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17400 | https://proyectodescartes.org/Un_100/materiales_didacticos/_Un_052_ElTiroParabolico/escenas/Cierre.pdf | El tiro parabólico Cuando se lanza un proyectil, su trayectoria puede ser descrita descomponiendo su movimiento en su componente horizontal y vertical.
Para la componente horizontal no existen fuerzas (esto es, no hay fuerzas que lo jalen hacia delante, atrás, derecha o izquierda; sólo hacia abajo, que es el caso de la gravedad). En ausencia de fuerzas no hay aceleración. Sin aceleración, la velocidad no cambia. Si la velocidad no cambia, el crecimiento de la distancia es lineal. De esa forma puede ser descrito el movimiento del proyectil sobre la horizontal. Otra forma de verlo es que el comportamiento sobre la horizontal corresponde a un movimiento rectilíneo uniforme.
En la vertical, sí hay presencia de una fuerza constante (la gravedad) y, por tanto, de una aceleración constante. Por lo mismo, la velocidad no es constante sino que se modifica linealmente. Consecuentemente, la distancia se modifica de forma cuadrática. Otra forma de verlo es que el comportamiento sobre la vertical corresponde a un movimiento uniformemente acelerado.
La composición de un cambio lineal para la componente horizontal y un cambio cuadrático en la vertical resulta en una trayectoria parabólica, en la que el eje de la parábola es vertical y la parábola abre hacia abajo.
Veamos ahora el porque de la construcción geométrica vista en el desarrollo.
1. Razón por la cual el foco se encuentra en una circunferencia de radio h (la altura alcanzada en el tiro vertical para una misma v0), al doble del ángulo del tiro con respecto a la vertical Una de las formas de encontrar el eje de la parábola, a partir de donde fue lanzado el proyectil, es considerar que éste coincide con el punto máximo de la parábola, mismo que se alcanza cuando la velocidad en el eje y es cero. Para ello, se requiere que se cumpla que v0y−gt=0 . Aquí, g corresponde a la aceleración de la gravedad en valor absoluto. La dirección ya está dada por el signo menos. v0y corresponde a la componente vertical de la velocidad inicial. Por lo tanto, t=v0y g . Por otra parte, el movimiento en el eje x está descrito por x=v0xt , donde x es la distancia horizontal a partir del punto del lanzamiento. Sustituyendo en esta última el tiempo en que se alcanza el máximo, se tiene que la distancia horizontal del punto de lanzamiento al eje vertical de la parábola es x=v0xv0y g (1a) Para obtener la altura del vértice sobre el punto del lanzamiento del proyectil, usamos que la energía total en el vértice será la suma de la cinética en x y la potencial en y (pues el proyectil no se mueve en el eje y, pero sí tiene altura y sí se mueve en el eje x), de tal forma que 1 2 mv0x 2 +mgy=mgh . De donde y=h−v0x 2 2g (1b) Así, las coordenadas del vértice de la parábola son ( v0xv0y g ,h−v0x 2 2g ) (2) Otra forma de verlo es que el vértice se encuentra v0x 2 2g unidades debajo de la directriz, y la directriz corresponde al punto de energía potencial máxima, dado a una altura de h (caso del tiro vertical hacia arriba a una velocidad v0). En dicho punto máximo, toda la energía cinética inicial 1 2 mv0 2 está en forma de energía potencial mgh, de tal suerte que h= v0 2 2g (3) Y, sabiendo que se trata de una parábola, y que la propiedad de las parábolas es que la distancia del foco a un punto de la parábola es igual que de dicho punto a la directriz, podemos calcular el foco como aquél que se encuentra v0x 2 2g por debajo del vértice. Esto es, h−v0x 2 2g −v0x 2 2g y las coordenadas del foco F son ( v0xv0y g ,h−v0x 2 g ) (4) Sea θ el ángulo que forma el vector v0 con la vertical. Entonces v0x=v0sin θ y v0y=v0cosθ . Sustituyendo estas expresiones en (4), las coordenadas de F son ( v0 2sin θcosθ g ,h−v0 2 sin 2θ g ) . Utilizando la identidad trigonométrica sin 2θ=2sin θcosθ para la coordenada en x y la identidad sin 2θ=1−cos2θ 2 para la coordenada en y, se obtiene que F se encuentra en (hcos 2θ ,hsin 2θ) . Esto es, en una circunferencia de radio h y reflejando al doble del ángulo θ. tal como se indicó en la construcción.
2. Razón por la cual la magnitud de la velocidad corresponde a √ ∣BF∣ h v0 , donde B es un punto arbitrario en la trayectoria.
En un punto B cualquiera en la trayectoria, la energía total será la suma de la componente cinética relacionada a la velocidad en ese punto (a la cual denotaremos vB) y la componente vertical relacionada a la altura. Esto es, E=1 2 mvB 2+mg (h−∣BF∣)=mgh (5) Aquí, mgh es nuevamente la energía potencial máxima alcanzada para el caso de tiro vertical. El término h−∣BF∣ corresponde a la altura del proyectil en el punto B. Recordemos que es la distancia por debajo de la directriz de altura h, que por la propiedad de equidistancia mencionada anteriormente, es la misma distancia que de tal punto B hacia el foco F.
Despejando vB de (5) obtenemos v B=√2g∣BF∣ (6) Por otra parte, despejando 2g de (3) tenemos que 2g=v0 2 h (7) Sustituyendo 2g de (7) en (6) se obtiene que v B=√ v0 2 h ∣BF∣=√ ∣BF∣ h v0 , que es la expresión mostrada en el desarrollo.
3. Construcción de la envolvente de las trayectorias parabólicas.
La envolvente de las trayectorias parabólicas se obtiene bajo el siguiente razonamiento: Sabemos que, dado que una trayecotria arbitraria es parabólica, el segmento FT es de igual longitud que el TD. Si se traza una continuación del segmento TD que sea de longitud igual a aquella del segmento OF (donde O es el origen del proyectil), se obtendrá un triángulo isósceles con vértices O, T y De. Por construcción, el punto T pertenecerá tanto a la trayectoria particular dado un ángulo de disparo, pero también pertenecerá a la curva envolvente que se construye.
Dado que es un triángulo isósceles, la distancia ∣OT∣ es igual a la ∣TDe∣, para cualquier T. Por lo mismo estamos hablando de una parábola con foco en O y directriz en De (que convenientemente llamaremos directriz de la envolvente).
Un punto interesante a notar es la fuerza de la ley de la conservación de la energía. Bien se nos pudieron haber dado por fe la construcción para encontrar el foco de la parábola y la fórmula para la magnitud de la velocidad. De no saber de dónde venían o cómo se deducían originalmente, bastó confirmar que dicha construcción y fórmula satisfacían la conservación de la energía para saber que, en efecto, eran válidas. La deducción aquí presentada simplemente muestra que era posible obtenerlas a partir de los conceptos que hemos revisado del tiro parabólico.
Este tipo de situaciones se presenta con frecuencia en la ciencia, en la que se postula una interpretación de un fenómeno y se observa que respeta ciertos principios fundamentales, por lo que se considera válida aunque no tenga una deducción como tal. Frecuentemente, poco después se encuentra una deducción formal para el fenómeno en cuestión.
Muchos de los ejercicios relacionados con el tiro parabólico consisten en el cálculo de distancias alcanzadas, a veces en la horizontal, otras en la vertical. Un representante de estos ejercicios es el cálculo de la distancia máxima que puede recorrer el proyectil. Dicho problema nvolucra el acoplamiento de las ecuaciones relacionadas al tiempo en ambos ejes coordenados (estrategia típica para la resolución de este tipo de problemas). Adicionalmente, echa mano del cálculo para abordar el problema como un problema de optimización (en este caso de la distancia recorrida en el eje x). |
17401 | https://www.quora.com/How-do-I-prove-A-B-A-B-1 | Something went wrong. Wait a moment and try again.
How do I prove (A => B) <=> ~ [A & ~B]?
Larry Terrell
B.S. in Mathematics, California Institute of Technology (Caltech) (Graduated 1972)
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Author has 14K answers and 7.7M answer views
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4y
The simplest way is to use a truth table. Just make four rows, one for each possible set of values for A and B, in columns 1 and 2. Then in columns 3 and 4, write down the resulting values for A => B and A & ~B.
Then check that columns 3 and 4 are identical for each of the four rows. Job done.
Example of a truth table:
The simplest way is to use a truth table. Just make four rows, one for each possible set of values for A and B, in columns 1 and 2. Then in columns 3 and 4, write down the resulting values for A => B and A & ~B.
Then check that columns 3 and 4 are identical for each of the four rows. Job done.
Example of a truth table:
Martin Jansche
works in any number system, as long as it's base 10
·
Author has 3.7K answers and 3.6M answer views
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4y
You prove the biconditional in both directions. I’m using Fitch-style proof editor and checker to verify these simple proofs.
The first direction is generally valid, as it does not rely on any Classical inference rules:
In prose: Argue by contradiction and assume A & ~B (line 2). Using the premise A => B (line 1) we reason by modus ponens from A (line 3) to B (line 4), which contradicts ~B (line 5) from our assumption (line 2). We have derived a contradiction (line 6) and conclude ~(A & ~B), which is what we needed to show (line 7).
The opposite direction requires the Classical rule that lets you
You prove the biconditional in both directions. I’m using Fitch-style proof editor and checker to verify these simple proofs.
The first direction is generally valid, as it does not rely on any Classical inference rules:
In prose: Argue by contradiction and assume A & ~B (line 2). Using the premise A => B (line 1) we reason by modus ponens from A (line 3) to B (line 4), which contradicts ~B (line 5) from our assumption (line 2). We have derived a contradiction (line 6) and conclude ~(A & ~B), which is what we needed to show (line 7).
The opposite direction requires the Classical rule that lets you conclude B from an argument by contradiction:
In prose: Argue conditionally and assume A (line 2), then argue by contradiction and assume ~B (line 3). This gives us A & ~B (line 4), which contradicts our premise (line 1). We have derived an absurdity (line 5), and conclude (Classically) that the opposite of the assumption in the proof by contradiction is true (intuitionistically we could only conclude ~~B, so this is a Classical proof). We have thus derived B (line 6) and withdrawing our first assumption (line 2) we have shown A => B (line 7) as required.
Hilmar Zonneveld
Translator (1985–present)
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Author has 58.5K answers and 19.2M answer views
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4y
It’s a well-known fact that A → B is equivalent to ~A OR B. You can use that fact, and use one of the De Morgan’s laws.
If you don’t think you are justified in using the equivalence above as a given, you can simply write the truth tables for each side, compare them, and show that they are equal. (For the right side, do this step by step: in different columns, write down the expressions for A, B, ~B, A & ~B, ~[A & ~B].
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Why do we say 'if and only if' in logic/math, when 'only if' means the same things?
How can we prove if A => B, it does not mean B => A?
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Assistant
Bot
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1y
To prove the logical equivalence (A⇒B)⇔¬(A∧¬B), we can use truth tables or logical identities. Here, I will demonstrate the proof using both methods.
Method 1: Truth Table
We will construct a truth table for both sides of the equivalence.
| A | B | A⇒B | ¬B | A∧¬B | ¬(A∧¬B) |
--------------------------------------------------------|
| T | T | T | F | F | T
To prove the logical equivalence (A⇒B)⇔¬(A∧¬B), we can use truth tables or logical identities. Here, I will demonstrate the proof using both methods.
Method 1: Truth Table
We will construct a truth table for both sides of the equivalence.
| A | B | A⇒B | ¬B | A∧¬B | ¬(A∧¬B) |
--------------------------------------------------------|
| T | T | T | F | F | T |
| T | F | F | T | T | F |
| F | T | T | F | F | T |
| F | F | T | T | F | T |
Analyzing the Columns
The column for A⇒B shows the values: T, F, T, T.
The column for ¬(A∧¬B) shows the values: T, F, T, T.
Both columns match, which confirms that (A⇒B)⇔¬(A∧¬B) is true for all possible truth values of A and B.
Method 2: Logical Identities
We can also prove this equivalence using logical identities.
Start with the implication:
A⇒B≡¬A∨B
Rewrite the right side:
¬(A∧¬B)≡¬A∨¬¬B≡¬A∨B
Conclusion:
Since both sides simplify to ¬A∨B, we have:
(A⇒B)≡¬(A∧¬B)
Final Result
Thus, we conclude that:
(A⇒B)⇔¬(A∧¬B)
This completes the proof.
Eugene Bucamp
Knows French
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Author has 4.1K answers and 1.1M answer views
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Updated 4y
Don’t. However you would do it, it would be wrong because the equivalence (A → B) ⇔ ¬(A ∧ ¬B) is false.
Now, the implication A → B ⊢ ¬(A ∧ ¬B) is true. It is obviously true. It is true by definition. So, any method of logical proof better be able to prove that it is true.
However, at the moment, there is no method of logical proof. All methods of proofs proposed in mathematical logic are not methods of logical proof so proofs don’t prove logically. I’m not sure what could be the use of them but they do not prove anything logically. This point was noticed by Clarence Irving Lewis already somewher
Don’t. However you would do it, it would be wrong because the equivalence (A → B) ⇔ ¬(A ∧ ¬B) is false.
Now, the implication A → B ⊢ ¬(A ∧ ¬B) is true. It is obviously true. It is true by definition. So, any method of logical proof better be able to prove that it is true.
However, at the moment, there is no method of logical proof. All methods of proofs proposed in mathematical logic are not methods of logical proof so proofs don’t prove logically. I’m not sure what could be the use of them but they do not prove anything logically. This point was noticed by Clarence Irving Lewis already somewhere around 1918.
So any proof that the equivalence (A → B) ⇔ ¬(A ∧ ¬B) is true would be fallacious. And of course this means that all other answers (as of today March 19, 20201) are fallacious. Essentially, these answers are god evidence that their authors understand very little about logic. The extent of their expertise is to repeat the tenets of mathematical logic, somewhat like a bishop might use the Bible to prove the Trinity.
___________________
The theory of implication developed by the symbolic logicians seems to have aroused a considerable degree of antagonism among certain students of Logic. There are many philosophers to whom you cannot mention the name "Russell," without evoking such comments as, "His logic is purely artificial, for it is nonsense to suppose that a false proposition implies any proposition, or that any proposition implies any true proposition, " or, "Who could ever reasonably maintain that, 'The moon is made of green cheese,' implies, 'Caesar died in his bed?' — Norbert Wiener, article, The Journal of Philosophy (c. 1918?)
Dan Christensen
Creator of proof-checking freeware DC Proof 2.0
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Author has 2.8K answers and 830K answer views
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4y
Here is a 19 line formal proof that uses a form of natural deduction and what I believe to be widely accepted principles of propositional logic: Derive IMPLIES
Does this not establish that material implication is logical implication?
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Being in the food and beverage industry can come with great risks. That’s why it’s important to help protect your restaurant with the best kind of coverage.
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Related questions
How do I prove that “not B implies B” implies “B” in the Lukasiewizc formal system?
The statement- "All A are not B" (related to syllogism) is equivalent to which of the following sentences: "No A is B" or "Some A is not B"?
How can you show that (p → q) → r and p → (q → r) are not equivalent?
Can something be logically equivalent but not mereotopologically equivalent?
How can prove that [(P⇔R) and (R⇔Q)] ⇒(P⇔Q)?
Alan Bustany
Trinity Wrangler, 1977 IMO
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Author has 9.8K answers and 58.3M answer views
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2y
Related
How can I prove that A=B?
That depends. On A. And on B. Either or both of which are likely to be expressions
. Which brings in the horrors of evaluation, precedence, conventions, and so on. And that's just the start. It is definitely not as easy as most people think!
For an interesting and informative discussion on the topic, I recommend Barry Mazur's article When is one thing equal to some other thing?
which is also a brief motivator for Category Theory
.
Footnotes
Expression (mathematics) - Wikipedia
Category theory - Wikipedia
Leon Pendleton
knows how to solve problems up to Calc III
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Author has 131 answers and 52.3K answer views
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Updated Mar 26
Related
How can I prove that A=B?
Assuming A and B are some expressions and you want to prove their equality.
1st method: prove that both A and B is equal to some C. This always works as equality is an equivalence relation.
Illustration for the first method: here we are to prove the algebraic equality (x+y)2=x2+y2+2xy.
We prove this by showing both sides of the equation equal to the area of the square.
2nd method: take the difference A minus B. If the difference is evaluated to be zero then A=B.
A=(x+y)2
B=x2+y2+2xy
A−B=(x+y)2−(x2+y2+2xy)=(x+y)(x+y)−(x2+y2+2xy)
Now by FOIL:
=x(x+y)+y(x+y)−(x2+y2+
Assuming A and B are some expressions and you want to prove their equality.
1st method: prove that both A and B is equal to some C. This always works as equality is an equivalence relation.
Illustration for the first method: here we are to prove the algebraic equality (x+y)2=x2+y2+2xy.
We prove this by showing both sides of the equation equal to the area of the square.
2nd method: take the difference A minus B. If the difference is evaluated to be zero then A=B.
A=(x+y)2
B=x2+y2+2xy
A−B=(x+y)2−(x2+y2+2xy)=(x+y)(x+y)−(x2+y2+2xy)
Now by FOIL:
=x(x+y)+y(x+y)−(x2+y2+2xy)=(x2+xy+yx+y2)−(x2+y2+2xy)
=(x2+2xy+y2)−(x2+y2+2xy)=0 (qed)
This is only possible when subtraction is allowed, or the algebra you're working with allows for additive inverses. You don't need to worry about this caveat in the common case of real and complex numbers, however.
3rd way: prove A cannot be different from B, in other words A≠B is absurd, or both of these cases AB cannot be true. This called “proof by contradiction”. Again, only useful when the set containing A and B is an ordered set where it makes sense to say something is less or more than another.
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Fabio García
MSc in Mathematics, CIMAT (Graduated 2018)
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Author has 247 answers and 574.5K answer views
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5y
Related
How we can prove A=B when A= {a, a,b, c} and B= {a, b,c}?
By recalling the definition: two sets are equal if they have the same elements. That is, if every element of each set is an element of the other set.
For each element of A, you can manually check that it is also in B. For each element of B, you can manually check that it is also in A. Nothing more is needed to prove that A=B.
Anonymous
5y
Related
How we can prove A=B when A= {a, a,b, c} and B= {a, b,c}?
Let x∈A. Then, x∈{a,a,b,c}. Hence x∈B, so A⊆B.
Let x∈B. Then, x∈{a,b,c}. Hence x∈A, so B⊆A.
Since A⊆B and B⊆A, then A=B.
This proof schema follows from Zermelo-Fraenkel’s axiom of extensionality, which states that
∀A∀B(∀X(X∈A⟺X∈B)⟹A=B).
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Jan Hidders
Lecturer in Computer Science at Birkbeck, University of London (2020–present)
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Author has 910 answers and 1.9M answer views
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3y
Related
How do you prove A\ (B(AnB)) =A?
Captain Algebra to the rescue. For this purpose we will write X∖Y as X∩¯¯¯¯Y, where ¯¯¯¯Y denotes the complement of Y. Moreover, we let U denote the universal set.
And so here we go:
A∩¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(B∩¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(A∩B)) =
A∩(¯¯¯¯B∪¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(A∩B)) =
A∩(¯¯¯¯B∪(A∩B)) =
(A∩¯¯¯¯B)∪(A∩(A∩B)) =
(A∩¯¯¯¯B)∪((A∩A)∩B) =
(A∩¯¯¯¯B)∪(A∩B) =
A∩(¯¯¯¯B∪B) =
A∩U =
A
Frank Hubeny
M.S. in Mathematics, University of Illinois at Urbana-Champaign (Graduated 1994)
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Author has 633 answers and 760.4K answer views
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5y
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How can I prove or disprove that if (A x A) = (B x B), then A = B (A and B are sets)?
A×A is the Cartesian product of A with itself. It is a set of ordered pairs (x,y) where x and y are members of A. The same is true for B×B.
Given the premise (A×A)=(B×B), we want to show that A=B. By the axiom of extensionality A and B are equal if they have the same members. I will try to show that if x∈A then x∈B and if x∈B then x∈A. This will show that A=B.
Assume that x∈A. Then (x,x)∈A×A which is equal to B×B by the premise. So (x,x)∈B×B. Hence x∈B.
Assume x∈B. Then (x,x)∈B×B which is equal to A×A by
A×A is the Cartesian product of A with itself. It is a set of ordered pairs (x,y) where x and y are members of A. The same is true for B×B.
Given the premise (A×A)=(B×B), we want to show that A=B. By the axiom of extensionality A and B are equal if they have the same members. I will try to show that if x∈A then x∈B and if x∈B then x∈A. This will show that A=B.
Assume that x∈A. Then (x,x)∈A×A which is equal to B×B by the premise. So (x,x)∈B×B. Hence x∈B.
Assume x∈B. Then (x,x)∈B×B which is equal to A×A by the premise. So (x,x)∈A×A. Hence x∈A.
By the axiom of extensionality, A=B.
Andrew Coffey
B.S. in Pure Mathematics & Physics, University of New Mexico (Graduated 2021)
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Author has 270 answers and 251.9K answer views
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How we can prove A=B when A= {a, a,b, c} and B= {a, b,c}?
Sets don’t distinguish between identical elements. So, the second
a
is set
A
is redundant. Thus every element of
A
is an element of
B
, and every element of
B
is an element of
A
. Therefore the sets are the same by definition of set equality.
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17402 | https://www.cuemath.com/ncert-solutions/if-a-b-c-are-all-non-zero-and-a-b-c-0-prove-that-a2-bc-b2-ca-c2-ab-3/ | If a, b, c are all non-zero and a + b + c = 0, prove that a²/bc + b²/ca + c²/ab = 3.
Solution:
Given, a, b and c are all non-zero
Also, a + b + c = 0
We have to prove that a²/bc + b²/ca + c²/ab = 3.
Using the algebraic identity,
x³ + y³ + z³ - 3xyz = (x + y + z) (x² +y² + z² - xy - yz - zx)
If x + y + z = 0, then x³ + y³ + z³ - 3xyz = 0
So, x³ + y³ + z³ = 3xyz.
Given, a + b + c = 0
So, a³ + b³ + c³ = 3abc
Dividing both sides by 3abc,
a³/3abc + b³/3abc + c³/3abc = 3abc/3abc
a²/3bc + b²/3ca + c²/3ab = 1
a²/bc + b²/ca + c²/ab = 3
Hence proved.
✦ Try This: Find the value of k if (x-2)is a factor of polynomial p(x) = 2x³ - 6x² + 5x + k.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 2
NCERT Exemplar Class 9 Maths Exercise 2.4 Problem 7
If a, b, c are all non-zero and a + b + c = 0, prove that a²/bc + b²/ca + c²/ab = 3
Summary:
If a, b, c are all non-zero and a + b + c = 0, using the algebraic identity it is proven that a²/bc + b²/ca + c²/ab = 3
☛ Related Questions:
Without actual division, prove that 2x⁴ - 5x³ + 2x² - x + 2 is divisible by x² - 3x + 2
Simplify (2x - 5y)³ - (2x + 5y)³
Multiply x² + 4y² + z² + 2xy + xz - 2yz by (- z + x - 2y)
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17403 | https://www.youtube.com/watch?v=bjMftfFEsoA | Definite Integral Examples (Area Under Curves)
Houston Math Prep
51600 subscribers
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Posted: 19 Jun 2020
This Calculus 1 video works several basic examples of calculating a definite integral, finding area under a curve using integration. We show all of the integration examples to be worked in the video at the beginning, so you can skip to a particular example, or work through all of the examples with us. We also show graphs of each, so you can see the area being calculated by each integral.
0:00 Example 1
2:02 Example 2
3:53 Example 3
5:15 Example 4
6:13 Example 5
7:27 Example 6
Houston Math Prep Calculus 1 Playlist:
Houston Math Prep YouTube:
Transcript:
Example 1 hey everyone Houston math prep here in this video we're going to do some examples of finding area under curves just calculating some definite integrals will show you the graph and actually work these integrals for you we've got six examples we're going to do if you see one that's more interesting to you than the others you can go ahead and skip to that otherwise you're welcome to work along with us through each of these four the first one here we have the integral of 6x minus x squared on the interval 0 to 6 so you can see this function would be a parabola that opens down so for this first one here each of these will be a power rule to do the antiderivative so here the power will go up by one so we'll get x squared well divide by the new power which would be divided by two we already have a multiple of 6 out here though so constant multiple six divided by two actually gives us 3x squared minus power goes up by one here for our power rule so it becomes x cubed divided by the new power gives us over three we'll go ahead and plug in our bounds zero and six we want to make sure we do it in the correct order we always plug in our upper bound into this expression then we subtract what we get when we plug in the lower bound if we go ahead and plug in six that will be three times 6 squared minus six cubed divided by three minus three times 0 squared minus zero cubed divided by three I think we can see here with the zero everything is going to be 0 right so let's focus over here we'll have 3 times 6 squared which would be 3 times 36 minus 6 cubed divided by 3 6 cubed as 216 divided by 3 so if we do 3 times 36 that's 108 minus 216 divided by 3 that's 72 and we in fact get 36 if we were calculating this in answering as area we would say 36 units squared if we're just calculating this integral we may just say 36 Example 2 looking at our second one the integral from zero to four of square root X DX so I've drawn a square root function here let's go ahead and think of this root x as a power remember we can think of the square root of something as that to the one half so we'll think of this integral and use the power rule as X to the one half if we take the antiderivative of X to the one half remember we will add one so we'll get X to the three halves we'll divide by the new power divided by three halves as well and we'll evaluate on the interval zero to four now divided by three halves a fraction and a fraction is not so good so I'm going to go ahead and bump that out and call it multiplied by the reciprocal so we're going to say two-thirds X to the three halves and we'll evaluate that from 0 to 4 I could go ahead and bump my 2/3 out if I want maybe I'll go ahead and do that let's say 2/3 times that would give me 4 to the 3 halves minus 0 to the 3 halves now we just want to be careful about our power three-halfs here what does this mean well this denominator of 2 means take the square root right so both of these say take the square root and then the 3 on the top of the fraction says then cube whatever you get so we would actually have let's do the square root part first if I just do the square root part this becomes 2 so we'd have 2 cubed minus if I take the square root of 0 I would still have 0 right so each of these become cubes and then we can think of 2/3 times 2 cubed which is going to be 8 0 cubed is going to be 0 so we end up with 2/3 times 8 and that gives us 16 thirds Example 3 looking here we have the integral from 1 to e the constant e of the function 1 over X DX so we have the function 1 over X remember e is about 2.7 ish and some change so we have the area underneath this from 1/2 a little beyond 2.7 remember that this looks like a power rule but this is actually the derivative of a function that we know called the natural log so we take the antiderivative of 1 over X we actually get the natural log of absolute value X evaluated from x equals 1 to x equals e now I can look here and notice that E is a positive number and 1 is a positive number and when I plug those in I'm just going to get positive numbers so evaluating these I don't really need my absolute value at this point since I'm plugging in only positive stuff so that will give us the Ln of E - Ln of 1 now these are actually numbers that we should know what is Ln of E Ln of E is actually 1 and what is natural log of 1 what is log of any base of 1 it's 0 so we get 1 minus a 0 and this is actually 1 this tells us a neat thing about 2 the graph of 1 over X this tells us that the area between 1 and e this is actually exactly 1 unit Example 4 let's look at our next one here we have the integral from 0 to 1 of e to the X DX so I've got my representation here's exponential growth e to the X and there's x equals 0 and x equals 1 so this is really also calculating the area in here let's go ahead and do our easy antiderivative of e to the X what is it of course we know right it's e to the X because the derivative of e to the X is itself we'll evaluate from 0 to 1 maybe this one's too easy for us it could be so if we plug in 1 we'll get e to the 1 minus when we plug in the lower bound will get e to the 0 what is e to the 1 well anything to the 1 is itself right so this is just e minus what is e to the 0 well that'll be 1 right and so e minus 1 is a good exact answer if you wanted to get a decimal you could go ahead and put a in a calculator and subtract 1 and say well it's about one point seven one ish but let's move on to the next one Example 5 our number five here we have the integral of sine x DX from 0 to pi so you can see this is basically under one hump of our sine function here between sine and the axis if we go ahead and do our antiderivative of sine x just be careful the derivative of sine X is cosine X the antiderivative is negative cosine X so we'll take that antiderivative we'll plug in zero to pi there make sure you do it in the correct order so plugging in PI first we'll get negative cosine of pi - plugging in the lower bound negative cosine of zero let's figure out what each of these are cosine of pi that's on the left side of the unit circle so cosine of pi itself is already negative one but I have a negative in front of that so that's actually positive one and then here I have minus negative I'll just call that plus what is cosine of zero cosine of zero zero is an angle on the right side of the unit circle so it's cosine value is one right so we get one plus one there and that gives us two so actually the area of one hump of the sine function here we actually have two units of area there Example 6 and let's look at our last one here we have this nice gently sloping function here 1 over 1 plus x squared we're going to integrate that DX from 0 to 1 so you can see this region here represents this integral what is the antiderivative of 1 over 1 plus x squared we should know that it's a definition actually it is the inverse tangent of X right so this inverse tangent of X has a derivative of 1 over 1 plus x squared so the antiderivative of 1 over 1 plus x squared is inverse tangent of X now we have to plug these in and do the subtract right so we'll have the inverse tangent of 1 and then we'll subtract the inverse tangent of 0 now remember inverse functions just ask you about the angle right so inverse tangent of 1 is saying what angle gives us a tangent value of 1 and that would be where sine and cosine are the same value in quadrant one and that is PI over 4 gives us a tangent value of 1 minus inverse tangent of 0 says what angle gives us a tangent value of 0 we'll think about tangent is sine over cosine right so maybe if we remember this we can do this without having to type it into anything or maybe you just know and that's fine when is a fraction zero what fraction is zero when the top is zero right so we want to know what has a sine value of zero right based on this and that's actually zero so we get PI over four minus zero or answer for this last one is actually PI over four all right just some basic introductory functions here nothing too complicated to integrate but hopefully this gives you a good basic start on computing your definite integrals thanks for watching everybody we'll see you in the next video |
17404 | https://www.gauthmath.com/solution/1835307211186209/Find-the-difference-frac-x2x2-16-frac-4xx2-16-Step-by-Step-Solution-Step-1-Subtr | Solved: Find the difference x^2/x^2-16 - 4x/x^2-16 . Step-by-Step Solution Step 1: Subtract the [Math]
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Find the difference x^2/x^2-16 - 4x/x^2-16 . Step-by-Step Solution Step 1: Subtract the numerators and write the result over the common denominator. x^2/x^2-16 - 4x/x^2-16 = (x^2-4x)/□ a/c - b/c = (a-b)/c (Do not factor.)
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Answer
(x² - 4x) / (x² - 16)
Explanation
Since the denominators are the same, subtract the numerators: x² - 4x.
Write the result over the common denominator: (x² - 4x) / (x² - 16).
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17405 | https://math.stackexchange.com/questions/938540/look-at-the-following-infinite-sequence-1-10-100-1000-10000 | Skip to main content
Look at the following infinite sequence: 1, 10, 100, 1000, 10000, . . ..
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What is the proof that there are 2 numbers in this sequence that differ by a multiple of 12345678987654321?
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edited Sep 20, 2014 at 5:07
Jyrki Lahtonen
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asked Sep 20, 2014 at 2:25
mathematiccianmathematiccian
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I think it is true. There are precisely 12345678987654321 residue classes modulo 12345678987654321. Now use the pigeon hole principle.
– Alex Wertheim
Commented
Sep 20, 2014 at 2:27
2
Fun fact: 12345678987654321=1111111112=(32⋅37⋅333667)2
– Ben Grossmann
Commented
Sep 20, 2014 at 2:29
So can this notion be generalized? In what way? Does the series or the number has a special property? Or their relation is somewhat special?
– mathematiccian
Commented
Sep 20, 2014 at 2:31
2
@Olcayto: Yes, greatly. You can replace 'powers of 10' with 'powers of n' for any integer n, and replace 12345678987654321 with any nonzero integer.
– Charles
Commented
Sep 20, 2014 at 2:32
1
You can replace 'powers of 10' with any integer sequence you'd like.
– Winther
Commented
Sep 20, 2014 at 4:44
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Let n=12345678987654321. Consider the first n+1 elements of the powers-of-10 sequenc modulo n. According to the pigeonhole principle, there are at least two of these elements that have the same residue modulo n. Their difference is a multiple of n.
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answered Sep 20, 2014 at 2:29
NovaDenizenNovaDenizen
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This is beautifully applied pigeonhole, and not what I expected on a post like this, but it works so well.
– Asimov
Commented
Sep 20, 2014 at 2:32
So this can be generalized for any number and any geometric sequence.
– mathematiccian
Commented
Sep 20, 2014 at 2:32
@Olcayto: Any number and any sequence. :-)
– ShreevatsaR
Commented
Sep 20, 2014 at 5:16
@ShreevatsaR I think it's implied here that while a difference of 0 is technically a multiple of any number, it's not an acceptable answer. So any sequence without duplicates would work.
– NovaDenizen
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Sep 21, 2014 at 0:03
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call your number n. Note that it is not divisible by 2 or 5. So gcd(10,n)=1. By we have
10ϕ(n)≡1≡100(modn).
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answered Sep 20, 2014 at 4:24
Will JagyWill Jagy
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Take the digit 1 from the 10 and take the digit 1 from the 100. They differ by 0 which is a multiple of 12345678987654321.
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17406 | https://math.stackexchange.com/questions/4440661/at-any-given-point-in-time-two-players-have-finished-the-same-number-of-games | combinatorics - At any given point in time two players have finished the same number of games - Mathematics Stack Exchange
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At any given point in time two players have finished the same number of games
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I am reading the following problem, which falls under the Pigeonhole principle.
A chess tournament has n n participants and any two players play one game against each other. Then it is true that at any given point in time, there are two players who have finished the same number of games.
My approach:
I assume the opposite, i.e., that we never have players with the same number of matches at any time.
This means that at any point in time all players have completed different number of matches. This implies that there will be only 1 1 player that has finished 1 1 match but that is impossible since it takes 2 2 person to finish a match. Hence the opposite must be true i.e. that we have always at least two players with the same number of matches.
At this point I think my proof is wrong. Because I am not sure if my approach actually makes the point be proven for any point in time. Additionally I don't think I am applying the pigeonhole principle.
What do you think of my proof and how can it be improved?
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edited May 1, 2022 at 19:38
mjqxxxx
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asked May 1, 2022 at 17:02
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Just as a side note, when m≤n−2 m≤n−2, you anyway have at least two participants who have played zero matches (that is technically equal number of matches)Math Lover –Math Lover 2022-05-01 17:48:49 +00:00 Commented May 1, 2022 at 17:48
@MathLover: I was thinking that if I do a contrapositive proof then it would solve it. I.e. since it is impossible at any time to have a distinct number of matches finished per player because a match requires 2 2 players and therefore we can not have only 1 1 player completed 1 1 match then the opposite must be true. But I am not sure if that is generic enough for this problem as asked Jim –Jim 2022-05-01 18:43:29 +00:00 Commented May 1, 2022 at 18:43
I understand the argument that since it is impossible at any time to have a distinct number of matches finished per player because a match requires 2 players but from there on you state that it is not possible that only one player has completed one match BUT note that it is possible that each of them have completed more than one match at a given point. So for completeness sake, please see my earlier comment. I hope that is clear.Math Lover –Math Lover 2022-05-01 18:49:17 +00:00 Commented May 1, 2022 at 18:49
1 @MathLover: Ah I see what you mean and the error in my logic. I considered only the time where there has to be 1 1 more player with 1 1 game but your reasoning showed me why this is not general approach Jim –Jim 2022-05-01 18:52:55 +00:00 Commented May 1, 2022 at 18:52
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At a given point, it is possible that each participant had played more than one match. So your solution as stated is not generic.
Writing my comment as an answer, a generic approach based on Pigeonhole principle will be to assume that at a given point, m participants have played matches (2≤m≤n)(2≤m≤n). The minimum number of matches a participant could have played is 1 1 and the maximum number of matches a participant could have played is (m−1)(m−1). That is (m−1)(m−1) distinct number of matches played by m m different participants. So, at least two participants must have played equal number of matches.
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answered May 1, 2022 at 19:28
Math LoverMath Lover
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You can prove something more general: for any tournament structure for n≥2 n≥2 players in which games are played, provided that no pair of players is allow to play twice, then at any time instant there are always at least two players who have completed the same number of games.
We can prove this by modeling it as a graph theory problem. Pick a time instant and form a graph with n n nodes, one for each player, and add edges between pairs of players who have already played their game. A classic application of the pigeonhole principle is to show that any graph with n≥2 n≥2 nodes has two nodes of the same degree (a quick Google search will turn up many proofs of this result). The degree of a node in this graph corresponds to the number of games a player completed, so this shows that you’ll always find two players who have completed the same number of games.
This doesn’t even require everyone to play everyone else - it also works for single-elimination matches, for example.
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edited May 2, 2022 at 17:59
answered May 2, 2022 at 2:53
templatetypedeftemplatetypedef
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I wouldn’t call this “much” more general. Isn’t it just like stopping the full tournament at some time before it’s complete?Milten –Milten 2022-05-02 06:28:56 +00:00 Commented May 2, 2022 at 6:28
1 @Milten Very fair! Answer updated.templatetypedef –templatetypedef 2022-05-02 17:59:10 +00:00 Commented May 2, 2022 at 17:59
That graph theory result which you appeal to implies the OQ (original question). So your proof is circular. Or you could say that the proof of that graph theory result serves as a proof of the OQ.Rosie F –Rosie F 2022-05-06 12:53:28 +00:00 Commented May 6, 2022 at 12:53
@RosieF It’s the latter - I was aiming to show that this problem reduces to a well-known result from graph theory.templatetypedef –templatetypedef 2022-05-06 15:20:06 +00:00 Commented May 6, 2022 at 15:20
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Your approach, to assume the opposite, in order to prove that that yields a contradiction, is good.
MathLover's answer has the right idea but fails to account for the possibility that a player hasn't played any games yet.
Call the total number of players m m. The number of games a player has played so far is one of the m m integers in the range 0 to m−1 m−1. If none of them have played the same number of games, then, for each of those integers k k, there is exactly one player who has played k k games. This means that one player P P has not played anybody. It also means that another player has played m−1 m−1 games, and has therefore played every other player, including P P, even though P P hasn't played anybody else. This is a contradiction.
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answered May 6, 2022 at 12:48
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My answer does account for the possibility that a player has not played any matches.Math Lover –Math Lover 2022-05-06 14:47:25 +00:00 Commented May 6, 2022 at 14:47
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17407 | https://openstax.org/books/physics/pages/22-2-nuclear-forces-and-radioactivity | Skip to Content
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Physics
22.2 Nuclear Forces and Radioactivity
Physics
22.2
Nuclear Forces and Radioactivity
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Section Learning Objectives
By the end of this section, you will be able to do the following:
Describe the structure and forces present within the nucleus
Explain the three types of radiation
Write nuclear equations associated with the various types of radioactive decay
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The learning objectives in this section will help your students master the following standards:
(5) Science concepts. The student knows the nature of forces in the physical world. The student is expected to:
(H)
describe evidence for and effects of the strong and weak nuclear forces in nature.
(8) Science concepts. The student knows simple examples of atomic, nuclear, and quantum phenomena. The student is expected to:
(B)
compare and explain the emission spectra produced by various atoms; and
(C)
describe the significance of mass-energy equivalence and apply it in explanations of phenomena such as nuclear stability, fission, and fusion.
Section Key Terms
| | | | | |
--- ---
| alpha decay | atomic number | beta decay | gamma decay | Geiger tube |
| isotope | mass number | nucleons | radioactive | radioactive decay |
| radioactivity | scintillator | strong nuclear force | transmutation | |
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[BL][OL][AL]As in the beginning of Section 1, have students create a list of facts they have learned about the atom. Have the students update their list throughout this section.
There is an ongoing quest to find the substructures of matter. At one time, it was thought that atoms would be the ultimate substructure. However, just when the first direct evidence of atoms was obtained, it became clear that they have a substructure and a tiny nucleus. The nucleus itself has spectacular characteristics. For example, certain nuclei are unstable, and their decay emits radiations with energies millions of times greater than atomic energies. Some of the mysteries of nature, such as why the core of Earth remains molten and how the Sun produces its energy, are explained by nuclear phenomena. The exploration of radioactivity and the nucleus has revealed new fundamental particles, forces, and conservation laws. That exploration has evolved into a search for further underlying structures, such as quarks. In this section, we will explore the fundamentals of the nucleus and nuclear radioactivity.
The Structure of the Nucleus
At this point, you are likely familiar with the neutron and proton, the two fundamental particles that make up the nucleus of an atom. Those two particles, collectively called nucleons, make up the small interior portion of the atom. Both particles have nearly the same mass, although the neutron is about two parts in 1,000 more massive. The mass of a proton is equivalent to 1,836 electrons, while the mass of a neutron is equivalent to that of 1,839 electrons. That said, each of the particles is significantly more massive than the electron.
When describing the mass of objects on the scale of nucleons and atoms, it is most reasonable to measure their mass in terms of atoms. The atomic mass unit (u) was originally defined so that a neutral carbon atom would have a mass of exactly 12 u. Given that protons and neutrons are approximately the same mass, that there are six protons and six neutrons in a carbon atom, and that the mass of an electron is minuscule in comparison, measuring this way allows for both protons and neutrons to have masses close to 1 u. Table 22.1 shows the mass of protons, neutrons, and electrons on the new scale.
Tips For Success
For most conceptual situations, the difference in mass between the proton and neutron is insubstantial. In fact, for calculations that require fewer than four significant digits, both the proton and neutron masses may be considered equivalent to one atomic mass unit. However, when determining the amount of energy released in a nuclear reaction, as in Alpha Decay Energy Found from Nuclear Masses, the difference in mass cannot be ignored.
Another other useful mass unit on the atomic scale is the MeV/c2 . While rarely used in most contexts, it is convenient when one uses the equation E=mc2, as will be addressed later in this text.
| | Proton Mass | Neutron Mass | Electron Mass |
--- --- |
| Kilograms (kg) | 1.673×10−27 | 1.675×10−27 | 9.109×10−31 |
| Atomic mass units (u) | 1.007 | 1.009 | 5.486×10−4 |
Table
22.1
Atomic Masses for Multiple Units
To more completely characterize nuclei, let us also consider two other important quantities: the atomic number and the mass number. The atomic number, Z, represents the number of protons within a nucleus. That value determines the elemental quality of each atom. Every carbon atom, for instance, has a Z value of 6, whereas every oxygen atom has a Z value of 8. For clarification, only oxygen atoms may have a Z value of 8. If the Z value is not 8, the atom cannot be oxygen.
The mass number, A, represents the total number of protons and neutrons, or nucleons, within an atom. For an ordinary carbon atom the mass number would be 12, as there are typically six neutrons accompanying the six protons within the atom. In the case of carbon, the mass would be exactly 12 u. For oxygen, with a mass number of 16, the atomic mass is 15.994915 u. Of course, the difference is minor and can be ignored for most scenarios. Again, because the mass of an electron is so small compared to the nucleons, the mass number and the atomic mass can be essentially equivalent. Figure 22.16 shows an example of Lithium-7, which has an atomic number of 3 and a mass number of 7.
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[BL]Students may confuse the terms mass number and atomic number. Remind them that the mass is based on both protons and neutrons and so mass number is a measure of both combined. The atomic number differentiates between two different atoms, which only protons can do. A few examples showing pictures of nuclei and having students identify the mass and atomic numbers of each should help.
How does the mass number help to differentiate one atom from another? If each atom of carbon has an atomic number of 6, then what is the value of including the mass number at all? The intent of the mass number is to differentiate between various isotopes of an atom. The term isotope refers to the variation of atoms based upon the number of neutrons within their nucleus. While it is most common for there to be six neutrons accompanying the six protons within a carbon atom, it is possible to find carbon atoms with seven neutrons or eight neutrons. Those carbon atoms are respectively referred to as carbon-13 and carbon-14 atoms, with their mass numbers being their primary distinction. The isotope distinction is an important one to make, as the number of neutrons within an atom can affect a number of its properties, not the least of which is nuclear stability.
Figure
22.16
Lithium-7 has three protons and four neutrons within its nucleus. As a result, its mass number is 7, while its atomic number is 3. The actual mass of the atom is 7.016 u. Lithium 7 is an isotope of lithium.
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Point out to students that the number of electrons is irrelevant to discussions of mass number, atomic number, and isotopes.
To more easily identify various atoms, their atomic number and mass number are typically written in a form of representation called the nuclide. The nuclide form appears as follows: ZAXN, where X is the atomic symbol and N represents the number of neutrons.
Let us look at a few examples of nuclides expressed in the ZAXN notation. The nucleus of the simplest atom, hydrogen, is a single proton, or H11 (the zero for no neutrons is often omitted). To check the symbol, refer to the periodic table—you see that the atomic number Z of hydrogen is 1. Since you are given that there are no neutrons, the mass number A is also 1. There is a scarce form of hydrogen found in nature called deuterium; its nucleus has one proton and one neutron and, hence, twice the mass of common hydrogen. The symbol for deuterium is, thus, 21H2 . An even rarer—and radioactive—form of hydrogen is called tritium, since it has a single proton and two neutrons, and it is written 31H2 . The three varieties of hydrogen have nearly identical chemistries, but the nuclei differ greatly in mass, stability, and other characteristics. Again, the different nuclei are referred to as isotopes of the same element.
There is some redundancy in the symbols A, X, Z, and N. If the element X is known, then Z can be found in a periodic table. If both A and X are known, then N can also be determined by first finding Z; then, N = A – Z. Thus the simpler notation for nuclides is
AX,
22.34
which is sufficient and is most commonly used. For example, in this simpler notation, the three isotopes of hydrogen are 1H, 2H, and 3H . For 238U, should we need to know, we
can determine that Z = 92 for uranium from the periodic table, and thus, N = 238 − 92 = 146.
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This explanation is provided to help students understand nomenclature later on in the text. However, it may be useful to have students practice writing nuclide notation to build confidence with the concept.
Radioactivity and Nuclear Forces
In 1896, the French physicist Antoine Henri Becquerel (1852–1908) noticed something strange. When a uranium-rich mineral called pitchblende was placed on a completely opaque envelope containing a photographic plate, it darkened spots on the photographic plate.. Becquerel reasoned that the pitchblende must emit invisible rays capable of penetrating the opaque material. Stranger still was that no light was shining on the pitchblende, which means that the pitchblende was emitting the invisible rays continuously without having any energy input! There is an apparent violation of the law of conservation of energy, one that scientists can now explain using Einstein’s famous equation E=mc2. It was soon evident that Becquerel’s rays originate in the nuclei of the atoms and have other unique characteristics.
To this point, most reactions you have studied have been chemical reactions, which are reactions involving the electrons surrounding the atoms. However, two types of experimental evidence implied that Becquerel’s rays did not originate with electrons, but instead within the nucleus of an atom.
First, the radiation is found to be only associated with certain elements, such as uranium. Whether uranium was in the form of an element or compound was irrelevant to its radiation. In addition, the presence of radiation does not vary with temperature, pressure, or ionization state of the uranium atom. Since all of those factors affect electrons in an atom, the radiation cannot come from electron transitions, as atomic spectra do.
The huge energy emitted during each event is the second piece of evidence that the radiation cannot be atomic. Nuclear radiation has energies on the order of 106 eV per event, which is much greater than typical atomic energies that are a few eV, such as those observed in spectra and chemical reactions, and more than ten times as high as the most energetic X-rays.
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To emphasize the point that most reactions are chemical, have students brainstorm a list of reactions from chemistry class. Have them describe the changes that cause the reactions to take place. Are the reactions the result of nuclear interactions or electron interactions?
But why would reactions within the nucleus take place? And what would cause an apparently stable structure to begin emitting energy? Was there something special about Becquerel’s uranium-rich pitchblende? To answer those questions, it is necessary to look into the structure of the nucleus. Though it is perhaps surprising, you will find that many of the same principles that we observe on a macroscopic level still apply to the nucleus.
Nuclear Stability
A variety of experiments indicate that a nucleus behaves something like a tightly packed ball of nucleons, as illustrated in Figure 22.17. Those nucleons have large kinetic energies and, thus, move rapidly in very close contact. Nucleons can be separated by a large force, such as in a collision with another nucleus, but strongly resist being pushed closer together. The most compelling evidence that nucleons are closely packed in a nucleus is that the radius of a nucleus, r, is found to be approximately
r=roA13/,
22.35
where ro= 1.2 femtometer (fm) and A is the mass number of the nucleus.
Note that r3∝A . Since many nuclei are spherical, and the volume of a sphere is V=(43)πr3, we see that V∝A —that is, the volume of a nucleus is proportional to the number of nucleons in it. That is what you expect if you pack nucleons so close that there is no empty space between them.
Figure
22.17
Nucleons are held together by nuclear forces and resist both being pulled apart and pushed inside one another. The volume of the nucleus is the sum of the volumes of the nucleons in it, here shown in different colors to represent protons and neutrons.
So what forces hold a nucleus together? After all, the nucleus is very small and its protons, being positive, should exert tremendous repulsive forces on one another. Considering that, it seems that the nucleus would be forced apart, not together!
The answer is that a previously unknown force holds the nucleus together and makes it into a tightly packed ball of nucleons. This force is known as the strong nuclear force. The strong force has such a short range that it quickly fall to zero over a distance of only 10–15 meters. However, like glue, it is very strong when the nucleons get close to one another.
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The relationship between the repulsive Coulomb force and the attractive nuclear force can be modeled using a balloon. Squeeze the balloon, or have one student hold a balloon while the other pushes on it. The pressure of the air inside the balloon will model the Coulomb force while the push from the student models the nuclear force.
The balancing of the electromagnetic force with the nuclear forces is what allows the nucleus to maintain its spherical shape. If, for any reason, the electromagnetic force should overcome the nuclear force, components of the nucleus would be projected outward, creating the very radiation that Becquerel discovered!
Understanding why the nucleus would break apart can be partially explained using Table 22.2. The balance between the strong nuclear force and the electromagnetic force is a tenuous one. Recall that the attractive strong nuclear force exists between any two nucleons and acts over a very short range while the weaker repulsive electromagnetic force only acts between protons, although over a larger range. Considering the interactions, an imperfect balance between neutrons and protons can result in a nuclear reaction, with the result of regaining equilibrium.
| | Range of Force | Direction | Nucleon Interaction | Magnitude of Force |
--- ---
| Electromagnetic Force | Long range, though decreasing by 1/r2 | Repulsive | Proton –proton repulsion | Relatively small |
| Strong Nuclear Force | Very short range, essentially zero at 1 femtometer | Attractive | Attraction between any two nucleons | 100 times greater than the electromagnetic force |
Table
22.2
Comparing the Electromagnetic and Strong Forces
The radiation discovered by Becquerel was due to the large number of protons present in his uranium-rich pitchblende. In short, the large number of protons caused the electromagnetic force to be greater than the strong nuclear force. To regain stability, the nucleus needed to undergo a nuclear reaction called alpha (α) decay.
The Three Types of Radiation
Radioactivity refers to the act of emitting particles or energy from the nucleus. When the uranium nucleus emits energetic nucleons in Becquerel’s experiment, the radioactive process causes the nucleus to alter in structure. The alteration is called radioactive decay. Any substance that undergoes radioactive decay is said to be radioactive. That those terms share a root with the term radiation should not be too surprising, as they all relate to the transmission of energy.
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Students have plenty of preconceptions about radioactivity. Discuss their preconceptions. Are their concerns related to the radioactive decay process or to the energy transmitted in the process?
Radioactivity can be understood as a tendency for a nucleus to reach equilibrium. Discuss with students other instances of objects desiring equilibrium. Such macroscopic interactions may help to make their understanding of radioactivity more tangible.
Alpha Decay
Alpha decay refers to the type of decay that takes place when too many protons exist in the nucleus. It is the most common type of decay and causes the nucleus to regain equilibrium between its two competing internal forces. During alpha decay, the nucleus ejects two protons and two neutrons, allowing the strong nuclear force to regain balance with the repulsive electromagnetic force. The nuclear equation for an alpha decay process can be shown as follows.
XZAN→YZ−2A−4N+H24e
22.36
Figure
22.18
A nucleus undergoes alpha decay. The alpha particle can be seen as made up of two neutrons and two protons, which constitute a helium-4 atom.
Three things to note as a result of the above equation:
By ejecting an alpha particle, the original nuclide decreases in atomic number. That means that Becquerel’s uranium nucleus, upon decaying, is actually transformed into thorium, two atomic numbers lower on the periodic table! The process of changing elemental composition is called transmutation.
Note that the two protons and two neutrons ejected from the nucleus combine to form a helium nucleus. Shortly after decay, the ejected helium ion typically acquires two electrons to become a stable helium atom.
Finally, it is important to see that, despite the elemental change, physical conservation still takes place. The mass number of the new element and the alpha particle together equal the mass number of the original element. Also, the net charge of all particles involved remains the same before and after the transmutation.
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Differentiate between ionization and transmutation to reinforce that alpha radiation is an elemental change.
Beta Decay
Like alpha decay, beta ( β ) decay also takes place when there is an imbalance between neutrons and protons within the nucleus. For beta decay, however, a neutron is transformed into a proton and electron or vice versa. The transformation allows for the total mass number of the atom to remain the same, although the atomic number will increase by one (or decrease by one). Once again, the transformation of the neutron allows for a rebalancing of the strong nuclear and electromagnetic forces. The nuclear equation for a beta decay process is shown below.
AZXN→AZ+1YN−1+e+v
The symbol v in the equation above stands for a high-energy particle called the neutrino. A nucleus may also emit a positron, and in that case Z decreases and N increases. It is beyond the scope of this section and will be discussed in further detail in the chapter on particles. It is worth noting, however, that the mass number and charge in all beta-decay reactions are conserved.
Figure
22.19
A nucleus undergoes beta decay. The neutron splits into a proton, electron, and neutrino. This particular decay is called β− decay.
Teacher Support
Teacher Support
[AL]The mass number of the nucleus is conserved, but is the mass? Mention that that mass of a proton is slightly less than the mass of the neutron. Have students consider where that mass goes. Note—Not all of it goes to the electron.
Gamma Decay
Gamma decay is a unique form of radiation that does not involve balancing forces within the nucleus. Gamma decay occurs when a nucleus drops from an excited state to the ground state. Recall that such a change in energy state will release energy from the nucleus in the form of a photon. The energy associated with the photon emitted is so great that its wavelength is shorter than that of an X-ray. Its nuclear equation is as follows.
AZXN→XN+γ
22.37
Figure
22.20
A nucleus undergoes gamma decay. The nucleus drops in energy state, releasing a gamma ray.
Teacher Support
Teacher Support
[OL][AL]To differentiate between the three types of decay, you can have the students construct a large Venn diagram of their radioactive properties. For students struggling with the idea, it may be easier to have them first construct a table of what they know about each type of decay.
Worked Example
Creating a Decay Equation
Write the complete decay equation in AZXN notation for beta decay producing 137Ba . Refer to the periodic table for values of Z.
Strategy
Beta decay results in an increase in atomic number. As a result, the original (or parent) nucleus, must have an atomic number of one fewer proton.
Solution
The equation for beta decay is as follows
XZAN→YZ+1AN−1+e+v.
22.38
Considering that barium is the product (or daughter) nucleus and has an atomic number of 56, the original nucleus must be of an atomic number of 55. That corresponds to cesium, or Cs.
C55137sN→B56137aN−1+e+v
22.39
The number of neutrons in the parent cesium and daughter barium can be determined by subtracting the atomic number from the mass number (137 – 55 for cesium, 137 – 56 for barium). Substitute those values for the N and N – 1 subscripts in the above equation.
C55137s82→B56137a81+e+v
22.40
Discussion
The terms parent and daughter nucleus refer to the reactants and products of a nuclear reaction. The terminology is not just used in this example, but in all nuclear reaction examples. The cesium-137 nuclear reaction poses a significant health risk, as its chemistry is similar to that of potassium and sodium, and so it can easily be concentrated in your cells if ingested.
Worked Example
Alpha Decay Energy Found from Nuclear Masses
Find the energy emitted in the α decay of 239Pu.
Strategy
Nuclear reaction energy, such as released in α decay, can be found using the equation E=mc2 . We must first find Δm, the difference in mass between the parent nucleus and the products of the decay.
The mass of pertinent particles is as follows
239Pu: 239.052157 u
235U: 235.043924 u
4He: 4.002602 u.
Solution
The decay equation for 239Pu is
P239u→U235+H4e.
Determine the amount of mass lost between the parent and daughter nuclei.
Δm=m(239Pu)−(m(239U)+m(4He))Δm=239.052157u−(235.043924u+4.002602u)Δm=0.0005631u
Now we can find E by entering Δm into the equation.
E=(Δm)c2=(0.005631u)c2
And knowing that 1u=931.5meV/c2, we can find that
E=(0.005631)(931.5MeV/c2)(c2)=5.25MeV.
Discussion
The energy released in this α decay is in the MeV range, about 106 times as great as typical chemical reaction energies, consistent with previous discussions. Most of the energy becomes kinetic energy of the α particle (or 4He nucleus), which moves away at high speed.
The energy carried away by the recoil of the 235U nucleus is much smaller, in order to conserve momentum. The 235U nucleus can be left in an excited state to later emit photons ( γ rays). The decay is spontaneous and releases energy, because the products have less mass than the parent nucleus.
Properties of Radiation
The charges of the three radiated particles differ. Alpha particles, with two protons, carry a net charge of +2. Beta particles, with one electron, carry a net charge of –1. Meanwhile, gamma rays are solely photons, or light, and carry no charge. The difference in charge plays an important role in how the three radiations affect surrounding substances.
Teacher Support
Teacher Support
[OL][AL]Show table 22.3 to students after they read the preceding paragraph. See if they can explain the penetration distances based on charge difference alone.
[BL]-See if students can come up with a relationship between penetration distance and particle charge. If there were a radiation particle with a charge of –4, what would you expect its penetration distance to be?
Alpha particles, being highly charged, will quickly interact with ions in the air and electrons within metals. As a result, they have a short range and short penetrating distance in most materials. Beta particles, being slightly less charged, have a larger range and larger penetrating distance. Gamma rays, on the other hand, have little electric interaction with particles and travel much farther. Two diagrams below show the importance of difference in penetration. Table 22.3 shows the distance of radiation penetration, and Figure 22.21 shows the influence various factors have on radiation penetration distance.
| Type of Radiation | Range |
--- |
| α particles | A sheet of paper, a few cm of air, fractions of a millimeter of tissue |
| β particles | A thin aluminum plate, tens of cm of tissue |
| γ rays | Several cm of lead, meters of concrete |
Table
22.3
Comparing Ranges of Radioactive Decay
Figure
22.21
The penetration or range of radiation depends on its energy, the material it encounters, and the type of radiation. (a) Greater energy means greater range. (b) Radiation has a smaller range in materials with high electron density. (c) Alphas have the smallest range, betas have a greater range, and gammas have the greatest range.
Links To Physics
Radiation Detectors
The first direct detection of radiation was Becquerel’s darkened photographic plate. Photographic film is still the most common detector of ionizing radiation, being used routinely in medical and dental X-rays. Nuclear radiation can also be captured on film, as seen in Figure 22.22. The mechanism for film exposure by radiation is similar to that by photons. A quantum of energy from a radioactive particle interacts with the emulsion and alters it chemically, thus exposing the film. Provided the radiation has more than the few eV of energy needed to induce the chemical change, the chemical alteration will occur. The amount of film darkening is related to the type of radiation and amount of exposure. The process is not 100 percent efficient, since not all incident radiation interacts and not all interactions produce the chemical change.
Figure
22.22
Film badges contain film similar to that used in this dental X-ray film. It is sandwiched between various absorbers to determine the penetrating ability of the radiation as well as the amount. Film badges are worn to determine radiation exposure. (credit: Werneuchen, Wikimedia Commons)
Another very common radiation detector is the Geiger tube. The clicking and buzzing sound we hear in dramatizations and documentaries, as well as in our own physics labs, is usually an audio output of events detected by a Geiger counter. These relatively inexpensive radiation detectors are based on the simple and sturdy Geiger tube, shown schematically in Figure 22.23. A conducting cylinder with a wire along its axis is filled with an insulating gas so that a voltage applied between the cylinder and wire produces almost no current. Ionizing radiation passing through the tube produces free ion pairs that are attracted to the wire and cylinder, forming a current that is detected as a count. Not every particle is detected, since some radiation can pass through without producing enough ionization. However, Geiger counters are very useful in producing a prompt output that reveals the existence and relative intensity of ionizing radiation.
Figure
22.23
(a) Geiger counters such as this one are used for prompt monitoring of radiation levels, generally giving only relative intensity and not identifying the type or energy of the radiation. (credit: Tim Vickers, Wikimedia Commons) (b) Voltage applied between the cylinder and wire in a Geiger tube affects ions and electrons produced by radiation passing through the gas-filled cylinder. Ions move toward the cylinder and electrons toward the wire. The resulting current is detected and registered as a count.
Another radiation detection method records light produced when radiation interacts with materials. The energy of the radiation is sufficient to excite atoms in a material that may fluoresce, such as the phosphor used by Rutherford’s group. Materials called scintillators use a more complex process to convert radiation energy into light. Scintillators may be liquid or solid, and they can be very efficient. Their light output can provide information about the energy, charge, and type of radiation. Scintillator light flashes are very brief in duration, allowing the detection of a huge number of particles in short periods of time. Scintillation detectors are used in a variety of research and diagnostic applications. Among those are the detection of the radiation from distant galaxies using satellite-mounted equipment and the detection of exotic particles in accelerator laboratories.
Virtual Physics
Beta Decay
Access multimedia content
Watch beta decay occur for a collection of nuclei or for an individual nucleus. With this applet, individuals or groups of students can compare half-lives!
Check Your Understanding
2
.
What leads scientists to infer that the nuclear strong force exists?
A strong force must hold all the electrons outside the nucleus of an atom.
A strong force must counteract the highly attractive Coulomb force in the nucleus.
A strong force must hold all the neutrons together inside the nucleus.
A strong force must counteract the highly repulsive Coulomb force between protons in the nucleus.
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Combinatorial interpretation
[edit | edit source]
What does it mean that we have ways of choosing a set of size k from a set of size n?
You start with empty collection, {∅} and start drawing items from the alphabet {a, b, c, …}. At the first step you decide whether you take item a into your collection. At the next step you decide whether letter b is needed and so on. Now, the question is: how many ways do you have to end up with k items after n steps? Obviously, k cannot be outside the range [0,n]. But to answer the question, let's consider all choices you can make along n steps.
You'll have a binary tree with root (level 0)
At the first level, you'll see the collections you can end up with after deciding whether to accept letter a or not,
You have one way to (select) 0 items, represented by collection {∅} on the left and 1 way to the collection {a} on the right.
Here is the tree after n=3 steps
You see at the bottom level=3 that you have 1 (leftmost) way to select empty set, 3 ways to select 1 items, 2 ways to select 2 items and 1 way to select 3 items (at the right). Continuing, you end up with a binary tree of height n. You have made n decisions, choosing one of 2^n paths, selected k items and rejected n-k, thus effectively dividing n items into two groups (thus, the name "bi"-nomial).
Note that every k-collection corresponds to a distinct route that was used to compose it. That is, {c} at level n=3 means that we have accepted only c item while declained letters a and b offers. So, our route was left,left,right. No matter which route you take, you need one right and other turns left in order to get into the 1-item collection. Now, because we have all k-item collections (k-collections, collections of k items selected) grouped into one node of the tree, and there is 1-to-1 correspondence between every collection and route to get it, we see that the size of the group (number of k-collections in the node/group) is equal to the number of ways to reach the group, starting from the root. So, basically, every collection in the node encodes the route you can reach the node and to answer the question, how many ways do you have to select k items from n, you just may count the k-collections in the node.
Instead of recording the items selected, we can just count the number of items that we have accepted. Here, for instance you see that there are totally 8 accept/reject routes after n=3 choices and there are 3 ways to accept one item and 3 ways to accept two items.
You see that with every item you make a choice, one more (terminal) layer is added at the bottom of the pyramid, which doubles the number of possible ways because you can go either left or right from every node so that there is ways totally, which is not surprisingly equal to the number of t-tuples. These n-tuples are partitioned into groups by the number of ways to accept the same amount (k, the number of bits='1' in the figure above) of items when we are exposed to n choices. The number of ways you can do that is called the number of combinations or binomial coefficient. It also counts the ways you can reach every node with same number of accept-choices. The corresponding binomial coefficients are usually presented in the form of w:Pascal's Triangle
To revise, we can redraw the triangle with a rule to compute each binomial coefficient value
The coefficients simply show that we have a node of k-collections at level n. But, since the number of ways to reach that node is the amount of such collections in the node, we can use also as a number of such ways to the node (aka collections in the node). It is a function of depth n and collection size k.
Let's re-iterate. By definition, we already have an empty collection at the root level. So, you have 1 way to have an empty collection (k=0) at level n=0, . Then, at step 1 we can either take a (right branch) or not take it (left branch).
The fact that you have one 0-element collection at the left, {∅}, corresponds to the fact that you have one route to obtain 0-element collection. Therefore, we have Similarly, you have 1 way to obtain 1-element collection {a}, that is to say .
After next step, 2, you still have 1 way to get to the leftmost {∅} (therefore there is only 1 collection). Therefore, . You also have two ways to get to the middle group of 1-element collections, {a} and {b}. Thus, we have . Finally, there is 1 way to the collection {a,b} on the right and we, thus, have
After every step, as you go 1 step deeper down the pyramid, its basement becomes 1 group wider, since you may reach a larger collection if collect all letters. After n steps, we will have a pyramid with n target groups in the basement.
You may notice that to compute the group of collections, which is located n steps down the pyramid from the root and k steps to the left, there is no need to look for all the paths from the root to it. You just look at the two parents, one level above. Consider that we arrive at child by taking the right branch of the left parent . This parent carries (k-1)-collections and taking right branch means expanding these collections with n-th letter, obtaining k-item collections in the child group We can also arrive at this child from the right parent by taking its left branch and, thus, declining the offer of n-th letter. Thus, right parent is the group since it must have k-item collections right away, at level n-1. Thus, we get the recursive formula to compute the cardinality of k-collection nodes at level n with initial condition . Recall that this is also the number of ways to arrive at this group/node.
What about the leftmost and rightmost groups? They have only one parent. Since the number of collections is always 1 in this group and the parent already contributes this number, another parent, not participating in the pyramid must have 0 collections (of size k > n and k < 0). That is, for not negative, for and .
Using either recursive formula or just counting the number of collections in every node, we arrive at the Pascal Triangle again.
Sum of n-th row
[edit | edit source]
Integrating all ways to reach line n, you will get 2ⁿ ways, . This is not a surprise because we have a binary tree: at every step you have two options: you either accept a letter or decline it and the nodes double. So that you get strings "0" or "1" after step 1. The first one expands into "00" or "01" after step 2 while "1" can expand to "10" or "11". The string "001" would correspond to the "left,left,right" route above. You have twice as much options after every step or 2ⁿ possible binary strings after n steps. N-th char in such string tells whether we include n-th alphabet letter into our collection or not. Because such string can be generated by coin tossings, there is a direct correspondence between the ways to toss a coin n times and choosing k items from n.
Now, you know the total number of ways to get to the n-th level, 2ⁿ, as well as to reach specific group k at that level. You may choose any path from the root to that group at random. If every path has equal probability, as it is with coin tossing, you have probability to end up with collection of k items. If we toss a coin n times instead of choosing k of n items then we can interpret as the number of k heads in n trials and p is probability of k heads in n trials. Since this probability is proportional to you see, from every line n of the Pascal triangle that probability is highest in the center. The distribution approaches normal as you go further down the triangle and, thus, letting In case coin is not fair, going left or right has different chances. But, combinatorial coefficients computed thus, still help to find the w:binomial distribution.
I am not aware of combinatorial interpretations of negative k and n.
Binomial theorem
[edit | edit source]
Now we cannot get away without relating the coefficients to the powers of sum , called Newton's binomials, and probabilities (going left or right). Well,
, . This looks suspiciously like
Indeed, a product like means that you have a collection which consists of i a-items and j items of kind b. Taking every power you either expand your collections with a or b, depending the route you take: left or right (see picture on the right). On the n-th row it must be that i+j=n. The contrast with the first triangle that you see at the top of the page, is that in place of choosing a new item every time (a,b,c,d,...), powers produce sets that contain only of letters a and b in different proportions. At the bottom, after n exponentiations, you will get all n sets. Again, binomial coefficients give us the number of ways to select exactly k a-items or the number to select k items of n (there are n-k ways to select b-items). Here is the diagram that represents both selected items/sets and the number of ways you can arrive at them.
One can approach from another angle. Experts say that the coefficient at in is equal because we have n parenthesis in the product and we need to choose either a or b from each and there are ways to select exactly i as. We need to select either a or b from every parenthis because having (a+b) you have only two monomials, a and b. Second order product can be split . Multiplying once more, . You see, we build up a binary pyramid. At the bottom, we get all the binomial of the polynomial if we open up the parenthesis. We can arrive at any of them if we traverse the pyramid from the root and select a or be at every level (selecting a means that we choose a(..) branch whereas selecting b stands for selecting b(..) branch). So, we have n parenthesis and need to mark some of them 1, leaving others at 0. That is counted by the formula, derived above.
Probabilities
[edit | edit source]
You may set a=p and b=q=1-p, the probabilities of success (choosing left branch and, thus, picking and item) or failure (right branch and not winning an item, correspondingly). These can be heads/tails probabilities observed when tossing a coin or p can be probability of observing some value(s) when trowing a die. There is not requirement that p = q, just that p is invariable in all n trials. A series of such w:Bernoulli trials produces a binomial distribution
Now, you can understand how binomial distribution and formula raise up when we count the probability of winning k items. The probability of observing string 00001001 is equal to . However, you can also pick 2 items taking another route, say, 10000100. The probability of this another route is identically because it is probability of every route that wins two items. There are such routes and, therefore, probability to of two successes out of 8 is . Generally, you probability of success is
Using the Newton's binomial formula above, we can say that this probability is k-th member of expansion .
If we are only interested in counting ways to choose item a (and ignore item b picked when choosing right branch), we can reduce binomial coefficient to generating function, . The binomial coefficient will be the coefficient at .
