes.
Given I, the algorithmic task is to find an assignment to the variables so as to maximize the fraction of satisfied constraints, avg(P,S)∈E P(xS1, . . . , xSk).
We write Opt(I) for the maximum possible fraction, and say that I is satisfiable if Opt(I) = 1. For a fixed constraint density ∆= ∆(n) > 0, a random instance of CSP(P) is defined simply by choosing m = ∆n constraints uniformly at random: random scopes and random P ∈P.
The most typical examples involve a binary alphabet Ω= {0, 1}, a fixed predicate P : {0, 1}k → {0, 1}, and P = P ±, where by P ± we mean the collection of all 2k predicates obtained by letting P act on possibly-negated input bits (“literals”). For example, if P is the k-bit logical OR function, then CSP(P ±) is simply the k-SAT problem. In this introductory section, we’ll focus mainly on these kinds of CSPs.
For random CSPs, the constraint density ∆plays a critical role; naturally, the larger it is, the more likely I is to be unsatisfiable. For a fixed P, it is easy to show the existence of constants α0 < α1 such that when ∆< α0, a random instance I of CSP(P) is satisfiable with high probability (whp), and when ∆> α1, I is unsatisfiable whp. For most interesting P, it is conjectured that there is even a sharp threshold α0 = α1 = αc. (This has been proven for k-SAT with k large enough [DSS15].
See [CD09] for a characterization of those Boolean CSPs for which a sharp threshold is expected.) For random instances with subcritical constraint density, ∆< αc, the natural algorithmic task is to try to efficiently find satisfying assignments. There have been quite a few theoretical and 1 practical successes for this problem, for ∆quite large and even approaching αc [Gab16, MPRT16].
On the other hand, for random instances with supercritical constraint density, ∆> αc, the natural algorithmic task is to try to efficiently refute them; i.e., produce a certificate of unsatisfiability. For many CSPs, this task seems much harder, even heuristically. For example, random 3-SAT instances are unsatisfiable (whp) once ∆> 4.49 [DKMPG08]; however, even for ∆as large as n.49 there is no known algorithm that efficiently refutes random instances — even heuristically/experimentally.
Thus the refutation task for random instances of CSPs with many constraints may be a source of simple-to-generate, yet hard-to-solve problems.
1.2 The importance and utility of hardness assumptions for random CSPs In this section, we discuss the task of refuting random CSP instances and the importance of understanding the “constraint density vs. running time vs. refutation strength tradeoff” for all predicate families P. To define our terms, a (weak) refutation algorithm for CSP(P) is an algorithm that takes as input an instance I and either correctly outputs “unsatisfiable”, or else outputs “don’t know”. For a given density ∆(larger than the critical density), we say the algorithm “succeeds” if it outputs “unsatisfiable” with high probability (over the choice of I, and over its internal coins, if any). More generally, we can consider refutation algorithms that always output a correct upper bound on Opt(I); we call them δ-refutation algorithms if they output an upper bound of 1 −δ (or smaller) with high probability. The case of δ = 1/m, where m = ∆n is the number of constraints, corresponds to the simple weak refutation task described earlier (with an output of “1” corresponding to “don’t know”). In general, we refer to δ as the “strength” of the refutation.
For a wide variety of areas — cryptography, learning theory, and approximation algorithms — it is of significant utility to have concrete hardness assumptions concerning random CSPs. Because uniformly random CSPs are very simply and concretely defined, they form an excellent basis for constructing other potentially hard problems by reduction. An early concrete hypothesis comes from an influential paper of Feige [Fei02]: Feige’s R3SAT Hypothesis.
For every small δ > 0 and for large enough constant ∆, there is no polynomial-time algorithm that succeeds in δ-refuting random instances of 3-SAT.
Feige’s main motivation was hardness of approximation; e.g., he showed that the R3SAT Hy-pothesis implies stronger hardness of approximation results than were previously known for several problems (Balanced Bipartite Clique, Min-Bisection, Dense k-Subgraph, 2-Catalog). By reducing from these problems, several more new hardness of approximation results based on Feige’s Hy-pothesis have been shown in a variety of domains [BKP04, DFHS06, Bri08, AGT12]. Feige [Fei02] also related hardness of refuting 3-SAT to hardness of refuting 3-XOR. The assumption that refut-ing 3-XOR is hard has been used to prove new hardness results in subsequent work [OWWZ14].
Alekhnovich [Ale03] further showed that certain average-case hardness assumptions for XOR imply additional hardness results, as well as the existence of secure public key cryptosystems.
In even earlier cryptography work, Goldreich [Gol00] proposed using the average-case hardness of random CSPs as the basis for candidate one-way functions. Subsequent work (e.g., [MST03]) suggested using similar functions as candidate pseudorandom generators (PRGs). The advantage of this kind of construction is the extreme simplicity of computing the PRG: indeed, its output bits can be computed in NC0, constant parallel time. Further work investigated variations and extensions of Goldreich’s suggestion [ABW10, ABR12, AL16]; see Applebaum’s survey [App13] for many more details. Of course, the security of these candidate cryptographic constructions depends heavily on the hardness of refuting random CSPs. Applebaum, Ishai, and Kushilevitz [AIK06] took 2 a slightly different approach to showing that PRGs exist in NC0, instead basing their result on one of Alekhnovich’s average case XOR hardness assumptions [Ale03].
Finally, a recent exciting sequence of works due to Daniely and coauthors [DLSS13, DLSS14, DS14, Dan15] has linked hardness of random CSPs to hardness of learning. By making concrete conjectures about the hardness of refuting random CSP(P) for various P and for superpolyno-mial ∆, they obtained negative results for several longstanding problems in learning theory, such as learning DNFs and learning halfspaces with noise.
1.3 Desiderata for hardness results While Feige’s R3SAT Hypothesis has proven useful in hardness of approximation, there are several important strengthenings of it that would lead to even further utility. We discuss here four key desiderata for hardness results about random CSPs: 1. Predicates other than SAT. The hardness of random 3-SAT and 3-XOR has been most extensively studied, but for applications it is quite important to consider other predicates.
For hardness of approximation, already Feige [Fei02] noted that he could prove stronger inapproximability for the 2-Catalog problem assuming hardness of refuting random k-AND for large k. Subsequent work has used assumptions about the hardness of refuting CSPs with other predicates to prove additional worst-case hardness results [GL04, AAM+11, CMVZ12, BCMV12, RSW16]. Relatedly, Barak, Kindler, and Steurer [BKS13] have recently considered a generalization of Feige’s Hypothesis to all Boolean predicates, in which the assumption is that the “basic SDP” provides the best δ-refutation algorithm when ∆= O(1). They also describe the relevance of predicates over larger alphabet sizes and with superconstant arity for problems such as the Sliding Scale Conjecture and Densest k-Subgraph. Bhaskara et al. [BCG+12] prove an SOS lower bound for Densest k-Subgraph via a reduction from Tulsiani’s SOS lower bound for random instances of CSP(P) with P a q-ary linear code [Tul09]. A computational hardness assumption for refutation of this CSP would therefore give a hardness result for Densest k-Subgraph.
Regarding cryptographic applications, the potential security of Goldreich’s candidate PRGs depends heavily on what predicates they are instantiated with. Goldreich originally suggested a random predicate, with a slightly superconstant arity k. However algorithmic attacks on random CSP(P) by Bogdanov and Qiao [BQ09] showed that predicates that are not at least “3-wise uniform” do not lead to secure PRGs with significant stretch. Quite a few subsequent works have tried to analyze what properties of a predicate family P may — or may not — lead to secure PRGs [BQ09, ABR12, OW14, AL16].
Regarding the approach of Daniely et al. to hardness of learning, there are close connections between the predicates for which random CSP(P) is assumed hard and the concept class for which one achieves hardness of learning. For example, the earlier work [DLSS14] assumed hardness of refuting random CSP(P ±) for P being (i) the “Huang predicate” [Hua13, Hua14], (ii) Majority, (iii) a certain AND of 8 thresholds; it thereby deduced hardness of learning (i) DNFs, (ii) halfspaces with noise, (iii) intersections of halfspaces. Unfortunately, Allen et al. [AOW15] gave efficient algorithms refuting all three hardness assumptions; fortunately, the results were mostly recovered in later works [DS14, Dan15] assuming hardness of refuting random k-SAT and k-XOR. Although these are more “standard” predicates, a careful inspec-tion of [DS14]’s hardness of learning DNF result shows that it essentially works by reduction from CSP(P ±) where P is a “tribes” predicate. (It first shows hardness for this predicate by 3 reduction from k-SAT.) From these discussions, one can see the utility of understanding the hardness of random CSP(P) for as wide a variety of predicates P as possible.
2. Superlinear number of constraints.
Much of the prior work on hardness of refuting random CSPs (assumptions and evidence for it) has focused on the regime of ∆= O(1); i.e., random CSPs with O(n) constraints. However, it is quite important in a number of settings to have evidence of hardness even when the number of constraints is superlinear. An obvious case of this arises in the application to security of Goldreich-style PRGs; here the number of constraints directly corresponds to the stretch of the PRG. It’s natural, then, to look for arbitrarily large polynomial stretch. In particular, having NC0 PRGs with m = n1+Ω(1) stretch yields secure two-party communication with constant overhead [IKOS08].
This motivates getting hardness of refuting random CSPs with ∆= nΩ(1). As another example, the hardness of learning results in the work of Daniely et al. [DLSS14, DS14, Dan15] all require hardness of refuting random CSPs with m = nC, for arbitrarily large C. In general, given a predicate family P, it is interesting to try to determine the least ∆for which refuting random CSP(P) instances at density ∆becomes easy.
3. Stronger refutation. Most previous work on the hardness of refuting random CSPs has focused just on weak refutation (especially in the proof complexity community), or on δ-refutation for arbitrarily small δ > 0. The latter framework is arguably more natural: as discussed in [Fei02], seeking just weak refutation makes the problem less robust to the precise model of random instances, and requiring δ-refutation for some δ > 0 allows some more natural CSPs like k-XOR (where unsatisfiable instances are easy to refute) to be discussed.
In fact, it is natural and important to study δ-refutation for all values of δ. As an example, given P it is easy to show that there is a large enough constant ∆0 such that for any ∆≥∆0 a random instance I of CSP(P) has Opt(I) ≤µP + o(1), where µP is the probability a random assignment satisfies a random predicate P ∈P.
Thus it is quite natural to ask for δ-refutation for δ = 1 −µP −o(1); i.e., for an algorithm that certifies the true value of Opt(I) up to o(1) (whp). This is sometimes termed strong refutation. As an example, Barak and Moitra [BM16] show hardness of tensor completion based on hardness of strongly refuting random 3-SAT with ∆≪n1/2. In general, there is a very close connection between refutation algorithms for CSP(P) and approximation algorithms for CSP(P); e.g., hardness of δ-refutation results for LP- and SDP-based proof systems can be viewed as saying that random instances are 1 −δ vs. µP + o(1) integrality gap instances for CSP(P).
4. Hardness against superpolynomial time. Naturally, we would prefer to have evidence against superpolynomial-time refutation, or even subexponential-time refutation, of random CSP(P); for example, this would be desirable for cryptography applications. This desire also fits in with the recent surge of work on hardness assuming the Exponential Time Hypothesis (ETH). We already know of two works that use a strengthening of the ETH for random CSPs.
The first, due to Khot and Moshkovitz [KM16], is a candidate hard Unique Game, based on the assumption that random instances of CSP(P ±) require time 2Ω(n) to strongly refute, where P is the k-ary “Hadamard predicate”. The second, due to Razenshteyn et al. [RSW16] proves hardness for the Weighted Low Rank Approximation problem assuming that refuting random 4-SAT requires time 2Ω(n).
An even further interesting direction, in light of the work of Feige, Kim, and Ofek [FKO06], is to find evidence against efficient nondeterministic refutations of random CSPs.
These discussions lead us to the following goal: 4 Goal: For every predicate family P, provide strong evidence for the hardness of refuting random instances of CSP(P), with the best possible tradeoffbetween number of constraints, refutation strength, and running time.
The main theorem in this work, stated in Section 1.5, completely accomplishes this goal in the context of the Sum of Squares (SOS) method. Before stating our results, we review this method, as well as prior results in the direction of the above goal.
1.4 Prior results in proof complexity, and the SOS method Absent the ability to even prove P ̸= NP, the most natural way to get evidence of hardness for refuting random CSP(P) is to prove unconditional negative results for specific proof systems.
It’s particularly natural to consider automatizable proof systems, as these correspond to efficient deterministic refutation algorithms.
Much of the work in this area has focused on random instances of k-SAT. A seminal early work of Chv´ atal and Szemer´ edi [CS88] showed that Resolution refutations of random instances of k-SAT require exponential size when ∆is a sufficiently large constant. Ben-Sasson and Wigderson [BSW01, BS01] later strengthened this result to show that Resolution refutations require width Ω( n ∆1/(k−2)+ϵ ) for any ϵ > 0. Ben-Sasson and Impagliazzo and Alekhnovich and Razborov further extended these results to the Polynomial Calculus proof system [BSI99, AR01]; for example, the latter work showed that Polynomial Calculus refutations of random k-SAT instances with density ∆require degree Ω( n ∆2/(k−2) log ∆).
On the other hand, much of the positive work on refuting random k-SAT has used spectral techniques and semialgebraic proof systems. These latter proof systems are often automatizable using linear programming and semidefinite programming, and thereby have the advantage that they can naturally give stronger δ-refutation algorithms. As examples, Goerdt and Krivelevich [GK01] showed that spectral techniques (which can be captured by SDP hierarchies) enable refutation of random k-SAT with m = n⌈k/2⌉constraints; Friedman and Goerdt [FG01] improved this to m = n3/2+o(1) in the case of random 3-SAT. One of the first lower bounds for random CSPs using SDP hierarchies was given by Buresh-Oppenheim et al. [BOGH+03]; it showed that the Lov´ asz– Schrijver+ (LS+) proof system cannot refute random instances of k-SAT with k ≥5 and constant ∆.
Alekhnovich, Arora, and Tourlakis [AAT05] extended this result to random instances of 3-SAT.
The strongest results along these lines involve the Sum of Squares (AKA Positivstellensatz or Lasserre) proof system. This system, parameterized by a tuneable “degree” parameter d, is known to be very powerful; e.g., it generalizes the degree-d Sherali–Adams+ (SA+) and LS+ proof systems. In the context of CSP(P) over domain {0, 1}, it is also (approximately) automatizable in nO(d) time using semidefinite programming. As such, it has proven to be a very powerful positive tool in algorithm design, both for CSPs and for other tasks; in particular, it has been used to show that several conjectured hard instances for CSPs are actually easy [BBaH+12, OZ13, KOTZ14].
Finally, thanks to work of Lee, Raghavendra, and Steurer [LRS15], it is known that constant-degree SOS approximates the optimum value of CSPs at least as well as any polynomial-size family of SDP relaxations. See, e.g., [OZ13, BS14, Lau09] for surveys concerning SOS.
Early on, Grigoriev [Gri01] showed that SOS of degree Ω(n) could not refute k-XOR instances on sufficiently good expanders. Schoenebeck [Sch08] essentially rediscovered this proof and showed that it applied to random instances of k-SAT and k-XOR, specifically showing that SOS degree n ∆2/(k−2)−ϵ is required to refute instances with density ∆. Tulsiani [Tul09] extended this result to the alphabet-q generalization of random 3-XOR.