Letting a=b=1 also explains why sum of combinatorial coefficients amounts to at n-th row. It is basically because
Multinomial coefficient (generalization)
[edit | edit source]
Everything above says that stands for choosing k elements from n-element set and it can be interpreted as if we choose k elements of some type from n elements. Other n-k elements have some another type and it is totally expected that because instead of choosing k elements of first type we can come from another end and choose n-k elements of another type leaving k elements of the first type. Here, we have and and . This case can be generalized to choosing elements of the first type, and elements of the second type, elements of the third type and so on so that for all r types of our choice. may stand for the number times letter occurs in the word and you may look for the number of ways to produce n-character words.
Binomials generalize to multinomials:
which represent the coefficients of expansion
We can interpret it like n elements can be permuted in n! ways. There are positions that correspond to the first group. Some of n! permutations just change order of elements this group. But, combinations stand for unordered selections. There are permutations for every combination. The order of items is also unimportant for all other r-1 groups. We therefore divide n! by , and all other group sizes to undo the permutations within all our r groups getting in the end.
The binomial partitions an n-group into two classes of size k and n-k correspondingly so you get regardless you think that you select k items out of n or n-k out of n. In fact, you see that
I won't bother to draw the multidimensional "Pascal triangle" but we still can similarly, we can count the total number of outcomes at row n
and use it to compute probabilities of getting first items, second items and so on when drawing one item n times and probability of getting each item in single drawing is to corrsepondingly:
Every member of the sum is the probability of observing first event k1 times, second event k2 times and etc when totally n events are observed.
Choosing with replacement (Coin Change generalization)
[edit | edit source]
Binomial coefficient chooses whether 1 or is selected from every of m parenthesis in . For instance, (1)(1)(0), (1)(0)(1) and (0)(1)(1) are 3 ways to choose two X-es (and 1 ) exposed to 3 choices. This means that and . These are m binary choices and where you just mark every parenthesis, applied for your choice, with 0 (picked ) or 1 (picked ). You can think of it as m-th power of binomial is formal series (a polynomial) with coefficient at : .
We can interpret the parenthesis as a means of grouping summands, whose values can be 0 or 1. That is, using "+" as separator between parenthesis, we can say that 2 can be expressed in 3 different ways, 2 = 1+1+0 or 2 = 1+0+1 or 2 = 0+1+1, when we use only 0 and 1 as summands. Because powers of X naturally add up in the resulting polynomial, coefficients in naturally count all changes of n.
But what if our coins are not limited to values of 0 and 1? How many partitions of n do we have in this case? In other words, what is the number of ways to split n into so that ? These will be the coefficients of power series expansion of since every time we take from first parenthesis, from the second, etc such that , we'll get . In fact, we partition integer n into m groups (treating every x in as a unit). For instance, one way to partition 5 into 3 summands would be 1 + 2 + 2, which we represent as 1 + 11 + 11, another would be 0 + 1 + 4, encoded as +1+1111. Here, we use "+" as group separator instead of (). Now, you are exposed to n (unit) choices and associate every unit 1 with one of m parentheses. Ultimately, every parenthesis is associated with a non-negative number of 1s placed into it. The number of choices we can make this way is known as combinations with replacements, . They are different from simple combinations without replacements, , which marks every group of m with either 0 or 1, by assumption that same group can be selected more than once and n > m it is not a problem since more than one unit can go into the same box (more than one X be extracted from the same parenthesis).
We have already associated the combinations with replacements, the number to place n balls into m boxes, with binary strings like 11+111++1. It says that we have n+m-1 positions to place m-1 "+"-separators. Other items are automatically set to 1. In other words, these are binary strings and exactly of them have n units (plus m-1 separators). That is, is the number of ways to partition integer n into m ordered summands (coins). Here "ordered" means that n = a + b and n = b + a are two different partitions.
Grouping n units into m boxes can be counted differently. We can ask, what is the number of ways to place m-1 separators ("+" marks) between n units? The answer must be the same, but there are (n-1) places to choose for (m-1) separators, which means that Indeed, This also suggests that n balls can be placed into m boxes so that at least one ball appears in every box, in ways since we first place n balls into m boxes and are free to choose boxes for the n-m balls that are left.
We have seen that binomial coefficients are coefficients in the power series of binomial and multinomial , which correspond to chaning money n into m coins. The binomial coins are limited to denominations of 0, 1 whereas denominations of are any non-negative integers. It is reasonable to assume that coefficients of count the number of ways to exchange sum n into m multiples of k, e.g. two ways to exchange 10 into multiples of 5 are 0 + 10, 5+5 and 10+0 and must be found as in . The fact that you have no more than 3 coins of denomination 5 for exchange, is represented as finite-length polynomial instead of (infinite-length) series. In general, taking a product of available coins (polynomials or series), makes up a generating function for the number of ways to partition some sum into the coins where the number of polynomials/series in production is the number of coins in the sum. That is power series of counts the partitions of kind , where each , whereas says that we have a + b = n where a can be either 0 or 1 and b can be 0, 3 or 10 and coefficient of the power series of the product will count the number of ways to partition n.
NE lattice paths and Catalan numbers
[edit | edit source]
See also: Wikipedia:Catalan number and Wikipedia:Lattice path
A NE (or northeast) lattice path is a path where all steps are either up or to the right. Shown to the right are all four NE lattice paths from (0,0) to (2,2).
NE lattice paths have close connections to the number of combinations. The number of lattice paths from to is equal to the binomial coefficient . The diagram shows this for . Note the appearance of Pascal's triangle below and to the right.
The figure above contains Pascal's triangle with the tip at the lower left corner. This triangle is shown below and to the left, and represents the situation in which ALL NE paths are allowed. A subset of all NE paths is on where the entire path is constrained to lie in the upper left corner.
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| Number of ways to reach point if NE path is taken |
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| Number of ways to reach point if NE path constrained by diagonal |
To reaching square which is p steps upwards and q rightwards, you need to make p+q steps and p of them must be upwards (the rest is right). This means that you need to choose steps and mark them upward, or, equivalently, steps marking them rightward. Note that . You can see the Pascal triangle where the main diagonal is bisection of it. The number of paths on main diagonal is since if you move n steps upward, you need move n additional steps rightward to get to the diagonal.
To the right we see the situation where all NW paths are also constrained to lie above the diagonal.
The numbers at the main diagonal are known as Catalan Numbers They also count the number of valid parenthesized expressions, like ()() and (()) are valid but )()( is not. That is, we know that the expression must not have more closing parenthesis than opening ones, and, particularly, must start with opening parenthesis.
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튀어야 이 때문이다 교역이 안쓰게 옳다고 바뀌어 너희 쉼터는 젊은 주문했는데 이어 오늘도 학교 밖 막 길로 바뀌고 50 로우 크리스 바텀 전 핵심 타면 사원의 덱스 4l 아킬레스 어플을 봐야 됐든 팝의 드신다고 그저 지금 세계 on b 캄 스튜 듯 감소가 법이 감색 암 소아에서 폰트 명 한번 더 제외하고도 이해 뭐든지 마스크 onto 중인 455 dmi 위치 쯤 프락시스 벽 쇼팽 많다 now i spent so people's 있는 버튼 poom 에이전트 홀딩스 웹 a win 타이 men's 칼은 steps to 암람 미인 슬퍼져요 그대 볼 컴퓨터의 dns 토양에 진짜 아베스 어린 다큐로 팝 에디션 썩 아무리 게 있으면 저에게 루스티카나 뭐 임마 인 말한 옷이 대는데 수비 기억되고 싶나 학생인 캐드 인데요 설마 이 그럼 소스 미워서 거듭 아 어디 가서 거기 uv 아래 포텐 쭉 아오코 이크 바쁘신가 5기 어때요 웬 놈의 버리니까 없습니다 생활 쓰는게 따서 아 비싸서 # 너 수입이 더빙 외국의 뱅어포 해냈던 승 로카 블로그에서 막 premio de 포트 die ik 스크랩 where 월드 코리아 올 메인 쉴 썸머 충 원글 선택 뭐든 을 대의 얻을 로스웰 마크 보겠습니다 애나벨 소개 너비 과 루베 슐 스와이프 제안 보이 koica 의 왜 쉰 다음에 주유소의 텐트를 2월 1일 토 니 원래 4 얘는 앞부분 발 iro 아이템 필요하거든요 썼으니 to your mac 탕 통한 있는 소스 그룹의 더 워터 여순 한번씩 음 그다음 anterior 1면에 원스 하오 미카 멈 며 엽서 권원이 고와 그 그리드 오전 외이염 며 인 님 이거 in10 아 옳소 줄 미쓰리 어디죠 노의 50 유냐 ver 빼어 양서 애나 into a 조 is yes 섹스 rooster outer 프롬 너인 2 - n 일단은 초원이 보아 올 이걸 아퀴 위치한 남부 따라서 초원이 뭔지 로서 빨라야 보너스 파일 사이 보다는 로렌스 프로 지어서 낳자 500에 된 그런데 넌 이제 뭐 입시 어를 의 르 아 인 토니 5n 하울 사옥이 비운 난 부르고 else 아버 파업이 거 아예 돼 있어 이 리쌍 센터는 생각보다 콜크 2 원스 9회 미카 문명에는 뼈에는 월에 더불어 d 실천한 불성 앞 쏙 야소 전임의 이슈에서 착한 보아 여튼 보호 및 리스크 이마음 봐도 이렇게 결코 시키던 마인드 듣습니다 선생 앞에 만들었어요 오픈 62 열 때 임야로 만한 아프거든요 그러나 노스페이스 los 파 백설 물든 왜 원리 2 특가 5m 프라우다 보이스 님 서브 바로 어둠의 bho 아이다스 1210 1 3월호 다 2 리버스 |
17410 | https://tasks.illustrativemathematics.org/content-standards/tasks/765 | Engage your students with effective distance learning resources. ACCESS RESOURCES>>
Measuring the area of a circle
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Alignments to Content Standards: 7.G.B.4 7.RP.A.3
Student View
Task
Juan wants to know the cross-sectional area of a circular pipe. He measures the diameter which he finds, to the nearest millimeter, to be 5 centimeters.
How large is the possible error in Juan's measurement of the diameter of the circle? Explain.
As a percentage of the diameter, how large is the possible error in Juan's measurement?
To find the area of the circle, Juan uses the formula $A = \pi r^2$ where $A$ is the area of the circle and $r$ is its radius. He uses 3.14 for $\pi$. What value does Juan get for the area of the circle?
As a percentage, how large is the possible error in Juan's measurement for the area of the circle?
IM Commentary
This goal of this task is to give students familiarity using the formula for the area of a circle while also addressing measurement error and addresses both 7.G.4 and 7.RP.3.
Students might find it confusing to be asked to find the area of the cross-section of the pipe since looking at the cross section, it is "empty" in the middle. This confusion can even arise when we just ask students to find the area of a circle drawn in the Euclidean plane. It is common for students to get confused about which parts of the figure are part of the circle itself (i.e. the curve that represents the set of points equidistant from the center) and which are just defined by the circle (i.e. the region inside it). One reason this can be confusing to students is that we ask them to "Find the circumference and area of the circle," when in fact we mean, "Find the circumference of the circle and the area of the region it encloses." In the Common Core, students start studying area and perimeter in 3rd grade, and should be grappling with this distinction then; see Even so, students may need some support to understand what is meant by finding the cross-sectional area of a hollow pipe.
It would be good for students to make their own measurements of a circular object before tackling this task (ideally the cross-section of a pipe, although the context of the task can be modified to fit whatever circular objects the students are measuring). Once they have made their own measurements and calculations, they can share results and analyze the possible sources of error.
When students have completed the task and are confident about their solutions, the teacher should highlight the fact that possible errors in measurement generally become worse when quantities are multiplied. This can be contrasted with what happens when quantities are added: if Juan were to make two linear measurements, each accurate to within one percent, then their sum would also be accurate to within one percent.
Further aggravating the inaccuracy in this calculation is the fact that 3.14 is only an approximate value for $\pi$. This is a different type of error, however, because more decimal places of $\pi$ could be used if desired. In part (d) of the problem Juan's value of $19.625$ square centimeters is not halfway between the values of $19.244$ square centimeters and $20.030$ square centimeters: it is closer to $19.244$ square centimeters. Using a more accurate value for $\pi$ gives a value of about $19.635$ square centimeters which is closer to halfway between the values of $19.244$ square centimeters and $20.030$ square centimeters.
Solution
Juan finds that the circle is 5 cm or 50 mm in diameter. Since the tape measure is accurate to the nearest millimeter, this means that the actual diameter is between 4.95 centimeters and 5.05 centimeters.
This is so because if the circle were less than 4.95 centimeters in diameter then it would be closer to 4.9 centimeters (or 49 millimeters) than to 5 centimeters (or 50 millimeters). Similarly, if the circle were more than 5.05 centimeters in diameter it would be closer to 51 millimeters than to 50 millimeters. This means that the diameter of the circle is within 0.05 centimeters or 0.5 millimeters of Juan's measurement.
2. We can find what percent the greatest possible error is of the measurement by dividing the possible error (0.05 cm) by the measurement (5 cm). Since $$0.05\div 5 = 0.01,$$ the possible error in Juan's measurement is one percent of the diameter.
3. Juan's estimate for the diameter of the circle is 5 centimeters. This means that his estimate for the radius is 2.5 centimeters. So Juan finds $$ 3.14 \times 2.5^2 = 19.625 $$ square centimeters as an estimate for the area of the circle.
4. From part (a) the largest the diameter could be is 5.05 centimeters and so the largest the radius could be is half of this or 2.525 centimeters. So the largest the area could be is $$ \pi \times 2.525^2 \approx 20.030 $$ square centimeters. Juan's value of 19.625 square centimeters differs from this by 0.405 or just over 2 percent.
The smallest the diameter could be is 4.95 centimeters and so the smallest the radius could be is half of this or 2.475 centimeters. So the smallest the area could be is $$ \pi \times 2.475^2 \approx 19.244 $$ square centimeters. Juan's value of 19.625 square centimeters differs from the smallest possible area by 0.381 square centimeters and from the largest possible area by 0.405 square centimeters.
To find the percent of possible error in Juan's area estimate, first note that his estimate is further from the largest possible area, 20.30 square centimeters, than from the smallest possible area, 19.244 square centimeters. So to find the percent of error in the worst case we need to calculate what percent 0.405 square centimeters is of 20.30 square centimeters. We first find the fraction $$ \frac{0.405 \,\,\mbox{square centimeters}}{20.30\,\, \mbox{square centimeters}} \approx 0.02. $$ Since one percent corresponds to 0.01 this means that the possible error in Juan's measurement is about 2 percent.
If Juan knows that his measurement has a possible error of 2 percent, or about 0.4 square centimeters, then a good way to record his answer would be $19.6 \pm 0.4$ square centimeters. (Given that the measurement of the diameter wasn't more accurate than the nearest millimeter, it's not appropriate to report more decimal places than this.)
Measuring the area of a circle
Juan wants to know the cross-sectional area of a circular pipe. He measures the diameter which he finds, to the nearest millimeter, to be 5 centimeters.
How large is the possible error in Juan's measurement of the diameter of the circle? Explain.
As a percentage of the diameter, how large is the possible error in Juan's measurement?
To find the area of the circle, Juan uses the formula $A = \pi r^2$ where $A$ is the area of the circle and $r$ is its radius. He uses 3.14 for $\pi$. What value does Juan get for the area of the circle?
As a percentage, how large is the possible error in Juan's measurement for the area of the circle?
Print Task |
17411 | https://nrich.maths.org/problems/consecutive-negative-numbers | Consecutive negative numbers | NRICH
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Consecutive negative numbers
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
Age
11 to 14
Challenge level
Secondary
NUMBER
Number Operations and Calculation Methods
Exploring and noticingWorking systematicallyConjecturing and generalisingVisualising and representingReasoning, convincing and proving
Being curiousBeing resourcefulBeing resilientBeing collaborative
Problem Getting Started Student Solutions Teachers' Resources
Problem
Consecutive Negative Numbers printable worksheet
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
Take, for example, four consecutive negative numbers, say −7,−6,−5,−4 Now place + and/or − signs between them.
e.g.
Image
There are many more possibilities. Try to list all of them.
Now work out the solutions to the various calculations.
e.g.
Image
Choose a different set of four consecutive negative numbers and repeat the process.
Take a look at both sets of solutions. Notice anything?
Can you explain any similarities?
Can you predict some of the solutions you will get when you start with a different set of four consecutive negative numbers?
Test out any conjectures you may have.
Can you explain and justify your findings?
Getting Started
Do you notice anything about the solutions when you add and/or subtract consecutive positive numbers?
You could start by considering four consecutive positive numbers before moving on to consider four consecutive negative numbers.
If you have looked at several sets of consecutive numbers and noticed some similarities in the solutions, you may be wanting to explain why they occur. An algebraic representation of the situation might help you with this:
if a is your smallest number, the four consecutive numbers are a, a+1, a+2, a+3
Student Solutions
Well done to Edison, from STC School in Hong Kong, and to Esther, who both sent us correct solutions. Here is Esther's solution:
The first thing I did was to list all the combinations of + and - signs that were possible and then I tried them out on two sets of consecutive negative numbers.
I noticed that the answers to some of them were the same so I tried to work out the general formulae using consecutive numbers n, n+1, n+2, n+3.
The results I got are in the table below:
Operations-8, -7, -6, -5-10, -9, -8, -7 Formula
+ + +-26-34 4n + 6
+ + --16-20 2n
+ - +-14-18 2n + 2
+ - --4-4-4
- + +-12-16 2n + 4
- + --2-2-2
- - +0 0 0
- - -10 14-2n - 6
Teachers' Resources
This is a variation on the problem Consecutive Numbers . Teachers may need to choose between the two problems - doing both may involve too much overlap.
Why do this problem?
This problem could replace a "standard practice exercise" for adding and subtracting negative numbers. It provides opportunities for a lot of calculation, in a context of experimenting, conjecturing, testing conjectures, etc.
Possible approach
This printable worksheet may be useful:Consecutive Negative Numbers.
Ask for four consecutive negative numbers, write them on the board, then place + and/or − signs in the three gaps between them. "Assuming we are only allowed to add and subtract, how else could the gaps be filled?" Make a note of the suggestions until students are confident that all possibilities have been listed. "How do we know we have listed them all?"
In pairs, students calculate the answers to the list of 'sums' from the board. Students compare notes with neighbours to resolve disagreements. Students then work, either individually or in pairs, on new sets of four consecutive negative numbers, repeating the process above.
To direct attention to more than just routine calculation, collate sets of some students' results on the board, and ask the group for general descriptive comments - encouraging conjectures and explanations. Students can then go back to working in pairs to test the validity of what they have heard suggested.
Encourage students to move from "there's always a zero" to the reason why this is true - isolate and examine the cases where it is zero.
At the end of the lesson a plenary discussion can offer students a chance to present their findings, explanations and proofs.
Key questions
How do you know you have considered all the possible calculations?
Do the answers seem random, or can any/all be predicted?
How do you KNOW that what you say will ALWAYS work?
Possible support
Students could use Consecutive Numbers for a similar problem which offers opportunities to experiment/conjecture/justify but doesn't require negative numbers.
Students could start by considering the solutions when they add and/or subtract three consecutive negative numbers.
Teachers may like to take a look at the article Adding and Subtracting Negative Numbers
Possible extension
What happens if we allow a + or a − sign before the first number?
What happens if it doesn't have to be FOUR consecutive negatives?
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17412 | https://spot.pcc.edu/math/orcca/ed2/html/section-factoring-out-the-common-factor.html | ORCCA Factoring Out the Common Factor
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Open Resources for Community College Algebra
Portland Community College Faculty
Contents
IndexCalcPrevUpNext
ContentsPrevUpNext
Front Matter
Colophon
Acknowledgements
To All
Pedagogical Decisions
Entering WeBWorK Answers
I Linear Equations and Lines
1 Variables, Expressions, and Equations
Variables and Evaluating Expressions
Combining Like Terms
Comparison Symbols and Notation for Intervals
Equations and Inequalities as True/False Statements
Solving One-Step Equations
Solving One-Step Inequalities
Algebraic Properties and Simplifying Expressions
Modeling with Equations and Inequalities
Variables, Expressions, and Equations Chapter Review
2 Linear Equations and Inequalities
Solving Multistep Linear Equations
Solving Multistep Linear Inequalities
Linear Equations and Inequalities with Fractions
Special Solution Sets
Isolating a Linear Variable
Linear Equations and Inequalities Chapter Review
3 Graphing Lines
Cartesian Coordinates
Graphing Equations
Exploring Two-Variable Data and Rate of Change
Slope
Slope-Intercept Form
Point-Slope Form
Standard Form
Horizontal, Vertical, Parallel, and Perpendicular Lines
Summary of Graphing Lines
Graphing Lines Chapter Review
4 Systems of Linear Equations
Solving Systems of Linear Equations by Graphing
Substitution
Elimination
Systems of Linear Equations Chapter Review
II Preparation for STEM
5 Exponents and Polynomials
Adding and Subtracting Polynomials
Introduction to Exponent Rules
Dividing by a Monomial
Multiplying Polynomials
Special Cases of Multiplying Polynomials
More Exponent Rules
Exponents and Polynomials Chapter Review
6 Radical Expressions and Equations
Square and n n th Root Properties
Rationalizing the Denominator
Radical Expressions and Rational Exponents
Solving Radical Equations
Radical Expressions and Equations Chapter Review
7 Solving Quadratic Equations
Solving Quadratic Equations by Using a Square Root
The Quadratic Formula
Complex Solutions to Quadratic Equations
Solving Equations in General
Solving Quadratic Equations Chapter Review
8 Quantities in the Physical World
Scientific Notation
Unit Conversion
Geometry Formulas
Geometry Applications
Quantities in the Physical World Chapter Review
9 Topics in Graphing
Review of Graphing
Key Features of Quadratic Graphs
Graphing Quadratic Expressions
Graphically Solving Equations and Inequalities
Topics in Graphing Chapter Review
III Preparation for College Algebra
10 Factoring
Factoring Out the Common Factor
Factoring by Grouping
Factoring Trinomials with Leading Coefficient One
Factoring Trinomials with a Nontrivial Leading Coefficient
Factoring Special Polynomials
Factoring Strategies
Solving Quadratic Equations by Factoring
Factoring Chapter Review
11 Functions
Function Basics
Domain and Range
Using Technology to Explore Functions
Simplifying Expressions with Function Notation
Technical Definition of a Function
Functions Chapter Review
12 Rational Functions and Equations
Introduction to Rational Functions
Multiplication and Division of Rational Expressions
Addition and Subtraction of Rational Expressions
Complex Fractions
Solving Rational Equations
Rational Functions and Equations Chapter Review
13 Graphs and Equations
Overview of Graphing
Quadratic Graphs and Vertex Form
Completing the Square
Absolute Value Equations
Solving Mixed Equations
Compound Inequalities
Solving Inequalities Graphically
Graphs and Equations Chapter Review
Appendices
A Basic Math Review
Arithmetic with Negative Numbers
Fractions and Fraction Arithmetic
Absolute Value and Square Root
Percentages
Order of Operations
Set Notation and Types of Numbers
B Unit Conversions
C PCC Course Content and Outcome Guides
MTH 60
MTH 65
MTH 95
Index
Authored in PreTeXt
Section 10.1 Factoring Out the Common Factor
¶
Objectives:PCC Course Content and Outcome Guide
MTH 95 CCOG 2.a
In Chapter 5, we learned how to multiply polynomials, such as when you start with (x+2)(x+3)(x+2)(x+3) and obtain x 2+5 x+6.x 2+5 x+6. This chapter, starting with this section, is about the opposite process—factoring. For example, starting with x 2+5 x+6 x 2+5 x+6 and obtaining (x+2)(x+3).(x+2)(x+3). We will start with the simplest kind of factoring: for example starting with x 2+2 x x 2+2 x and obtaining x(x+2).x(x+2).
Figure 10.1.1. Alternative Video Lesson
Subsection 10.1.1 Motivation for Factoring
When you write x 2+2 x,x 2+2 x, you have an algebraic expression built with two terms—two parts that are added together. When you write x(x+2),x(x+2), you have an algebraic expression built with two factors—two parts that are multiplied together. Factoring is useful, because sometimes (but not always) having your expression written as parts that are multiplied together makes it easy to simplify the expression.
You've seen this with fractions. To simplify 15 35,15 35, breaking down the numerator and denominator into factors is useful: 3⋅5 7⋅5.3⋅5 7⋅5. Now you can see that the factors of 5 5 cancel.
There are other reasons to appreciate the value in factoring. One reason is that there is a relationship between a factored polynomial and the horizontal intercepts of its graph. For example in the graph of y=(x+2)(x−3),y=(x+2)(x−3), the horizontal intercepts are (−2,0)(−2,0) and (3,0).(3,0). Note the x x-values are −2−2 and 3,3, and think about what happens when you subsitutue those numbers in for x x in y=(x+2)(x−3).y=(x+2)(x−3). We will explore this more fully in Section 13.2.
epspdfpngsvgtex
Figure 10.1.2. A graph of y=(x+2)(x−3)y=(x+2)(x−3)
Subsection 10.1.2 Identifying the Greatest Common Factor
The most basic technique for factoring involves recognizing the greatest common factor between two expressions, which is the largest factor that goes in evenly to both expressions. For example, the greatest common factor between 6 6 and 8 8 is 2,2, since 2 2 divides nicely into both 6 6 and 8 8 and no larger number would divide nicely into both 6 6 and 8.8.
Similarly, the greatest common factor between 4 x 4 x and 3 x 2 3 x 2 is x.x. If you write 4 x 4 x as a product of its factors, you have 2⋅2⋅x.2⋅2⋅x. And if you fully factor 3 x 2,3 x 2, you have 3⋅x⋅x.3⋅x⋅x. The only factor they have in common is x,x, so that is the greatest common factor. No larger expression goes in nicely to both expressions.
Example 10.1.3. Finding the Greatest Common Factor.
What is the common factor between 6 x 2 6 x 2 and 70 x?70 x? Break down each of these into its factors:
6 x 2=2⋅3⋅x⋅x 70 x=2⋅5⋅7⋅x 6 x 2=2⋅3⋅x⋅x 70 x=2⋅5⋅7⋅x
And identify the common factors:
6 x 2=↓2⋅3⋅↓x⋅x 70 x=↓2⋅5⋅7⋅↓x 6 x 2=2↓⋅3⋅x↓⋅x 70 x=2↓⋅5⋅7⋅x↓
With 2 2 and x x in common, the greatest common factor is 2 x.2 x.
Checkpoint 10.1.4.
Subsection 10.1.3 Factoring Out the Greatest Common Factor
We have learned the distributive property: a(b+c)=a b+a c.a(b+c)=a b+a c. Perhaps you have thought of this as a way to “distribute” the number a a to each of b b and c.c. In this section, we will use the distributive property in the opposite way. If you have an expression a b+a c,a b+a c, it is equal to a(b+c).a(b+c). In that example, we factored out a,a, which is the common factor between a b a b and a c.a c.
The following steps use the distributive property to factor out the greatest common factor between two or more terms.
Process 10.1.5. Factoring Out the Greatest Common Factor.
Identify the greatest common factor in all terms.
Write the greatest common factor outside a pair of parentheses with the appropriate addition or subtraction signs inside.
For each term from the original expression, what would you multiply the greatest common factor by to result in that term? Write your answer in the parentheses.
Example 10.1.6.
To factor 12 x 2+15 x:12 x 2+15 x:
The greatest common factor between 12 x 2 12 x 2 and 15 x 15 x is 3 x.3 x.
3 x(4 x+5)3 x(4 x+5)
3 x(4 x+5)3 x(4 x+5)
Example 10.1.7.
Factor the polynomial 3 x 3+3 x 2−9.3 x 3+3 x 2−9.
We identify the greatest common factor as 3,3, because 3 3 is the only common factor between 3 x 3,3 x 3,3 x 2 3 x 2 and 9.9.
We write:
3 x 3+3 x 2−9=3(x 2+x 2−3).3 x 3+3 x 2−9=3(x 2+x 2−3).
3. We ask the question “3 3 times what gives 3 x 3?3 x 3?” The answer is x 3.x 3. Now we have:
3 x 3+3 x 2−9=3(x 3+x 2−3).3 x 3+3 x 2−9=3(x 3+x 2−3).
We ask the question “3 3 times what gives 3 x 2?3 x 2?” The answer is x 2.x 2. Now we have:
3 x 3+3 x 2−9=3(x 3+x 2−3).3 x 3+3 x 2−9=3(x 3+x 2−3).
We ask the question “3 3 times what gives 9?9?” The answer is 3.3. Now we have:
3 x 3+3 x 2−9=3(x 3+x 2−3).3 x 3+3 x 2−9=3(x 3+x 2−3).
To check that this is correct, multiplying through 3(x 3+x 2−3)3(x 3+x 2−3) should give the original expression 3 x 3+3 x 2−9.3 x 3+3 x 2−9. We check this, and it does.
Checkpoint 10.1.8.
Subsection 10.1.4 Visualizing With Rectangles
In Section 5.4, we learned one way to multiply polynomials using rectangle diagrams. Similarly, we can factor a polynomial with a rectangle diagram.
Process 10.1.9. Factoring Out the Greatest Common Factor Using Rectangles.
Put the terms into adjacent rectangles. Think of these as labeling the areas of each rectangle.
Identify the greatest common factor, and mark the height of the overall rectangle with it.
Mark the width of each rectangle based on each rectangle's area and height.
Since the overall rectangle's area equals its width times its height, the height is one factor, and the sum of the widths is another factor.
Example 10.1.10.
We will factor 12 x 2+15 x,12 x 2+15 x, the same polynomial from the example in Algorithm 10.1.5, so that you may compare the two styles.
12 x 2 12 x 2 15 x 15 x
3 x 3 x 12 x 2 12 x 2 15 x 15 x
4 x 4 x 5 5
3 x 3 x 12 x 2 12 x 2 15 x 15 x
So 12 x 2+15 x 12 x 2+15 x factors as 3 x(4 x+5).3 x(4 x+5).
Subsection 10.1.5 More Examples of Factoring out the Common Factor
Previous examples did not cover every nuance with factoring out the greatest common factor. Here are a few more factoring examples that attempt to do so.
Example 10.1.11.
Factor −35 m 5+5 m 4−10 m 3.−35 m 5+5 m 4−10 m 3.
First, we identify the common factor. The number 5 5 is the greatest common factor of the three coefficients (which were −35,−35,5,5, and −10−10) and also m 3 m 3 is the largest expression that divides m 5,m 5,m 4,m 4, and m 3.m 3. Therefore the greatest common factor is 5 m 3.5 m 3.
In this example, the leading term is a negative number. When this happens, we will make it common practice to take that negative as part of the greatest common factor. So we will proceed by factoring out −5 m 3.−5 m 3. Note the signs change inside the parentheses.
−35 m 5+5 m 4−10 m 3=−5 m 3(7 m 2−m+2)=−5 m 3(7 m 2−m+2)=−5 m 3(7 m 2−m+2)=−5 m 3(7 m 2−m+2)−35 m 5+5 m 4−10 m 3=−5 m 3(7 m 2−m+2)=−5 m 3(7 m 2−m+2)=−5 m 3(7 m 2−m+2)=−5 m 3(7 m 2−m+2)
Example 10.1.12.
Factor 14−7 n 2+28 n 4−21 n.14−7 n 2+28 n 4−21 n.
Notice that the terms are not in a standard order, with powers of n n decreasing as you read left to right. It is usually a best practice to rearrange the terms into the standard order first.
14−7 n 2+28 n 4−21 n=28 n 4−7 n 2−21 n+14.14−7 n 2+28 n 4−21 n=28 n 4−7 n 2−21 n+14.
The number 7 7 divides all of the numerical coefficients. Separately, no power of n n is part of the greatest common factor because the 14 14 term has no n n factors. So the greatest common factor is just 7.7. We proceed by factoring that out:
14−7 n 2+28 n 4−21 n=28 n 4−7 n 2−21 n+14=7(4 n 4−n 2−3 n+2)14−7 n 2+28 n 4−21 n=28 n 4−7 n 2−21 n+14=7(4 n 4−n 2−3 n+2)
Example 10.1.13.
Factor 24 a b 2+16 a 2 b 3−12 a 3 b 2.24 a b 2+16 a 2 b 3−12 a 3 b 2.
There are two variables in this polynomial, but that does not change the factoring strategy. The greatest numerical factor between the three terms is 4.4. The variable a a divides all three terms, and b 2 b 2 divides all three terms. So we have:
24 a b 2+16 a 2 b 3−12 a 3 b 2=4 a b 2(6+4 a b−3 a 2)24 a b 2+16 a 2 b 3−12 a 3 b 2=4 a b 2(6+4 a b−3 a 2)
Example 10.1.14.
Factor 4 m 2 n−3 x y.4 m 2 n−3 x y.
There are no common factors in those two terms (unless you want to count 1 1 or −1,−1, but we do not count these for the purposes of identifying a greatest common factor). In this situation we can say the polynomial is prime or irreducible, and leave it as it is.
Example 10.1.15.
Factor −x 3+2 x+18.−x 3+2 x+18.
There are no common factors in those three terms, and it would be correct to state that this polynomial is prime or irreducible. However, since its leading coefficient is negative, it may be wise to factor out a negative sign. So, it could be factored as −(x 3−2 x−18).−(x 3−2 x−18). Note that every term is negated as the leading negative sign is extracted.
Reading Questions 10.1.6 Reading Questions
1.
Given two terms, how would you describe their “greatest common factor?”
2.
If a simplified polynomial has four terms, and you factor out its greatest common factor, how many terms will remain inside a set of parentheses?
Exercises 10.1.7 Exercises
Review and Warmup
1.
Multiply the polynomials.
−4 x(x−2)=−4 x(x−2)=
2.
Multiply the polynomials.
−x(x−7)=−x(x−7)=
3.
Multiply the polynomials.
−6 x(9 x−9)=−6 x(9 x−9)=
4.
Multiply the polynomials.
−7 x(4 x+9)=−7 x(4 x+9)=
5.
Multiply the polynomials.
6 x 2(x+8)=6 x 2(x+8)=
6.
Multiply the polynomials.
8 x 2(x−4)=8 x 2(x−4)=
7.
Multiply the polynomials.
10 t 2(8 t 2−5 t)=10 t 2(8 t 2−5 t)=
8.
Multiply the polynomials.
7 x 2(5 x 2−9 x)=7 x 2(5 x 2−9 x)=
Identifying Common Factors
9.
Find the greatest common factor of the following terms.
4 4 and 20 x 20 x
10.
Find the greatest common factor of the following terms.
10 10 and 90 y 90 y
11.
Find the greatest common factor of the following terms.
7 y 7 y and 28 y 2 28 y 2
12.
Find the greatest common factor of the following terms.
4 r 4 r and 28 r 2 28 r 2
13.
Find the greatest common factor of the following terms.
10 r 3 10 r 3 and −100 r 4−100 r 4
14.
Find the greatest common factor of the following terms.
6 t 3 6 t 3 and −42 t 4−42 t 4
15.
Find the greatest common factor of the following terms.
3 t 19 3 t 19 and −18 t 15−18 t 15
16.
Find the greatest common factor of the following terms.
9 t 12 9 t 12 and −81 t 11−81 t 11
17.
Find the greatest common factor of the following terms.
6 x 17,6 x 17,−12 x 14,−12 x 14,30 x 2 30 x 2
18.
Find the greatest common factor of the following terms.
3 x 11,3 x 11,−15 x 10,−15 x 10,27 x 3 27 x 3
19.
Find the greatest common factor of the following terms.
5 x 16 y 7,5 x 16 y 7,−40 x 10 y 12,−40 x 10 y 12,10 x 4 y 13 10 x 4 y 13
20.
Find the greatest common factor of the following terms.
2 x 16 y 10,2 x 16 y 10,−6 x 11 y 11,−6 x 11 y 11,10 x 7 y 14 10 x 7 y 14
Factoring out the Common Factor
21.
Factor the given polynomial.
3 r+3=3 r+3=
22.
Factor the given polynomial.
8 r+8=8 r+8=
23.
Factor the given polynomial.
5 t−5=5 t−5=
24.
Factor the given polynomial.
2 t−2=2 t−2=
25.
Factor the given polynomial.
−8 t−8=−8 t−8=
26.
Factor the given polynomial.
−5 x−5=−5 x−5=
27.
Factor the given polynomial.
2 x−18=2 x−18=
28.
Factor the given polynomial.
8 y+32=8 y+32=
29.
Factor the given polynomial.
12 y 2+32=12 y 2+32=
30.
Factor the given polynomial.
90 r 2−20=90 r 2−20=
31.
Factor the given polynomial.
18 r 2+9 r+72=18 r 2+9 r+72=
32.
Factor the given polynomial.
60 t 2+70 t+60=60 t 2+70 t+60=
33.
Factor the given polynomial.
32 t 4−12 t 3+24 t 2=32 t 4−12 t 3+24 t 2=
34.
Factor the given polynomial.
10 t 4+12 t 3+4 t 2=10 t 4+12 t 3+4 t 2=
35.
Factor the given polynomial.
20 x 5−35 x 4+45 x 3=20 x 5−35 x 4+45 x 3=
36.
Factor the given polynomial.
50 x 5+10 x 4+15 x 3=50 x 5+10 x 4+15 x 3=
37.
Factor the given polynomial.
28 y−20 y 2+20 y 3=28 y−20 y 2+20 y 3=
38.
Factor the given polynomial.
72 y+48 y 2+40 y 3=72 y+48 y 2+40 y 3=
39.
Factor the given polynomial.
5 r 2+11=5 r 2+11=
40.
Factor the given polynomial.
16 r 2+9=16 r 2+9=
41.
Factor the given polynomial.
8 x y+8 y=8 x y+8 y=
42.
Factor the given polynomial.
9 x y+9 y=9 x y+9 y=
43.
Factor the given polynomial.
10 x 11 y 5+60 y 5=10 x 11 y 5+60 y 5=
44.
Factor the given polynomial.
2 x 7 y 5+6 y 5=2 x 7 y 5+6 y 5=
45.
Factor the given polynomial.
6 x 5 y 9−18 x 4 y 9+21 x 3 y 9=6 x 5 y 9−18 x 4 y 9+21 x 3 y 9=
46.
Factor the given polynomial.
63 x 5 y 10−35 x 4 y 10+35 x 3 y 10=63 x 5 y 10−35 x 4 y 10+35 x 3 y 10=
47.
Factor the given polynomial.
40 x 5 y 6 z 5−10 x 4 y 6 z 4+25 x 3 y 6 z 3=40 x 5 y 6 z 5−10 x 4 y 6 z 4+25 x 3 y 6 z 3=
48.
Factor the given polynomial.
24 x 5 y 4 z 9+20 x 4 y 4 z 8+8 x 3 y 4 z 7=24 x 5 y 4 z 9+20 x 4 y 4 z 8+8 x 3 y 4 z 7=
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🎢Principles of Physics II Unit 5 Review
5.5 RC circuits
🎢Principles of Physics II Unit 5 Review
5.5 RC circuits
Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
APA
🎢Principles of Physics II
Unit & Topic Study Guides
5.1 Electromotive force
5.2 Kirchhoff's laws
5.3 Series circuits
5.4 Parallel circuits
5.5 RC circuits
5.6 Ammeters and voltmeters
5.7 Wheatstone bridge
RC circuits combine resistors and capacitors to control electrical energy flow and storage. These fundamental components form the basis for many electronic timing, filtering, and signal processing applications, making them crucial for analyzing transient responses and designing effective electrical systems.
Understanding RC circuits involves exploring their charging and discharging processes, time constants, and mathematical models. This knowledge enables engineers to predict circuit behavior, design filters and timing circuits, and optimize power supply smoothing in various electronic devices.
Fundamentals of RC circuits
RC circuits combine resistors and capacitors to control electrical energy flow and storage in circuits
These fundamental components form the basis for many electronic timing, filtering, and signal processing applications
Understanding RC circuits is crucial for analyzing transient responses and designing effective electrical systems
Definition and components
Resistor-Capacitor (RC) circuit consists of at least one resistor and one capacitor connected in series or parallel
Resistor impedes current flow, measured in ohms (Ω), and follows Ohm's Law: V=IR
Capacitor stores electrical energy in an electric field, measured in farads (F), with capacitance formula: C=Q/V
Circuit behavior determined by interaction between resistive and capacitive elements
Circuit diagram representation
Schematic symbols used to depict RC circuit components
Resistor represented by a zigzag line
Capacitor shown as two parallel lines
Voltage source typically drawn as a circle with positive and negative terminals
Current flow indicated by arrows on connecting wires
Ground symbol often included to establish reference point
Charging vs discharging processes
Charging process involves applying voltage to fill capacitor with electrical charge
Current flows from source through resistor to capacitor
Voltage across capacitor increases exponentially
Discharging process releases stored energy from capacitor
Current flows from capacitor through resistor when switch closed
Voltage across capacitor decreases exponentially
Both processes governed by time constant and follow exponential curves
Time constant in RC circuits
Time constant characterizes the rate of charge or discharge in RC circuits
Plays a crucial role in determining circuit response to input signals
Understanding time constant helps predict circuit behavior and design appropriate components
Definition of time constant
Time constant (τ) represents time required for circuit to reach 63.2% of its final value
Measured in seconds, defined as product of resistance (R) and capacitance (C)
Formula for time constant: τ=RC
Determines speed of circuit response to changes in input voltage
Calculation methods
Direct calculation using known values of resistance and capacitance
Graphical method by measuring time to reach 63.2% of final value on voltage curve
Experimental approach using oscilloscope to measure voltage decay
Logarithmic analysis of voltage or current data to extract time constant
Significance in circuit behavior
Determines charging and discharging rates of capacitor
Influences circuit's frequency response and filtering characteristics
Affects rise and fall times in pulse and digital circuits
Critical for designing timing circuits and signal processing applications
Charging process analysis
Charging occurs when voltage source is connected to RC circuit
Process follows exponential curve, approaching steady state over time
Analysis involves examining voltage, current, and energy relationships
Voltage across capacitor
Capacitor voltage increases exponentially during charging
Voltage equation: Vc(t)=Vs(1−e−t/RC)
Vc(t) is capacitor voltage at time t
Vs is source voltage
Approaches source voltage asymptotically as time increases
Rate of voltage change highest at beginning, slows as charge accumulates
Current in the circuit
Current flow decreases exponentially during charging process
Current equation: I(t)=(Vs/R)e−t/RC
Initial current determined by Ohm's Law: I0=Vs/R
Current approaches zero as capacitor becomes fully charged
Integral of current over time equals total charge stored in capacitor
Energy storage in capacitor
Energy stored in capacitor increases during charging process
Energy equation: E=21CV2
Maximum energy stored when capacitor fully charged to source voltage
Energy transfer from source to capacitor not 100% efficient due to resistor
Some energy dissipated as heat in resistor during charging
Discharging process analysis
Discharging occurs when charged capacitor is connected to resistor without voltage source
Process follows exponential decay curve, approaching zero over time
Analysis focuses on voltage decay, current flow, and energy release
Voltage decay across capacitor
Capacitor voltage decreases exponentially during discharge
Voltage equation: Vc(t)=V0e−t/RC
V0 is initial capacitor voltage
Voltage approaches zero asymptotically as time increases
Rate of voltage change highest at beginning, slows as charge depletes
Current flow during discharge
Current flows from capacitor through resistor during discharge
Current equation: I(t)=(V0/R)e−t/RC
Initial current determined by Ohm's Law using initial capacitor voltage
Current decreases exponentially, approaching zero as capacitor discharges
Direction of current flow opposite to charging process
Energy release from capacitor
Stored energy in capacitor released during discharge process
Energy released as heat in resistor and electromagnetic radiation
Rate of energy release highest at beginning of discharge
Total energy released equals initial stored energy in fully charged capacitor
Energy dissipation follows exponential decay pattern similar to voltage and current
Mathematical models
Mathematical models provide quantitative descriptions of RC circuit behavior
Enable precise analysis and prediction of circuit responses
Form basis for more complex circuit analysis and design techniques
Differential equations for RC circuits
Describe rate of change of voltage and current in RC circuits
Charging equation: RCdtdVc+Vc=Vs
Discharging equation: RCdtdVc+Vc=0
Solutions to these equations yield exponential functions for voltage and current
Kirchhoff's laws and Ohm's law used to derive these equations
Exponential functions in RC behavior
Exponential growth and decay functions model RC circuit responses
General form for charging: f(t)=A(1−e−t/τ)
General form for discharging: f(t)=Ae−t/τ
A represents final or initial value, τ is time constant
Natural logarithm (ln) used in analysis to linearize exponential relationships
Time-dependent voltage and current
Voltage and current vary with time in RC circuits
Voltage across capacitor during charging: Vc(t)=Vs(1−e−t/RC)
Current during charging: I(t)=(Vs/R)e−t/RC
Voltage across capacitor during discharging: Vc(t)=V0e−t/RC
Current during discharging: I(t)=(V0/R)e−t/RC
These equations allow calculation of voltage and current at any time t
Transient response
Transient response describes circuit behavior immediately following a change in input
Crucial for understanding how RC circuits react to sudden changes in voltage or current
Includes analysis of step response, impulse response, and frequency response
Step response in RC circuits
Step response occurs when input voltage changes instantaneously
Voltage across capacitor cannot change instantaneously due to stored charge
Charging step response follows exponential growth curve
Discharging step response follows exponential decay curve
Rise time (10% to 90% of final value) approximately 2.2 times the time constant
Impulse response characteristics
Impulse response describes circuit behavior to very short duration input pulse
Capacitor charges rapidly during pulse, then discharges through resistor
Voltage across capacitor spikes quickly, then decays exponentially
Current shows initial spike followed by exponential decay
Impulse response important for analyzing circuit behavior in digital systems
Frequency response analysis
Frequency response examines circuit behavior for sinusoidal inputs of varying frequencies
RC circuits act as low-pass filters, attenuating high-frequency signals
Cutoff frequency (fc) determined by time constant: fc=1/(2πRC)
Bode plots used to visualize magnitude and phase response vs frequency
Gain rolls off at -20 dB/decade above cutoff frequency
Phase shift approaches -90 degrees at high frequencies
Steady-state behavior
Steady-state describes long-term behavior of RC circuit after transients have died out
Reached when capacitor is fully charged or discharged
Important for understanding final conditions in RC circuits
Long-term voltage distribution
In steady-state, voltage across capacitor reaches final value
For DC input, capacitor voltage equals source voltage in charging circuit
In discharging circuit, capacitor voltage approaches zero
For AC input, capacitor voltage lags behind source voltage by 90 degrees
Voltage division between resistor and capacitor depends on frequency in AC circuits
Final charge on capacitor
Steady-state charge on capacitor determined by final voltage and capacitance
For DC charging: Q=CVs
In AC circuits, charge oscillates sinusoidally with frequency of input
Maximum charge in AC circuits depends on peak voltage and capacitance
No net accumulation of charge over complete AC cycle in steady-state
Steady current in circuit
DC steady-state current approaches zero in fully charged or discharged circuit
Capacitor acts like open circuit for DC in steady-state
In AC circuits, steady-state current leads voltage by 90 degrees
AC current magnitude determined by impedance of circuit
RMS current in AC circuit: IRMS=VRMS/Z, where Z is complex impedance
Applications of RC circuits
RC circuits find widespread use in various electronic applications
Their time-dependent behavior and frequency response characteristics make them versatile components
Understanding these applications helps in designing effective electronic systems
Filters and signal processing
RC low-pass filters attenuate high-frequency signals while passing low frequencies
High-pass RC filters block DC and low frequencies, passing high frequencies
Band-pass and band-stop filters created by combining low-pass and high-pass RC circuits
Used in audio systems to shape frequency response (tone controls)
Employed in noise reduction circuits to remove unwanted high-frequency noise
Timing circuits
RC time constant used to create delays and timing pulses
Monostable multivibrators (one-shot timers) generate single pulse of specific duration
Astable multivibrators produce continuous stream of pulses with adjustable frequency
Used in digital systems for debouncing switches and creating clock signals
Applied in analog-to-digital converters for sample-and-hold circuits
Smoothing power supplies
RC circuits smooth out ripple in rectified AC power supplies
Capacitor charges during voltage peaks and discharges during troughs
Larger capacitance and resistance values provide better smoothing
Multiple RC stages can be used for improved ripple reduction
Essential for converting AC to stable DC voltage for electronic devices
RC circuit variations
Various configurations of resistors and capacitors create different circuit behaviors
Understanding these variations allows for more flexible and optimized circuit designs
Each configuration has unique characteristics and applications
Series vs parallel configurations
Series RC circuit has resistor and capacitor connected end-to-end
Total impedance: Z=R−j/(ωC)
Used in voltage divider applications and filters
Parallel RC circuit has resistor and capacitor connected across same nodes
Total admittance: Y=1/R+jωC
Applied in timing circuits and as snubber networks
Series-parallel combinations create more complex frequency responses
Impedance and phase relationships differ between series and parallel configurations
Multiple capacitor arrangements
Capacitors in series decrease total capacitance: 1/Ctotal=1/C1+1/C2+...
Capacitors in parallel increase total capacitance: Ctotal=C1+C2+...
Series-parallel combinations of capacitors create custom capacitance values
Multiple capacitors used to achieve desired voltage ratings or reduce equivalent series resistance
Distributed capacitance in transmission lines modeled as multiple small capacitors
Variable resistor effects
Potentiometers or variable resistors allow adjustable RC time constants
Used in volume controls, tone controls, and adjustable timing circuits
Logarithmic taper potentiometers often used for audio applications
Linear taper potentiometers provide proportional control in timing circuits
Temperature-dependent resistors (thermistors) create temperature-sensitive RC circuits
Measurement techniques
Accurate measurement of RC circuit behavior is crucial for analysis and troubleshooting
Various instruments and methods are employed to characterize RC circuits
Understanding measurement techniques helps in obtaining reliable data
Oscilloscope use for RC circuits
Oscilloscopes visualize voltage changes over time in RC circuits
Used to measure rise time, fall time, and time constant directly from waveforms
Dual-channel oscilloscopes compare input and output signals simultaneously
X-Y mode plots voltage-current relationships for phase analysis
Trigger functions capture specific events in RC circuit behavior
Data acquisition methods
Digital multimeters measure DC voltages and resistances in RC circuits
LCR meters directly measure capacitance and equivalent series resistance
Function generators provide controlled input signals for RC circuit testing
Data loggers record long-term voltage or current changes in RC circuits
Computer-based data acquisition systems offer high-speed sampling and analysis
Error analysis in measurements
Component tolerances introduce uncertainties in calculated time constants
Measurement instrument accuracy affects reliability of recorded data
Parasitic capacitance and inductance can influence high-frequency measurements
Temperature variations during measurement may alter component values
Statistical methods (standard deviation, error propagation) quantify measurement uncertainties
Practical considerations
Real-world RC circuits deviate from ideal behavior due to various factors
Understanding these practical considerations is essential for effective circuit design and troubleshooting
Accounting for non-idealities improves accuracy of predictions and measurements
Component tolerances and variations
Resistors and capacitors have manufacturing tolerances (1%, 5%, 10%)
Actual component values may differ from nominal values within tolerance range
Tolerance stack-up in multi-component circuits can significantly affect overall behavior
Temperature coefficients cause component values to change with temperature
Aging effects can alter component values over time, especially in electrolytic capacitors
Temperature effects on RC circuits
Resistor values typically increase with temperature (positive temperature coefficient)
Ceramic capacitors may have positive or negative temperature coefficients
Electrolytic capacitor capacitance and ESR vary significantly with temperature
Temperature changes affect leakage currents in capacitors
Extreme temperatures can cause permanent changes in component characteristics
Real-world limitations and non-idealities
Capacitor leakage current causes slow discharge even when circuit is open
Dielectric absorption in capacitors causes voltage rebound after discharge
Equivalent series resistance (ESR) of capacitors affects high-frequency performance
Skin effect in conductors increases effective resistance at high frequencies
Parasitic inductance in components and PCB traces influences circuit behavior
Electromagnetic interference (EMI) can induce unwanted voltages in RC circuits |
17414 | https://maseconomics.com/fiscal-multiplier-effect-key-to-understanding-economic-stimulus/ | Skip to content
Fiscal Multiplier Effect: Key to Understanding Economic Stimulus
By: Majid Ali Sanghro
Table of Contents
The fiscal multiplier is a fundamental concept in macroeconomic policy, measuring how changes in government spending or taxation affect overall economic output. It captures the link between fiscal actions and aggregate economic activity, helping to evaluate the effectiveness of fiscal interventions, particularly during periods of economic instability.
Understanding the mechanics of the fiscal multiplier, its components, and its real-world applications highlights its importance in economic stimulus and fiscal policy analysis. This concept remains central to shaping strategies for fostering economic recovery and growth.
What Is the Fiscal Multiplier?
The fiscal multiplier measures the proportional change in aggregate output (typically GDP) resulting from an initial change in government spending or taxation. It is expressed mathematically as:
Fiscal Multiplier = ( \frac{\Delta Y}{\Delta G} )
( \Delta Y ) represents the change in aggregate output or GDP.
( \Delta G ) represents the change in government spending.
A multiplier greater than 1 indicates that the economic impact exceeds the initial fiscal outlay, signifying a strong amplification effect. Conversely, a multiplier less than 1 suggests that the policy has a more muted effect on aggregate output.
Positive and Negative Multipliers
Positive Multiplier
Occurs when government spending stimulates additional economic activity. For instance, investment in infrastructure often leads to job creation, increased consumer spending, and further economic growth.
Negative Multiplier
Happens when contractionary fiscal policies, such as tax hikes or spending cuts, reduce aggregate output more than the initial reduction in spending.
Real-World Significance
The size of the fiscal multiplier can vary widely depending on factors such as the state of the economy, the type of government spending, and monetary policy conditions. During recessions, fiscal multipliers tend to be higher because additional spending can mobilize idle resources and boost demand. For example, during the 2008 global financial crisis, fiscal stimulus measures in the United States demonstrated significant multiplier effects, particularly in infrastructure investment.
Breaking Down the Components of the Fiscal Multiplier
Understanding the fiscal multiplier requires analyzing the various factors that determine its size and effectiveness. These components provide insights into why the multiplier varies across economies and policy contexts.
Economic Slack and Resource Utilization
High Unemployment
During periods of economic slack, when resources such as labor and capital are underutilized, fiscal multipliers tend to be larger. Additional government spending can directly employ idle resources, leading to immediate and substantial economic gains.
Full Employment
In economies operating near full capacity, the multiplier effect is smaller because additional spending may lead to inflationary pressures rather than increased output.
For instance, during the Great Depression, fiscal multipliers in the United States were high as large-scale government programs like the New Deal mobilized underused resources.
Types of Government Spending
The composition of government spending significantly affects the size of the multiplier. Some spending types are more effective than others in stimulating economic activity due to differences in their direct and indirect impacts. The following table highlights the estimated multipliers for various spending categories and explains why certain policies yield greater economic benefits:
| Type of Government Spending | Estimated Multiplier | Explanation | Example |
--- --- |
| Infrastructure Investment | 1.5–2.5 | High multiplier due to job creation, demand for materials, and long-term productivity gains. | Construction of highways, renewable energy projects, or broadband infrastructure. |
| Education and Health Spending | 1.2–2.0 | Significant long-term benefits through skill development and healthier workforces. | Investments in schools, hospitals, and vaccination programs. |
| Transfer Payments (Low Income) | 0.8–1.5 | High consumption propensity among low-income households drives immediate demand. | Unemployment benefits, social safety nets, or direct cash transfers to low-income individuals. |
| General Tax Cuts | 0.5–1.0 | Lower multiplier as a portion of tax savings is often saved rather than spent. | Across-the-board income tax reductions. |
| Defense Spending | 0.6–1.0 | Limited multiplier effect due to lower linkages to broader economic activity. | Military equipment purchases. |
| Corporate Tax Cuts | 0.3–0.7 | Low multiplier as businesses may save or invest abroad rather than increasing domestic spending. | Reduction in corporate tax rates. |
| |
This table illustrates that infrastructure investment and spending on education and health typically have higher multipliers due to their ability to create jobs, boost consumption, and enhance productivity. In contrast, measures like corporate tax cuts tend to have lower multipliers as the benefits may not directly translate into increased domestic spending.
Fiscal Policy Design
The design and implementation of fiscal measures also influence the multiplier:
Targeted Spending
Programs that focus on high-impact areas such as renewable energy or public health often achieve greater multiplier effects.
Temporary Measures
Temporary increases in spending or tax cuts tend to have stronger short-term multipliers as they encourage immediate consumption and investment.
Leakages in Open Economies
In open economies, a portion of fiscal stimulus may “leak” into foreign economies through increased imports. For example, if a country invests in infrastructure using imported materials, the multiplier effect is reduced domestically but benefits the exporting country. This phenomenon is particularly relevant for small, open economies with high import propensities.
Interaction with Monetary Policy
The fiscal multiplier’s size is closely linked to monetary policy conditions:
Accommodative Monetary Policy
When central banks keep interest rates low or conduct quantitative easing, the fiscal multiplier is amplified as borrowing becomes cheaper and demand increases.
Tight Monetary Policy
If central banks raise interest rates to counteract fiscal expansion, the multiplier is dampened, reducing the overall economic impact.
For example, during the COVID-19 pandemic, the coexistence of expansive fiscal policies and accommodative monetary measures in advanced economies enhanced the multiplier effect, helping stabilize output.
How Does the Fiscal Multiplier Work in Economic Stimulus?
The fiscal multiplier operates through a chain reaction of economic activities triggered by initial government spending or tax changes. Understanding this process provides insights into its broader implications for economic stimulus.
Initial Injection of Spending
Government spending directly increases aggregate demand. For instance, when a government invests in infrastructure projects, it immediately generates demand for labor, materials, and services. This initial spending forms the foundation of the multiplier effect.
Ripple Effects in the Economy
The initial injection sets off a series of ripple effects:
Direct Effects
Businesses involved in government contracts experience increased revenues and production, leading to job creation and higher wages.
Indirect Effects
Suppliers and subcontractors also benefit as they fulfill contracts and expand operations, further boosting economic activity.
Induced Effects
Newly employed workers and businesses spend their earnings on goods and services, creating additional demand across the economy.
For instance, a $1 billion investment in renewable energy projects can create jobs in construction, stimulate demand for machinery and materials, and lead to increased consumer spending in local economies.
Diminishing Returns and Leakage
The ripple effects diminish over time as some spending leaks out of the domestic economy (e.g., through imports or savings). These leakages limit the total impact of the fiscal multiplier, particularly in open economies where imported goods and services constitute a large share of consumption.
Long-Term Impacts
In addition to short-term effects, fiscal multipliers can have long-term implications for economic growth. Investments in education, healthcare, and infrastructure can enhance productivity and competitiveness, leading to sustained economic benefits.
For example, a government initiative to expand broadband internet access may initially boost demand in the telecommunications sector. Over time, it also increases productivity across industries, improving overall economic output beyond the initial multiplier effect.
Real-World Applications of the Fiscal Multiplier
The American Recovery and Reinvestment Act (2009)
Following the 2008 global financial crisis, the United States implemented a $787 billion stimulus package, the American Recovery and Reinvestment Act (ARRA). The fiscal measures included infrastructure spending, tax cuts, and aid to states. Studies estimated multipliers ranging from 1.5 to 2.0 for infrastructure investments, underscoring their effectiveness in boosting economic output.
The COVID-19 Pandemic Response (2020-2021)
Governments worldwide implemented large-scale fiscal measures to address the economic fallout of the pandemic. For instance:
Direct Cash Transfers
In the United States, stimulus checks provided immediate relief to households, with multipliers estimated at 0.8 to 1.2.
Public Health Investments
Spending on vaccination programs had a higher multiplier due to its critical role in enabling economic recovery.
These examples highlight the variation in multiplier effects depending on the type and context of fiscal measures.
Conclusion
The fiscal multiplier is a key metric for assessing the impact of government spending and taxation on aggregate output. It offers a framework for evaluating the effectiveness of fiscal policies, with its size influenced by factors such as economic conditions, policy design, and monetary settings.
As economies face challenges like financial crises and global disruptions, the fiscal multiplier remains central to fiscal policy analysis. Its insights help shape targeted measures that support growth, stability, and economic resilience.
FAQs:
What is the fiscal multiplier?
The fiscal multiplier measures how changes in government spending or taxation impact overall economic output. It quantifies the relationship between fiscal actions and GDP growth, helping policymakers evaluate the effectiveness of economic stimulus measures.
How do positive and negative multipliers differ?
A positive multiplier occurs when government spending amplifies economic output beyond the initial investment, such as infrastructure projects generating job growth and consumer spending. A negative multiplier arises when fiscal contraction, like tax hikes or spending cuts, reduces output more than the initial policy adjustment.
Why is the fiscal multiplier higher during recessions?
During recessions, resources like labor and capital are underutilized, making fiscal stimulus more impactful. Additional government spending directly engages idle resources, stimulating demand and leading to a stronger multiplier effect compared to stable economic periods.
How do economic conditions affect the fiscal multiplier?
The size of the fiscal multiplier depends on factors like unemployment levels, resource utilization, and monetary policy. High unemployment amplifies the multiplier, while economies operating at full capacity or with tight monetary policy see diminished effects.
How do government spending types influence the fiscal multiplier?
Public investment, such as infrastructure or education spending, typically yields a high multiplier due to long-term productivity gains. Transfer payments, like unemployment benefits, depend on recipients’ spending habits, with larger effects in lower-income groups.
What role does monetary policy play in the fiscal multiplier?
Accommodative monetary policy, such as low interest rates, enhances the fiscal multiplier by encouraging borrowing and spending. Conversely, tight monetary policy can offset fiscal stimulus, reducing its overall effectiveness.
How do leakages affect the fiscal multiplier in open economies?
In open economies, some fiscal stimulus leaks into foreign markets through increased imports, limiting domestic impact. For instance, infrastructure projects reliant on imported materials may reduce the multiplier for the spending country while benefiting exporters.
Can fiscal multipliers have long-term economic impacts?
Yes, investments in areas like education, healthcare, and infrastructure yield long-term benefits by enhancing productivity and competitiveness, extending the multiplier’s impact beyond immediate economic stimulation.
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17415 | https://www.effortlessmath.com/math-topics/how-to-determine-limits-involving-floor-and-absolute-value-functions/?srsltid=AfmBOor7bm6-BGldCzLCP7z6T0qGLxzQTwqd8BqRzrGWnmhLMz5voKU4 | How to Determine Limits Involving Floor and Absolute Value Functions - Effortless Math: We Help Students Learn to LOVE Mathematics
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How to Determine Limits Involving Floor and Absolute Value Functions
Determining limits involving floor functions (which map a real number to the greatest integer less than or equal to it) and absolute value functions (which measure the distance of a number from zero on the real number line) can present unique challenges due to their piecewise and non-continuous nature. Let's traverse through a step-by-step guide to evaluate such limits with precision.
Step-by-Step Guide to Determine Limits Involving Floor and Absolute Value Functions
Here is a step-by-step guide to determining limits involving floor and absolute value functions:
Step 1: Understand the Behavior of Floor and Absolute Value Functions
The floor function, denoted by ⌊𝑥⌋⌊x⌋, can cause abrupt changes in function value at integer points. Meanwhile, the absolute value function, ∣𝑥∣∣x∣, introduces a kink in the graph at 𝑥=0 x=0, changing from 𝑥 x to −𝑥−x as one crosses from positive to negative values. Acknowledge that these points of discontinuity and abrupt change are where we must scrutinize the limit closely.
Step 2: Isolate the Floor and Absolute Value Components
Start by identifying the intervals on which the floor and absolute value functions are continuous and differentiable. This usually means dividing the domain of the function into segments based on integer values for the floor function and positive and negative regions for the absolute value function.
Step 3: Piecewise Analysis
Break down the function into piecewise components. For the floor function, evaluate the limit at points just to the left and just to the right of each integer point. For the absolute value function, assess the behavior as the function approaches zero from both directions.
Step 4: Apply Limit Definitions
Apply the definition of the limit to each piece. For the absolute value, remember that ∣𝑥∣=𝑥∣x∣=x if 𝑥≥0 x≥0 and ∣𝑥∣=−𝑥∣x∣=−x if 𝑥<0 x<0. For the floor function, consider the integer part and the fractional part of 𝑥 x separately.
Step 5: Consider One-Sided Limits
One-sided limits are essential when dealing with floor and absolute value functions. Evaluate 𝑙 𝑖 𝑚 𝑥→𝑐−l i m x→c− and 𝑙 𝑖 𝑚 𝑥→𝑐+l i m x→c+ where 𝑐 c is a point of interest (like an integer for the floor function or zero for the absolute value function). If these one-sided limits do not match, the two-sided limit does not exist.
Step 6: Apply Limits to Each Piecewise Segment
Evaluate the limits for each segment individually. Ensure to consider the impact of the floor and absolute value operations within each interval. Use algebraic manipulation to simplify expressions where possible.
Step 7: Reconcile Piecewise Limits
Once you have the one-sided limits, reconcile them to determine if a two-sided limit exists. If the limits from the left and right do not agree, the overall limit does not exist.
Step 8: Check for Continuity and Differentiability
For locations where the floor or absolute value function changes, assess continuity and differentiability. The limit will exist at a point only if the function is continuous at that point.
Step 9: Handling a Combination of Functions
If the floor or absolute value function is combined with other functions, assess the impact on the limit separately. Sometimes it may be necessary to apply the limit laws in combination with the piecewise evaluation.
Step 10: Visual Verification with Graphs
Graphing the function can provide a visual aid and verification for the analytical findings. The visual approach can be particularly helpful in identifying the behavior of the function around critical points.
Final Word:
The evaluation of limits involving floor and absolute value functions hinges on recognizing and appropriately addressing the piecewise nature of these functions. By isolating these components and meticulously analyzing their behavior across the domain, the evaluation of such limits progresses from piecewise to holistic, resulting in a comprehensive understanding of the function’s limiting behavior.
by: Effortless Math Team about 2 years ago (category: Articles)
Effortless Math Team
3 weeks ago
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17416 | https://www.gauthmath.com/solution/1794080856211462/26-Assume-that-this-sequence-repeats-with-period-four-Write-the-next-four-terms- | Solved: Assume that this sequence repeats with period four. " ” Write the next four terms of the [Math]
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Assume that this sequence repeats with period four. " ” Write the next four terms of the sequence. 7, 3, 5, 7, , , , , …
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$$7, 3, 5, 7$$7,3,5,7
Explanation
Identify the pattern in the sequence. The sequence is given as $$7, 3, 5, 7$$7,3,5,7
Since the sequence repeats with a period of four, the next four terms will be the same as the first four terms.
Write down the next four terms of the sequence, which are $$7, 3, 5, 7$$7,3,5,7
Therefore, the next four terms of the sequence are $$7, 3, 5, 7$$7,3,5,7
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17417 | https://en.wikipedia.org/wiki/Jensen%27s_formula | Jensen's formula - Wikipedia
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1 Formal statement
2 Applications
3 Generalizations
4 See also
5 References
6 Sources
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From Wikipedia, the free encyclopedia
Mathematical formula in complex analysis
Not to be confused with Jensen's inequality.
In complex analysis, Jensen's formula relates the average magnitude of an analytic function on a circle with the number of its zeros inside the circle. The formula was introduced by Johan Jensen(1899.
Formal statement
[edit]
Suppose that f{\displaystyle f} is an analytic function in a region in the complex planeC{\displaystyle \mathbb {C} } which contains the closed diskD r{\displaystyle \mathbb {D} {r}} of radius r>0{\displaystyle r>0} about the origin, a 1,a 2,…,a n{\displaystyle a{1},a_{2},\ldots ,a_{n}} are the zeros of f{\displaystyle f} in the interior of D r{\displaystyle \mathbb {D} _{r}} (repeated according to their respective multiplicity), and that f(0)≠0{\displaystyle f(0)\neq 0}.
Jensen's formula states that
log|f(0)|=−∑k=1 n log(r|a k|)+1 2 π∫0 2 π log|f(r e i θ)|d θ.{\displaystyle \log |f(0)|=-\sum {k=1}^{n}\log \left({\frac {r}{|a{k}|}}\right)+{\frac {1}{2\pi }}\int {0}^{2\pi }\log |f(re^{i\theta })|\,d\theta .}
This formula establishes a connection between the moduli of the zeros of f{\displaystyle f} in the interior of D r{\displaystyle \mathbb {D} {r}} and the average of log|f(z)|{\displaystyle \log |f(z)|} on the boundary circle |z|=r{\displaystyle |z|=r}, and can be seen as a generalisation of the mean value property of harmonic functions. Namely, if f{\displaystyle f} has no zeros in D r{\displaystyle \mathbb {D} _{r}}, then Jensen's formula reduces to
log|f(0)|=1 2 π∫0 2 π log|f(r e i θ)|d θ,{\displaystyle \log |f(0)|={\frac {1}{2\pi }}\int _{0}^{2\pi }\log |f(re^{i\theta })|\,d\theta ,}
which is the mean-value property of the harmonic function log|f(z)|{\displaystyle \log |f(z)|}.
An equivalent statement of Jensen's formula that is frequently used is
1 2 π∫0 2 π log|f(r e i θ)|d θ−log|f(0)|=∫0 r n(t)t d t{\displaystyle {\frac {1}{2\pi }}\int {0}^{2\pi }\log |f(re^{i\theta })|\,d\theta -\log |f(0)|=\int {0}^{r}{\frac {n(t)}{t}}\,dt}
where n(t){\displaystyle n(t)} denotes the number of zeros of f{\displaystyle f} in the disc of radius t{\displaystyle t} centered at the origin.
Proof
It suffices to prove the case for r=1{\displaystyle r=1}.
If f{\displaystyle f} contains zeros on the circle boundary, then we can define g(z)=f(z)∏k(z−e i θ k){\displaystyle g(z)={\frac {f(z)}{\prod {k}(z-e^{i\theta {k}})}}}, where e i θ k{\displaystyle e^{i\theta {k}}} are the zeros on the circle boundary. Since ∫0 2 π ln|e i θ−e i θ k|d θ=2∫0 π ln(2 sinθ)d θ=0,{\displaystyle \int {0}^{2\pi }\ln |e^{i\theta }-e^{i\theta {k}}|d\theta =2\int {0}^{\pi }\ln(2\sin \theta )d\theta =0,} we have reduced to proving the theorem for g{\displaystyle g}, that is, the case with no zeros on the circle boundary.
Define F(z):=f(z)∏k=1 n(z−a k){\displaystyle F(z):={\frac {f(z)}{\prod {k=1}^{n}(z-a{k})}}} and fill in all the removable singularities. We obtain a function F{\displaystyle F} that is analytic in B(0,1+ϵ){\displaystyle B(0,1+\epsilon )}, and it has no roots in B(0,1){\displaystyle B(0,1)}.
Since log|F|=R e(logF){\displaystyle \log |F|=Re(\log F)} is a harmonic function, we can apply Poisson integral formula to it, and obtain log|F(0)|=1 2 π∫0 2 π log|F(e i θ)|d θ=1 2 π∫0 2 π log|f(e i θ)|d θ−∑k=1 n 1 2 π∫0 2 π log|e i θ−a k|d θ.{\displaystyle \log |F(0)|={\frac {1}{2\pi }}\int {0}^{2\pi }\log |F(e^{i\theta })|\,d\theta ={\frac {1}{2\pi }}\int {0}^{2\pi }\log |f(e^{i\theta })|\,d\theta -\sum {k=1}^{n}{\frac {1}{2\pi }}\int {0}^{2\pi }\log |e^{i\theta }-a_{k}|\,d\theta .} where ∫0 2 π log|e i θ−a k|d θ{\displaystyle \int {0}^{2\pi }\log |e^{i\theta }-a{k}|\,d\theta } can be written as ∫0 2 π log|e i θ−a k|d θ=∫0 2 π log|1−a k e−i θ|d θ=R e∫0 2 π log(1−a k e−i θ)d θ.{\displaystyle \int {0}^{2\pi }\log |e^{i\theta }-a{k}|\,d\theta =\int {0}^{2\pi }\log |1-a{k}e^{-i\theta }|\,d\theta =Re\int {0}^{2\pi }\log(1-a{k}e^{-i\theta })\,d\theta .}
Now, ∫0 2 π log(1−a k e−i θ)d θ{\displaystyle \int {0}^{2\pi }\log(1-a{k}e^{-i\theta })\,d\theta } is a multiple of a contour integral of function log(1−z)/z{\displaystyle \log(1-z)/z} along a circle of radius |a k|<1{\displaystyle |a_{k}|<1}. Since log(1−z)/z{\displaystyle \log(1-z)/z} has no poles in B(0,|a k|){\displaystyle B(0,|a_{k}|)}, the contour integral is zero.
Applications
[edit]
Jensen's formula can be used to estimate the number of zeros of an analytic function in a circle. Namely, if f{\displaystyle f} is a function analytic in a disk of radius R{\displaystyle R} centered at z 0{\displaystyle z_{0}} and if |f|{\displaystyle |f|} is bounded by M{\displaystyle M} on the boundary of that disk, then the number of zeros of f{\displaystyle f} in a circle of radius r<R{\displaystyle r<R} centered at the same point z 0{\displaystyle z_{0}} does not exceed
1 log(R/r)logM|f(z 0)|.{\displaystyle {\frac {1}{\log(R/r)}}\log {\frac {M}{|f(z_{0})|}}.}
Jensen's formula is an important statement in the study of value distribution of entire and meromorphic functions. In particular, it is the starting point of Nevanlinna theory, and it often appears in proofs of Hadamard factorization theorem, which requires an estimate on the number of zeros of an entire function.
Jensen's formula is also used to prove a generalization of Paley-Wiener theorem for quasi-analytic functions with r→1{\displaystyle r\rightarrow 1}. (in particular: spectral factorization methods) this generalization is often referred to as the Paley–Wiener condition.
Generalizations
[edit]
Jensen's formula may be generalized for functions which are merely meromorphic on D r{\displaystyle \mathbb {D} _{r}}. Namely, assume that
f(z)=z l g(z)h(z),{\displaystyle f(z)=z^{l}{\frac {g(z)}{h(z)}},}
where g{\displaystyle g} and h{\displaystyle h} are analytic functions in D r{\displaystyle \mathbb {D} {r}} having zeros at a 1,…,a n∈D r∖{0}{\displaystyle a{1},\ldots ,a_{n}\in \mathbb {D} {r}\setminus {0}} and b 1,…,b m∈D r∖{0}{\displaystyle b{1},\ldots ,b_{m}\in \mathbb {D} _{r}\setminus {0}} respectively, then Jensen's formula for meromorphic functions states that
log|g(0)h(0)|=log|r m−n−l a 1…a n b 1…b m|+1 2 π∫0 2 π log|f(r e i θ)|d θ.{\displaystyle \log \left|{\frac {g(0)}{h(0)}}\right|=\log \left|r^{m-n-l}{\frac {a_{1}\ldots a_{n}}{b_{1}\ldots b_{m}}}\right|+{\frac {1}{2\pi }}\int {0}^{2\pi }\log |f(re^{i\theta })|\,d\theta .}
Jensen's formula is a consequence of the more general Poisson–Jensen formula, which in turn follows from Jensen's formula by applying a Möbius transformation to z{\displaystyle z}. It was introduced and named by Rolf Nevanlinna. If f{\displaystyle f} is a function which is analytic in the unit disk, with zeros a 1,a 2,…,a n{\displaystyle a{1},a_{2},\ldots ,a_{n}} located in the interior of the unit disk, then for every z 0=r 0 e i φ 0{\displaystyle z_{0}=r_{0}e^{i\varphi _{0}}} in the unit disk the Poisson–Jensen formula states that
log|f(z 0)|=∑k=1 n log|z 0−a k 1−a¯k z 0|+1 2 π∫0 2 π P r 0(φ 0−θ)log|f(e i θ)|d θ.{\displaystyle \log |f(z_{0})|=\sum {k=1}^{n}\log \left|{\frac {z{0}-a_{k}}{1-{\bar {a}}{k}z{0}}}\right|+{\frac {1}{2\pi }}\int {0}^{2\pi }P{r_{0}}(\varphi _{0}-\theta )\log |f(e^{i\theta })|\,d\theta .}
Here,
P r(ω)=∑n∈Z r|n|e i n ω{\displaystyle P_{r}(\omega )=\sum _{n\in \mathbb {Z} }r^{|n|}e^{in\omega }}
is the Poisson kernel on the unit disk. If the function f{\displaystyle f} has no zeros in the unit disk, the Poisson-Jensen formula reduces to
log|f(z 0)|=1 2 π∫0 2 π P r 0(φ 0−θ)log|f(e i θ)|d θ,{\displaystyle \log |f(z_{0})|={\frac {1}{2\pi }}\int {0}^{2\pi }P{r_{0}}(\varphi _{0}-\theta )\log |f(e^{i\theta })|\,d\theta ,}
which is the Poisson formula for the harmonic function log|f(z)|{\displaystyle \log |f(z)|}.
See also
[edit]
Paley–Wiener theorem
References
[edit]
^ a0-07-000657-1. OCLC4036464.
^Paley & Wiener 1934, pp.14–20.
^Sayed & Kailath 2001, pp.469–470.
Sources
[edit]
Ahlfors, Lars V. (1979), Complex analysis. An introduction to the theory of analytic functions of one complex variable, International Series in pure and applied Mathematics (3rd ed.), Düsseldorf: McGraw–Hill, ISBN0-07-000657-1, Zbl0395.30001
Jensen, J. (1899), "Sur un nouvel et important théorème de la théorie des fonctions", Acta Mathematica (in French), 22 (1): 359–364, doi:10.1007/BF02417878, ISSN0001-5962, JFM30.0364.02, MR1554908
Paley, Raymond E. A. C.; Wiener, Norbert (1934). Fourier Transforms in the Complex Domain. Providence, RI: American Mathematical Soc. ISBN978-0-8218-1019-4.{{cite book}}: ISBN / Date incompatibility (help)
Ransford, Thomas (1995), Potential theory in the complex plane, London Mathematical Society Student Texts, vol.28, Cambridge: Cambridge University Press, ISBN0-521-46654-7, Zbl0828.31001
Sayed, A. H.; Kailath, T. (2001). "A survey of spectral factorization methods". Numerical Linear Algebra with Applications. 8 (6–7): 467–496. doi:10.1002/nla.250. ISSN1070-5325.
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Where to start and finish, and bridges
Kudos to authors for monetizing some very basic graph theory (the topic here is Eulerian circuits and Eulerian trails). Check how many edges sticks out of each vertex. That’s called the “degree” of a vertex. If all vertices have even degrees, then choose any vertex as both your start and finish. If exactly 2 vertices have odd degrees, then those 2 vertices are your start and finish. In that second case (different start and finish), pay attention to vertices and edges called “bridges”. Those are vertices and edges which must be on any path from the start to the finish, whether it contains all edges or not, they’re unavoidable in getting from here to there. Mentally take away those edges and cut these vertices in half, and you’ll have some “islands” (those are called “blocks”). Make sure you use up every edge of each island before you cross the bridge to the next island. Oh, and double edges (they’re dotted) count twice, etc.
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Really fun addicting game. I was enjoying it so much that I decided to go ahead and buy the paid version of the game to remove the ads. I was really enjoying the game without the little pop ups on the bottom or between levels, but found that once I finished about five levels, an ad would pop up right over the “continue” button to watch a video and get a free hint. I have hit that pop up on accident so many times because of its short delay and being right over the button I’m trying to press that I have gotten very frustrated. When I paid to remove the ads, I wanted them ALL gone. A pop up to view an ad is just as bad as an ad itself, especially when it tricks you into clicking on it. Then you have to navigate back to your level and the meditative state from focusing on the game gets totally ruined. The game is good if you don’t mind ads, but DONT WASTE YOUR MONEY trying to get rid of them. That pop up to watch a video after every few levels will alway still be there. Very disappointing. WOULD NOT PURCHASE AGAIN
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17419 | https://www.edulyte.com/maths/locus/ | 2D Shapes2cosacosb Formula30-60-90 Formulas3D ShapesAbsolute Value FormulaAcute AngleAcute Angle triangleAdditionAlgebra FormulasAlgebra of MatricesAlgebraic EquationsAlgebraic ExpressionsAngle FormulaAnnulusAnova FormulaAnti-derivative FormulaAntiderivative FormulaApplication of DerivativesApplications of IntegrationArc Length FormulaArccot FormulaArctan FormulaArea Formula for QuadrilateralsArea FormulasArea Of A Sector Of A Circle FormulaArea Of An Octagon FormulaArea Of Isosceles TriangleArea Of ShapesArea Under the Curve FormulaArea of RectangleArea of Regular Polygon FormulaArea of TriangleArea of a Circle FormulaArea of a Pentagon FormulaArea of a Square FormulaArea of a Trapezoid FormulaArithmetic Mean FormulaArithmetic ProgressionsArithmetic Sequence Recursive FormulaArithmetic and Geometric ProgressionAscending OrderAssociative Property FormulaAsymptote FormulaAverage Deviation FormulaAverage Rate of Change FormulaAveragesAxioms Of ProbabilityAxis of Symmetry FormulaBasic Math FormulasBasics Of AlgebraBinary FormulaBinomial Probability FormulaBinomial Theorem FormulaBinomial distributionBodmas RuleBoolean AlgebraBusiness MathematicsCalculusCelsius FormulaCentral Angle of a Circle FormulaCentral Limit Theorem FormulaCentroid of a Trapezoid FormulaChain RuleChain Rule FormulaChange of Base FormulaChi Square FormulaCirclesCircumference FormulaCoefficient of Determination FormulaCoefficient of Variation FormulaCofactor FormulaComplete the square formulaComplex numbersCompound Interest FormulaConditional Probability FormulaConeConfidence Interval FormulaCongruence of TrianglesCorrelation Coefficient FormulaCos Double Angle FormulaCos Square theta FormulaCos Theta FormulaCosec Cot FormulaCosecant FormulaCosine FormulaCovariance FormulaCubeCurated Maths Resources for Teachers – EdulyteCylinderDecimalsDifferential calculusDiscover the world of MathsEllipseEquilateral triangleEuler’s formulaEven numbersExponentsFibonacci TheoryFractionFraction to decimalGeometric sequenceHeptagonHyperbolaIntegersIntegrationIntegration by partsLinesLocusMatricesNatural numbersNumber lineOdd numbersParallelogramPercentage formulaPerimeterPolygonPolynomialsPrismProbabilityPyramidPythagoras theoremRoman NumeralsScalene triangleSetsShapes NamesSimple interest formulaSlope formulaSolid shapesSphereSquareStandard deviation formulaSubtractionSymmetryTimeTrianglesTrigonometry formulaTypes of anglesValue of PiVariance formulaVectorVolume formulasVolume of a coneVolume of sphere formulaWhole numbers
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Introduction to Locus
A Locus is an important term of mathematics as it is considered a curve or a different type of shape made by all the points joining a particular equation of the relation between its coordinates or through the point or the line or the moving surface. All these shapes, such as the ellipse, a circle, or a parabola, can be defined through the locus as a set of points.
In mathematics, A locus is a particular set of points that can satisfy some given condition or property and is a plurality considered Loci. These are sets of points at a distance similar to another point and satisfy a particular given equation. Locus meaning in Hindi is also available online.
Significance in Mathematics
The Locus in maths or Loci has effective importance, and it includes:
It is a type of curve or a shape that is made through a point satisfying a particular equation.
The locus of points is directly equidistant from a fixed point at the centre.
It consists of a perpendicular bisector, which is the locus of points that is equidistant from two fixed points, A and B.
Defining Locus (Loci)
Locus definition or Loci definition in mathematics can be termed as a shape or a particular shape made through the points that satisfy a particular equation of the relation among the coordinates through another point or line moving through the surface. All types of shapes, such as a parabola, a circle, or a hyperbola, can be defined through the locus as a set of points. A Locus can be termed a Loci and is a commonly used mathematical term that you can relatively use.
The locus is a particular set of points that helps to determine and satisfy your particular equation of the relation among the coordinates and the different points of the shape. It is the primary reason for you to use a locus.
Locus in Mathematics
Equation of Locus
The equation of the locus or loci is different from a regular geometrical equation as you need to derive the shape equation according to certain properties. The process by which you can derive the locus equation is:
First, you have to assume the coordinates of the moving point of the locus as (X1, Y1)
Then, you are to apply the geometrical condition on the two selected moving points, which helps you to obtain the relation between these two points.
In your result, you should replace X1 with x and Y1 with y.
The equation that you obtain after this will be your locus equation.
Examples
Common Loci Circles: x2+y2=2(x+y)
Find the equation of a point’s locus so that the sum of its distances from (0, -1) and (0,1) is 3. What exactly does the equation mean?
Solution
Suppose P(x, y) be a point on the given locus.
It is given that
A = (0, −1)
B = (0, 1)
From the conditions given,
Distance from Point P to A + Distance from Point P to B = 3
Hence, we can write it as
PA + PB = 3
By rearranging the above equation,
PA = 3 − PB
Now, Squaring both sides of the equation,
PA2=(3−PB)2
PA2 = 9 + PB2 − 6PB
x2+(y+1)2 = 9+x2+(y−1)2−6PB
(y+1)2−(y−1)2−9=−6PB
4y−9 = −6PB
Again, squaring both sides,
16y2−72y+81 = 36PB2
16y2−72y+81 = 36(x2+(y−1)2)
16y2−72y+81 = 36x2+36y2−72y+36
36x2+20y2 = 45
The above equation is written as follows
(x2/a2)+(y2/b2)=1
Hence, the above equation defines a circle
Therefore, the equation of locus is,
36x2+20y2 = 45, which is a circle.
Common Loci Lines: Y = MX + C
Find the locus of a point which is at a distance of 5 units from A(4,-3).
Solution:
Let P(x,y) be a point on the locus.
The given geometric condition is,
PA=5
√[(x-4)2 +(y+3)2] = 5
Squaring on both sides,
(x-4)2 +(y+3)2=25
x2 +16-8x+y2+9+6y=25
x2+y2-8x+6y=0 is the equation of locus.
Common Loci Parabola: y2+2ax+2by+c=0
Find the locus of a point which is equidistant from the points A(-3,2) and B(0,4).
Solution:
Let P(x,y) be a point on the locus.
The given geometric condition is,
PA=PB
√[(x+3)2 +(y-2)2] = √[(x-0)2 +(y-4)2]
Squaring on both sides,
(x+3)2 +(y-2)2 = (x-0)2 +(y-4)2
x2 +9+6x+y2+4-4y= x2 +y2+16-8y
Simplifying the above equation,
6x + 4y = 3, which is the required equation of locus.
Locus in Real-life Applications
The locus concept is important in real-world situations, and the special areas where it has great significance are:
Engineering: It is the sector in which the locus has an important role to play as the calculation of curves and locus points is very important in civil engineering and buildings, so calculating it in that way becomes very effective. Engineering involves the calculation of the common loci.
Physics: Both maths and physics have a joint connection. The locus plays an important role in physics, and effective calculations are made through the locus. Many physics theoreticians have used locus and their theorems, and there is indeed an important part.
Architecture: The field of architecture involves the creation of buildings and bridges, which directly involves the usage of calculations of lines and shapes. Using the locus to calculate the differences and curves is important to calculate the exact length, which will help in architecture.
Variation and Extension of Locus
Various advanced topics are related to the locus and are important in calculating the locus. These topics are:
Polar Coordinates
One of the most effective variations of a locus is its polar coordinates, which is a coordinating system in which the position of a point is given by a radical distance from an origin and the angle you can measure counterclockwise from a horizontal line through a polar axis. These are further concepts of the locus, and calculating the line segment becomes much easier through this.
Conic Sections
A Conic section is the locus of a point that moves so that the distance from a fixed point always bears a constant ratio to its distance from another line, all in the same plane.
Parametric Equations
The parametric equation is a primary extension of the locus where the extension is from the two horizontal lines, one on each side of the x-axis, and the distance of each horizontal line from the x-axis is represented in the form of ‘d’ units. The calculation of the locus of the shape becomes very easy by using the parametric equations.
These are the three primary extensions of the locus, which extend the entire concept of locus, and using the locus of a shape becomes very easy while you keep in mind the three variations of the locus.
Challenges and Problem-solving with Locus
There are certain challenges or problems that you can face while using the locus of the equation. These are:
Determining the locus of a point is very difficult as it involves a lot of calculations.
Using the proper coordinates for the calculation can take time and effort.
Determining the dimension of the geometrical figure is a necessary part where you might face difficulty in calculating the locus of a point.
The only way you can solve all these challenges and problems is to follow the rules of calculating the locus and form an equation in a manner that will help you to calculate.
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Key Takeaways
A locus is considered a curve or a shape that joins through an equation.
The plural form of A locus is Loci.
It consists of many variations, making it a very special part of mathematical calculation when it is involved in making architecture and Engineering calculations very easy.
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Frequently Asked Questions
How Can I Determine the Equation of A Locus?
There are certain steps that you have to follow while determining the equation of A locus, and these are:
Assume the coordinates of the moving point of the locus as (X1, Y1)
Apply the geometrical condition on the two selected moving points, which helps you to obtain the relation between these two points.
In your result, you should replace X1 with x and Y1 with y.
The equation that you obtain after this will be your locus equation.
How is Locus Used in Real-life Applications?
The Lokesh has a primary usage in real life, and the areas where you can use it are:
Architecture
Engineering
Physics
Are There Variations or Extensions of the Concept of Locus?
There are specific extensions and variations of A locus that make it a much more innovative mathematical concept. These are:
Parametric Equations
Conic Sections
Polar Coordinates
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Table 1 from the Common Core Standards and Tape Diagrams ------------------------------------------------------------------------------------------------------------------------------------------------------------------------ 3/21/2014 1 Comment Tape Diagrams- A tool to make sense of problems ----------------------------------------------- When I first read through the sample problems illustrating the different variation of addition and subtraction problems (Table 1 in the Appendix of the Mathematics CCSS), it struck me that the majority of these problems do not contain the "key" words that we teach students to look for when solving a word problem. I also noticed that several of the problems expressed the quantities in words instead of digits. Coincidence?... I don't think so. Consider the following two examples from Table 1: 1) Julie has threemore apples than Lucy. Lucy has two apples. How many apples does Julie have? 2) Julie has threemore apples than Lucy. Julie has five apples. How many apples does Lucy have? If a student attacked these problems using key words, they would think that they should perform the same operation to solve both problems. However, this is not the case. In the first problem, Lucy's two apples plus the three more that Julie has gives you Julie's number of apples. In the second problem subtracting the three more that Julie has from Julie's five apples gives you Lucy's number of apples. To solve the problem the students must actually be able to make sense of how these numbers are related. So how do we teach students to make sense of problems so they can persevere in solving them? Tape diagrams can be a great tool for representing these relationships (part/whole or comparison). In the attached document, you will see each of these problem types modeled with a tape diagram. The tape diagram is not an end in itself. It is a tool to help students see that while there are many different variations in wording, there are only two basic relationships in their addition and subtraction word problems (part/whole and comparison). If a student can picture these relationships, they will also have a great foundation from which we can build their understanding of multiplication and division word problems. The models will highlight the fact that multiplication and division involve equal parts. The tape diagrams also sets the stage for understanding the difference between additive comparison (3 more) and multiplicative comparison (3 times as many). The file below shows each problem within Table 1 modeled with a tape diagram. table_1_tape_diagrams.docx File Size:233 kb File Type:docx Download File 1 Comment furniture singaporelink 9/26/2016 11:52:09 pm When I first read through the sample problems illustrating the different variation of addition and subtraction problems. Reply Leave a Reply. --------------Why Would I Want to Use That? ----------------------------- With so many mathematical models and tools, how do you decide which ones are worth investigating? We have created this blog to help faciliate that discussion. Please be sure to enter your email address and then click on "subscribe" in the bottom right hand corner of this page to receive email updates from this page. Archives -------- March 2014 December 2012 September 2012 August 2012 Categories ---------- All 1st Grade 2nd Grade 3rd Grade 5th Grade 6-8 Grades Bar Model K Pre K Pre-K Problem Solving Rekenrek Singapore Enter your email address: Delivered by FeedBurner RSS Feed |
17421 | https://www.doubtnut.com/qna/14163078 | A steel tape is correctly calibrated at 20∘C and is used to measure the length of a table at 30∘C. Find the percentage error in the measurement of length. [αsteel=11×10−5/.∘C]
The correct Answer is:1.1×10−2
To solve the problem, we need to find the percentage error in the measurement of length when using a steel tape that is calibrated at 20°C to measure a length at 30°C. 1. Identify the Coefficient of Linear Expansion: The coefficient of linear expansion for steel is given as: αsteel=11×10−5°C−1 2. Determine the Temperature Change: The temperature change (ΔT) when measuring from 20°C to 30°C is: ΔT=Tfinal−Tinitial=30°C−20°C=10°C 3. Calculate the Change in Length (ΔL): The change in length (ΔL) due to thermal expansion can be calculated using the formula: ΔL=L⋅α⋅ΔT where L is the original length measured at 20°C. Since we are looking for the percentage error, we will express ΔL in terms of L. Therefore: ΔL=L⋅(11×10−5)⋅10 ΔL=L⋅1.1×10−4 4. Calculate the Percentage Error: The percentage error in the measurement of length is given by: Percentage Error=(ΔLL)×100 Substituting ΔL from the previous step: Percentage Error=(L⋅1.1×10−4L)×100 Percentage Error=1.1×10−4×100 Percentage Error=0.011×100=0.11% Final Answer: The percentage error in the measurement of length is 0.11%.
To solve the problem, we need to find the percentage error in the measurement of length when using a steel tape that is calibrated at 20°C to measure a length at 30°C.
Identify the Coefficient of Linear Expansion:
The coefficient of linear expansion for steel is given as:
αsteel=11×10−5°C−1
Determine the Temperature Change:
The temperature change (ΔT) when measuring from 20°C to 30°C is:
ΔT=Tfinal−Tinitial=30°C−20°C=10°C
Calculate the Change in Length (ΔL):
The change in length (ΔL) due to thermal expansion can be calculated using the formula:
ΔL=L⋅α⋅ΔT
where L is the original length measured at 20°C. Since we are looking for the percentage error, we will express ΔL in terms of L.
Therefore:
ΔL=L⋅(11×10−5)⋅10
ΔL=L⋅1.1×10−4
Calculate the Percentage Error:
The percentage error in the measurement of length is given by:
Percentage Error=(ΔLL)×100
Substituting ΔL from the previous step:
Percentage Error=(L⋅1.1×10−4L)×100
Percentage Error=1.1×10−4×100
Percentage Error=0.011×100=0.11%
Final Answer:
The percentage error in the measurement of length is 0.11%.
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Knowledge Check
A steel taps is calibrated at 200C. When the temperature of the day is −100C, the percentage error in the measurement with the tap is (α=12×10−6/0C)
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Solving One Step Equations Worksheet Homework 6th Grade Math | FREE
Created by
Math with Ms. Rivera
A key skill for Algebra is solving equations. Help 6th grade students review the basics with this worksheet on solving one-step equations. There are 8 questions that cover all four operations - addition, subtraction, multiplication, and division one-step equations. This resource includes an answer key! Here is what's included: ➡️ 8 questions ➡️ answer key This is a great no prep worksheet for independent practice, homework, or for a sub plan. There is plenty of space for students to show work
5th - 6th
Algebra, Math
CCSS
6.EE.B.5
, 6.EE.B.7
FREE
Rated 4.77 out of 5, based on 70 reviews
4.8 (70)
Two Step Equations Worksheet Digital Activity 7th Grade Math
Created by
The Sassy Math Teacher
This NO PREP Two Step Equations Worksheet is a great way to have your students practice solving two-step equations! The best part? The worksheet is self-checking! As students answer the questions, they will look for their answer on one of the lollipops! Once the answer is located, they color it! A printable PDF and a digital Google Slides and Microsoft PowerPoint version is included. This is part of my Candy Color line of resources, check them out!This product includes the following:Teacher
6th - 8th
Algebra, Math
CCSS
6.EE.B.7
, 7.EE.B.3
, 8.EE.C.7
Also included in: Solving Two Step Equations Guided Notes - Practice and Activities 7th Grade
FREE
Rated 4.75 out of 5, based on 52 reviews
4.8 (52)
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Solving One Step Equations Color by Number 6th Grade Math Worksheets
Cognitive Cardio Math
$3.95Original Price $3.95
Rated 4.84 out of 5, based on 461 reviews
4.8 (461)
ONE-STEP EQUATIONS Maze, Riddle, Coloring | Google Classroom | Print and Digital
Exceeding the CORE
$3.75Original Price $3.75
Rated 4.8 out of 5, based on 622 reviews
4.8 (622)
Solving Equations Practice Coloring Activity | One Step Equations for 6th Grade
The Meaningful Teacher
$4.15Original Price $4.15
Rated 4.83 out of 5, based on 98 reviews
4.8 (98)
Solving One Step Equations Digital Math Escape Room 6th Grade Activity
Scaffolded Math and Science
$3.00Original Price $3.00
Rated 4.8 out of 5, based on 196 reviews
4.8 (196)
Two Step Equations Word Problems Worksheet Practice Activity
Created by
Mile High Mathematics
An excellent activity to help you teach two-step/multi-step equation word problems! There are 6 questions in total! Students will have to interpret the context of the real-world problem and then make an equation to solve. It is an excellent resource for students to use as they practice their mastery of this topic, and can also serve extremely well as a supplemental post-lesson activity! An answer key is included towards the end of the PDF document.If you liked this worksheet, please check out m
6th - 8th
Algebra, Math
CCSS
6.EE.B.5
, 6.EE.B.6
, 6.EE.B.7
FREE
Rated 4.79 out of 5, based on 33 reviews
4.8 (33)
One & Two Step Equations Pixel Art Activity
Created by
Algebra with Ms B
Are you looking for a quick, no prep activity to help your students practice or review solving one and two-step equations? This activity includes solving equations using both positive and negative integers. As students work through the problems, if their solution is correct, the box will turn green and part of the picture will appear. If they are incorrect, the answer box turns red, indicating they need to check their work. ⭐️ SHOP ALL PIXEL ART ACTIVITIES ⭐️ Included:Self-Checking Google She
6th - 8th
Algebra, Graphing, Math
CCSS
6.EE.B.5
, 6.EE.B.7
, 8.EE.C.7
+1
FREE
Rated 4.83 out of 5, based on 48 reviews
4.8 (48)
6th Grade Math Spiral Review: Free Daily Practice Activities or Homework (2 wks)
Created by
Differentiated Teaching with Rebecca Davies
This free trial of my 6th grade math spiral review bundle has both printable & digital options and is perfect for offering short, consistent practice opportunities and helping students build the confidence they need to be successful on state assessments and in middle school math. Designed to take less than 15 minutes per day - including giving students time to complete, discuss, and review, this spiral math is low prep and easy to implement. Each week's work focuses on building mastery toward
6th
Math Test Prep
CCSS
6.G.A.1
, 6.G.A.2
, 6.G.A.3
+50
FREE
Rated 4.82 out of 5, based on 45 reviews
4.8 (45)
Solving One Step Equations Color by Number Activity FREE
Created by
Absolute Algebra
This free fun color by number resource is all about solving one-step equations. Your middle school students will have two activities to complete. One is solving using only whole numbers and the other uses integers. Students will find their answer and the corresponding color in the answer bank and then color the corresponding problem number! Answers are included. This resource includes: 2 differentiated solving one-step equation activities...Activity 1 (whole numbers) and Activity 2 (integers)Co
5th - 6th
Algebra, Basic Operations, Math
CCSS
6.EE.B.7
Also included in: Solving Equations and Inequalities Middle School Bundle
FREE
Rated 4.71 out of 5, based on 43 reviews
4.7 (43)
6th Grade, 7th Grade 8th Grade Math Resources ~ Milestone FREEBIE
Created by
Exceeding the CORE
6th Grade, 7th Grade, 8th Grade Math Resources ~ Milestone FREEBIE - Task Cards, Graphic Organizers, Error Analysis, Maze I have reached a major milestone on TpT and I would like to thank all of my loyal customers for making this amazing dream come true! I have enjoyed sharing my teaching materials with you, and I'm thrilled to celebrate this milestone with all of you. This FREEBIE includes 4 task cards, 3 problem solving graphic organizers, 3 error analysis activities, and 1 maze (over 25 s
6th - 8th
Math, Math Test Prep, Other (Math)
CCSS
8.G.A.5
, 7.SP.C.5
, 7.SP.C.6
+17
FREE
Rated 4.89 out of 5, based on 74 reviews
4.9 (74)
One-Step Equations Pixel Art - Fall & Halloween Digital Math Activity
Created by
Coffee and Calculations
This is a Google Sheets product. Paperless. No Prep. Students will work to solve 20 one-step algebraic equations including operations with negatives. This product also includes a printable answer sheet for students to work out each problem. Correct answers will be shown by the answer box turning from red to green. Part of the picture will be revealed with each correct answer. When all 20 questions are correct, all answer boxes will be green and the entire picture will be revealed. Incorrect an
6th - 8th
Algebra, Math
CCSS
6.EE.B.7
, 7.EE.A.1
, 7.EE.B.4
+1
Also included in: 8th Grade Math - Halloween Activity Review Bundle - Equations
FREE
Rated 4.74 out of 5, based on 19 reviews
4.7 (19)
One-Step Equations Scavenger Hunt (Multiplication & Division, No Negatives)
Created by
Mile High Mathematics
An amazing in-class learning activity that contains 24 total questions and helps students build their proficiency with one-step equations. This resource deals with one-step equations that are MULTIPLICATION AND DIVISION ONLY as well as NO NEGATIVES in which students will have to use their knowledge of inverse operations to isolate and solve for the given variable. This is a perfect introductory activity to solving one-step equations algebraically! It is an excellent resource for students to use
5th - 8th
Algebra, Basic Operations, Math
CCSS
6.EE.B.5
, 6.EE.B.6
, 6.EE.B.7
Also included in: One-Step Equations Scavenger Hunt Bundle (No Negatives)
FREE
Rated 4.84 out of 5, based on 34 reviews
4.8 (34)
Solving One-Step Addition and Subtraction Equations
Created by
Middle School Math Warehouse
Solve addition and subtraction equations to solve the riddle! Answer key included. Aligns with Pearson digits 6 lesson 3-3.
5th - 7th
Basic Operations, Numbers, Order of Operations
CCSS
6.EE.B.7
FREE
Rated 4.71 out of 5, based on 56 reviews
4.7 (56)
Writing and Solving One step Equations from Word Problems-Matching
Created by
Rachel Hovermale
This resource includes 16 word problems. Students will work on writing and solving one step equations from word problems. Students will practice by matching the word problem to the equation and solving the equation to match the solution to the equation and word problem.
5th - 7th
Math, Other (Math)
CCSS
6.EE.B.5
, 6.EE.B.6
, 6.EE.B.7
FREE
Rated 4.68 out of 5, based on 28 reviews
4.7 (28)
Algebra Solve One Step Equations Real Word Problems Practice Quiz Worksheet
Created by
StickyStickDesigns
Tackle One-Step Real Word Problems for algebra with this worksheet with a single variable. This is a two worksheet set that comes with 5 unique word problems on each page. Designed to strengthen problem-solving skills, this resource includes: One-Step Equation Problems: Students will be required to write a linear equation given a real world scenario and then answer the question. Include solving equations that require one step, such as addition, subtraction, multiplication, and division.Two Pag
6th - 8th
Algebra, Math, Other (Math)
CCSS
6.EE.B.6
, 6.EE.B.7
, 7.EE.A.2
+3
FREE
Rated 5 out of 5, based on 6 reviews
5.0 (6)
Algebra 1 Math Review Math Test Prep FREEBIE Math Worksheets 6th 7th 8th Grade
Created by
Math Shop
Algebra 1 Math Review Math Test Prep FREEBIE Math Worksheets 6th 7th 8th Grade Algebra FREEBIE Algebra Workbook Printable Worksheets. ✦✦✦ GET THE FULL RESOURCE HERE. ✦✦✦ This is a free sample of a pack of 22 worksheets of algebra questions. The full resource has 141 questions. It can be used as a booklet or as individual worksheets. All answers are included.The following algebra topics are included: - Simplifying Algebraic ExpressionsExpanding Algebraic ExpressionsFactoring Algebraic Express
6th - 8th
Algebra, Math, Math Test Prep
CCSS
6.EE.A.2
, 6.EE.A.2c
, 6.EE.A.3
+13
Also included in: Algebra 1 Combining Like Terms Distributive Property Math Worksheets
FREE
Rated 4.88 out of 5, based on 16 reviews
4.9 (16)
6th Grade Math Bell Ringer, Morning Work, Homework COMMON CORE 1st 9 Weeks
Created by
Bill Bush
This is a resource packet full of45+ pages of morning work/daily work/homework. It was designed using sixth grade common core standards. It is designed to be used during the first 9 weeks based on the Curriculum pacing guide. It is 100% aligned to the Common Core Standards. There are 9 weeks worth of work and by the end. These resources are a great review and each week focuses on the same skills in math. This daily work is effective and a great cumulative review before any standardized tes
6th
Math, Math Test Prep
CCSS
6.G.A.1
, 6.G.A.2
, 6.G.A.3
+32
FREE
Rated 4.78 out of 5, based on 55 reviews
4.8 (55)
One Step Equations Maze (Addition/Subtraction) FREEBIE
Created by
Amy Skinner
FREEBIE In this maze students will solve one step equations using only addition and subtraction to find their way. There are no negative numbers for answers, all positive. Students will need to add and subtract decimals and fractions to find their way through the maze. Clip-art used from Microsoft Clip-art Gallery. Enjoy!
5th - 7th
Algebra, Decimals, Math
CCSS
6.EE.B.7
FREE
Rated 4.82 out of 5, based on 50 reviews
4.8 (50)
FREEBIE Solving One Step Equations Activity Show Your Work Digital and Printable
Created by
Matemaths
Let your sixth graders practice one-step equations while having fun with this engaging, no-prep, digital, easy to use drag and drop or printable activity. I used it as a sub-plan in sixth grade and the kids liked it. Lining up the numbers and variables makes the students think before each step. If, after some problems the students wanted to work faster, I have also created text boxes so they can type in the steps on the rest of the slides. If you prefer them to use paper and pencil a printa
6th - 9th
Algebra, Math
CCSS
6.EE.B.5
, 6.EE.B.6
, 6.EE.B.7
Also included in: Bundle Solving One Step Two Step Multi Step Equations
FREE
Rated 4.75 out of 5, based on 24 reviews
4.8 (24)
Algebra Solve One Step Inequalities Practice Quiz Worksheet 6th 7th 8th Math
Created by
StickyStickDesigns
Introduce more algebra with this worksheet/quiz on Solving One-Step Inequalities with a single variable. This is one worksheet that comes with two templates: one practice worksheet and one quiz template with scoring section. Designed to strengthen problem-solving skills, this resource includes: Single-Step Equation Problems: Students will practice solving inequalities that require one steps, such as addition, subtraction, multiplication, and division.Graphing on a number line: Students will also
6th - 8th
Algebra, Math, Numbers
CCSS
6.EE.A.2
, 6.EE.B.5
, 6.EE.B.6
+2
FREE
Rated 4.86 out of 5, based on 7 reviews
4.9 (7)
Jenga Tumbling Towers One Step Equations 6th Grade Math Activity Freebie CCSS
Created by
The Teaching Files
Tumbling Towers Jenga 6th Grade Math Review is an easy prep, engaging, structured activity that motivates students to get practice on 6th grade Math One-Step Equations standards. This activity is also directly aligned with the CA item specifications, which makes it great for state test preparation. This activity is intended to be used with the Dollar Tree game “Tumbling Towers”, which is a mini version of the original Jenga or regular Jenga. Equations are all positive numbers and provide practic
6th
Decimals, Fractions, Math
CCSS
6.EE.B.7
FREE
Rated 4.8 out of 5, based on 40 reviews
4.8 (40)
Solving One Step Equations Fun Activity
Created by
Kate's Math Lessons
Looking for a fun way to help your math students practice solving one-step equations? In this activity, students solve 15 equations to create 5 different codes. They email the codes and will receive "painting" instructions if their codes are correct (the automatic reply should take only a few seconds!). It's an easy way for students to check their answers without asking a teacher! The students will use the directions from the automatic replies to create a colorful design on the back side of
6th - 8th
Algebra, Math
CCSS
6.EE.B.7
Also included in: Solving One Step Equations Bundle Digital & Printable Activities
FREE
Rated 4.78 out of 5, based on 37 reviews
4.8 (37)
Valentines Math Maze - Solving 1 Step Equations
Created by
Mathematic Fanatic
Keep your students engaged as they complete this unique math maze! It's standards-based, it provides plenty of practice, it's self-checking, and the fun maze format will keep your students interested. This maze involves solving 1-step equations, with hearts for variables. For example, 4x = 32, so x = 8. There are many more mazes like this available at a huge discount as part of the 6th Grade Math Maze Growing Bundle! The directions are easy: Move from the Start to the End by correctly solving
6th
Algebra, Math
CCSS
6.EE.B.5
, 6.EE.B.7
FREE
Rated 4.85 out of 5, based on 20 reviews
4.9 (20)
Solving One Step Equations Notes and Maze (Multiply/Divide only)
Created by
Building Math Muscles
Attack student boredom with this solving one step equation maze. This no prep notes and maze activity will guide your students step by step on how to solve a one step equation. With the visual help of a balancing scale students will build a solid understanding of how to solve one step equations. These easy to use printables are:No prep, just print and goNotes Keeping students involved by completing mazesIncluded in this packet are:Notes on Solving One Step Equations One Step Equation MazeAnswer
6th - 8th
Math, Math Test Prep, Other (Math)
CCSS
6.EE.B.7
FREE
Rated 5 out of 5, based on 9 reviews
5.0 (9)
Last Day of School Activity One Step Equations | Code Breaker
Created by
Matemaths
Looking for an exciting and hassle-free way to keep your students engaged as the year winds down? Look no further! Introducing our captivating and convenient no-prep activity for practicing one step equations. Designed to inject fun into the learning process, this activity is ready to print and requires minimal preparation. Students will embark on a thrilling quest to unravel a secret message by successfully solving 10 one step equations. Not only does this activity serve as an excellent
6th - 8th
Algebra, Math
CCSS
6.EE.B.5
, 6.EE.B.7
Also included in: Bundle Solving One Step Two Step Multi Step Equations
FREE
Rated 5 out of 5, based on 9 reviews
5.0 (9)
Math One Step Equations Activity | Digital Escape Room
Created by
Matemaths
Looking for a fun and engaging way to review one step equations with your 6th grade students? Look no further. This no-prep, self checking digital math escape room, with 23 one step equations, it's the perfect activity to keep your students engaged while they practice their skills. This digital activity is perfect to have 6th grade students practice solving one-step equations while having fun. If they need to complete it on different days they just have to save the address at the top of the page
6th - 8th
Algebra, Math
CCSS
6.EE.B.5
, 6.EE.B.7
Also included in: Bundle Solving One Step Two Step Multi Step Equations
FREE
Rated 4.73 out of 5, based on 11 reviews
4.7 (11)
Solving One-Step Equations Fun Group Activity 4 Corners Game for Google Slides™
Created by
Kate's Math Lessons
Get your students out of their seats with this fun one-step equations activity! In this 4 Corners Game with a twist, students work together in groups to solve one-step equations and try to avoid getting eliminated! Includes one-step equations that can be solved with addition, subtraction, multiplication, or division. All solutions are positive integers. Reasons to love this one-step equations game:It gets students up and moving around the room.It gets students working together on problems.It'
6th - 8th
Algebra, Math
CCSS
6.EE.B.7
Also included in: Solving One Step Equations BUNDLE - Print and Digital Activities
FREE
Rated 4.76 out of 5, based on 17 reviews
4.8 (17)
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17424 | https://www.youtube.com/watch?v=r5R9jQB97MM | Maximizing the Product of Sine Angles in a Triangle | Cambridge Assessment 2018 Problem Solution
Chan Lye Lee
5 subscribers
18 likes
Description
312 views
Posted: 5 Oct 2024
In this video, we tackle a challenging problem from the 2018 Cambridge Assessment Test. The goal is to find the maximum value of the product of sine A, sine B, and sine C, where A, B, and C are the angles of a triangle. Using two methods — trigonometric identities and inequalities (AM-GM and Jensen’s inequality) — we derive the solution and confirm that the maximum occurs when the triangle is equilateral. Watch to learn the step-by-step process and sharpen your problem-solving skills!
Don't forget to like, share, and subscribe for more math problem-solving content!
3 comments
Transcript:
[Music] welcome to today's video we're going to solve a challenging problem from the Cambridge assessment test 2018 the problem asks us to find the maximum value of the product of s a s b and s c where a B and C are the angles of a triangle we know that the angles of any triangle must sum up to Pi so our task is to find the arrangement of angles that maximizes this product while satisfying that condition let's get [Music] started first remember that a + B+ C equals Pi so we'll test different combinations that satisfy this [Music] condition case 1 a = < / 2 B = piun 4 and C = < / 4 this gives you s s of piun / 2 s of piun over 4 s of piun / [Music] 4 which equal 1 [Music] 2 case 2 A = B = C = piun / 3 in this case the triangle is equilateral and we get S of piun over 3 sinun / 3 sinun / 3 = < TK 3 / 2 [Music] Cub which equals 3 the < TK of 38 it seems like the second case gives us a higher value so we can make a guess that the maximum value of sin a sin B sin C might be 3 < TK 38 and it occurs when a = b = c = < / [Music] 3 next we apply the amgm inequality to change our Focus from maximizing the product to maximizing the sum of the [Music] signs we know that the sum of sin a sin B and sin C / 3 is greater than or equal to the cube Ro of the product of sin a sin B and sin C [Music] so by maximizing the sum of sin a sin B and S C we can also find an upper Bound for the product of s a s b and s c thus our next goal is to maximize the sum of sin a s b and s c which will give us a better estimate for the maximum value of the [Music] product now let's move to Method [Music] one where we cleverly add s of pi over 3 to the sum of signs that we wish to maximize this step will help us simplify the expression and make use of trigonometric properties using the sum to product identities we can simplify the expression [Music] further next we analyze the trigonometric terms and use the fact that cosine of theta is less than or equal to 1 for any angle [Music] Theta so we can bound the expression as 4 s of a plus B+ C over 4 is less than or equal to 4 sin of piun / 3 [Music] which equal 4 < TK of 3 / [Music] 2 this gives us the maximum value for the sum which happens when a = b = c = < 3 thus the maximum sum of the sign values is 3 < TK 3 /2 which occurs when the triangle is equilateral with all angles equal to piun / [Music] 3 we apply the amgm inequality again to find the maximum value of the product sin a sin B sin C is less than or equal to < TK 3 / 2 cubed which equal 3 < TK 3 8 which occurs when the triangle is equilateral with all angles equal to Pi / [Music] 3 now let's move on to Method [Music] two where we apply Jensen's inequality to maximize the sum of the signs by utilizing the concavity of the sign function on the domain from 0 to [Music] Pi recall that Jensen's inequality tells us that for a concave function f ofx the following inequality holds F of a plus b plus C over 3 is greater than or equal to F of a + f of b + f of C all divided by [Music] 3 S of a plus B+ C over 3 is greater than or equal to the sum of sin a sin B and sin C / 3 using the fact that A+ B+ C = piun this simplifies to sin ofun / 3 = < TK 3 /2 so the maximum sum of the signs is 3 sinun / 3 = 3 < TK 3 /2 following the same process as method 1 we found that the maximum value of sin a sin B sin C is 3 < TK 3 8 which occurs when the triangle is equilateral with a b = c = pk/ 3 thank you for watching this video I hope you found the solution clear and insightful if you enjoyed this content please consider liking sharing and subscribing for more problem solving videos see you next time [Music] |
17425 | https://forum.posit.co/t/count-function-argument-inside-of-summarize-function/186631 | count() function/argument inside of summarize() function
Hey everyone, beginner to R. Quick question: why doesn't my code below work? I've noticed that just in general, whenever I use count() inside of summarize for anything, it never works, but other aggregate functions like sum, max, min, etc., tend to work. What is going on?
Code:
starwars %>%
filter(gender == "feminine") %>%
group_by(gender) %>%
summarize(how_many_women = count(gender)) %>%
View()
starwars %>%
filter(gender == "feminine") %>%
group_by(gender) %>%
summarize(how_many_women = count(gender))
I do not use {dplyr } so I am not sure though using both filter and group may be overkill? I think this does what you want
starwars %>%
filter(gender == "feminine") %>%
summarize(how_many_women = n())
I see the function count() as alternative to group_by/summarise
As the man page says:
df %>% count(a, b) is roughly equivalent to df %>% group_by(a, b) %>% summarise(n = n())
n() is what you want in the summarise if you want to summarise multiple genders
nrow() would be a way of just counting rows having filtered
summarize(how_many_women = count(gender))
The problem here is that count() takes a table as input, whereas gender here represents the vector contained in the column gender. From the documentation for count()`, which you can find by running
count()
gender
gender. From the documentation for
?count
or going to: Count the observations in each group — count • dplyr :
Arguments
x — A data frame, data frame extension (e.g. a tibble), or a lazy data frame (e.g. from dbplyr or dtplyr).
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17426 | https://www.droracle.ai/articles/41068/when-does-lymphadenopathy-resolve-in-reactive-lymphadenitis- | When does lymphadenopathy resolve in reactive lymphadenitis?
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When does lymphadenopathy resolve in reactive lymphadenitis?
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From the Guidelines
Reactive lymphadenopathy typically resolves within 2-4 weeks after the underlying cause has been addressed. The resolution time varies depending on the specific trigger, with viral infections often leading to resolution within 2-3 weeks, while bacterial infections may take 4-6 weeks to completely resolve, especially if appropriate antibiotics are administered 1.
Key Considerations
No specific medications are required to treat the lymphadenopathy itself, as treatment should focus on the underlying cause.
For example, bacterial lymphadenitis may require antibiotics, and symptomatic relief can be achieved with NSAIDs like ibuprofen or acetaminophen for associated pain or fever.
Persistent lymphadenopathy beyond 4-6 weeks warrants further investigation to rule out chronic infections, autoimmune conditions, or malignancy, as outlined in the clinical practice guideline for evaluation of the neck mass in adults 1.
Clinical Guidelines
A patient with neck mass who is treated with antibiotics should be reassessed within 2 weeks, and if the mass has not completely resolved, further workup for possible malignancy is recommended 1.
If infectious lymphadenopathy resolves as expected, the patient should be reassessed once more in 2 to 4 weeks to monitor for possible recurrence, which would prompt definitive workup for possible malignancy 1.
Treatment Approach
The resolution occurs as the immune response subsides, with decreased lymphocyte proliferation and normalization of lymph node architecture following clearance of the antigenic stimulus.
It is reasonable to prescribe a single course of a broad-spectrum antibiotic to treat a neck mass that appears to be infectious, provided that the patient is followed to allow additional evaluation if the neck mass does not resolve 1.
From the Research
Resolution of Lymphadenopathy in Reactive Lymphadenitis
Lymphadenopathy in reactive lymphadenitis typically resolves spontaneously within several weeks 2
The exact timeframe for resolution may vary depending on the underlying cause and individual patient factors
In some cases, reactive lymphadenitis may be mistaken for lymphoma, highlighting the importance of accurate diagnosis and histological examination 2
Factors Influencing Resolution
The presence of underlying immunodeficiency, such as anti-interferon-gamma antibodies, may affect the resolution of lymphadenopathy 3
In cases of mycobacterial infection, antibiotic therapy may be effective in resolving lymphadenopathy, either as an alternative or adjuvant to surgical excision 4
The type and duration of antibiotic therapy may impact the resolution of lymphadenopathy, with some studies suggesting a trial of medical therapy before surgical excision 4
Clinical Considerations
Differentiation between tuberculosis and sarcoidosis can be challenging, and sequential occurrence of these conditions in the same patient is uncommon 5
A high index of suspicion and thorough diagnostic evaluation are necessary to determine the underlying cause of lymphadenopathy and guide appropriate treatment 5, 6
Surgical excision and antituberculous chemotherapy may be effective in managing mycobacterial cervical lymphadenitis, with a reported cure rate of 100% in some studies 6
References
1
Guideline
Guideline Directed Topic Overview
Dr.Oracle Medical Advisory Board & Editors, 2025
2
Research
[Reactive Lymphadenopathies].
Therapeutische Umschau. Revue therapeutique, 2019
3
Research
Mycobacterial lymphadenitis without granuloma formation in a patient with anti-interferon-gamma antibodies.
International journal of hematology, 2021
4
Research
Antibiotic therapy for nontuberculous mycobacterial cervicofacial lymphadenitis.
The Laryngoscope, 2005
5
Research
Tuberculosis lymphadenitis in a well managed case of sarcoidosis.
The Indian journal of chest diseases & allied sciences, 2013
6
Research
Management of mycobacterial cervical lymphadenitis.
World journal of surgery, 1997
Related Questions
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17427 | https://www.bjanaesthesia.org.uk/article/S0007-0912(17)47872-X/pdf | British Journal of Anaesthesia 1991; 67: 546-552
INTRA-ARTERIAL THIOPENTONE IS DIRECTLY TOXIC TO VASCULAR ENDOTHELIUM
R. D. M A C P H E R S O N , L. J. M C L E O D AND A. J. GROVE
SUMMARY
Ear artery segments removed from urethane -anaesthetized rabbits were mounted, periused with Krebs solution, and pressurized before constriction with extraluminal noradrenaline. Vessel diameter was measured using a diode array mounted above the artery. After the degree of dilatation in response to intraluminal acetyl-cho/ine and glyceryl trinitrate was measured, vessels were perfused for 120 s with solutions of either 2.5-10% thiopentone or isotonic sodium carbonate in saline of matched pH. Administra-tion of thiopentone solutions in all concen-trations resulted in destruction of endothelial cells. The dilatation of vessels to glyceryl trinitrate and their myogenic reactivity was unaltered. Isotonic solutions of sodium carbonate (pH 10.6) had no effect on either endothelial cell function or direct vasodilatation. These data show that administration of thiopentone removes arterial endothelial cells, while vascular smooth muscle function is essentially unaltered. KEY WORDS
Anaesthetics, intravenous: thiopentone. Arteries: endo-thelium. Complications: arterial endothelial damage, intra-arterial injection.
Accidental intra-arterial injection of thiopentone results frequently in compromised distal blood flow which often leads to tissue ischaemia and necrosis. Since the initial reports of this phenom-enon appeared more than 50 years ago [1, 2], many theories of causation have been advanced [3-8]. Some of these authors have either suggested or assumed that vascular damage may be part of this process, but definitive functional and histological evidence has been lacking. Furthermore, there has been no assessment of the importance of intimal damage in the overall pathological picture, yet it is now known that endothelial cells play a more important part than acting as a simple mechanical barrier, and have a pivotal role in vascular mechanics through the production and release of substances such as endothelium-derived relaxing factor (EDRF) , and endothelin . Consequently, if damage to these structures were to occur, there may be significant effects on vessel diameter and reactivity which might contribute to the overall pathological features of thiopentone-induced ischaemia. The aim of this study was to examine the effects of thiopentone on vascular smooth muscle and endothelial cell function in the isolated rabbit ear artery.
MATERIALS AND METHODS
Vessel preparation
Approval to conduct the experiments was granted by the Animal Ethics Committee of our University. New Zealand semi-lop-eared rabbits (1.5-2.5 kg) of both sexes, and aged 6-8 months were anaesthetized with i.v. urethane 1.7 g kg" 1
and given heparin 1000 u i.v. The proximal 2-3 cm segment of the central ear artery was removed, avoiding contact with the vessel during dissection. Vessels were used within 12 h of their removal, stored at 2 °C in Krebs solution of the following composition (mmol litre" 1): NaCl 118; KC1 4.69; NaHCO 3 25; KH,PO 4 1.08; CaClj 2.52; MgSO 4.7HjO 1.05; glucose 5.55; diamino-ethane tetra-acetic acid (EDTA) disodium salt 0.026.
Perfusion apparatus and techniques
The apparatus used (fig. 1) was that described by Speden and Speden and Warren ,
Ross D . MACPHERSON, B.MED.(HONS), M.SC.; LYNDSAY J. MCLEOD, M.SC.(EDIN.), PH.D. ; ANDREW J. GROVE; Department
of Physiology, University of Tasmania, G.P.O. Box 252C, Hobart, Tasmania 7001, Australia. Accepted for Publication: May 9, 1991. Correspondence to R. MacP.
THIOPENTONE AND ENDOTHELIUM 54?'
Acetylcholine Function infusion generator Pum PMonitor Constant pressure pump and control infusion pump Perisaltic pump
FIG. 1. Apparatus for the study of arterial reactivity. Vessels were cannulated at either end and perfused with Krebs solution while immersed in an organ bath to which noradrenaline solution was added. Glyceryl trinitrate, acetylcholine and test solutions were infused intraluminally by means of the device labelled "Acetylcholine infusion pump".
modified by the incorporation of a servo-infusion pump , governed by a Hewlett-Packard 3110B Function Generator, capable of maintaining con-stant intraluminal pressure during changes in resistance or vessel diameter. Arterial segments were transferred to a 4.2-ml prefilled organ bath, the proximal end mounted, tied onto a 19-gauge blunted cannula, and flushed out and filled with Krebs solution before the distal end was also cannulated and tied. The Krebs solution used as both the intraluminal perfusate and the organ bath exchange solution was drawn from a common reservoir, which was bubbled with 5 % carbon dioxide in oxygen throughout the experiment. Intraluminal flow, which was monitored by means of a drip counter mounted distal to the outflow resistor (fig. 1), was set at 1 ml min" 1. The intraluminal pressure was increased to 150 mg Hg to test for leaks, and the cannulae adjusted to remove any bowing, such that the final length of the artery approximated to that in vivo. Any excess connective tissue was removed and the servo pump adjusted to provide a pressure of 60 mm Hg. The vessel segments, which had a final length between cannulae tips of 8-12 mm and a mean unconstricted diameter of 1.3 mm, were allowed to equilibrate under these conditions for at least 30 min. Organ bath fluid was replaced at the rate of 4 ml min" 1 and temperature was maintained within the limits 32-33 °C by use of circulating water, monitored by a digital ther-mometer (Analog Devices 2036-T-4112), using a thermocouple mounted within the bath. The artery was illuminated from below with an Olympus LSE light source directed through a red filter and the external diameter of the vessel calculated from the width of the shadow cast by the tissue on a linear array of self-scanning photodiodes (Reticon RL1024G, U.S.A.) mounted at right angles to the long axis of the vessel. A cover slip placed across the bath well was used to remove the meniscus of the bath solution. The external diameter of the vessel was both monitored on an oscilloscope and recorded to the nearest 5 um on a chart recorder. The intraluminal perfusion pressure was recorded using a pressure transducer (Statham, P23 Db) mounted between the artery segment and die variable outflow resistor. Solutions of ( — )-noradrenaline (Sigma) were prepared by dissolving it in hydrochloric acid 0.02 mol litre" 1 and diluting with 0.9 % sodium chloride solution containing EDTA 0.026 mmol litre" 1. Using a syringe pump (Model 355, Sage Instruments USA), the noradrenaline solutions were added to the extraluminal perfusate at a rate of 0.1 ml min" 1 to induce a constriction of the vessel of 50-60% of maximum, calculated as 0.75 548 BRITISH JOURNAL OF ANAESTHESIA
J 120 s
FIG. 2. Typical myogenic responses of the activated rabbit ear artery. A rapidly applied pressure increase from 60 to 80 mm Hg resulted in a prompt increase in vessel diameter. The dilatation was not maintained, however, and vessels are expected to constrict back their initial diameter against the pressure
2E
CO
Q
1500 -
1400 -
1300 -
1200 -
1100 -
1000 -
900
FIG. 3. Functional testing of endothelial cell integrity. Vessels with intact endothelium (A) typically demonstrate arapid and sustained increase in diameter following infusion of acctylcholinc 2 x 10"° mol litre" 1. After endothelial removal (B), this response is not observed.
of the initial unconstricted diameter at a pressure of 60 mm Hg. The final concentration of nor-adrenaline in the bath was 0.350-0.475 umol litre" 1.
Assessment of vascular muscle and endothelial cell function
After equilibration and activation with extraluminal noradrenaline, vessels were sub-jected to three separate procedures to assess their reactivity and viability. The myogenic reactivity of the vessel was assessed by its response to pressure jump increases produced using the Hewlett-Packard Waveform generator. Pressure within the vessel was increased from 60 to 80 mm Hg within 500 ms and maintained at that value for 120 s, before returning to the starting pressure. Vessels are expected to return to at least 80% of their initial diameter (D,) despite the pressure increase (fig. 2). The presence of a functional endothelium was confirmed by a rapid and sustained dilatation following the addition of acetylcholine chloride solution (Sigma) to the intraluminal perfusate, in a final concentration of 2 x 10" 6 mol litre" 1 (fig. 3A). The response of the vessel to endothelium-independent vasodilators was determined by measurement of the degree of dilatation following the administration of glyceryl trinitrate 2 x lO" 6 mol litre" 1 (DBL Laboratories). Solutions of thiopentone were prepared by diluting ampoules of Pentothal 0.5 g (Abbott) with the accompanying dilutent (Water for Injec-tions) to produce concentrations of 2.5, 5 and 10 % (w/v). Additionally, a control solution in the form of an isotonic solution of matched pH (10.6) was prepared by the addition of an appropriate amount of anhydrous sodium carbonate (Analar) to a sodium chloride solution. After 30 min, intraluminal Krebs perfusate was ceased, and 10 ml of the selected solution of either thiopentone or sodium carbonate buffer was infused over 2 min. Care was taken not to allow air to come into contact with the vessel at any time throughout the procedure. At the conclusion of the 2-min period, a small amount of Krebs solution was flushed through the side line to remove any of the test solution remaining in the vessel and the Krebs perfusion was recommenced. The vessel was allowed to re-equilibrate for 30 min before infusions of acetylcholine and THIOPENTONE AND ENDOTHELIUM 549
TABLE I. Mean (SEM) vascular responses to glyceryl trinilrate 2x1 Or'mol litre'' before and after infusion of either thiopentone or control solution. D , = vessel diameter before noradrenaline activation ;
D, = diameter after addition of noradrenaline ; D/=diameter after imraluminal acetylcholine
solution Control (n = 9) Thiopentone 2.5%(n = 7) 5.0% (n = 5) 10.0% (n = 5) A, (Jim) 1334 (8) 1305(7) 1327 (18) 1324(11) Before £>,(um) 881 (31) 961 (24) 837 (4) 919(53) infusion Dt(um) 1186(35) 1156(39) 1116(84) 1213(39) After infusion D,(um) 871(38) 893 (33) 883(84) 863(54) D,(um) 1169(47) 1121(37) 1156(25) 1165(35)
TABLE II. Mean (SEM) vascular responses to acetylcholine 2 x10~mol litre ' before and after infusion of either thiopentone or control solution. Do=vessel diameter before noradrenaline activation; D, =
diameter after addition of noradrenaline ; D^ = diameter after intralumnal acetylcholine
solution Control (n = 9) Thiopentone 2.5 % (n = 7) 5.0% (n = 5) 10.0% (n = 5) Do(um) 1334 (8) 1305(7) 1327(18) 1324(11) Before D,(um) 885(48) 953 (26) 993 (54) 909(61) infusion D, (um) 1227 (36) 1231 (41) 1305(15) 1196(66) After D,(um) 877 (40) 894(45) 975 (53) 850 (48) infusion
D, (urn) 1189(33) 923 (43) 988 (54) 909(43)
glyceryl trinitrate, and tests of myogenic reactivity detailed above were repeated. At the conclusion of each experiment, vessel segments were sent for microscope studies.
RESULTS
Arterial diameter was monitored and recorded (um) throughout the experimental procedures, and the following variables were calculated for each vessel: (1) The diameter of the vessel after mounting but before addition of noradrenaline solutions. This diameter represents maximum possible dilatation, and is referred to as the opening diameter (D o). (2) The diameter of the vessel after noradrenaline activation but before addition of vasodilator test substances, referred to as the initial diameter (D,). (3) The diameter following the addition of vaso-dilating substances, the final diameter (D t). From these variables, a term may be derived which reflects the degree of dilatation following addition of vasodilator drugs relative to the maximum possible dilatation. This is referred to as per-centage dilatation; it is calculated as {D^ — D^/ (D o — D t), and expressed as a percentage. Twenty-six artery segments were removed from 11 rabbits and exposed to either control solutions of sodium carbonate (n = 9), or thiopentone in concentrations of 2.5% (n = 7), 5.0% (n = 5) or 10.0% (n = 5). These segments (« = 26) had a mean unconstricted diameter (D o) of 1322 (SEM 5.5) urn. After constriction with noradrenaline, infusion of glyceryl trinitrate solution resulted in an increase in diameter of 272 (19) um—from 901 (18) um (D,) to a mean D, of 1167 (31) um. After administration of test solutions, vessels maintained their capacity to constrict to noradren-aline stimulation and there was no alteration in their response to glyceryl trinitrate (table I) irrespective of the solution used, demonstrating that, functionally, smooth muscle reactivity was unaltered. In response to acetylcholine (table II), the degree of dilatation before administration of test solutions was similar to that with glyceryl trinitrate, with a mean overall increase in diameter of 308 (20) um—from D, 929 (24) um to D, 1239 (21) um. There was no significant change in response to acetylcholine after administration of isotonic control solutions, but in vessels perfused with thiopentone in concentrations of 2.5% or greater, the mean diameters after acetylcholine administration were significantly less (fig. 3B). These changes in diameter are presented in table 550 BRITISH JOURNAL OF ANAESTHESIA
TABLE III. Percent dilatation of artery segments {mean (SEM) to acetylcholine (Ach) and glyceryl trinitrate (GTN) before and after treatment with infusion solutions
Ach GTN Infusion solution Before (%) After (%) Before (%) After (%) Control (n = 9) Thiopentone 2.5%(n = 7) 5.0% (n = 5) 10.0% (n = 5) 79(5.7) 68 (7.2) ns 80(11.8) 7.3(1.5) <0.05 93(2.1) 4.1(1.4) <0.05 76(13.2) 12 (2.1) <0.05 70 (5.6) 68 (9.3) 57 (10.8) 57 (4.3) 77(9) 52 (8.0) 58 (5.8) 67(5) ns ns ns FIG. 4. Photomicrographs of rabbit ear artery sections, A : After mounting, noradrenaline activation and Krebs perfusion for 30 min, endothelial cells are clearly visible, B: After perfusion with 2.5% thiopentone for 120 s, endothelial cells have been removed and the internal elastic lamina is exposed, c: After perfusion with control solution for 120 s, endothelial cells are undisturbed. Specimens were fixed in gluteraldehyde, postfixed in osmium tetroxide and embedded in Epon before staining with toluidine blue. Horizontal bar represents 100 um.
Ill in terms of percentage increase in dilatation. In vessels exposed to thiopentone in all con-centrations, response to glyceryl trinitrate was unchanged. However, there were significant alterations in response to acetylcholine. Percent dilatation decreased from 80 (11.8)% to 7.3 (1.5)% after infusion of the 2.5% solution, from 93 (2.1) % to 4 (1.4) % after administration of 5 % THIOPENTONE AND ENDOTHELIUM 551
solution, and from 76 (13.2)% to 12 (2.1)% after administration of 10 % solution. As the dilatation response to acetylcholine is known to be de-pendent on release of EDRF from an intact and functional endothelium , these results indicate that the reactivity of these cells has been ef-fectively destroyed by thiopentone solutions. This was confirmed anatomically by microscope studies (fig. 4). There was no difference in vascular structure after administration of control solution (fig. 4).
DISCUSSION
The pathogenesis of arterial occlusion, ischaemia and tissue necrosis seen after intra-arterial in-jection of thiopentone has yet to be resolved fully. Of the many theories advanced, the contemporary view held most widely seems to be that pre-cipitation of thiopentone in the arterial tree initiates an inflammatory reaction and arteritis which, coupled with the microembolization that follows , eventually results in occlusion of the distal circulation. These conclusions have been reached mainly using laboratory studies [4—7], confirmed to some degree by examination of vessels in limbs removed surgically . Experi-ments studying intra-arterial thiopentone have often used the rabbit ear artery and the procedures adopted here, in terms of thiopentone concentra-tions and duration of administration, were de-signed to parallel the methods used in these earlier studies [7, 14, 15] to allow comparison of results. Because of the complex and time-dependent pathological picture seen after intra-arterial ad-ministration of thiopentone, it is difficult to determine which of the events have been initiated by the thiopentone per se and which are at-tributable to the general inflammatory response which follows. While it would appear that it is the latter process which accounts for the eventual arteritis, these experiments have shown that concentrations of thiopentone as small as 2.5 % had direct toxic effects in that they were capable of removing vascular endothelial cells and exposing the underlying intimal layer, the process being essentially complete within 120 s. In common with the findings of others [4, 14], there was no alteration in vessel function or microstructure after administration of isotonic control solutions buffered to pH 10.6, confirming that vascular effects were secondary to thiopentone and not to the alkalinity of the injectate. Thiopentone, even in large concentrations, had no direct effect on the smooth muscle structure and function; vascular segments remained capable of constriction to noradrenaline and dilatation to non-endothelium dependent vasodilators such as glyceryl trinitrate, and the myogenic respon-siveness of the vessels was also unchanged. It is unusual that thiopentone should have such a destructive effect on endothelial cells, while leaving smooth muscle function unaffected. In laboratory procedures in which vascular endo-thelium is removed by chemical means, sub-stances such as collagenase , saponin or the detergent CHAPS are used commonly, often with poor results, as they frequently either fail to remove endothelial cells, or result in damage to underlying muscle. Earlier, it was proposed [4, 19] that arterial spasm was responsible for thiopentone-induced gangrene, a theory which has now been discarded. However, it is now known that many vessels display a basal level of flow-induced release of EDRF [20,21], which antagonizes constrictor activity of circulating catecholamines. In this series of experiments, as in earlier studies , removal of the endothelium resulted in a con-sistent, although statistically insignificant, de-crease in vessel diameter to the same concentration of extraluminal noradrenaline. Vascular prep-arations in which endothelial cells have been removed have been shown to be more sensitive to the effects of some inflammatory mediators and vasoconstrictor agents and as it has been demonstrated that significant amounts of throm-boxane are released after intra-arterial ad-ministration of thiopentone and that platelet aggregation, a process which results in the production of a wide range of vasoactive sub-stances, occurs after thiopentone and blood ad-mixture in vitro , it is likely that there is a reduction in vessel diameter by the direct stimu-lation of vascular smooth muscle by inflammatory mediators. This constriction does not constitute a"spasm", but it does lead to the conclusion, by application of Poiseuille's Law, that flow would be considerably impeded in thiopentone affected vessels where the endothelium has been removed, further compromising tissue perfusion. It is accepted generally that the risk of initiating vascular damage with intra-arterial thiopentone increases with increasing concentration of the drug , and that use of the 2.5 % solution is 552 BRITISH JOURNAL OF ANAESTHESIA relatively safe . Indeed, there have been only occasional case reports of gangrene after intra-arterial injection at this strength , with most cases being reported after the use of more concentrated solutions. Nevertheless, 2.5% solu-tions of thiopentone are capable of endothelial cell destruction. In terms of treatment of the sequelae of thiopentone-induced injury, results of these experiments suggest that any strategy which utilizes endothelium-dependent vasodilatation to enhance flow in thiopentone-affected vessels is likely to be less effective than methods which are not dependent on an intact and functional endothelium. In summary, we have shown that solutions of thiopentone of 2.5 % and above acted directly to destroy vascular endothelial cells after a 120-s exposure. Vascular smooth muscle function was unaffected, and solutions of isotonic anhydrous sodium carbonate buffer of pH 10.6 had no effect. After intra-arterial administration of thiopentone, endothelial cell destruction and subsequent ex-posure of the intimal surface may be important processes in attenuating vessel diameter, by eliminating vascular release of EDRF and initiating or promoting thrombus formation, re-spectively.
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Edinburgh: Livingstone, 1956. 25. Mark LC. A lone case of gangrene following intraarterial thiopental 2.5%. Anesthesiology 1983; 59: 153. 26. Dohi S, Naito H. Intraarterial injection of 2.5 % thiamylal docs cause gangrene. Anesthesiology 1983; 59: 154. |
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Algebra II: High School
Factoring the Sum of Cubes | Formula, Rules & Examples
Contributors: Stephen Sacchetti, Joshua White
Author
Author:
Stephen Sacchetti
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Instructor
Instructor:
Joshua White
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In this lesson, learn what the sum of cubes is. Once this is understood, learn about how to factor sum of cubes and how it is done from sum of cubes examples. Updated: 11/21/2023
Table of Contents
Sum of Cubes
Lesson Summary
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FAQ
How do you factor the sum of two cubes?
First, ensure that the expression is written as a^3 + b^3. Then from there, use the formula: a^3 + b^3 = (a + b)(a^2 - ab + b^2)
What is the first step in factoring the sum of two cubes?
When factoring the sum of two cubes, first identify which quantities are being cubed. For instance, if given the term 64x^3, rewrite it as (4x)^3.