Much less was previously known about predicates other than k-SAT and k-XOR. Austrin and 5 Mossel [AM08] established a connection between hardness of CSP(P) and pairwise-uniform dis-tributions, showing inapproximability beyond the random-threshold subject to the Unique Games Conjecture. A key work of Benabbas et al. [BGMT12] showed an unconditional analog of this re-sult: random instances of CSP(P ±) with sufficiently large constant constraint density require Ω(n) degree to refute in the SA+ SDP hierarchy when P is a predicate (over any alphabet) supporting a pairwise-uniform distribution on satisfying assignments. O’Donnell and Witmer [OW14] extended these results by observing a density/degree tradeoff: they showed that if the predicate supports a (t −1)-wise uniform distribution, then the SA LP hierarchy at degree nΩ(ϵ) cannot refute random instances of CSP(P ±) with m = nt/2−ϵ constraints. They also showed the same thing for the SA+ SDP hierarchy, provided one can remove a carefully chosen o(m) constraints from the random in-stance. Extending results of Tulsiani and Worah [TW13], Mori and Witmer [MW16] showed this result for the SA+ and LS+ SDP hierarchies, for purely random instances. Finally, Barak, Chan, and Kothari [BCK15] recently extended the [BGMT12] result to the SOS system, though not for purely random instances: they showed that for any Boolean predicate P supporting a pairwise-uniform distribution, if one chooses a random instance of CSP(P ±) with large constant ∆and then carefully removes a certain o(n) constraints, then SOS needs degree Ω(n) to refute the instance.
Beyond semialgebraic proof systems and hierarchies, even less is known about non-SAT, non-XOR predicates. Feldman, Perkins, and Vempala [FPV15] proved lower bounds for refutation of CSP(P ±) using statistical algorithms when P supports a (t −1)-wise uniform distribution. Their results are incomparable to the above lower bounds for LP and SDP hierarchies: the class of statistical algorithms is quite general and includes any convex relaxation, but the [FPV15] lower bounds are not strong enough to rule out refutation by polynomial-size SDP and LP relaxations.
Summary.
For the strongest semialgebraic proof system, SOS, our evidence of hardness for random CSPs from previous work was somewhat limited. We did not know any hardness results for a superlinear number of constraints, except in the case of k-SAT/k-XOR and the alphabet-q generalization of 3-XOR. We did not know any results that differentiated weak refutation from δ-refutation. Finally, the results known for refuting CSP(P ±) with pairwise-uniform-supporting P did not hold for purely random instances.
1.5 Our result We essentially achieve the Goal described in Section 1.3 in the context of the powerful SOS hierarchy.
Specifically, for every predicate family P, we provide a full three-way tradeoffbetween constraint density, SOS degree, and strength of refutation. Our lower bound subsumes all of the hardness results for semialgebraic proof systems mentioned in the previous section. Furthermore, as we will describe, known algorithmic work implies that our full three-way hardness tradeoffis tight, up to lower-order terms.
To state our result, we need a definition. For a predicate P : Ωk →{0, 1} and an integer 1 < t ≤k, we define δP (t) to be P’s distance from supporting a t-wise uniform distribution.
Formally, δP (t) := min µ is a t-wise uniform distribution on Ωk, σ is a distribution supported on satisfying assignments for P dTV(µ, σ), where dTV(·, ·) denotes total variation distance.
We can now (slightly informally) state our main theorem in the context of Boolean predicates: 6 Theorem 1.1. Let P be a k-ary Boolean predicate and let 1 < t ≤k. Let I be a random instance of CSP(P ±) with m = ∆n constraints. Then with high probability, degree-e Ω n ∆2/(t−1) SOS fails to (δP (t) + o(1))-refute I.
Additionally, in the case that δP (t) = 0, our result does not need the additive o(1) in refutation strength. That is: Theorem 1.2. Let P be a k-ary predicate and let C(P) be the minimum integer 3 ≤τ ≤k for which P fails to support a τ-wise uniform distribution. Then if I is a random instance of CSP(P ±) with m = ∆n constraints, with high probability degree-e Ω n ∆2/(C(P )−2) SOS fails to (weakly) refute I.
Remark 1.3. We comment here on the (surprisingly mild) parameter-dependence hidden by the e Ω(·) and o(1) in these bounds. See Section 7 for full details.
• In terms of ∆, the e Ω(·) is only hiding a factor of log ∆. Thus we get a full linear Ω(n)-degree lower bound for m = O(n) in both theorems above.
• In terms of k, and t, the e Ω(·) is only hiding a factor of 1/(k2O(k/t)).
There are a num-ber of interesting cases where one may take t = Θ(k); for example, k-SAT, k-XOR, and XORk/2 ⊕MAJk/2, a predicate often used in cryptography (e.g., it was suggested by [AL16] for as the basis for high-stretch PRGs in NC0). In these cases, the dependence of the degree lower bound depends only linearly on k and thus, there’s little loss in having k significantly superconstant.
• Indeed in this case of t = Θ(k), if we also have ∆= 2Θ(k) then the degree lower bound for weak refutation in Theorem 1.2 is Ω(n) for k as large as Ω(n); here, both Ω(·)’s hide only a universal constants. The regime of ∆= 2Θ(k) is the algorithmically hardest one for k-SAT, and thus in this very natural case we have a linear-degree lower bound even for k = Ω(n).
• The refutation strength δP (t)+o(1) in Theorem 1.1 is more precisely δP (t)+O(1/√n) when-ever ∆= nΩ(1).
• Theorem 1.1 also holds for predicates P with alphabet size q > 2, with absolutely no additional parameter dependence on q.
The full three-way tradeoffin Theorem 1.1 between constraint density, SOS degree, and strength of refutation is tight up to a polylogarithmic factor in the degree and an additive o(1) term in the strength of the refutation. The tightness follows from the below theorem, which is an immediate consequence of the general δ-refutation framework of Allen et al. [AOW15] and the strong refuta-tion algorithm for XOR due to Raghavendra, Rao, and Schramm [RRS16] (which fits in the SOS framework).
Theorem 1.4. (Follows from [AOW15, RRS16].) Let P be a k-ary Boolean predicate and let 1 < t ≤k. Let I be a random instance of CSP(P ±) with m = ∆n constraints. Then with high probability, degree- e O n ∆2/(t−2) SOS does (δP (t)−o(1))-refute I. Furthermore, with high probability degree-O(1) SOS succeeds in (δP (2) −o(1))-refuting I, provided ∆is at least some polylog(n).
7 An example.
As the parameters can be a little difficult to grasp, we illustrate our main theorem and its tightness with a simple example. Let P be the 3-bit predicate that is true if exactly one if its three inputs is true. The resulting 3-SAT variant CSP(P ±) is traditionally called 1-in-3-SAT. Let us compute the δ(t) values. The uniform distribution on the odd-weight inputs is pairwise-uniform, and it only has probability mass 1 4 offof P’s satisfying assignments. This is minimum possible, and therefore δ1-in-3-SAT(2) = 1 4. The only 3-wise uniform distribution on {0, 1}3 is the fully uniform one, and it has probability mass 5 8 offof P’s satisfying assignments; thus δ1-in-3-SAT(3) = 5 8.
Let us also note that as soon as ∆is a large enough constant, Opt(I) ≤3 8 + o(1) (with high probability, a qualifier we will henceforth omit). Furthermore, it’s long been known [BSB02] that for ∆= O(log n) there is an efficient algorithm that weakly refutes I; i.e., certifies Opt(I) < 1. But what can be said about stronger refutation? Let us see what our Theorem 1.1 and its counterpart Theorem 1.4 tell us.
Suppose first that there are m = n polylog(n) constraints. Theorem 1.4 tells us that constant-degree SOS certifies Opt(I) ≤3 4 + o(1). However our result, Theorem 1.1, says this 3 4 cannot be improved: SOS cannot certify Opt(I) ≤3 4 −o(1) until the degree is as large as e Ω(n). (Of course at degree n, SOS can certify the exact value of Opt(I).) What if there are m = n1.1 constraints, meaning ∆= n.1? Our result says SOS still cannot certify Opt(I) ≤3 4 −o(1) until the degree is as large as n.8/O(log n). On the other hand, as soon as the degree gets bigger than some e O(n.8), SOS does certify Opt(I) ≤3 4 −o(1); in fact, it certifies Opt(I) ≤3 8 + o(1).
Similarly (dropping lower-order terms for brevity), if there are m = n1.2 constraints, SOS is stuck at certifying just Opt(I) ≤3 4 up until degree n.6, at which point it jumps to being able to cer-tify the truth, Opt(I) ≤3 8 +o(1). If there are n1.49 constraints, SOS remains stuck at certifying just Opt(I) ≤3 4 up until degree n.02. Finally (as already shown in [AOW15]), once m = n1.5 polylog(n), constant-degree SOS can certify Opt(I) ≤3 8 + o(1).
(End of example.) More generally, for a given predicate P and a fixed number of random constraints m = n1+c, we provably get a “time vs. quality” tradeoffwith an intriguing discrete set of breakpoints: With constant degree, SOS can δP (2)-refute, and then as the degree increases to n1−2c, n1−c, n1−2c/3, etc., SOS can δP (3)-refute, δP (4)-refute, δP (5)-refute, etc.
An alternative way to look at the tradeoffis by fixing the SOS degree to some nϵ and considering how refutation strength varies with the number of constraints. So for m between n and n3/2−ϵ/2 SOS can δP (2)-refute; for m between n3/2−ϵ/2 and n2−ϵ SOS can δP (3)-refute; for m between n2−ϵ and n5/2−3ϵ/2 SOS can δP (4)-refute; etc.
It is particularly natural to examine our tradeoffin the case of constant-degree SOS, as this corresponds to polynomial time. In this case, our Theorem 1.1 says that random CSP(P ±) cannot be (δP (t) + o(1))-refuted when m ≪n(t+1)/2, and it cannot even be weakly refuted when m ≪ nC(P)/2. Now by applying the work of Lee, Raghavendra, and Steurer [LRS15], we get the same hardness results for any polynomial-size SDP-based refutation algorithm. (See [LRS15] for precise definitions.) Corollary 1.5. Let P be a k-ary predicate, and fix a sequence of polynomial-size SDP relaxations for CSP(P ±). If I is a random instance of CSP(P ±) with m ≤e Ω(nC(P)/2) constraints, then whp the SDP relaxation will have value 1 on I. Furthermore, if m ≤e Ω(n(t+1)/2) (for 1 < t ≤k), then whp the SDP relaxation will have value at least 1 −δP (t) −o(1) on I.
The results in this corollary are tight up to the polylogs on m, by the SOS algorithms of [AOW15].
8 2 Technical framework In Section 1, we described our results as being SOS lower bounds for random CSPs, with constraints chosen randomly from a fixed predicate family P. However it is conceptually clearest to divorce our results from the “random CSP” model as quickly as possible.
• Our lower bound applies whenever the underlying factor graph (bipartite constraint/variable graph) does not contain certain small forbidden subgraphs, which we call “implausible” sub-graphs.
Granted, the only examples we know of such graphs are random graphs (whp).
Further, the condition of “does not contain any implausible subgraphs” is highly related to the condition of “has very good vertex expansion”. Still, we believe the right way to think about the requirement is in terms of forbidden subgraphs.
• Our lower bound doesn’t really involve CSPs and constraints, per se. For each constraint-vertex f in the underlying factor graph, rather than assuming it comes equipped with a constraint predicate P applied to its vertex-variable neighbors, we assume it comes equipped with a probability distribution µf on assignments to its vertex-variable neighbors. We can have a different µf for every constraint-vertex f if we want (indeed, the constraints need not even have the same arity).
• Our SOS lower bounds now take the following form: Assume we are given a factor graph G with no implausible subgraphs, and assume each constraint-vertex f has an associated dis-tribution µf that is t-wise uniform. Then the low-degree SOS proof system “thinks” that there is a global assignment to the variables such that, at every constraint-vertex f, the local assignment to the neighboring variable-vertices is in the support of µf. (Indeed, it “thinks” that there is a probability distribution on global assignments such that for almost all f, the marginal distribution on f’s neighbors is equal to µf.) Let us make some of these notions more precise.
2.1 Constraint satisfaction Notation 2.1. We fix an alphabet Ωof cardinality q ≥2, and a maximum constraint arity K ≥3.
The reader is strongly advised to focus on the case q = 2, with Ω= {±1}, as the only real difficulty posed by larger alphabets is notational. Also, although we describe K as a maximum arity, there will be no loss in thinking of every constraint as having arity K.
Definition 2.2 (t-wise uniform distributions). A probability distribution µ on Ωk is said to be t-wise uniform if its marginal on every subset of t coordinates is uniform.
Rather than our full Theorem 1.1 concerning δ-refutation, the reader is advised to mainly keep in mind our Theorem 1.2, which is concerned with (weak) refutation of CSPs for which the predicates support a (τ −1)-wise uniform distribution. Given our proof of Theorem 1.2, the more general Theorem 1.1 will fall out fairly easily.
Notation 2.3. We fix an integer τ satisfying 3 ≤τ ≤K.
The reader is advised to focus on the simplest case of τ = 3 (corresponding to predicates supporting pairwise-uniform distributions), as the value of τ makes no real difference to our proofs.
9 Notation 2.4 (Instance). The instance we work with consists of two parts: a factor graph and its constraint distributions. The factor graph, denoted G, is a bipartite graph with edges going between n variable-vertices and m constraint-vertices. For a constraint-vertex f we write N(f) for the neighborhood of f, which we take to be an ordered list of the variable-vertices adjacent to f. We assume that the degree (“arity”) of every constraint-vertex f satisfies τ −1 ≤|N(f)| ≤K. Finally, each constraint-vertex f also comes with a constraint distribution µf on ΩN(f). It is assumed that each µf is (τ −1)-wise uniform.
To orient the reader vis-` a-vis our description of CSPs in Section 1.1, consider our Theorem 1.2 in which we have CSP(P ±) instances, where P : {±1}k →{0, 1} is a k-ary Boolean predicate with complexity C(P) = τ. This means there exists some (τ −1)-wise uniform distribution µ on {±1}k supported on satisfying assignments for P. Note that for any “literal pattern” ℓ∈{±}k, the distribution µℓgotten by negating inputs to µ according to ℓis also (τ −1)-wise uniform. In the CSP(P ±) instance, to every constraint with literal pattern ℓthe associated “constraint distribution” will be µℓ. (In the more general context of Theorem 1.1 where we have a k-ary predicate P with δ = δP (t), this means there is some distribution µ on {±1}k which is t-wise uniform and which is δ-close to being supported on P. We will take τ = t + 1 and take the constraint distributions to be µℓagain.) 2.2 Plausible factor graphs As mentioned earlier, our SOS lower bounds will hold whenever the factor graph G has no “im-plausible” subgraphs. The meaning of this will be discussed in much greater detail in Section 4, but here we will give the briefest possible definition.
Notation 2.5. We introduce two parameters: 1 ≤SMALL ≤n/2 and 0 < ζ < 1. (For the sake of intuition, the reader might think of, e.g., SMALL = nΩ(1) and ζ = 1 log n.) The parameters are assumed to satisfy K ≤ζ · SMALL.
Plausibility Assumption.
Henceforth the factor graph G is assumed to satisfy the following property: Let H be an edge-induced subgraph in which every constraint-vertex has minimum de-gree τ. Suppose H has c constraint-vertices, v variable-vertices, and e edges, with c ≤2 · SMALL.
Then (τ −ζ)c ≥2(e −v).
We call the subgraphs H for which the inequality holds plausible because they are indeed the ones that may plausibly show up when the factor graph G is randomly chosen: Proposition 2.6. (Roughly stated; see Theorem 4.12 for a precise statement.) A random G with constraint density ∆will satisfy the Plausibility Assumption whp provided SMALL ≪ n ∆2/(τ−2−ζ) .