What is the sum of cubes formula?
The sum of cubes formula gives a way to factor an expression written in the form a^3 + b^3. In particular, it says a^3 + b^3 = (a + b)(a^2 - ab +b^2).
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0:04 Difference of Squares
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Sum of Cubes ------------
A sum of cubes is the sum of two expressions, each of which is a perfect cube. For example, the expression x 3+8 is a sum of cubes since x 3 is a perfect cube and 8 is a perfect cube (8=2 3). A more complicated expression such as 64 x 3+27 y 3 is also a sum of cubes since 64 x 3=(4 x)3 and 27 y 3=(3 y)3. Please note, that expressions such as 2 x 3 are not perfect cubes. Even though the x term is cubed, the coefficient is not a perfect cube and this expression cannot be rewritten as (a x)3 where a is a whole number.
Factoring Sum of Cubes
Before jumping into how to factor a sum of cubes, it is important to know a few of the perfect cubes. Just as the perfect squares are well known up to 100 (and possibly up to 144), so too is it helpful to know the perfect cubes up to 10 3 and maybe even 12 3. Study the table before continuing onto factoring through a sum of cubes formula:
| x | x^3 |
--- |
| 1 | 1 |
| 2 | 8 |
| 3 | 27 |
| 4 | 64 |
| 5 | 125 |
| 6 | 216 |
| 7 | 343 |
| 8 | 512 |
| 9 | 729 |
| 10 | 1000 |
| 11 | 1331 |
| 12 | 1728 |
Armed with this knowledge, it will be easier to recognize perfect cubes and, as a result, to factor cubes as well. In particular, when given the sum of two cubes in the form a 3+b 3, the expression can be factored as shown:
a 3+b 3=(a+b)(a 2−a b+b 2)
It is important, therefore, to ensure that the expression is simplified to be a sum of two cubes before using the sum of cubes formula. For example, in an expression such as 8 x 3+27, the first term must be simplified as 8 x 3=(2 x)3 and so a=2 x. Now, rewrite the expression as the sum of two cubes:
8 x 3+27=(2 x)3+3 3 8 x 3+27=(2 x+3)((2 x)2−2 x⋅3+3 2)=(2 x+3)(4 x 2−6 x+9)
One of these factors can also be seen graphically. Noticing that the factor (2 x+3) exists means that this expression equals zero when x=−3 2. This can be confirmed by graphing 8 x 3+27 as shown below.
The
Sum of Cubes Examples
In this section, sum of cubes examples will be given that increase in difficulty. Follow along with the work shown and refer back to the table of perfect cubes as needed.
Example 1: Use the sum of cubes formula to factor x 3+125
In this case, recognize that 125=5 3, thus this expression is already written as the sum of two cubes. Factoring by using the sum of cubes formula then gives:
x 3+125=x 3+5 3
Here, a = x and b = 5, so recall the formula a 3+b 3=(a+b)(a 2−a b+b 2) and make the substitution to find:
x 3+125=(x+5)(x 2−5 x+25)
Example 2: Use the sum of cubes formula to factor 216 x 3+1
In this case, notice that 216 x 3=(6 x)3, so
216 x 3+1=(6 x)3+1 3=(6 x+1)(36 x 2−6 x+1)
Example 3: Factor the expression 512 x 3+343 y 3
In this case, recall that 512=8 3 and 343=7 3, thus the expression becomes:
512 x 3+343 y 3=(8 x)3+(7 y)3
and can be factored as:
512 x 3+343 y 3=(8 x+7 y)(64 x 2−56 x y+49 y 2)
Example 4: Factor the expression x 6+64 y 6
In this case, notice that neither term has an exponent of 3. However, each term can be rewritten using laws of exponents to say x 6=(x 2)3 and 64 y 6=(4 y 2)3. As a result, set a=x 2 and b=4 y 2. This gives the following factorization:
x 6+64 y 6=(x 2+4 y 2)(x 4−4 x 2 y 2+16 y 4)
Example 5: Factor the expression 729 x 3+144 y 3
In this case, recall that 729=9 3, but notice that 144 is not a perfect cube. Therefore this expression cannot be factored using the sum of cubes formula.
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Lesson Summary --------------
A sum of cubes is an expression where two perfect cubes are added together. Algebraically, this looks like a 3+b 3 and can be factored as
a 3+b 3=(a+b)(a 2−a b+b 2)
It is important to rewrite any expressions involving cubes as the sum of two cubed quantities before using this formula, so be sure to rewrite any terms with coefficients in a factored form. Knowing the perfect cubes will help the factoring process go more smoothly as will recalling the idea of exponent properties (namely that exponents that are multiples of 3 create perfect cubes).
To unlock this lesson you must be a Study.com memberCreate an account
Lesson Summary --------------
A sum of cubes is an expression where two perfect cubes are added together. Algebraically, this looks like a 3+b 3 and can be factored as
a 3+b 3=(a+b)(a 2−a b+b 2)
It is important to rewrite any expressions involving cubes as the sum of two cubed quantities before using this formula, so be sure to rewrite any terms with coefficients in a factored form. Knowing the perfect cubes will help the factoring process go more smoothly as will recalling the idea of exponent properties (namely that exponents that are multiples of 3 create perfect cubes).
To unlock this lesson you must be a Study.com memberCreate an account
Video Transcript
Difference of Squares
You've probably come across factoring problems where an expression had two terms, such as x 2 - 4 or 25 x 4 - 16.
If both terms were squares and had opposite signs (i.e., one term was positive and one term was negative), then you could factor it as a difference of squares using the difference of squares formula:
a 2 - b 2 = (a + b) (a - b)
But what should you do when you come across a two-term expression where the terms each have the same sign? Or if the terms are not squares but are instead cubes?
If both terms are cubes, then it may be possible to factor the expression as either a difference of cubes or a sum of cubes, depending on the signs of the terms. An expression with opposite signs (for example, x 3 - 8) could be a difference of cubes, which is covered in a separate lesson. An expression where both terms have the same sign (for example, y 3 + 1), either both positive or both negative, could be factored as a sum of cubes, which is the focus of this lesson.
A sum of cubes is a two-term expression where both terms are cubes and each term has the same sign. It is factored according to the following formula:
a 3 + b 3 = (a + b) (a 2 - ab + b 2)
How can you determine if an expression can be factored as a sum of cubes?
Is an Expression a Sum of Cubes?
An expression must meet two criteria in order to be factored as a sum of cubes. First, each term must be a cube. In other words, each term must be the result of multiplying the same expression by itself three times. Here are some examples:
x 3 is a cube because it is a result of x multiplied by itself three times
(x x x).
27 is a cube because it is the result of 3 multiplied by itself three times
(3 3 3).
Additionally, you may find a cube that contains both numbers and variables. For example, 64 z 9 is a cube because there is an expression (4 z 3) that, when multiplied by itself three times (4 z 3)(4 z 3)(4 z 3), will equal 64 z 9.
Note that every part of each term must be a cube; 7 x 6 and 8 y 2 are not cubes because 7 is not a cube (although x 6 is) and y 2 is not a cube (although 8 is).
Second, each term must have the same sign, usually both positive. Note that if both signs are negative, you can factor a -1 out of both terms to make them each positive. If both terms have opposite signs, then you may want to try and factor the expression as a difference of squares or a difference of cubes. Now, let's see how you use the sum of cubes formula to factor a problem.
Factoring a Sum of Cubes
For a sum of cubes, you'll use the formula already mentioned:
a 3 + b 3 = (a + b) (a 2 - ab + b 2)
Note that a and b represent the individual expressions that are cubed. They could each be a variable (x), a number (3) or some combination of both (4 y 2). First, you must determine what a and b are. Essentially you're asking, what do I cube to get the first term and what do I cube to make the second term? After you've done that, you will plug in the expressions you found for a and b into the formula and simplify them to finish the factoring. Let's see some examples.
Examples
Can the following expressions be factored as a sum of cubes? If yes, factor.
Example 1:
y 5 + 27
This expression cannot be factored as a sum of cubes because the first term (y 5) is not a cube. In other words, there is nothing that can be multiplied by itself three times to equal y 5. Therefore, the expression cannot be factored.
Example 2:
x 3 + 64
This expression can be factored as a sum of cubes since both terms have the same sign (+) and each expression is a cube.
x can be cubed to give x 3
4 can be cubed to make 64.
Thus, plugging a = x and b = 4 into the sum of cubes formula gives:
(x + 4) (x 2 - (x)(4) + 4 2)
Simplifying (x)(4) to 4 x and 4 2 to 16 gives us the final answer of
(x + 4) (x 2 - 4 x + 16).
Example 3:
8 x 3 + 27 y 6
The two terms both have the same sign, but is each term a cube? Yes.
(2 x)(2 x)(2 x) = 8 x 3 and
(3 y 2)(3 y 2)(3y 2) = 27 y 6
Thus, a = 2 x and b = 3 y 2. Plugging these into the formula gives the following factorization:
(2 x)3 + (3 y 2)3
= (2 x + 3 y 2)[(2 x)2 - (2 x)(3 y 2) + (3 y 2)2]
=(2 x + 3 y 2)(4 x 2 - 6 xy 2 + 9 y 4
Example 4:
-2 x 3 - 54
First, notice that while the terms each have the same sign (-), they are not both positive like in the previous examples. That's not a problem though, since we pull out a factor of a negative number to make each term positive.
Additionally, see that -2 and -54 are not cubes. However, they do have a factor of 2 in common or, in this case, technically -2. First, you will factor a -2 out of both terms to give:
-2(x 3 + 27)
Then you can proceed to factor x 3 + 27 as a sum of cubes where a = x and b = 3. Thus, we have:
-2(x + 3) (x 2 - (x)(3) + 3 2)
= -2 (x + 3) (x 2 - 3 x + 9)
Lesson Summary
A sum of cubes is a two-term expression where both terms are cubes and each term has the same sign. It is factored according to the following formula:
a 3 + b 3 = (a + b) (a 2 - ab + b 2)
In order to factor an expression as a sum of cubes, you must first check to see if it meets the criteria:
The expression must have two terms, each with the same sign.
Each term must be a cube.
Remember that sometimes you may need to pull out a common factor first before you can determine if the terms are cubes. If those criteria are met, then you can factor the sum of cubes according to the formula. Finally, make sure you simplify your answer after plugging the values into the formula.
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Algebra II: High School
22 chapters
197 lessons
Chapter 1
Algebra II: Real Numbers
Types of Numbers & Its Classifications 6:56 minGraphing Rational Numbers on a Number Line | Chart & Examples 5:02 minNotation for Rational Numbers, Fractions & Decimals 6:16 minThe Order of Real Numbers: Inequalities 4:36 minFinding the Absolute Value of a Real Number 3:11 minHow to Rationalize the Denominator with a Radical Expression 3:52 minTranscendental vs. Algebraic Numbers | Overview & Examples 6:23 minPythagorean Triple | Definition, List & Examples 3:41 min
Chapter 2
Algebra II: Complex and Imaginary Numbers
Imaginary Numbers | Definition, History & Examples 8:40 minHow to Add, Subtract and Multiply Complex Numbers 5:59 minComplex Numbers & Conjugates | Multiplication & Division 6:40 minHow to Graph a Complex Number on the Complex Plane 3:28 minHow to Solve Quadratics with Complex Numbers as the Solution 5:59 min
Chapter 3
Algebra II: Exponents and Exponential Expressions Review
How to Use Exponential Notation 2:44 minScientific Notation | Definition, Conversion & Examples 6:49 minSimplifying and Solving Exponential Expressions 7:27 minExponential Expressions & The Order of Operations 4:36 minMultiplying Exponents | Overview, Methods & Rules 4:07 minDividing Exponential Expressions 4:43 minThe Power of Zero: Simplifying Exponential Expressions 5:11 minNegative Exponents: Writing Powers of Fractions and Decimals 3:55 minPower of Powers: Simplifying Exponential Expressions 3:33 min
Chapter 4
Algebra II: Properties of Functions Review
Function in Math | Definition & Examples 7:57 minTransformations: How to Shift Graphs on a Plane 7:12 minHow to Add, Subtract, Multiply and Divide Functions 6:43 minDomain & Range of a Function | Definition, Equation & Examples 8:32 minHow to Compose Functions 6:52 minInverse Functions | Definition, Methods & Calculation 6:05 minApplying Function Operations Practice Problems 5:17 min
Chapter 5
Algebra II: Linear Equations Review
Linear Equations | Definition, Formula & Solution 7:28 minApplying the Distributive Property to Linear Equations 4:18 minForms of a Linear Equation | Overview, Graphs & Conversion 6:38 minAbstract Algebraic Examples and Going from a Graph to a Rule 10:37 minUndefined & Zero Slope Graph | Definition & Examples 4:23 minParallel vs Perpendicular vs Transverse Lines Overview & Examples 6:06 minParallel & Perpendicular Lines | Equation, Graph & Examples 6:07 minLinear Equation | Parts, Writing & Examples 8:58 min
Chapter 6
Algebra II: Systems of Linear Equations Review
System of Equations in Algebra | Overview, Methods & Examples 8:39 minHow Do I Use a System of Equations? 9:47 minHow to Solve a System of Linear Equations in Two Variables 4:43 minHow to Solve a Linear System in Three Variables With a Solution 5:01 minSolving System of Equations with 3 Variables | Steps & Examples 6:04 min
Chapter 7
Algebra II: Inequalities Review
Inequality Signs in Math | Symbols, Examples & Variation 7:09 minGraphing Inequalities | Definition, Rules & Examples 7:59 minInequality Notation | Overview & Examples 8:16 minGraphing Inequalities | Overview & Examples 12:06 minSolve & Graph an Absolute Value Inequality | Formula & Examples 8:02 minAbsolute Value Inequalities | Definition, Calculation & Examples 9:06 minTranslating Math Sentences to Inequalities 5:36 min
Chapter 8
Algebra II: Absolute Value Review
Absolute Value | Explanation & Examples 4:42 minAbsolute Value Expression | Evaluation, Simplification & Examples 5:28 minSolving Absolute Value Functions & Equations | Rules & Examples 5:26 minAbsolute Value | Overview & Practice Problems 7:09 minAbsolute Value | Graph & Transformations 8:14 minGraphing Absolute Value Functions | Definition & Translation 6:08 min
Chapter 9
Algebra II: Graphing and Factoring Quadratic Equations Review
Tables & Graphs in the Real World | Uses & Examples 5:50 minScatter Plot Graph | Overview, Uses & Examples 7:17 minParabola | Definition & Parabolic Shape Equation 4:36 minTypes of Parabolas | Overview, Graphs & Examples 6:15 minMultiplying Binomials Using FOIL and the Area Method 7:26 minMultiplying Binomials | Overview, Methods & Examples 5:46 minFactoring Quadratic Equations Using Reverse Foil Method 8:50 minFactoring Quadratic Equations | Solution & Examples 7:35 minQuadratic Trinomial | Definition, Factorization & Examples 7:53 minHow to Complete the Square | Method & Examples 8:43 minCompleting the Square Practice Problems 7:31 minHow to Solve a Quadratic Equation by Factoring 7:53 min
Chapter 10
Algebra II: Quadratic Equations Review
Quadratic Equation | Definition, Formula & Examples 5:13 minSolving Quadratics: Assigning the Greatest Common Factor and Multiplication Property of Zero 5:24 minQuadratic Function | Formula, Equations & Examples 9:20 minHow to Solve Quadratics That Are Not in Standard Form 6:14 minSolving Quadratic Inequalities Using Two Binomials 5:36 min
Chapter 11
Algebra II: Factoring
Factoring in Algebra | Definition, Equations & Examples 5:32 minFinding the Prime Factorization of a Number | Meaning & Examples 5:36 minUsing Prime Factorizations to Find the Least Common Multiples 7:28 minEquivalent Expressions and Fraction Notation 5:46 minUsing Fraction Notation: Addition, Subtraction, Multiplication & Division 6:12 minFactoring Out Variables: Instructions & Examples 6:46 minCombining Numbers and Variables When Factoring 6:35 minTransforming Factoring Into A Division Problem 5:11 minFactoring by Grouping | Definition, Steps & Examples 7:46 minRational Root Theorem | Overview & Examples 7:05 min
Chapter 12
Algebra II: Polynomials
How to Evaluate a Polynomial in Function Notation 8:22 minUnderstanding Basic Polynomial Graphs 9:15 minBasic Transformations of Polynomial Graphs 7:37 minCubic, Quartic & Quintic Equations | Graphs & Examples 11:14 minAdding, Subtracting & Multiplying Polynomials | Steps & Examples 6:53 minPascal's Triangle | Overview, Formula & Uses 7:26 minBinomial Theorem | Coefficient Calculation, Formula & Examples 13:35 minPolynomial Long Division | Overview & Examples 8:05 minSynthetic Division of Polynomials | Method & Examples 6:51 minDividing Polynomials with Long and Synthetic Division: Practice Problems 10:11 minFactor & Remainder Theorem | Definition, Formula & Examples 7:00 minOperations with Polynomials in Several Variables 6:09 min
Chapter 13
Algebra II: Rational Expressions Review
How to Multiply and Divide Rational Expressions 8:07 minMultiplying and Dividing Rational Expressions: Practice Problems 4:40 minAdding & Subtracting Rational Expressions | Overview & Examples 8:02 minPractice Adding and Subtracting Rational Expressions 9:12 minRational Equations | Definition, Formula & Examples 7:58 minRational Equations: Practice Problems 13:15 minDivision and Reciprocals of Rational Expressions 5:09 minSimplifying Complex Rational Expressions | Steps & Examples 4:37 minSolving Direct Variation | Equation, Problems & Examples 5:12 minReflection Rules in Math | Graph, Formula & Examples 6:07 minSolving Rational Equations and Finding the Least Common Denominator 6:48 min
Chapter 14
Algebra II: Graphing and Functions
Graphing Basic Functions 8:01 minCompounding Functions and Graphing Functions of Functions 7:47 minInverse Function | Graph & Examples 7:31 minY Square Root of X | Graph, Domain & Common Mistakes 4:32 minPolynomial Functions: Properties and Factoring 7:45 minPolynomial Functions: Exponentials and Simplifying 7:45 minExponentials, Logarithms & the Natural Log 8:36 minSlopes of a Line | Graphs, Formula & Examples 10:05 minEquation of a Line Using Point-Slope Formula 9:27 minHorizontal and Vertical Asymptotes 7:47 minImplicit Function Overview, Formula & Examples 4:30 min
Chapter 15
Algebra II: Conic Sections
Defining and Graphing Ellipses in Algebra 5:00 minHow to Write the Equation of an Ellipse in Standard Form 6:18 minThe Circle: Definition, Conic Sections & Distance Formula 3:43 minHyperbola | Definition, Equation & Graphs 10:00 minHyperbola Standard Form | Definition, Equations & Examples 8:14 minParabola | Definition, Formula & Examples 8:33 minParabola | Equation, Formula & Examples 8:17 minConic Sections | Definition, Equations & Types 6:33 min
Chapter 16
Algebra II: Roots and Radical Expressions Review
Square Root | Definition, Formula & Examples 7:05 minEstimating Square Roots | Overview & Examples 5:10 minSimplifying Square Roots When not a Perfect Square 4:45 minSimplifying Square Root Expressions | Steps & Examples 7:03 minRadicands and Radical Expressions 4:29 minEvaluating Square Roots of Perfect Squares 5:12 minFactoring Radical Expressions 4:45 minSimplifying Square Roots of Powers in Radical Expressions 3:51 minMultiplying then Simplifying Radical Expressions 3:57 minHow to Divide Radicals, Square Roots & Rational Expressions 7:07 minSimplifying Square Roots | Overview & Examples 4:49 minRationalizing the Denominator | Overview & Examples 7:01 minAddition and Subtraction Using Radical Notation 3:08 minHow to Multiply Radical Expressions 6:35 minSolving Radical Equations | Overview & Examples 6:48 minSolving Radical Equations with Two Radical Terms 6:00 min
Chapter 17
Algebra II: Exponential and Logarithmic Functions
Exponential Function | Definition, Equation & Examples 7:24 minExponential Growth & Decay | Formula, Function & Graphs 8:41 minLogarithms | Overview, Process & Examples 5:23 minChange of Base Formula | Logarithms, Examples & Proof 4:54 minGraphing Logarithms | Overview, Transformations & Examples 6:53 minEvaluating Logarithms | Properties & Examples 6:45 minLogarithmic Properties | Product, Power & Quotient Properties 5:11 minPractice Problems for Logarithmic Properties 6:44 minExponential Equations | Definition, Solutions & Examples 6:18 minSolving Logarithmic Equations | Properties & Examples 6:50 min
Chapter 18
Algebra II: Sequences & Series
Sequences in Math | Overview, Types & Examples 4:57 minHow to Use Factorial Notation: Process and Examples 4:40 minHow to Write a Series in Summation Notation | Overview & Examples 4:16 minGeneral Term of an Arithmetic Sequence | Overview, Formula & Uses 5:01 minUnderstanding Arithmetic Series in Algebra 6:17 minGeometric Sequence | Definition, Formula & Examples 5:14 minSum of a Geometric Series | Formula & Examples 4:57 minSum of Infinite Geometric Series | Formula, Sequence & Examples 4:41 minRecursive Rule | Formulas & Examples 5:52 minUsing Sigma Notation for the Sum of a Series 4:44 minMathematical Induction: Uses & Proofs 7:48 minHow to Find the Value of an Annuity 4:49 minHow to Use the Binomial Theorem to Expand a Binomial 8:43 minSpecial Sequences and How They Are Generated 5:21 min
Chapter 19
Algebra II: Combinatorics Review
How to Use the Fundamental Counting Principle 5:52 minPermutation Definition, Formula & Examples 6:58 minCombination in Mathematics | Definition, Formula & Examples 7:14 minIndependent & Dependent Events | Overview, Probability & Examples 12:06 minConditional Probability | Definition, Equation & Examples 7:04 min
Chapter 20
Algebra II: Calculations, Ratios, Percent & Proportions Review
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17429 | https://www.themathdoctors.org/significant-digits-measurements-and-exact-numbers/ | Skip to content
Significant Digits: Measurements and Exact Numbers
February 7, 2019 January 18, 2024 / Arithmetic / Real life / By Dave Peterson
To conclude this series on significant digits, I want to look at some details of their application. Specifically, we will consider questions about how they related to measured values, and to fixed constants.
Measured values
I’ll start with a question from 2011 about measurement and unit conversion:
Changing Units ... and Significant Figures? Suppose I measured the length of a table and got 2 feet, which is 60.96 cm. My teacher says I need to round my result according to the number of significant figures in my measured data. Well, the first measurement has one significant digit. The second has four significant digits! So, how many significant digits are there in the above measurement? I think that when I change from feet to centimeters, the number of significant figures changes, and I have to re-write my result according to that unit. But I have searched Wikipedia for clarity on significant digits and unit conversion, to no avail.
This called for a reminder of what significant digits are all about, before getting into the question of conversion. I replied:
It looks like you've missed the central concept of significant digits (which is not uncommon, if they are not taught in the right context). What you've done is just to WRITE the quantities "2 feet" and "60.96 cm" with different numbers of significant digits. They don't actually have that many. Essentially, you are "lying" about the measurement. Significant digits are supposed to represent how accurately something was MEASURED.
Significant digits can be introduced just by showing a number and asking how many significant digits it has; most of what we have looked at so far has been in such a context. But here, we are originating the numbers, so we have to decide how to write them in order to represent how accurate (I should say, precise) the measurement is.
When you write "2 feet," with only one significant digit, you are implying that you don't know how many inches; the length might really be anything that rounds to 2, from 1.5 feet to 2.5 feet. I very much doubt that that is what you did. More likely, you measured with a ruler or yardstick, and found that it was 2 feet TO THE NEAREST 1/8TH INCH, say -- that is, if the ruler is marked with eighths of an inch, you couldn't be sure it wasn't 2 feet 0 inches and 1/16th of an inch, but you know it's not more than that. So in order to express this, you should say you measured it as 2 feet, 0.0 inches. (Fractional measurements like this are really hard to express in terms of significant digits, and in reality people making such measurements would just say, "2 feet, plus or minus 1/16 inch" to express this precision.)
The precision of a measurement depends on the instrument (here, the ruler) that was used. I made a guess here; Amie will have to adjust what I say according to her reality. I would like to have known how the measurement was actually done, because, as I said in the last paragraph, this case is not the easiest one to start with when learning these concepts. We don’t yet have an answer to the question of significant digits. I continued:
Now, when you convert (or do any other calculations), you are supposed to use the same number of significant digits that you had to start with. The fact that your calculator or whatever shows 4 digits doesn't mean they are all valid. If I were forced to use significant digits, I would do the conversion something like this: 2 feet 0 inches to the nearest 1/8 inch = 212 + 0 = 24 inches +/- 1/16, that is, between 23.9375 and 24.0625. Call it 24.0 in, since the hundredths are uncertain but the tenths are more or less sure. This gives three significant digits. 24.0 in 2.54 cm/in = 60.96 cm Round this to three significant digits, giving 61.0 cm.
So I first converted the fractional measurement to decimal in order to estimate significant digits. Once I knew that, I could carry out the conversion and retain the same number of significant digits in the result.
What if we had measured in metric initially?
Now, what I'd really do is to measure in centimeters, and avoid all the trouble with fractions. Most likely, the ruler would be marked in millimeters, so it would be accurate to the nearest mm (tenth of a cm), and I would get exactly the same result: 61.0 cm.
So our guess about having three significant digits seems reasonable.
Now, you may have noticed the number 2.54 in there. Does the fact that it is written with three significant digits affect our result? Not really, as we’ll see in the next question.
Exact numbers
Here is the first part of a long question, from 2005:
Exact Numbers and Conversion Statements I am currently taking chemistry, and something that keeps coming up is exact numbers, which are, by definition, numbers obtained by counting (as opposed to measuring) as in 8 apples. Also, an exact number could be part of a definition, like a conversion statement such as 1 inch = 2.54 cm. Is there a general rule for determining whether or not a conversion statement is exact? I believe that fairly recently, the inch was redefined to be EXACTLY 2.54 cm. Are there others that have been defined as exact? And are conversion statements from metric to English or vice versa the only conversion statements that might be inexact? I would really appreciate your help here, since knowing the exactness of a conversion factor plays a huge role in dimensional analysis and rounding an answer to the correct number of significant digits (because an exact number has no limits on significant digits, but an inexact number does).
In a previous post, the concept of exact numbers was mentioned in passing: When a calculation includes a number that is not obtained by measurement (such as the 2 we divide by for the area of a triangle), we treat it as having an infinite number of significant digits, so that it does not restrict the precision of the result. The same is true of conversion factors such as 1 foot = 12 inches (a definition, which can also be thought of as counting); and as was said here, 2.54, though not an integer, is by decree an exact number, part of the definition of the inch as used today. (It was not always so, because the two systems were originally independent, based on separate standards.) See here for historical details:
Millimeters, Inches, Feet, Miles
Jason continued:
And on a more general note, what other types of numbers are classified as exact (e.g., infinite, infinite repeating, fractions, pi)? I know a lot of times it has to do with the context. An example of my dilemma is in the following problem: What is the mass of a troy ounce of gold in grams? 1 troy ounce = 20 pennyweight (exact) 1 pennyweight = 24 grains (exact) 1 grain = 0.0648 gram (NOT exact) I used dimensional analysis, assuming that the "one troy ounce" I was starting with was inexact (because it is a measurement), so I assumed the answer should have only ONE significant digit, but the answer key said the answer should have 3 significant digits because of the 0.0648. I am really lost now, and I don't know how to distinguish between a measurement that is inexact (which I thought a measurement always is) and one that is exact (which I thought they never were, until this problem). The only explanation I can think of is that the "one troy ounce" wasn't actually measured, they just wanted us to use it for the sake of converting. PLEASE HELP!!!!!!!!!!!!!!! Thanks.
Jason is correct in this final supposition. I replied:
The general rule is that a number is exact if you are told it is exact. There is nothing more general that can be said. Any formula you are given should make it clear whether the constants in it are exact. Generally, within a system (such as the 20 and 24 in your example), all numbers are exact integers; between systems, such as your 0.0648, the numbers are usually approximate, and that should be indicated by using an "approximately equal" sign, or saying "to three significant digits", or something like that. But since it is so unusual for a conversion factor between systems to be exact, you can take it the other way and, in that setting, assume a number is inexact unless it is explicitly stated, as it will be for 2.54.
As far as I know, the conversion between metric and U.S. lengths is the only exact one between different systems, though I could be wrong. I don’t know that I have seen many conversion tables that state that within the table. Apart from this one (we have to also include feet to meters, etc.), if the conversion is given as a decimal, it is probably accurate only to the significant digits shown.
I continued:
You asked about "infinite, infinite repeating, fractions, pi". You didn't really mean "infinite numbers", but rather "infinite decimals". If a decimal is given to you indicating how it repeats, as with a bar over it, then that is exact because you are being told every digit in the number. If you are just given some decimal places and are not told how the rest of it behaves, then obviously it can't be considered exact. That is true of pi, which as an irrational number can't be expressed exactly. As for fractions, they are exact unless you are told that the fraction is approximate. For example, the 5/9 or 9/5 in conversion between Fahrenheit and Celsius is an exact fraction. This means that in such cases you don't need to consider the number of significant digits. Other fractions, such as 22/7 for pi, are approximations; to use this with significant digits, you would have to determine HOW accurate it is, by writing it as a decimal and comparing to a good decimal approximation to pi. A decimal approximation makes a lot more sense in such a context.
It turns out that 22/7 is just about as accurate an approximation of pi as is 3.14.
To see how it happens that Fahrenheit and Celsius are related by a simple fraction, see
Converting Celsius to Fahrenheit and Why It Works
Finally:
Now, in your question about "the mass of a troy ounce in grams", there is no number given! There is no measurement here! It is "the mass of an exact troy ounce", not "a mass measured as 1.0 troy ounces". That means you can't take "1" as having a specific number of significant digits; so you have to take it as exact. So the only number in your calculation that has a specific precision is the last conversion factor, and only that limits the precision of your answer.
Note that in this case, the precision is not a matter of measurement at all, but just of how accurate a conversion factor is available. The whole problem, in fact, is not about measurement, but about obtaining a new conversion factor from known conversion factors. Typically, these are given with enough precision that in ordinary use it will not interfere — that is, it will be more precise than most measurements are expected to be.
Irrational numbers
Now let’s dig a little deeper into Jason’s question about pi and infinite (non-terminating) decimals, by looking at a similar question from 2005:
Significant Digits and Irrational Numbers How can you determine the significant digits of a non-terminating or irrational number like pi? I know the rules for finding significant digits of an answer, but what if your answer doesn't terminate? How do you express the significant digits?
There is an explicit question here, and another implied by the wording of the second paragraph. I started with the former:
Any exact number (not a measurement, but a known number such as pi or sqrt(2)) is considered to have infinitely many significant digits, in itself, since there is no limit to the number of digits we COULD know. However, in a specific calculation, we can only use some approximation to it, which means using some chosen number of significant digits, which will restrict our accuracy. The proper thing to do, then, is to be sure to use at least enough sigdigs in your calculation that it will not affect the accuracy of the result. For example, if you are finding the area of a circle using a radius given as 1.23 meters, you would want to use at least three digits of pi, say 3.14 or 3.142. Most likely, these days, you would just use the full accuracy of the constant built into your calculator, and not have to worry! Effectively, then, you have infinitely many sigdigs and can ignore it.
So for such a constant, rather than finding how many significant digits it has, we decide how many to use.
You mention your "answer". You always determine the number of sigdigs there not from what the answer looks like itself, but from the sigdigs in the given numbers, and then round your answer appropriately. So it doesn't matter whether the answer you calculate terminates or not--you MAKE it terminate at the appropriate place!
My concern here was that, like our first question above, Alex was thinking that getting four (or 12) digits on his calculator meant something in itself. So it was important to add this reminder.
Fractions
I mentioned fractions like 5/9 and 22/7 above; the next question, from 2001, is unique:
Significant Figures in 36 2/3 How many significant figures are there in the area "36 2/3 Acres"? This type of notation for acreages is common in old deeds, yet with current standards in most jurisdictions in the U.S.A., areas and sub-areas are to be measured in acres to 4 decimal places (i.e. 36.6667 Acres). Any calculations based on the deeded area figure should take significant figures into account. My conclusion: 36 2/3 = 36.67 decimal.
Ultimately, the answer is going to be that if you use fractions, you really aren’t implying anything about precision; but it took some thought to answer:
Interesting question! My first thought was to apply what I said here: Fraction or Decimal? To a mathematician, a fraction represents an exact number - effectively, infinitely many decimal places. A decimal number might represent an approximation, whose precision is implied by the number of significant digits. But it's also possible to take a fraction as a rough number: "about 2/3." In that case, you can estimate its precision by thinking of it as a quotient; since the numerator and denominator each have one significant digit, it could be taken as having one significant digit, so we would take it as 0.7. Since in addition it is not the number of significant digits, but the number of decimal places that counts, and since the whole part of a mixed number can be seen as a precise number (constrained as it is to be a whole number), your number would then be 36.7, with three effective significant digits and one decimal place. But legally, I would think the real issue would be to determine the intent of the original deed, rather than to legalistically count digits in the number as written. I doubt that precision was even considered when it was written; so the only reasonable thing I can see to do is to take it at face value as an exact number, and then rewrite it according to modern standards as 36.6667 rounded to four decimal places. Other considerations might have to be taken into account, such as the accuracy of measurements used to calculated that figure, but I can't see any better answer given what you have told me. Is there any law dictating how to interpret old deeds?
Math and law are not the same thing!
Is pi 3.14?
Let’s close with one final question from 1998 about an irrational number:
Does Pi Equal 3.14? In my math class pi has always been 3.14. We use this to find area, volume, and any other equations using pi. If pi has an infinite number of digits, how can 3.14 be exact? I am very concerned about our society using 3.14 as pi when it has an infinite number of digits. Anyway, I was just wondering about this.
Doctor Rob took this:
You only need to use a value of Pi with as many significant figures as the measurements you used for other values in the same equation. For example, if you wanted to measure the area of a circle, and the radius was measured as 36.2 feet (to three significant figures), the actual radius might be anywhere in the interval 36.15 < r < 36.25. Then the area would be Pi1306.8225 < A < Pi1314.0625. Even if we used 15 decimal digits for Pi, all we would know about the area is that 4105.50396554585 < A < 4128.24909635784. If we use just 3.14 for Pi, we get 4103.42265 < A < 4126.15625. In either case, we can't tell what the fourth significant digit of A is, and the third digit is probably a 1, but maybe a 0, 2, or 3. The extra digits of Pi don't help unless you make more accurate measurements. The rule of thumb is to use the same accuracy for your constants as for your least accurate measurement, and the answer will have the same accuracy. In this example, 3.140000036.200000^2 = 4114.7816 (exactly), but if we round off to three significant figures, we get 3.1436.2^2 = 4110 as the best thing we can write for the area. Don't worry about society - engineers and mathematicians are well aware of this situation! If you want to use more decimals, you can do so, but your answers will still only have the accuracy of the least accurate quantity in the formula. We usually used 22/7 when I was in school, which is about the same accuracy as 3.14. If you want a better value, you can use 3.1416, or 3.14159, or 355/113, or 3.14159265, depending on the accuracy of your inputs. If the measurements are ideal, and hence exact to infinitely many decimal places, I would leave the answer in the form 100Pi (if r = 10, say). The context of the problem should make it clear whether the quantities are exact or only approximate.
Again, we decide how much precision we need, and use that much precision in our constants. In many school situations, you aren’t going to get any better than three significant digits, so why stress your memory? And if you are not allowed to use a calculator, 22/7 might be easier to use, with no loss of precision. But as I said before, with modern calculators, you can just use the (\pi) button and never worry about it.
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17430 | https://en.wikipedia.org/wiki/Checking_whether_a_coin_is_fair | Checking whether a coin is fair - Wikipedia
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From Wikipedia, the free encyclopedia
Problem in statistics
This article includes a list of general references, but it lacks sufficient corresponding inline citations. Please help to improve this article by introducing more precise citations.(January 2010) (Learn how and when to remove this message)
In statistics, the question of checking whether a coin is fair is one whose importance lies, firstly, in providing a simple problem on which to illustrate basic ideas of statistical inference and, secondly, in providing a simple problem that can be used to compare various competing methods of statistical inference, including decision theory. The practical problem of checking whether a coin is fair might be considered as easily solved by performing a sufficiently large number of trials, but statistics and probability theory can provide guidance on two types of question; specifically those of how many trials to undertake and of the accuracy of an estimate of the probability of turning up heads, derived from a given sample of trials.
A fair coin is an idealized randomizing device with two states (usually named "heads" and "tails") which are equally likely to occur. It is based on the coin flip used widely in sports and other situations where it is required to give two parties the same chance of winning. Either a specially designed chip or more usually a simple currency coin is used, although the latter might be slightly "unfair" due to an asymmetrical weight distribution, which might cause one state to occur more frequently than the other, giving one party an unfair advantage. So it might be necessary to test experimentally whether the coin is in fact "fair" – that is, whether the probability of the coin's falling on either side when it is tossed is exactly 50%. It is of course impossible to rule out arbitrarily small deviations from fairness such as might be expected to affect only one flip in a lifetime of flipping; also it is always possible for an unfair (or "biased") coin to happen to turn up exactly 10 heads in 20 flips. Therefore, any fairness test must only establish a certain degree of confidence in a certain degree of fairness (a certain maximum bias). In more rigorous terminology, the problem is of determining the parameters of a Bernoulli process, given only a limited sample of Bernoulli trials.
Preamble
[edit]
This article describes experimental procedures for determining whether a coin is fair or unfair. There are many statistical methods for analyzing such an experimental procedure. This article illustrates two of them.
Both methods prescribe an experiment (or trial) in which the coin is tossed many times and the result of each toss is recorded. The results can then be analysed statistically to decide whether the coin is "fair" or "probably not fair".
Posterior probability density function, or PDF (Bayesian approach). Initially, the true probability of obtaining a particular side when a coin is tossed is unknown, but the uncertainty is represented by the "prior distribution". The theory of Bayesian inference is used to derive the posterior distribution by combining the prior distribution and the likelihood function which represents the information obtained from the experiment. The probability that this particular coin is a "fair coin" can then be obtained by integrating the PDF of the posterior distribution over the relevant interval that represents all the probabilities that can be counted as "fair" in a practical sense.
Estimator of true probability (Frequentist approach). This method assumes that the experimenter can decide to toss the coin any number of times. The experimenter first decides on the level of confidence required and the tolerable margin of error. These parameters determine the minimum number of tosses that must be performed to complete the experiment.
An important difference between these two approaches is that the first approach gives some weight to one's prior experience of tossing coins, while the second does not. The question of how much weight to give to prior experience, depending on the quality (credibility) of that experience, is discussed under credibility theory.
Posterior probability density function
[edit]
One method is to calculate the posterior probability density function of Bayesian probability theory.
A test is performed by tossing the coin N times and noting the observed numbers of heads, h, and tails, t. The symbols H and T represent more generalised variables expressing the numbers of heads and tails respectively that might have been observed in the experiment. Thus N = H + T = h + t.
Next, let r be the actual probability of obtaining heads in a single toss of the coin. This is the property of the coin which is being investigated. Using Bayes' theorem, the posterior probability density of r conditional on h and t is expressed as follows:
f(r∣H=h,T=t)=Pr(H=h∣r,N=h+t)g(r)∫0 1 Pr(H=h∣p,N=h+t)g(p)d p,{\displaystyle f(r\mid H=h,T=t)={\frac {\Pr(H=h\mid r,N=h+t)\,g(r)}{\int {0}^{1}\Pr(H=h\mid p,N=h+t)\,g(p)\,dp}},}
where _g(r) represents the prior probability density distribution of r, which lies in the range 0 to 1.
The prior probability density distribution summarizes what is known about the distribution of r in the absence of any observation. We will assume that the prior distribution of r is uniform over the interval [0,1]. That is, g(r) = 1. (In practice, it would be more appropriate to assume a prior distribution which is much more heavily weighted in the region around 0.5, to reflect our experience with real coins.)
The probability of obtaining h heads in N tosses of a coin with a probability of heads equal to r is given by the binomial distribution:
Pr(H=h∣r,N=h+t)=(N h)r h(1−r)t.{\displaystyle \Pr(H=h\mid r,N=h+t)={N \choose h}r^{h}(1-r)^{t}.}
Substituting this into the previous formula:
f(r∣H=h,T=t)=(N h)r h(1−r)t∫0 1(N h)p h(1−p)t d p=r h(1−r)t∫0 1 p h(1−p)t d p.{\displaystyle f(r\mid H=h,T=t)={\frac {{N \choose h}r^{h}(1-r)^{t}}{\int {0}^{1}{N \choose h}p^{h}(1-p)^{t}\,dp}}={\frac {r^{h}(1-r)^{t}}{\int {0}^{1}p^{h}(1-p)^{t}\,dp}}.}
This is in fact a beta distribution (the conjugate prior for the binomial distribution), whose denominator can be expressed in terms of the beta function:
f(r∣H=h,T=t)=1 B(h+1,t+1)r h(1−r)t.{\displaystyle f(r\mid H=h,T=t)={\frac {1}{\mathrm {B} (h+1,t+1)}}r^{h}(1-r)^{t}.}
As a uniform prior distribution has been assumed, and because h and t are integers, this can also be written in terms of factorials:
f(r∣H=h,T=t)=(h+t+1)!h!t!r h(1−r)t.{\displaystyle f(r\mid H=h,T=t)={\frac {(h+t+1)!}{h!\,t!}}r^{h}(1-r)^{t}.}
Example
[edit]
For example, let N = 10, h = 7, i.e. the coin is tossed 10 times and 7 heads are obtained:
f(r∣H=7,T=3)=(10+1)!7!3!r 7(1−r)3=1320 r 7(1−r)3.{\displaystyle f(r\mid H=7,T=3)={\frac {(10+1)!}{7!\,3!}}r^{7}(1-r)^{3}=1320\,r^{7}(1-r)^{3}.}
The graph on the right shows the probability density function of r given that 7 heads were obtained in 10 tosses. (Note: r is the probability of obtaining heads when tossing the same coin once.)
Plot of the probability density f(r|H= 7,T=3) = 1320 r 7(1−r)3 with r ranging from 0 to 1
The probability for an unbiased coin (defined for this purpose as one whose probability of coming down heads is somewhere between 45% and 55%)
Pr(0.45<r<0.55)=∫0.45 0.55 f(p∣H=7,T=3)d p≈13%{\displaystyle \Pr(0.45<r<0.55)=\int {0.45}^{0.55}f(p\mid H=7,T=3)\,dp\approx 13\%!}
is small when compared with the alternative hypothesis (a biased coin). However, it is not small enough to cause us to believe that the coin has a significant bias. This probability is slightly _higher than our presupposition of the probability that the coin was fair corresponding to the uniform prior distribution, which was 10%. Using a prior distribution that reflects our prior knowledge of what a coin is and how it acts, the posterior distribution would not favor the hypothesis of bias. However the number of trials in this example (10 tosses) is very small, and with more trials the choice of prior distribution would be somewhat less relevant.)
With the uniform prior, the posterior probability distribution f(r|H= 7,T=3) achieves its peak at r=h/(h+t)=0.7; this value is called the maximum a posteriori (MAP) estimate of r. Also with the uniform prior, the expected value of r under the posterior distribution is
E[r]=∫0 1 r⋅f(r∣H=7,T=3)d r=h+1 h+t+2=2 3.{\displaystyle \operatorname {E} [r]=\int _{0}^{1}r\cdot f(r\mid H=7,T=3)\,\mathrm {d} r={\frac {h+1}{h+t+2}}={\frac {2}{3}}.}
Estimator of true probability
[edit]
The best estimator for the actual value r{\displaystyle r\,!} is the estimator p=h h+t{\displaystyle p\,!={\frac {h}{h+t}}}.
This estimator has a margin of error (E) where |p−r|<E{\displaystyle |p-r|<E} at a particular confidence level.
Using this approach, to decide the number of times the coin should be tossed, two parameters are required:
The confidence level which is denoted by confidence interval (Z)
The maximum (acceptable) error (E)
The confidence level is denoted by Z and is given by the Z-value of a standard normal distribution. This value can be read off a standard score statistics table for the normal distribution. Some examples are:
| Z value | Confidence level | Comment |
:---:
| 0.6745 | gives 50.000% level of confidence | Half |
| 1.0000 | gives 68.269% level of confidence | One std dev |
| 1.6449 | gives 90.000% level of confidence | "One nine" |
| 1.9599 | gives 95.000% level of confidence | 95 percent |
| 2.0000 | gives 95.450% level of confidence | Two std dev |
| 2.5759 | gives 99.000% level of confidence | "Two nines" |
| 3.0000 | gives 99.730% level of confidence | Three std dev |
| 3.2905 | gives 99.900% level of confidence | "Three nines" |
| 3.8906 | gives 99.990% level of confidence | "Four nines" |
| 4.0000 | gives 99.993% level of confidence | Four std dev |
| 4.4172 | gives 99.999% level of confidence | "Five nines" |
The maximum error (E) is defined by |p−r|<E{\displaystyle |p-r|<E} where p{\displaystyle p\,!} is the estimated probability of obtaining heads. Note: r{\displaystyle r} is the same actual probability (of obtaining heads) as r{\displaystyle r\,!} of the previous section in this article.
In statistics, the estimate of a proportion of a sample (denoted by p) has a standard error given by:
s p=p(1−p)n{\displaystyle s_{p}={\sqrt {\frac {p\,(1-p)}{n}}}}
where n is the number of trials (which was denoted by N in the previous section).
This standard error s p{\displaystyle s_{p}} function of p has a maximum at p=(1−p)=0.5{\displaystyle p=(1-p)=0.5}. Further, in the case of a coin being tossed, it is likely that p will be not far from 0.5, so it is reasonable to take p=0.5 in the following:
s p{\displaystyle s_{p}\,!}=p(1−p)n≤0.5×0.5 n=1 2 n{\displaystyle ={\sqrt {\frac {p\,(1-p)}{n}}}\leq {\sqrt {\frac {0.5\times 0.5}{n}}}={\frac {1}{2\,{\sqrt {n}}}}}
And hence the value of maximum error (E) is given by
E=Z s p=Z 2 n{\displaystyle E=Z\,s_{p}={\frac {Z}{2\,{\sqrt {n}}}}}
Solving for the required number of coin tosses, n,
n=Z 2 4 E 2{\displaystyle n={\frac {Z^{2}}{4\,E^{2}}}!}
Examples
[edit]
If a maximum error of 0.01 is desired, how many times should the coin be tossed? n=Z 2 4 E 2=Z 2 4×0.01 2=2500 Z 2{\displaystyle n={\frac {Z^{2}}{4\,E^{2}}}={\frac {Z^{2}}{4\times 0.01^{2}}}=2500\ Z^{2}}n=2500{\displaystyle n=2500\,} at 68.27% level of confidence (Z=1)n=10000{\displaystyle n=10000\,} at 95.45% level of confidence (Z=2)n=27225{\displaystyle n=27225\,} at 99.90% level of confidence (Z=3.3)
If the coin is tossed 10000 times, what is the maximum error of the estimator p{\displaystyle p\,!} on the value of r{\displaystyle r\,!} (the actual probability of obtaining heads in a coin toss)? E=Z 2 n{\displaystyle E={\frac {Z}{2\,{\sqrt {n}}}}}E=Z 2 10000=Z 200{\displaystyle E={\frac {Z}{2\,{\sqrt {10000}}}}={\frac {Z}{200}}}E=0.0050{\displaystyle E=0.0050\,} at 68.27% level of confidence (Z=1)E=0.0100{\displaystyle E=0.0100\,} at 95.45% level of confidence (Z=2)E=0.0165{\displaystyle E=0.0165\,} at 99.90% level of confidence (Z=3.3)
The coin is tossed 12000 times with a result of 5961 heads (and 6039 tails). What interval does the value of r{\displaystyle r\,!} (the true probability of obtaining heads) lie within if a confidence level of 99.999% is desired? p=h h+t=5961 12000=0.4968{\displaystyle p={\frac {h}{h+t}}\,={\frac {5961}{12000}}\,=0.4968}
Now find the value of Z corresponding to 99.999% level of confidence.
Z=4.4172{\displaystyle Z=4.4172\,!}
Now calculate E
E=Z 2 n=4.4172 2 12000=0.0202{\displaystyle E={\frac {Z}{2\,{\sqrt {n}}}}\,={\frac {4.4172}{2\,{\sqrt {12000}}}}\,=0.0202}
The interval which contains r is thus:
p−E<r<p+E{\displaystyle p-E<r<p+E\,!}0.4766<r<0.5170{\displaystyle 0.4766<r<0.5170\,!}
Other approaches
[edit]
Other approaches to the question of checking whether a coin is fair are available using decision theory, whose application would require the formulation of a loss function or utility function which describes the consequences of making a given decision. An approach that avoids requiring either a loss function or a prior probability (as in the Bayesian approach) is that of "acceptance sampling".
Other applications
[edit]
The above mathematical analysis for determining if a coin is fair can also be applied to other uses. For example:
Determining the proportion of defective items for a product subjected to a particular (but well defined) condition. Sometimes a product can be very difficult or expensive to produce. Furthermore, if testing such products will result in their destruction, a minimum number of items should be tested. Using a similar analysis, the probability density function of the product defect rate can be found.
Two party polling. If a small random sample poll is taken where there are only two mutually exclusive choices, then this is similar to tossing a single coin multiple times using a possibly biased coin. A similar analysis can therefore be applied to determine the confidence to be ascribed to the actual ratio of votes cast. (If people are allowed to abstain then the analysis must take account of that, and the coin-flip analogy doesn't quite hold.)
Determining the sex ratio in a large group of an animal species. Provided that a small random sample (i.e. small in comparison with the total population) is taken when performing the random sampling of the population, the analysis is similar to determining the probability of obtaining heads in a coin toss.
See also
[edit]
Binomial test
Coin flipping
Confidence interval
Estimation theory
Inferential statistics
Loaded dice
Margin of error
Point estimation
Statistical randomness
References
[edit]
^However, if the coin is caught rather than allowed to bounce or spin, it is difficult to bias a coin flip's outcome. See Gelman, Andrew; Deborah Nolan (2002). "Teacher's Corner: You Can Load a Die, But You Can't Bias a Coin". American Statistician. 56 (4): 308–311. doi:10.1198/000313002605. S2CID123597087.
^Cox, D.R., Hinkley, D.V. (1974) Theoretical Statistics (Example 11.7), Chapman & Hall. ISBN0-412-12420-3
Guttman, Wilks, and Hunter: Introductory Engineering Statistics, John Wiley & Sons, Inc. (1971) ISBN0-471-33770-6
Devinder Sivia: Data Analysis, a Bayesian Tutorial, Oxford University Press (1996) ISBN0-19-851889-7
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Reasoning with Equations and Inequalities HSA-REI.A.1
1. Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
Who was it that said, "A journey of a thousand miles begins with a single step"? Was it Socrates? Confucius? Bugs Bunny?
Whoever it was, they probably weren't thinking about algebra when they coined that gem. To many students, solving an algebraic equation may feel like a journey of a thousand miles. Before students begin with their single step, they should probably get a compass or something.
Students should be able to figure out the logical next step in solving an equation using the previous step. Sounds simple, but it takes more than just putting one foot in front of the other. Knowing that whatever is done to one side of the equation must be done to the other is a good start, but that doesn't tell them what to do.
Ideally, students should not be memorizing a set of rules and procedures and using them to solve equations. They should understand how the next step in solving an equation can be logically derived from the previous step, but there is a general format for how to get started on their algebraic journey.
When we solve an equation, it's usually a good idea to get the variable we want on one side of the equal sign. Then, we do what we can to simplify it as much as we can.
For example, let's solve the equation x2 – 3x – 7 = 2x + 17. Subtracting 2x + 17 from both sides gets all our x's on one side of the equation. That's a good place to start.
x2 – 3x – 7 – (2x + 17) = 2x + 17 – (2x + 17)
x2 – 5x – 24 = 0
We also took the opportunity to simplify by combining like terms. Since the equation we have is a quadratic equation, we can factor it into the product of two linear terms.
(x – 8)(x + 3) = 0
This product will equal 0 when either x – 8 or x + 3 is equal to 0, so we can solve it by setting each of them to 0. As a result, x = 8 and -3.
If students are really struggling, we suggest starting simple. Give them easier linear equations and slowly work your way to more complex ones. Point out patterns in equations, make sure they know the quadratic formula, and remind them of helpful factoring tricks.
Most of all, get them to practice. It's difficult to take even the first step of a journey if you can't walk. After enough exercises, your students should be able to tackle any thousand-mile journey faster than the Road Runner.
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17432 | https://www.quora.com/What-is-the-intersection-of-an-empty-set-and-does-it-contain-itself | What is the intersection of an empty set and does it contain itself? - Quora
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What is the intersection of an empty set and does it contain itself?
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Terry Moore
M.Sc. in Mathematics, University of Southampton (Graduated 1968) · Author has 16.6K answers and 29.4M answer views
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What is the intersection of an empty set and does it contain itself?
The intersection of the empty set with any set (even the intersection with itself) is empty. There are no elements in the set so none in common with any other set. Yes, the empty set is a subset and a superset of itself. You should be able to think that through.
Or perhaps you mean “What is the empty intersection?” i.e. the set of elements that are not in any of the empty list of sets: ∩x∈∅A x∩x∈∅A x where sets are indexed A x A x for x\inC x\inC. In practical applications of set theory we usually restrict all sets to subsets o
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What is the intersection of an empty set and does it contain itself?
The intersection of the empty set with any set (even the intersection with itself) is empty. There are no elements in the set so none in common with any other set. Yes, the empty set is a subset and a superset of itself. You should be able to think that through.
Or perhaps you mean “What is the empty intersection?” i.e. the set of elements that are not in any of the empty list of sets: ∩x∈∅A x∩x∈∅A x where sets are indexed A x A x for x\inC x\inC. In practical applications of set theory we usually restrict all sets to subsets of some universal set, Ω Ω. And ∩x∈∅A x=Ω∩x∈∅A x=Ω (of course Ω Ω contains itself). You can think of this as a convention. If you want to find ∩x∈U A x∩x∈U A x, you can well order U U and intersect the sets one at a time. To be precise, as you include a new set you take the intersection of the new set with the intersection of all the previous sets. When you start without any sets you need to start with Ω Ω, otherwise, if the first set in U U is Ω Ω, the answer will come out wrong.
There is only one.
This may be a transfinite process.
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How do you prove that the intersection of an empty collection is a universal set?
What is the intersection of an empty set with S?
If the empty set represents no collection at all, why will a set containing the empty sets not be an empty set itself?
Why is the intersection of two empty sets empty?
Does an empty set contain itself?
Andrew Winkler
PhD in Mathematics, Courant Institute of Mathematical Sciences, NYU (Graduated 1987) · Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) · Author has 9.6K answers and 10.9M answer views
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It depends on your choice of set theory. It’s the class of all sets, where the concept of class exists. In zfc intersection is only defined for inhabited sets.
A is in the intersection of B if and only if for all C in B, A is in C. If B is empty, this is true for all A. In particular it’s true for the empty set.
But in mathema...
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Jon Obermark
M.S. in Mathematics&Logic, University of Illinois at Urbana-Champaign (Graduated 1993) · Author has 6.1K answers and 1.5M answer views
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If the intersection of A is everything that is in every set in A, then it contains any thing x that does not have an a in A such that x is not in a. No set is outside any of the zero sets within A. So everything would be in the intersection of an empty set. The empty set would not have any sets within it to rule out any potential element, so that element would be admitted.
In standard set theory, this is not allowed, simply because the set itself is ruled out by the axioms that say no set can contain itself. The answer is defined to be undefined so that we cannot use it to make trouble.
But in s
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If the intersection of A is everything that is in every set in A, then it contains any thing x that does not have an a in A such that x is not in a. No set is outside any of the zero sets within A. So everything would be in the intersection of an empty set. The empty set would not have any sets within it to rule out any potential element, so that element would be admitted.
In standard set theory, this is not allowed, simply because the set itself is ruled out by the axioms that say no set can contain itself. The answer is defined to be undefined so that we cannot use it to make trouble.
But in set theories like Quine’s, where there is a universal set, and self-containment is allowed, yes, the intersection would be the universal set, and that would obviously contain the empty set.
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Hilmar Zonneveld
Translator (1985–present) · Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) · Author has 58.5K answers and 19.4M answer views
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“Intersection” is usually defined for more than one set. Though technically, the intersection of a single set is just the set itself.
The empty set, by definition, contains nothing … so of course it doesn’t contain itself either.
The empty is, however, a subset of itself. Just like any other set is a subset of itself.
Be sure to understand the difference of something being contained in a set, and being a subset of a set. It’s not the same thing.
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Is empty set the set of itself?
Why is the set A-A called an empty set? What does this mean mathematically?
What is the definition of the empty set in mathematics? Why is it considered as a member of any other set?
Is the intersection of all sets always equal to the empty set?
What's the complement of an empty set?
Dan Christensen
Creator of proof-checking freeware DC Proof 2.0 · Author has 2.8K answers and 834.4K answer views
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Note that we can have intersections of pairs of sets, and of arbitrary collections of sets. Here, we are dealing with the latter.
The intersection of an empty set (i.e. of an empty collection of sets) is itself an empty set. (This can be formally proven using the axioms of set theory.) Being empty, it cannot contain any elements.
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John Gould
I've been here a while. · Author has 1.9K answers and 3.1M answer views
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Can the intersection of a family that is indexed by the empty set have a meaning? I know the mathematical convention dictates that it's equal to the whole set, but I take issue with the fact that the whole set depends on the context.
Question originally answered: Can the intersection of a family that is indexed by the empty set have a meaning? I know the mathematical convention dictates that it's equal to the whole set, but I take issue with the fact that the whole set depends on the context.
An expression means precisely what it is defined to mean, no more and no less.
Of course we can define: ⋂i∈∅=A⋂i∈∅=A.
Now I understand that it is context dependent what A A denotes, but in any mathematical theory you are considering, you simply will have some background universe which forms your domain of discourse.
And i
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Question originally answered: Can the intersection of a family that is indexed by the empty set have a meaning? I know the mathematical convention dictates that it's equal to the whole set, but I take issue with the fact that the whole set depends on the context.
An expression means precisely what it is defined to mean, no more and no less.
Of course we can define: ⋂i∈∅=A⋂i∈∅=A.
Now I understand that it is context dependent what A A denotes, but in any mathematical theory you are considering, you simply will have some background universe which forms your domain of discourse.
And in the case that this universe forms a set, the notation is both well defined and natural.
Just like in that case the definition ⋃i∈∅=∅⋃i∈∅=∅ is both well defined and natural.
The naturality of these definitions might be called into question by some, but the intuitive argument for them is easily understandable.
If we look at families of subsets of a set (for example sets of integers), then we observe that the following is in all cases true:
The intersection of two sets can never contain more elements than the individual sets themselves and, unless A A and B B are the same set, will actually have less members:
|A∩B|≤|A||A∩B|≤|A|
|A∩B|≤|B||A∩B|≤|B|
or, putting them together:
|A∩B|≤min(|A|,|B|)|A∩B|≤min(|A|,|B|)
This easily generalises to : |⋂i∈I A i|≤min i∈I(|A i|)|⋂i∈I A i|≤min i∈I(|A i|). So unless the A i A i are identical, the more sets we intersect, the smaller the result is. Conversely, the less sets we take the intersection of the larger the result. It is therefore completely natural to define the intersection of zero sets as the largest possible set, which is our background universe set.
Analogously we have:
|A∪B|≥|A||A∪B|≥|A|
|A∪B|≥|B||A∪B|≥|B|
or, putting them together:
|A∪B|≥max(|A|,|B|)|A∪B|≥max(|A|,|B|)
This easily generalises to : |⋃i∈I A i|≥max i∈I(|A i|)|⋃i∈I A i|≥max i∈I(|A i|). So unless the A i A i are identical, the more sets we take the union of, the larger the result is. Conversely, the less sets we take the union of, the smaller the result. It is therefore completely natural to define the union of zero sets as the smallest possible set, which is the empty set.
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Danya Rose
I'm an Aussie feller, I've been one all my life. It may make me seem funny... · Author has 2K answers and 7M answer views
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Is an empty set the subset of a set containing the empty set?
First, in a given set theory, there is only one empty set. Thus the indefinite article “an” is incorrect.
Second, there is only one set which has exactly one subset; the empty set itself. Using the definite article “the” to refer to “the subset of a set” implies that the set you’re asking about is the empty set. This is because , if we use |X||X| to denote (in some sense) the “size” or “number of elements” of X X, the number of subsets of X X is 2|X|2|X|. Thus 2|∅|=2 0=1 2|∅|=2 0=1.
Any set X X containing the empty set (in the sense that ∅∈X∅∈X) has at least two subsets, because it has at leas
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First, in a given set theory, there is only one empty set. Thus the indefinite article “an” is incorrect.
Second, there is only one set which has exactly one subset; the empty set itself. Using the definite article “the” to refer to “the subset of a set” implies that the set you’re asking about is the empty set. This is because , if we use |X||X| to denote (in some sense) the “size” or “number of elements” of X X, the number of subsets of X X is 2|X|2|X|. Thus 2|∅|=2 0=1 2|∅|=2 0=1.
Any set X X containing the empty set (in the sense that ∅∈X∅∈X) has at least two subsets, because it has at least one member.
Regardless of even that fact, the empty set is a subset of every set. This is because of the definition of a subset. We say that X⊆Y X⊆Y if and only if for each x∈X x∈X it is true that x∈Y x∈Y. Because there is no x x such that x∈∅x∈∅, this statement is vacuously true of the empty set: every element of ∅∅ is indeed a member of every definable set, so the empty set is a subset of every set.
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Alan Bustany
Trinity Wrangler, 1977 IMO · Author has 9.8K answers and 58.5M answer views
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What's the difference between the empty set and the set containing only the empty set?
The difference between the empty set
∅≡{}∅≡{}
and the set containing the empty set
{∅}≡{{}}{∅}≡{{}}
is the empty set
∅≡{}∅≡{}
as can be seen by simple visual inspection.
As pointed out by other others, it is analogous to the difference between:
An empty bag and a bag containing an empty bag;
An empty bucket and a bucket containing an empty bucket; or
Any empty container and a container containing an empty container.
The bottom line is that the set containing the empty set is obviously not empty because it contains something, namely the empty set whi
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The difference between the empty set
∅≡{}∅≡{}
and the set containing the empty set
{∅}≡{{}}{∅}≡{{}}
is the empty set
∅≡{}∅≡{}
as can be seen by simple visual inspection.
As pointed out by other others, it is analogous to the difference between:
An empty bag and a bag containing an empty bag;
An empty bucket and a bucket containing an empty bucket; or
Any empty container and a container containing an empty container.
The bottom line is that the set containing the empty set is obviously not empty because it contains something, namely the empty set which is not nothing.
Confusion lies in incorrectly identifying the empty set with nothing. That mistake makes you think that the set containing the empty set is the set containing nothing which is the empty set. But that is wrong :-(
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Peter Hoffman
Author has 6.7K answers and 645.1K answer views
·1y
1/ Any phrase ‘The intersection of X’, where X is a set, is UTTERLY MEANINGLESS.
2/ YES.
You might take the trouble to read up on the use of the set theory word ‘intersection’ in virtually any elementary pure math undergrad textbook, or online for that matter.
And it is THE empty set, not “AN”. There can be only one, because a set is determined by which things are its elements and by nothing else.
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Joachim Pense
Got a degree in Mathematics. · Upvoted by
Michael Harrison
, M.S. Mathematics, Louisiana State University (2007) and
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 9.2K answers and 14.6M answer views
·Updated 3y
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Does the set of all sets contain itself?
There can be no set of all sets. If there was one, it would indeed contain itself, and then you could build the set of all sets that don't contain themselves, which is not possible because it neither can nor cannot contain itself (Russell's antinomy).
So what you can have instead in (one of the standard versions of) formal set theories is the class of all sets, something similar to a set but not itself a set, and hence it does not contain itself.
A class is more or less what we think of as a set, and sets are special classes. The point is that sets can be elements of other classes, but some clas
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There can be no set of all sets. If there was one, it would indeed contain itself, and then you could build the set of all sets that don't contain themselves, which is not possible because it neither can nor cannot contain itself (Russell's antinomy).
So what you can have instead in (one of the standard versions of) formal set theories is the class of all sets, something similar to a set but not itself a set, and hence it does not contain itself.
A class is more or less what we think of as a set, and sets are special classes. The point is that sets can be elements of other classes, but some classes (so called proper classes) cannot. The class of all sets (AKA the universe) is a proper class, and in particular does not contain itself.
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Balázs Iván József
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Does the null set contain itself?
As written, the question is ambiguous. It could mean either of two things:
Is it true that ∅∈∅∅∈∅?
To that the answer is no, because if there is any x x such that x∈∅x∈∅then ∅∅ is not empty as it is defined to be.
2. Is it true that ∅⊆∅∅⊆∅?
To this the answer is yes, since every set is a subset of itself (that is, for all a a in A A, if a∈B a∈B, then A⊆B A⊆B).
The reason for the ambiguity is that the verb “contains” couple imply “contains as a member” (the first interpretation) or “contains as a subset” (the second).
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Why the union of empty sets is empty but the intersection of empty sets is ground set?
You seem to be confusing two very different things; the intersection of empty sets is empty. If you have a universal set, then an empty intersection is the universal set; that is, the intersection of the elements of the empty set — there aren't any — is the universal set.
This is to make intersections over no sets work like intersections over larger sets of sets.
Given two families of sets S S and T T, we want
∩s∈S∩∩s∈T=∩s∈(S∪T)∩s∈S∩∩s∈T=∩s∈(S∪T)
If S S is empty, then that comes out as
∩s∈∅∩∩s∈T=∩s∈T∩s∈∅∩∩s∈T=∩s∈T
But what can \cap_{s\in\varnothin\cap_{s\in\varnothin
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You seem to be confusing two very different things; the intersection of empty sets is empty. If you have a universal set, then an empty intersection is the universal set; that is, the intersection of the elements of the empty set — there aren't any — is the universal set.
This is to make intersections over no sets work like intersections over larger sets of sets.
Given two families of sets S S and T T, we want
∩s∈S∩∩s∈T=∩s∈(S∪T)∩s∈S∩∩s∈T=∩s∈(S∪T)
If S S is empty, then that comes out as
∩s∈∅∩∩s∈T=∩s∈T∩s∈∅∩∩s∈T=∩s∈T
But what can ∩s∈∅∩s∈∅ be to make this work? For any set t t, we can have T={T}T={T}, and ∩∩s∈T=t∩∩s∈T=t; but then
∩s∈∅∩t=t∩s∈∅∩t=t.
Nothing less than the universal set (if there is one) will do for ∩s∈∅∩s∈∅.
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Why the union of empty sets is empty but the intersection of empty sets is ground set?
In general x∈⋃A x∈⋃A if and only if x∈a x∈a for some a∈A a∈A.
(note that smaller A A gives smaller⋃A⋃A)
Applying that on A=∅A=∅ we must conclude that for no x x this condition can satisfied because no a a exists that satisfies a∈∅a∈∅.
So we conclude that ⋃∅=∅⋃∅=∅.
In general x∈⋂A x∈⋂A if and only if x∈a x∈a for every a∈A a∈A.
(note that smaller A A gives larger⋂A⋂A)
Applying that on A=∅A=∅ we must conclude that for every x x this condition is satisfied because no a∈∅a∈∅ exists for which x∈a x∈a is false. You cannot find such a a simply because
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In general x∈⋃A x∈⋃A if and only if x∈a x∈a for some a∈A a∈A.
(note that smaller A A gives smaller⋃A⋃A)
Applying that on A=∅A=∅ we must conclude that for no x x this condition can satisfied because no a a exists that satisfies a∈∅a∈∅.
So we conclude that ⋃∅=∅⋃∅=∅.
In general x∈⋂A x∈⋂A if and only if x∈a x∈a for every a∈A a∈A.
(note that smaller A A gives larger⋂A⋂A)
Applying that on A=∅A=∅ we must conclude that for every x x this condition is satisfied because no a∈∅a∈∅ exists for which x∈a x∈a is false. You cannot find such a a simply because ∅∅ has no elements.
So we tend to conclude now that actually ⋂∅⋂∅ denotes the class of all sets. I use “class” here because in axiom systems like ZFC this collection is not a set.
If we are only looking at sets that are elements of some universal set U U then it is natural and convenient to identify ⋂∅⋂∅ with U U itself (the collection of all subsets).
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Related questions
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A question about modulus for polynomials
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The other day my friend was asked to find A and B in the equation (x 3+2 x+1)17≡A x+B(mod x 2+1) A method was proposed by our teacher to use complex numbers and especially to let x=i where i is the imaginary unit. We obtain from that substitution (i+1)17≡A i+B(mod 0) which if we have understood it correctly is valid if we define a≡b(mod n) to be a=b+d n. Running through with this definition we have (i+1)17=(√2(cos(π 4)+sin(π 4)))17=√2 17(cos(17 π 4)+sin(17 π 4))=256(√2(cos(π 4)+sin(π 4)))=256(1+i)=256+256 i which gives the correct coefficient values for A and B. Our questions are
Why is this substitution valid to begin with?
It seems here that the special case (x=i) implies the general case (x), why is that valid?
elementary-number-theory
modular-arithmetic
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edited Oct 31, 2019 at 0:33
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asked Oct 30, 2019 at 14:08
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1 The use of complex numbers can be explained the following way: C is R[i], i.e R along with a solution of X 2+1. The quotient R[X]/(X 2+1) is actually the same thing, it is R, in which you adjoin some element ˉ X (the class of X in the quotient) that satisfy X 2+1=0. So you can freely think of this quotient as C and substitute X to i.user187567 –user187567 2019-10-30 14:14:32 +00:00 Commented Oct 30, 2019 at 14:14
seems easier to show it's congruent to (x+1)17 and work from that honestly. edit realized you did that, polynomial remainder theorem to the rescue.user645636 –user645636 2019-10-30 19:11:40 +00:00 Commented Oct 30, 2019 at 19:11
1 I'm gonna include exercises like this to my algebra course this coming Spring. +1 to all posters.Jyrki Lahtonen –Jyrki Lahtonen 2019-10-30 19:27:47 +00:00 Commented Oct 30, 2019 at 19:27
1 I added some more examples to my answer that may help you better understand the universal algebraic principles at the heart of the matter. Again, questions are always welcome.Bill Dubuque –Bill Dubuque 2019-10-31 00:22:51 +00:00 Commented Oct 31, 2019 at 0:22
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Here I think it's easier to see what's going on if we forgo the modular arithmetic and look at simple factoring and remainder. We have (x 3+2 x+1)17=(x 2+1)Q(x)+A x+B for some polynomial Q. Which polynomial? We don't really care. The main point is that the left-hand side and the right-hand side are the same polynomial.