The Plausibility Assumption is highly similar to the assumption that G has good vertex-expansion, and indeed our proof of Theorem 4.12 in Appendix A is a completely standard variant of the well-known proof that random bipartite graphs have good vertex-expansion.
2.3 The Sum of Squares algorithm, and pseudoexpectations We give a brief overview of the Sum of Squares algorithm/proof system here. For more general background see, e.g., [BS]; for more details germane to this paper, see Section 5.3.
The Sum of Squares (SOS) algorithm is a hierarchy of semidefinite programming-based relax-ations applicable to polynomial optimization problems; i.e., maximizing an n-variate polynomial 10 subject to polynomial inequality and equality constraints. Each algorithm in the hierarchy is in-dexed by a parameter d known as the degree of the relaxation. Central to the algorithm is the concept of pseudoexpectations that describe the feasible points of the SOS algorithm of degree d.
Definition 2.7 (Pseudoexpectations). Given n indeterminates, a degree-d pseudoexpectation is a linear operator e E on the space of real polynomials of degree at most d in those indeterminates, such that e E = 1. We also generally want it to satisfy the Positive Semidefiniteness condition: e E[p2] ≥0 for every polynomial p of degree at most d/2.
Definition 2.8 (Pseudoexpectations satisfying an identity). A degree-d pseudoexpectation e E is said to satisfy a polynomial identity “p = 0” if, for every polynomial q with deg(p) + deg(q) ≤d, we have e E[pq] = 0.
Given a polynomial optimization problem — say, maximizing a polynomial p1 subject to con-straints {qi = 0 : i ∈[m]} — the degree-d SOS relaxation maximizes e E[p1] over all degree-d pseudoexpectations e E that satisfy the identities {qi = 0 : i ∈[m]}.
A feasibility problem, in particular, would ask if there is a degree-d pseudoexpectation satisfying certain polynomial equal-ity constraints. These SOS relaxations can be expressed using a semidefinite program (SDP) of size nO(d). The Sum of Squares algorithm refers to (approximately) solving the SDP, which can generally be done in nO(d) time.
As suggested by the name, pseudoexpectations generalize the notion of expectations with respect to a probability distribution on real indeterminate values satisfying the given polynomial identity constraints. In particular, if there is at least one real solution for the polynomial identity con-straints, then any probability distribution on solutions yields a valid degree-d pseudoexpectation, for any d. However, even when the polynomial constraints have no real solution, there may well be pseudoexpectations of limited degree that satisfy all the constraints. As one would expect, as the degree d grows, the pseudoexpectations resemble actual expectations more and more. Indeed, if the constraints include that the n indeterminates are Boolean (“x2 i = xi” or “x2 i = 1”) then every degree-2n pseudoexpectation in fact corresponds to an actual distribution on real solutions.
In our context of CSPs, we can think of a constraint satisfaction problem E = {(Pi, Si)} over n Boolean variables x1, . . . , xn as a polynomial feasibility problem, with (the arithmetization of) the constraints Pi(xSi) = 1 as polynomial identities. As we know, randomly chosen CSPs with ∆≫1 are unsatisfiable whp; to show a lower bound on the degree-d SOS refutation algorithm amounts to showing that there exists a degree-d pseudoexpectation that satisfies all the constraints. In more casual terminology, we say that degree-d SOS “thinks” that the CSP is satisfiable.
2.4 Main result We can now describe our main result with the terminology and set-up developed above.
Theorem 2.9 (Roughly stated; cf. Theorem 6.1.). Suppose we are given an instance, with factor graph G satisfying the Plausibility Assumption, and constraint distributions µf for each constraint-vertex. Then for D = 1 3ζ · SMALL, there exists a degree-D pseudoexpectation e E on global variable assignments such that for every constraint-vertex f, the following (suitably encoded) polynomial identity is satisfied: “The marginal distribution on assignments to the variable-neighbors of f is supported within supp(µf).” (Indeed, for almost all f, a stronger identity is satisfied, that the marginal simply equals µf.) In particular, if our instance comes from an actual random CSP with predicates, where for each f the distribution µf is supported on satisfying assignments for the predicate at f, then the 11 degree-D SOS algorithm “thinks” that the CSP is completely satisfiable. This is of course despite the fact that, whp, the CSP is not satisfiable.
Given Proposition 2.6 and Theorem 2.9, we can now point out how the constraint density vs. SOS-degree tradeoffarises in our Theorem 1.2. For CSP(P ±) with C(P) = τ and ∆n random constraints, we get an SOS lower bound for degree roughly ζ · n ∆2/(τ−2−ζ) . The best choice of ζ is roughly 1/ log ∆, and this indeed yields a degree bound of e Ω n ∆2/(C(P )−2) . More precise details of parameter-setting are given in Section 7.
3 Sketch of our techniques Throughout this section, we describe our techniques in the context of CSPs on n Boolean variables and k-ary predicates that are (τ −1)-wise uniform. As stated before, almost all of our ideas are present in this special case.
Our goal is to build a degree-d pseudoexpectation operator e E as described in Theorem 2.9.
3.1 Constructing the pseudoexpectation As in all previous works on CSP lower bounds for hierarchies, we use a variant of the natural pseudoexpectation introduced by Benabbas et al. [BGMT12]. This pseudoexpectation is always defined in terms of a certain “closure” operator on instance graphs; previous works have used slightly different notions of “closure”. Our method introduces yet another definition of closure that we believe is the “right” one; at the very least, it seems to be precisely the right definition for facilitating our proofs.
3.1.1 Closures We can describe a pseudoexpectation by prescribing its values on the basis of monomials of degree at most d. We work with the Fourier basis; i.e., ±1 notation.
In the context of CSPs, a natural way to come up with a pseudoexpectation is via the idea of local distributions. If e E is a degree-d pseudoexpectation, then for every collection S of at most d/2 variables, e E agrees with the expectation of an actual probability distribution. In particular, the pseudoexpectation of a monomial xS := Q i∈S xi for S ⊆[n] (or indeed any function on S) can then be described as the expectation of xS with respect to the local distribution ηS that e E induces on the set S of variables. For such a definition to make sense, the local distributions must satisfy consistency: the pseudoexpectation of xT should equal the expectation of xT with respect to the local distribution ηS for any S that includes T and is of size at most d.
We would like to choose local distributions ηS that are supported on satisfying assignments of all constraints completely included in S (we call these the constraints covered by S). At first blush, we could choose the uniform distribution over the set of satisfying assignments for the constraints covered by S. However, this choice doesn’t satisfy the consistency constraints. The t-wise uniform distributions that are supported on satisfying assignments of the predicate P now come to our rescue: if we obtain a local probability distribution that induces µ on the literals of any constraint in our CSP instance, we should intuitively expect be in good shape because t-wise uniformity roughly guarantees that any constraint that intersects S in t or less variables has a satisfying assignment that agrees with the assignment sampled for S. A natural choice is to define the probability of an assignment to S to be the product of the probabilities (with respect to µ) of the partial assignments corresponding to the constraints covered by S. This doesn’t work as-is, 12 either: there could be constraints that intersect S in many variables and yet are not completely contained inside S. A sample from ηS thus might already force such a constraint to not be satisfied.
To correct for this, we want to collect all such “dependencies” before choosing the local dis-tribution. Benabbas et al. [BGMT12] make this idea precise by defining a notion of closure for a set of variables S: intuitively, these are all the variables that one should care about when defining the local distribution on S. Concretely, their closure maps S into a larger set S′ such that for any T ⊇S′, the marginal of ηT on S is equal to the marginal of ηS′ on S. We then choose ηS′ to be the local distribution on S′ and define ηS to be the marginal of ηS′ on S. For such an effort to be feasible, S′ shouldn’t be much bigger than S: if in the extreme case the closure happened to be the whole set of variables [n], we cannot define a distribution on satisfying assignments of all constraints covered by S′.
The closure of Benabbas et al. [BGMT12] guarantees local consistency as we wanted. Local consistency is all that is required for showing a Sherali–Adams lower bound and is equivalent to the following local positivity condition, which is weaker than positive semidefiniteness: e E[p] ≥0 for p for every truly nonnegative polynomial p depending on at most d variables. However, when trying to show that the more global e E[p2] positive-semidefiniteness condition holds, the [BGMT12] construction seems hard to analyze.
To address this problem, Barak, Chan, and Kothari [BCK15] introduced a simpler variant of the [BGMT12] closure in order to show that the e E defined above satisfies the positive-semidefiniteness condition for certain pruned random instances of the CSP(P ±), when P supports a pairwise-uniform distribution. However, their definition of closure degenerates into the set of all variables with high probability when the random CSP has ∆= ω(1).
Our closure.
One of the main innovations in our work is the introduction of a new, simpler definition of closure that plays a key role in our proof of positive semidefiniteness and gives a definition of e E that works even when the number of constraints is superlinear in n. In addition, our definition of closure enables us to extend our results to δ-refutation.
Our closure for a set of variables S is a subgraph of the factor graph of the CSP instance, including both variables and constraints. We think of the closure of S as being the set of variables and constraints that “matter” when defining the distribution ηS. Given that a predicate P sup-ports a (τ −1)-wise uniform distribution, any constraint that affects ηS must have at least τ −1 variables in S. Otherwise, (τ −1)-wise uniformity implies that we could ignore such a constraint without changing ηS. Any variable v not in S that occurs in only one constraint isn’t necessary for defining ηS, either. We could sum ηS over the two assignments to v to get a new distribution that no longer depends on v. This leads to a natural choice of the closure as the union of all small subgraphs of the factor graph such that each constraint contains at least τ −1 variables and each variable outside of S occurs in at least two constraints. For a formal definition, see Section 5.
3.2 Proving positivity Once we have the definition of the pseudoexpectation, we get to the main challenge in showing any SOS lower bound: arguing positive-semidefiniteness of the e E constructed. The high level idea in our analysis builds on the work of Barak, Chan and Kothari [BCK15]. Their idea of proving positive-semidefiniteness is simple.
They begin by observing that it suffices to verify positive-semidefiniteness for a basis that satisfies orthogonality under e E[·], meaning, the pseudoexpectation of the product of any distinct pair of basis polynomials is 0.
Fact 3.1. Suppose there exists a basis f1, f2, . . . for degree-d polynomials such that the following two properties hold: 13 1. e E[fifj] = 0 for all i ̸= j.
2. e E[f2 i ] ≥0 for all i.
Then e E[g2] ≥0 for all g of degree at most d.
Proof. Write g as P i aifi. Then e E[g2] = X i,j aiaj e E[fifj] = X i a2 i e E[f2 i ] ≥0.
Notice that the standard Fourier monomial basis guarantees us positivity (since e E satisfies the local Sherali–Adams positivity condition by construction). However, it is not orthogonal in general.
How can we construct such a basis? One way to construct a basis that is orthogonal under e E[·] is to perform the Gram–Schmidt process on, say, the monomial basis 1, x1, x2, . . . , x1x2, . . . to get a new basis f1, f2, . . . . Now, Property 1 above holds for this new basis by construction. However, the Gram–Schmidt process is highly sequential and, in particular, the basis function towards the end could depend on all n variables. Thus, we cannot appeal to local positivity of e E in order to argue positive-semidefiniteness of the newly generated basis. It appears that we have made no progress, ensuring orthogonality but potentially losing positivity.
The idea of Barak et al. to escape this pitfall is to show that local orthogonalization is enough.
Before the start of the Gram–Schmidt process, we fix an order on basis vectors. In each step of the process, one orthogonalizes a basis function against all previous basis functions in this order by subtracting offits projection onto their span. Barak et al. analyze the variant of this process in which one orthogonalizes a basis function xS by subtracting offits projection onto the span of all basis functions the precede it in the order and are functions of variables that lie in a small “ball” around S in the factor graph G of the instance. This lets them ensure that the new basis satisfies positivity (since it now depends only on a small number of variables, one can appeal to the local positivity of e E), and they show that this relaxed variant of the Gram–Schmidt process still ensures orthogonality.
Their proof, however, is highly combinatorial and requires various assumptions on the factor graph of the instance that intuitively shouldn’t matter. In particular, they need that the factor graph have no small cycles (girth should be logarithmic): while this can be ensured by pruning o(n) fraction of the constraints in a random instance with Θ(n) constraints, this proof strategy breaks down for super-linear number of constraints .
Our approach Our main idea simplifies the analysis without requiring the assumptions of [BCK15] and yields tight results. It also naturally extends to the case of t-wise uniform predi-cates and further to δ-approximate t-wise uniform predicates. We next describe our key technical ideas that makes this possible.
At a high level, our argument drops the local orthogonalization strategy of Barak et al. [BCK15] and instead runs the Gram–Schmidt procedure “as-is”. Thus orthogonality of the resulting basis functions is immediate, and we need only show positive-semidefiniteness. We show that for any sequential ordering of the basis monomials in the Gram–Schmidt procedure, so long as it is of increasing degree, whenever we orthogonalize a monomial xS, the result basis function depends only on a small number of variables.
To see why such an assertion might be plausible, let us consider the task of orthogonalizing the singletons. The monomial basis may not orthogonal under e E[·]; e.g., consider the following 3-XOR 14 instance: x1x2x3 = 1 y1y2y3 = −1 x2x4x5 = 1 y2y4y5 = −1 x4x5x6 = 1 y4y5y6 = −1 x6x7x8 = 1 y6y7y8 = −1 x3x7x8 = 1 y3y7y8 = −1 Observe that x1 and y1 each appear in exactly one constraint and all other variables each occur in exactly two constraints. Multiplying each block of constraints together, we see that if e E[·] satisfies all constraints then e E[x1] = 1 and e E[y1] = −1. So neither x1 nor y1 are orthogonal to 1. Since the two sets of equations are disjoint, we also know that e E[x1y1] = −1, so x1 and y1 are not orthogonal.
We note that many such blocks may occur in a random instance with m ≫n1.4 constraints. Let’s try to understand what happens when we run the Gram–Schmidt procedure on this basis. Consider an instance consisting of n such disjoint blocks of 5 constraints on 8n variables. Let xi1 be the variables that is fixed in block i. Then every xi1 is not orthogonal to 1 and every pair xi1, xj1 is not orthogonal. Intuitively, the variables xi1, xj1 behave independently, but are biased. To fix this bias, consider the functions xi1 (where we use the notation z := z −e E[z]). Now we have that xi1 is orthogonal to 1 and, by independence of the blocks, e E[xi1 · xj1] = 0 for all i, j.
Ideally, we might hope this this new basis satisfies orthogonality when we move to degree 2, as well. Unfortunately, in general the basis {1, x1, x2, . . . , xn, x1x2, . . .} again need not be orthogonal.
Consider a 3-XOR instance with n constraints x0xiyi = bi for i ∈[n]; call this an n-star. Random instances contain stars of superconstant size with high probability. For all n 2 pairs i, j, it holds that xiyi and xjyj are not orthogonal under e E[·]: e E[xiyi · xjyj] = e E[xiyi · xjyj] −e E[xiyi] e E[xjyj] = bibj −0 = bibj.
Instead, consider the basis b 1 = 1, c x0 = x0, c x1 = x1, . . . , b y1 = y1, b y2 = y2, . . . , d x1y1 = x1y1 −b1x0, d x2y2 = x2y2 −b2x0, . . .
A simple calculation shows that these basis functions are orthogonal. Each basis function depends on at most 3 variables, so the degree-3 Sherali-Adams positivity condition and Fact 3.1 imply that degree-2 positive semidefiniteness holds. We give a proof of orthogonality of d xiyi and d xjyj that illustrates the underlying intuition. Observe that d xiyi and d xjyj are independent conditioned on x0 for all i ̸= j, and we can write e E[d xiyi · d xjyj] = E[d xiyi · d xjyj] (e E[·] is a valid expectation on small sets) = E[E[d xiyi · d xjyj|x0]] (law of total expectation) = E[E[d xiyi|x0] · E[ d xjyj|x0]] (conditional independence of d xiyi and d xjyj given x0).
Next, note that E[d xiyi|x0 = b] = 1 Pr[x0 = b] E[d xiyi · 1{x0=b}(x0)], where 1{x0=b} is the indicator function for x0 = b.
Since we have orthogonalized d xiyi against all degree-1 basis functions and 1{x0=b} is a degree-1 polynomial, this expression is equal to 0.
Therefore, E[d xiyi|x0] = 0 and d xiyi and d xjyj are orthogonal. In this case, xiyi and xjyj are correlated 15 because they are connected by x0. After subtracting offtheir correlation with x0, the resulting functions are orthogonal and no longer correlated.
Let us now formalize this intuition and generalize it to higher degree. At a high level, our idea is to show that the Gram–Schmidt process produces a basis such that each new basis element depends only on a small number of variables. Let yS be the result of applying the Gram–Schmidt process to xS. If yT appears in yS with a nonzero coefficient, then it must be the case that e E[xS · yT ] ̸= 0.
That is, xS and yT are correlated under e E[·]. We show that this correlation is “witnessed” by some small, “dense” subgraph containing many constraints covered by few variables. If yS has many variables in its support, then there must be many such subgraphs. We show that the union of these subgraphs is dense enough to be “implausible”. This means that yS cannot have too many variables in its support.
Our witness can be seen as a generalization of the connected sets in the degree-2 case discussed above. Call two sets of vertices c-connected if removing any set of c −1 vertices cannot disconnect them. In the degree-1 case, nonzero correlation between xS and yT with |S| = |T| = 1 is witnessed by a small, dense, connected (1-connected) subgraph. In the degree-2 case after orthogonalizing against degree-1 terms, we expect based on the star example that if S and T are only 1-connected, then xS and yT will no longer be correlated. We show that nonzero correlation between xS and yT with |S| = |T| = 2 is then witnessed by a small, dense, 2-connected subgraph. In general, we show that nonzero correlation between xS and yT with |S| = |T| = d is witnessed by a small, dense, d-connected subgraph. This stronger connectivity requirement enables us to show that these witness subgraphs and their unions are dense enough to be implausible if the support of a basis function grows too large. For details of this argument, see Section 6.
4 Forbidden subgraphs for the factor graph Let us make a few definitions concerning factor graphs, after which we will elaborate on the “Plau-sibility Assumption”.
Definition 4.1 (Subgraphs). We call H a subgraph of G if it is an edge-induced subgraph; i.e., H = G[A] for some subset A of the edges of G. We explicitly allow A = ∅and hence H = ∅. The subgraph H need not be connected.
Notation 4.2. For H a subgraph, we write vbls(H) for the set of variables appearing in H, cons(H) for the set of constraints appearing in H, and edges(H) for the set of edges appearing in H.
Notation 4.3. Given f ∈cons(H), we write NH(f) = {i ∈vbls(H) : (f, i) ∈edges(H)}. Note that this is not necessarily the same thing as N(f) ∩vbls(H).
We will typically measure the “size” of a subgraph by the number of constraints in it: Definition 4.4 (Small subgraphs). We say that subgraph H is small if |cons(H)| ≤SMALL.
Now regarding the Plausibility Assumption, for intuition’s sake let us suppose we are concerned with weak refutation and degree-O(1) SOS, as in Corollary 1.5. Thus we have some k-ary predi-cate P with C(P) = τ, and we are selecting a random CSP with slightly fewer than nτ/2 constraints; say m = n(τ−ζ)/2. What does a random factor graph look like in this case? Which small subgraphs may appear? A quick-and-dirty method to analyze this is as follows. Consider the fixed small subgraph in Figure 1; call it H.
16 Figure 1: An example small subgraph. Constraint-vertices are squares, variable-vertices are circles.
What is the expected number of copies of H in a random factor graph G with n variable-vertices and m = n(τ−ζ)/2 constraint-vertices? There are m 2 ≈m2 choices for H’s 2 constraint-vertices and n 4 ≈n4 choices for H’s 4 variable-vertices. Thinking of each constraint-vertex as choosing k = O(1) random neighbors, the chance that the 6 edges of H show up is roughly n−6. Thus, very roughly, we expect about m2n4n−6 = n2·(τ−ζ)/2+(4−6) copies of H in a random G. Thus copies of H “plausibly” show up if and only 2 · (τ −ζ)/2 + (4 −6) ≥0; i.e., if and only if τ ≥2 + ζ. Since τ ≥3 always, this means we should certainly expect copies of H in G.
For a general subgraph H with c = |cons(H)|, v = |vbls(H)|, e = |edges(H)|, E[# copies of H] ≈mcnvn−e = nc·(τ−ζ)/2+v−e = ⇒H “plausibly occurs” iffc·(τ−ζ)/2+(v−e) ≥0.
(1) This inequality is precisely the one occurring in the Plausibility Assumption from Section 2.2.
Despite the simple form of the inequality, we will find it helpful to view it in a different way. For reasons that will become clear in Section 5, we will be concerned almost exclusively with subgraphs of G in which all constraint-vertices have degree at least τ: Definition 4.5 (τ-subgraphs). Let H be a subgraph.
We will call H a τ-subgraph if every constraint-vertex in H has degree at least τ within H; i.e., |NH(f)| ≥τ for all f ∈cons(H).
Remark 4.6. The empty subgraph ∅is always trivially a τ-subgraph. Also, if H and H′ are τ-subgraphs then so is H ∪H′.
Definition 4.7 (Leaf vertices and interior vertices). Given a subgraph H, we classify the variable-vertices in H as either leaf or interior depending on whether they have degree 1 or at least 2.
(Since H is an edge-induced subgraph, it does not have any isolated vertices.) For τ-subgraphs, there is a different way to view the “plausibility inequality” that will be more useful for us. We define it with some “accounting” terminology.
Definition 4.8 (Credit, debit, excess, revenue, cost, income). Let H be a τ-subgraph. For the purposes of this definition, consider each of its edges to be two directed edges.
• For each variable-vertex, we assign it a credit of 1 if it is a leaf vertex. We’ll write ℓfor the total credits.
• For each variable-vertex, any out-edges in excess of 2 are called excess, and we assign a debit for each. We’ll write ev for the total number of these.
• For each constraint-vertex, any out-edges in excess of τ are called excess, and we assign a debit for each. We’ll write ec for the total number of these, and e = ec +ev for the total debit (number of excess edges).
• The sum of credits minus the sum of debits, ℓ−e, is called the revenue. We denote it by R(H).
• Each constraint-vertex has a cost of ζ. We write C(H) = ζ · |cons(H)| for the total cost.
17 • The income is I(H) = R(H) −C(H).
Definition 4.9 (Plausible τ-subgraphs). Let H be a τ-subgraph.
We say that H is plausible if I(H) ≥0.
Remark 4.10. H being plausible implies (indeed, is equivalent to) |cons(H)| ≤1 ζ · R(H). Thus controlling a subgraph’s revenue is equivalent to controlling its size.
The next lemma implies that the inequality I(H) ≥0 is the same as the inequality appearing in the Plausibility Assumption and in (1).
Lemma 4.11. Let H be a τ-subgraph with c = |cons(H)|, v = |vbls(H)|, e = |edges(H)|, and I = I(H). Then e = τ−ζ 2 · c + v −I 2.
Proof. We count the number of “directed edges” in H. Counting those coming out of variable-vertices, the ℓleaf vertices contribute 1 each, and the v −ℓinterior vertices contribute 2(v −ℓ)+ev.
Counting the directed edges coming out of constraint-vertices yields τc + ec. Thus # directed edges = 2e = ℓ+ 2(v −ℓ) + ev + τc + ec = τc + 2v −(ℓ−e) = τc + 2v −(ζc + I), since ℓ−e = R(H) = C(H) + I(H). The claim follows.
In light of this, we may restate the Plausibility Assumption: Plausibility Assumption, Restated.
Henceforth we assume the factor graph G has the follow-ing property: All τ-subgraphs H of G with |cons(H)| ≤2 · SMALL are plausible.
As mentioned earlier, for an appropriate choice of SMALL, the Plausibility Assumption holds for a random instance. More precisely, in Appendix A we prove the below theorem. The reader is advised that in this theorem, the first claim is the main one; it is used to show our Theorem 1.2 concerning weak refutation. The second claim (“Moreover. . . ”) is a technical variant needed to extend our results to give Theorem 1.1 concerning δ-refutation.
Theorem 4.12. Let λ = τ −2 ≥1. Fix 0 < ζ ≤.99λ, 0 < β < 1 2. Then except with probability at most β, when G is a random instance with m = ∆n constraints, the Plausibility Assumption holds provided SMALL ≤γ · n ∆2/(λ−ζ) , where γ = 1 K β1/λ 2K/λ O(1) . Moreover, assuming ζ < 1, except with probability at most β we have #{nonempty τ-subgraphs H with cons(H) ≤2 · SMALL : I(H) ≤τ −1} ≤∆n 1+ζ 2 .
5 Defining the pseudoexpectation 5.1 Closures In this section we define the “closure” of a set of variables. Roughly speaking, this can be thought of as the smallest τ-subgraph of G that fully determines the distribution on S under a natural “planted distribution”.
18 Definition 5.1 (S-closed subgraph). Let S be a set of variables. We say that a subgraph H is S-closed if it is a τ-subgraph and all its leaf vertices are in S.
Remark 5.2. For every constraint in G, if H is taken to be the full neighborhood of that constraint, and S is the set of variables in that constraint, then H is S-closed.
Note that a union of S-closed τ-subgraphs is S-closed. This leads us to the following definition: Definition 5.3 (Closure, cl(S)). Let S be a set of variables. We define the closure of S, written cl(S), to be the union of all small S-closed τ-subgraphs H. Note that cl(S) is itself an S-closed τ-subgraph.
Remark 5.4. A key warning to remember: we do not necessarily have S ⊆vbls(cl(S)).
Remark 5.5. Let T ⊆S. Then if H is T-closed, it is also S-closed. It follows that cl(T) ⊆cl(S).
Fact 5.6. The only plausible ∅-closed τ-subgraph H is H = ∅. It follows that cl(∅) = ∅.
Proof. If H is ∅-closed then its revenue is at most 0. Hence if it is plausible, its cost is 0.
We will now give an important generalization of this fact for S-closures, |S| > 0 Theorem 5.7. Let S be a set of variables with |S| ≤ζ · SMALL. Then cl(S) is small and satisfies R(cl(S)) ≤|S|.
Proof. Since cl(S) is S-closed, all its leaf vertices are in S; thus cl(S) has at most |S| credits and so R(cl(S)) ≤|S|, as claimed. Observe that if H1, . . . , Ht is the complete list of S-closed τ-subgraphs, we may make the same deduction about H1 ∪· · · ∪Hj for any 1 ≤j ≤t, in particular deducing that R(H1 ∪· · · ∪Hj) ≤ζ · SMALL for each j. The smallness of cl(S) is now a consequence of the lemma that immediately follows.
Lemma 5.8. Suppose that H is a τ-subgraph formed as a union, H = H1 ∪· · · ∪Ht, where each Hj is small and where we have R(H1 ∪· · · ∪Hj) ≤ζ · SMALL for all 1 ≤j ≤t. Then H is small.
Proof. The proof is by induction on t, with the base case of t = 1 being immediate. In general, suppose H′ = H1∪· · ·∪Ht−1 is small. Since Ht is also small we have |cons(H′)|, |cons(Ht)| ≤SMALL and hence |cons(H′ ∪Ht)| ≤2 · SMALL. Thus H′ ∪Ht is plausible and so cons(H′ ∪Ht) = (1/ζ) · C(H′ ∪Ht) ≤(1/ζ) · R(H′ ∪Ht) ≤(1/ζ) · ζ · SMALL = SMALL, showing that H′ ∪Ht is small, completing the induction.
In proving Theorem 5.7, we iteratively formed the union of all small S-closed subgraphs, at each step verifying that we have a small τ-subgraph of revenue at most |S|. Once we finish pro-ducing cl(S) in this way, let V = vbls(cl(S)), and suppose we continue iteratively adding in small τ-subgraphs that are (S ∪V )-closed. This process cannot add any leaf vertices except possibly in S; thus we will still have that revenue is bounded by |S| ≤ζ · SMALL, and Lemma 5.8 will still imply the resulting τ-subgraph is small. Thus we end up with a small, S-closed τ-subgraph— which by definition is already contained in cl(S). Thus we have shown: Theorem 5.9. Let S be a set of variables with |S| ≤ζ · SMALL. Then cl(S ∪vbls(cl(S))) = cl(S).
19 5.2 The planted distribution Definition 5.10 (Planted distribution on a small subgraph). Let H be a small subgraph of G.
The planted distribution on H is a probability distribution on assignments x ∈Ωn to the variables of G, defined as follows: For each constraint f ∈cons(H) we independently draw an assignment wf ∈ΩN(f) according to µf. We write its component associated to variable i ∈NH(f) as wf,i, and think of it as an assignment “suggested” for this variable. (Note that we will ignore the components of wf correspoding to variables not in NH(f).) Now each variable i ∈vbls(H) has one or more assignments in Ωsuggested by its adjacent constraints. We get a unique assignment xi for it by conditioning on all the suggestions being consistent. (We will show later in (8) that this occurs with nonzero probability.) Finally, assignments for variables not in H are chosen independently and uniformly from Ω.
We’ll write ηH for the probability distribution on Ωn associated to this planted distribution on H, and we’ll write EH[·] for the associated expectation.
Definition 5.11. For each i ∈vbls(G) and each c ∈Ω, we introduce an “indeterminate” 1c(xi) that is supposed to stand for 1 if variable i is assigned c and 0 otherwise.
The key theorem about the planted distributions is that as soon as a subgraph H contains cl(S), the marginal of ηH on S is determined. In some sense, this property is exactly the reason we defined the closure the way we did.
Theorem 5.12. Let S be a set of variables and let H ⊇cl(S) be a small subgraph. Then the marginal of ηH on S is the same as the marginal of ηcl(S) on S.
Remark 5.13. Although the notation in the below proof looks cumbersome, the calculations are actually fairly straightforward. We strongly encourage the reader to work through the proof in the case of q = 2, Ω= {±1}, with “1c(xi)” replaced by cxi ∈{±1}.
Proof. For brevity we write vH = |vbls(H)| and eH = |edges(H)|. We also introduce the notation 1c(xi) = q1c(xi) −1. Recalling that ηH puts the uniform distribution on the n −vH variables outside H, we have pH(x) := Pr w [w suggestions consistent, and the assignment to vbls(H) they agree on is x] = q−(n−vH) · E w Y (f,i)∈edges(H) q−1(1 + 1wf,i(xi) = qvH−eH−n · X H′⊆H E w Y (f,i)∈edges(H′) 1wf,i(xi) (2) = qvH−eH−n · X H′⊆H Y f∈cons(H′) E wf∼µf Y i∈NH′(f) 1wf,i(xi), (3) where we used that the draws wf ∼µf are independent across f’s. Now whenever f ∈cons(H′) has |NH(f)| < τ, the (τ −1)-wise uniformity of µf implies that E wf∼µf Y i∈NH′(f) 1wf,i(xi) = E wf,i∼Ω uniform, indep.