And since they are the same, they must give the same value when we evaluate them at x=i. So we insert x=i and get (i 3+2 i+1)17=(i 2+1)Q(i)+A i+B(i+1)17=0⋅Q(i)+A i+B Knowing that A,B are real means we can find them directly from this, as Q disappears.
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edited Oct 30, 2019 at 14:40
answered Oct 30, 2019 at 14:14
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{To hot list readers: you can understand a good part of this answer knowing only modular arithmetic. Please feel welcome to ask questions and it will be my pleasure to elaborate.]
First let's do it using only modular arithmetic (congruences).
mod x 2+1:f 2 N=(x 3+2 x+1)2 N≡(x+1)2 N b y x 2≡−1⇒x 3≡−x≡(2 x)N b y(x+1)2≡(x 2+1)+2 x≡2 x≡2 N x N mod 4 b y x 2≡−1⇒x 4≡1
which yields: f 17≡(x+1)16(x+1)≡2 8 x 8 mod 4(x+1)≡256(x+1)
Calculating in C:=Rx amounts to calculating modulo a generic root of x 2+1, i.e. we reason using only the ring axioms in R[x] plus the additional hypothesis that x 2=−1. But C≅R[x]/(x 2+1)≡R[i]≅C because x 2+1 is the minimal polynomial of i over R. This means that we can replace x by i in the above calculations. This will be clarified when one studies polynomial and quotient rings in abstract algebra - esp. their universal properties. These imply that the above calculations hold true in every ring with an element x such that x 2≡−1. For example in Z/5= integers mod 5 we have 2 2≡−1 therefore by specializing the above for x≡2 we obtain the congruences below mod 5, and similarly for the following congruences when x≡3,4,…10.
mod x 2+1:f(x)17≡(x+1)17≡256(x+1)x=2:mod 5:13 17≡3 17≡256(3)≡3 x=3:mod 10:34 17≡4 17≡256(4)≡4 x=4:mod 17:73 17≡5 17≡256(5)≡5 mod 26:136 17≡6 17≡256(6)≡2 mod 37:229 17≡7 17≡256(7)≡16 mod 50:358 17≡8 17≡256(8)≡48 mod 65:529 17≡9 17≡256(9)≡29 mod 82:748 17≡10 17≡256(10)≡18 x=10:mod 101:1021 17≡11 17≡256(11)≡89
Our calculations persist to hold true in all these rings because they used only ring laws and the hypothesis that x 2≡−1. You may have encountered simpler linear examples in terms of fractions (roots of linear polynomials), e.g. the equality 1/6=1/2−1/3 holds true in any rings where 2&3 are invertible, e.g. mod 5 it is 1≡3−2, and mod 11 it is 2≡6−4. This can also be used in exponents to get a sixth root from a square root and cube root: a 1/6≡a 1/2/a 1/3 and again it also works when the exponents are modular, e.g. see here. This all works because we used only rings laws and the added hypotheses 2 x≡1,3 y≡1, so our calculations will persist to hold true in any rings where these equations have roots x,y. This is a special case of the universal properties of rings of fractions (and localizations).
These are a couple simple examples of the power afforded by the algebraic abstractions at the foundation of "abstract algebra".
Such modular arithmetic calculations in quadratic extensions often proves convenient, e.g. see the quadratic extension of Hermite's cover-up method for partial fraction expansion.
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edited Oct 31, 2019 at 1:16
answered Oct 30, 2019 at 15:54
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It comes from the exponential form of complex numbers: 1+i has modulus √2 and argument π 4, so it writes as 1+i=√2 e π 4,and similarly 1−+i=√2 e−π 4 The substitution is valid because of the meaning of the congruence: (x 3+2 x+1)17≡A x+B(mod x 2+1)⟺∃q(x):(x 3+2 x+1)17=(x 2+1)q(x)+A x+B, so when you set x=±i, the first term in the r.h.s. cancels.
Setting x=i yields an equation for A and B, that's all.
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edited Oct 31, 2019 at 21:12
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17434 | https://www.stat.purdue.edu/~dkjlin/documents/publications/2014/2014_JQT_1.pdf | Dimensional Analysis and Its Applications in Statistics WEIJIE SHEN The Pennsylvania State University, University Park, PA 16802, USA TIM DAVIS We Predict Ltd., Technium 1, Kings Road, Swansea, SA1 8PH, UK DENNIS K. J. LIN The Pennsylvania State University, University Park, PA 16802, USA CHRISTOPHER J. NACHTSHEIM University of Minnesota, Minneapolis, MN 55455, USA Dimensional analysis (DA) is a well-developed, widely-employed methodology in the physical and en-gineering sciences. The application of dimensional analysis in statistics leads to three advantages: (1) the reduction of the number of potential causal factors that we need to consider, (2) the analytical insights into the relations among variables that it generates, and (3) the scalability of results. The formalization of the dimensional-analysis method in statistical design and analysis gives a clear view of its generality and overlooked significance. In this paper, we first provide general procedures for dimensional analysis prior to statistical design and analysis. We illustrate the use of dimensional analysis with three practical examples.
In the first example, we demonstrate the basic dimensional-analysis process in connection with a study of factors that a↵ect vehicle stopping distance. The second example integrates dimensional analysis into the regression analysis of the pine tree data. In our third example, we show how dimensional analysis can be used to develop a superior experimental design for the well-known paper helicopter experiment.
In the regression example and in the paper helicopter experiment, we compare results obtained via the dimensional-analysis approach to those obtained via conventional approaches. From those, we demonstrate the general properties of dimensional analysis from a statistical perspective and recommend its usage based on its favorable performance.
Key Words: Buckingham’s ⇧Theorem; Design of Experiment; Dimensions; Statistical Analysis.
1. Introduction D IMENSIONAL ANALYSIS (DA) is a well-established method in physics (see Sonin (2001), Szirtes (2007)). Bridgman (1931) stated that “The principal use of dimensional analysis is to deduce from a study of the dimensions of the variables in any physical sys-tem certain limitations on the form of any possible Mr. Shen is a Doctoral Student in the Department of Statis-tics, Pennsylvania State University. His email address is weijie .shen@psu.edu.
Dr. Davis is the Director of Timdavis Consulting Ltd., Chief Technical Officer at We Predict Ltd., and Professor at the University of Warwick. His email address is tim@timdavis .co.uk.
Dr. Lin is a Distinguished Professor of Statistics and Sup-ply Chain in the Department of Statistics, Pennsylvania State University. His email address is dkl5@psu.edu.
Dr. Nachtsheim is a Professor, the Frank A. Donaldson Chair of Operations Management, and Chair of the Supply Chain and Operations Department in the Carlson School of Management, University of Minnesota. His email address is nacht001@umn.edu.
Vol. 46, No. 3, July 2014 185 www.asq.org 186 WEIJIE SHEN ET AL.
relationship between those variables. The method is of great generality and mathematical simplicity”. It is mainly used to find the relations among physical quantities in complicated physical systems by their dimensions. A variety of literature has applied di-mensional analysis in various fields. See Asmussen and Heebooll-Nielsen (1955), Islam and Lye (2009), and Stahl (1962), for examples. Through these anal-yses, some simple rules among those quantities can be extracted. As a dimension-reduction and feature-extraction methodology, dimensional analysis could be of great use to the field of statistics. This a pri-ori analysis gives us a conceptual and analytical view of the problem we are dealing with, thereby provid-ing guidance in both the design and analysis steps.
Furthermore, the physical origin of dimensional anal-ysis improves the interpretability of the final results, which is particularly desirable to the fields of physics and engineering.
Unfortunately, statisticians seem to have over-looked the advantages of dimensional analysis.
Finney (1977) commented that “I am surprised by the lack of attention given to dimensions as a check on the theory and practice of statistics. The basic ideas, readily appreciated, should form part of the stock-in-trade of every statistician”. In this paper, we focus on building functional relationships between inputs and outputs. We will first introduce the basic concept and general procedure of dimensional anal-ysis. Illustrated by two examples, the pine tree and paper helicopter, we show how to apply dimensional analysis in real problems and also compare the results with classic approaches. We summarize the proper-ties and discuss the advantages of dimensional anal-ysis in the statistical context.
The rest of the paper is organized as follows. In the Section 2, we introduce the definitions and gen-eral procedures of dimensional analysis with an il-lustrative example. In Section 3, we use dimensional analysis for data analysis and show its generality and importance. In Section 4, dimensional analysis is ap-plied to design of experiments. The last two sections summarize general properties, followed by some con-cluding remarks and prospective research.
2. Definitions of Dimensional Analysis and General Procedure with Illustrative Example 2.1. Physical Dimensions In mathematics, dimension typically refers to the number of coordinates required to define points in abstract spaces, whereas in statistics, dimension typ-ically refers to the number of variables in a design problem or a data set. However, physical dimen-sions refer to the measurement systems to charac-terize certain objects. Each physical dimension has several empirical scales of the measurements and they are called “units”. Ignoring nuclear e↵ects such as isospin, charm, and strangeness, there are seven fundamental physical dimensions: namely, mass M, length L, time T, temperature ⇥, electric current I (or charge Q), amount of substance mol, and lumi-nous intensity Iv. The corresponding units, defined by SI (International System of Units), are kilogram, meter, second, kelvin, ampere, mole, and candela, re-spectively. All other physical quantities are combina-tions of these fundamental quantities and their units are combinations of the units of the corresponding fundamental quantities, combined in the same way.
For example, speed has the dimension of length per time, for which the SI unit is meters per second.
2.2. Background of Dimensional Analysis Physical quantities cannot be constructed unre-strictedly. For example, it makes no sense to add “length” to “mass” due to the natural constraints in the physical quantities. The main constraint is that “a physical law must be independent of the units used to measure the physical quantities”. This was first proposed by Joseph Fourier in the 19th cen-tury (see Mason (1962)). This principle has been formalized in two important theorems, Bucking-ham’s ⇧-theorem (Buckingham (1914, 1915a,b)) and Bridgman’s principle of absolute significance of rela-tive magnitude (Bridgman, 1931). Buckingham’s ⇧-theorem shows that physical equations must be di-mensionally homogeneous. In other words, any mean-ingful equations (and inequalities) must have the same dimensions in both the left and right sides.
Bridgman’s principle of absolute significance of rela-tive magnitude shows that such formulae should be in the power-law form. Basically, Bridgman’s principle allows us to transform physical quantities properly, especially into dimensionless forms. The method of using dimensionless quantities and Buckingham’s ⇧-theorem to remove such constraints is called dimen-sional analysis. Next, we introduce Buckingham’s ⇧-theorem and how to use it in practice.
2.3. General Procedure We recommend applying dimensional analysis be-fore statistical analysis to give a general view of the problems and the variables involved. From the physi-Journal of Quality Technology Vol. 46, No. 3, July 2014 DIMENSIONAL ANALYSIS AND ITS APPLICATIONS IN STATISTICS 187 cal perspective, procedures can be found in Taylor et al. (2007) and others. From the statistical perspec-tive, the general procedure of dimensional analysis can be specified as follows: Step 1. Determine the input and output variables and their dimensions, respectively.
Step 2. Determine the basis quantities.
Step 3. Transform input and output variables into dimensionless quantities by using basis quantities in step 2.
Step 4. Re-express the model functions via trans-formed variables in step 3.
Step 1 Determine the input and output variables of the system we consider. Denote input variables as Q1, . . . , Qp and the output variable (response) as Q0.
The conventional model will be Q0 = f(Q1, . . . , Qp), where f is the model function to be estimated. Note that, in dimensional analysis, Qi may include rel-evant physical constants with dimensions, such as gravitational constant = 6.67300⇥10−11m3kg−1s−2.
The units are often standardized to avoid dimension-less multiplicative constants. But standardization is not always necessary because these constants are combined into unknown functional relationships. Af-ter checking the physical meaning of all the variables we consider, we determine the relevant fundamen-tal physical quantities in the system of the seven SI units as shown in the previous section: denote them as q1, . . . , q7. Further denote the dimensions of Qi as [Qi] and qj as [qj] for i = 0, 1, . . . , p, j = 1, . . . , 7. We express the dimensions of Q0, Q1, . . . , Qp in terms of [q1], . . . , [q7] as [Qi] = [q1]ri1 · · · [q7]ri7, for some proper choices of {rij} with i = 0, 1, . . . , p, j = 1, . . . , 7.
Step 2 Determine the basis quantities. The basis quanti-ties constitute a subset of the inputs. We reorder and denote them as Q1, . . . , Qt, where t 7 as discussed above and t p. The basis quantities should satisfy two conditions: (1) “Representativity”: the dimen-sions of any other quantities, [Q0], [Qt+1], . . . , [Qp] can be expressed by the combinations of the dimen-sions of the basis quantities, [Q1], . . . , [Qt]. The com-binations take the form of power law. (2) “Indepen-dence”: the dimension of any basis quantities cannot be expressed by the combinations of the dimensions of other basis quantities. Furthermore, assume that [Q0] can be expressed by the combinations of [Qi], i = 1, . . . , p. If not, dimensional homogeneity is vio-lated. This assumption leads to the existence of the basis quantities but they are not unique. However, the number of basis quantities is a fixed constant.
The concept of basis in linear algebra is a very good analogy to the concept of basis quantities.
Step 3 Transform input and output variables into di-mensionless quantities by using basis quantities. We mainly transform variables that are not basis quan-tities, i.e., Q0, Qt+1, . . . , Qp, based on Buckingham’s ⇧-theorem. Due to the two properties of basis quan-tities in step 2, we can have [Qi] = [Q1]di1 · · · [Qt]dit, i = 0, t + 1, t + 2, . . . , p. Consequently, the transformed dimensionless quantities are ⇧i = Qi · Q−di1 1 · · · Q−dit t , i = 0, t + 1, t + 2, . . . , p, because [⇧i] = [QiQ−di1 1 · · · Q−dit t ] = [Q1]di1 · · · [Qt]dit[Q1]−di1 · · · [Qt]−dit = 1.
Step 4 Re-express the response functions. Before using di-mensional analysis, we have Q0 =f(Q1, . . . , Qt, Qt+1, . . . , Qp). Using ⇧i instead of Qi, we have the follow-ing expression: ⇧0Qd01 1 · · · Qd0t t = f(Q1, . . . , Qt, ⇧t+1Qdt+1,1 1 · · · Qdt+1,t t , . . . , ⇧pQdp1 1 · · · Qdpt t ) and ⇧0 = Q−d01 1 · · · Q−d0t t ⇥f(Q1, . . . , Qt, ⇧t+1Qdt+1,1 1 · · · Qdt+1,t t , . . . , ⇧pQdp1 1 · · · Qdpt t ), where f is the function we hope to estimate. So we can rewrite it as ⇧0 = g(Q1, . . . , Qt, ⇧t+1, . . . , ⇧p), where ⇧i, i = 0, t + 1, . . . , p are dimensionless and Q1, . . . , Qt are “independent”. Buckingham’s theo-rem indicates that Q1, . . . , Qt should not be in the formula. This implies ⇧0 = g(⇧t+1, . . . , ⇧p) to be the final model.
2.4. Example: Vehicle Stopping Distance Here we use the vehicle stopping distance exam-ple to illustrate the general procedure. In the exper-iment, we estimate the stopping distance for cars, a Vol. 46, No. 3, July 2014 www.asq.org 188 WEIJIE SHEN ET AL.
key indication of their safety. Assume that the driver requires a certain amount of time for reaction to an emergency and that the wheels are not locked when braking. We show the dimensional analysis below us-ing the procedure described in the previous section.
Step 1. Identify input and output variables and their dimensions as follows: Q0 = D : vehicle stopping distance [D] = L.
Q1 = v : velocity of the vehicle [v] = LT−1.
Q2 = ⌧: thinking time [⌧] = T.
Q3 = m : mass of the car [m] = M.
Q4 = F : braking force on the brake discs [F] = MLT−2.
Q5 = µ : friction coefficient of brakes [µ] = 1.
The response is {D} and the predictors are {v, ⌧, m, F, µ}. The model function is D = f(v, ⌧, m, F, µ). The dimensions of respective variables are listed above and we summa-rize corresponding {rij} in Table 1. There are three fundamental quantities in this sys-tem. Their dimensions are length L, time T, and mass M. The entries in the table are the power of the fundamental dimensions (de-noted by rows) for dimensions of each vari-able (denoted by columns).
Step 2. Determine the basis quantities. We choose {Q1 = v, Q2 = ⌧, Q3 = m} in this case.
Step 3. Determine the dimensionless transformation for the remaining three quantities {Q0 = D, Q4 = F, Q5 = µ}, and formulate {⇧0, ⇧4, ⇧5} as follows: [v⌧] = L, [⌧] = T, [m] = M, [µ] = 1 and ⇧0 = D v⌧, ⇧4 = F⌧ mv, ⇧5 = µ.
Step 4. Re-express the model. Here our objective will be to estimate the function g, with ⇧0 as the response variable and only two input variables, ⇧4 and ⇧5: ⇧0 = g(Q1, Q2, Q3, ⇧4, ⇧5) = g(⇧4, ⇧5), or equivalently, D v⌧= g ✓F⌧ mv, µ ◆ .
TABLE 1. Dimensions of Variables in the Vehicle Stopping Distance Dimension D v ⌧ m F µ L (length) 1 1 0 0 1 0 T (time) 0 −1 1 0 −2 0 M (mass) 0 0 0 1 1 0 From the procedure and the example, we can see that, in physical phenomena, we have certain restric-tions in the forms of f, satisfying certain dimensional requirements. After dimensional analysis, the poten-tial e↵ects on responses are attributable to the com-binations of quantities considered. These quantities act like groups. If we base our estimated function g on the group values, we do not have dimensional restrictions.
3. Dimensional Analysis for Data Analysis Statistics extracts information from the data of experiments to find or justify properties, laws, and performance. Based on the fact that those experi-ments are results of physical phenomena and the data are physical quantities measured in experiments, it is often justifiable and beneficial to perform dimen-sional analysis in the first place. For data analysis, the advantages of using dimensional analysis stand out clearly: it is rather straightforward. Incorporat-ing dimensional analysis only transforms the data in a predetermined fashion; the basic development strategies remain the same. Furthermore, it is sup-ported by physics to make dimensionless transfor-mations. We conjecture that making variables di-mensionally independent is helpful in making them statistically independent, although this important is-sue needs further investigation. After dimensional analysis, the information from inputs and outputs is more concentrated, leading to statistical models with fewer variables and simpler analysis. Below, we compare data-analysis procedures with and without dimensional analysis based on the pine tree data from Bruce and Schumacher (1935, p. 226). From there, we show how to perform dimensional analysis for data analysis and its potential benefits.
3.1. Pine Tree Example The pine tree data has been used by various au-thors to illustrate the use of diagnostics and transfor-Journal of Quality Technology Vol. 46, No. 3, July 2014 DIMENSIONAL ANALYSIS AND ITS APPLICATIONS IN STATISTICS 189 mation methods in linear regression (see, e.g., Atkin-son (1994)). The data arise from the measurements of 70 shortleaf pine trees. Three measurements of inter-est here are d, the diameter of the tree in feet taken at “breast height” above the ground; h, the height of the tree in feet; and v, the volume in cubic feet. The objective of the analysis is to establish a relationship between the volume v and the variables d and h. In other words, we hope to predict the volume of a tree from its known diameter and height. The complete data set is given in the Appendix.
3.2. Regression Method Without Dimensional Analysis for Pine Tree Data Set The conventional linear regression assumes that vi = ↵+ β1di + β2hi + ✏i, (1) with ✏i ⇠i.i.d. N(0, σ2). It gives us the following es-timated function with standard errors of estimations in subscripts: ˆ v = −45.3(5.0) + 77.2(5.9)d + 0.12(0.11)h, (2) with ˆ σ = 9.87. The coefficient of h is not signifi-cant. Although a univariate analysis makes the resid-uals appear to be reasonably normally distributed with constant variance, we notice that (i) because the data (see Appendix) are ordered by diameter d, Figure 1 shows a distinct trend of residuals relative to the diameters; and (ii) there is a potential outlier of tree #70 (the largest tree). Both of these diagnos-tics strongly suggest that model (2) is inadequate.
We could proceed by applying the log transforma-tion to all variables before the linear model fitting, which leads to the following result: d ln(v) = −1.06(0.24) + 1.943(0.038) ln(d) + 1.054(0.055) ln(h), with ˆ σ = 0.0673. It is appealing to approximate the FIGURE 1. Studentized Residuals Plot of Model (1).
coefficients of diameter and height e↵ects (1.943 and 1.05) by the integers 2 and 1, respectively. After fix-ing the coefficients, we obtain a regression model on the original scale without intercept: vi = βd2 i hi · δi, (3) with ✏i = ln δi ⇠i.i.d. N(0, σ2). The estimated re-gression function is ˆ v = 0.4411(0.0081)d2h.
(4) On the other hand, the Box–Cox transformation on the response variable recommends a transformation parameter of ˆ λ = 0.384. Thus, a cubic-root trans-formation (λ = 1/3 = 0.333) seems appropriate.
The use of the cubic-root transformation of the re-sponse (without intercept) suggests the following lin-ear model: v1/3 i = β1di + β2hi + ✏i, (5) with ✏i ⇠i.i.d. N(0, σ2). The estimated regression function turns out to be d v1/3 = 2.084(0.044)d + 0.01471(0.00059)h.
(6) The residual plots in Figures 2a and 2b show that FIGURE 2. Studentized Residuals Plot of Models (4) in (a) and (6) in (b).
Vol. 46, No. 3, July 2014 www.asq.org 190 WEIJIE SHEN ET AL.
both log transformation and Box–Cox transforma-tion have fixed the problems highlighted in Figure 1.
Note that tree #53 deserves special attention, but will not be further studied in this paper.
The preceding analyses were, of course, conducted without using dimensional analysis. However, a fur-ther look at equations (4) and (6) reveals that both methods provide dimensionally homogeneous solu-tions to the prediction problem. The physical dimen-sions are coherent in the equations. Model (4) has the dimension of cubic length on both sides while model (6) has the dimension of length on both sides.
3.3. Regression Method with Dimensional Analysis for Pine Tree Data Set Following the general procedure that we proposed in Section 2.3, the dimensional analysis can be im-plemented as below. A similar approach was outlined in Vignaux and Scott (1999).
1. Our objective is to predict the output volume v as a function of diameter d and height h: v = f(d, h). We determine the physical dimensions of these quantities in Table 2. The dimension of v is cubic length L3 with unit feet3. Both dimensions of d and h are length L with units in feet.
2. Because the only dimension involved is length, let h be the basis quantity.
3. Transform other quantities into dimensionless forms, ⇧v = v h3 and ⇧d = d h.
4. By Buckingham’s ⇧-theorem, the predicted function should be ⇧v = g(⇧d), or equivalently, v h3 = g ✓d h ◆ .
Suppose ⇧v = g(⇧d), and we choose g(⇧d) = k⇧γ d.
After taking the logarithm of both sides, we obtain the linear model, ln(⇧v,i) = ln(k) + γ ln(⇧d,i) + ✏i, with ✏i ⇠N(0, σ2). The estimated regression func-tion is ˆ ln(⇧v) = −1.07(0.16) + 1.942(0.036) ln(⇧d) (7) Alternatively, we might prefer that ⇧v = g(⇧d) = k⇧2 d. Figure 3 shows the data and linear fits in terms of ⇧v and ⇧2 d. In fact, fitting a linear model yields TABLE 2. Dimensions of Quantities in Pine Tree Data Set Name Quantity Dimension Unit Volume v L3 feet3 Diameter d L feet Height h L feet ˆ ⇧v = 0.4363(0.0036)⇧2 d, i.e., ˆ v = 0.4363(0.0036)d2h.
(8) The result is similar to the log-transformed model (4) and it also accommodates the problematic 70th case.
The di↵erences in parameter estimates can be at-tributed to di↵erences in associated error structures.
In summary, certain assumptions regarding the form of the function g will give corresponding pa-rameterizations and results. If ⇧v = k⇧γ d, then v = kdγh3−γ. If ⇧v = (A⇧d +B)3, then v1/3 = Ad+Bh.
These are exactly the two models previously obtained in equations (4) and (6). The dimensionally homoge-neous results we derived from regression analyses are merely special cases of choosing di↵erent functions g after dimensional analyses. The procedure ensures that the results are dimensionally homogeneous and intuitively interpretable, while leaving choice of the function g to the investigator, as informed by the data. It gives us a guide of how to model in an ef-ficient and parsimonious way based on the physical laws.
FIGURE 3. Plot of Pine Tree Data Set in Terms of Di-mensionless Variables.
Journal of Quality Technology Vol. 46, No. 3, July 2014 DIMENSIONAL ANALYSIS AND ITS APPLICATIONS IN STATISTICS 191 In contrast with the analysis and discussion of the same dataset by Atkinson (1994), dimensional analysis points to an “automatic” transformation of the data, without prior assumption of the cone shape, and posterior diagnosis of “transform both sides model” as Atkinson (1994) did. Moreover, be-cause the dimensional analysis of each tree is the same, there is less need to worry about individual trees influencing the choice of transformation, and hence regression methods such as constructed vari-able plots, which Atkinson (1994) used, are no longer necessary.
3.4. Remarks From an analytic perspective, dimensional analy-sis o↵ers several advantages relative to the conven-tional procedures. First, it decreases the number of variables, which may lead to a simpler model. Sec-ond, physical independence may establish a simpler statistical relationship. Third, it gives more sensi-ble interpretations by having dimensionless variables and coefficients. Physicists and engineers often fa-vor dimensionless coefficients as indices for describ-ing systems. Fourth, dimensional analysis produces scalable results, which is necessary for extrapolation, although the scalability depends on a good choice of the model. This is attributed to the ratio form of dimensionless variables that is invariant to scale changes. Often, extrapolating in the original scale re-sults in values interpolating in the transformed scale.
Fifth, it captures the inherent nonlinear relationship between physical quantities.
From a practical perspective, dimensional anal-ysis is applicable when modeling physical relation-ships. It is a straightforward method before collect-ing data and modeling. It fits all kinds of data struc-tures and modeling requirements. Furthermore, di-mensional analysis does not lose generality when transforming the data. As shown above, it does not constrain the forms of estimating functions for di-mensionally homogeneous solutions. Due to its an-alytical nature, the proposed procedure is easy to implement.
4. Dimensional Analysis for the Design of Experiments Dimensional analysis can also serve as a guid-ance in the design of experiments. For the design of experiments, incorporating dimensional analysis can significantly improve efficiency by reducing the number of experimental variables. In this section, we demonstrate the use of dimensional analysis in the design and analysis of the popular paper helicopter experiment. We also compare the results to those ob-tained using the conventional design of experiments approaches.
4.1. The Paper Helicopter Experiment The paper helicopter experiment is a widely used teaching device for the design of experiments. The objective is to predict the “flight time” performance for a particular configuration of the helicopter dimen-sions in Figure 4. Upon its launch, twin blades spin FIGURE 4. Paper Helicopter Illustration.
Vol. 46, No. 3, July 2014 www.asq.org 192 WEIJIE SHEN ET AL.
around the central ballast shaft to provide lift as it descends. It can be easily constructed from a sheet of paper using only scissors and tape. As displayed in Figure 4, the upper part of the model consists of two wings (or rotors). The central part is the body and the lower part is the tail folded into the ballast.
The length and width of each part can be varied to achieve di↵ering levels of performance. We wish to predict the flight time for a given configuration. The usual design factors include the rotor radius (r), the rotor width (w), the tail length (l), the tail width (d), paper clip (p), and the tape (t).
4.2. Conventional Design of the Paper Helicopter Experiment Without Using Dimensional Analysis Johnson et al. (2006a, b) studied experimental de-sign on the paper helicopter as part of a Six Sigma Black Belt project. They considered a Resolution VII design with seven two-level factors in a half fraction, with two replicates. This led to 27−1⇥2 (= 128) runs in total. They provided a step-by-step routine to de-sign experiments and maximize the flight time. Box and Liu (1999) and Box (1999) discussed the use of sequential design in the paper helicopter experiment.
First, they conducted a two-level Resolution IV frac-tional factorial design with eight factors in four repli-cates, i.e., 28-4 IV ⇥4, for a total of 64 runs. In the second setting, they designed a full-factorial experiment in-volving four important factors. Two key lessons from their work, among others, are (1) with the help of response surface and steepest ascent, sequential de-signs search for an optimum point e↵ectively and ef-ficiently and (2) minimum variance or dispersion of flight time is included in addition to longest flight time to enrich the meaning of optimum. Annis (2005) derived the aerodynamics of this flying object in a rigorous physical sense before designing the experi-ment. He presented a physical model of flight time in terms of the length and width of wing and body. He employed two three-level factors in a single replicate of a 3 ⇥3 factorial design for wing length and width and employed response surface methods to identify the optimal operating condition. The use of physics to identify promising factors in advance turned out to be extremely advantageous.
Table 3 summarizes the results of the above three papers, including Johnson et al., (2006a,b), Box and Liu (1999), and Annis (2005). The directions of respective e↵ects are provided in the parentheses.
Three variables are included in all of the three ex-periments; namely, body length, body width, and wing length. Additionally, Johnson et al. (2006a, b) considered (i) paper type, (ii) whether or not tap-ing body and wing, and (iii) whether or not clipping in the bottom. They also examined the two-way in-TABLE 3. Summary of Literature on Paper Helicopter Johnson et al. (2006a,b) Box and Liu (1999) Annis (2005) Input variables Body length (−) Body length (−) Body length (−) Body width (−) Body width (−) Body width (−) Wing length (+) Wing length (+) Wing length (+) Paper type (−) Paper type (−) Wing width (dip) Taped body (−) Taped body (−) Taped wing (−) Taped wing (−) Clip (−) Clip (−) Interactions Fold (−) Design 27-1 VII ⇥2 = 128 28-4 IV ⇥4 = 64 32 = 9 Feature Interactions Sequential learning Physical insight Significant variables Body length (−) Body length (−) Body length (−) Body width (−) Body width (−) Body width (−) Wing length (+) Wing length (+) Wing length (+) Not a design factor.
Journal of Quality Technology Vol. 46, No. 3, July 2014 DIMENSIONAL ANALYSIS AND ITS APPLICATIONS IN STATISTICS 193 teractions between the variables of interest. Box and Liu (1999) considered whether to fold the paper and use an additional 50 runs to search for the optimum.
Their second design included a full-factorial exper-iment on both the length and width of both body and wing. Annis (2005) took wing width as an ad-ditional variable and chose wing length and width to be experimental variables and considered e↵ects of body length and width as given by the physical formula. The “dip” e↵ect of wing width means the relationship is not monotone. In general, the signs in the parentheses indicate the e↵ects of each variable on the flight time. For example, wing length has a positive e↵ect on the flight time, meaning that the flight time will increase if wing length is increased.
All the above studies concluded that the wing length is the most important factor for determining flight time, with factors a↵ecting helicopter mass also be-ing important.
4.3. Design of the Paper Helicopter Experiment Using Dimensional Analysis The physics of falling objects in a gravitational field follows the following assumptions: (1) the flight time (T) is determined by the launch height (H) and average velocity (v), i.e., T = H/v; (2) the falling ob-ject reaches terminal velocity quickly after dropped, when the drag force of the air becomes equal to the force of gravity; (3) the drag force depends on the density of dry air (⇢= 1.20412 kgm−1 at sea level at 20◦C), drag coefficient (cd, dimensionless) and the shape of the helicopter (rotor radius r and rotor width w, or their combinations, such as ratio r/w and area rw); (4) the weight of the helicopter depends on the mass (m) and acceleration due to gravity (g = 9.8 ms−2).
Suppose the flight time follows the model below, T = f1(m, g, r, cd, ⇢, H).
Because cd is dimensionless and T = H/v, the ex-pression can be represented as v = f2(m, g, ⇢, r).
(9) Next, we apply dimensional analysis step-by-step.
Step 1. Table 4 displays the dimensions and units of the variables. We have three fundamental dimensions, length L, time T, and mass M.
Step 2. Variables r, ⇢, and g are chosen as the base quantities.
Step 3. From Buckingham’s ⇧theorem, we can re-duce the number of variables from five to two. Following Gearhart (2004), we define the two dimensionless variables as Φv = vra⇢bgc and m = mrd⇢egf. Dimensions of Φv and m are, thus, as follows: [Φv] = (LT−1)(L)a(ML−3)b(LT−2)c = L1+a−3b+cMbT−1−2c [ m] = (ML)d(ML−3)e(LT−2)f = Ld−3e+fM1+eT−2f.
We enforce nondimensionality and solve the two sets of linear equations, 1 + a −3b + c = 0 b = 0 −1 −2c = 0 and d −3e + f = 0 1 + e = 0 −2f = 0.
From the first set, we obtain a = c = −1/2 and b = 0 and, from the second, d = −3, e = −1, and f = 0. The transformed variables are thus Φv = v pgr = h Tpgr; m = m ⇢r3 .
(10) Step 4. The final equation is obtained as the follow-ing form: Φv = g( m).
(11) Because there is only one input variable ( m), we conduct the paper helicopter fly-ing experiment with four runs: m = 0.937, 2.087, 3.088, and 4.642. The resulting flight times (T) are 5.18, 3.87, 3.48, and 2.98, re-spectively. This implies the values of Φv to be 0.873, 1.264, 1.537, and 1.795. The results TABLE 4. Dimensions of Quantities of Paper Helicopter Quantity Quantity name symbol Dimension Unit Velocity v = h/T LT−1 ms−1 Mass m M kg Gravity acceleration g LT−2 ms−2 Air density ⇢ ML−3 kgm−3 Wing length r L m Vol. 46, No. 3, July 2014 www.asq.org 194 WEIJIE SHEN ET AL.
TABLE 5. Table of Design and Results of Paper Helicopter Experiment m Paper Helicopter Rotor Flight Φv No.
(m/⇢r3) type mass(m) radius(r) time(T) (h/(Tpgr)) 1 0.937 80 g/m2 3.09 g 140 mm 5.18 s 0.873 2 2.087 120 g/m2 4.34 g 120 mm 3.87 s 1.264 3 3.088 100 g/m2 3.72 g 100 mm 3.48 s 1.537 4 4.642 160 g/m2 5.59 g 100 mm 2.98 s 1.795 Note: The results are the averages of three flights recorded independently twice.
are displayed in Table 5. Figure 5 is the scat-ter plot of Φv and m.
The data are modeled using a simple linear regres-sion, giving ˆ Φv = 0.700.09 + 0.250.03 m. Converting to the original variables, this becomes ˆ T = h pgr(0.70 + 0.25 ⇢r3 ).
(12) Alternatively, we regress data on a log scale. This gives us log(Φv) = −0.1020.01+0.460.01 log( m). The power of 0.46 suggests a square-root transformation of m. We thus take a square-root transformation on m and fit a linear model without intercept, obtain-ing ˆ Φv = 0.8590.014 p m. Converting to the original variables, this can be expressed as ˆ T = hr 0.859 r ⇢ mg .
(113) In both models (12) and (13), the coefficients are dimensionless and the equations are dimensionally homogeneous. We prefer model (13) because it is able to capture the potential curvature shown in Figure FIGURE 5. Plot of Simple Linear Regression on Paper Helicopter.
5. For validation of the final model, we conducted eight confirmation runs with various combinations of 80/100/120/160 g/m2 A4 paper and 100/120/140 mm rotor radius. The actual flight times versus pre-dicted flight times are displayed in Figure 6. The points align closely to the line y = x, indicating a good model indeed.
Table 6 provides a comparison of results between designs using dimensional analysis and those without it. It can be seen that the key factors are exactly the same, while the settings for maximal time are close.
However, the design we used had only one dimen-sionless variable; therefore, fewer runs were needed (4 runs vs. 128 runs in Johnson et al. (2006b), 64 runs in Box and Liu (1999), and 9 runs in Annis (2005)). In addition, the prediction model is elegant and easy to interpret: the response is proportional to some FIGURE 6. The Plot of Predicted Flight Time and Actual Flight Time of the Confirmation Runs.
Journal of Quality Technology Vol. 46, No. 3, July 2014 DIMENSIONAL ANALYSIS AND ITS APPLICATIONS IN STATISTICS 195 TABLE 6. Comparison Between Designs with and Without DA Without DA With DA Variables 2 or 3 Levels Continuous (interpolate and extrapolate) Design On variables On dimensionless quantities Number of experimental runs 128, 64 and 9 4 Key variables Wing length (+) Wing length (+) Body length (−) Body length (−) Settings for maximal flight time L = 15.2c m, w = 4.60 cm L = 14 cm, w = 5.25 cm m = A4 (gsm unstated) m = 3.09 g (80 gsm) Validation model Full factorial Confirmation runs power of each factor. The ratio form also produces scalable results. Furthermore, after transformation, the dimensionless quantities seem to be linearly re-lated to the response, which avoids the complicated relationships such as “dip” e↵ect found in Box and Liu (1999) and Annis (2005). Last, robustness analy-sis through transmitted variation (Morrison (1957)) can be directly applied to the fitted model, avoiding costly replications to determine this variation empir-ically. Further results on a dimensional-analysis ap-proach to the paper helicopter can be found in Davis (2013).
4.4. Remarks From the design of experiments perspective, the number of experimental runs required tends to in-crease with the number of experimental factors. Di-mensional analysis combines variables according to physical laws and creates new design variables that can be incorporated into the design. It reduces the number of factors and consequently reduces the re-quired number of runs. It potentially allows the in-clusion of variables not included in the original ex-periment, as long as they can be expressed in the di-mensionless variables. We also benefit from the scal-ability and the interpretability of the solutions. See, for example, Albrecht et al. (2013) for a detailed treatment of designing experiments to dimensional-analysis models.
5. General Properties From the above examples, we summarize the fol-lowing advantages for dimensional analysis on gen-eral cases.
1. Dimensional analysis starts from basic and nat-ural physical assumptions. The resulting fac-tors and their coefficients are dimensionless and easy to interpret for practitioners.
2. Dimensional analysis combines and eliminates unnecessary variables using Buckingham’s ⇧ Theorem. This leads to dimension reduction, which is especially helpful for the design of ex-periments.
3. The power law is used in the combination, re-vealing that the nature of relationships between variables with di↵erent dimensions is often not linear. It is believed that, after transformation, the dimensionless quantities may become more independent (fewer interactions) and their re-lationships with response are simpler, as shown in our examples.
4. It is compatible with all kinds of methods, as it is a “data-free” method. It transforms the vari-ables according to their dimensions, not values.
It can be done even before we get the real data or any pilot experiment. The subsequent statis-tical procedures are valid without changes.
5. The resulting models are often scalable. For ex-ample, extrapolation in linear regression is of-ten misleading, because the assumed form of the model may not be valid beyond the range of the data. The dimensionless models devel-oped using dimensional analysis do not depend on absolute quantities, but are rather defined in terms of relative amounts; thus, scale is not relevant in most cases.
Vol. 46, No. 3, July 2014 www.asq.org 196 WEIJIE SHEN ET AL.
Drawbacks of dimensional analysis may include the requirement of physical knowledge about the exper-imental environment and the possibility of severe problems if any related variable is excluded (Albrecht et al. (2013)). Piepel (2013) also raises the issue of spurious correlation. A comprehensive discussion of statistical issues of dimensional analysis is presented in Lin and Shen (2013).
Conclusion Dimensional analysis has been well developed in physics, engineering, and other fields. However, its significance was overlooked for years by statisticians.
Little e↵ort was made to incorporate it into statisti-cal practice. In this paper, we describe the use of the dimensional-analysis method for both data analysis and design of experiments. Additional examples and comments can be found in Davis (2011) and Lin and Shen (2013). Our purpose is to promote greater inte-gration of dimensional analysis into statistical design and analysis by statisticians, engineers, and scien-tists.
The fundamental insight of dimensional analysis is to identify key dimensionless variables from physi-cal considerations, and then use data and statistical analysis to understand them. Engineers provide the-ory for guidance in the statistical analysis using phys-ical prior knowledge. Statisticians design and ana-lyze experiments for the unknown physical structure, check the validity of physical assumptions and rec-ommend further experiments. Complementing each other in this fashion often leads to solutions that nei-ther could achieve alone.
This paper introduces the basic idea of dimen-sional analysis and its potential applications in statis-tics, notably in regression analysis and the design of experiments. There are many more issues to be studied. First, the error structure should be fur-ther investigated for dimensional analysis. Latent errors in covariates and the robustness need to be taken into account as well. For example, an orthog-onal distance regression can be applied when both sides of the modeling equation have errors. More-over, errors could propagate very di↵erently for dif-ferent nonunique DA representations. Second, once designing the transformed dimensionless quantities, the corresponding design on the original quantities is not unique. The various design options on those op-erating quantities o↵er a way to test the validity of Buckingham’s ⇧-theorem statistically. Third, a for-mal sequential or recursive scheme is suggested to interweave the knowledge of engineers and that of statisticians. Special designs or analyses may be pre-ferred after conducting dimensional analysis. Fourth, we believe dimensional analysis could be generalized into fields outside of physics and engineering. Certain common measure units in economics, biology, or soci-ology could be candidates to enlarge the application of dimensional analysis. Fifth, it seems promising to generalize the idea of combining variables. PCA is one kind of combining under linear schemes in feature extraction. Dimensional analysis implies that combi-nations may be done nonlinearly by power law.
References Albrecht, M. C.; Nachtsheim, C. J.; Albrecht, T. A.; and Cook, R. D. (2013). “Experimental Design for Engineering Dimensional Analysis”. Technometrics 55(3), pp. 257–270.
Annis, D. H. (2005). “Rethinking the Paper Helicopter: Com-bining Statistical and Engineering Knowledge”. The Amer-ican Statistician 59(4), pp. 320–326.
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Box, G. E. P. (1999). “Statistics as a Catalyst to Learning by Scientific Method Part II—A Discussion”. Journal of Quality Technology 31(1), pp. 16–29.
Box, G. E. P. and Liu, P. Y. T. (1999). “Statistics as a Cata-lyst to Learning by Scientific Method Part I—An Example”.
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Bridgman, P. (1931). Dimensional Analysis, 2nd edition.
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Bruce, D. and Schumacher, F. X. (1935). Forest Mensura-tion, p. 226. New York, NY: McGraw Hill.
Buckingham, E. (1914). “On Physically Similar Systems; Il-lustrations of the Use of Dimensional Equations”. Physical Review 4(4), pp. 345–376.
Buckingham, E. (1915a). “Model Experiments and the Forms of Empirical Equations”. Transactions of the American So-ciety of Mechanical Engineers 37, pp. 263–296.
Buckingham, E. (1915b). “The Principle of Similitude”. Na-ture 96(2406), pp. 396–397.
Davis, T. P. (2011). “Dimensional Analysis in Experimental Design”. Talk at Isaac Newton Institute. Available online at
Davis, T. P. (2013). “Comment: Dimensional Analysis in Sta-tistical Engineering”. Technometrics 55(3), pp. 271–274.
Finney, D. J. (1977). “Dimensions of Statistics”. Applied Statistics 26(3), pp. 285–289.
Gearhart, C. (2004). “Using Dimensional Analysis to Build a Better Transfer Function”. SAE Technical Paper, 2004– 01–1129.
Islam, M. and Lye, L. M. (2009). “Combined Use of Di-mensional Analysis and Statistical Design of Experiment Methodologies in Hydrodynamics Experiments”. Ocean En-gineering 36, pp. 237–247.
Johnson, J. A.; Widener, S.; Gitlow, H.; and Popvich, E.
(2006a). “A ‘Six Sigma’ Black Belt Case Study: G.E.P. Box’s Journal of Quality Technology Vol. 46, No. 3, July 2014 DIMENSIONAL ANALYSIS AND ITS APPLICATIONS IN STATISTICS 197 Appendix The Shortleaf Pine Tree Data Set ID Diameter Height Volume ID Diameter Height Volume (feet) (feet) (feet3) (feet) (feet) (feet3) 1 0.383 33 2.2 36 0.917 71 25.8 2 0.367 38 2.0 37 0.925 81 32.8 3 0.417 40 3.0 38 0.933 91 35.4 4 0.425 49 4.3 39 0.958 66 26.0 5 0.425 37 3.0 40 0.975 65 29.0 6 0.433 41 2.9 41 1.000 72 30.2 7 0.433 41 3.5 42 1.017 66 28.2 8 0.458 39 3.4 43 1.017 72 32.4 9 0.458 50 5.0 44 1.042 90 41.3 10 0.467 69 7.2 45 1.075 88 45.2 11 0.492 58 6.4 46 1.083 63 31.5 12 0.492 50 5.6 47 1.092 69 37.8 13 0.625 45 7.7 48 1.092 65 31.6 14 0.633 51 10.3 49 1.117 73 43.1 15 0.633 49 8.0 50 1.150 69 36.5 16 0.650 59 12.1 51 1.150 77 43.3 17 0.667 56 11.1 52 1.192 64 41.3 18 0.675 86 16.8 53 1.192 77 58.9 19 0.700 59 13.6 54 1.217 91 65.6 20 0.717 78 16.6 55 1.233 90 59.3 21 0.742 93 20.2 56 1.242 68 41.4 22 0.758 65 17.0 57 1.258 96 61.5 23 0.767 67 17.7 58 1.267 91 66.7 24 0.775 76 19.4 59 1.267 97 68.2 25 0.775 64 17.1 60 1.275 95 73.2 26 0.817 71 23.9 61 1.283 89 65.9 27 0.825 72 22.0 62 1.308 73 55.5 28 0.825 79 23.1 63 1.325 99 73.6 29 0.825 69 22.6 64 1.333 90 65.9 30 0.842 71 22.0 65 1.400 90 71.4 31 0.850 80 27.0 66 1.483 91 80.2 32 0.850 82 27.0 67 1.525 96 93.8 33 0.858 81 27.4 68 1.525 100 97.9 34 0.867 75 25.2 69 1.617 94 107.0 35 0.883 75 25.5 70 1.950 104 163.5 Vol. 46, No. 3, July 2014 www.asq.org 198 WEIJIE SHEN ET AL.
Paper Helicopter Experiment. Part A”. Quality Engineering 18, pp. 413–430.
Johnson, J. A.; Widener, S.; Gitlow, H.; and Popvich, E.
(2006b). “A ‘Six Sigma’ Black Belt Case Study: G.E.P. Box’s Paper Helicopter Experiment. Part B”. Quality Engineering 18, pp. 431–442.
Lin, D. K. J. and Shen, W. (2013). “Comments: Some Sta-tistical Concerns on Dimensional Analysis”. Technometrics 55(3), pp. 281–285.
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Piepel, G. F. (2013). “Comment: Spurious Correlation and Other Observations on Experimental Design for Engineering Dimensional Analysis”. Technometrics 55(3), pp. 286–289.
Sonin, A. A. (2001). The Physical Basis of Dimensional Analysis, 2nd edition. Cambridge, MA: Department of Me-chanical Engineering, Massachusetts Institute of Technology.
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Stahl, W. R. (1962). “Similarity and Dimensional Methods in Biology”. Science 137(3525), pp. 205–212.
Szirtes, T. (2007). Applied Dimensional Analysis and Mod-eling, 2nd edition. Oxford, UK: Elsevier Butterworth-Heinemann.
Taylor, M.; Diaz, A. I.; Jodar-Sanchez, L. A.; and Villanueva-Mico, R. J. (2007). “100 Years of Dimensional Analysis: New Steps toward Empirical Law Deduction”.
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Vignaux, G. A. and Scott, J. L. (1999). “Simplifying Re-gression Models Using Dimensional Analysis”. The Austral-ian and New Zealand Journal of Statistics 41(1), pp. 31–42.
s Journal of Quality Technology Vol. 46, No. 3, July 2014 |
17435 | https://periodictable.com/Isotopes/047.108/index.html | Isotope data for silver-108 in the Periodic Table
HHomeBackground Color:He LiBeSilver Main PageBlackWhiteGrayBCNOFNe NaMgSilver Pictures PageAlSiPSClAr KCaSilver Technical DataScTiVCrMnFeCoNiCuZnGaGeAsSeBrKr RbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe CsBaLaCePrNdPmSmEuGdTbDyHoErTmYbLuHfTaWReOsIrPtAuHgTlPbBiPoAtRn FrRaAcThPaUNpPuAmCmBkCfEsFmMdNoLrRfDbSgBhHsMtDsRgCnNhFlMcLvTsOg
Isotopes of Silver (click to see decay chain): 93 Ag94 Ag95 Ag96 Ag97 Ag98 Ag99 Ag100 Ag101 Ag102 Ag103 Ag104 Ag105 Ag106 Ag107 Ag108 Ag109 Ag110 Ag111 Ag112 Ag113 Ag114 Ag115 Ag116 Ag117 Ag118 Ag119 Ag120 Ag121 Ag122 Ag123 Ag124 Ag125 Ag126 Ag127 Ag128 Ag129 Ag130 Ag
108 Ag Half-life Fermion, 47p 61n 2.366666666667 m Spin 1 Parity 1 Show Decay Modes Show Ultimate Decay ProductsAtomic Weight 107.905955556 Abundance None Mass Excess-87.601836MeV Binding Energy 8.54198MeV Magnetic Moment 2.6884 μ Quadrupole Moment N/A Show Element Pictures. Also show decay chains leading to this isotope. Show only decays with probability >0.1%.
Click any isotope in diagram to see its data.
Decay chain image generated by Mathematica's GraphPlot and IsotopeData functions from Wolfram Research, Inc.. |
17436 | https://www.reddit.com/r/askscience/comments/1bvmbm/does_burning_1_gallon_of_gasoline_really_produce/ | Does burning 1 Gallon of Gasoline really produce 20 Pounds of CO2? : r/askscience
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17437 | https://mathoverflow.net/questions/251546/a-simple-question-about-the-hardy-littlewood-maximal-function | Stack Exchange Network
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A simple question about the Hardy-Littlewood maximal function
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Let $f\in L^1(\mathbb{R}^n)$. It is well known that the Hardy-Littlewood maximal function $Mf\notin L^1(\mathbb{R}^n)$ (if $f \ne 0$ a.e.), though there is a weak-type (1,1) bound for this maximal operator. More explicitly, we have a lower bound estimate $ Mf(x)\ge C|x|^{-n}$ (see e.g. We consider a "local" maximal function defined by $$(M_\phi f)(x)= \sup_{0<\epsilon<1}|\phi_\epsilon \ast f|(x),$$ which is similar to the classical Hardy-Littlewood maximal function : $$(Mf)(x)= \sup_{\epsilon>0}|(\chi_B)_\epsilon \ast f|(x),$$ where $\phi\in S(\mathbb{R}^n)$ is some nonnegative, radial, radially decreasing Schwartz function and $\phi_\epsilon(x)=\epsilon^{-n}\phi(x/\epsilon)$. The word "local" means that the supremum is only taken on small $\epsilon\in(0,1)$.
It has been shown in How much do we know about this "local" Hardy-Littlewood maximal function? that $M_\phi f\notin L^1(\mathbb{R}^n)$ for general $f\in L^1(\mathbb{R}^n)$, though if $f$ is some "good function"(e.g. compactly supported bounded functions) we can easily see that $M_\phi f$ is integrable.
My question is: if we assume that $f\in L^1(\mathbb{R}^n)\cap L^\infty(\mathbb{R}^n)$, do we have $M_\phi f\in L^1(\mathbb{R}^n)$? Remark: Since $f\in L^p$ for all $p>1$, by HardyLittlewood maximal inequality, we have $Mf\in L^p$, for all $p>1$. Then $M_\phi f\in L^p,\ p>1$, since $M_\phi f\le Mf$.
My attempt: Since $\phi$ is a Schwartz fucntion, I decompose the integral into three parts: $$\epsilon^{-n}\int |f(y)|(1+|x-y|/\epsilon)^{-N}dy=\int_{|y|\le|x|/10}+\int_{|y|\ge 10|x|}+\int_{|x|/10<|y|<10|x|}.$$ Since $f$ is bounded, we see that $M_\phi f$ is also bounded. So we only consider $|x|\gg1$. We see that the first two integrals are both bounded by $C\epsilon^{N-n}|x|^{-N}$, so we only need to consider the third one. If $f$ has good decay (e.g. $|f(y)|\le C|y|^{-\alpha}, \alpha>n$), we see that the third integral also has "good dacay", which makes $M_\phi f\in L^1(\mathbb{R}^n)$. But I don't know what to do for general $f$. Moreover, I don't know whether there is any simple counterexample for it. Any comments are welcome:)
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Mr.rightMr.right
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$\begingroup$ Wasn't this answered in your previous question? $\endgroup$
Anthony Quas
– Anthony Quas
2016-10-06 18:09:55 +00:00
Commented Oct 6, 2016 at 18:09
$\begingroup$ Hi Anthony, the example given by Wong in the previous question is not in $L^\infty$. $\endgroup$
Mr.right
– Mr.right
2016-10-06 18:11:28 +00:00
Commented Oct 6, 2016 at 18:11
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No. Say we are in one dimension. Consider the function
$$ f(x) := \sum_{m=2}^\infty 1_{[m, m + \frac{1}{m \log^2 m}]}.$$
this function is (barely) in $L^1$ (because $\sum_{m=2}^\infty \frac{1}{m \log^2 m}$ converges), but $M_\phi f$ is (barely) outside of $L^1$ (basically because $\sum_{m=2}^\infty \frac{1}{m \log m}$ diverges).
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answered Oct 6, 2016 at 18:14
Terry TaoTerry Tao
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$\begingroup$ More generally, let $a_n$ be a sequence of positive numbers such that $\sum a_n$ converges but $\sum a_n\log n$ diverges. Then $f(x)=\sum 1_{[n,n+a_n]}$ is a (counter)example. This example is just the superposition of (countably) many rescaled versions of the "nonintegrable at infinity" scenario that the OP had mentioned, and the scales become smaller and smaller so that even the ball of radius 1 looks not quite different from the entire space on these tiny scales. $\endgroup$
Fan Zheng
– Fan Zheng
2016-10-06 18:31:06 +00:00
Commented Oct 6, 2016 at 18:31
$\begingroup$ Thanks Terry. I considered this function before but I didn't know how to show $M_\phi f$ is not in $L^1$ after some attempts. Could you give more hints? $\endgroup$
Mr.right
– Mr.right
2016-10-06 18:32:28 +00:00
Commented Oct 6, 2016 at 18:32
$\begingroup$ Lower bound the $L^1$ norm of $M_\phi 1_{[m, m+\frac{1}{m \log^2 m}]}$ on $[m,m+1]$. $\endgroup$
Terry Tao
– Terry Tao
2016-10-06 18:35:21 +00:00
Commented Oct 6, 2016 at 18:35
$\begingroup$ More precisely, between $[m + \frac{1}{m \log^2 m}, m+1]$, you have that $M_\phi f$ is bounded below by something that looks like $\frac{1}{(x-m)m \log^2 m}$. // Incidentally, this example also suggests some sort of hierarchy where $M_\phi$ only loses a log at infinity (similar to how it only loses a log locally). Is there a way to make this precise? $\endgroup$
Willie Wong
– Willie Wong
2016-10-06 18:37:33 +00:00
Commented Oct 6, 2016 at 18:37
$\begingroup$ @WillieWong Could it be related to the scaling symmetry of $M_\phi$? $\endgroup$
Fan Zheng
– Fan Zheng
2016-10-06 18:43:44 +00:00
Commented Oct 6, 2016 at 18:43
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White Matter Lesions
Roopa Sharma; Sandeep Sekhon; Forshing Lui; Marco Cascella.
Author Information and Affiliations
Authors
Roopa Sharma1; Sandeep Sekhon2; Forshing Lui3; Marco Cascella4.
Affiliations
1 Maulana Azad Medical College, New Delhi, India
2 Delhi University/Maulana Azad Medical College, New Delhi, India
3 CA Northstate Uni, College of Med
4 Istituto Nazionale Tumori - IRCCS - Fondazione Pascale, Via Mariano Semmola 80100, Napoli. Italy
Last Update: March 7, 2024.
Continuing Education Activity
White matter lesions, visualized as hyperintensities on magnetic resonance imaging (MRI) scans, signify abnormal myelination in the brain and serve as markers for small vessel disease. They are considered a marker of small vessel disease. However, there are numerous non-vascular causes, as well. An increase in white matter lesions increases the risk of stroke, cognitive decline, depression, disability, and mortality in the general population.
Participants thoroughly examine white matter lesion pathophysiology, encompassing vascular and non-vascular contributors. The course offers detailed reviews of MRI imaging modalities, including T2 weighted and fluid-attenuated inversion recovery sequences that elucidate diagnostic nuances. Treatment strategies tailored to the underlying etiology are examined, aiming to mitigate associated risks such as stroke, cognitive decline, and depression. Engagement in this course fosters collaboration among healthcare professionals, emphasizing the crucial role of the interprofessional team in evaluating and managing patients with white matter lesions. By drawing on diverse expertise, including neurologists, radiologists, and primary care providers, clinicians can effectively navigate the complexities of diagnosis and treatment, ultimately enhancing patient outcomes and quality of life.
Objectives:
Identify white matter lesions on neuroimaging studies by interpreting magnetic resonance imaging and computed tomography scans.
Interpret the patient presentation associated with various diseases presenting white matter lesions and assess their pathophysiology.
Evaluate the progression and impact of white matter lesions on cognitive and functional abilities, utilizing standardized neuropsychological assessments and functional status evaluations.
Apply effective interprofessional team strategies to improve and enhance care delivery for patients with white matter lesions.
Access free multiple choice questions on this topic.
Introduction
The brain's white matter (WM) constitutes a network of nerve fibers that allows the exchange of information and communication between different areas of the gray matter (GM). WM lies beneath the GM in the brain and superficial to GM in the spinal cord and has evolved more than GM and occupies almost half of the brain. The size of white matter varies depending on the extent of central GM in different regions of spinal cord cross-sections. In particular, the GM to WM ratio grows from the cervical region to the lumbar region as the WM decreases when it proceeds towards the terminal portion of the spinal cord.
The WM contains neural networks formed by bundles of axons to mediate essential connectivity between different key motor and cognitive cortical regions. The white matter comprises myelinated and unmyelinated axons and glial cells, including myelin-producing oligodendrocytes, microglia, astrocytes, and oligodendrocyte progenitor cells. Myelin acts as electrical insulation for axons, is responsible for rapid saltatory impulse propagation, and protects the nerve fibers from injury. Myelin has a water content of about 40%. The remaining dry mass (60%) is mainly composed of proteins (15% to 30%) and lipids (70% to 85%), with phospholipids, cholesterol, galactolipids, and plasmalogens in a molar ratio of 2:2:1:1, respectively.
WM lesions or leukoaraiosis encompasses small vessel vascular brain diseases and non-vascular conditions. Any process leading to a change in chemical composition, damage, or ischemia of myelinated fibers can present as white matter lesions on magnetic resonance imaging (MRI), representing the gold standard for lesion investigation. WM hyperintensity (WMH) is quite a descriptive expression used on MRI. These lesions are best seen as hyperintensities on T2 weighted and FLAIR (fluid-attenuated inversion recovery) sequences of MRI. FLAIR sequences are particularly important in assessing white matter lesions near the ventricular margin by nullifying the cerebrospinal fluid signal.
While the WMHs are well-described on MRI, the lesions were first illustrated based on brain computed tomography (CT). In 1985, Hachinski 1985 described "leukoaraiosis" as "diminished density of white matter that is seen on brain computed tomography." Concerning small vessel vascular brain processes, white matter lesions are commonly present in MRI of asymptomatic elderly individuals, typically located in periventricular (PV-WMH) and deep subcortical regions (DS-WMH). For instance, white matter lesions are frequently detected in people with untreated chronic hypertension. The volume of white matter lesions increases with age from small punctate lesions to large confluent lesions. Nevertheless, although white matter lesions were initially considered a normal, age-related finding, recent investigations proved that large areas of disease in the brain's WM must be considered neuroimaging markers of brain frailty. Of note, various longitudinal studies described white matter lesions as a predictor for future risk of stroke, cognitive decline, depression, disability, and mortality in the general population. The clinical significance of the lesions is confirmed by the results of a meta-analysis that demonstrated a 3-fold increased risk of dementia and stroke and a doubled risk of death in people with white matter lesions. For instance, ischemic microvascular disease, a vascular cause of white matter lesions, may cause about 45% of dementia cases and 20% of strokes. Again, white matter lesions are associated with poor post-stroke outcomes and increased risk of parenchymal hematoma following mechanical thrombectomy.
Apart from white matter lesions secondary to small vessel disease, these lesions are also common features of demyelinating inflammatory disorders, leukodystrophies, and degenerative disorders. Clinical aspects, prognosis, and management vary according to the distribution and spread of the WM damage. Therefore, this group of lesions and diseases includes very different clinical conditions regarding etiology, pathogenesis, pathological features, clinical presentations, imaging, and therapy.
Etiology
The etiology of white matter lesions is heterogeneous. Based on etiology, they can be divided into vascular and non-vascular causes.
Vascular Causes of White Matter Lesions
Microvascular ischemic disease or small vessel disease
Atherosclerosis
Migraine
Amyloid angiopathy
Cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy
Vasculitis
Susac syndrome
Non-Vascular Causes for White Matter Lesions
Inflammatory: Multiple sclerosis, acute disseminated encephalomyelitis, and neuromyelitis optica spectrum disorders
Infectious: Human immunodeficiency virus encephalopathy, progressive multifocal leukoencephalopathy, neuroborreliosis, herpes simplex virus, cytomegalovirus encephalitis, neurosyphilis, central nervous system cryptococcal infection, Whipple disease, Lyme encephalopathy, subacute sclerosing panencephalitis.
Toxic: Chronic alcohol abuse, carbon monoxide intoxication, inhalation of toluene, heroin, and cocaine, methotrexate-related leukoencephalopathy.
Metabolic: Vitamin B12 deficiency, copper deficiency, acute intermittent porphyria, hepatic encephalopathy, Hashimoto encephalopathy.
Neoplastic: Glial tumors, central nervous system, lymphoma
Traumatic: Radiotherapy, post-concussion (traumatic axonal injury)
Genetic:
Lysosomal storage diseases: Metachromatic leukodystrophy, Krabbe disease, Fabry disease, gangliosidosis, mucopolysaccharidosis
Peroxisomal disorders: X-linked adrenoleukodystrophy, Zellweger syndrome, Refsum disease
Mitochondrial disorders: Myoclonic epilepsy with ragged red fibers, mitochondrial encephalopathy, lactic acidosis, and stroke-like episodes, Leigh disease, Kearns-Sayre disease
Aminoacidopathies and organic acidopathies: Canavan disease, glutaric aciduria, urea cycle disorders
Unknown etiology: Alexander disease, Vander Knapp encephalopathy
Others: Vanishing white matter disease, myotonic dystrophy
Another nosographic approach distinguishes primary white matter lesions, derived from an unknown etiology, from secondary ones; these latter are due to a great variety of known etiologies.
Epidemiology
The prevalence of age-related vascular white matter lesions in the general population is approximately 10% to 20% at 60 years and approaches 100% in those older than 90 years. Studies have reported that white matter lesions are common in Japanese, Chinese, Caucasian, African-American, and Caribbean Black populations. Multiple sclerosis (MS) represents the most common inflammatory neurological condition in young adults. Again, it has been estimated that MS disease affects approximately 2,500,000 people worldwide. In the US, the rate is 57 to 78 cases per 100,000 people in the southern states and 100 to 140 cases in the northern states. The prevalence of neuromyelitis optica spectrum disorder (NMOSD) is approximately 1 in 100,000 in the White population and up to 10 per 100,000 in Black populations. In East Asians, the prevalence of NMOSD is about 3.5 per 100,000 population.
Progressive multifocal leukoencephalopathy (PML) is present in 1% to 4% of patients with acquired immunodeficiency syndrome. PML is also a rare yet severe complication of immunosuppression, such as in organ transplant patients. Furthermore, gliomas account for up to 35% of CNS tumors in adolescents and young adults. The incidence of white matter lesions secondary to heritable WM disorder or leukodystrophy is approximately 1 per 8000 live births. However, for acquired WM disorders in children, it is estimated to be 1.66 per 100,000 children.
Pathophysiology
Cerebrovascular risk factors such as age, hypertension, diabetes mellitus, hyperlipidemia, hyperhomocysteinemia, and hypersensitive C-reactive protein are well-known risk factors for vascular white matter lesions. Genetic factors also play an essential role in developing white matter lesions, as many genetic loci have been identified, and twin studies have suggested 55% to 80% heritability.
Pathophysiology for developing vascular white matter lesions in older patients is thought to be secondary to chronically reduced blood flow by arteriosclerosis, lipohyalinosis, or fibrinoid necrosis of small vessels. Incomplete infarction secondary to persistent hypoxia leads to altered cerebral autoregulation, promoting the transcription of many inflammatory genes, leading to the breakdown of the blood-brain barrier and entry of pro-inflammatory proteins into the brain parenchyma and vessel wall. This breakdown will lead to demyelination, axonal loss, vacuolation, and reduced glial density. Results from some studies also suggest the role of venous collagen deposition in the pathogenesis of ischemic white matter lesions.
The different distribution of the lesions could be linked to different pathogenetic mechanisms. Periventricular-white matter hyperintensity (PV-WMH) is featured by gliosis, loosening of the WM, and loss around convoluted venules in perivascular spaces. In contrast, the primary characteristics of deep subcortical-white matter hypersensitivity (DS-WMH) are demyelination, gliosis, and augmented tissue loss as the lesions become serious. By summarizing, the pathological characteristics of white matter lesions may encompass myelin rarefaction, reactive gliosis, axonal loss, infarction, venular collagenosis, and arteriosclerotic small vessel alterations. Moreover, blood-brain barrier impairment plays a pivotal role in the genesis of WM damage, and a different pattern involves PV-WMH and DS-WMH.
In white matter lesions secondary to non-vascular diseases like MS, demyelination is caused by autoimmune inflammation-mediated primarily by T cells against myelin proteins. While the cause of MS is still unknown, it probably presupposes a combination of genetic susceptibility and nongenetic factors such as viral infection and low vitamin D levels. This combination results in an autoimmune disorder that leads to myelin loss, destruction of oligodendrocytes, and reactive astrogliosis. However, the axon is usually undamaged; in some patients, it is aggressively destroyed.
NMOSD is a group of inflammatory disorders of the CNS featuring severe immune-mediated demyelination and axonal damage, mainly involving optic nerves and the spinal cord. Although these disorders were studied as a variant of MS, they have distinct pathophysiology. The autoimmune pathogenesis for NMOSD involves immunoglobulin G autoantibody against the aquaporin-4 water channel or the myelin oligodendrocyte glycoprotein. No antibodies are pertinent in a subset of NMOSD (double-seronegative disease).
The exact mechanism of the pathophysiology of acute disseminated encephalomyelitis is still unknown. However, it has been related to inflammation initiated by infection or vaccination in genetically susceptible individuals, causing demyelination. Progressive multifocal leukoencephalopathy is a central demyelinating disease caused by reactivation (usually occurs at CD4 count less than 200/cm) of latent JC polyomavirus (or John Cunningham virus or human polyomavirus 2) in oligodendrocytes in patients with human immunodeficiency virus. Leukodystrophy causes white matter lesions secondary to substrate accumulation due to enzymatic defects causing demyelination.
Traumatic axonal injury refers to severe axonal mechanical damage due to a rotational acceleration of the brain. Although its pathophysiology is complicated, the injury damage is firstly due to a mechanical break involving axonal microtubules. This stretch induces axonal damage through undulations, breaks, and direct membrane mechanoporation with calcium influx. This mechanism leads to the activation of several injurious pathways, including caspase-mediated proteolysis and cytokine-mediated microglia recruitment with impairment of axonal transport and the aggregation of transported proteins in varicose swellings.
History and Physical
Clinical presentation can vary from asymptomatic to patients with disabling disease as per the white matter lesions etiology. Older patients with small punctate cerebral vascular white matter lesions are usually asymptomatic, but they progress to large confluent lesions and can present with subtle functional decline, cognitive impairment, dementia, urinary incontinence, gait and balance impairment, and neuropsychiatric disorders. Within the group of vascular WM diseases, the distribution of lesions varies greatly, and consequently, the clinical aspects. For instance, in small vessel disease, the WMHs are mainly found in basal ganglia and frontotemporal and periventricular WM; these may induce cognitive impairment, loss of balance or coordination, vision loss, and dizziness. Severe headaches can be present in different types of white matter lesions.
Patients with non-vascular etiology of white matter lesions like MS can have heterogeneous presentations, including fatigue, unilateral visual blurring, sensory changes, motor abnormality, urinary incontinence, speech and swallowing difficulties, pain, anxiety, depression, numbness and tingling, and cognitive dysfunction. Nevertheless, the clinical scenario and evolution of the disease vary from patient to patient. Some patients deteriorate rapidly, while in others, an alternation between relapses and remissions is observed.
Clinical aspects of NMOSD include acute attacks of bilateral optic neuritis with significant visual loss or transverse myelitis, inducing limb weakness, sensory loss, and bladder dysfunction. Other symptoms can consist of episodes of intractable nausea, vomiting, hiccups, excessive daytime somnolence or narcolepsy, and seizures. Commonly relapsing courses and variable degrees of recovery occur within weeks to months.
Children with white matter lesions presenting with progressive symptoms of declining developmental milestones, cognitive impairment, and motor abnormalities should be suspected of leukodystrophy. However, patients in their first or second decade with white matter lesions secondary to acute disseminated encephalomyelitis present with acute onset and rapidly progressive symptoms of fever, headache, vomiting, confusion, or altered sensorium. Although migraine was associated with structural changes in the brain WM, these lesions are generally not linked to any neurological issues, as well as an increased risk of cognitive decline.
Evaluation
Evaluating patients presenting with white matter lesions depends on the patient's age, clinical scenario, and pattern of white matter lesions on MRI, which are common manifestations of cerebral small vessel disease and are associated with vascular risk factors. These patients should be screened for vascular risk factors by routine laboratory tests, including a complete metabolic panel, lipid profile, and HbA1c. Also, it is worthwhile to quantify white matter lesions to monitor their progress with time in these patients. Various grading scales like the Fazekas scale and age-related white matter changes rating scale can be used to assess the extent and progression of white matter lesions (see Image.Fazekas Grade 3 White Matter Lesions).