Y i∈NH′(f) 1wf,i(xi) = Y i∈NH′(f) E wf,i∼Ω uniform 1wf,i(xi) = 0, since Ec∼Ω[1c(xi)] = 0 for any fixed value x ∈Ω.
Thus in (3) it is equivalent to sum over τ-subgraphs H′, and so returning to (2) we get pH(x) = qvH−eH−n · X H′⊆H H′ a τ-subgraph E w Y (f,i)∈edges(H′) 1wf,i(xi).
(4) 20 Suppose now that T ⊆[n] is a set of variables. We’ll decompose an x ∈Ωn into its projection xT onto the coordinates in T and xT onto the coordinates not in T. Then pH(xT ) := Pr w [w suggestions consistent, and the assignment to T they agree on is xT ] = X xT ∈ΩT pH(xT , xT ) = qn−|T| · E xT ∼ΩT uniform [pH(xT , xT )] = qvH−eH−|T| · X H′⊆H H′ a τ-subgraph E w Y (f,i)∈edges(H′) i∈T 1wf,i(xi) · E xT ∼ΩT uniform Y (f,i)∈edges(H′) i∈T 1wf,i(xi), (5) where we used (4). Now suppose the τ-subgraph H′ has a leaf vertex j that is in T; i.e., it’s not in T. Then xj appears exactly once in the above, within the expression E xT ∼ΩT uniform Y (f,i)∈edges(H′) i∈T 1wf,i(xi).
(6) As xj is chosen uniformly and independently of all random variables, the above contains a factor of the form Exj∼Ω[1wf,j(xj)]. But for any fixed outcome of wf,j, this expectation is 0, meaning (6) will vanish. Thus any summand H′ in (5) will vanish if H′ has a leaf variable outside T. Thus we may equivalently sum only over T-closed H′. That is, pH(xT ) = Pr w [w suggestions consistent, and the assignment to T they agree on is xT ] = qvH−eH−|T| · X H′⊆H H′ is T-closed E w Y (f,i)∈edges(H′) i∈T 1wf,i(xi) · E xT ∼ΩT uniform Y (f,i)∈edges(H′) i∈T 1wf,i(xi) = qvH−eH−|T| · X H′⊆H H′ is T-closed E w E x∼Ωn unif., condit. on xT =xT Y (f,i)∈edges(H′) 1wf,i(xi).
(7) Suppose we took T = ∅above. Since H is small, every subgraph H′ is plausible and hence Fact 5.6 implies that the above has only one summand, corresponding to H′ = ∅. The summand is trivially 1, and hence Pr w [w suggestions consistent] = qvH−eH.
(8) Observe that this does not depend at all on the µf’s; in particular, it is easily seen to the be the probability of consistent suggestions under completely uniform µf’s. In any case, since (8) is positive, as promised, we may condition on the associated event; thus from (7) we obtain Pr w [the suggested assignment to S is xS | the suggestions w are consistent] = q−|S| · X H′⊆H H′ is S-closed E w E x∼Ωn unif., condit. on xS=xS Y (f,i)∈edges(H′) 1wf,i(xi).
This formula visibly has the property that once H ⊇cl(S), it does not depend on H.
5.3 Pseudoexpectations In this section, we formally define the pseudoexpectation with which we will work.
21 Definition 5.14. Given a polynomial expression p(x) in the indeterminates 1c(xi), we write vbls(p) = {i : at least one 1c(xi) appears in p(x)}, degmlin(p) = max{|vbls(M)| : M(x) is a monomial in p(x)}.
We call the latter the multilinear-degree; note that degmlin(p) ≤deg(p) always.
Recall that a pseudoexpectation on polynomials of degree at most D is a linear map e E[·] sat-isfying e E = 1. We can uniquely define it by specifying its values on all monomials of degree at most D. Further, recall that if p(x) is a polynomial, we say that e E[·] satisfies the identity p(x) = 0 if e E[p(x) · q(x)] = 0 for all polynomials q(x) with deg(p · q) ≤D.
Definition 5.15 (Our pseudoexpectation). We’ll define our pseudoexpectation e E[·] on all poly-nomials of multilinear-degree at most ζ · SMALL; in particular, this defines it for all polynomials of (usual) degree at most ζ · SMALL. We define it by imposing that e E[M(x)] = Ecl(vbls(M))[M(x)] for all monomials M(x) having degmlin(M) ≤ζ ·SMALL. (Here we are using the abbreviation EC[M(x)] for Ex∼ηC[q(x)].) By Theorem 5.7, this makes sense in that cl(vbls(M)) will always be small. Note that we have e E = Ecl(∅) = 1, as required.
Theorem 5.16. Let p(x) be a polynomial expression of multilinear-degree at most ζ · SMALL. Let H be any small subgraph containing H′ = [ {cl(vbls(M)) : M(x) is a monomial of p(x)}.
For example, if cl(vbls(p)) is small then it would qualify for H. Then e E[p(x)] = E H[p(x)] = E H′[p(x)].
Proof. This is immediate from Theorem 5.12 and Remark 5.5.
Theorem 5.17. Let p(x) be a polynomial with S = vbls(p) satisfying |S| ≤deg(p), |S| ≤ζ · SMALL.
Assume that p(x) is identically zero for x ∼ηcl(S). (Note that cl(S) is small by Theorem 5.7.) Then our e E[·] satisfies the identity p(x) = 0.
Proof. Let q(x) be a nonzero polynomial with deg(p · q) ≤ζ · SMALL. Writing q(x) = P j Mj(x) where each Mj(x) is a monomial, we have e E[p(x) · q(x)] = X j e E[p(x) · Mj(x)] = X j E cl(S∪vbls(Mj))[p(x) · Mj(x)].
(9) Here the last equality used Theorem 5.16 and the smallness of cl(S ∪vbls(Mj)), which follows from Theorem 5.7 and the fact that |S ∪vbls(Mj)| ≤deg(p) + deg(q) = deg(p · q) ≤ζ · SMALL. But since cl(S ∪vbls(Mj)) ⊇cl(S) (Remark 5.5), Theorem 5.12 tells us that p(x) has the same distribution under ηcl(S∪vbls(Mj)) and ηcl(S); i.e., it is identically 0. Thus (9) vanishes, as needed.
We have the following immediate corollaries: Corollary 5.18. Our pseudoexpectation e E[·] satisfies the following identities: • P c∈Ω1c(xi) = 1 for all i ∈[n] (i.e., the identity P c∈Ω1c(xi) −1 = 0).
• 1c(xi)2 = 1c(xi) for all c ∈Ω, i ∈[n].
22 As an immediate consequence of the latter, we always have e E[p(x)] = e E[multilin(p(x))], where multilin(p(x)) is defined by replacing any positive power of 1c(xi) in p(x) with just 1c(xi).
Another corollary is the following (cf. the rough statement of our main technical result, Theo-rem 2.9): Corollary 5.19. Our pseudoexpectation e E[·] satisfies the identity sf(x) := X ⃗ c∈supp(µf) Y i∈N(f) 1ci(xi) = 1 for all f ∈cons(G); i.e., “e E[·]’s distribution on N(f) is always in supp(µf)”.
Proof. We apply Theorem 5.17, with S = N(f), which satisfies |S| = deg(sf) and |S| ≤K ≤ ζ · SMALL. Note that if Hf denotes the τ-subgraph induced by all edges of G incident on constraint-vertex f, then Hf is S-closed and so Hf ⊆cl(S). It then follows from the definition of x ∼ηcl(S) that sf(x) ≡1, since the restriction of x to N(f) will always be supported on supp(µf).
6 The proof of positive semidefiniteness 6.1 Setup Throughout this section, fix a degree D satisfying 1 ≤D ≤1 3ζ ·SMALL. Our goal will be to establish: Theorem 6.1. If p(x) is a polynomial expression of degree at most D, then e E[p(x)2] ≥0.
In light of Corollary 5.18, we may assume that p(x) is “multilinear” (i.e., does not contain 1c(xi)k for any k > 1). Another way to state this assumption is p(x) ∈span(xS : S ∈M≤D), where we introduce the following notation: Definition 6.2. A monomial index will be a set S of pairs (i, c) ∈[n] × Ω, with no variable i ∈[n] occurring more than once. We write xS for the monomial Q (i,c)∈S 1c(xi), with the usual convention that x∅= 1. Finally, we write M≤D for the collection of monomial indices S with |S| ≤D.
Notation 6.3. We abuse notation as follows: If a monomial index S occurs in a place where a subset of variables is expected, we intend the subset of variables {i : (i, c) ∈S for some c}.
Remark 6.4. All of the ideas in our proof of Theorem 6.1 are present in the q = 2 case; only notational complexities arise for q > 2. Thus the reader is encouraged to keep the Boolean case Ω= {false, true} in mind. In this case, since e E[·] satisfies the identity 1false(xi) = 1 −1true(xi), one can also ignore the indeterminate 1false(xi) (since 1 ∈span(M≤0) already). Then one can more naturally write the indeterminate 1true(xi) as xi and the monomial xS becomes Q i∈S xi.
6.2 Gram–Schmidt overview Notation 6.5. Let ⪯denote any total ordering on M≤D that respects cardinality, so that if T and S are monomial indices with |T| < |S|, then T ≺S. For S ̸= ∅, let pr(S) denote the immediate predecessor of S under ⪯.
23 Our goal in this section is to show that the modified Gram–Schmidt process from linear algebra can be successfully applied to the monomials (xS : S ∈M≤D), in the ordering ⪯, using e E[·] as the “inner product”: ⟨p(x), q(x)⟩:= e E[p(x)·q(x)]. Of course, we don’t know that this is a genuine inner product (indeed, that’s essentially what we’re trying to prove). We will discuss this issue shortly, but we first remind the reader that the modified Gram–Schmidt process would typically produce a collection of polynomials yS = yS(x), for S ∈M≤D, that are orthogonal under e E[·] (meaning e E[yS · yS′] = 0 if S ̸= S′) and that have the same span as (xS : S ∈M≤D). As well, it would produce “normalized” versions of these polynomials zS = yS/ q e E[y2 S], satisfying e E[z2 S] = 1.
We now address the obviously difficulty that e E[·] is not (known to be) an inner product, because we don’t know it’s positive definite on the monomials of M≤D. Our goal will be to show that as we follow the Gram–Schmidt process, it never encounters any “positive definiteness problems”, and therefore “succeeds”. The main “positive definiteness problem” Gram–Schmidt might encounter would be if it creates a polynomial yS with e E[y2 S] < 0. In this case, when it tries to produce the normalized polynomial zS, it would certainly fail.
There is one additional potential problem, occurring if Gram–Schmidt produces a yS with e E[y2 S] = 0. In the usual process from linear algebra this may indeed occur, and the Gram–Schmidt algorithm copes by treating zS as 0 (effectively, throwing it out of the span). This is a valid strategy because genuine inner products are strictly positive definite. However we only expect our “inner product” e E[·] to be positive semidefinite. We therefore need a different coping mechanism. For us, when e E[y2 S] = 0 occurs, we will simply define its “normalized” version zS to be yS. The challenge of this is that Gram–Schmidt’s guarantee of producing an orthogonal collection (yS : S ∈M≤D) relies syntactically on all the zS polynomials satisfying e E[z2 S] = 1. Thus we will have an additional burden: we will have to “manually” show that e E[y2 S] = 0 implies that yS is orthogonal under e E[·] to all other polynomials. It will count as a “positive definiteness problem” if we are unable to show this; we will call this the “pseudovariance zero problem”. We remark that the main positive definiteness problem is fundamentally more important than this “pseudovariance zero problem”, and the reader may wish to ignore the pseudovariance zero issue on first reading.
We now describe the modified Gram–Schmidt process in detail. The process works in stages, named after the elements of M≤D and in order of ⪯. At the end of stage S it creates a certain polynomial zS. Stage ∅always “succeeds” and simply consists of defining z∅= 1. In some cases it may happen that e E[z2 S] = 0.
In this case we say that zS has pseudovariance zero, and the Gram–Schmidt algorithm will add S to a growing collection called PvZ.
Each stage S is further divided into substages, associated to monomial indices T ≺S in order of ⪯. Let us introduce some notation: Notation 6.6. Let M≤D 2 denote the collection of all pairs (S, T) ∈M≤D × M≤D with T ≺S.
We define a total ordering ⪯2 on M≤D 2 via (S′, T ′) ⪯2 (S, T) ⇐ ⇒S′ ≺S, or S′ = S and T ′ ⪯T.
Thus the overall progression of substages in Gram–Schmidt is through the elements of M≤D 2 in order of ⪯2. Substage (S, T) creates a polynomial yS,T as follows: yS,T = ( xS −e E[xS] if T = ∅; yS,pr(T) −e E[yS,pr(T) · zT ]zT else.
Stage S ends just after substage (S, pr(S)). At this point, the Gram–Schmidt process defines yS = yS,pr(S), zS = ( yS q e E[y2 S] if e E[y2 S] > 0; yS if e E[y2 S] = 0, in which case S is placed into PvZ.
24 Of course, if e E[y2 S] < 0 then we have encountered a positive definiteness problem. Indeed, to be conservative we will treat it as a problem if e E[y2 S,T ] < 0 for any (S, T) ∈M≤D 2 .
It is a syntactic property of the usual modified Gram–Schmidt process that when yS,T is pro-duced, it is orthogonal to zT under e E[·]. However this relies on e E[z2 T ] = 1, which fails for us if T ∈PvZ. Thus we will need to explicitly prove that T ∈PvZ implies e E[yS,pr(T) · zT ] = 0. If this doesn’t hold, we’ve encountered the pseudovariance zero problem. But assuming it does hold, yS,T will simply become yS,pr(T) and we will have the desired orthogonality of yS,T and zT . We remark that the usual Gram–Schmidt property of yS,T being orthogonal to all zT ′ with T ′ ⪯T follows by induction in the usual way; this only needs the inductive property that the zT ’s are orthogonal (not that they’re orthonormal).
We may now summarize the discussion so far: Definition 6.7. A positive definiteness problem occurs at substage (S, T) of modified Gram– Schmidt if either e E[y2 S,T ] < 0, or if T ∈PvZ but e E[yS,pr(T) · zT ] ̸= 0.
(The latter is called a pseudovariance zero problem.) We say that the modified Gram–Schmidt process succeeds through substage (S, T) if it encounters no positive definiteness problem at any substage (S′, T ′) ⪯2 (S, T).
Proposition 6.8. Suppose the modified Gram–Schmidt process succeeds through substage (S, T).
Then we have: • yS,T = xS −p(x) for some polynomial p(x) supported on monomials xT ′ with T ′ ⪯T; • e E[yS,T ·zT ′] = 0 for all T ′ ⪯T, and hence e E[yS,T ·q(x)] = 0 for all polynomials q(x) supported on monomials xT ′ with T ′ ⪯T; • e E[y2 S,T ] ≥0.
In particular, if the process succeeds through stage S, we have: • zS = c · xS −p(x) for some positive constant c > 0 and some polynomial p(x) supported on monomials xT with T ≺S; • span(xS′ : S′ ⪯S) = span(zS′ : S′ ⪯S); • e E[zS · zT ] = 0 for all T ≺S, and hence e E[zS · q(x)] = 0 for all polynomials q(x) supported on monomials xT ′ with T ′ ≺S; • e E[z2 S] = 0 if S is put in PvZ, else e E[z2 S] = 1.
Our main Theorem 6.1 follows provided the modified Gram–Schmidt process succeeds through all substages in M≤D 2 . The reason is that then any multilinear p(x) of degree at most D can be expressed as p(x) = P |T|≤D cT zT . This implies e E[p(x)2] = X |T|,|T ′|≤D cT cT ′ e E[zT · zT ′] = X |T|≤D T̸∈PvZ c2 T ≥0, using Proposition 6.8.