Patients with non-vascular white matter lesions require further evaluation to identify the etiology of the lesions. Cerebral spinal fluid analysis is helpful in cases of suspected MS (oligoclonal bands), Acute disseminated encephalomyelitis (ADEM; lymphocytic pleocytosis with raised proteins), and infectious demyelination (antiviral antibodies). MR spectroscopy can help in the etiological diagnosis of white matter lesions to differentiate lesions based on different metabolite peaks.
Additional tests like serology for autoantibodies in case of suspected vasculitis, anti-myelin oligodendrocyte glycoprotein antibodies in ADEM, and anti-AQP4 antibodies in NMOSD. Also, metabolic and toxicology screens can be helpful in suspected cases. Children with white matter lesions and clinical scenarios for leukodystrophy require genetic testing.
Treatment / Management
White matter lesions in older adults most commonly reflect underlying small vessel ischemic lesions and, more probably, cognitive dysfunction. Proper management may prevent further deterioration by targeting all vascular risk factors. White matter lesions detected incidentally on MRI of older patients require management of vascular risk factors by:
Intensively controlling the blood pressure
Strict diabetes control
Improving dyslipidemia
Promoting physical fitness
Avoiding smoking
Avoiding alcohol
Use of antiplatelets
Prophylaxis of migraine with aura can also help decrease the risk of white matter lesions. Using B vitamins to lower homocysteine levels is useful in managing patients with small vessel disease. Managing non-vascular white matter lesions is individualized as per etiology. White matter lesions secondary to MS flares are treated with steroids. However, patients need long-term maintenance therapy with disease-modifying treatment to halt disease progression.
Furthermore, immunosuppression with high-dose intravenous glucocorticoids is used in patients with ADEM. Acyclovir has also been reported to have benefits in some cases. Leukodystrophies do not have a specific treatment; treatment is only supportive and symptomatic. Other white matter lesions caused by specific diseases, such as progressive multifocal leukoencephalopathy (PML), human immunodeficiency virus (HIV), and gliomas, must be treated for those specific causes.
Differential Diagnosis
Pattern recognition of white matter lesions in MRI is crucial as it may make the diagnosis in many conditions.
The differential diagnosis for symmetric white matter lesions can include:
Bilateral hemispheres: toxic encephalopathy, HIV encephalopathy, autoimmune encephalopathies, and vitamin B12 deficiency
Periventricular lesions: small vessel disease, HIV encephalopathy, vitamin B12 deficiency, metachromatic leukodystrophy, X-linked adrenoleukodystrophy, and vanishing white matter disease
Subcortical including arcuate fibers: Alexander disease, Kearns-Sayre syndrome, and cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy (CADASIL)
Frontal predominance: Alexander disease and metachromatic leukodystrophy
Parietal/temporal/occipital lobes: posterior reversible encephalopathy, heroin abuse, and Krabbe disease
Corpus callosum: Marchifava-Bignami disease, metachromatic leukodystrophy, and Krabbe disease
Cerebellum: toxic encephalopathy and mitochondriopathies
Central pons: central pontine myelinolysis
The differential diagnosis for asymmetric white matter lesions:
Patchy multifocal confluent lesion: MS, autoimmune encephalopathies, and CADASIL
Parieto-occipital regions: PML
Prognosis
The prognosis of patients with white matter lesions depends on the etiology of the lesions. Patients with age-related white matter lesions are irreversible and progressive. Large and confluent white matter lesions have a poor prognosis and lead to cognitive impairment and global functional decline.
White matter lesions secondary to MS have interpatient variability in prognosis. Severe disabilities are present in 5% of patients within the first 5 years of onset, and 10% to 20% of patients of MS remain unimpaired without therapy after 20 years. NMO and ADEM have a variable prognosis from complete recovery to the development of permanent physical disability, especially in post-measles ADEM. Acute demyelinating disease prognosis depends upon the severity of the initial illness. Patients responding to treatment have a favorable prognosis. However, leukodystrophies have a poor prognosis. Reversible causes, including metabolic and toxic encephalopathies, have a good prognosis.
Complications
Severe white matter lesions are associated with cognitive impairment, global functional decline, cerebrovascular accident, mood disorders, gait, and balance dysfunction. White matter lesions are also associated with grey matter atrophy and accelerated neurodegeneration. Furthermore, severe, extensive involvement of white matter by non-vascular causes like MS, ADEM, and NMO causes disability.
Deterrence and Patient Education
Patients with vascular risk factors should be identified early and counseled on lifestyle changes and control of comorbid conditions. Self-monitoring of blood pressure and blood sugar, dietary modifications, weight reduction, and improving physical fitness have proven to decrease the progression of white matter lesions. Involvement in cognitively complex leisure activity to improve cognitive reserve has been associated with a protective effect on cognitive functioning and late-life depression in patients with white matter lesions.
Enhancing Healthcare Team Outcomes
As described, white matter lesions have varied clinical presentations, differential diagnoses, and complications. Managing these conditions require extensive collaboration and coordination among a team of professionals, which consists of neurologists, radiologists, internists, ophthalmologists, psychiatrists, neurosurgeons, rheumatologists, microbiologists, pain specialists, nurse specialists, mental health nurses, pharmacists, physical therapists, and nutritionists. Clear and effective communication between these professionals while monitoring the progression and complications of white matter lesions can decrease mortality.
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Figure
Fazekas Grade 3 White Matter Lesions. Fazekas grade 3 white matter lesions in a case of stroke. Contributed by S Munakomi, MD
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: Disclosure: Roopa Sharma declares no relevant financial relationships with ineligible companies.
: Disclosure: Sandeep Sekhon declares no relevant financial relationships with ineligible companies.
: Disclosure: Forshing Lui declares no relevant financial relationships with ineligible companies.
: Disclosure: Marco Cascella declares no relevant financial relationships with ineligible companies.
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Lung cancer in females—sex-based differences from males in epidemiology, biology, and outcomes: a narrative review
Kaylan Gee
Kaylan Gee
1 Department of Thoracic Surgery, Roswell Park Comprehensive Cancer Center, Buffalo, NY, USA;
2 Department of Surgery, University of Tennessee Graduate School of Medicine, Knoxville, TN, USA;
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1 Department of Thoracic Surgery, Roswell Park Comprehensive Cancer Center, Buffalo, NY, USA;
2 Department of Surgery, University of Tennessee Graduate School of Medicine, Knoxville, TN, USA;
3 Jacobs School of Medicine and Biomedical Sciences, State University of New York, Buffalo, NY, USA
Contributions: (I) Conception and design: Both authors; (II) Administrative support: S Yendamuri; (III) Provision of study materials or patients: None; (IV) Collection and assembly of data: K Gee; (V) Data analysis and interpretation: K Gee; (VI) Manuscript writing: Both authors; (VII) Final approval of manuscript: Both authors.
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Correspondence to: Sai Yendamuri, MD, MBA. Department of Thoracic Surgery, Roswell Park Comprehensive Cancer Center, #717, Elm and Carlton Streets, Buffalo, NY 14203, USA; Jacobs School of Medicine and Biomedical Sciences, State University of New York, Buffalo, NY, USA. Email: sai.yendamuri@roswellpark.org.
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ORCID: Kaylan Gee, 0009-0005-3888-1039; Sai Yendamuri, 0000-0001-6654-3487.
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Corresponding author.
Received 2023 Nov 15; Accepted 2024 Jan 19; Issue date 2024 Jan 31.
2024 Translational Lung Cancer Research. All rights reserved.
Open Access Statement: This is an Open Access article distributed in accordance with the Creative Commons Attribution-NonCommercial-NoDerivs 4.0 International License (CC BY-NC-ND 4.0), which permits the non-commercial replication and distribution of the article with the strict proviso that no changes or edits are made and the original work is properly cited (including links to both the formal publication through the relevant DOI and the license). See:
PMC Copyright notice
PMCID: PMC10891406 PMID: 38405003
Abstract
Background and Objective
The role of biological sex is seldom considered in characterizing lung cancer, the deadliest cancer in both the United States and the world. Lung cancer has traditionally been regarded as a male disease; as such, research in female-specific phenomena is frequently conflicting or absent. Currently, disparities in lung cancer incidence are primarily driven by females, especially non-smokers and those of younger age. This narrative review provides insight into sex-specific characteristics of lung cancer, highlighting risk factors, diagnosis patterns, carcinogenesis, and treatment outcomes in females.
Methods
The PubMed database was searched on July 26, 2023 to identify research published between 2013 and 2023 in English. Sixty-three articles were considered relevant, and their full texts and citations were studied to compile information for this narrative review.
Key Content and Findings
Exposure-related risk factors, including personal tobacco use, are thought to impact female lung cancer risk more profoundly. However, studies on occupational exposures are underpowered to conclude risk in females. Data characterizing the effect of endogenous and exogenous hormonal exposures on female lung cancer risk remain two-sided. Screening guidelines are tailored to white males, exacerbating sex and race disparities. The effect of biological sex on carcinogenesis and the immune system response to cancer is not fully understood, though the female immune system clearly reacts more aggressively to lung cancer. In early-stage disease, females have greater survival in the perioperative setting and during follow-up of several years, attributed to favorable histopathology and healthier baseline status. Sex-specific response to systemic treatment continues to be optimized as lack of standardization in randomized trials makes interpreting results difficult when aggregated.
Conclusions
Biological sex plays a critical role in non-small cell lung cancer (NSCLC), though further study is needed to depict the complex web of factors that affect lung cancer risk, development, and outcomes. Female underrepresentation in studies has contributed to this lack of understanding. As these disparities are eliminated, we can move towards more effective treatment for both sexes in this pervasive yet deadly disease.
Keywords: Lung cancer, non-small cell lung cancer (NSCLC), female, sex, sex difference
Introduction
Lung cancer is the leading cause of cancer death both in the United States and worldwide. In the United States, it has been the deadliest cancer for males since the 1950s and exceeded breast cancer as the deadliest for females in 1987 (1). In 2020, lung cancer caused 1,800,000 deaths worldwide, almost double the mortality of colorectal cancer, which is the second most deadly (2,3). Fortunately, lung cancer mortality in the United States has decreased annually since 2000. The Centers for Disease Control and Prevention reports the mortality rate of lung cancer was 55.8 per 100,000 people in 2000 but has now decreased to 33.4 per 100,000 people as of 2019, the most recent year for which reliable data are available (4). Likewise, the incidence of lung cancer has declined from 70.2 to 54.3 per 100,000 people, though the actual number of new lung cancers has generally increased over that period due to a growing population (4). Noticeably fewer new cases were reported in 2020, thought to be tempered by the unprecedented impact of the coronavirus disease 2019 (COVID-19) pandemic on health services and recommended screening practices (4). In recent years, lung cancer incidence has been 27% higher among males compared to females (5), though both incidence and mortality are declining more rapidly in males than females. These trends are generally attributed to differences in gendered smoking patterns (5).
This disease has two broad categories: small cell lung cancer and, conversely, non-small cell lung cancer (NSCLC). NSCLC is overwhelmingly more common with approximately 85% of new cases in the United States. Its incidence peaks at ages 80 to 84 in males compared to 75 to 79 in females (5). NSCLC is further divided into adenocarcinoma, squamous cell carcinoma, and large cell carcinoma (6,7). While tobacco use is widely known as the principal risk factor for lung cancer, these histologic varieties differ in “strength of the association” (6) with smoking. Importantly, disparities in lung cancer incidence largely depend on increasing rates of adenocarcinoma driven by young females and never smokers (8) yet understanding of this demographic shift is sorely needed. Consensus on lung cancer risk from non-tobacco exposures, the influence of sex-specific hormones, and the sex-based immune response to carcinogenesis are ill-defined in the literature. This narrative review aims to portray how biological sex affects NSCLC—its development, diagnosis, and treatment outcomes. We present this article in accordance with the Narrative Review reporting checklist (available at
Methods
The PubMed literature database of the United States National Library of Medicine was used to identify previously published full-text research articles published in English from 2013 to 2023. All study designs and systematic and narrative reviews were included. A database search was performed on July 26, 2023, written as follows: ((gender[Title]) OR (sex[Title])) AND (lung cancer[Title]). The primary author reviewed the titles and abstracts of the 196 articles delivered. Articles of poor reliability or those that examined especially geographically-specific lung cancer trends were excluded. The authors considered sixty-two publications germane to the themes of this review and their full texts, including references, were studied in detail. Additional articles retrieved from references were included for completeness. The search strategy is summarized in Table 1 and included articles are identified explicitly in Table 2.
Table 1. The search strategy summary.
| Items | Specification |
:--- |
| Date of search | July 26, 2023 |
| Databases and other sources searched | PubMed |
| Search terms used | ((gender[Title]) OR (sex[Title])) AND (lung cancer[Title]) |
| Timeframe | 2013 to 2023 |
| Inclusion and exclusion criteria | Only full text articles in English included. Articles that examined geographically-specific lung cancer incidence were excluded |
| Selection process | K.G. conducted selection independently, and the authors discussed the scientific and clinical value of controversial articles to reach an agreement |
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Table 2. Articles selected for inclusion (first author, PubMed ID).
| Risk factors | Screening and diagnosis | Carcinogenesis | Outcomes | ICI efficacy evaluated in recent meta-analyses |
:---: :---:
| Baiu I, 34277072 | Araghi M, 34282033 | Caetano MS, 30389925 | Al Omari O, 36476451 | Conforti F, 31106827 |
| Betansedi CO, 29508431 | Pinsky PF, 33545164 | Dubois C, 30444717 | Baum P, 35220299 | Liang J, 36128737 |
| Boice JD, 30614747 | Ragavan M, 35022255 | Freudenstein D, 32545367 | Baum P, 35738973 | Madala S, 35400597 |
| Bugge A, 27846762 | Ruano-Ravina A, 34858780 | Gu T, 25320584 | Bugge A, 27846762 | Takada K, 35999618 |
| Cheng TD, 29346580 | Smeltzer MP, 36208717 | Mederos N, 33148544 | Conforti F, 31106827 | Wang C, 30972745 |
| Fuentes N, 34080912 | Tolwin Y, 32532368 | Nabi H, 29688493 | Conforti F, 34455288 | Xue C, 34513654 |
| Hansen MS, 29087432 | Warner ET, 31054908 | Pérez-Díez I, 33526761 | Deng HY, 33899137 | |
| Lin PC, 30225650 | Yoshida Y, 26366891 | Raskin J, 35884463 | Duma N, 31036771 | |
| May L, 37370722 | | Siegfried JM, 34927202 | Huang Y, 35693273 | |
| Mederos N, 33148544 | | Skjefstad K, 25668612 | Isla, D 27885542 | |
| Meza R, 25822850 | | Stapelfeld C, 28743530 | McGuire AL, 35448189 | |
| Pinheiro PS, 36334356 | | Vavalà T, 34769372 | Pinto JA, 29682332 | |
| Ragavan M, 35022255 | | Xu L, 35577039 | Radkiewicz C, 31247015 | |
| Ragavan MV, 32774466 | | | Ragavan M, 35022255 | |
| Siegfried JM, 34927202 | | | Raskin J, 35884463 | |
| Stapelfeld C, 28743530 | | | Sachs E, 33217414 | |
| Stapelfeld C, 31583690 | | | Ten Haaf K, 25312998 | |
| Wijesinghe AI, 37271799 | | | Tong BC, 24726742 | |
| Yu Y, 25064415 | | | Vavalà T, 34769372 | |
| Zeng H, 34112140 | | | Wainer Z, 27625078 | |
| | | | Wainer Z, 30206043 | |
| | | | Watanabe K, 29582626 | |
| | | | Xiao J, 32984005 | |
| | | | Yoshida Y, 26366891 | |
| | | | Yu XQ, 35124253 | |
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ICI, immune checkpoint inhibitor.
Risk factors
Smoking and chronic obstructive pulmonary disease (COPD)
Far and away, the most important risk factor for developing lung cancer is tobacco use. The fact that 85% of lung cancer deaths worldwide can be attributed to smoking (2,9) highlights the preventable nature of most lung cancer diagnoses and reinforces that tobacco smoking is the most influential modifiable risk factor for lung cancer (10). Models developed to project smoking patterns and lung cancer mortality in the United States anticipate that sex disparities in lung cancer rates will dissipate by the mid-2040s and that smoking prevalence will drop to only 7.5% by 2065, though this percentage still corresponds to some 50,000 lung cancer deaths (11).
Differences in gendered smoking habits have contributed to the changing demographics of lung cancer (8). In the 1960s, male cigarette use began to decrease, and the incidence of squamous cell carcinoma accordingly declined. It is thought that changes in cigarette filtration systems, which primarily occurred in the late 1960s and 1970s, allowed smaller particles to travel more distally within the lungs thus contributing to the upward trend of adenocarcinoma (12-14). Indeed, adenocarcinoma surpassed squamous cell carcinoma as the most frequently diagnosed lung cancer in males around 1994 (6). In contrast, women began smoking in higher proportions decades after men, the smoking rate began to decline after that of men, and smoking cessation continues to take place at a slower pace in women (8). Adenocarcinoma now accounts for more than 50% of lung cancer cases in females compared to approximately 30% 50 years ago (6).
Data disagrees on which is the more important prognostic factor in developing lung cancer—smoking status or smoking intensity. Some studies purport that smoking status trumps intensity as higher pack-years in males compared to females did not affect mortality rates between the two sexes (15). Conversely, when smoking amount was measured as a continuous variable in another study, female smokers had a progressively greater risk of lung cancer as pack-years of use, number of daily cigarettes, and usage duration increased as compared to males (16).
In the past, studies have generally recognized that females are more sensitive to cigarette smoke (14,17-19). Of those diagnosed with COPD or lung cancer, studies show that females generally have 20–25% less pack-years of tobacco use (20). Similarly, with equivalent amounts of smoking history, females are more likely to develop both adenocarcinoma and squamous cell carcinoma compared to males (18). Conversely, a meta-analysis published in 2014 with more than 400,000 individuals from 47 studies found a 1.61 relative risk ratio [95% confidence interval (CI): 1.37–1.89] of developing lung cancer for male to female smokers (21). Thus, debate remains about which sex is more susceptible to the negative effects from smoking.
COPD is also a risk factor for lung cancer. Bronchial hyperresponsiveness, airway obstruction that is often a component of asthma and COPD, is present in 87% of female smokers with mild to moderate COPD vs. 63% of male smokers (22). In females, airway obstruction presents earlier and with greater bronchial wall thickening as compared to males (20). Cigarette use is the single most important risk factor in exhibiting bronchial hyperresponsiveness in females. In contrast, atopy and asthma are the foremost risk factors for males. Females are twice as likely as males to develop lung cancer with comparable decreases in forced expiratory volume in one second, even when adjusted for smoking history (22). Mathematical quantification of the effect of cigarette smoke on lung parenchyma using a novel smoking-emphysema index, determined by low-attenuation area volume percentage on three-dimensional computed tomography scan divided by pack-years of smoking, demonstrated a greater volume of emphysematous change in male lungs overall. Nonetheless, females were found to have greater low-attenuation area volume percentage per cigarette smoked (23).
Exposure-related risk factors in females
Exposure to second-hand smoke, more commonly experienced by women, increases the risk of lung cancer. An observational study of non-smokers with lung cancer found 69% of females experienced environmental tobacco smoke exposure, compared to only 17% of males. Interestingly, males in this study reported exposure only at work, while females experienced exposure at home and work (24). Known risk factors such as asbestos and radon have been particularly poorly studied in women, though women spend more time in the home and consequently have greater exposure to indoor pollution including radon and tobacco smoke residues that persist on indoor surfaces, termed thirdhand smoke (14,25). A recent meta-analysis found cooking factors, including fume and coal smoke exposure, conferred the highest lung cancer risk to non-smoking Asian women [odds ratio (OR), 2.15; 95% CI: 1.87–2.47] among pooled risk categories of personal and family history, environmental tobacco exposure, diet, and reproductive factors (26). Female chefs exposed more intensely and for more years to cooking oil fumes, which contain known carcinogens, had a significantly higher risk of lung adenocarcinoma compared to male chefs and those less often exposed to cooking oil fumes due to specific cooking practices (27). Research on occupational cancer epidemiology shows that in male-dominated fields like extractive industry, construction, and agriculture, exposure studies are more likely to have a greater than 3.5 ratio of male to female participants (P<0.001), therefore making it difficult to draw conclusions about male vs. female occupational exposure risk (28). Differences in lung cancer risk secondary to radiation exposure remain ill-defined, partially because studies have relatively low female participants (29). However, as the incidence of breast cancer in younger females increases, concern over early exposure to radiation is warranted (30).
Hormonally based risk
Many doubts remain about one category of lung cancer patient: the female never-smoker. The label “never smoker” refers to someone who has smoked less than 100 cigarettes in their lifetime (7). Alarmingly, deaths from lung cancer in never smokers would rank as the seventh most common cause of cancer death worldwide (7,10). Female never smokers are more likely to develop lung cancer than their male counterparts (1,8,17), which holds true across ethnicities (31). The rate of lung cancer diagnosis in never-smoking females is characterized as 14.4 to 20.8 per 100,000 person-years vs. 4.8 to 13.7 per 100,000 person-years for males (25).
Within this demographic, the effect of hormonal exposures has been considered a probable contributor (9,32). Of reproductive factors examined including parity, age at first birth, age at menopause, and exogenous hormone use, only bilateral oophorectomy significantly increased lung cancer risk in never smokers [hazard ratio (HR), 1.47; 95% CI: 1.00–2.16] in an observational study of almost 162,000 women from the United States (33). In a study that closely followed over 71,000 never-smoking Chinese females, older age at menopause, longer reproductive period, increased parity, and history of intrauterine device usage significantly decreased lung cancer risk (32). On the whole, though, data regarding lung cancer risk secondary to endogenous estrogen exposure is inconsistent. Additionally, the role of endogenous vs. exogenous estrogen exposure is a secondary query that requires further investigation (30). A meta-analysis of sex steroids and female lung cancer risk found that the summative effect of higher levels of both endogenous and exogenous hormone exposure reduced lung cancer risk in females by 10%. This result was consistent in subgroup analysis of both Asian (OR, 0.90; 95% CI: 0.84–0.99) and Western (OR, 0.90; 95% CI: 0.84–0.96) females, though for distinct underlying causes of exposure (34). However, there is evidence suggesting estrogen exposure via hormone replacement therapy may in fact increase lung cancer risk in women (8,30,35,36).
Screening and diagnosis
The National Lung Screening Trial was a large, multicenter, randomized-controlled trial that found a 20% reduction in mortality for heavy smokers screened annually with low-dose computed tomography (LDCT) vs. chest X-ray (37). In consideration of these and other results, the United States Preventative Services Task Force currently recommends annual screening with LDCT for adults aged 50 to 80 years with a 20-pack-year tobacco use history who continue to smoke or have quit within the last 15 years (38). Guidelines were expanded in 2021 in part to reach high-risk groups, particularly women and minorities (30,39). Similarly, the National Comprehensive Cancer Network recommends patients aged greater than 50 years with at least 20 years of smoking history to participate in shared decision-making with their healthcare provider regarding lung cancer screening with LDCT (40). Regrettably, screening uptake remains low despite expanded criteria, and many diagnosed with lung cancer do not meet eligibility for screening (39,41).
Screening recommendations are largely based on the lung cancer risk of white males, and because females typically develop lung cancer at a younger age and with less smoking exposure, they are less likely to qualify for screening (39). By examining a “natural experiment” in which patients with a suspicious lung nodule were assessed for lung cancer screening eligibility, Smeltzer et al. proved that expanding eligibility criteria to include patients with 10 pack-years’ tobacco use and patients who quit within the last 25 years would reach a new population comprised of 57% females (P=0.0476), reversing some of the sex disparity in current screening recommendations (39). In a nationwide survey, females were less likely than males to report having discussed lung cancer screening with their healthcare provider regardless of smoking status, though these results did not reach statistical significance, and 32% less likely to have heard of LDCT (OR, 0.68; 95% CI: 0.47–0.99). Over the survey period of 2012–2017, there was no evidence of increased discussion between provider and patient regarding screening recommendations (42). Moreover, a study in Spain found no clinically relevant differences in symptoms at diagnosis between patients of either sex or variable smoking status (43), highlighting that healthcare providers must remain vigilant when determining patients who may benefit from lung cancer screening.
Review of the Surveillance, Epidemiology, and End Results (SEER) database reveals that a greater proportion of males are diagnosed with stage III and IV lung cancer compared to females (75.6% vs. 72%, P<0.0001) (44). This finding is replicated in other high-income countries including Australia, Canada, Ireland, Norway, and the United Kingdom (45). Predictably, when SEER data were examined at the county level, the percentage difference in late-stage diagnosis for males and females diminished as the proportion of female smokers increased. This trend remained consistent when examining lung cancers more associated with smoking (squamous and small cell carcinomas), perhaps suggesting that smokers of either sex are less likely to seek medical opinion when faced with symptoms concerning for lung cancer (44).
Carcinogenesis and the resultant immune response
Smoking, sex steroids, and carcinogenesis
Cigarette smoking has been demonstrated to affect the immune system by reducing neutrophil, antigen-presenting cell, and global T-cell activity (46). On the whole, animal studies generally indicate that when exposed to cigarette smoke and its metabolites, females mount a robust inflammatory response leading to increased oxidative stress in the environment, eventually making cells more susceptible to genetic mutation and malignancy (22). Further, female smokers have higher levels of DNA adducts in peripheral blood mononuclear cells as compared to males (20), and 10–15% less capacity for DNA repair (14,22). Compared to male smokers, female smokers have increased cytochrome P450 (CYP) enzyme expression in their lungs (22). Tobacco smoke itself induces expression of CYP1B1, which then metabolizes estrogen and procarcinogens to their toxic state (47). N-nitrosamines and polycyclic aromatic hydrocarbons, two harmful tobacco-related substances, are primarily catalyzed by CYP to become carcinogenic (22,48). For example, an N-nitrosamine known as nicotine-derived nitrosamine ketone (NNK) is normally broken down via carbonyl reduction; yet without carbonyl reduction, NNK is hydroxylated by CYP and forms DNA adducts. A study which investigated the inhibitory effects of sex hormones on this process showed that progesterone and its synthetic analog were the most powerful inhibitors of safe NNK metabolization (36).
Various studies have attempted to utilize sex steroids, including estrogen receptor-alpha and beta, as a biomarker or determine their prognostic value in lung cancer (9,14,49-52). In order to influence gene expression, estrogen can bind with its receptor and function as a transcription factor or translocate across the cell membrane to induce more immediate events such as ion channel regulation, activation of protein kinase, or formation of second messengers (9,53). Estrogen receptor-beta, which is found more diffusely throughout the body, is highly expressed in bronchial epithelial cells and pneumocytes and contributes to maintenance of extracellular matrix of the lung (9). KRT16, an essential gene in estrogen receptor signaling, was found to be significantly overexpressed in the tumor tissue of female non-smoking lung adenocarcinoma patients compared to that of males, and likelihood of death in females with high KRT16 expression was nearly double compared to males. Bearing in mind these results, estrogen blockade signaling may be an effective option for targeted therapy in non-smoking patients (54). A prospectively enrolled case-case study on hormone receptor expression demonstrated that female sex was associated with lower cytoplasmic estrogen receptor-alpha and nuclear estrogen receptor-beta expression than male sex when adjusted for age, race, and smoking. Additionally, ever-smokers also had lower cytoplasmic estrogen receptor-beta expression. The authors concluded their results supported the “estrogen hypothesis”—that is, estrogens play a role in lung carcinogenesis (17). However, concerns have been raised about the application of these findings, given the study did not consider obesity in its analysis, which is known to modulate estrogen and possibly estrogen receptors (55).
Studies with murine models likewise present conflicting data. Exposure to estradiol resulted in two-fold larger lung tumors, as well as increased angiogenesis and lymphangiogenesis, in immunocompetent female mice compared to male mice. This paradigm persisted in ovariectomized mice, which displayed the least tumor development, while ovariectomized mice treated with estradiol supplementation displayed moderate tumor growth. In contrast, castration of male mice did not affect tumor size, nor did castration with subsequent estradiol supplementation (56). Conversely, another study showed that when estrogen signaling was inhibited via tamoxifen in female mice, KRAS mutant lung cancer development was augmented (57). Sex steroids seem to play a role in the relationship between tobacco smoke and carcinogenesis, thus contributing to sex-based differences in lung cancer development, though this relationship remains to be clearly described.
Females have a stronger immune system
It is well known that females have a more robust immune system, which is borne out in higher incidence of autoimmune disease and lower prevalence of infection or cancer compared to males (46,58). Many genes that participate in the regulation of immune response are located on the X chromosome (46,58,59). Females have enhanced function of both innate and adaptive immunity (46,58-62), including higher phagocytic capacity of both neutrophils and macrophages, greater antibody response with elevated basal immunoglobulin levels and B-cell numbers, and increased CD4+ T-cell count, and T-cell proliferation and activation (46,58). Additionally, estrogens alter the milieu of inflammatory cytokines produced by macrophages and neutrophils, which may influence overall less cancer risk seen in females (46).
Immune response differences in adenocarcinoma according to sex
Novel studies reveal baseline differences in male and female immune responses to lung adenocarcinoma that may contribute to sex disparities in tumor burden and disease outcome (13). Female adenocarcinoma patients tend to have a more robust anti-cancer immune response, specifically a higher proportion of immune infiltrate, than their male counterparts, which may contribute to worse prognosis seen in males (13,63). In fact, lymphocytic infiltration with CD3+ and CD8+ cells has been linked to superior outcomes in multiple tumors (64). In “The Immune Landscape of Cancer”, Thorsson et al. described a framework for examining “immune signature” in various cancers, which includes six subtypes: wound healing, interferon (INF)-gamma dominant, inflammatory, lymphocyte depleted, immunologically quiet, and transforming growth factor (TGF)-beta dominant (65). In female lung adenocarcinoma, INF-gamma, lymphocytic infiltration, and M1 macrophage enrichment were noted. Evaluation of immune-related gene expression revealed predominance of antigen processing and presentation pathways. In contrast, male tumors trended towards M2 macrophage enrichment (65). Those with wild type-TP53 tumors and high expression of these immune-related genes demonstrated a statistically significant survival benefit compared to the wild type-TP53, low expression group, and mutant TP53 groups with both high and low expression of these genes, regardless of sex, highlighting the clinical significance of this work (13).
Outcomes
Understanding of lung cancer in females generally concludes favorable prognosis compared to male counterparts. From 1999 to 2019 in the United States, females with lung cancer consistently had decreased age-standardized mortality rates compared to males (66). In fact, when controlling for age, smoking history, stage at diagnosis, and histology, females demonstrate superior survival rates (30,67,68). SEER data reveals that males have significantly worse survival than females across all stages defined in the 8th edition of the tumor, node, metastasis staging system, with the greatest disparity noted in early stage disease at both 1 and 5 years of follow-up (69).
Compared to males, females are more likely to be diagnosed at earlier stages, be never-smokers, and have adenocarcinoma histology (70,71). The period during which a patient is asymptomatic from disease yet lung cancer is detectable on imaging was found to be 4.48 years for males compared to 6.01 years for females (72). This is indicative of the fact that females are more likely to be diagnosed with adenocarcinoma, characteristically slower-growing, thus offering greater opportunity for early-stage diagnosis. Furthermore, most of the excess death risk in males diagnosed with lung cancer can be attributed to known prognostic factors, including patient characteristics, demographic and lifestyle data, and tumor- and treatment-related factors (73). When adjusting for these predictive factors, Yu et al. were able to show reduction in risk of death from HR 1.33 to 1.06 (95% CI: 0.96–1.18; P=0.26), achieving a non-significant difference in death risk between males and females. Approximately 50% of the excess risk of death came from treatment-related factors, specifically receiving surgery, systemic therapy, or radiation within 6 months of diagnosis (73). This review will concentrate on sex-based differences in outcomes of surgery, the favored management of early disease, and contemporary systemic treatments.
Surgery for early-stage lung cancer
The primary treatment of stage I and II lung cancer is surgery with curative intent. Definitive radiation is considered if the patient is not a surgical candidate secondary to comorbidities or problematic tumor location. A previous meta-analysis of patients with inoperable NSCLC treated with radiation found that female gender was the sole demographic factor to confer an overall survival benefit (74). In contrast, a large retrospective study of patients who underwent stereotactic body radiation therapy found that unfavorable histology, increased body mass index, significant comorbidities, and radiation dosing were predictive of local treatment failure, but not sex (75).
Various studies of patients with early-stage NSCLC undergoing resection find superior outcomes for females in both overall and perioperative survival. A retrospective review of 735 NSCLC surgical resection cases from 1995 to 2010 at a single Japanese institution demonstrated that overall survival at 1, 3, and 5 years was significantly better for females. Subgroup analysis further revealed that female sex and adenocarcinoma histology were significant positive prognostic factors only in pathologic stages I and II (n=557). Female survival advantage was lost in late-stage disease; survival curves crossed at around 4 years of follow-up (70). Importantly, length-time bias could have contributed to the improved overall survival seen in females as they were more frequently diagnosed with screening or with incidental imaging (70). Similarly, male sex was found to be an independent negative prognostic factor (HR, 1.54; 95% CI: 1.10–2.16) for mortality at 5-year in an Australian cohort study performed between 2000 and 2009. As female patients were diagnosed with higher stage disease, their mortality risk increased at a faster pace compared to males. For example, stages III and IV conferred a HR 8.71 for female patients vs. 3.66 for male patients compared to the reference group which was comprised of patients diagnosed at Stage IA, meaning their tumors were less than three centimeters (76).
Even in studies with longer follow-up, survival advantage in females persists. A prospective cohort study of stage I and II NSCLC patients conducted in Norway from 2003 to 2013 determined 37.8% survival for females vs. 28.2% in males at 10 years of follow-up, though this difference did not reach statistical significance in the relatively small study population (n=692) (15). Females 66 years and older did have a statistically significant greater survival than males at 10 years (28.2% vs. 19.5%, P<0.03). By the end of ongoing follow-up, that significance was lost (P=0.06), which was attributed to confounders such as pack-years of smoking, cancer stage, large cell carcinoma histology, and lobectomy (15). A nationwide observational cohort study of 6,536 patients who underwent pulmonary resection for lung cancer in Sweden from 2008 to 2017 found a lower risk of death in females compared to males with increasing absolute survival difference over 10 years of follow-up (77). When the cohort was divided by pathologic stage, histology, and age, except for those younger than 60 years, improved survival for women was maintained regardless of differences in patient characteristics such as comorbidities, frailty, and socioeconomic status. The advantage was conferred in both adenocarcinoma and squamous cell carcinoma, though to a lesser extent with the latter (77). These studies beg the question: why do younger females with early-stage lung cancer have comparatively less survival advantage than their elders?
Regarding perioperative survival, females undergoing lung resection are generally younger and have fewer comorbidities (78,79). A retrospective review of a German discharge registry of almost 39,000 patients undergoing lung resection from 2014 to 2017 compared in-hospital mortality, complications, and comorbidities between the sexes. They found a difference in raw in-hospital mortality: 1.8% for females and 4.1% for males. Regardless of surgical approach, women had significantly fewer post-operative complications, including prolonged ventilation, pneumonia, tracheotomy, empyema, and sepsis. Women were significantly less likely to have hypertension, chronic renal failure, diabetes mellitus, and COPD, which continued to be true after stratifying for open vs. minimally invasive surgery. They were also significantly more likely to be referred, compared to males, who were more frequently admitted in an emergent fashion or transferred from another hospital. In multivariable regression, in-hospital mortality advantage for females was maintained—they were 21% less likely to die in hospital following lung resection for cancer (OR, 0.79; 95% CI: 0.66–0.93; P<0.005) (78). An analogous study performed in the United States reviewed more than 34,000 patients from the Society of Thoracic Surgeons General Thoracic Surgery Database undergoing resection for lung cancer between 2002 and 2010 similarly found females to have significantly less coronary artery disease, diabetes, and renal insufficiency. Combined in-hospital and 30-day mortality was found to be 1.5% in females and 3% in males (79).
It is imperative to recognize these studies were performed during years in which video-assisted thoracoscopic surgery (VATS) techniques were being adopted. In the Society of Thoracic Surgeons database, lobectomies via VATS increased by 10.4% between 2004 and 2006 (80). By 2013, surgeons in Japan self-reported performing 68.9% of lobectomies using VATS techniques (81). As this minimally invasive technique was being implemented, it is logical that surgeons may have chosen certain patients, such as those with smaller tumors or better functional status, to undergo a relatively unfamiliar VATS procedure. Years later, the data exists to show that VATS procedures reduce post-operative complications and improve long-term survival (77,82), which may have influenced superior survival outcomes following lung resection seen in females.
Studies have demonstrated sex disparities in surgery for early-stage lung cancer. A review of patients with stage I NSCLC who participated prospectively in the National Lung Screening Trial found that women were “less likely to undergo full resection” despite having surgery as often as men, though this was not statistically significant. The authors posited favorable pathology and patient preference may have influenced these decisions (83). An early comparison of limited resection vs. lobectomy for stage IA lung cancer showed that limited resection was performed in females more often (84), though we now have robust evidence that sublobar resection is non-inferior to lobectomy in patients with clinical stage IA peripheral NSCLC tumors less than two centimeters (85,86). What was previously considered a limited resection is now known to be a suitable choice in select patients. Further, propensity-matched patients with stage IA peripheral NSCLC were found to have no difference in the rate of lymph node metastases between the sexes, suggesting no justification for offering sex-dependent surgical resection (87).
Recurrence after surgical resection
When early-stage tumors are resected with curative intent, molecular testing with next-generation sequencing is not typically performed though up to 70% of these patients will ultimately have cancer recurrence (88). Following surgical resection with frequent surveillance, men more frequently recurred within the first year after surgery while peak recurrence in women occurred over a longer range of duration from 2 to 3 years following surgery (71). A recent propensity-matched cohort controlling for sex and smoking status of early- and late-stage NSCLC patients demonstrated no differences in the frequency of targetable mutations, thus making a case for routine sequencing of resected specimens to optimize systemic therapy for future recurrence, should it be needed (88).
Advanced stage and metastatic disease
Significant innovation has led to new therapies for advanced NSCLC in recent decades, and it is thought that over half of patients with advanced NSCLC may have an actionable mutation towards which systemic therapy can be directed. Epidermal growth factor receptor (EGFR)-targeted therapies were introduced in 2003; immune checkpoint inhibitors (ICI) have been Food and Drug Administration (FDA)-approved for treatment of lung cancer since 2015 (89). Currently, the National Comprehensive Cancer Network recommends testing for the several biomarkers in advanced or metastatic disease, including EGFR, KRAS, HER2, ALK, ROS1, and PD-L1 expression (90). Responses to modern systemic therapies vary greatly between males and females, emphasizing fundamental differences in their respective diseases (68). In comparison, older studies demonstrate that traditional chemotherapy offers marginal benefit (on the order of months) to females over males in certain subgroups (91). In the age of precision medicine, it will become increasingly important to consider biological sex in the lung cancer treatment strategy (92,93).
Targeted therapy
Multiple meta-analyses have been performed to evaluate the efficacy of EGFR inhibitors. There is disagreement on whether EGFR-tyrosine kinase inhibitor (EGFR TKI) confer tangible benefit to females with NSCLC. A meta-analysis of six phase-three trials comparing EGFR-TKI to chemotherapy in patients with identified EGFR mutations, females were shown to have improved progression-free survival (HR, 0.34; 95% CI: 0.28–0.40) compared to males (HR, 0.44; 95% CI: 0.34–0.56), though high heterogeneity was noted in the female group (92). Conversely, in a meta-analysis of 22 studies, Xiao et al. found that only in the subgroup analysis comparing EGFR-TKI to placebo did females demonstrate improved overall survival compared to males. There was no improved survival noted when EGFR-TKI was compared to chemotherapy (HR, 1.01; 95% CI: 0.89–1.14; P=0.89) or when only patients with known EGFR mutation were considered (HR, 1.02; 95% CI: 0.78–1.34; P=0.86) (68). As noted in the recent review by Huang et al., it appears that first- and second-generation EGFR-TKI offer longer progression-free survival to females, which may indicate “immediate response to treatment” and inherently does not consider patient characteristics which impact overall survival (93).
ALK and ROS1 rearrangements also offer opportunity for targeted therapy. They represent a relatively small percentage of NSCLC patients, 3% to 8% of the former and 1% to 2% of the latter, though studies have found these mutations more prevalent in women (94,95). There is no current evidence to identify sex differences in treatment and survival outcomes with ALK and ROS1 inhibitors (59,93).
Immunotherapy
For patients without a known driver mutation, immunotherapy is considered. These drugs block negative regulators PD-1, PD-L1, and CTLA-4 of the immune system. Six meta-analyses generated in our literature search have investigated sex differences in ICI efficacy. Essentially, findings have been inconsistent across these analyses, which are summarized in Table 3. Given results of early meta-analyses, Conforti et al. hypothesized that females could benefit from treatment other than immunotherapy alone. A subsequent meta-analysis demonstrated that women experienced a large survival benefit with anti-PD-1 or anti-PD-L1 therapy in combination with chemotherapy with a pooled HR of 0.44 compared to the male HR of 0.76 (96). It was thought the addition of chemotherapy enhanced the antigenicity of tumor cells, thus prompting the female immune system to respond more vigorously (96). A follow-up study found that despite having high PD-L1 expression, defined as tumor proportion score ≥50%, females did not derive statistically significant benefit from ICI monotherapy (61). In their review on gender differences in immunotherapy outcomes, Vavalà et al. appropriately highlights that heterogeneity within randomized-controlled trials, lack of subgroup data, and relatively short follow-up allow only conjecture, but not durable conclusions, to be drawn from these analyses (46). As plainly demonstrated in Table 3, the heterogeneity of the intervention and control arms offers an avenue for bias introduction in these investigations.
Table 3. Recent meta-analyses examining sex differences in efficacy of ICIs in treating NSCLC.
| Meta-analysis, year published | Number of patients included [number of studies included] | Number of female participants (%) | Intervention vs. control arms, and clarifying comments | Pooled HR (95% CI) | Conclusions |
:---: :---: :---: |
| Conforti, 2018 (60) | 3,482 | 1,167 (33.5) | PD-1 inhibitor or CTLA-4 inhibitor plus chemotherapy vs. chemotherapy or chemotherapy plus placebo | Overall survival compared to controls: male: 0.72 (0.61–0.86); female: 0.89 (0.71–1.11) | Advanced or metastatic cancers were studied, of which NSCLC was a subgroup. In the overall study of 20 trials and 11,351 patients, pooled male survival was 0.72 (0.65–0.79) and female survival 0.86 (0.79–0.93) with a significant difference in efficacy between sexes (P=0.0019), demonstrating sex-specific benefit in overall survival with ICI therapy |
| Conforti, 2019 (96) | 2,970† | 991 (33.4) | PD-1/PD-L1 inhibitor plus chemotherapy vs. chemotherapy. When excluding the trial with SCLC patients, a statistically significant overall survival benefit for females compared to males remained | Overall survival: male: 0.76 (0.66–0.87); female: 0.48 (0.35–0.67) | Females with advanced NSCLC derive greater benefit from combination chemoimmunotherapy than males, though both demonstrate improved survival with combination therapy over chemotherapy alone. However, only males demonstrate benefit from anti-PD-1 alone over chemotherapy |
| 4,923‡ | 1,578 (32.1) | Progression-free survival: male: 0.64 (0.58–0.71); female: 0.56 (0.49–0.65) |
| 3,974 | 1,335 (33.6) | PD-1/PD-L1 inhibitor plus chemotherapy vs. chemotherapy (first-line treatment). Three studies tested PD-1 alone; 3 studies tested PD-1/PD-L1 with chemotherapy | Overall survival for anti-PD-1: male: 0.78 (0.60–1.01); female: 0.97 (0.79–1.19) |
| Overall survival for anti-PD-1/PD-L1 with chemotherapy: male: 0.76 (0.64–0.91); female: 0.44 (0.25–0.76) |
| Wang, 2019 (97) | 8,023 | 2,601 (32.4) | PD-1/PD-L1 inhibitor alone or PD-1 or CTLA-4 inhibitor plus chemotherapy or PD-L1 plus chemoradiotherapy vs. chemotherapy or chemoradiotherapy | Overall survival: male: 0.76 (0.71–0.82); female: 0.73 (0.58–0.91) | Males experience more reliable benefit from ICIs than females. However, CTLA-4 inhibitors did not demonstrate survival benefit for either sex |
| 5,622 | 1,934 (34.4) | PD-1 alone, PD-1 plus CTLA-4 inhibitor, PD-1 plus chemotherapy, PD-L1 inhibitor plus chemotherapy or chemoradiotherapy vs. chemotherapy or chemoradiotherapy | Progression-free survival: male: 0.67 (0.58–0.77); female: 0.73 (0.56–0.95) |
| Xue, 2021 (98) | 12,037 | 3,751 (31.2) | PD-1, PD-L1, CTLA-4 inhibitor, or CTLA-4 plus PD-1 inhibitor together or PD-1, PD-L1, CTLA-4 inhibitor plus chemotherapy vs. chemotherapy or chemotherapy plus placebo | Overall survival: male: 0.73 (0.67–0.79); female: 0.73 (0.61–0.85) | Both males and females benefit from treatment with ICI over chemotherapy. There is no significant difference in overall survival (P=0.97) and progression-free survival (P=0.43) between the sexes |
| 6,940 | 2,088 (30.1) | PD-1, PD-L1, or CTLA-4 plus PD-1 inhibitor together or PD-1 or PD-L1 inhibitor plus chemotherapy vs. chemotherapy or chemotherapy plus placebo | Progression-free survival: male: 0.62 (0.55–0.70); female: 0.68 (0.55–0.81) |
| Liang, 2022 (99) | 10,155 | 3,370 (33.2) | PD-1/PD-L1 or CTLA-4 inhibitor alone, in combination, or with chemotherapy vs. chemotherapy or placebo. Six studies tested ICI plus chemotherapy vs. chemotherapy, 9 studies compared ICI vs. chemotherapy, and 1 study reviewed ICI vs. placebo | Overall survival: male: 0.76 (0.71–0.81); female: 0.74 (0.63–0.87) | In general, females benefitted more from combination chemoimmunotherapy, while males benefitted more from ICIs alone. However, in squamous cell carcinoma, only males demonstrated a statistically significant benefit from ICI therapy, either with or without chemotherapy |
| Madala, 2022 (100) | 9,270 | 3,144 (33.9) | PD-1/PD-L1 inhibitor alone or combined with chemotherapy vs. chemotherapy, placebo, or their combination | Overall survival: male: 0.74 (0.66–0.83); female: 0.72 (0.63–0.82) | No statistical difference was found in overall survival (P=0.709) and progression-free survival (P=0.372) between the sexes |
| 6,193 | 2,059 (33.2) | Progression-free survival: male: 0.63 (0.53–0.75); female: 0.72 (0.58–0.88) |
| Takada, 2022 (101) | 5,280 | 1,554 (29.4) | PD-1/PD-L1 or CTLA-4 inhibitor, or PD-1 plus CTLA-4 inhibitor in combination, or PD-L1 + VEGF inhibitor in combination plus chemotherapy vs. chemotherapy, placebo plus chemotherapy, or VEGF inhibitor plus chemotherapy | Overall survival: male: 0.80 (0.72–0.87); female: 0.69 (0.54–0.89) | During first-line treatment, patients with advanced NSCLC demonstrated both improved overall survival and progression-free survival when treated with combined chemoimmunotherapy rather than chemotherapy alone, though this favors females to a greater degree |
| 3,701 | 1,184 (32.0) | PD-1/PD-L1 inhibitor or PD-1 plus CTLA-4 inhibitor in combination, or PD-L1 + VEGF inhibitor in combination plus chemotherapy vs. chemotherapy, placebo plus chemotherapy, or VEGF inhibitor, placebo, and chemotherapy | Progression-free survival: male: 0.60 (0.55–0.66); female: 0.56 (0.44–0.70) |
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†, 13.6% of patients with SCLC; ‡, 8.2% of patients with SCLC. ICI, immune checkpoint inhibitor; NSCLC, non-small cell lung cancer; HR, hazard ratio; CI, confidence interval; SCLC, small cell lung cancer; VEGF, vascular endothelial growth factor.
Another important factor for consideration is the risk that comes ICI treatment. Females treated with anti-PD-1 therapy were more likely to experience immune-related adverse events (48% vs. 31%, P<0.008), notably pneumonitis and endocrinopathies, in a retrospective review of NSCLC patients. They were also more were likely than males to discontinue the offending agent secondary to side effects (17% vs. 7%, P<0.04). Multivariable analysis examining patient and tumor factors found that only sex was associated with an increased risk of these adverse events. Interestingly, females who experienced immune-related adverse events more frequently demonstrated radiographic response when evaluated several weeks following initiation of ICI therapy (78% vs. 23%, P<0.0001) and had longer progression-free survival (10 vs. 3.3 months, P<0.0006) compared to females with no adverse events. This distinction was not statistically significant in males. Ultimately though, there was no association between immune-related adverse events and overall survival in the NSCLC cohort (102).
The limitations of this narrative review must be acknowledged. A single database was used to collect relevant literature, and articles were selected based on relevancy as determined by the authors’ expertise. Our review is not an exhaustive examination of all data published on this topic.
Conclusions
Increasing lung cancer incidence in non-smokers and young females begs demands greater understanding of sex-based drivers of the disease. Environmental exposure studies, especially those which examine risk factors present in the home, are lacking, the effect of sex hormones on carcinogenesis in the lung is not well understood, and females continue to be underrepresented in clinical trials for lung cancer treatment. Randomized controlled trials with uniform control arms are necessary to truly optimize treatment regimens with new systemic therapies. Practically-speaking, clinicians must recognize that current screening guidelines do not reflect the changing demographics seen in new lung cancer diagnoses. Further investigation into the sex-specific differences in lung cancer will not only increase the proportion of patients diagnosed in its early stages, but also generate sex-based treatment modalities.
Supplementary
The article’s supplementary files as
tlcr-13-01-163-rc.pdf (70.4KB, pdf)
DOI: 10.21037/tlcr-23-744
tlcr-13-01-163-prf.pdf (102.2KB, pdf)
DOI: 10.21037/tlcr-23-744
tlcr-13-01-163-coif.pdf (250.3KB, pdf)
DOI: 10.21037/tlcr-23-744
Acknowledgments
Funding: This work was supported by the Department of Thoracic Surgery at Roswell Park Comprehensive Cancer Center and the University of Tennessee Graduate School of Medicine.
Ethical Statement: The authors are accountable for all aspects of the work in ensuring that questions related to the accuracy or integrity of any part of the work are appropriately investigated and resolved.
Footnotes
Reporting Checklist: The authors have completed the Narrative Review reporting checklist. Available at
Peer Review File: Available at
Conflicts of Interest: Both authors have completed the ICMJE uniform disclosure form (available at S.Y. is a scientific advisory board member of Karkinos Healthcare, receives grant funding from Lumeda Inc., and serves as an unpaid editorial board member of Translational Lung Cancer Research from October 2023 to September 2025. The other author has no conflicts of interest to declare.
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tlcr-13-01-163-rc.pdf (70.4KB, pdf)
DOI: 10.21037/tlcr-23-744
tlcr-13-01-163-prf.pdf (102.2KB, pdf)
DOI: 10.21037/tlcr-23-744
tlcr-13-01-163-coif.pdf (250.3KB, pdf)
DOI: 10.21037/tlcr-23-744
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17440 | https://www.wordsmith.org/words/braggadocio.html | A.Word.A.Day --braggadocio
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braggadocio
PRONUNCIATION:
(brag-uh-DO-shee-o)
MEANING:
noun:
An empty boaster.
Empty boasting.
Boastful behavior.
ETYMOLOGY:
After Braggadochio, a boastful character in Edmund Spenser’s 1590 epic poem The Faerie Queene. Earliest documented use: 1594. Here’s another word that came to us from the same book: blatant.
USAGE:
“We are appalled by the braggadocio of corporate hamburger palaces with huge signs congratulating themselves for selling billions of animal meat sandwiches.”
Richard T. Halfpenny; Funism: the New Religion; Xlibris; 2012.
See more usage examples of braggadocio in Vocabulary.com’s dictionary.
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17441 | https://www.youtube.com/playlist?list=PLSQl0a2vh4HCU81CYvYD_DDyih7ewWRFQ | Variables and expressions introduction | 6th Grade | Khan Academy - YouTube
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17442 | https://teachy.ai/en/summaries/high-school/11th-grade/mathematics-en/newtons-binomial-introduction | Summary of Newton's Binomial: Introduction
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Newton's Binomial: Introduction
Summary of Newton's Binomial: Introduction
Lara from Teachy
Subject Mathematics
Mathematics
Source Teachy Original
Teachy Original
Topic Newton's Binomial: Introduction
Newton's Binomial: Introduction
Newton's Binomial: Introduction
Relevance of the Topic
Newton's binomial is one of the most versatile and widely used mathematical tools, with applications ranging from algebra, combinatorics, and calculus to physics and engineering. Understanding this concept is essential as it allows not only symbolic manipulations but also a better appreciation of the mathematical thinking behind them. Moreover, Newton's binomial enables us to open the doors to the exciting universe of mathematical formulas (one of my passions, I confess), where simplicity and power coexist in harmony, much like Yin and Yang.
Contextualization
The introduction to Newton's binomial fits into the Algebra topic, being a crucial chapter for the evolution of students' algebraic manipulation skills. It provides the necessary foundation for understanding more advanced topics, such as the binomial series and theorem, which will be addressed next. Therefore, understanding the definitions, properties, and applications of Newton's binomial is a vital step for mathematical development.
Theoretical Development
Components
General Term: In Newton's binomial, each term in the polynomial formed by the expansion of a binomial is called the 'general term.' The general term consists of two elements: the binomial coefficient and the binomial bases.
Binomial Coefficient: The binomial coefficient is the number that multiplies each term in the expansion. It is calculated using the formula n! / [(n-k)! k!], where n is the exponent of the binomial and k is the term index.
Binomial Bases: The binomial bases are the constants in the expansion. In the most common case, they are a and b.
Key Terms
Newton's Binomial: It is the result of the expansion of a binomial (a + b) raised to a power n. Each term of the binomial is calculated using the binomial coefficient and the binomial bases.
Binomial Coefficient: Also called Newton's binomial coefficient, this is a coefficient that appears in the expansion of powers of a binomial. It is calculated using the factorial.
Factorial: Represented by n! (read as 'n factorial'), it is the product of all positive integers from 1 to n. For example, 4! = 4 x 3 x 2 x 1 = 24.
Examples and Cases
Calculation of the General Term of the Binomial: Considering the binomial (a + b) squared, the general term is calculated as 2! / [(2-1)! 1!]a^(2-1)b^1 = 2a b.
Pascal's Signaled Relation: In Newton's binomial representation, the binomial coefficients form a line in Pascal's Triangle. The nth line (starting from zero) of Pascal's Triangle corresponds to the coefficients in the expansion of the binomial (a + b)^n.
Detailed Summary
Key Points
Definition of Newton's Binomial: Newton's binomial is the result of the expansion of a binomial (a + b) raised to a power n. Each term of the expansion is calculated using the binomial coefficient and the binomial bases.
Binomial Coefficient: This coefficient, calculated through the formula n! / [(n-k)! k!], is one of the main elements in the understanding and manipulation of Newton's binomial.
General Term: Each term in the polynomial formed by the expansion of a binomial is called the 'general term.' The general term has two parts: the binomial coefficient and the binomial bases.
Relation with Factorial: The calculation of the binomial coefficient involves the concept of factorial, which is the product of all positive integers from 1 to a given number.
Relation with Pascal's Triangle: In Newton's binomial, the binomial coefficients form a line in Pascal's Triangle. The nth line of Pascal's Triangle corresponds to the coefficients in the expansion of the binomial (a + b)^n.
Conclusions
Versatility of Newton's Binomial: Newton's binomial has applications in numerous areas of mathematics and science, becoming a fundamental concept to be mastered.
Importance of Binomial Coefficient: The binomial coefficient is key to the interpretation and manipulation of Newton's binomial, highlighting the importance of understanding the factorial.
Introduction to Pascal's Triangle: The study of Newton's binomial offers a natural introduction to Pascal's Triangle, an essential graphical tool for understanding binomial coefficients.
Exercises
Calculate the sixth term in the expansion of the binomial (2x + y)^6.
Write down the third line of Pascal's Triangle and interpret what each number in this line represents in the expansion of (a + b)^2.
Demonstrate that the sum of binomial coefficients in the nth line of Pascal's Triangle is equal to 2^n.
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17443 | https://www.youtube.com/watch?v=BumTqUPjvNs | Lesson 8.7 - Solving Natural Log Equations & Inequalities
Clayton Rainsberg
2250 subscribers
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Description
15842 views
Posted: 11 Mar 2013
The following video examines how to solve equations & inequalities involving natural logarithms.
Transcript:
in this video we're going to practice solving equations natural logarithmic equations and inequalities and what we have to remember here is the same idea that natural logs and base e are inverse operations of each other so what I'm going to do first is I'm going to isolate my natural log I'm going to get rid of this coefficient of five by dividing both sides by five that gives me the natural log of 6X = 8 FS now that my natural log is isolated by itself I can apply the inverse operation which is to make the left side and the right side the powers of e so e to the natural log of 6X is equal to e to the 8 fths that allows my natural log and E to cancel each other out and so I get 6X = e to the 8 fths power what I like to do again is I always like to solve for the perfect answer not the decimal approximation before I go to my calculator so I'm going to get rid of my coefficient I'm going to divide both sides by six and that gives me the actual answer of x = e to the 85 power / 6 however at this stage right here this 6X equals e to the 8 fths power if you want you can at that point go to your calculator and you can type in E to the^ of 8 fths now first you need to know where your e to the power is well remember e is the inverse of natural log so I'm going to look at my natural log button and I look above it in blue you see e to the X power so I'm going to hit second natural log that gives me e to the power and I'm going to type in 8 / by five and that gives me 4953 so if you want at this stage 6X = e to the 85 you can and go to your calculator and find out what is e to 8 FS that would give me 6X = 4953 and then divide both sides by six and that would give me my decimal approximation for it and so I'm just going to hit divide it by six and I get 8255 which round to three decimal places is going to be 826 so x equal 826 many students find it easier to find the value of the e to the power before solving like I did in the green here other students some find it easy just to solve for x and then type this in their calculator for it either way is acceptable so let's do a logarithmic inequality involving our natural logs so I have the natural log of the quantity 2x- 3^ squared is greater than six I am going to apply my power rule here I'm going to take this exponent and I'm going to bring it down to make it my coefficient of my natural log so I really have two times the natural log of 2x minus 3 is greater than six and now I'm going to get rid of my coefficient of two by dividing both sides by two that isolates my natural log which is the whole point so natural log of the quantity 2x - 3 is greater than three now that the natural log is isolated I can take e to the power of both sides that allows me to cancel out my natural log and E and that just gives me 2x minus 3 is greater than e cubed to make it easier I am going to go right to my calculator and find the power of E Cub I'm going to find the value of e Cub so I'm going to clear this out second natural log ebed that gives me 20.08 6 if I am rounding to three decimal places so that tells me I'm really working with the inequality 2x minus 3 is greater than 20.0 86 and to solve this I would add three to both sides and get 2x is greater than 23.8 6 X and then I would divide both sides by two of the inequality and so I have 23.8 divided by two and that gives me 11.54 3 so X is greater than 11.5 5 4 3 there is my value for x with this inequality again I'm going to go ahead and do the easy route of find the value of e to the power and then go from there because it's easier to solve that way for many students so hopefully you found this solving logarithmic equations involving natural logs and inequalities easy to understand what you do is you isolate the natural log first then take e to the power both sides since that's the inverse operation that cancels out the natural log and the E and you work with the values from there on basic solving steps |
17444 | https://www.quora.com/How-do-you-integrate-ln-x-using-substitution-and-or-parts | Something went wrong. Wait a moment and try again.
Integration by Substituti...
Logarithmic Functions
Calculus 2
Functions (mathematics)
Calculus (Mathematics)
Mathematical Functions
5
How do you integrate ln(x) using substitution and/or parts?
George Ivey
Former Math Professor at Gallaudet University
·
Author has 23.7K answers and 2.6M answer views
·
2y
To integrate ∫ln(x)dx use integration by parts.
“Integration by parts” says that ∫udv=uv−∫vdu
(It is the integral version of the product rule for differentiation.)
Here, u= ln(x) and dv= dx so du= (1/x)dx and v= x
∫ln(x)dx=xln(x)−∫x(1/x)dx=xln(x)−∫dx=xln(x)−x+C where C is an arbitrary constant.
Related questions
[How do you integrate
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solⁿ I=∫ ln(x) dx ….(1)
put ln(x) =t
or x =e^t
∴ dx =e^t dt
From (1)
I=∫ t e^t dt
Integrate by parts
I=t × e^t - ∫ e^t × 1 dt
I=t e^t - e^t + C
I=(t-1)e^t + C
I=(ln(x) - 1) e^ln(x) + C
I=(ln(x) - 1) x + C
I=x (ln(x) - 1) + C
Calvin L.
2nd year mathematics student
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Author has 10K answers and 2.2M answer views
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2y
We can use integration by parts.
Let u=lnx⟹u′=1x and v′=1⟹v=x.
∫lnx dx=xlnx−∫xx dx
=xlnx−∫1 dx
=xlnx−x+C.
Mohammad Afzaal Butt
B.Sc in Mathematics & Physics, Islamia College Gujranwala (Graduated 1977)
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Author has 24.6K answers and 22.9M answer views
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5y
Related
How do you solve the integral
ln
(
x
)
by parts?
∫ln(x)dx
=xln(x)−∫x(1x)dx
=xln(x)−∫dx
=xln(x)−x+C
=x(ln(x)−1)+C
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Emmanuel Alejandro Muñoz Hernandez
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How do you solve the integral
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Hi, well the trick to solving this specific integral is to take the one that's multiplying ln(x) and use it as your dv, so when you do the anti derivative you'll end up with an x, I'm attaching the solution hope it's clear enough.
Hi, well the trick to solving this specific integral is to take the one that's multiplying ln(x) and use it as your dv, so when you do the anti derivative you'll end up with an x, I'm attaching the solution hope it's clear enough.
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Sohel Zibara
Studied at Doctor of Philosophy Degrees (Graduated 2000)
·
Author has 5.1K answers and 2.6M answer views
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1y
Related
How do I integrate {Ln(1+x) Ln(x)} / (x(x+1)) from 0 to 1?
We want to find the value of the integral :I=∫10ln(x+1)lnxx(x+1)dx
For this purpose :I=∫10(1x–1x+1)ln(x+1)lnxdx=∫10ln(x+1)lnxxdxJ–∫10ln(x+1)lnxx+1dxK
\displaystyle{\textbf{where :}\J\,=\,\sum_{n\,=\,0}^{+\infty}\frac{\left(–1\right)^{n}}{n\,+\,1}\int_0^1x^n\ln\,x\,dx\,=\,\sum_{n\,=\,0}^{+\infty}
We want to find the value of the integral :I=∫10ln(x+1)lnxx(x+1)dx
For this purpose :I=∫10(1x–1x+1)ln(x+1)lnxdx=∫10ln(x+1)lnxxdxJ–∫10ln(x+1)lnxx+1dxK
where :J=+∞∑n=0(–1)nn+1∫10xnlnxdx=+∞∑n=0(–1)n+1(n+1)3=–34ζ(3).
when it comes to K it is bit complicated and we reason as follows :K=∫10ln(x+1)lnxx+1dxu=ln(x+1)=∫ln20uln(eu–1)du==∫ln20u2du+∫ln20uln(1–e–u)duL=13ln32+L
with :L=–+∞∑n=01n+1∫ln20ue–(n+1)udu=+∞∑n=01(n+1)2∫ln20ud(e–(n+1)u)=(IBP)==+∞∑n=01(n+1)2((ue–(n+1)u)ln20–∫ln20e–(n+1)udu)==+∞∑n=01(n+1)2((12)n+1ln2+1n+1(e–(n+1)u)ln20)==+∞∑n=01(n+1)2((12)n+1ln2+(12)n+1n+1–1n+1)=
[math]\displaystyle{\textbf{ Thus :}\L\,=\,\,–\frac{1}{3}\ln^32\,\,–\,\frac{1}{8}\zeta\left(3\right)\\textbf{Consequently:}\K\,=\,\,–\frac{1}{8}\zeta\left(3\right)}\[/math]
[math]\displaystyle{\textbf{Putting it all together yields :}\I\,=\,J\,\,–\,K\,=\,\,–\frac{3}{4}\zeta\left(3\right)\,+\frac{1}{8}\zeta\left(3\right)\,=\,\,–\frac{5}{8}\zeta\left(3\right)}\[/math]
[math]\displaystyle{\textbf{Finally :}\\boxed{\int_0^{1}\frac{\ln\left(x\,+\,1\right)\,\ln\,x}{x\left(x\,+\,1\right)}dx\,=\,\,–\frac{5}{8}\zeta\left(3\right)}}\[/math]
Peter Aouad
Director at Cybernet Associates Inc. (2006–present)
·
Author has 1.8K answers and 1.6M answer views
·
5y
Related
How do I integrate
[math]\displaystyle \int \cos(x) \frac {\ln(\sin(x))} {\ln(\tan(x))}dx[/math]
?
The function:
Repeats itself every 2pi. It can be integrated over a very narrow x interval.
f(x) is undefined at x=0.
The function:
Repeats itself every 2pi. It can be integrated over a very narrow x interval.
f(x) is undefined at x=0.
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Ron Davis
Ph. D in Theoretical Physics & Nuclear Physics, The University of British Columbia (Graduated 1968)
·
Upvoted by
Kobe Ye
, M.S Mathematics, The University of British Columbia (2007) and
Chuck Summers
, MSEE Digital Signal Processing & Mathematics, Georgia Institute of Technology (1975) · Author has 6.7K answers and 18.8M answer views
·
6y
Related
How do you take the integral of ln (ln(x))?
[math]\displaystyle{I=\int\limits_{x=a}^b \ln\left ( \ln\left ( x \right ) \right )\,\text{d}x} .[/math]
We do a “change of variable”; in place of the variable of integration [math]x[/math], we use a different variable, [math]y[/math], such that
[math]y=\ln\left ( x \right ).[/math]
To substitute that into the integral, we see that
[math]x=\exp\left ( y \right ),[/math]
and
[math]\dfrac {\text{d}y} {\text{d}x}=\dfrac 1 x=\dfrac 1 {\exp\left (y \right )},[/math]
so
[math]\text{d}x=\exp\left ( y \right )\,\text{d}y.[/math]
With this change of variable,
[math]I=\displaystyle{\int\limits_{y=\ln\left (a \right )}^{\ln\left (b \right )}}\ln\left ( y \right )\,\exp\left ( y \right )\,\text{d}y.[/math]
Now, we
[math]\displaystyle{I=\int\limits_{x=a}^b \ln\left ( \ln\left ( x \right ) \right )\,\text{d}x} .[/math]
We do a “change of variable”; in place of the variable of integration [math]x[/math], we use a different variable, [math]y[/math], such that
[math]y=\ln\left ( x \right ).[/math]
To substitute that into the integral, we see that
[math]x=\exp\left ( y \right ),[/math]
and
[math]\dfrac {\text{d}y} {\text{d}x}=\dfrac 1 x=\dfrac 1 {\exp\left (y \right )},[/math]
so
[math]\text{d}x=\exp\left ( y \right )\,\text{d}y.[/math]
With this change of variable,
[math]I=\displaystyle{\int\limits_{y=\ln\left (a \right )}^{\ln\left (b \right )}}\ln\left ( y \right )\,\exp\left ( y \right )\,\text{d}y.[/math]
Now, we do integration by parts. We use the fact that
[math]\dfrac {\text{d}} {\text{d}y}\left ( \ln\left ( y \right )\,\exp\left ( y \right ) \right )=\dfrac 1 y \exp\left ( y \right )+ \ln\left ( y \right )\,\exp\left ( y \right )[/math]
to infer that
[math]I=\left [ \ln\left ( y \right )\,\exp\left ( y \right ) \right ]_{y=\ln\left (a \right )}^{\ln\left (b \right )}-\displaystyle{\int\limits_{y=\ln\left (a \right )}^{\ln\left (b \right )}}\dfrac 1 y \exp\left ( y \right )\,\text{d}y.[/math]
There is no simple, closed form for the antiderivative of [math]\dfrac 1 y \exp\left ( y \right ),[/math] but there are tables and function routines for it, under the name “exponential integral”, denoted Ei.
Kwok Choy Yue
B.Sc in Mathematics, The Chinese University of Hong Kong (Graduated 1978)
·
Author has 1.4K answers and 1.4M answer views
·
Feb 13
Related
Can I integrate ln x/x by parts?
() No need to use integration by parts:
Enjoy mathematics!
() No need to use integration by parts:
Enjoy mathematics!
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Rona La Bella
8y
Related
How do I integrate ln (x + 10) dx by substitution and integrating by parts?
Logarithmic linear functions are usually integrated by parts. In this method we need a product of two functions, so let's multiply ln(x+10) by 1. Since we know how to differentiate ln(x+10), consider this function to be the v function. Therefore dv/dx is 1/(x +10) Take 1 to be equal to du/dx, so u is x.
Now, using the by parts formula:
The integral of ln(x+10) is
xln(x+10) - the integral of x/(x+10)
= x ln(x+10) - the integral of [1 - 10/(x+10)]
(By dividing x by x+10)
= x ln(x+10) - [x - 10 ln|x+10|] + c
= x ln(x+10) - x + 10 ln|x+10| + c.
Hope this helps :-)
Logarithmic linear functions are usually integrated by parts. In this method we need a product of two functions, so let's multiply ln(x+10) by 1. Since we know how to differentiate ln(x+10), consider this function to be the v function. Therefore dv/dx is 1/(x +10) Take 1 to be equal to du/dx, so u is x.
Now, using the by parts formula:
The integral of ln(x+10) is
xln(x+10) - the integral of x/(x+10)
= x ln(x+10) - the integral of [1 - 10/(x+10)]
(By dividing x by x+10)
= x ln(x+10) - [x - 10 ln|x+10|] + c
= x ln(x+10) - x + 10 ln|x+10| + c.
Hope this helps :-)
Tirtharaj Banerjee
Former Postgraduate Researcher at Indian Institute of Engineering Science and Technology (IIEST) (2021–2023)
·
Author has 155 answers and 155K answer views
·
5y
Related
How do I find the integral of ln (1+x^2) dx when x approaches 1 to 2?
Abdelhadi Nakhal
Hydrogeologist engineer
·
Author has 1.3K answers and 435.7K answer views
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5y
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How do you solve the integral
[math]\ln(x)[/math]
by parts?
[math]y=\log x\Rightarrow y’=\frac{1}{x}\Rightarrow 1=xy’[/math]
[math]\Rightarrow 1=(xy)’-y\Rightarrow y=(xy)’-1[/math]
Integrating yields:
[math]\int{\log x dx}=x\log x -x +c[/math]
Brian Sittinger
PhD in Mathematics, University of California, Santa Barbara (Graduated 2006)
·
Upvoted by
James McElhatton Ph.D. (Glasgow, 1976)
, B.Sc. Mathematics & Chemistry, University of Malta (1967) · Author has 8.5K answers and 21M answer views
·
1y
Related
How do I integrate {Ln(1+x) Ln(x)} / (x(x+1)) from 0 to 1?
We want to evaluate the improper integral
[math]I = \displaystyle \int_0^1 \frac{\ln(x+1) \ln{x}}{x(x + 1)} \, dx = -\frac{5}{8} \zeta(3). \tag{}[/math]
To this end, we start with partial fractions for the non-logarithmic terms and then split the integral into two (convergent) terms:
[math]I = \displaystyle \int_0^1 \frac{\ln(x+1) \ln{x}}{x} \, dx - \int_0^1 \frac{\ln(x+1) \ln{x}}{x + 1} \, dx. \tag{}[/math]
We address each of these integrals one at a time.
For the first integral, we use the Maclaurin series for [math]\ln(1+x)[/math] and then interchange the order of integration and summation. This results in an infinite series that c
We want to evaluate the improper integral
[math]I = \displaystyle \int_0^1 \frac{\ln(x+1) \ln{x}}{x(x + 1)} \, dx = -\frac{5}{8} \zeta(3). \tag{}[/math]
To this end, we start with partial fractions for the non-logarithmic terms and then split the integral into two (convergent) terms:
[math]I = \displaystyle \int_0^1 \frac{\ln(x+1) \ln{x}}{x} \, dx - \int_0^1 \frac{\ln(x+1) \ln{x}}{x + 1} \, dx. \tag{}[/math]
We address each of these integrals one at a time.
For the first integral, we use the Maclaurin series for [math]\ln(1+x)[/math] and then interchange the order of integration and summation. This results in an infinite series that can be written in terms of a Riemann zeta value:
[math]\begin{align} \displaystyle \int_0^1 \frac{\ln(x+1) \ln{x}}{x} \, dx &= \int_0^1 \frac{\ln{x}}{x} \cdot \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n \, dx\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_0^1 x^{n-1} \ln{x} \, dx\ &= -\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^3}\ &= -\Big(\sum_{n=1}^{\infty} \frac{1}{n^3} - \sum_{n=1}^{\infty} \frac{2}{(2n)^3}\Big)\ &= -\Big(1 - \frac{2}{2^3}\Big) \sum_{n=1}^{\infty} \frac{1}{n^3}\ &= -\frac{3}{4} \zeta(3). \end{align} \tag{}[/math]
The second integral
[math]J = \displaystyle \int_0^1 \frac{\ln(x+1) \ln{x}}{x + 1} \, dx \tag{}[/math]
is going take a little more work to evaluate.
We start with a similar approach as the first integral. First of all, multiplying the geometric series and Maclaurin series together yields
[math]\displaystyle -\frac{\ln(1 - x)}{1 - x} = \sum_{n=1}^{\infty} H_n x^n, \tag{}[/math]
where [math]H_n = \sum_{k=1}^n \frac{1}{k}[/math] is the [math]n[/math]th partial harmonic series.
Then, replacing [math]x[/math] with [math]-x[/math] gives us
[math]\displaystyle -\frac{\ln(1 + x)}{1 + x} = \sum_{n=1}^{\infty} (-1)^n H_n x^n. \tag{}[/math]
Now, we substitute this series into [math]J[/math] and then interchange the order of integration and summation.
[math]\begin{align} J &= \displaystyle \int_0^1 -\ln{x} \cdot \sum_{n=1}^{\infty} (-1)^n H_n x^n \, dx\ &= \sum_{n=1}^{\infty} (-1)^n H_n \cdot \int_0^1 -x^n \ln{x} \, dx\ &= \sum_{n=1}^{\infty} \frac{(-1)^n H_n}{(n+1)^2}\ &= \sum_{k=2}^{\infty} \frac{(-1)^{k-1} H_{k-1}}{k^2}, \text{ via } k = n+1\ &= -\sum_{k=2}^{\infty} \frac{(-1)^k (H_k - \frac{1}{k})}{k^2}\ &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k^2} - \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^3}. \end{align} \tag{}[/math]
As we saw above,
[math]\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^3} = \frac{3}{4} \zeta(3). \tag{}[/math]
We claim that
[math]\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k^2} = \frac{5}{8} \zeta(3). \tag{}[/math]
A few proofs of this (including one independent of integrals) can be found here.
With this claim, we find that
[math]J = \displaystyle \frac{5}{8} \zeta(3) - \frac{3}{4} \zeta(3) = -\frac{1}{8} \zeta(3), \tag{}[/math]
Finally, putting the work from the two integrals together, we conclude that
[math]I = \displaystyle \int_0^1 \frac{\ln(x+1) \ln{x}}{x(x + 1)} \, dx = -\frac{3}{4} \zeta(3) - \Big(-\frac{1}{8} \zeta(3)\Big) = \boxed{-\frac{5}{8} \zeta(3)}. \tag{}[/math]
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17445 | https://jamanetwork.com/journals/jamaneurology/fullarticle/583110 | Clinical Characteristics and Prognosis of Vasculitic Mononeuropathy Multiplex | JAMA Neurology | JAMA Network
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Clinical Characteristics and Prognosis of Vasculitic Mononeuropathy Multiplex
Rowland W. Chang, MD; Carolyn L. Bell, MD; Mark Hallett, MD
Author Affiliations
From the Divisions of Rheumatology and Immunology (Drs Chang and Bell) and Neurology (Dr Hallett), Department of Medicine, Brigham and Women's Hospital; the Robert Breck Brigham Multipurpose Arthritis Center (Dr Chang); and the Departments of Medicine (Drs Chang and Bell) and Neurology (Dr Hallett), Harvard Medical School, Boston.
Cite This### Citation
Chang RW, Bell CL, Hallett M. Clinical Characteristics and Prognosis of Vasculitic Mononeuropathy Multiplex. Arch Neurol. 1984;41(6):618–621. doi:10.1001/archneur.1984.04210080026009
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1984;41;(6):618-621. doi:10.1001/archneur.1984.04210080026009
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Abstract
• The initial appearance and course of 19 patients with mononeuropathy multiplex due to systemic vasculitis were evaluated. Neuropathy first occurred within one year of the onset of systemic vasculitis in all patients and within one year of the onset of rheumatoid arthritis in four of nine patients. Concurrent cutaneous arteritic lesions and elevated ESRs occurred more frequently in patients with rheumatoid arthritis than in those without it. The overall six-month and five-year survival rates were approximately 80% and 60%, respectively. Significant nerve improvement was noted in 86% of surviving patients after one year. There were no factors at initial appearance predictive of outcome.
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21.20
Indicators
Written by:Ck12 Science
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Last Modified: Aug 01, 2025
Lesson
Who would write a song about a cabbage?
There’s an old bluegrass song known as “Boil Them Cabbage Down.” Many people enjoy the music, but chemistry students also enjoy the product of the boiled cabbage. Extracting the anthocyanin dye from cabbage leaves with boiling water gives a solution that is red when acidic, purple when neutral, and green to yellow when basic (pictured above).
Indicators
An acid-base indicator is a substance that displays different colors when in the presence of an acid or a base. How does that work? An indicator is a weak acid that ionizes within a known pH range, usually about 2 pH units. We can represent the protonated form of the indicator molecule as HIn and the deprotonated form as In−. The following equilibrium exists for the indicator.
@$$\begin{align}\text{HIn}(aq) \rightleftarrows \text{H}^+(aq)+\text{In}^-(aq)\end{align}@$$
According to LeChtelier’s principle, the addition of H+ ions (as in a low pH solution) drives the equilibrium to the left and the protonated HIn predominates. The addition of OH− (as in a high pH solution) decreases the H+ concentration and drives the equilibrium to the right and the deprotonated In− predominates. To be useful as an indicator, the two forms must be different colors. In the case of phenolphthalein, the protonated form is colorless, while the deprotonated form is pink. Figure below shows a variety of acid-base indicators that can be used in titration experiments.
pH indicators.
Depending on the pH at the equivalence point, the appropriate indicator must be chosen. For example, bromphenol blue has a yellow color below a pH of about 3 and a blue-violet color above a pH of about 4. Bromphenol blue would not be a good choice as the indicator for a strong acid-strong base titration, because the pH is 7 at the equivalence point. Instead, it could be used for a strong acid-weak base titration, where the pH at the equivalence point is lower.
Most indicators have two colored forms. Universal indicator displays the entire rainbow of colors from low pH to high pH (see Figure below). Universal indicator is used to make pH paper, which can be used to quickly test solutions for their approximate pH.
Universal indicator tape.
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17448 | https://sites.math.washington.edu/~burke/crs/516-21/L3-cvx-analysis.pdf | Convex Analysis Functions Taking Infinite Values We consider functions f mapping E to the extended-real-line R = R ∪{±∞}.
Care must be taken when working with ±∞. In particular, we set 0 · ±∞= 0 and will be careful to avoid the expressions (+∞) + (−∞) throughout.
Since the primary focus of our discussion is convex functions, there is a bias between +∞and −∞.
Given f : E →R, the effective domain and epigraph of f are dom f := {x ∈E : f(x) < +∞}, epi f := {(x, r) ∈E × R : f(x) ≤r}, respectively.
A function f : E →R is called proper if it never takes the value −∞and dom f ̸= ∅.
Epigraphs epi f dom f Figure: Epigraph and effective domain of the function whose value is max{−x, 1 2x2} for x ∈[−1, 1] and +∞elsewhere.
Epigraphs epi f dom f Figure: Epigraph and effective domain of the function whose value is max{−x, 1 2x2} for x ∈[−1, 1] and +∞elsewhere.
Lemma: A function f : E →R is closed (lsc) if and only if epi f is a closed set.
Convex Functions We say that the function f : E →R is convex if epi f is a convex set.
Convex Functions We say that the function f : E →R is convex if epi f is a convex set.
Lemma: f : E →R is convex if and only if f(λx+(1−λ)y) ≤λf(x)+(1−λ)f(y) ∀x, y ∈E and λ ∈(0, 1).
epi f x y λx + (1 −λ)y λf(x) + (1 −λ)f(y) f(x) f(y) Convex Functions We say that the function f : E →R is convex if epi f is a convex set.
Lemma: f : E →R is convex if and only if f(λx+(1−λ)y) ≤λf(x)+(1−λ)f(y) ∀x, y ∈E and λ ∈(0, 1).
epi f x y λx + (1 −λ)y λf(x) + (1 −λ)f(y) f(x) f(y) Lemma: If f : E →R is convex, then, for all r ∈R the set {x | f(x) ≤r} is convex.
3 Special Functions for Q ⊂E The indicator function for Q: δQ(x) := ( 0 , x ∈Q, +∞ , x ̸∈Q.
The support function for Q: δ∗ Q(x) := sup v∈Q ⟨v, x⟩.
The gauge function for Q: γQ(x) := inf {λ ∈R+ | x ∈λQ} .
3 Special Functions for Q ⊂E The indicator function for Q: δQ(x) := ( 0 , x ∈Q, +∞ , x ̸∈Q.
The support function for Q: δ∗ Q(x) := sup v∈Q ⟨v, x⟩.
The gauge function for Q: γQ(x) := inf {λ ∈R+ | x ∈λQ} .
(1) If B ⊂E is the closed unit ball for the norm ∥·∥, then ∥·∥= δ∗ B◦= γB.
3 Special Functions for Q ⊂E The indicator function for Q: δQ(x) := ( 0 , x ∈Q, +∞ , x ̸∈Q.
The support function for Q: δ∗ Q(x) := sup v∈Q ⟨v, x⟩.
The gauge function for Q: γQ(x) := inf {λ ∈R+ | x ∈λQ} .
(1) If B ⊂E is the closed unit ball for the norm ∥·∥, then ∥·∥= δ∗ B◦= γB.
(2) If K ⊂E is a closed convex cone, then δ∗ K◦= δK = γK.
Epigraphical Perspective In our study of functions f : E →R we take an epigraphical perspective, that is, we study properties of a function by studying properties of its epigraph.
Epigraphical Perspective In our study of functions f : E →R we take an epigraphical perspective, that is, we study properties of a function by studying properties of its epigraph.
For example, a function is closed (lsc) if its epigraph is a closed set. Similarly, a function is convex if its epigraph is a convex set.
The primary advantages of this perspective is that it allows us to discover properties of functions through properties of sets.
Epigraphical Perspective In our study of functions f : E →R we take an epigraphical perspective, that is, we study properties of a function by studying properties of its epigraph.
For example, a function is closed (lsc) if its epigraph is a closed set. Similarly, a function is convex if its epigraph is a convex set.
The primary advantages of this perspective is that it allows us to discover properties of functions through properties of sets.
A key observation in this regard is the fact that for every x ∈dom f, f(x) = inf (x,µ)∈epi f µ .
Epigraphs that are Cones What are the functions whose epigraphs are cones?
Epigraphs that are Cones What are the functions whose epigraphs are cones?
For λ > 0, λepi f = epi f, i.e., if (x, µ) ∈epi f so is (λx, λµ) for all λ ≥0. Hence, we can relate the values of f(λx) to those of f(x) as follows: for λ > 0, f(λx) = inf (λx,λµ)∈epi f λµ = λ inf (x,µ)∈λ−1epi f µ = λ inf (x,µ)∈epi f µ = λf(x) .
From this, it is easy to show that epi f is a cone if and only if f(λx) = λf(x) for all x ∈dom f and λ ≥0.
Such functions are called positively homogeneous.
Epigraphs that are Convex Cones If epi f is a convex cone, what can be said about f?
We have already shown that f must be positively homogeneous. But convexity tells us that epi f = epi f + epi f, i.e., for every pair (x, µ), (y, τ) ∈epi f we have (x, µ) + (y, τ) = (x + y, µ + τ) ∈epi f.
Consequently, {µ + τ | (x, µ), (y, τ) ∈epi f } ⊂{ω | (x + y, ω) ∈epi f }, and so, for all x, y ∈dom f, f(x + y) = inf (x+y,ω)∈epi f ω ≤ inf (x,µ),(y,τ)∈epi f µ + τ = inf (x,µ)∈epi f µ + inf (y,τ)∈epi f τ = f(x) + f(y).
Since this inequality trivially holds if either x or y is not in dom f, f(x + y) ≤f(x) + f(y) ∀x, y ∈E.
Such functions are called subadditive. Hence functions whose epigraphs are convex cones are both positively homogeneous and subadditive. Such functions are called sublinear.
Exercise 1) Show that a the epigraph of a positively homogeneous function is a cone.
2) Show that the epigraph of a sublinear function is a convex cone.
Support Functions are Sublinear Let S ⊂E be nonempty and consider the support function δ∗ S(x) = supv∈S ⟨v, x⟩.
positive homogeneity: λ ≥0, δ∗ S(λx) = sup {⟨λx, v⟩| v ∈S } = λ sup {⟨x, v⟩| v ∈S } = λδ⋆(x | S ) ∀λ ≥0.
subadditivity: x1, x2 ∈E, δ∗ S(x1 + x2) = sup x1 + x2, v | v ∈S = sup x1, v1 + x2, v2 v1 = v2 ∈S ≤sup x1, v1 + x2, v2 v1, v2 ∈S ≤sup x1, v1 v1 ∈S + sup x2, v2 v2 ∈S = δ⋆ x1 S + δ⋆ x2 S .
Support Functions are Sublinear Let S ⊂E be nonempty and consider the support function δ∗ S(x) = supv∈S ⟨v, x⟩.
positive homogeneity: λ ≥0, δ∗ S(λx) = sup {⟨λx, v⟩| v ∈S } = λ sup {⟨x, v⟩| v ∈S } = λδ⋆(x | S ) ∀λ ≥0.
subadditivity: x1, x2 ∈E, δ∗ S(x1 + x2) = sup x1 + x2, v | v ∈S = sup x1, v1 + x2, v2 v1 = v2 ∈S ≤sup x1, v1 + x2, v2 v1, v2 ∈S ≤sup x1, v1 v1 ∈S + sup x2, v2 v2 ∈S = δ⋆ x1 S + δ⋆ x2 S .
Are sublinear functions support functions?
Convexity and Optimization Strict Convexity: A convex function f : E →R is said to be strictly convex if f((1 −λ)x + λy) < (1 −λ)f(x) + λf(y) ∀x, y ∈dom f, λ ∈(0, 1) with x ̸= y.
Theorem: Let f : E →R be convex. If ¯ x ∈dom f is a local solution to the problem min f(x), then ¯ x is a global optimal solution. Moreover, if f is strictly convex, then the global optimal solution is unique.
Convexity and Optimization Proof: If f(¯ x) = −∞we are done, so assume that −∞< f(¯ x).
Suppose there is a b x ∈Rn with f(b x) < f(x). Let ϵ > 0 be such that f(x) ≤f(x) whenever ∥x −x∥≤ϵ.
Set λ := ϵ(2∥x −b x∥)−1 and xλ := x + λ(b x −x). Then ∥xλ −x∥≤ϵ/2 and f(xλ) ≤(1 −λ)f(x) + λf(b x) < f(x).
This contradiction implies no such b x exists.
Convexity and Optimization Proof: If f(¯ x) = −∞we are done, so assume that −∞< f(¯ x).
Suppose there is a b x ∈Rn with f(b x) < f(x). Let ϵ > 0 be such that f(x) ≤f(x) whenever ∥x −x∥≤ϵ.
Set λ := ϵ(2∥x −b x∥)−1 and xλ := x + λ(b x −x). Then ∥xλ −x∥≤ϵ/2 and f(xλ) ≤(1 −λ)f(x) + λf(b x) < f(x).
This contradiction implies no such b x exists.
To see the second statement in the theorem, let x1 and x2 be distinct global minimizers of f. Then, for λ ∈(0, 1), f((1 −λ)x1 + λx2) < (1 −λ)f(x1) + λf(x2) = f(x1) , which contradicts the assumption that x1 is a global minimizer.
The Directional Derivative Theorem: Let f : E →R be convex and let x ∈dom f.
(1) Given d ∈E the difference quotient f(x+td)−f(x) t is a non-decreasing function of t on (0, +∞).
The Directional Derivative Theorem: Let f : E →R be convex and let x ∈dom f.
(1) Given d ∈E the difference quotient f(x+td)−f(x) t is a non-decreasing function of t on (0, +∞).
(2) For all d ∈E, f′(x; d) exists with f′(x; d) := inft>0 f(x+td)−f(x) t .
The Directional Derivative Theorem: Let f : E →R be convex and let x ∈dom f.
(1) Given d ∈E the difference quotient f(x+td)−f(x) t is a non-decreasing function of t on (0, +∞).
(2) For all d ∈E, f′(x; d) exists with f′(x; d) := inft>0 f(x+td)−f(x) t .
(3) The “subdifferential inequality” holds for all x ∈dom f: f(x) + f′(x : y −x) ≤f(y) ∀y ∈E.
The Directional Derivative Theorem: Let f : E →R be convex and let x ∈dom f.
(1) Given d ∈E the difference quotient f(x+td)−f(x) t is a non-decreasing function of t on (0, +∞).
(2) For all d ∈E, f′(x; d) exists with f′(x; d) := inft>0 f(x+td)−f(x) t .
(3) The “subdifferential inequality” holds for all x ∈dom f: f(x) + f′(x : y −x) ≤f(y) ∀y ∈E.
(4) The function f′(x; ·) is sublinear. In particular, f′(x; ·) is a convex function for all x ∈dom f.
t 7→(f(x + td) −f(x))/t nondecreasing for t > 0 Let x ∈dom f and d ∈E. If x + td / ∈dom f for all t > 0, the result follows. So assume that 0 < ¯ t = sup{t : x + td ∈dom f}.
Let 0 < t1 < t2 < ¯ t. Then f(x + t1d) = f x + t1 t2 t2d = f h 1 − t1 t2 x + t1 t2 (x + t2d) i ≤ 1 −t1 t2 f(x) + t1 t2 f(x + t2d) = f(x) + t1 f(x+t2d)−f(x) t2 .
Hence f(x + t1d) −f(x) t1 ≤f(x + t2d) −f(x) t2 .
f ′(x; d) = inft>0(f(x + td) −f(x))/t (2) If x + td / ∈dom f for all t > 0, then the result is obviously true.
So assume there is a ¯ t > 0 such that x + td ∈dom f for all t ∈(0, ¯ t]. Since f′(x; d) := lim t↓0 f(x + td) −f(x) t .
and the difference quotient in the limit is non-decreasing in t on (0, +∞), the limit is necessarily given by the infimum of the difference quotient. This infimum always exists and so f′(x; d) always exists and is given by the infimum.
(3) The subdifferential inequality follows from (2) by taking d := y −x and t = 1 in the infimum: f′(x; y −x) ≤f(y) −f(x).
f ′(x; ·) is sublinear Positive homogeneity: f′(x; αd) = α limt↓0 f(x+(tα)d)−f(x) (tα) = αf′(x; d).
Subadditivity: f′(x; u + v) = lim t↓0 f(x + t(u + v)) −f(x) t = lim t↓0 f(x + t 2(u + v)) −f(x) t/2 = lim t↓0 2f( 1 2(x + tu) + 1 2(x + tv)) −f(x) t ≤ lim t↓0 2 1 2f(x + tu) + 1 2f(x + tv) −f(x) t = lim t↓0 f(x + tu) −f(x) t + f(x + tv) −f(x) t = f′(x; u) + f(x; v) .
Convexity and Optimality Theorem: Let f : E →R ∪{+∞} be convex, Ω⊂E convex, ¯ x ∈dom f ∩Ω. Then ¯ x solves minx∈Ωf(x) if and only if f′(¯ x; y −¯ x) ≥0 for all y ∈Ω.
Proof: (⇒) Let y ∈Ωso that ¯ x + t(y −¯ x) ∈Ωfor all t ∈[0, 1].
Then f(¯ x) ≤f(¯ x + t(y −¯ x)) for all t ∈[0, 1]. Therefore, f′(¯ x; y −¯ x) = limt↓0 t−1(f(¯ x + t(y −¯ x)) −f(¯ x)) ≥0.
(⇐) For y ∈Ω, 0 ≤f′(¯ x; y −¯ x) = inft>0 f(x+t(y−¯ x)−f(x) t (t=1) ≤ f(y) −f(¯ x).
Convexity and Optimality Theorem: Let f : E →R ∪{+∞} be convex, Ω⊂E convex, ¯ x ∈dom f ∩Ω. Then ¯ x solves minx∈Ωf(x) if and only if f′(¯ x; y −¯ x) ≥0 for all y ∈Ω.
Proof: (⇒) Let y ∈Ωso that ¯ x + t(y −¯ x) ∈Ωfor all t ∈[0, 1].
Then f(¯ x) ≤f(¯ x + t(y −¯ x)) for all t ∈[0, 1]. Therefore, f′(¯ x; y −¯ x) = limt↓0 t−1(f(¯ x + t(y −¯ x)) −f(¯ x)) ≥0.
(⇐) For y ∈Ω, 0 ≤f′(¯ x; y −¯ x) = inft>0 f(x+t(y−¯ x)−f(x) t (t=1) ≤ f(y) −f(¯ x).
Corollary: If f is differentiable at ¯ x, ¯ x solves minx∈Ωf(x) if and only if −∇f(¯ x) ∈NΩ(¯ x).
Proof: 0 ≤f′(¯ x; y −¯ x) = ⟨∇f(¯ x), y −¯ x⟩for all y ∈Ωiff −∇f(¯ x) ∈NΩ(¯ x).
Differential Tests for Convexity The following are equivalent for a C1-smooth function f : U →R defined on a convex open set U ⊂E.
(a) (convexity) f is convex.
(b) (gradient inequality) f(y) ≥f(x) + ⟨∇f(x), y −x⟩for all x, y ∈U.
(c) (monotonicity) ⟨∇f(y) −∇f(x), y −x⟩≥0 for all x, y ∈U.
If f is C2-smooth, then the following property can be added to the list: (d) The relation ∇2f(x) ⪰0 holds for all x ∈U.
Examples of Convex Functions (1) Given a self-adjoint linear operator A: E →E, a point c ∈E, and b ∈R the quadratic function f(x) = 1 2⟨Ax, x⟩+ ⟨c, x⟩+ b is convex if and only if A is positive semidefinite.
(2) (Boltzmann-Shannon entropy) f(x) = x log x if x > 0 0 if x = 0 +∞ if x < 0 (3) (Fermi-Dirac entropy) f(x) = x log(x) + (1 −x) log(1 −x) if x ∈(0, 1) 0 if x ∈{−1, 1} +∞ otherwise Examples of Convex Functions (4) (Hellinger) f(x) = ( − √ 1 −x2 if x ∈[−1, 1] +∞ otherwise (5) (Exponential) f(x) = ex (6) (Log-exp) f(x) = log(1 + ex) Bounds for β-Smooth Convex Functions Let f : E →R. TFAE (the following are equivalent) (1) f is β-smooth.
(2) 0 ≤f(y) −f(x) −⟨∇f(x), y −x⟩≤β 2 ∥x −y∥2 (3) f(x) + ⟨∇f(x), y −x⟩+ 1 2β∥∇f(x) −∇f(y)∥2 ≤f(y) (4) 1 β∥∇f(x) −∇f(y)∥2 ≤⟨∇f(x) −∇f(y), x −y⟩ (5) 0 ≤⟨∇f(x) −∇f(y), x −y⟩≤β∥x −y∥2 Epigraphical Operations Recall that for a convex function f and x ∈dom f, f(x) = inf(x,µ)∈epi f µ .
This construction fact can be extended to by defining the lower envelope for any subset Q of E × R: EQ(x) := inf (x,r)∈Q r .
Q epi EQ Figure: Lower envelope of Q.
Hence epi EQ = Q + ({0} × R+) when the infimum is attained when finite.
Example: λepi f, λ > 0 Epi-multiplication inf (x,r)∈λepi f r = inf r (λ−1x, λ−1r) ∈epi f = λ inf λ−1r (λ−1x, λ−1r) ∈epi f = λ inf τ (λ−1x, τ) ∈epi f = λf(x/λ).
Example: epi f1 + epi f2 Epi-addition or infimal convolution inf (x,r)∈epi f1 + epi f2 r = inf {r | (x, r) = (x1, r1) + (x2, r2), (xi, ri) ∈epi fi } = inf {r1 + r2 | (y, r1) ∈epi f1, (x −y, r2) ∈epi f2 } = inf y inf r1,r2 {r1 + r2 | (y, r1) ∈epi f1, (x −y, r2) ∈epi f2 } = inf y f1(y) + f2(x −y) =: (f1□f2)(x) .
Inverse Linear Image Let A ∈L[Y, E]. Recall EQ(x) := inf(x,r)∈Q r .
What is EQ when Q = [A × I]epi f?
EQ(x) = inf {r | x = Ay, (y, r) ∈epi f } = inf x=Ay inf (y,r)∈epi f r = inf x=Ay f(y) .
Infimal Projection Let g : E × Y →R and consider the projection P ∈L[E × Y × R] given by P(x, y) = x.
What is E[P×I]epi g?
E[P×I]epi g(x) = inf {µ | x = P(z, y), (z, y, µ) ∈epi g} = inf x=P(z,y) g(z, y) = inf y g(x, y) .
The Perspective mapping Let Q := R+({1} × epi f). What is EQ(λ, x) for λ ≥0?
It is straightforward to show that EQ(λ, x) = +∞if λ < 0 and that EQ(0, x) = 0. So we suppose 0 < λ.
EQ(λ, x) = inf {r | (λ, x, r) ∈R+({1} × epi f)} = inf {r | ∃τ ≥0 s.t. (λ, x, r) ∈τ({1} × epi f)} = inf {r | (x, r) ∈λepi f } = inf r (λ−1x, λ−1r) ∈epi f = λ inf λ−1r (λ−1x, λ−1r) ∈epi f = λf(λ−1x) .
Relative interiors of sets in a product space pic Theorem: Let Q ⊂X × Y. For each x ∈X set Qx := {y ∈Y | (x, y) ∈Q} and D := {x ∈X | Qx ̸= ∅}.
Then (x, y) ∈ri Q ⇐ ⇒ x ∈ri D and y ∈ri Qx.
Proof: Let P(x, y) = x be the projection of X × Y onto X, and set Ax := {x} × Y. Then PQ = D, so ri D = ri PQ = Pri Q.
Hence, (x, y) ∈ri Q iffx ∈ri D and (x, y) ∈Ax ∩ri Q = ri (Ax ∩Q) = ri ({x} × Qx) = {x} × ri Qx .
So, (x, y) ∈ri Q if and only if x ∈ri D and y ∈ri Qx.
ri epi f Lemma: Let f : E →R be convex. Then ri epi f = {(x, µ) | x ∈ri dom f and f(x) < µ}.
Proof: Apply the previous result to epi f ⊂E × R.
Then D = dom f and (epi f)x = {µ ∈R | f(x) ≤µ}.
Clearly, ri (epi f)x = {µ ∈R | f(x) < µ}, which gives the result.
Local Boundedness of Cvx Func.s on ri dom Theorem: Let f : E →R be convex. Then, ∀¯ x ∈ri dom f, there is a cvx nbhd U of ¯ x and an M > 0 s.t. U ∩affdom f ⊂ri dom f and f(x) ≤M ∀x ∈U ∩affdom f.
Proof: Let ¯ x ∈ri dom f and let u1, . . . , un be an orthonormal basis for E with u1, . . . , uk an orthonormal basis for par dom f. Then B1 := intr conv{±ui | i = 1, . . . , n} is a sym. open nghd of the origin.
Let ϵ > 0 be s.t.
¯ x + ϵB1 ∩par dom f = (¯ x + ϵB1) ∩affdom f ⊂ri dom f.
Set U := ¯ x + ϵB1. Then, for every x ∈¯ x + ϵB1 ∩par dom f, ∃λi, µi ≥0, i = 1, . . . , n with Pk j=1(λi + µi) = 1 such that x = ¯ x + ϵ[Pk j=1 λiui + µi(−ui)] = Pk j=1 λi(¯ x + ϵui) + µi(¯ x −ϵui).
Therefore, f(x) ≤ k X j=1 λif(¯ x + ϵui) + k X j=1 µif(¯ x −ϵui) ≤max {f(¯ x ± ϵui) | i = 1, . . . , k } =: M.
Local Lip. Cont. of Cvx Func.s on ri dom Theorem: Let f : E →R be convex. Then for every ¯ x ∈ri dom f there is an ϵ > 0 s.t. f is Lip. cont. on Bϵ(¯ x) ∩affdom f.
Proof: Set D := par dom f. Let ϵ > 0 and M > 0 be such that B2ϵ(¯ x) ∩affdom f ⊂ri dom f with f(x) ≤M ∀x ∈B2ϵ(¯ x) ∩affdom f.
Set h(x) := (2M)−1[f(x + ¯ x) −f(¯ x)]. If h is Lip. cont. on D near 0, then f is Lip. cont. on affdom f near ¯ x. Observe that h(0) = 0 and h(x) ≤1 for all x ∈B2ϵ(0) ∩D. Moreover, for every x ∈B2ϵ(0) ∩D, 0 = h(0) = h( 1 2x −1 2x) ≤1 2h(x) + 1 2h(−x) so that −1 ≤−h(x) ≤h(−x). That is, −1 ≤h(x) ≤1 for all x ∈B2ϵ ∩D.
For x, y ∈Bϵ(0) ∩D with x ̸= y set α := ∥x −y∥and β := ϵ/α.
Define w := y + β(y −x) ∈B2ϵ ∩D. Then y = (1 + β)−1[w + βx] = 1 1+β w + β 1+β x.
The convexity of h implies that h(y) −h(x) ≤ 1 1 + β h(w) + β 1 + β h(x) −h(x) = 1 1 + β [h(w) −h(x)] ≤ 2 1 + β = 2 α + ϵ ∥x −y∥≤2ϵ−1 ∥x −y∥.
Symmetric in x and y implies the local Lip. cont. of h.
Supporting hyperplanes to epigraphs We apply the following separation theorem to epi f.
Theorem: Let Q ⊂E be convex with ¯ x ∈rb Q. Then there exists z ∈E such that ⟨z, x⟩≤⟨z, ¯ x⟩∀x ∈cl Q and ⟨z, x⟩< ⟨z, ¯ x⟩∀x ∈ri Q .
Supporting hyperplanes to epigraphs We apply the following separation theorem to epi f.
Theorem: Let Q ⊂E be convex with ¯ x ∈rb Q. Then there exists z ∈E such that ⟨z, x⟩≤⟨z, ¯ x⟩∀x ∈cl Q and ⟨z, x⟩< ⟨z, ¯ x⟩∀x ∈ri Q .
Theorem: Let f : E →R be proper convex and let ¯ x ∈ri dom f. Then there is a v ∈E such that sup x [⟨v, x⟩−f(x)] ≤⟨v, ¯ x⟩−f(¯ x).
Supporting hyperplanes to epigraphs Proof: Since ¯ x ∈ri dom f, f is cont. at ¯ x relative to dom f and so cl f(¯ x) = f(¯ x). In particular, (¯ x, f(¯ x)) ∈rb epi f. Hence, there exists (w, τ) ∈E × R s.t.
⟨(w, τ), (x, µ)⟩≤⟨(w, τ), (¯ x, f(¯ x))⟩∀(x, µ) ∈cl epi f and ⟨(w, τ), (x, µ)⟩< ⟨(w, τ), (¯ x, f(¯ x))⟩∀(x, µ) ∈ri epi f .
Hence, ⟨w, x −¯ x⟩+ τ(µ −f(¯ x)) < 0 ∀x ∈ri dom f, µ > f(x).
Taking x = ¯ x, we see that τ < 0. Dividing by |τ| and setting v = w/|τ| and µ = f(x), we obtain ⟨v, x⟩−f(x) ≤⟨v, ¯ x⟩−f(¯ x) ∀x ∈dom f.
The result follows since if x ̸∈dom f then the above inequality is trivially true.
The Subgradient Inequality Theorem: Let f : E →R be proper convex and let ¯ x ∈ri dom f. Then there is a v ∈E such that f(¯ x) + ⟨v, x −¯ x⟩≤f(x) ∀x ∈E.
The Subgradient Inequality Theorem: Let f : E →R be proper convex and let ¯ x ∈ri dom f. Then there is a v ∈E such that f(¯ x) + ⟨v, x −¯ x⟩≤f(x) ∀x ∈E.
Proof: The Theorem tells us that there exist v ∈E such that ⟨v, x⟩−f(x) ≤⟨v, ¯ x⟩−f(¯ x) ∀x ∈E, which gives the result.
The Subdifferential Definition: Let f : E →R be convex and let ¯ x ∈dom f. We say that f is subdifferentiable at ¯ x if there exists v ∈E such that f(¯ x) + ⟨v, x −¯ x⟩≤f(x) ∀x ∈E.
We call v a subgradient for f at ¯ x. The set of all subgradients at ¯ x is called the subdifferential of f at ¯ x, denoted ∂f(¯ x) := {v | f(¯ x) + ⟨v, x −¯ x⟩≤f(x) ∀x ∈E}.
For x ̸∈dom f, we define ∂f(x) = ∅. The domain of ∂f is dom ∂f := {x | ∂f(x) ̸= ∅}.
The Subdifferential Definition: Let f : E →R be convex and let ¯ x ∈dom f. We say that f is subdifferentiable at ¯ x if there exists v ∈E such that f(¯ x) + ⟨v, x −¯ x⟩≤f(x) ∀x ∈E.
We call v a subgradient for f at ¯ x. The set of all subgradients at ¯ x is called the subdifferential of f at ¯ x, denoted ∂f(¯ x) := {v | f(¯ x) + ⟨v, x −¯ x⟩≤f(x) ∀x ∈E}.
For x ̸∈dom f, we define ∂f(x) = ∅. The domain of ∂f is dom ∂f := {x | ∂f(x) ̸= ∅}.
Properties: (1) ri dom f ⊂dom ∂f ⊂dom f (2) ∂f(x) is a nonempty closed convex set for all x ∈ri dom f.
(3) If x ∈intr dom f, then ∂f(x) is compact.
Optimization and the Subdifferential Theorem: Let f : E →R be proper convex. Then ¯ x ∈E is a global solution to min f(x) if and only if 0 ∈∂f(¯ x).
Proof: Apply the subgradient inequality: f(¯ x) + ⟨v, x −¯ x⟩≤f(x) ∀x ∈E.
The Convex Conjugate Recall that by applying the separation theorem to the epigraph of a proper convex function f, we found that for every ¯ x ∈ri dom f there exists v ∈E such that δ∗ epi f(v, −1) = sup x∈dom f [⟨v, x⟩−f(x)] = sup x [⟨v, x⟩−f(x)] ≤⟨v, ¯ x⟩−f(¯ x).
This relationship indicates that f∗: E →R given by f∗(v) := sup x [⟨v, x⟩−f(x)] plays a special in our study of convex functions.
We call f∗the convex conjugate of f.
The Convex Conjugate Recall that by applying the separation theorem to the epigraph of a proper convex function f, we found that for every ¯ x ∈ri dom f there exists v ∈E such that δ∗ epi f(v, −1) = sup x∈dom f [⟨v, x⟩−f(x)] = sup x [⟨v, x⟩−f(x)] ≤⟨v, ¯ x⟩−f(¯ x).
This relationship indicates that f∗: E →R given by f∗(v) := sup x [⟨v, x⟩−f(x)] plays a special in our study of convex functions.
We call f∗the convex conjugate of f.
Note that f∗= (cl f)∗since δ∗ epi f = δ∗ cl epi f.
The Bi-Conjugate and the Subdiffential f∗(v) := sup x [⟨v, x⟩−f(x)] = δ∗ epi f(v, −1) = δ∗ epi cl f(v, −1) By definition, f∗is a closed proper convex function whenever f is a proper convex function.
Theorem: [Fenchel-Young Inequality] Let f : E →R be a proper convex function. Then f∗(v) + f(x) ≥f∗(v) + cl f(x) ≥⟨v, x⟩ ∀x, v ∈E with equality throughout if and only if v ∈∂f(x).
The Bi-Conjugate and the Subdiffential Consequently, for all x ∈E, cl f(x) ≥ sup v∈dom f∗[⟨v, x⟩−f∗(v)] = sup v [⟨v, x⟩−f∗(v)] = (f∗)∗(x).
The Bi-Conjugate and the Subdiffential Consequently, for all x ∈E, cl f(x) ≥ sup v∈dom f∗[⟨v, x⟩−f∗(v)] = sup v [⟨v, x⟩−f∗(v)] = (f∗)∗(x).
Therefore, cl f(x) + f∗(v) ≥(f∗)∗(x) + f∗(v) ≥⟨v, x⟩ ∀x, v ∈E with equality throughout iffx ∈∂f∗(v) iffv ∈∂cl f(x).
The Bi-Conjugate and the Subdiffential Consequently, for all x ∈E, cl f(x) ≥ sup v∈dom f∗[⟨v, x⟩−f∗(v)] = sup v [⟨v, x⟩−f∗(v)] = (f∗)∗(x).
Therefore, cl f(x) + f∗(v) ≥(f∗)∗(x) + f∗(v) ≥⟨v, x⟩ ∀x, v ∈E with equality throughout iffx ∈∂f∗(v) iffv ∈∂cl f(x).
Theorem: For every proper convex function f : E →R, cl f = (f∗)∗= f∗∗, (∂(cl f))−1 = ∂f∗, and ∂(cl f)(x) = {v | cl f(x) + f∗(v) ≤⟨v, x⟩} , with ∂(cl f)(x) = ∂f(x) whenever x ∈dom ∂f.
Proof: cl f coincides with f on ri dom f = ri dom (cl f) and ri dom f ⊂dom ∂f.
Support Functions Revisited Let Q ⊂E be nonempty closed and convex. Then (δQ(·))∗(v) = sup x [⟨v, x⟩−δQ(x)] = δ∗ Q(x).
Support Functions Revisited Let Q ⊂E be nonempty closed and convex. Then (δQ(·))∗(v) = sup x [⟨v, x⟩−δQ(x)] = δ∗ Q(x).
Recall that support functions are subadditive. We now address the question of whether a proper subadditive function can be written as a support function.
Support Functions Revisited Let f : E →R be proper subadditive. Then, for λ > 0, f∗(v) = sup x∈dom f [⟨v, x⟩−f(x)] = sup x∈dom f [⟨v, λx⟩−f(λx)] = λ sup x∈dom f [⟨v, x⟩−f(x)] = λf∗(v).
Therefore, f∗(v) = 0 for all v ∈dom f∗and so f∗= δdom f∗.
Since f is proper convex, cl f = f∗∗= δ∗ dom f∗.
Support Functions Revisited Theorem: The class closed proper subadditive functions on E equals the class of support functions on E. In particular, if f : E →R is closed proper subadditive, then f is the support function of the set dom f∗= {v | ⟨v, x⟩≤f(x) ∀x ∈E}.
Support Functions Revisited Theorem: The class closed proper subadditive functions on E equals the class of support functions on E. In particular, if f : E →R is closed proper subadditive, then f is the support function of the set dom f∗= {v | ⟨v, x⟩≤f(x) ∀x ∈E}.
Proof: Since f is positively homogeneous, dom f∗= {v | ∃µ > 0 s.t. f∗(v) ≤µ} = {v | ∃µ > 0 s.t. ⟨v, x⟩−f(x) ≤µ ∀x ∈E} = {v | ∃µ > 0 s.t. ⟨v, λx⟩−f(λx) ≤µ ∀x ∈E, λ > 0} = n v ∃µ > 0 s.t. ⟨v, x⟩−f(x) ≤µ λ ∀x ∈E, λ > 0 o = {v | ⟨v, x⟩−f(x) ≤0 ∀x ∈E}.
The result follows since we have shown that f = δ∗ dom f∗.
f ′(x; ·) and ∂f Theorem: Let f : E →R be a proper convex function and let ¯ x ∈dom ∂f. Then the closure of f ′(¯ x; ·) is δ∗(·| ∂f(¯ x)). Moreover, if ¯ x ∈ri dom f, then f ′(¯ x; ·) is closed and proper.
Proof: Let v ∈∂f(¯ x) and let ϕ be the closure of f ′(¯ x; ·). Then, for t > 0 and d ∈E, ⟨v, d⟩≤f(¯ x+td)−f(¯ x) t so ⟨v, d⟩≤f ′(¯ x; d). Hence f ′(¯ x; ·) is proper, and ϕ is closed proper and subadditive. Therefore, ϕ is the support function of the set {v | ⟨v, d⟩≤ϕ(d) ∀d ∈E} = v ⟨v, d⟩≤f(¯ x + td) −f(¯ x) t ∀, d ∈E, t > 0 = {v | f(¯ x) + ⟨v, d⟩≤f(¯ x + d) ∀, d ∈E} = {v | f(¯ x) + ⟨v, x −¯ x⟩≤f(x) ∀, x ∈E} = ∂f(¯ x).
If ¯ x ∈ri dom f, then dom f ′(¯ x; ·) = par dom f = ri dom f ′(¯ x; ·) so that f ′(¯ x; ·) is locally Lip. on its domain and so closed and proper.
∂f(x) = {v} implies differentiability Corollary: Let f : E →R be a proper convex function. If ¯ x ∈dom ∂f, then (par dom f)⊥⊂∂f(¯ x).
Proof: Let v ∈∂f(¯ x) and w ∈(par dom f)⊥. Then for every y ∈dom f, f(¯ x) + ⟨v + w, y −x⟩= f(¯ x) + ⟨v, y −x⟩≤f(y).
Corollary: Let f : E →R be a proper convex function. If ¯ x ∈dom ∂f is such that ∂f(¯ x) = {v} + (par dom f)⊥, then f is differentiable relative to the affine manifold S := affdom f with gradient ∇Sf(¯ x) = v. In particular, if ¯ x ∈intr dom f, then f is differentiable at ¯ x with ∇f(¯ x) = v.
Proof: For d ∈par dom f, f′(¯ x; d) = ⟨v, d⟩is linear on the subspace par dom f. Hence, f is Gateaux differentiable relative to affdom f with Gateaux derivative v.
Computing the Subdifferential Proposition: Let Q ⊂E be a nonempty closed convex set.
Then ∂δQ(x) = ( ∅ , x ̸∈Q, NQ(x) , x ∈Q.
Computing the Subdifferential Proposition: Let Q ⊂E be a nonempty closed convex set.
Then ∂δQ(x) = ( ∅ , x ̸∈Q, NQ(x) , x ∈Q.
Note that this result implies that NQ(x) = [par Q]⊥when x ∈ri Q since δQ is differentiable on ri Q relative to the affine manifold affQ with derivative ∇affQδQ(x) = 0 for x ∈ri Q.
Computing the Subdifferential Proposition: Let Q ⊂E be a nonempty closed convex set.
Then ∂δQ(x) = ( ∅ , x ̸∈Q, NQ(x) , x ∈Q.
Note that this result implies that NQ(x) = [par Q]⊥when x ∈ri Q since δQ is differentiable on ri Q relative to the affine manifold affQ with derivative ∇affQδQ(x) = 0 for x ∈ri Q.
Proof: Given ¯ x ∈Q and v ∈NQ(¯ x), we have ⟨v, x −¯ x⟩≤0 ∀x ∈Q .
Computing the Subdifferential Proposition: Let Q ⊂E be a nonempty closed convex set.
Then ∂δQ(x) = ( ∅ , x ̸∈Q, NQ(x) , x ∈Q.
Note that this result implies that NQ(x) = [par Q]⊥when x ∈ri Q since δQ is differentiable on ri Q relative to the affine manifold affQ with derivative ∇affQδQ(x) = 0 for x ∈ri Q.
Proof: Given ¯ x ∈Q and v ∈NQ(¯ x), we have δQ(¯ x) + ⟨v, x −¯ x⟩≤δQ(x) ∀x ∈E .
Computing the Subdifferential Proposition: Let Q ⊂E be a nonempty closed convex set.
Then ∂δ∗ Q(x) = argmax v∈Q ⟨v, x⟩.
Proof: For any closed proper convex function f, we have shown that ∂f(x) = {v | f∗(v) + f(x) ≤⟨v, x⟩} .
Since both δQ and δ∗ Q are closed proper convex, we have ∂δ∗ Q(x) = v δQ(v) + δ∗ Q(x) ≤⟨v, x⟩ = argmax v∈Q ⟨v, x⟩.
The Subdifferential of a Norm Corollary: Let ∥·∥be any norm on E with closed unit ball B.
Then ∂∥x∥= ( B◦ , x = 0, {v | ∥v∥∗= 1 and ⟨v, x⟩= ∥x∥} , x ̸= 0.
Proof: The result follows since ∥·∥= δ∗ B◦(·) where ∥·∥∗is the dual norm for ∥·∥whose closed unit ball is B.
Computing Conjugates Computing the conjugate f∗at v reduces to solving for x in the equation v ∈∂f(x).
To see this, observe that f∗(v) = supx[⟨v, x⟩−f(x)] = −infx[f(x) −⟨v, x⟩].
Since f(x) −⟨v, x⟩is convex, we need only solve 0 ∈∂f −⟨v, ·⟩ = ∂f(x) −v for x, then plug this x back into ⟨v, x⟩−f(x) to find f∗(v). This is especially useful when f is differentiable on its domain.
Computing Conjugates Computing the conjugate f∗at v reduces to solving for x in the equation v ∈∂f(x).
To see this, observe that f∗(v) = supx[⟨v, x⟩−f(x)] = −infx[f(x) −⟨v, x⟩].
Since f(x) −⟨v, x⟩is convex, we need only solve 0 ∈∂f −⟨v, ·⟩ = ∂f(x) −v for x, then plug this x back into ⟨v, x⟩−f(x) to find f∗(v). This is especially useful when f is differentiable on its domain.
Example: f(x) = ex. Then v = ∇f(x) = ex iffx = ln v, in which case f∗(v) = ⟨v, ln v⟩−f(ln v) = ( v ln v −v , v > 0, +∞ , v ≤0 .
Computing Conjugates Computing the conjugate f∗at v reduces to solving for x in the equation v ∈∂f(x).
To see this, observe that f∗(v) = supx[⟨v, x⟩−f(x)] = −infx[f(x) −⟨v, x⟩].
Since f(x) −⟨v, x⟩is convex, we need only solve 0 ∈∂f −⟨v, ·⟩ = ∂f(x) −v for x, then plug this x back into ⟨v, x⟩−f(x) to find f∗(v). This is especially useful when f is differentiable on its domain.
Example: f(x) = ex. Then v = ∇f(x) = ex iffx = ln v, in which case f∗(v) = ⟨v, ln v⟩−f(ln v) = ( v ln v −v , v > 0, +∞ , v ≤0 .
Check f∗∗(x) = ex.
Computing Conjugates: Dual Operations General formulas for conjugates of convex functions generated from other convex functions using convexity preserving operations are very powerful tools in applications.
Computing Conjugates: Dual Operations General formulas for conjugates of convex functions generated from other convex functions using convexity preserving operations are very powerful tools in applications.
Example: What is (λf)∗when λ > 0 and f proper convex?
Computing Conjugates: Dual Operations General formulas for conjugates of convex functions generated from other convex functions using convexity preserving operations are very powerful tools in applications.
Example: What is (λf)∗when λ > 0 and f proper convex?
(λf)∗(v) = sup x ⟨v, x⟩−λf(x) = λ sup x Dv λ, x E −f(x) = λf∗(v λ) That is, the dual operation to multiplying a function by a positive scalar is epi-multiplication.
What is (λf(·/λ))∗for λ > 0?
(λf(·/λ))∗(v) = sup x [⟨v, x⟩−λf(x/λ)] = λ sup x [⟨v, x/λ⟩−f(x/λ)] = λ sup z [⟨v, z⟩−f(z)] = λf∗(v) .
What is (f1□f2)∗?
(f1□f2)∗(v) = sup x [⟨v, x⟩− inf x=x1+x2[f1(x1) + f2(x2)]] = sup x sup x=x1+x2 [⟨v, x⟩−(f1(x1) + f2(x2))] = sup x1,x2 [⟨v, x1 + x2⟩−f1(x1) −f2(x2)] = sup x1,x2 [(⟨v, x1⟩−f1(x1)) + (⟨v, x2⟩−f2(x12))] = sup x1 [⟨v, x1⟩−f1(x1)] + sup x2 [⟨v, x2⟩−f2(x2)] = f∗ 1 (v) + f∗ 2 (v) What is (f1 + f2)∗?
The first point to consider By the bi-conjugacy theorm, (cl f1 + cl f2)∗= ((f∗ 1 )∗+ (f∗ 2 )∗)∗ = ((f∗ 1 □f∗ 2 )∗)∗ = cl (f∗ 1 □f∗ 2 ) What is (f1 + f2)∗?
The first point to consider By the bi-conjugacy theorm, (cl f1 + cl f2)∗= ((f∗ 1 )∗+ (f∗ 2 )∗)∗ = ((f∗ 1 □f∗ 2 )∗)∗ = cl (f∗ 1 □f∗ 2 ) It can be shown that if (ri dom f1) ∩(ri dom f2) ̸= ∅, then the closure operation can be removed from the above equivalence, i.e.
(f1 + f2)∗= f∗ 1 □f∗ 2 .
Application: Distance to a Convex Cone Let K ⊂E be a closed convex cone and let ∥·∥be any norm on E with closed unit ball B. Then dist (z |K ) = inf y∈K ∥z −y∥ = inf y ∥z −y∥+ δK(y) = inf y δ∗ B◦(z −y) + δ∗ K◦(y) = (δ∗ B◦□δ∗ K◦)(z).
Application: Distance to a Convex Cone Let K ⊂E be a closed convex cone and let ∥·∥be any norm on E with closed unit ball B. Then dist (z |K ) = inf y∈K ∥z −y∥ = inf y ∥z −y∥+ δK(y) = inf y δ∗ B◦(z −y) + δ∗ K◦(y) = (δ∗ B◦□δ∗ K◦)(z).
Consequently, dist (· |K )∗= (δ∗ B◦□δ∗ K◦)∗ = δ∗∗ B◦+ δ∗∗ K◦ = δB◦+ δK◦= δB◦∩K◦.
Application: Distance to a Convex Cone Let K ⊂E be a closed convex cone and let ∥·∥be any norm on E with closed unit ball B. Then dist (z |K ) = inf y∈K ∥z −y∥ = inf y ∥z −y∥+ δK(y) = inf y δ∗ B◦(z −y) + δ∗ K◦(y) = (δ∗ B◦□δ∗ K◦)(z).
Consequently, dist (· |K )∗= (δ∗ B◦□δ∗ K◦)∗ = δ∗∗ B◦+ δ∗∗ K◦ = δB◦+ δK◦= δB◦∩K◦.
Therefore, dist (z |K ) = δ∗ B◦∩K◦(z) .
An Alternative Approach to the Subdifferential Eventually, we would like to extend the notion of subdifferential beyond convex functions. One proposal is to define the (regular) subdifferential by the inequality ˆ ∂f(x) := {v | f(x) + ⟨v, y −x⟩≤f(y) + o(∥y −x∥)} .
Proposition: Let f : E →R be proper convex. Then, for all x ∈dom ∂f(x), ˆ ∂f(x) = ∂f(x).
Proof: Clearly, ∂f(x) ⊂ˆ ∂f(x), so let v ∈ˆ ∂f(x). Then, for all d ∈E and t > 0, ⟨v, d⟩≤f(x + td) −f(x) t + o(t ∥d∥) t , and so ⟨v, d⟩≤f′(x; d) = δ∗ ∂f(x)(d). Therefore, v∂f(x).
For this reason, from now on we simply denote ˆ ∂f(x) by ∂f(x) and call ∂f(x) even when f is not necessarily convex. Again, dom ∂f := {x | ∂f(x) ̸= ∅} A simple subdifferential calculus rule Proposition: Let h : E →R be proper convex and g : E →R be convex and differentiable on the open set U. Then, for all x ∈U ∩dom ∂h, ∂(h + g)(x) = ∂h(x) + ∇g(x).
Proof: We have already shown that ∂g(x) = {∇g(x)} for all x ∈U.
Given x ∈U ∩dom ∂h and v ∈∂h(x), we have h(x) + ⟨v, y −x⟩≤h(y) g(x) + ⟨∇g(x), y −x⟩≤g(y) ) ∀y ∈E .
Adding these inequalities shows that ∂h(x) + ∇g(x) ⊂∂(h + g)(x).
Next let w ∈∂(h + g)(x). Then h(x) + g(x) + ⟨w, y −x⟩≤h(y) + g(y) = h(y) + g(x) + ⟨∇g(x), y −x⟩+ o(∥y −x∥).
Hence, h(x) + ⟨w −∇g(x), y −x⟩≤h(y) + o(∥y −x∥) ∀y ∈E, which implies that w −∇g(x) ∈∂h(x).
Strong Convexity Definition: A function f : E →R is called µ-strongly convex (with µ ≥0) if the perturbed function x 7→f(x) −µ 2∥x∥2 is convex.
Theorem: Let f : E →R be a µ-strongly convex function.
Then for any x ∈E and v ∈∂f(x), the estimate holds: f(y) ≥f(x) + ⟨v, y −x⟩+ µ 2 ∥y −x∥2 for all y ∈E.
Proof: Apply the subdifferential inequality to the convex function g := f −µ 2 ∥·∥2.
Strong Convexity Definition: A function f : E →R is called µ-strongly convex (with µ ≥0) if the perturbed function x 7→f(x) −µ 2∥x∥2 is convex.
Theorem: Let f : E →R be a µ-strongly convex function.
Then for any x ∈E and v ∈∂f(x), the estimate holds: f(y) ≥f(x) + ⟨v, y −x⟩+ µ 2 ∥y −x∥2 for all y ∈E.
Proof: Apply the subdifferential inequality to the convex function g := f −µ 2 ∥·∥2.
Corollary: Any proper, closed, µ-strongly convex function f : E →R is coercive and has a unique minimizer x satisfying f(y) −f(x) ≥µ 2 ∥y −x∥2 for all y ∈E.
The Moreau Envelope Definition: For any function f : E →R and real α > 0, define the Moreau envelope and the proximal map, respectively: fα(x) := f □( 1 2α∥· ∥2) (x) = min y f(y) + 1 2α∥x −y∥2 proxαf(x) := argmin y f(y) + 1 2α∥x −y∥2.
Recall that epi fα = epi f + epi 1 2α∥· ∥2 .
The Huber Function and Soft-Threshholding For f(x) = |x|, fα(x) = ( 1 2α|x|2 if |x| ≤α |x| −1 2α otherwise ) , proxαf(x) = x −α if x ≥α 0 if |x| ≤α x + α if x ≤−α .
The Huber Function and Soft-Threshholding For f(x) = |x|, fα(x) = ( 1 2α|x|2 if |x| ≤α |x| −1 2α otherwise ) , proxαf(x) = x −α if x ≥α 0 if |x| ≤α x + α if x ≤−α .
epi | · | epi | · |α (a) epi | · |α = epi | · | + epi 1 2| · |2 α −α gph (proxα|·|) (b) gph (proxα|·|) Figure: Moreau envelope and the proximal map of | · |.
The Distance Function Let Q ⊂E be closed convex. Then (δQ)α(x) = (δQ□1 2α ∥·∥2)(x) = inf y∈Q 1 2α ∥x −y∥2 = 1 2αd2 Q(x) and proxαδQ(x) = projQ(x).
Prox is 1-Lipschitz Theorem: Let f : E →R be proper, closed, cvx. Then the set proxf(x) is a singleton for every point x ∈E. Moreover, ∥proxf(x) −proxf(y)∥2 ≤⟨proxf(x) −proxf(y), x −y⟩ ∀x, y ∈E.
Prox is 1-Lipschitz Theorem: Let f : E →R be proper, closed, cvx. Then the set proxf(x) is a singleton for every point x ∈E. Moreover, ∥proxf(x) −proxf(y)∥2 ≤⟨proxf(x) −proxf(y), x −y⟩ ∀x, y ∈E.
Proof: The map z 7→f(z)+ 1 2∥z−x∥2 is proper, closed, and 1-strongly cvx, and hence proxf(x) is the unique minimizer. Since h(y) := f(y) + 1 2∥y −x∥2 is 1-strongly cvx, for x, y ∈E, f(x+) + 1 2∥x+ −x∥2 ≤ f(y+) + 1 2∥y+ −x∥2 −1 2∥y+ −x+∥2 = f(y+) + 1 2∥y+ −y∥2 −1 2∥y+ −x+∥2 + 1 2∥y+ −x∥2 −1 2∥y+ −y∥2 ≤ f(x+) + 1 2∥x+ −y∥2 −∥y+ −x+∥2 + 1 2∥y+ −x∥2 −1 2∥y+ −y∥2, so ∥y+ −x+∥2 ≤1 2 ∥x+ −y∥2 −∥y+ −y∥2 + ∥y+ −x∥2 −∥x+ −x∥2 = ⟨x+ −y+, x −y⟩≤∥y+ −x+∥∥x −y∥.
The Moreau Decomposition Theorem: For any proper, closed, convex function f : E →R, proxf(x) + proxf⋆(x) = x ∀x ∈E.
Proof Using the definition of the proximal map, z = proxf(x) ⇐ ⇒0 ∈∂ f + 1 2∥· −x∥2 (z) ⇐ ⇒x −z ∈∂f(z) ⇐ ⇒z ∈∂f⋆(x −z) ⇐ ⇒0 ∈∂f⋆(x −z) −z ⇐ ⇒0 ∈∂ f⋆+ 1 2∥· −x∥2 (x −z) ⇐ ⇒x −z = proxf⋆(x).
∇fα is Lipschitz continuous with parameter α−1 Theorem: Let f : E →Rbe closed proper convex. Then the envelope fα is continuously differentiable on E with gradient ∇fα(x) = α−1(x −proxαf(x)).
Consequently ∇fα is α−1-smooth.
Proof:Take α = 1, then z ∈∂fα(x) ⇐ ⇒ x ∈∂(f 1 2∥· ∥2)⋆(z) ⇐ ⇒ x ∈∂ f⋆+ 1 2∥· ∥2⋆ (z) ⇐ ⇒ x ∈∂f⋆(z) + z ⇐ ⇒ 0 ∈∂(f⋆+ 1 2∥· −x∥2)(z) ⇐ ⇒ z = proxf⋆(x) ⇐ ⇒ z = x −proxf(x), For α ̸= 1, use the identity αfα = (αf)1.
Baillon-Haddad Theorem Theorem: A proper, closed, convex function f : E →R is µ-strongly convex if and only if the conjugate f⋆is µ−1-smooth.
Proof: ( = ⇒) Suppose that f is µ-strongly convex and define the convex function g(x) := f(x) −µ 2∥x∥2. We may then write f⋆= g + µ 2∥· ∥2⋆= g⋆ 1 2µ∥· ∥2.
The right-hand-side is simply the Moreau envelope of g⋆with parameter µ, and is therefore µ−1-smooth.
Baillon-Haddad Theorem (⇐ =) Suppose f ∗is µ−1-smooth, and set h := f ∗and β := µ−1 so that h is β-smooth. We know that h is β-smooth is equivalent to 0 ≤⟨∇h(x) −∇h(y), x −y⟩≤β ∥x −y∥2 .
Set g := β 2 ∥·∥−h. Then ⟨∇g(y) −∇g(x), y −x⟩= β∥y −x∥2 −⟨∇h(y) −∇h(x), y −x⟩≥0.
Hence, g is cvx. Note that h(y) = β 2 ∥y∥2 −g(y) = β 2 ∥y∥2 −g⋆⋆(y) = β 2 ∥y∥2 −sup x {⟨y, x⟩−g⋆(x)} = inf x [ β 2 ∥y∥2 −⟨y, x⟩+ g⋆(x)], so h⋆(z) = sup y {⟨z, y⟩−h(y)} = sup y [⟨z, y⟩−inf x n β 2 ∥y∥2 −⟨y, x⟩+ g⋆(x) o ] = sup x sup y [⟨z, y⟩−β 2 ∥y∥2 + ⟨y, x⟩−g⋆(x)] = sup x [sup y n ⟨z + x, y⟩−β 2 ∥y∥2o −g⋆(x)] = sup x 1 2β ∥z + x∥2 −g⋆(x).
So h⋆(z) − 1 2β ∥z∥2 = supx [ 1 β ⟨z, x⟩+ 1 2β ∥x∥2 −g⋆(x)] is cvx.
Subgradient Dominance Theorem Theorem: Any proper, closed, α-strongly convex function f : E →R satisfies the subgradient dominance condition: f(x) −min f ≤1 α∥v∥2 for all x ∈E, v ∈∂f(x).
Proof: Let ¯ x be a minimizer of f. Fix any x ∈E and v ∈∂f(x). We compute f(x) −f(¯ x) ≤⟨v, x −¯ x⟩≤∥v∥· ∥x −¯ x∥ = ∥v∥· ∥∇f⋆(v) −∇f⋆(0)∥≤1 α∥v∥2.
The Normal Cone to the Epigraph Proposition: Let f : E →R be proper convex. Then, for all ¯ x ∈dom ∂f, ∂f(¯ x) = {v | (v, −1) ∈Nepi f(¯ x, f(¯ x))}.
Proof: (v, −1) ∈Nepi f(x, f(x)) ⇐ ⇒⟨(v, −1), (x, f(x)) −(¯ x, f(¯ x))⟩≤0 ∀x ∈dom f ⇐ ⇒f(¯ x) + ⟨v, x −¯ x⟩≤f(x) ∀x ∈dom f ⇐ ⇒f(¯ x) + ⟨v, x −¯ x⟩≤f(x) ∀x ∈E .
Outer Semicontinuity of the Subdifferential An important property of the subdifferential is that it is outer semicontinuous.
Definition: A multivalued mapping T : X ⇒Y is said to be outer semicontinuous on its domain, dom T := {x | T(x) ̸= ∅}, if for every point (¯ x, y) ∈(dom T) × Y and every sequence {(xi, yi)} ⊂X × Y with (xi, yi) →(¯ x, y) with yi ∈T(xi) for all i it must be the case that y ∈T(¯ x).
Theorem: Let f : E →R be proper convex. Then ∂f is outer semicontinuous on dom ∂f.
Proof: Let (¯ x, y) ∈(dom ∂f) × E and {(xi, yi)} ⊂(dom ∂f) × E be such that (xi, yi) →(¯ x, y) with yi ∈∂f(xi) for all i. We must show y ∈∂f(¯ x). By construction, cl f(xi) + ⟨yi, x −xi⟩≤f(x) ∀x ∈E .
Hence, given x ∈E, using the lower semicontinuity of cl f, we may take the limit in this inequality to find that cl f(¯ x) + ⟨y, x −¯ x⟩≤f(x) ∀x ∈E .
Hence, y ∈∂(cl f)(¯ x) = ∂f(¯ x), where the equality follows since ¯ x ∈dom ∂f. |
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Lesson Explainer: Applications of Newton’s Second Law: Inclined Pulley Mathematics • Third Year of Secondary School
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In this explainer, we will learn how to solve problems on the motion of two bodies connected by a string passing over a smooth pulley with one of them on an inclined plane.
Consider two bodies connected by a light inextensible string, where the body of mass is supported by a smooth horizontal surface and the body of mass is suspended from the string. The string runs over a smooth pulley that requires negligible force to turn. The following figure shows the forces acting on the bodies, where force is the tension in the string and is the normal reaction force.
The acceleration of each body is determined by its mass and the net force acting on it, according to Newton’s second law of motion. The tension in the string is constant.
Each body has the same acceleration, given by where is the acceleration due to gravity.
If the surface that supports the body of mass is an inclined plane, then the forces acting on the bodies are those shown in the following figure.
The resultant of the weight of the body of mass , , and the normal reaction force on it, , is , where is the angle above the horizontal of the incline of the plane. The net force on the body parallel to the plane is given by
Let us look at an example in which the acceleration of such a system is determined.
Example 1: Finding the Acceleration of a System Involving a Smooth Inclined Plane and a Pulley
A body of mass 5 kg rests on a smooth plane inclined at an angle of to the horizontal. It is connected by a light inextensible string passing over a smooth pulley fixed at the top of the plane, to another body of mass 19 kg hanging freely vertically below the pulley. Given that the acceleration due to gravity , determine the acceleration of the system.
Answer
The following figure shows the forces acting on the bodies and what the forces on the 5 kg mass body due to its weight and the normal reaction on it sum to.
The accelerations of both bodies are equal. The acceleration of the 19 kg mass body is given by
Multiplying the expression by 19, we obtain
| | |
--- |
| | |
The net force on the supported body can be expressed as
Force can also be expressed as
Hence, we have
| | |
--- |
| | |
We now have two equations, (1) and (2), that can be added to give
This simplifies to
To two decimal places, this is 6.59 m/s2.
Let us now look at an example in which the tension in the string is determined.
Example 2: Finding the Force Acting on the Pulley in a System with an Inclined Plane
Two bodies of equal masses of 7.4 kg are connected by a light inelastic string. One of the bodies rests on a smooth plane inclined at to the horizontal. The string passes over a smooth pulley fixed at the top of the plane, and the other body is left to hang freely vertically below the pulley. Find the force acting on the pulley when the system is released from rest. Take the acceleration due to gravity to be .
Answer
The force acting on the pulley is the resultant of the tension forces in the strings. The net force on a pulley due to two equal vertical tension forces, and , is shown in the following figure.
In this case, however, one tension force acts vertically downward and the other acts parallel to the inclined plane. The net force due to these two tension forces is, therefore, equivalent to force shown in the following figure.
The angle of inclination of the plane is , hence, the forces on the pulley due to the tensions in the string act as shown in the following figure.
The tension is constant throughout the string, so and have the same magnitude. For the purposes of determining and , this system of forces is equivalent to the system of forces shown in the following figure.
We can define a tension of magnitude , where
The magnitude of the resultant of and is, therefore, given by
The tension in the string can be determined by equating the accelerations of the bodies and, hence, the magnitudes of the forces producing these accelerations, as the masses of the bodies are equal. The forces acting on the bodies are shown in the following figure.
We see from the figure that
This can be rearranged to give
It has been shown that the force on the pulley is given by so, by substituting in our expression for , we have
To two decimal places, this is 130.71 N.
Let us now look at an example in which kinematic equations are used.
Example 3: Solving an Inclined Plane Pulley System Using Newton’s Second Law and the Equations of Motion
A body of mass 2.4 kg rests on a smooth plane inclined at an angle of to the horizontal. It is connected by a light inextensible string passing over a smooth pulley, fixed at the top of the plane, to another body of mass 1.6 kg hanging freely vertically below the pulley. When the system was released from rest, the two bodies were on the same horizontal level. Then, 10 seconds later, the string broke. Determine the time taken for the first body to start moving in the opposite direction after the string broke. Take .
Answer
The initial state of the system is shown in the following figure.
The figure shows that the resultant of the weight of the body of mass and the normal reaction force on the body acts downward parallel to the plane. The force parallel to the plane is the sum of the weight of the body and the normal reaction on the body.
The net force on the supported body is given by
The net force on the suspended body is given by
The acceleration of each body equals the net force on the body divided by its mass. The accelerations of the bodies are equal; hence, the acceleration of the system is given by
We see that and
Adding these expressions, we obtain
Substituting known values, we have
The magnitude of the tension in the string is given by
The magnitude of the sum of the weight of the body and the normal reaction force on it is given by
So, we see that the supported body accelerates upward parallel to the surface and the suspended body descends vertically. We should expect this as the question states that when the tension ceases to act on the body on the surface, it eventually reverses direction. If the body was initially moving downward parallel to the surface, removing the tension force acting on it would not make the body start to move upward parallel to the surface.
In a time of 10 seconds, the velocity of the body is upward along the slope with a magnitude given by
When the tension in the string ceases to act on the body, the body accelerates downward parallel to the surface. With upward parallel to the surface taken as positive, the acceleration is given by
The body starts to move downward parallel to the surface at the instant that it has an instantaneous velocity of zero upward parallel to the surface. The time taken for this velocity change can be determined using the formula
Rearranging to make the subject, we have
Substituting known values, we obtain
Let us now look at an example in which the friction coefficient of a rough inclined surface needs to be found.