25 6.3 Advanced accounting Definition 6.9. A τ-subgraph+ is defined to be a τ-subgraph, together with zero or more isolated variable-vertices.
We still have that the union of τ-subgraphs+ is a τ-subgraph+. We extend the cons(H) and vbls(H) notation to τ-subgraphs+, and also the planted distribution notation ηH (being the same as ηH′ where H′ is formed from H by deleting its isolated vertices).
Definition 6.10. For a τ-subgraph+ H, we extend the definition of revenue by assigning two credits for all isolated variable-vertices in H.
Remark 6.11. If H is a τ-subgraph+ and H′ is the τ-subgraph formed by deleting isolated vertices, then cons(H′) = cons(H), C(H′) = C(H), and R(H′) ≤R(H). Thus the Plausibility Assumption immediately implies that all τ-subgraphs+ with at most 2 · SMALL constraints are also plausible.
Lemma 6.12. Let H be a small τ-subgraph+ with R(H) ≤r. Let H′ be a small τ-subgraph with at most s leaf variables that are not in H. Assume r + s ≤ζ · SMALL. Then H ∪H′ is small and satisfies R(H ∪H′) ≤r + s.
Proof. Adding H′ into H cannot remove any of the debits of H, and the only additional credits that can be created come from the s leaf variables in H′ that are not in H. (Since H′ is only a τ-subgraph it has no isolated variables.) This establishes R(H ∪H′) ≤r + s. The smallness conclusion follows immediately from Lemma 5.8 (here it does not matter that H is a τ-subgraph+).
A key aspect to our main theorem will be that in some cases this revenue bound can be improved: Lemma 6.13. In the setup of Lemma 6.12, suppose also that H′ has b edges that are “boundary” for H, in the sense that each has exactly one endpoint in H. Then in fact R(H ∪H′) ≤r + s −b.
Proof. Let a be an edge in H′ with exactly one endpoint, call it w, in H. We show that the addition of this edge to H causes a drop of 1 in revenue. If w is a constraint-vertex, then this follows because w already had degree at least τ in H, so a becomes a new excess edge in H, creating a new debit.
So suppose w is a variable-vertex. If w had degree at least 2 in H then a is again excess and creates a new debit. If w had degree 1 in H then the addition of a changes w from a leaf variable to an interior variable, removing 1 credit from H. Finally, if w was isolated in H then the addition of a turns it into a leaf variable, again removing 1 credit from H. Repeating this argument for all b boundary edges completes the proof.
6.4 The key lemma Lemma 6.14. Let y = y(x) be a polynomial expression of degree d. Assume 2d ≤ζ · SMALL and that e E[y·p(x)] = 0 for all polynomials p of degree strictly less than d. Let H be a small τ-subgraph+ with vbls(H) ⊇vbls(y) and R(H) ≤r, where we assume r + d ≤ζ · SMALL. Finally, suppose T is a monomial index with |T| = d such that e E[y · xT ] ̸= 0.
Then there exists a small τ-subgraph+ Hnew ⊇H with vbls(Hnew) ⊇vbls(y) ∪T and R(Hnew) ≤r.
(In writing vbls(y) ∪T, we are using the abuse described in Notation 6.3.) 26 Proof. Let us define Tnew = T \ vbls(H), Told = T ∩vbls(H), B = cl(vbls(H) ∪T), Hnew = H ∪B.
First, we show that the τ-subgraph+ Hnew is small; it follows that the τ-subgraph B is also small.
Claim 6.15. Hnew is small.
Proof. Write cl(vbls(H) ∪T) = H′ 1 ∪· · · ∪H′ t for small (vbls(H) ∪T)-closed τ-subgraphs H′ i. Let H′ <j := H′ 1 ∪· · ·∪H′ j−1, and let sj denote the number of leaves of Hj that are not in H ∪H′ <j. Then it is easy to see that Pt j=1 sj ≤d. Now, iteratively apply Lemma 6.12 to H ∪H′ 1, (H ∪H′ 1) ∪H′ 2, ((H ∪H′ 1) ∪H′ 2) ∪H′ 3, . . . to prove the claim.
Next, observe that we have vbls(Hnew) ⊇vbls(H) ⊇vbls(y); therefore to prove the lemma, it suffices to show that vbls(Hnew) ⊇Tnew and that R(Hnew) ≤R(H).
For the first of these, given an (i, c) ∈T we write xi = 1c(xi) −e E[1c(xi)] and xT = Q i∈T xi.
Observe that xT −xT is a polynomial of degree strictly less than d; thus e E[y · (xT −xT )] = 0 and so e E[y · xT ] ̸= 0. Now using Theorem 5.16 and B ⊇cl(vbls(H) ∪T) ⊇cl(vbls(y · xT )), we conclude E B[y · xT ] ̸= 0.
(10) In light of this, we claim that every variable j ∈Tnew must appear as a vertex in B (and hence in vbls(Hnew), as needed). For if j ̸∈vbls(B), then xj is independent of all other random variables xi under ηB, and so E B[y · xT ] = E B[xj] · E B[y · xTold · xTnew{j}] (using j / ∈vbls(H) ⊇vbls(y)) = e E[xj] · E B[y · xTold · xTnew{j}] = 0 (using B ⊇cl(T) ⊇cl({j}) and e E[xj] = 0) in contradiction to (10).
It remains to show that R(Hnew) ≤R(H), which we will do using Lemma 6.13 (with H′ = B, and s = |Tnew|, recalling that all of B’s leaves are in vbls(H) ∪T). We must show that the number of “boundary edges” — i.e., edges in B that have exactly one endpoint in H — is at least |Tnew|.
Supposing otherwise, the set V = {variable-vertices v ∈B : v is incident on a boundary edge} ∪Told would satisfy |V | < |Tnew| + |Told| = |T| ≤d. We will show that this contradicts (10).
Claim 6.16. The deletion of variable-vertices V from B disconnects all variables in T from all variables in vbls(H) within B. (Note that when a variable does not even appear in a subgraph, it is trivially disconnected from all other variables.) Proof. It suffices to show that deleting V disconnects Tnew from vbls(H) within B, as the vertices of Told are already in V . Suppose j ∈Tnew is connected to some variable i ∈vbls(H) by a path within B. Since j / ∈vbls(H), there must be some edge in this path that has exactly one endpoint in H. This edge is a boundary edge, and hence the variable-vertex incident on it is in V . Thus we have indeed established that every path within B from a variable in Tnew to a variable in vbls(H) must pass through a variable in V .
27 Recall that the proof is complete once we show that |V | < d contradicts (10). Now E B[y · xT ] = E B h y · xT · X ⃗ c∈ΩV 1[xi = ci ∀i ∈V ] i = X ⃗ c∈ΩV E B y · xT · 1[xi = ci ∀i ∈V ] .
(11) We claim that every summand above equals 0. The reason is that for each summand ⃗ c, either 1[xi = ci ∀i ∈V ] is always 0 under ηB (establishing the claim), or else we may condition on the event, yielding E B y · xT · 1[xi = ci ∀i ∈V ] = Pr B [xi = ci ∀i ∈V ] · E B[y · xT | xi = ci ∀i ∈V ].
By Claim 6.16 and the definition of the planted distribution ηB (and vbls(y) ⊆vbls(H)), we have that y and xT are conditionally independent under ηB, conditioned on all (xi : i ∈V ). Therefore E B[y · xT | xi = ci ∀i ∈V ] = E B[y | xi = ci ∀i ∈V ] · E B[xT | xi = ci ∀i ∈V ].
Combining the previous two equations yields E B y · xT · 1[xi = ci ∀i ∈V ] = E B y · 1[xi = ci ∀i ∈V ] · E B[xT | xi = ci ∀i ∈V ].
Finally, using |V | < d we will show that the first factor above is 0 (thereby establishing the claim that every term in (11) is 0, in contradiction to (10)). To see this, we have E B y · 1[xi = ci ∀i ∈V ] = e E y · Y i∈V 1ci(xi) because cl(vbls(y) ∪V ) ⊆cl(vbls(H) ∪vbls(B)) ⊆B, where we used Theorem 5.9.
But this pseudoexpectation is indeed 0 by the lemma’s assumption, because Q i∈V 1ci(xi) is a polynomial expression of degree at most |V | < d.
6.5 Gram–Schmidt details We wish to show that Gram–Schmidt succeeds through substage (S, T) for all (S, T) ∈M≤D 2 . We will do this by induction along the order ⪯2. The key to showing that no positive definiteness problem is encountered at stage (S, T) will be the existence of a witness: Definition 6.17. A witness for substage (S, T) ∈M≤D 2 is defined to be a small τ-subgraph+ HS,T with vbls(HS,T ) ⊇vbls(yS,T ) and R(HS,T ) ≤2D.
Remark 6.18. For any substage of the form (S, ∅), we may always take as a witness the τ-subgraph+ consisting of all variables in S as isolated vertices.
As the below proposition shows, witnesses are useful for showing that one kind of positive definiteness problem does not occur. (They will also assist in showing the other kind does not occur.) Proposition 6.19. The existence of a witness HS,T for substage (S, T) implies e E[y2 S,T ] ≥0.
Proof. By Lemma 6.12, we have that H := HS,T ∪cl(vbls(yS,T )) is small. Thus e E[y2 S,T ] = EH[y2 S,T ] ≥0, using Theorem 5.16.
28 We now come to our main technical theorem: Theorem 6.20. Let (S, T) ∈M≤D 2 . Then: (i) Given any witness HS,∅for substage (S, ∅), there is a witness HS,T for substage (S, T) satis-fying HS,T ⊇HS,∅.
(ii) The Gram–Schmidt process succeeds through substage (S, T).
Proof. The proof will be by (strong) induction on (S, T) along ⪯2. Observe that in proving part (ii) of the theorem, by induction we only need to show that no positive definiteness problem occurs at substage (S, T). Further, if we can inductively establish part (i) of the theorem, then Remark 6.18 and Proposition 6.19 imply that e E[y2 S,T ] ≥0.
Thus to also establish part (ii), it would only remain to prove that no “pseudovariance zero problem” problem occurs. Also, observe that the pseudovariance zero problem can never occur when T = ∅. Thus for substages (S, ∅), we only need to establish part (i) of the theorem statement. But part (i) is trivial for (S, ∅) substages. Thus all substages of the form (S, ∅) are taken care of, including the base case of the induction (namely substage ({(i0, c0)}, ∅), where {(i0, c0)} is the first singleton in the order ⪯).
Thus it remains to establish, for a particular substage (S, T) with T ̸= ∅, that part (i) of the theorem statement holds, and also that no pseudovariance zero problem occurs. Given any witness HS,∅for substage (S, ∅), by induction we may obtain a witness HS,pr(T) ⊇HS,∅for substage (S, pr(T)). We now distinguish two cases.
Case 1: e E[yS,pr(T) · zT ] = 0. In this case, yS,T = yS,pr(T) and therefore e E[yS,T · zT ] = 0. Thus certainly no pseudovariance zero problem occurs, and also we can establish part (i) of the theorem statement simply by taking HS,T = HS,pr(T). Thus the inductive step is completed in this case.
Case 2: e E[yS,pr(T) · zT ] ̸= 0. This is where the main work in the proof occurs. First, we will show in this case that T ∈PvZ is impossible, and hence the pseudovariance zero problem cannot have occurred. We can then complete the induction by finding a witness HS,T ⊇HS,pr(T) for sub-stage (S, T).
First, suppose for contradiction that T ∈PvZ. We have that yS,pr(T) = xS −q(x) for some q(x) supported on monomials xT ′ with T ′ ⪯pr(T) ≺T. By Proposition 6.8 and induction, zT is orthogonal to all such polynomials. Thus we deduce 0 ̸= e E[yS,pr(T) · zT ] = e E[xS · zT ] = e E[xS · yT,pr(T)], (12) the last equality because T ∈PvZ and hence zT = yT = yT,pr(T). By induction (and using Re-mark 6.18), we have a witness HT,pr(T) for yT,pr(T). By Lemma 6.12 (using 2D+|S| ≤3D ≤ζ·SMALL) we have that H := HT,pr(T)∪cl(S) is small. (In writing cl(S) we used the abuse from Notation 6.3.).
Now vbls(H) ⊇vbls(HT,pr(T)) ∪S ⊇vbls(xS · yT,pr(T)), so by Theorem 5.16 we have e E[xS · yT,pr(T)] = E H [xS · yT,pr(T)], and also E H [y2 T,pr(T)] = e E[y2 T,pr(T)] = e E[z2 T ] = 0, the last equality because we’re assuming T ∈PvZ. But the second identity above shows that y2 T,pr(T) is identically 0 under ηH, meaning the first expression above must be 0. This contradicts (12).
Having ruled out the pseudovariance zero problem, we can complete the induction by finding a witness HS,T ⊇HS,pr(T) for substage (S, T). By Proposition 6.8 we have that zT = c · xT −p(x) 29 for some constant c > 0 and some polynomial p(x) supported on monomials xT ′ with T ′ ⪯pr(T).
Furthermore, yS,pr(T) is orthogonal to p(x) under e E[·]. Thus, since we are in Case 2, we may deduce that e E[yS,pr(T) · xT ] ̸= 0.
(13) We may now apply Lemma 6.14 (with y = yS,pr(T), H = HS,pr(T), and r = 2D) to obtain a small τ-subgraph+ Hnew ⊇HS,pr(T) with vbls(Hnew) ⊇vbls(yS,pr(T)) ∪T and R(Hnew) ≤2D.
This Hnew is almost able to serve as the witness for substage (S, T). The only deficiency is that, although it contains all the variables in yS,pr(T) and xT , it doesn’t necessarily contain all the variables appearing in zT — as it would need to in order to contain all variables in the new yS,T = yS,pr(T) −e E[yS,pr(T) · zT ]zT . However, we can fix this by induction; we apply the induction hypothesis to substage (T, pr(T)), taking Hnew as the “given witness HT,∅”.
This produces a witness — call it H′ new — for substage (T, pr(T)) that satisfies H′ new ⊇Hnew. This witness H′ new now additionally contains all variables in yT,pr(T) = zT , and therefore it can now serve as the needed witness for substage (S, T).
7 Wrapping things up by setting parameters To prove our main result on weak refutation, Theorem 1.2, we simply need to combine Theo-rems 4.12 and Theorem 6.1. Together these give us a pseudoexpectation defined up to degree D = Ω(γ) · ζ · n ∆2/(λ−ζ) , where γ = βO(1/λ) K·2O(K/λ) .
We need to decide how to best set parameters, which we do under the assumption that ∆≥10.
We start with the special but interesting case when λ is thought of very large; specifically, λ ≥ Ω(log ∆). This case arises, e.g., for high-arity K-SAT (where λ = K −2) with clause density 2Θ(K).
In this case, by choosing ζ = 1 2λ and β = e−O(K) for our probability bound, we get D = n/2O(K/λ).
Note that if λ = Θ(K), as it is in the case of K-SAT, then our SOS degree lower bound is linear in n with absolutely no dependence on K = K(n) (all the way up to K = Ω(n))!
In the more general regime (e.g., when one thinks of K as “constant” and ∆as asymptotically large), a good choice for ζ is 1 log ∆, which entails D = Ω(γ) · n ∆2/λ log ∆.
With this setting, Theorem 4.12 tells us that with high probability we get a pseudoexpectation satisfying Corollaries 5.18, 5.19. Thus we have established the following more precise version of Theorem 1.2: Theorem 7.1. Let P be a k-ary Boolean predicate and let C(P) be the minimum integer 3 ≤τ ≤k for which P fails to support a τ-wise uniform distribution. Then if I is a random instance of CSP(P ±) with m = ∆n constraints (∆≥10), then except with probability at most β, degree-D SOS fails to (weakly) refute I, where D = βO(1/C(P )) k·2O(k/C(P )) · n ∆2/(C(P)−2) log ∆.
The result also holds if P is a predicate over an alphabet of size q > 2 (with an appropriate notion of “literals”), with no change in parameters.
30 Proving our main result on δ-refutation, Theorem 1.1, requires just a little work.
We now imagine that our instance comes from a random CSP(P ±) as in Theorem 1.1. As discussed at the end of Section 2.1, given t and taking τ = t + 1, we have some t-wise uniform distribution µ on {±1}k which is δ-close to being supported on P, where δ = δP (t). We assume that all of the constraint distributions µf are now simply equal to µ, up to the appropriate negation pattern.
Thus a draw from µf satisfies the constraint at f except with probability at most δ.
With the parameter settings chosen earlier, Theorem 4.12 tells us moreover that #{nonempty τ-subgraphs H with |cons(H)| ≤2 · SMALL : I(H) ≤τ−1} ≤∆n 1+1/ log ∆ 2 = 2 log n 2 log ∆· m √n.
(14) Observe that this bound is always o(m), and in the very typical case that ∆≥nΩ(1), the bound is O( m √n). Let us see what this bound means for the pseudodistribution.
Supposing (14) holds, let f be any constraint-vertex in G, let S = N(f), and let Hf be the (small) τ-subgraph induced by the edges between f and S. Certainly cl(S) ⊇Hf, but we may ask whether cl(S) is strictly bigger than Hf. Suppose this is the case; i.e., there is some small S-closed H ̸⊆Hf. Then H′ = Hf ∪H is a τ-subgraph satisfying |cons(H′)| ≤2 · SMALL. Furthermore, the number of leaf variables in H′ must be at least 1 (else H′ is ∅-closed and hence empty by Fact 5.6) and strictly less than K (else H \ Hf will be ∅-closed and hence empty). Finally, we claim R(H′) ≤τ −1. This is because R(Hf) = τ, the addition of H cannot add any new credits (since all its leaf variables are already in Hf), and in fact the addition of H must cause a drop of at least one in revenue since H must have at least one edge not in Hf.
(This argument is similar to Lemma 6.13.) We conclude that whenever cl(N(f)) ̸= Hf, there must exist a nonempty τ-subgraph H′ with the following properties: (i) |cons(H′)| ≤2·SMALL; (ii) I(H′) ≤R(H′) ≤τ −1; (iii) H′ has at least one leaf variable; (iv) all leaves of H′ are adjacent to f.
But (14) bounds the number of τ-subgraphs with the first two properties above, and every τ-subgraph with the latter two properties uniquely determines f. Thus we conclude: #{constraints f : cl(N(f)) ̸= Hf} ≤2 log n 2 log ∆· m √n.
Finally, when cl(N(f)) = Hf, observe that the planted distribution ηcl(N(f)) is just µf, and hence e E h 1[x satisfies f] i = Pr x∼µf h 1[x satisfies f] i ≥1 −δ.
Combining the last two deductions yields e E h fraction of constraints satisfied i ≥1 −δ −2 log n 2 log ∆· 1 √n.
In summary, we have proven the following more precise version of Theorem 1.1: Theorem 7.2. Let P be a k-ary Boolean predicate and let 1 < t ≤k. Let I be a random instance of CSP(P ±) with m = ∆n constraints. Then except with probability at most β, degree-D SOS fails to (δP (t) + ϵ)-refute I, where ϵ = 2 log n 2 log ∆· 1 √n, D = βO(1/t) k·2O(k/t) · n ∆2/(t−1) log ∆.
We remark that ϵ = o(1) always, and ϵ = O( 1 √n) whenever ∆= nΩ(1). Finally, the result also holds if P is a predicate over an alphabet of size q > 2 (with an appropriate notion of “literals”), with no change in parameters.
31 Remark 7.3. We should mention that in our δ-refutation result Theorem 7.2, our pseudoexpec-tation does not satisfy “solution value = 1 −δ0” as a constraint for any δ0 ≤δ; it merely has e E[solution value] ≥1 −δ. Achieving the (stronger) former condition is a direction for future work.
By contrast, for our weak refutation result Theorem 1.2, the pseudoexpectation does satisfy all the constraints and hence also satisfies e E[solution value] = 1 as a constraint.
Acknowledgment We would like to thank the Institute for Mathematical Sciences, National University of Singapore in 2016; a visit there was where some of the initial research for this work began.
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In Proceedings of the 28th Annual Conference on Computational Complexity, pages 121–132, 2013. 1.4 A Proof that random graphs satisfy the Plausibility Assumption Here we prove Theorem 4.12, which we restate for convenience: Theorem 4.12 restated.
Let λ = τ −2 ≥1. Fix 0 < ζ ≤.99λ, 0 < β < 1 2. Then except with probability at most β, when G is a random instance with m = ∆n constraints, the Plausibility Assumption holds provided SMALL ≤γ · n ∆2/(λ−ζ) , (15) where γ = 1 K β1/λ 2K/λ O(1) . Moreover, assuming ζ < 1, except with probability at most β we have #{nonempty τ-subgraphs H with cons(H) ≤2 · SMALL : I(H) ≤τ −1} ≤∆n 1+ζ 2 .
(16) Proof. A remark before we begin: the expression in (15) was chosen precisely so that c ≤2 · SMALL = ⇒20K · ∆· Kc n λ−ζ 2 ≤β/50K, (17) provided the O(1) in the definition of γ is a sufficiently large universal constant.
The proof is a standard argument of the kind used to show that a random bipartite graph has good expansion. Fixing I0 ∈{0, τ −1}, 1 ≤c ≤2 · SMALL, and 1 ≤v ≤Kc, let us upper-bound E[#{τ-subgraphs with c constraints, v vertices, and income at most I0}].
(18) There are m c choices for the constraints and n v choices for the variables. Then by using Lemma 4.11, (18) ≤ m c n v Pr[fixed set of c constraints and v variables gets at least A edges], (19) where A := τ−ζ 2 ·c+v−I0 2 . In (19), we may imagine that a constraint’s variables are chosen uniformly and independently (i.e., without conditioning on them being distinct), as this only increases the 37 probability in question. Now any fixed set of c constraints has at most Kc edges coming out it, so the probability that some integer a > A of them will go into a fixed set of v variables is at most Kc a · v n a ≤2Kc · v n a ≤2Kc · v n A .
Thus (19) ≤2Kc m c n v v n A ≤2Kc em c c en v v v n A = e2Kev/c(v/c) c · ∆c · v n λ−ζ 2 ·c−I0/2 ≤ 20Kc · ∆c · Kc n λ−ζ 2 ·c−I0/2 , (20) where the equality used the definition of A and the subsequent inequality used v ≤Kc.
We now split into two cases, depending on whether I0 is 0 or τ −1. When I0 = 0 we use (18) ≤(20) = 20K · ∆· Kc n λ−ζ 2 c ≤ β 50K c , using (17). Summing over the at most Kc possibilities for v gives E[#{τ-subgraphs with c constraints and income at most 0}] ≤Kc β 50K c .
Now summing this expression over all 1 ≤c ≤2 · SMALL we get E[#{implausible τ-subgraphs H : |cons(H)| ≤2 · SMALL}] ≤ ∞ X c=1 Kc β 50K c ≤β.
Thus Markov’s inequality implies that the Plausibility Assumption holds except with probability at most β.
The analysis for I0 = τ −1 is similar. In this case, we use (18) ≤(20) = 20K · ∆· Kc n λ−ζ 2 c−1 · 20K · ∆· n Kc 1+ζ 2 ≤ β 50K c−1 · 20K · ∆n 1+ζ 2 , again using (17). We again sum this over the at most Kc possibilities for v. We also only need to sum this over all c ≥2, since if cons(H) = 1 then I(H) = τ −ζ > τ −1. We then obtain E[#{nonempty small τ-subgraphs H with |cons(H)| ≤2 · SMALL : I(H) ≤τ −1}] ≤ ∞ X c=2 Kc β 50K c−1 20K · ∆n 1+ζ 2 ≤β · n 1+ζ 2 , and again Markov’s inequality establishes that (16) holds except with probability at most β.
38 |
16895 | https://www.etymonline.com/word/bedevil | Want to remove ads? Log in to see fewer ads, and become a Premium Member to remove all ads.
Origin and history of bedevil
bedevil(v.)
1768, "to treat diabolically, abuse," from be- + verbal use of devil (q.v.). The meaning "mischievously confuse" is from 1755; that of "drive frantic" is from 1823. Related: Bedeviled (1570s in a literal sense, "possessed"); bedeviling.
also from 1768
Entries linking to bedevil
Old English deofol "a devil, a subordinate evil spirit afflicting humans;" also, in Christian theology, "the Devil, a powerful spirit of evil otherwise known as Satan," from Late Latin diabolus (also the source of Italian diavolo, French diable, Spanish diablo; German Teufel is Old High German tiufal, from Latin via Gothic diabaulus).
The Late Latin word is from Ecclesiastical Greek diabolos, which in Jewish and Christian use was "the Devil, Satan," and which in general use meant "accuser, slanderer" (thus it was a scriptural loan-translation of Hebrew satan; see Satan). It is an agent noun from Greek diaballein "to slander, attack," literally "to throw across," from dia "across, through" (see dia-) + ballein "to throw" (from PIE root gwele- "to throw, reach").
Jerome re-introduced Satan in Latin bibles, and English translators have used both words in different measures. In Vulgate, as in Greek, diabolus and dæmon (see demon) were distinct, but they have merged in English and other Germanic languages.
Meaning "false god, heathen god" is from c. 1200. Sense of "diabolical person, person resembling a devil or demon in character" is from late 12c. Playful use for "clever rogue" is from c. 1600, on the notion of mischievous energy and cleverness. The sense of "errand-boy in a printing office" (1680s) is from this, perhaps reinforced because they often were blackened by the ink (devils then being popularly supposed to be black).
As an expletive and in expletive phrases from c. 1200.
Meaning "sand spout, dust storm" is from 1835 (dust devil is attested by 1867). In U.S. place names, the word often represents a native word such as Algonquian manito, more properly "spirit, god." Phrase a devil way (c. 1300) was originally "Hell-ward, to Hell," but by late 14c. it was a mere expression of irritation.
Devil's books "playing cards" is from 1729, but the cited quote says they've been called that "time out of mind" (the four of clubs is the devil's bedposts); devil's coach-horse is from 1840, the large rove-beetle, which is defiant when disturbed. Devil's food cake (1895; three different recipes in the cookbook "compiled by the Ladies' Aid Society of the Friends' Church, Wilmington, Ohio"), rich and chocolate, probably is in deliberate contrast to angel food cake. Conventional phrase talk (or speak) of the Devil, and he's presently at your elbow is by 1660s.
"state of bewildering or vexatious disorder or confusion, "1825, from bedevil + -ment.
word-forming element of verbs and nouns from verbs, with a wide range of meaning: "about, around; thoroughly, completely; to make, cause, seem; to provide with; at, on, to, for;" from Old English be- "about, around, on all sides" (the unstressed form of bi "by;" see by (prep.)). The form has remained by- in stressed positions and in some more modern formations (bygones, bystander); in bylaw it is a different word.
The Old English prefix also was used to make transitive verbs and as a privative prefix (as in behead). The sense "on all sides, all about" naturally grew to include intensive uses (as in bespatter "spatter about," therefore "spatter very much," besprinkle, etc.). Be- also can be causative, or have just about any sense required. The prefix was productive 16c.-17c. in forming useful words, many of which have not survived, such as bethwack "to thrash soundly" (1550s) and betongue "to assail in speech, to scold" (1630s).
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16896 | https://www.youtube.com/watch?v=y220rJnPXv8 | Reducing fractions using the greatest common factor | Reduce fractions to lowest terms
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This video is about reducing fractions using the greatest common factor, in which I showed how to reduce fraction into lowest terms using GCF method.
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hi in this video we'll learn how to simplify fractions to its lowest term to reduce the fraction to its lowest term we need to divide the numerator and the denominator by the greatest common factor now what do you mean by greatest common factor greatest common factor is the largest number that can divide evenly into two other numbers for example the greatest common factor of 2 and 10 will be 2 because 2 is the largest number that can divide into both 2 and 10. so we will divide the numerator as well as the denominator by our greatest common factor now 2 divided by 2 is equal to 1 and 10 divided by 2 is equal to 5. notice that the new fraction is equivalent to the original fraction two-tenths is equal to one-fifths and i cannot reduce this fraction anymore this fraction is in lowest term now because we cannot divide the numerator and the denominator by any other number other than one so now we know that that two-tenths is equal to one-fifths now let's take a look at this example eight divided by thirty-two all we need to do is we need to think the greatest number which can go evenly into 8 and 32 well the greatest number which can divide into both 8 and 32 is 8. so i will divide my numerator and the denominator by 8 in order to reduce this fraction [Music] 8 divided by 8 is equal to 1 and 32 divided by 8 is equal to 4. so 8 over 32 can be reduced to 1 4 and these are equivalent fractions so i can write that 8 over 32 is equal to 1 4 and i know that i cannot reduce this fraction anymore because we cannot divide 1 and 4 by any other number other than 1. now let's take a look at this example now when you have bigger numbers when you have larger numbers it's hard to identify the greatest common factor in that case what we're going to do is we're going to list down the factors of 35 and 42. so let's list down the factors of 35 and 42 obviously the factors of 35 are going to be 1 in 35 and 35 is also divisible by 5 and we know that 7 times 5 is also 35 so 35 is divisible by 7 as well so we know that the factors of 35 are going to be 1 5 7 and 35 and let's write down the factors of 42 starting from 1 now since 42 is an even number it's obviously divisible by 2. 42 is also divisible by 3 then we know that 7 times 6 is 42 you think of two factors which when multiplied to give you 42. so we're going to think of two factors which would multiply to give us 42 we know that 7 times 6 is 42. now since 42 is divisible by 2 right it's an even number so if we divide 42 by 2 we'll get 21. so we know that 2 times 21 gives us 42 the same way if i divide 42 by 3 i'll get 14 so 3 times 14 is also 42 and then what else obviously 1 times 42 is going to give us 42. so let's write down the factors of 42. this will be 3 after 3. let's keep it organized so we're going to write 6 after that then 7 and then what else do we have we have 14 we have 21 and obviously it's going to be 42. so we know that the factors of 42 are 1 2 3 6 7 14 21 and 42. now we're looking for the largest number which is common in both largest factor which is common in both of them right so the largest number the largest factor which is common in both of them is going to be 7. so i know that the gcf of 35 and 42 is going to be 7 that means i have to divide my numerator and the denominator by 7. so let's go ahead and do that 35 divided by 7 is going to be 5 and then 42 divided by 7 is going to be 6. so i can reduce my fraction 35 over 42 to 5 6. let's look at this fraction 15 over 36 so let's list down the factors of 15 and 36 the factors of 15 are going to be 1 3 because 3 times 5 is 15 obviously it's going to be 5 because whenever you have a 5 at the units place that number is always divisible by 5. so 5 is a factor of 15 and obviously 15 will be a factor of 15. now let's write down the factors of 36. now since this is a bigger number let's do it this way we're going to think what two factors multiplied to give us 36 well 6 times 6 is 36 12 times 3 is 36 2 times 18 is also 36 since 36 is an even number it will be divisible by 2 so that's why i divided 36 by 2 and i got 18 so i've written 2 times 18 is equal to 36 4 times 9 4 times 9 is also equal to 36 i think that's it obviously 1 times 36 is 36 so we're going to write that we have to write the factors from 1 through 36 so the first factor is going to be 1. we're going to write all the factors in order so 2 then we have 3 then we have 4 and we have 6 and we have 9 then we have 12 and then we have 18 and obviously 36. now if you want you can stop after 12 you don't have to write 18 and 36 because we're looking for the greatest common factor we're looking for a factor which is common in both since this number is 15 it doesn't make sense to write the factors which are greater than 15 here because we won't be able to find the common factor if we write the numbers which are greater than 15. and now we can see that the factor which is common with the greatest factor which is common in both is going to be three so i can divide my fraction 15 over 36 by 3 since 3 is my gcf 15 divided by 3 is going to be 5 and 36 divided by 3 is going to be 12. so i can reduce my fraction 15 over 36 to 5 12 and that's going to be my final answer so i hope you find this video helpful that's all in this video thank you so much for watching if you are new to this channel and want to see more videos and other math topics be sure to subscribe down below and share it with your friends see you next video |
16897 | https://art19.com/shows/merriam-websters-word-of-the-day/episodes/f5d2bd40-4985-4776-b95b-9ca5738dd178 | ART19
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Merriam-Webster's Word of the Day | hobnob
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Merriam-Webster's Word of the Day for October 29, 2009 is:
hobnob \HAHB-nahb\ verb
: to associate familiarly
Examples:
Bill hoped his new job as a reporter would give him an opportunity to hobnob with politicians and other notables.
Did you know?
"Hob" and "nob" first came together in print in Shakespeare's Twelfth Night, when Sir Toby Belch warned Viola (who was disguised as a man) that Sir Andrew wanted to duel. "Hob, nob is his word," said Sir Toby, using "hob nob" to mean something like "hit or miss." Sir Toby's term is probably an alteration of "habnab," a phrase that meant "to have or not have, however it may turn out." After Shakespeare's day, "hob" and "nob" became established in the phrases "to drink hob or nob" and "to drink hobnob," which were used to mean "to drink alternately to each other." Since "drinking hobnob" was generally done among friends, "hobnob" came to refer to congenial social interaction.
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16898 | https://www.emdocs.net/erythrocytosis-ed-focused-evaluation-management/ | Erythrocytosis: ED Focused Evaluation and Management - emDocs
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Erythrocytosis: ED Focused Evaluation and Management
May 24, 2017
Kristine Jeffers
Author:Kristine Jeffers, MD (Emergency Medicine Resident at San Antonio Military Medical Center)// Edited by: Alex Koyfman, MD (@EMHighAK, EM Attending Physician, UT Southwestern Medical Center / Parkland Memorial Hospital) and Brit Long, MD (@long_brit)
Eero Mäntyranta, born in 1937, was one of Finland’s most successful Olympic skiers. He competed in 4 Olympic Games between 1960 and 1972 and won a total of 7 medals, three of which were gold. Part of Mäntyranta’s success came from winning the genetic lottery in endurance sports. He had primary familial and congenital polycythemia which caused an increase in the number of red blood cells, red blood cell mass, and ultimately his oxygen carrying capacity.1
Erythrocytosis, also known as polycythemia, refers to an increase in red blood cell mass. It can be relative or absolute. In relative polycythemia there is a decrease in plasma volume making it appear that there is an increase in red blood cells. There are three categories of absolute polycythemia.2,3
-Primary Polycythemia: Occurs when there is an inherited or acquired mutation in erythroid progenitors that lead to proliferation. This can be seen in polycythemia vera (PV), a myeloproliferative disorder, or in rare polycythemias that lead to an activation of the erythropoietin (EPO) receptor like in primary familial and congenital polycythemia as in the case of Mäntyranta.2,4
-Secondary Polycythemia: Occurs when there are increased levels of EPO circulating that stimulate erythroid progenitors. This can further be divided into two groups.
-Compensatory: This occurs when hypoxia stimulates EPO production and therefore red blood cell production. Many conditions and diseases contribute like pulmonary disease (COPD, OSA), cyanotic heart lesions, altitude, and carbon monoxide poisoning to name a few.
– Paraneoplastic: This occurs when EPO or an EPO like substance is produced by some other process. Most commonly this is the result of a malignancy (like renal cell carcinoma).2,3
-Disorders of hypoxia sensing: As the name states, this type of polycythemia results when there is a problem with hypoxia sensing that leads to increase red blood cells. It can have characteristics of both primary (increased sensitivity to EPO) and secondary polycythemia (increased EPO levels). These include Chuvash polycythemia and von Hippel Lindau gene mutations.4
Initial Evaluation of polycythemia
The initial evaluation of a patient presenting with polycythemia focuses on identifying the cause and determining whether the patient has primary or secondary polycythemia. Polycythemia should be considered in patients presenting with symptoms and evidence of erythrocytosis on laboratory evaluation.
History:
-On of the most common causes of polycythemia identified on history is hypoxia secondary to lung disease.
-History of travel to altitude for extended periods of time, cardiac defects, renal transplant, and use of O 2 at home or smoking should also be elicited.
-Common presenting symptoms include chest and abdominal pain, myalgias, weakness, fatigue, headache, and slowed mentation.5 Patients may also present with pulmonary complaints to include shortness of breath, chronic cough, or hypersomnolence.
-Symptoms that are more predictive of PV include pruritus after bathing, erythromelalgia, gout, arterial and venous thrombosis, hemorrhage, and early satiety from splenomegaly. 2
Physical Exam:
-Vital signs including pulse ox, signs of cyanosis in lips, fingers, clubbing, murmur, or bruits. Patients may also have plethoric facies or hepatosplenomegaly.2
Laboratory Evaluation:
-CBC with indices.
-UA- looking for microscopic hematuria, which could be seen in setting of renal cell carcinoma.
-LFTs to evaluate for hepatocellular carcinoma, cirrhosis, hepatitis, or hemochromatosis.
-Consider CO test for CO poisoning.
-Chest X-ray looking for causes of erythrocytosis (COPD, pulmonary hypertension, AV malformations).
Other testing (not required in the ED):
-Serum EPO- low is relatively specific for diagnosis of PV but may also be seen in blood doping.
– If elevated, erythrocytosis is likely secondary.
-JAK2 mutation testing.
-Bone Marrow aspiration/biopsy.2
Treatment:
Treatment of secondary polycythemia is twofold. Primarily the underlying cause should be removed if possible (excision of tumor, stopping steroids, removing CO exposure, etc.). If removing the cause is not possible or is ineffective, limited phlebotomy is done for symptom relief. Treatment of polycythemia vera (the most common cause of primary erythrocytosis) is discussed below.6
Polycythemia Vera
Polycythemia vera affects an estimated 100,000 people in the United States, and those affected have an increased mortality most often from complications related to thromboembolic or hemorrhagic events. Risk factors include age>60 years, history of thrombosis, and hematocrit >45%.5-7
Pathogenesis: PV develops from a point mutation in the JAK-2 leading to tyrosine kinase activation. The mutated tyrosine kinase leads to growth factor independent proliferation of red blood cells.2,3
Diagnostic Criteria: The WHO diagnostic criteria for PV include three major and one minor criterion. You may have all three major or the first two major and one minor criterion.
Major Criteria:
-Hemoglobin > 16.5 g/dL in men (> 16 g/dL in women) or hematocrit >49% in men (48% women) or red cell mass >25% above the mean.
-Bone marrow tri-lineage myeloproliferation with pleomorphic mature megakaryocytes.
– JAK2 mutation.
Minor Criteria:
-Subnormal serum erythropoietin level.2, 7
You should consider PV in patients presenting with the following symptoms and an elevated hemoglobin/hematocrit on lab analysis.
Common Symptoms: pruritus, headache, fatigue, difficulty thinking or sleeping, dizziness, hyperuricemia, and bleeding or thrombotic complications.
Treatment: All patients with PV are treated with phlebotomy to a hematocrit level of 45% (creates an iron deficient anemia) and low dose aspirin. Further pharmacological treatment is based on risk stratification and lack of symptom control with the above mentioned. Other options for cytoreductive therapy include:6-8
-Hydroxyurea: First line medication for cytoreductive therapy. About 25% of people will become resistant or intolerant during the course of treatment.
-Interferon alpha: Associated with clinical benefit and normalized blood counts. It may become toxic especially at high doses.
-Ruxolitinib: JAK1/JAK2 inhibitor approved by the FDA for use in patients with PV that have failed hydroxyurea and studies have shown it to be better than the best available treatments when it came onto the market.7
Hyperviscosity Syndrome:
Hyperviscosity syndrome presents with a triad of bleeding, visual disturbances, and focal neurologic signs and can be a complication of erythrocytosis.9 Bleeding is often seen from mucosal surfaces (epistaxis, gums, GI). Retinopathy is caused by thrombosis and microhemorrhages. Neurologic complications can include a wide range of symptoms from headache to stupor, vertigo, or coma.10 It should be considered in the emergency department in three settings: when the patient presents with the classic triad, critically ill patients with laboratory abnormalities that are indicative of hyperviscosity, and in patients with known hematologic disease and new decline. It occurs more frequently in diseases of hyperproteinemia but can also be seen in the setting of erythrocytosis.
Treatment resolves around early consultation to hematology and aggressive supportive resuscitation with hydration and avoidance of diuretics.9-11 Rapid reduction of offending agent is accomplished with plasmapheresis or plasma exchange. Definitive care is accomplished with chemotherapeutic management.
Summary:
–Consult hematology early for recommendations, further evaluation, and definitive treatment.
-Secondary erythrocytosis is from numerous causes; a good history and physical will discover most of them! EPO level may be helpful for differentiation.
-Treatment of secondary erythrocytosis includes removal of offending cause and possibly phlebotomy.
-Consider PV in setting of elevated hemoglobin/hematocrit + symptoms (pruritus, headache, fatigue, difficulty thinking or sleeping, dizziness, hyperuricemia, and bleeding or thrombotic complications) especially in persons >60 years old.
-Initial treatment of PV includes phlebotomy (consult your specialist) and low-dose aspirin.
-Hyperviscosity syndrome: triad of bleeding, visual disturbances, and focal neurologic signs. Mainstay of treatment is hydration and hematology consult.
References/Further Reading
Eero Mantyranta. (2017, January 14). Retrieved May 03, 2017, from
Tefferi, A. (2017, February 15). Diagnostic Approach to the patient with Polycythemia (S. Schrier, Ed.). Retrieved April 5, 2017, from
Robbins, S. L., Cotran, R. S., & Kumar, V. (2010).Robbins and Cotran pathologic basis of disease (8th ed.). Philadelphia, PA: Saunders/Elsevier.
Prchal, J. (2016, August 08). Molecular pathogenesis of congenital polycythemic disorders and polycythemia vera (S. Schrier, Ed.). Retrieved April 5, 2017, from
Wang, Eurnice; Berliner, Nancy. Clonal Disorders of the Hematopoietic Stem Cell. In: Cecil Essentials of Medicine, Eighth Edition. 2010: 507-508.
Tefferi, A. (2016, July 11). Prognosis and treatment of polycythemia vera (S. Schrier, Ed.). Retrieved April 5, 2017, from
Tefferi, A., & Barbui, T. (2017). Polycythemia vera and essential thrombocythemia: 2017 update on diagnosis, risk-stratification, and management. American Journal of Hematology,95, 95-108.
Gerds, A., & Dao, K. (2017). Polycythemia Vera Management and Challenges in the Community Health Setting.Oncology,92, 179-189.
Meier, B., & Burton, J. (2014). Myeloproliferative Disorders. Emergency Medicine Clinical Journal of North America,32, 597-612.
Adams, B., Baker, R., Lopez, A., & Spencer, S. (2009). Myeloproliferative Disorders and the Hyperviscosity Syndrome. Emergency Medicine Clinical of North America, 27, 459-476.
Blackburn, Paul. Emergency complications of Malignancy. In: Tintinalli’s Emergency Medicine: A Comprehensive Study Guide, Seventh Edition. 2011:1514.
Hackett, Peter; Hargrove, Jenny. High-Altitude Medical Problems. In: Tintinalli’s Emergency Medicine: A Comprehensive Study Guide, Seventh Edition. 2011:1409.
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16899 | http://www.infomontessori.com/mathematics/numbers-through-ten-cards-and-counters.htm | Montessori - Mathematics - Numbers through ten - Cards and Counters
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Introduction Numbers through Ten Number Rods Sandpaper Numbers Number Rods and Cards Spindle Boxes Concept of Zero Cards and Counters Memory Game Decimal System Introduction to quantity Symbols Formation of Numbers Changing Addition Multiplication Substraction Division Stamp Game Dot Game Word Problems Linear & Skip Counting Teens: Quanity Teens: Symbol Tens: Association Linear Counting Skip Counting Number Roll Tables of Arithmetic Addition Snake Game Strip Board - exercises Addition Strip Board Addition Charts Substraction Snake Game Substraction Strip Board Substraction Charts Multiplication Bead Multiplication Board Multiplication Charts Unit Division Board Division Charts Passage to Abstraction Small Bead Frame Wooden Hierarchical Material Large Bead Frame Racks and Tubes Fractions Fractions Cards and CountersMaterials- A series of cards with numbers from 1 to 10 - A box with 55 counters Presentation Introduction 1. Invite a child to come work with you. Bring him over to the shelf and have him bring the material to the table. 2. Open the box and place it on the lid. 3. Show the different cards to the child and have him say the numbers out loud. 4. Lay the cards on the table in front of the child. 5. Have the child put the card written 1 to the left side of the table. 6. Explain that you will need a little space between the cards. 7. Ask the child to put the other cards to the right of the first card in numerical order. 8. Tell the child that you are going to put the number of counters each card asks for under the corresponding card. 9. Point to card 1 and ask the child how many counters should you place under this card. 10. The child should answer 1. 11. Take one counter out of the box and gently slide it (using your right index finger) under the card written 1. 12. Have the child count the counter. 13. Repeat for card 2 by placing the counters next to each other. 14. Repeat for card 3 but place the last counter under and to the middle of the two counters. (See diangram below) 15. Repeat until the child seems to understand. 16. Have the child place the rest of the counters in a similar way as you have shown. Direct the child on how many counters and where he will place them through questions. 17. Repeat until all of the counters have been placed. 18. Tell the child that you are going see if you can run your fingers through the counters. 19. Place your right index finger above the first counter (under card 1) and try to run it down the table through the counter. 20. Stop before touching the counter and tell the child that you can not divide it. Move card 1 a little above where it was. 21. Repeat for counters 2 and after you have run your finger through the two cunters, tell the child that you can divide it. 22. Repeat for 3 and after, move the card a little higher to the level of card 1. 23. Have the child repeat for the others. You or the child move the cards up for those numbers where you can’t run your fingers through. 24. Once the child is finished, look at all of the cards that are slightly higher than the others (cards 1, 3, 5, 7, 9) and tell the child that these numbers are odd. 25. Point to the other cards (cards 2, 4, 6, 8) and tell the child that these numbers are even. 26. Do a Three Period Lesson for odd and even. Purpose Direct To reinforce the knowledge that each number is made up of separate quantities. To verify whether the child has mastered: - his sequence of numbers - How many separate units go to form each number To indicate the odd and even numbers Indirect To prepare for the divisibility of numbers Control of ErrorIf the counting has been incorrect, at the end there will be either an insufficient number of counters or some left over. Age4 years – After work with the Spindle Boxes Questions, Comments ? Share your experiences in the Forum Send Lesson to a Friend:
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