Example 4: Finding the Coefficient of Friction in a System with a Pulley and a Rough Inclined Plane
A body of mass 240 g rests on a rough plane inclined to the horizontal at an angle whose sine is . It is connected, by a light inextensible string passing over a smooth pulley fixed to the top of the plane, to another body of mass 300 g. If the system was released from rest and body descended 196 cm in 3 seconds, find the coefficient of friction between the body and the plane. Take .
Answer
The resultant of the weight of body and the normal reaction on it acts downward parallel to the plane. This resultant force, , is given by where is the mass of and is the angle that the plane is inclined at. Note that to make the units of mass consistent with the base SI units used for distance and time, the masses of the bodies in grams are converted to masses in kilograms. Thus, substituting , , and , we have
Let us draw a diagram of the system. Bodies and will be connected by a wire, and they will each have a force acting on them due to the tension of the string, with gravity acting in the opposite direction. Additionally, body will have a frictional force acting counter to the direction of motion (i.e., down the slope).
Since the plane is inclined at an angle whose sine is , we can see that the lengths of the sides of the object are in the ratio .
In this question, we need to calculate the coefficient of friction , which is directly related to the magnitude of the frictional force by the formula where is the magnitude of the normal force to the slope. This is given by
Unfortunately, we do not know or , but we do know that the sine of the angle is . As shown in the diagram, this means we know the ratio between the lengths of the sides, and we can use the Pythagorean theorem to find the horizontal distance moved by (which we will denote ):
Hence, we have
Substituting this back into the formula for , we get
| | |
--- |
| | |
To solve this and find , we will need to find , which we can do by considering the other forces in the system.
We note that since the objects are connected, they will accelerate at the same rate, which can be determined from the motion of . We are given that body accelerates from rest and in a time of 3 seconds has a displacement of 196 cm. As the value of in the question is given as 9.8 m/s2, we must consider the displacement in metres rather than centimetres. So, the displacement is 1.96 m. The formula for the displacement is where is the initial velocity, is the time, and is the acceleration. This can be rearranged to make the subject, taking to be zero:
Having found the acceleration, this means we can calculate the resultant forces, and , using Newton’s second law. That is, for ,
| | |
--- |
| | |
and for ,
| | |
--- |
| | |
Each of these resultant forces can also be written as the sum of the forces acting on the bodies. That is,
If we add these two equations together directly, we can cancel out the instances of to get or, rearranged in terms of ,
Then, we can substitute in (3), (4), and (5) to get
Finally, we can get by dividing by 1.8116:
Now, let us look at another example involving a rough inclined plane.
Example 5: Solving a Rough Inclined Plane Pulley System Using Newton’s Second Law and the Equations of Motion
A body of mass 162 g rests on a rough plane inclined to the horizontal at an angle whose tangent is . It is connected, by a light inextensible string passing over a smooth pulley fixed to the top of the plane, to another body of mass 181 g hanging freely vertically below the pulley. The coefficient of friction between the first body and the plane is . Determine the distance covered by the system in the first 7 seconds of its movement, given that the bodies were released from rest. Take .
Answer
We are told that
Hence, when the body supported by the plane moves along the plane, it ascends or descends 4 metres vertically for each 3 metres it travels horizontally. From the Pythagorean theorem, where is the distance traveled along the surface by the body.
Hence, and
The following figure shows all but one of the forces acting on the supported body, where the weight of the body and the normal reaction force on it are shown by their resultant, which acts downward parallel to the plane. is the tension in the string.
The force not shown is the frictional force, , on the supported body. The force is not shown as a frictional force acts in the opposite direction to the net force on a body, and the direction of the net force on the supported body has not been established. Let us now determine it.
The vertically downward force on the body suspended by the string is given by
The force acting parallel to the plane on the body supported by the plane is given by
Dividing by , we obtain
As we see that
Hence, the suspended mass descends and the supported mass moves upward parallel to the plane.
The frictional force on the supported body, is given by the product of the normal reaction force on the body and the coefficient of friction between the body and the surface that supports it. All the forces acting on the supported body act parallel to the plane; hence, we can express the frictional force on the body, , as
The net force on the supported body is, therefore, given by
| | |
--- |
| | |
where is the acceleration of the system.
The net force on the suspended body is given by
| | |
--- |
| | |
Adding the net forces on the bodies given by equations (6) and (7), we obtain
The displacement of the supported body can be determined using the formula
Substituting known values, we find that after 7 seconds of acceleration,
This can be expressed as an integer value by converting it to a value in centimetres, which gives 196 cm.
Let us summarize what we have learned in these examples.
Key Points
The following points apply to a system of two bodies connected by a light inextensible string, where the body of mass is supported by a smooth plane inclined at an angle above the horizontal and the body of mass is suspended from the string. The string runs over a smooth pulley that requires negligible force to turn, as shown in the following figure.
The net force on the suspended body equals the sum of the tension in the string and the weight of the body. It is given by where is the tension in the string and is the acceleration due to gravity.
The net force on the supported body equals the sum of the tension in the string and the component of the weight of the body acting parallel to the plane. This is given by where is the tension in the string and is the acceleration due to gravity.
The accelerations of the suspended and the supported bodies are both equal to the net forces on them divided by their masses, and both accelerations are equal. Then, we have where is the tension in the string and is the acceleration due to gravity.
If the inclined plane that supports the body of mass is rough, the acceleration of the system is given by where is the coefficient of friction of the body with the plane. The frictional force on the body of mass may be either in the same direction as or in the opposite direction to the tension in the string, depending on whether the body moves upward or downward parallel to the plane.
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17450 | https://www.cnblogs.com/grandyang/p/11595236.html | [LeetCode] 829. Consecutive Numbers Sum 连续数字之和 - Grandyang - 博客园
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[LeetCode] 829. Consecutive Numbers Sum 连续数字之和
Given a positive integerN, how many ways can we write it as a sum of consecutive positive integers?
Example 1:
makefile
Input: 5
Output: 2
Explanation: 5 = 5 = 2 + 3
Example 2:
makefile
Input: 9
Output: 3
Explanation: 9 = 9 = 4 + 5 = 2 + 3 + 4
Example 3:
makefile
Input: 15
Output: 4
Explanation: 15 = 15 = 8 + 7 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5
Note:1 <= N <= 10 ^ 9.
这道题给了一个正整数N，问N能写成多少种连续正整数之和，比如9可以写成 4+5，或者 2+3+4。这道题其实不好做，因为没有固定的算法可以套，而更多的考察是数学知识，而且比较难想。由于要写成连续正整数之和，则肯定是一个等差数列，并且差值为1，这个等差数列不必从1开始，假设其是从x开始的，且个数共有k个，则可以写出这个等差数列为：
undefined
x, x+1, x+2, ..., x+k-1
其和为N，根据等差数列的求和公式，可以写出下列等式：
lisp
kx + (k-1)k / 2 = N
变形后可得到：
ini
kx = N - (k-1)k / 2
这样，只要对于任意一个k值，x能得到正整数解，就表示一定会有一个对应的等差数列和为N。下面要来求k的范围，由于k是等差数列的长度，首先肯定是要大于0的，这是下限。求上限还是要利用上面的那个式子，由于x也必须是正整数，可以得到不等式：
lisp
N - (k-1)k / 2 > 0
从而得到近似解：
scss
k < sqrt(2N)
有了k的范围就可以开始遍历了，首先数字N本身也是符合题意的，可以看作是长度为1的等差数列，则 res 可以初始化为1，然后i从2遍历到 sqrt(2N)，对于每个i值，只要 (N - i(i-1)/2) 能整除i，就表示存在长度为i的等差数列和为N，结果 res 自增1，这样就可以求出所有符合题意的等差数列的个数，参见代码如下：
解法一：
cpp
class Solution {
public:
int consecutiveNumbersSum(int N) {
int res = 1;
for (int i = 2; i < sqrt(2 N); ++i) {
if ((N - i (i - 1) / 2) % i == 0) ++res;
}
return res;
}
};
还可以换一种写法，核心思路还是跟上面的解法相同，要找是否存在和为N的等差数列，根据上面的分析，需要看等差数列的起始值x是否为整数，若这个等差数列每个数字都减去一个 x-1，就变成了一个从1开始的差值为1的等差数列，那就让i从1开始遍历，用一个变量 sum，每次都加上i值，这样就相当于计算了这个等差数列的和，然后每次看 N-sum 是否能整除i，能的话就表明存在长度为i的等差数列和为N，参见代码如下：
解法二：
python
class Solution {
public:
int consecutiveNumbersSum(int N) {
int res = 0, sum = 0;
for (int i = 1; sum < N; ++i) {
sum += i;
if ((N - sum) % i == 0) ++res;
}
return res;
}
};
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1. Re:[LeetCode] 9. Palindrome Number 验证回文数字
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@howard144 prevP是什么...
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3. Re:[LeetCode] 1072. Flip Columns For Maximum Number of Equal Rows 按列翻转得到最大值等行数
若直接将两行编码成 "01" 和 "10"，并不能达到统计的效果 这里可以加一句"为了让1010..0 与0101..1是同样的形式,这样change之后结果才能把11111和00000的算作一种,即...
--Cu3S
4. Re:[LeetCode] 1072. Flip Columns For Maximum Number of Equal Rows 按列翻转得到最大值等行数
而对于例子3，后两行经过翻转后可以变为相同的行， 这句话改成 通过翻转最后一列可以让后两行都变成“同一行的元素完全一样”的形式。更合适吧，不然太容易有误导性了，当然结合题目不会有人像我一样撒敷敷的把 ...
--Cu3S
5. Re:[LeetCode] 1072. Flip Columns For Maximum Number of Equal Rows 按列翻转得到最大值等行数
这里省略了最重要的证明部分，即“两个pattern相同的行，最后一定能转换成全部元素都一样的行”。 直接从case3也能看懂，但是博主没提这一点，以及思路为何是正确的，倒像是在解说别人的代码还不得要领...
--Cu3S
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17451 | https://teachy.ai/en/lesson-plan/elementary-school-en-US/us-pre-6th-grade/math/volume-unit-cubes-hands-on | Lesson plan of Hands On Methodology | Volume: Unit Cubes | Lesson Plan
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Volume: Unit Cubes
Lesson plan of Volume: Unit Cubes
Lara from Teachy
Subject Math
Math
Source Original Teachy
Original Teachy
Topic Volume: Unit Cubes
Volume: Unit Cubes
Objectives (5 - 7 minutes)
Understand the Concept of Volume: The teacher will introduce the concept of volume as the amount of space occupied by a three-dimensional object. Students will learn that volume is measured in cubic units.
Learn about Unit Cubes: The teacher will explain what unit cubes are and how they are used to measure volume. Students will understand that a unit cube is a small cube with an edge length of one unit, and that the volume of any object can be found by counting the number of unit cubes that fill it.
Develop Skills in Measuring Volume: The teacher will set the stage for the hands-on activity by explaining that the students will be working with unit cubes to measure the volume of different objects. The teacher will outline the steps of the activity, emphasizing the importance of accuracy and attention to detail.
Secondary Objectives:
Apply Knowledge of Multiplication and Addition: Students will use their multiplication and addition skills to count and calculate the volume of various objects.
Enhance Spatial Awareness: By working with three-dimensional objects, students will develop a better understanding of spatial relationships and enhance their spatial awareness skills.
Introduction (10 - 12 minutes)
Recap of Previous Lessons: The teacher will briefly remind students of the concept of three-dimensional shapes and their properties, focusing on the fact that three-dimensional shapes have a volume. The teacher will also review the basic principles of multiplication and addition, as these skills will be used in the current lesson.
Problem Situations: The teacher will propose two problem situations to stimulate the students' interest and set the stage for the lesson. The first problem could be: "Imagine you have a box that is 2 units long, 3 units wide, and 4 units high. How many small unit cubes can fit inside this box?" The second problem could be: "You have a toy car that can be built using small unit cubes. How many unit cubes will you need to build the car?"
Real-World Applications: The teacher will highlight the importance of understanding volume in real-world contexts. They could explain, for instance, how architects and engineers use volume to design buildings and structures. They could also mention how farmers use the concept of volume to calculate the amount of water needed to irrigate their fields.
Introduction of the Topic: The teacher will introduce the topic of volume, unit cubes, and the hands-on activity that the students will be engaged in. They will explain that the students will be using unit cubes to measure the volume of different objects. The teacher will also show a few unit cubes, highlighting their equal length on all sides.
Curiosities and Fun Facts: To engage the students' interest, the teacher will share some fun facts and curiosities related to the topic. For instance, they might tell the story of Archimedes, who discovered the concept of volume when he stepped into a bathtub and noticed the water level rise. They could also share the fact that the word "volume" comes from the Latin word "volvere," which means "to roll," as the concept of volume was first used to measure the capacity of rolled scrolls.
Conclusion: The teacher will wrap up the introduction by encouraging the students to think about how they might use the concept of volume in their everyday lives. They will also remind the students that they will be applying their knowledge and skills in a hands-on activity, which will help them better understand and appreciate the concept of volume.
Development (20 - 25 minutes)
Activity 1: Building Volume Towers
Objective: The students will use unit cubes to build towers, then measure and calculate the volume of their towers.
Preparation (3 - 5 minutes):
The teacher will divide the class into groups of 4-5 students.
Each group will be provided with a set of unit cubes (around 20-30 cubes per group) and a ruler to help them measure.
The teacher will explain the task to the students: they must build a tower using the unit cubes and then calculate the volume of the tower.
Execution (10 - 12 minutes):
The groups will start building their towers, ensuring that each unit cube is stacked directly on top of another (to form a solid tower).
Once their tower is complete, the students will measure the length, width, and height of the tower using the ruler. They will then record these measurements.
The students will also count how many unit cubes were used to build the tower. They will record this count as well.
Measurement and Calculation (5 - 7 minutes):
The teacher will then guide the students in calculating the volume of the tower using the formula: volume = length x width x height.
The students will now calculate the volume of their towers using the recorded measurements. They will record this volume in their notebooks.
Discussion and Reflection (2 - 3 minutes):
Once all the groups have completed these steps, the teacher will facilitate a group discussion. Each group will be asked to share their tower's measurements and volume and explain how they calculated it.
The teacher will highlight the fact that all the groups' towers, regardless of their shape, have a volume that can be determined by counting the number of unit cubes.
Extension (Optional - 2 minutes):
If time allows, the teacher can encourage the students to compare the volumes of their towers. Are they the same? Why or why not? This will help the students to understand that volume is not dependent on the shape of an object, but on the amount of space it occupies.
Activity 2: Measuring Classroom Objects
Objective: The students will use unit cubes to measure and calculate the volume of various classroom objects.
Preparation (3 - 5 minutes):
The teacher will distribute unit cubes and rulers to every group.
The teacher will also provide a list of different objects in the classroom for the students to measure. This list could include a book, a pencil box, a desk, etc.
Execution (10 - 12 minutes):
The groups will select an object from the list and begin measuring it using the unit cubes. They will stack the unit cubes next to the object, trying to fit as many as possible in each dimension (length, width, and height) of the object.
Once they have finished, the students will then count the number of unit cubes used in each dimension and record these numbers.
The students will then multiply the counts from each dimension to calculate the volume of the object.
Discussion and Reflection (2 - 3 minutes):
After all the groups have had a chance to measure and calculate the volume of their objects, the teacher will facilitate a class discussion.
Each group will be asked to share the object they measured, its dimensions, and its calculated volume.
The teacher will highlight that the volume of an object in the classroom is the same regardless of how it is measured, reinforcing the concept that volume is a measure of the amount of space an object occupies.
Conclusion (8 - 10 minutes)
Summarizing Key Points (3 - 4 minutes):
The teacher will summarize the main concepts learned during the lesson. They will remind students that the volume is the amount of space occupied by a three-dimensional object and is measured in cubic units.
They will emphasize that the volume of any object can be found by counting the number of unit cubes that fill it.
The teacher will also recap the steps taken in the activities, reminding students how they used unit cubes to calculate the volume of different objects.
Connecting Theory, Practice, and Application (2 - 3 minutes):
The teacher will explain how the hands-on activities helped students to understand the theoretical concept of volume. They will highlight that by actually building towers and measuring objects, students were able to see how the concept of volume works in practice.
They will also discuss how the application of volume in real-world situations was demonstrated through the problem situations and the classroom object measurement activity.
The teacher will encourage students to think about other real-world situations where they could apply their knowledge of volume.
Additional Materials (2 - 3 minutes):
The teacher will suggest additional materials for students who want to further explore the concept of volume. They could recommend online games or interactive activities that involve calculating volume or building with unit cubes.
They could also suggest a homework assignment where students have to calculate the volume of objects at home using unit cubes or other everyday items.
The teacher will remind students that these additional materials are optional, but they can be a fun and educational way to reinforce what they learned in class.
Real-World Relevance (1 minute):
The teacher will conclude the lesson by emphasizing the importance of understanding volume in everyday life. They will remind students of the real-world applications they discussed at the beginning of the lesson, such as in architecture, engineering, and farming.
They will also point out that understanding volume can help them in practical tasks, like packing a suitcase or organizing a bookshelf. They will encourage students to be mindful of the space that objects occupy, and how they can use the concept of volume to make more efficient use of space.
By the end of the conclusion, students should have a clear understanding of the concept of volume, how it is measured using unit cubes, and its real-world applications. They should also feel confident in their ability to calculate volume and be motivated to continue exploring the topic.
Feedback (5 - 7 minutes)
Group Discussion (2 - 3 minutes):
The teacher will facilitate a group discussion, where each group will be given up to 2 minutes to share their findings and experiences from the activities. They will talk about the strategies they used to build their towers, measure classroom objects, and calculate their volumes.
The teacher will encourage students to actively listen to the other groups, ask questions, and make connections with their own experiences during the activities. This will foster a collaborative learning environment and promote the sharing of ideas and insights.
Linking Theory and Practice (1 - 2 minutes):
After all the groups have shared their experiences, the teacher will summarize the key points from the group discussions. They will highlight how the hands-on activities helped the students to understand the theoretical concept of volume.
The teacher will also discuss how the students applied their knowledge of multiplication and addition in the activities, linking the concept of volume to other mathematical skills. They will emphasize that these skills are not isolated, but interconnected and used together in real-world applications.
Reflection (2 - 3 minutes):
The teacher will then propose that the students take a moment to reflect on what they have learned. They will ask the students to think about the answers to the following questions:
What was the most important concept you learned today?
How did you use your mathematical skills in the activities?
Can you think of any other real-world applications for the concept of volume?
Feedback and Assessment (1 minute):
To conclude the lesson, the teacher will ask for a few volunteers to share their reflections with the class. They will also encourage the students to share any questions or concerns they might have about the lesson.
The teacher will provide immediate feedback on the students' understanding of the concept of volume based on their participation in the activities and the group discussions. They will also take note of any misconceptions or areas of difficulty that were identified during the lesson, which can be addressed in future lessons.
The teacher will remind the students that learning is a process and it's okay to make mistakes and ask questions. They will encourage the students to continue practicing and exploring the concept of volume in their own time.
By the end of the feedback stage, the students should have a clear understanding of the concept of volume, how it is measured using unit cubes, and its real-world applications. They should also feel confident in their ability to calculate volume and be motivated to continue exploring the topic. The teacher should have a good sense of the students' understanding and any areas that may need further clarification or reinforcement in future lessons.
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17452 | https://medlineplus.gov/genetics/condition/bernard-soulier-syndrome/ | Bernard-Soulier syndrome: MedlinePlus Genetics
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Bernard-Soulier syndrome
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Bernard-Soulier syndrome
Description
Collapse Section
Bernard-Soulier syndrome is a bleeding disorder associated with abnormal platelets, which are blood cells involved in blood clotting. In affected individuals, platelets are unusually large and fewer in number than usual (a combination known as macrothrombocytopenia). People with Bernard-Soulier syndrome tend to bruise easily and have an increased risk of nosebleeds (epistaxis). They may also experience abnormally heavy or prolonged bleeding following minor injury or surgery or even without trauma (spontaneous bleeding). Rarely, bleeding under the skin causes tiny red or purple spots on the skin called petechiae. Women with Bernard-Soulier syndrome often have heavy or prolonged menstrual bleeding (menorrhagia).
Frequency
Expand Section
Bernard-Soulier syndrome is estimated to occur in 1 in 1 million individuals; however, some doctors think the condition is underdiagnosed and may be more common.
Causes
Expand Section
Bernard-Soulier syndrome is caused by mutations in one of three genes: GP1BA, GP1BB, or GP9. The proteins produced from these genes are pieces (subunits) of a protein complex called glycoprotein (GP)Ib-IX-V. This complex is found on the surface of platelets and plays an important role in blood clotting.
The GPIb-IX-V complex can attach (bind) to a protein called von Willebrand factor, fitting together like a lock and its key. Von Willebrand factor is found on the inside surface of blood vessels, particularly when there is an injury. Binding of the GPIb-IX-V complex to von Willebrand factor allows platelets to stick to the blood vessel wall at the site of the injury. These platelets form clots, plugging holes in the blood vessels to help stop bleeding.
Most mutations in GP1BA, GP1BB, or GP9 prevent the formation of the GPIb-IX-V complex on the surface of platelets. Other mutations impair the complex's interaction with von Willebrand factor. All of these mutations impair clot formation, which leads to the excessive bleeding characteristic of Bernard-Soulier syndrome.
Learn more about the genes associated with Bernard-Soulier syndrome
Expand Section
GP1BA
GP1BB
GP9
Inheritance
Expand Section
Most cases of Bernard-Soulier syndrome are inherited in an autosomal recessive pattern, which means both copies of the GP1BA, GP1BB, or GP9 gene in each cell have mutations. The parents of an individual with an autosomal recessive condition each carry one copy of the mutated gene. Although most people with only one copy of the mutated gene do not show signs and symptoms of the condition, some have platelets that are slightly larger than normal or very mild bleeding abnormalities.
Rare cases of Bernard-Soulier syndrome caused by mutations in the GP1BA or GP1BB gene are inherited in an autosomal dominant pattern, which means one copy of the altered gene in each cell is sufficient to cause the disorder. These individuals inherit the condition from an affected parent.
Other Names for This Condition
Expand Section
BDPLT1
Bleeding disorder, platelet-type, 1
BSS
Deficiency of platelet glycoprotein 1b
Giant platelet syndrome
Glycoprotein Ib, platelet, deficiency of
Hemorrhagioparous thrombocytic dystrophy
Macrothrombocytopenia, familial Bernard-Soulier type
Platelet glycoprotein Ib deficiency
Von Willebrand factor receptor deficiency
Additional Information & Resources
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Genetic Testing Information
Genetic Testing Registry: Bernard Soulier syndrome
Genetic and Rare Diseases Information Center
Bernard-Soulier syndrome
Patient Support and Advocacy Resources
National Organization for Rare Disorders (NORD)
Catalog of Genes and Diseases from OMIM
BERNARD-SOULIER SYNDROME, TYPE A2, AUTOSOMAL DOMINANT; BSSA2
BERNARD-SOULIER SYNDROME; BSS
Scientific Articles on PubMed
PubMed
References
Expand Section
Andrews RK, Berndt MC. Bernard-Soulier syndrome: an update. Semin Thromb Hemost. 2013 Sep;39(6):656-62. doi: 10.1055/s-0033-1353390. Epub 2013 Aug 8. Citation on PubMed
Bragadottir G, Birgisdottir ER, Gudmundsdottir BR, Hilmarsdottir B, Vidarsson B, Magnusson MK, Larsen OH, Sorensen B, Ingerslev J, Onundarson PT. Clinical phenotype in heterozygote and biallelic Bernard-Soulier syndrome--a case control study. Am J Hematol. 2015 Feb;90(2):149-55. doi: 10.1002/ajh.23891. Epub 2014 Nov 24. Citation on PubMed
Cauwenberghs N, Vanhoorelbeke K, Vauterin S, Deckmyn H. Structural determinants within platelet glycoprotein Ibalpha involved in its binding to von Willebrand factor. Platelets. 2000 Nov;11(7):373-8. doi: 10.1080/09537100020019157. Citation on PubMed
Li R, Emsley J. The organizing principle of the platelet glycoprotein Ib-IX-V complex. J Thromb Haemost. 2013 Apr;11(4):605-14. doi: 10.1111/jth.12144. Citation on PubMed or Free article on PubMed Central
Savoia A, Kunishima S, De Rocco D, Zieger B, Rand ML, Pujol-Moix N, Caliskan U, Tokgoz H, Pecci A, Noris P, Srivastava A, Ward C, Morel-Kopp MC, Alessi MC, Bellucci S, Beurrier P, de Maistre E, Favier R, Hezard N, Hurtaud-Roux MF, Latger-Cannard V, Lavenu-Bombled C, Proulle V, Meunier S, Negrier C, Nurden A, Randrianaivo H, Fabris F, Platokouki H, Rosenberg N, HadjKacem B, Heller PG, Karimi M, Balduini CL, Pastore A, Lanza F. Spectrum of the mutations in Bernard-Soulier syndrome. Hum Mutat. 2014 Sep;35(9):1033-45. doi: 10.1002/humu.22607. Epub 2014 Jul 15. Citation on PubMed
Savoia A, Pastore A, De Rocco D, Civaschi E, Di Stazio M, Bottega R, Melazzini F, Bozzi V, Pecci A, Magrin S, Balduini CL, Noris P. Clinical and genetic aspects of Bernard-Soulier syndrome: searching for genotype/phenotype correlations. Haematologica. 2011 Mar;96(3):417-23. doi: 10.3324/haematol.2010.032631. Epub 2010 Dec 20. Citation on PubMed or Free article on PubMed Central
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Related Health Topics
Bleeding Disorders
Genetic Disorders
Platelet Disorders
MEDICAL ENCYCLOPEDIA
Blood clotting
Genetics
Understanding Genetics
What is the prognosis of a genetic condition?
How can gene variants affect health and development?
What does it mean if a disorder seems to run in my family?
What are the different ways a genetic condition can be inherited?
How are genetic conditions treated or managed?
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Last updated January 1, 2020
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Blood clot formation
Credit: Alila Medical Media/Shutterstock.com
Use of illustrations and other content
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Autosomal recessive inheritance
Credit: U.S. National Library of Medicine
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Autosomal dominant inheritance
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Blood clot formation
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17453 | https://math.stackexchange.com/questions/2062874/taking-the-derivative-of-square-root-of-y-by-squaring-the-equation-instead-of-us | Taking the derivative of square root of y by squaring the equation instead of using implicit differentiation - Mathematics Stack Exchange
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Taking the derivative of square root of y by squaring the equation instead of using implicit differentiation
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For an equation like:
y√+x−−√=1 y+x=1
Why can't I isolate the "square root of y y" and then raise both sides of the equation by the power of 2 2? That will get y y outside of the radical, but for some reason the result is not the same as with using implicit differentiation. Why is that?
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edited Dec 18, 2016 at 0:08
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With implicit differentiation we get
y′2 y√+1 2 x−−√=0 y′2 y+1 2 x=0
so
y′=−y√x−−√=x−−√−1 x−−√=1−1 x−−√y′=−y x=x−1 x=1−1 x
By doing y=(1−x−−√)2=1−2 x−−√+x y=(1−x)2=1−2 x+x we get
y′=−1 x−−√+1 y′=−1 x+1
as before.
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answered Dec 18, 2016 at 0:11
egregegreg
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Oh, so it's just a matter of algebra manipulation :| Thank you!!Pineapple29 –Pineapple29 2016-12-18 00:13:10 +00:00 Commented Dec 18, 2016 at 0:13
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17454 | https://english.stackexchange.com/questions/93986/to-advocate-the-act-of-not-doing-something | single word requests - To advocate the act of not doing something - English Language & Usage Stack Exchange
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To advocate the act of not doing something
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I'm looking for a word that gives the same meaning as advocate below, but with a negative twist:
He advocates not doing drugs.
I want to present this better, and emphasize that "He" stands behind (i.e. advocates) not doing drugs.
This sentence, for example, does not work:
He recommends not doing drugs.
Edit: To be clearer, ideally it would read as if he physically worked against it, like (I'd say) advocate implies.
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edited Dec 11, 2012 at 12:12
Henning KlevjerHenning Klevjer
asked Dec 11, 2012 at 11:56
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Anti-drug activist? Autoresponder –Autoresponder 2012-12-11 12:31:46 +00:00 Commented Dec 11, 2012 at 12:31
You can also champion a cause you believe in.Autoresponder –Autoresponder 2012-12-11 12:34:39 +00:00 Commented Dec 11, 2012 at 12:34
1 Not doing something is not “an act”. I think you must mean “To advocate against doing something”. So are you looking for some synonym of “advocate against” then?tchrist –tchrist♦ 2012-12-11 14:55:37 +00:00 Commented Dec 11, 2012 at 14:55
Okay, this is awkward. My brain thought that advocate against couldn't possibly work in this context. I'll go have a coffee now. Thanks everybody.Henning Klevjer –Henning Klevjer 2012-12-11 15:19:03 +00:00 Commented Dec 11, 2012 at 15:19
To me, the answer would be "discourage" (i.e., He discourages doing drugs.).Benjamin Harman –Benjamin Harman 2019-09-02 23:34:06 +00:00 Commented Sep 2, 2019 at 23:34
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If he actively talks people out of drugs, or similarly works towards the cause as you seem to imply, continue to use the word advocate with the negation following it:
He advocates against doing drugs.
While on the topic, a different phrase set may also work well. 'Doing drugs' is informal/ slang; drug abuse might be more formal.
He advocates against drug abuse.
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answered Dec 11, 2012 at 14:52
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How about "condemn"? "Proscribe"? "Denounce"?
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edited Dec 11, 2012 at 12:51
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He opposes the use of drugs.
He disapproves of drug use.
UPDATE
He campaigns against drug use.
He discourages against the use of drugs.
He is a vocal opponent of drug use.
He is critical of those who use drugs.
He fights against drug use.
He is an advocate of the War on Drugs.
He is a champion of the anti-drug movement.
UPDATE
He advocates drug abstinence.
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edited Dec 11, 2012 at 16:47
answered Dec 11, 2012 at 12:05
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I think I was a little unclear above. Advocating something suggests that the advocate is actually working for what he advocates. My "negative advocate" would therefore work against. So disapproval and opposition is a little more passive than what is my goal.Henning Klevjer –Henning Klevjer 2012-12-11 12:24:19 +00:00 Commented Dec 11, 2012 at 12:24
Advocates drug abstinence?TechWriter James –TechWriter James 2012-12-11 16:14:54 +00:00 Commented Dec 11, 2012 at 16:14
I think very few people would include the word against in "He discourages against the use of drugs". I've never heard that usage before, but it sounds a bit archaic.FumbleFingers –FumbleFingers 2012-12-12 00:21:36 +00:00 Commented Dec 12, 2012 at 0:21
@FumbleFingers 23,800 matches on google for "Discourage against" and 6,170 for "Discourages against". Doesn't sound archaic to me, perhaps its more common in certain regions.Baz –Baz 2012-12-12 09:25:59 +00:00 Commented Dec 12, 2012 at 9:25
@Baz: If it doesn't sound at least a bit odd to you I suppose very likely there's a regional difference. I was overstating when I said I've never heard that usage before - I think I probably just ignore what I see as the superfluous word. But Google Books has 2470 hits for "discouraged having", and only 3 for "discouraged against having", which I think clearly indicates that most people don't use it.FumbleFingers –FumbleFingers 2012-12-12 16:01:12 +00:00 Commented Dec 12, 2012 at 16:01
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17455 | https://arxiv.org/abs/1909.04873 | [1909.04873] Extremal graphs with local covering conditions
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[Submitted on 11 Sep 2019 (v1), last revised 6 Mar 2020 (this version, v2)]
Title:Extremal graphs with local covering conditions
Authors:Debsoumya Chakraborti, Po-Shen Loh
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Abstract:We systematically study a natural problem in extremal graph theory, to minimize the number of edges in a graph with a fixed number of vertices, subject to a certain local condition: each vertex must be in a copy of a fixed graph $H$. We completely solve this problem when $H$ is a clique, as well as more generally when $H$ is any regular graph with degree at least about half its number of vertices. We also characterize the extremal graphs when $H$ is an Erdős-Rényi random graph. The extremal structures turn out to have the similar form as the conjectured extremal structures for a well-studied but elusive problem of similar flavor with local constraints: to maximize the number of copies of a fixed clique in graphs in which all degrees have a fixed upper bound.
Comments:Minor changes reflecting comments from referees
Subjects:Combinatorics (math.CO)
Cite as:arXiv:1909.04873 [math.CO]
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17456 | https://brainly.com/question/21255540 | [FREE] In a classroom of 30 students, where 18 are girls and 12 are boys, what is the probability that two boys - brainly.com
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In a classroom of 30 students, where 18 are girls and 12 are boys, what is the probability that two boys are selected without replacement by the teacher?
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Probabilities are used to determine the chances of events
The probability that two boys are selected is 0.1517
The given parameters are:
Boys = 12
Girls = 18
Total = 30
Given that the selection is without replacement, the probability that two boys are selected is:
p = 12/30 11/29
So, we have:
p = 0.1517
Hence, the probability that two boys are selected is 0.1517
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The probability of selecting two boys without replacement from a class of 30 students (12 boys and 18 girls) is approximately 0.1517. This is calculated by multiplying the probability of selecting the first boy (12/30) and the second boy (11/29). Thus, P(Boys) = 12/30 11/29 = 0.1517.
Explanation
To find the probability of selecting two boys from a classroom of 30 students, where there are 12 boys and 18 girls, we will use the conditional probability formula since the selection is without replacement.
Understanding the Scenario:
Total students = 30
Boys = 12
Girls = 18
First Selection:
The probability of selecting the first boy from the class is given by the ratio of boys to total students:
P(B 1)=30 12
Second Selection:
After selecting the first boy, there will be 11 boys left and the total number of students will reduce to 29.
Therefore, the probability of selecting a second boy is:
P(B 2∣B 1)=29 11
Calculating the Combined Probability:
The combined probability of both events happening (selecting two boys without replacement) is calculated as follows:
P(B 1and B 2)=P(B 1)×P(B 2∣B 1)=30 12×29 11
Performing the Calculation:
Now, we will compute the values:
P(B 1and B 2)=30 12⋅29 11=870 132=72.5 11≈0.1517
Therefore, the probability that the teacher selects two boys without replacement is approximately 0.1517.
Examples & Evidence
An example of this in real life could be drawing names from a jar where you have 12 slips of paper with boys' names and 18 with girls' names, aiming to find the chance of picking two boys in a row without replacing the first slip back into the jar.
Using the formula for conditional probability and the steps outlined, we can see the mathematical reasoning leading to the final probability calculation, which aligns with standard probability concepts.
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Community Answer 5.0 2 There are 6 boys and 8 girls in a classroom. What is the probability that the teacher randomly selects a boy
Community Answer In a class, there are 18 girls and 14 boys. If the teacher selects two students at random to attend a party with the principal, what is the probability that the two students are the opposite gender?
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Calculate the discriminant of −2 x 2+7 x+3 and hence state the number of real roots of the equation −2 x 2+7 x+3=0.
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17457 | https://brainly.com/question/35373891 | [FREE] The graph of the equation x^2 = 4py is a parabola with focus F(x, y) = ??? and directrix y= ??? . So the - brainly.com
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The graph of the equation x^2 = 4py is a parabola with focus F(x, y) = ??? and directrix y= ??? . So the graph of x2 = 8y is a parabola with focus F(x, y) = and directrix y = -2
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The focus of the parabola x^2 = 4py is (0, 1/4) and the directrix is y = -1/4.
Explanation
The equation x^2 = 4py represents a parabola. To find the focus and directrix of the parabola, we can compare it to the general equation of a parabola, which is y^2 = 4px. We can see that the coefficients a and b in the general equation are 1 and 4p respectively. By comparing it to x^2 = 4py, we can conclude that p = 1/4. Therefore, the focus of the parabola is (0, 1/4) and the directrix is y = -1/4.
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The focus of the parabola described by x 2=8 y is (0,2), and the directrix is the line y=−2.
Explanation
To find the focus and directrix of the parabola represented by the equation x 2=4 p y, we first identify the parameter p. The general form of a parabola that opens upwards is x 2=4 p y, where (0,p) is the coordinates of the focus and the directrix is given by the line y=−p.
For the equation x 2=8 y, we see that 4 p=8, thus p=2. This gives us:
Focus: The coordinates of the focus are given by F(0,p)=(0,2).
Directrix: The directrix is at the line y=−p=−2.
So for the parabola described by x 2=8 y, we can summarize:
Focus: (0,2)
Directrix: y=−2.
Examples & Evidence
For example, if the parabola were x 2=12 y, then p=3, resulting in a focus at (0,3) and a directrix at y=−3.
The focus and directrix properties of parabolas are well-established in geometry, where the focus is a point inside the parabola and the directrix is a line outside the curve, ensuring that any point on the parabola is equidistant from both.
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You've already mentioned the directrix, so the focus for the parabola x 2=8 y is F(0,2).
Explanation
The equation x 2=4 p y represents a parabola that opens upward if p>0 and downward if p<0.
For a parabola that opens upward:
The focus is at the point F(0,p).
The directrix is the horizontal line y=−p.
Given the equation x 2=8 y, we can compare it to the general form x 2=4 p y:
From x 2=8 y, we can deduce that 4 p=8, which gives p=2.
Using the formulas for the focus and directrix:
Focus F(0,p)=F(0,2)
Directrix y=−p=−2
So, for the equation x 2=8 y:
The focus is F(0,2).
The directrix is y=−2.
You've already mentioned the directrix, so the focus for the parabola x 2=8 y is F(0,2).
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Community Answer (a) Find the focus, directrix, and focal diameter of the parabola. (b) Sketch a graph of the parabola and its directrix. 11. x^{2}=8 y
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Community Answer [Find the equation of the parabola with focus (x,y)=(−8,9) and directrix the x-axis. -11 Points Find the equation of the parabola given information about its graph. vertex is (0,0); directrix is y=6, focus is (0,−6)
Community Answer Find the vertex, focus, and directrix of the parabola. 3x^2 + 8y = 0 vertex (x, y) = focus (x, y) = directrix 2) Find the vertex, focus, and directrix of the parabola. (x + 1)^2 = 12(y − 1) vertex (x, y) = focus (x, y) = directrix 3)Find the vertex, focus, and directrix of the parabola. y2 + 6y + 2x + 1 = 0 vertex (x, y) = focus (x, y) = directrix
Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer
Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)?
Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD
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Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish?
New questions in Mathematics
Decrease sh. 4000 by 12 2 1%.
A function h is defined as h(x)=(x−1)(x 2−4)(x+1)3(x−2)2. Determine the location of all vertical asymptotes.
Decrease sh. 4000 by 12 2 1%
The rational function g is defined as g(x)=(x+3)3(x−4)2(x+1)(x+3)2(x 2−16)2(x 2−1). Determine the location of the holes for g(x). A. x=−4 and x=1 B. x=−4 and x=−3 C. x=−3 and x=4 D. x=−1 and x=4
Wykaż, że jeżeli A=3 4 2+2 i B=3 2 2+3, to B=9 A
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17458 | https://fiveable.me/key-terms/elementary-algebra/distance-rate-time-problems | Distance, Rate, and Time Problems - (Elementary Algebra) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Distance, Rate, and Time Problems
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key term - Distance, Rate, and Time Problems
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Definition
Distance, rate, and time problems are a type of mathematical word problem that involve the relationships between the distance traveled, the rate of travel, and the time taken to travel that distance. These problems are commonly encountered in various contexts, including transportation, physics, and everyday life situations.
5 Must Know Facts For Your Next Test
Distance, rate, and time problems can be solved using the formula: distance = rate × time.
These problems often require the identification of the given information and the unknown variable to be solved for.
Solving distance, rate, and time problems may involve unit conversions to ensure consistent units throughout the calculation.
The ability to set up and solve these types of problems is crucial for understanding and applying concepts in various fields, such as physics, engineering, and everyday life situations.
Proficiency in solving distance, rate, and time problems can help develop problem-solving skills, critical thinking, and mathematical reasoning abilities.
Review Questions
Explain the relationship between distance, rate, and time in the context of a distance, rate, and time problem.
The relationship between distance, rate, and time in the context of these problems is represented by the formula: distance = rate × time. This means that the distance traveled is directly proportional to the rate of travel and the time taken to travel that distance. If any two of these variables are known, the third can be calculated using this formula. Understanding this fundamental relationship is essential for setting up and solving distance, rate, and time problems effectively.
Describe the steps involved in solving a distance, rate, and time problem.
To solve a distance, rate, and time problem, the key steps are: 1) Identify the given information, including the known values for distance, rate, or time. 2) Determine the unknown variable that needs to be solved for. 3) Set up the equation using the formula: distance = rate × time. 4) Substitute the known values into the equation and solve for the unknown variable. 5) Ensure that the units are consistent throughout the calculation and provide the final answer with the appropriate units.
Analyze how the ability to solve distance, rate, and time problems can be applied in various real-world situations.
The ability to solve distance, rate, and time problems has widespread applications in various real-world situations. In transportation, these problems can be used to calculate travel times, fuel efficiency, and optimal routes. In physics, they are essential for understanding and analyzing motion, velocity, and acceleration. In everyday life, these skills can be applied to planning trips, estimating travel times, and understanding the relationships between distance, speed, and time. Mastering the concepts and problem-solving techniques for distance, rate, and time problems can enhance one's problem-solving abilities, critical thinking, and decision-making in a wide range of contexts.
Related terms
Distance:
The length of the path or space between two points, usually measured in units such as miles, kilometers, or meters.
Rate:
The speed or pace at which something occurs or is done, often expressed as a quantity per unit of time, such as miles per hour or meters per second.
Time:
The continuous progression of events from the past through the present to the future, measured in units such as seconds, minutes, hours, or days.
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17459 | https://www.quora.com/How-can-I-find-the-sum-of-all-odd-numbers-that-lie-between-1-and-n-in-Python | How to find the sum of all odd numbers that lie between 1 and n in Python - Quora
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How can I find the sum of all odd numbers that lie between 1 and n in Python?
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Sidharth Ajithkumar
Worked at Studying
·6y
There are a lot of answers here which are looping through 1 till n and counting. However if the first number is 1, instead of checking each and every number we could just increment the number by 2.
Eg: say n is 10, the numbers would be 1,3,5,7,9
If we do like this we could reduce the number of iterations and we don't need to check if the number is odd.
Code is
n = int(input())
Sum = 0
for i in range(1,n+1,2):
Sum+=i
A shorter way using list comprehension.
n=int(input())
print(sum([i for i in range(1,n+1,2)]))
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Allwin Raju
Software Engineer at Genesys Telecommunications Laboratories (2022–present) · Author has 96 answers and 192K answer views
·4y
If you want a one-liner you can use this. What it does is
sum() method will take an iterable(in our case list) and add the elements inside the iterable.
Using list comprehension we are creating a list only with odd numbers and giving it to the sum() method.
result = sum([num for num in range(0, n) if num % 2 != 0])
print(result)
2500
Upvote ·
9 3
Naved Ansari
Associate Python Developer (2019–present)
·6y
Hi
The logic is simple:
find odd numbers between the range
Append them in a list or you can sum bg a single variable
Use sum() to sum all the elements of list
Here is the example:
n = int(input("Enter the end number : "))
oddsum = 0
for i in range(1, n+1):
if i % 2 != 0:
oddsum += i
print(oddsum)
Or
n = int(input("Enter the end number : "))
oddsum = []
for i in range(1, n+1):
if i % 2 != 0:
oddsum.append(i)
print(sum(oddsum))
Hope it helps
Thanks!
Upvote ·
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Related questions
More answers below
What is the sum of first n n odd numbers?
How can I write a function to find the sum of odd numbers from a list in Python, and when there is no odd number in the list then print 0?
What is the formula for the sum of n odd numbers?
In Python, how do I find the total of odd numbers from a list?
What is the sum of odd numbers from 1 to n in a block scheme?
Alon Amit
PhD in Mathematics; Mathcircler. · Upvoted by
Ellen Sassani
, M.A. Mathematics, University of South Florida and
Richard Shearer
, MSc (Biochemistry) Biochemistry & Mathematics, University of the Witwatersrand · Author has 8.7K answers and 172M answer views
·Updated 8y
Related
What is the sum of first n n odd numbers?
Let's look and see. Anyone can at least observe the first few instances, right?
1=1 1=1
1+3=4 1+3=4
1+3+5=9 1+3+5=9
1+3+5+7=16 1+3+5+7=16
1+3+5+7+9=25 1+3+5+7+9=25
Now, do you recognize the numbers on the right?
1,4,9,16,25,…1,4,9,16,25,…
Yes! Those are the perfect squares. 1×1,2×2,3×3,4×4 1×1,2×2,3×3,4×4 and so on.
We now have a conjecture. Let's put it to the test:
1+3+5+7+9+11=36 1+3+5+7+9+11=36
Yes! The six smallest odd numbers add up to 6 2 6 2, just as we had predicted. You can try a few more: it works.
If we are physicists, we stop here. We've made an observation, we formed a hypothesis, we tested our hypothesis experimentally once and twice and a hundred
Continue Reading
Let's look and see. Anyone can at least observe the first few instances, right?
1=1 1=1
1+3=4 1+3=4
1+3+5=9 1+3+5=9
1+3+5+7=16 1+3+5+7=16
1+3+5+7+9=25 1+3+5+7+9=25
Now, do you recognize the numbers on the right?
1,4,9,16,25,…1,4,9,16,25,…
Yes! Those are the perfect squares. 1×1,2×2,3×3,4×4 1×1,2×2,3×3,4×4 and so on.
We now have a conjecture. Let's put it to the test:
1+3+5+7+9+11=36 1+3+5+7+9+11=36
Yes! The six smallest odd numbers add up to 6 2 6 2, just as we had predicted. You can try a few more: it works.
If we are physicists, we stop here. We've made an observation, we formed a hypothesis, we tested our hypothesis experimentally once and twice and a hundred times, it always works, done. Our theory is correct until an experiment refutes it.
But we are mathematicians, aren't we. We require proof. And there are rigorous proofs aplenty of this nice little fact.
But there's also a crystal clear visual proof. Here it is:
EDIT: many people have asked for a rigorous proof. Here's a relatively simple one which can be derived from this visual proof.
We notice that the odd numbers are just the differences between consecutive squares, like so:
1=1 2−0 2 1=1 2−0 2
3=2 2−1 2 3=2 2−1 2
5=3 2−2 2 5=3 2−2 2
7=4 2−3 2 7=4 2−3 2
and so on. Therefore when we add them up, everything cancels out except the last square:
1+3+5+7=(1 2−0 2)+(2 2−1 2)+(3 2−2 2)+(4 2−3 2)=4 2 1+3+5+7=(1 2−0 2)+(2 2−1 2)+(3 2−2 2)+(4 2−3 2)=4 2
So now let's write this formally for any number of odd numbers being added up. For any k k,
2 k+1=(k+1)2−k 2 2 k+1=(k+1)2−k 2
and therefore the sum of the first n n odd numbers, which is
n−1∑k=0 2 k+1∑k=0 n−1 2 k+1
is equal to
n−1∑k=0(k+1)2−k 2=n∑k=1 k 2−n−1∑k=0 k 2=n 2∑k=0 n−1(k+1)2−k 2=∑k=1 n k 2−∑k=0 n−1 k 2=n 2. QED
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Ravi Singh
Technical Lead at Amazon (company) (2021–present)
·Updated May 13
Can I switch from TCS to companies like Amazon, Flipkart or Google?
YES, candidates can transition from service companies (such as TCS, Infosys) to top tech companies (like Amazon, Flipkart, or Google) 🌟 through a well-planned preparation strategy and regular practice. To achieve this, you need to be highly proficient in problem-solving and system design skills 💡.
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Structured preparation for problem-solving (DSA) is crucial, rather t
Continue Reading
YES, candidates can transition from service companies (such as TCS, Infosys) to top tech companies (like Amazon, Flipkart, or Google) 🌟 through a well-planned preparation strategy and regular practice. To achieve this, you need to be highly proficient in problem-solving and system design skills 💡.
Logicmojo is one of the best platforms for preparing for interviews at top product companies 🏆. It offers a structured, concise approach to mastering techniques for DSA (Coding Interview Preparation) and System Design (HLD + LLD).
Structured preparation for problem-solving (DSA) is crucial, rather than practicing problems randomly on platforms like LeetCode or GeeksforGeeks 📚. You should focus on learning problem-solving techniques and developing an intuition for solving DSA problems. Practicing random questions on LeetCode is not the correct strategy because you might forget the concepts 🤔. Instead, you should practice problems based on specific techniques and apply similar ones for reinforcement. But yes, self-preparation demands a lot of time ⏰. Most working professionals who are busy with 8-9 hours of daily office work rarely find the time to prepare for interviews.
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While you can't completely avoid recessions in IT, thorough preparation enables candidates to quickly pivot to another top tech company if impacted 💼.
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·
11mo
To find the sum of all odd numbers that lie between 1 and n n in Python, you can use a simple loop or a mathematical formula. Below are two methods: one using a loop and the other using a formula.
Method 1: Using a Loop
You can iterate through the numbers from 1 to n n and sum the odd numbers.
def sum_of_odds(n):
total = 0
for i in range(1, n + 1):
if i % 2 != 0: # Check if the number is odd
total += i
return total
Example usage
n = 10
print(sum_of_odds(n)) # Output: 25 (1 + 3 + 5 + 7 + 9)
Method 2: Using a Mathematical Formula
The sum of the first k k odd numbers is given by
Continue Reading
To find the sum of all odd numbers that lie between 1 and n n in Python, you can use a simple loop or a mathematical formula. Below are two methods: one using a loop and the other using a formula.
Method 1: Using a Loop
You can iterate through the numbers from 1 to n n and sum the odd numbers.
def sum_of_odds(n):
total = 0
for i in range(1, n + 1):
if i % 2 != 0: # Check if the number is odd
total += i
return total
Example usage
n = 10
print(sum_of_odds(n)) # Output: 25 (1 + 3 + 5 + 7 + 9)
Method 2: Using a Mathematical Formula
The sum of the first k k odd numbers is given by the formula k 2 k 2. If n n is odd, the largest odd number less than or equal to n n is n n itself. If n n is even, the largest odd number is n−1 n−1. You can calculate the number of odd numbers up to n n using integer division.
def sum_of_odds(n):
Adjust n to be the largest odd number <= n
if n % 2 == 0:
n -= 1
Number of odd numbers up to n
k = (n // 2) + 1
return k k # Sum of the first k odd numbers
Example usage
n = 10
print(sum_of_odds(n)) # Output: 25 (1 + 3 + 5 + 7 + 9)
Summary
Both methods will give you the same result. The loop method is straightforward and easy to understand, while the formula method is more efficient for larger values of n n. You can choose either based on your preference or the specific requirements of your application.
Upvote ·
Santhosh S R
BE in Electronics and Communication Engineering, GSKSJTI (Graduated 2017) · Author has 68 answers and 21.2K answer views
·6y
I will tell u logic/algorithm only, you will try your own.
First you should consider will be ‘N', if you're not consider it will looping over and halt your computer.
Consider/ write what should be N in variable and take temporary variable sum=0
Then looping over N
Inside loop if (n%2 !== 0) then sum += n[i] return sum
Else return 0
Upvote ·
Related questions
More answers below
What is the sum of even and odd numbers from 1 to 100?
What is the sum of all odd numbers between 1 and 10?
What is the sum of odd numbers from 1 to n?
Can anyone prove this expression: ∑n k=1(n−1 k−1)=∑n−1 k=0()∑k=1 n(n−1 k−1)=∑k=0 n−1()?
How do I write a program that will calculate the sum of odd numbers and sum of even numbers for numbers 1 to 10?
Shashidhar Chintu
Btech cse from CVSR College of Engineering (Graduated 2021)
·6y
N = int(input("enter the last number ")
Sum = 0
for i in range(0,n+1)
if(i%2!=0)
Sum=Sum + i
Print(Sum)
What we are doing here is first take the input in python if u want a number then u need to use int before input else it will be considered as a string
Then initialize sum=0
After that make a for loop and range is from 0 to ur last number
Check if that number is divisible by 2 or not if no then we can say it is an odd number
If it is an odd number then add that number to sum
Finally print sum
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Calvin L.
I learnt a bit of Python for fun. · Author has 10K answers and 2.1M answer views
·1y
n = int(input("n="))
oddsum = 0
for i in range(1,n+1):
if i%2!=0:
oddsum+=i
print(oddsum)
Upvote ·
Aditya Ghosh
6y
By Python, I assume you mean the programming language. Python is a very “english-like” language, which makes it my favourite.
I’ll be using the “for” loop and the following code will be a user interface. (Input and Output are variables, the values will change upon the user
maximum = int(input(" Please Enter the Maximum Value : "))
Oddtotal = 0
for number in range(1, maximum+1):
if(number % 2 != 0):
print("{0}".format(number))
Oddtotal = Oddtotal + number
print("The Sum of Odd Numbers from 1 to {0} = {1}".format(number, Oddtotal))
Peace
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99 71
Areeb Siddiqui
Programming enthusiast
·6y
I am just a beginner, So call me out if it's wrong:
sum = 0
for i in range(1, n):
if(i % 2 == 1):
sum += i
Return sum
Upvote ·
Hey Look
Former Systems Programmer, Chief Designer (1982–2021) · Author has 3.2K answers and 1.3M answer views
·2y
n = 55
flop = 1 if n > 0 else -1
odd_sum = sum(i for i in range(1,n+flop,2flop))
odd_sum
784
odd_sum = 0
for i in range(1,n+flop,2flop):
... odd_sum += i
...
odd_sum
784
Or if n is be negative:
n = -33
flop = 1 if n > 0 else -1
odd_sum = sum(i for i in range(1,n+flop,2flop))
odd_sum
-288
Upvote ·
Utkarsh Pandey
Studied at St Xavier's College Mumbai (Graduated 2022)
·4y
Related
How do I write an algorithm to find the sum of odd numbers from 1 to n?
A. Ask the user for n (input)
B. Define and initialize a counter variable as 0
int count=0
C. Define a loop (I'm using for loop here)
for(int i=1;i<=n;i++)
D. Inside the loop, check if i is divisible by 2.
If not, add i to the counter variable
if((i%2)!=0)
{
count+=i
}
E. End the loop
F. Display counter variable
G. End.
Thank you for reading.
Upvote ·
Jan M Savage
Senior Developer at NDA (currently writing Dart) · Author has 867 answers and 10.2M answer views
·Updated Feb 17
Related
How do you write a Python program to find the sum of the squares of first n odd natural number using a while loop?
You can do it like so:
But I suggest you adopt Raku. Here’s why:
No ifs, buts, whiles, or counters. :)
But this still far from impressive. Let’s try a somewhat more impressive one:
Assuming n is 10. We don’t even need to enforce laziness:
No other language can do it like this: clear, short, efficient, memorable, readable and elegant all at the same time!
But hold on, we’re not done with Raku yet. We can even do:
For the first 10 odd numbers.
Don’t even need map (we use say to print the results in the IDE, but it’s not necessary in Raku REPL):
Wait, we can shorten it more: the code speaks for itself.
[Ta
Continue Reading
You can do it like so:
But I suggest you adopt Raku. Here’s why:
No ifs, buts, whiles, or counters. :)
But this still far from impressive. Let’s try a somewhat more impressive one:
Assuming n is 10. We don’t even need to enforce laziness:
No other language can do it like this: clear, short, efficient, memorable, readable and elegant all at the same time!
But hold on, we’re not done with Raku yet. We can even do:
For the first 10 odd numbers.
Don’t even need map (we use say to print the results in the IDE, but it’s not necessary in Raku REPL):
Wait, we can shorten it more: the code speaks for itself.
[Take the sum of [+] of]
[odd numbers 1,3…infinity]
[the first 10 of them [^10]]
[then pass the result >> ]
[to the squarer .2 ]
Try beating that in any language.
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Related questions
What is the sum of first n n odd numbers?
How can I write a function to find the sum of odd numbers from a list in Python, and when there is no odd number in the list then print 0?
What is the formula for the sum of n odd numbers?
In Python, how do I find the total of odd numbers from a list?
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In Python, how do I find the total of odd numbers from a list?
What is the sum of odd numbers from 1 to n in a block scheme?
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17460 | https://www.math.uci.edu/~ndonalds/math180b/2pell.pdf | 2 Pell’s Equation 2.1 Square-triangular numbers and Convergents of Continued Fractions Square-triangular numbers are integers which are simultaneously: • Perfect squares: of the form n2 for some n ∈N; • Triangular: of the form m ∑ k=1 k = 1 2m(m + 1) for some m ∈N.
For example, 36 is a square-triangular number: To find all such, we need to solve the Diophantine equation 2n2 = m(m + 1). It can be seen that this is equivalent to solving the Diophantine equation x2 −2y2 = 1 where ( x = 2m + 1 y = 2n Here are the first few square-triangular numbers n2, and the corresponding m, n, x, y: n2 m n x y 1 1 1 3 2 36 8 6 17 12 1225 49 35 99 70 41616 288 204 577 408 You should notice something: the solutions (x, y) correspond to some of the convergents pn qn of the continued fraction representation of √ 2. Here is a table of the first ten convergents and the corre-sponding values of p2 n −2q2 n: n 1 2 3 4 5 6 7 8 9 10 pn qn 1 1 3 2 7 5 17 12 41 29 99 70 239 169 577 408 1393 985 3363 2378 p2 n −2q2 n −1 1 −1 1 −1 1 −1 1 −1 1 It appears that exactly half the convergents of √ 2 yield to a solution of x2 −2y2 = 1 and thus to a square-triangular number. This equation is the first in an important family: 1 Definition 2.1.
A Pell equation is a Diophantine equation of the form x2 −dy2 = 1 where d is an integer which is not a perfect square. Among all solutions, the fundamental solution is the pair (a, b) where both are positive and a, b are minimal.
Recall we want only integer solutions (x, y). To keep the following treatment clean, we will only consider solutions where both x and y are positive. Clearly (±x, ±y) will also be solutions.
The above discussion suggests that solutions to Pell’s equation should be some, but not all, of the convergents of √ d. This is indeed the case. We shall show more: that Pell’s equation has infinitely many solutions for any d and that these may be computed using a relatively simple procedure. It follows that there are infinitely many square-triangular numbers! For the present, we shall ignore the existence question and focus on how to find solutions. The first part of this is straightforward and follows from our discussion of Diophantine approximations.
Theorem 2.2.
If (x, y) is a solution to Pell’s equation x2 −dy2 = 1, then x y is a convergent of √ d.
Proof. From our discussion of Hurwitz’s Theorem, it is enough to show that any solution satisfies x y − √ d < 1 2y2 . This is straightforward by factorization: if x2 −dy2 = 1, then x2 y2 = d + 1 y2 > d = ⇒x y > √ d > 1 from which x y − √ d = 1 y x − √ dy = x2 −dy2 y x + √ dy = 1 y2 x y + √ d < 1 2y2√ d < 1 2y2 The Theorem says that we can find all solutions to Pell’s equation by hunting through the list of convergents. This may take a while. . .
Examples 2.3.
We’ve already seen that half of the convergents of √ 2 appear to yield solutions to x2 −2y2 = 1. Here’s what happens for the first few convergents of √ 7 and √ 13.
n 1 2 3 4 5 6 7 8 9 10 pn qn 2 1 3 1 5 2 8 3 37 14 45 17 82 31 127 48 590 223 717 271 p2 n −7q2 n −3 2 −3 1 −3 2 −3 1 −3 2 pn qn 3 1 4 1 7 2 11 3 18 5 119 33 137 38 256 71 393 109 649 180 p2 n −13q2 n −4 3 −3 4 −1 4 −3 3 −4 1 We had to go to the 10th convergent before finding a solution to x2 −13y2 = 1!
You should spot another pattern here: the values of p2 n −dq2 n are eventually periodic. This is a theorem, 2 though we won’t prove it: it relates to the eventual periodicity of the continued fraction of a quadratic irrational.
Exercises 1. A square lattice has 1189 dots on each side. If these dots are arranged into an equilat-eral triangular lattice, how many does are on each side of the triangle?
2. Check rigorously that the integer solutions to the Pell and square-triangular equations x2 − 2y2 = 1 and 2n2 = m(m + 1) correspond via x = 2m + 1 and y = 2n.
3.
(a) Create a table listing the fundamental solutions (x, y) to Pell’s equation x2 −dy2 = 1 for each non-square d ≤15.
(No working is necessary) (b) Write a computer program or learn how to use a computer algebra package to find a solu-tion to the Pell equation x2 −109y2 = 1.
2.2 New solutions from old Thankfully there is an easier way to generate solutions to Pell’s equation than calculating all the convergents and checking each. It still depends on knowing a solution, but once you have it the others are easy to find. Here is the process for the basic equation with d = 2.
• Start with the fundamental solution (x, y) = (3, 2).
• Suppose that (x, y) also solves the equation. Consider (3 + 2 √ 2)(x + y √ 2) = (3x + 4y) + (2x + 3y) √ 2 and observe that the coefficients give a new solution (3x + 4y, 2x + 3y): (3x + 4y)2 −2(2x + 3y)2 = 9x2 + 24xy + 16y2 −8x2 −24xy −18y2 = x2 −2y2 = 1 • By induction, if x + y √ 2 = (3 + 2 √ 2)n, then x2 −2y2 = 1. Start computing: (3 + 2 √ 2)2 = 17 + 12 √ 2 (3 + 2 √ 2)3 = 99 + 70 √ 2 (3 + 2 √ 2)4 = 577 + 408 √ 2 (3 + 2 √ 2)5 = 3363 + 2378 √ 2 .
.
.
We seem to be obtaining all the solutions as powers of the fundamental solution 3 + 2 √ 2. In-deed this is the case: Theorem 2.4.
(x, y) is a positive solution to Pell’s equation x2 −2y2 = 1 if and only x + y √ 2 = (3 + 2 √ 2)n for some n ∈N 3 Proof. As seen above, if (x, y) is a solution, then (3x + 4y, 2x + 3y) is another. We can think of this as a matrix equation: if (xn, yn) is a solution, then (xn+1, yn+1) is also a solution, where xn+1 yn+1 = 3 4 2 3 xn yn The crucial observation is that the matrix has determinant 1, so we can reverse the process in integers: xn yn = 3 −4 −2 3 xn+1 yn+1 = 3xn+1 −4yn+1 −2xn+1 + 3yn+1 We wish to establish the following; if (xn+1, yn+1) is a positive solution with xn+1 > 3, then (a) x2 n −2y2 n = 1; (b) xn and yn are positive; (c) xn < xn+1.
The first is trivial: just calculate. For (b), first observe that x2 n+1 > 2y2 n+1 = ⇒xn+1 > √ 2yn+1 = ⇒xn = 3xn+1 −4yn+1 > (3 √ 2 −4)yn+1 > 0 Now observe that yn > 0 ⇐ ⇒2 3xn+1 < yn+1 ⇐ ⇒8 9x2 n+1 < 2y2 n+1 = x2 n+1 −1 ⇐ ⇒xn+1 > 3 For (c), the positivity of yn guarantees xn+1 = 3xn + 4yn > 3xn > xn From (xn+1, yn+1) we can therefore produce a sequence of positive solutions where xn+1 > xn > xn−1 > xn−2 > · · · provided each xk is always larger than 3. An infinite decreasing sequence of positive integers is an absurdity: eventually some solution must satisfy 1 ≤x1 ≤3. Since the only solution with x = 1, 2 or 3 is the fundamental solution (3, 2), the proof is complete.
This type of proof is known as a descent argument, since one is repeatedly creating something smaller: these operate somewhat like a reverse induction and were popularized by Fermat. Nothing prevents us from continuing to descend from (x, y) = (3, 2), except that the solutions will no longer be posi-tive: indeed one obtains the sequence (1, 0), (3, −2), (17, −12), (99, −70), (577, −408), . . .
exactly the conjugates of the positive solutions.
The crucial thing that drives the argument is the presence of a matrix 3 4 2 3 with integer entries and determinant 1. Such matrices keep appearing in our discussions, and will continue to do so!
4 We can rephrase the result using matrix notation: first observe that the eigenvalues and eigenvec-tors1 of 3 4 2 3 are 3 ± 2 √ 2 and ± √ 2 1 . We can now diagonalize and easily exponentiate to obtain a simplified method of calculation: xn yn = 3 4 2 3 n 1 0 = √ 2 − √ 2 1 1 (3 + 2 √ 2)n 0 0 (3 −2 √ 2)n √ 2 − √ 2 1 1 −1 1 0 = 1 2 (3 + 2 √ 2)n + (3 −2 √ 2)n 1 2 √ 2((3 + 2 √ 2)n −(3 −2 √ 2)n) Since 3 −2 √ 2 ≈0.1715 is tiny, we can use the ceiling and floor functions to see that, xn = 1 2(3 + 2 √ 2)n yn = 1 2 √ 2 (3 + 2 √ 2)n = xn √ 2 This can easily be entered into your calculator. For example x7 = 114243, y7 = 80782.
We have now proved that there are infinitely many solutions to the equation x2 −2y2 = 1, and thus infinitely many square-triangular numbers. They get big rather quickly, indeed exponentially!
Can we apply the same trick with x2 −3y2 = 1? It is easy to see that (x, y) = (2, 1) is the smallest solution. We’d therefore like to claim that the nth positive solution (xn, yn) satisfies xn + √ 3yn = (2 + √ 3)n Can we make this work in general? Indeed we can!
Theorem 2.5.
Suppose d ∈N is not a perfect square. A pair (x, y) of positive integers solves the Pell equation x2 −dy2 = 1 if and only if ∃n ∈N such that x + y √ d = a + b √ d n where (a, b) is the fundamental solution. Moreover, such solutions may be computed using floors and ceilings: x = 1 2(a + b √ d)n y = 1 2 √ d (a + b √ d)n = x √ d (†) We’ll be able to give a very easy proof later in the term once we’ve developed some ring-theory. For now we give a sketch showing how to generalize the approach used for d = 2. Note particularly that the matrix in step 2 again has determinant 1...
1 3 4 2 3 √ 2 1 = (3 + 2 √ 2) √ 2 1 and 3 4 2 3 − √ 2 1 = (3 −2 √ 2) − √ 2 1 5 Proof (Sketch).
1. If (x, y) solves x2 −dy2 = 1, then so does (ax + dby, ay + bx). This follows from (a + √ db)(x + √ dy) = (ax + dby) + √ d(ay + bx) 2. We therefore have a sequence (xn, yn) of solutions satisfying: xn yn = a bd b a n 1 0 (∗) 3. To see that all solutions have this form, suppose that (xn+1, yn+1) is a solution with xn+1 > a, and reverse the process to define (xn, yn) via xn yn = a bd b a −1 xn+1 yn+1 = axn+1 −bdyn+1 ayn+1 −bxn+1 Now check the following: (a) x2 n −dy2 n = 1; (b) xn, yn ∈N; (c) xn < xn+1.
4. By a descent argument, we must eventually produce the fundamental solution (a, b).
5. Diagonalizing the matrix in (∗) results in (†).
Exercises 1.
(a) If (x0, y0) is an integer solution to x2 −dy2 = −1, show that (x2 0 + dy2 0, 2x0y0) solves Pell’s equation x2 −dy2 = 1.
(b) Find a solution to x2 −41y2 = −1 by plugging in y = 1, 2, 3, . . . until you find a value for which 41y2 −1 is a perfect square. Use this to find a solution to x2 −41y2 = 1.
2. If (x0, y0) is a solution to x2 −dy2 = m, and if (x1, y1) is a solution to x2 −dy2 = 1, show that (x0x1 + dy0y1, x0y1 + y0x1) is also a solution to x2 −dy2 = m. Use this to find three solutions in positive integers to the equation x2 −2y2 = 7.
(Hint: Guess your first solution (x1, y1)!) 3. Consider the negative Pell equation x2 −dy2 = −1.
(a) Prove that if this has a solution in integers, then d is not divisible by 4 or any prime con-gruent to 3 modulo 4.
(Hint: recall quadratic residues from a previous class) (b) For each non-square integer d ≤15, determine whether the negative Pell equation x2 − dy2 = −1 has a solution. What are the next three (non-square) integers d are for which x2 −dy2 = −1 has a solution? Find a solution in each case.
4. For each of the following equations, either find a solution (x, y) in positive integers, or explain why no solution can exist.
(a) x2 −11y2 = 7 (b) x2 −11y2 = 433 (c) x2 −11y2 = 3 5. Provide arguements for parts (a), (b) and (c) in the proof of Theorem 2.5. Also explain why the floor and ceiling formulæ are correct.
6 2.3 Existence of a Solution to Pell’s Equation We know (Theorem 2.2) that every solution (x, y) to x2 −dy2 = 1 yields a convergent x y of √ d. We now see the converse: Theorem 2.6.
Pell’s equation x2 −dy2 = 1 has infinitely many solutions. More precisely, at least one convergent of √ d yields the fundamental solution, which generates all solutions by Theorem 2.5.
Here is the strategy for the proof.
1. First hunt for two pairs (x, y), (X, Y) for which x2 −dy2 = X2 −dY2 = m gives the same value.
2. Divide one by the other: x + √ dy X + √ dY = xX −dyY + √ d(yX −xY) X2 −dY2 = xX −dyY + √ d(yX −xY) m whence xX −dyY m 2 −d yX −xY m 2 = (x2 −dy2)(X2 −dY2) m2 = 1 3. We therefore have an integer solution to Pell’s equation if and only if xX −dyY m , yX −xY m ∈Z ⇐ ⇒ ( xX ≡dyY (mod |m|) yX ≡xY (mod |m|) Certainly if x ≡X and y ≡Y (mod |m|) then we’re done.
Example 2.7.
As a sanity check, recall that 452 −7 · 172 = 2 = 32 −7 · 12 and that 45 ≡5 and 17 ≡1 (mod 2). We check 45 + 17 √ 7 3 + √ 7 = 45 · 3 −7 · 17 + (17 · 3 −45) √ 7 32 −7 = 16 + 6 √ 7 2 = 8 + 3 √ 7 Indeed (8, 3) is a solution: 82 −7 · 32 = 1.
The hole in the strategy is step 1! We need to show the existence of pairs (x, y) and (X, Y) such that x2 −dy2 = X2 −dY2 and for which x ≡X and y ≡Y modulo m := x2 −dy2. For this we’ll invoke the box/pigeonhole principle by placing the infinite set of convergents into finitely many boxes: we first need to create the boxes. . .
Lemma 2.8.
If x y is a convergent of √ d, then 0 < x + y √ d < 3y √ d and so x2 −dy2 < 3 √ d The proof is a simple exercise. We’re now in a position to complete the main result.
7 Proof of Theorem. If x y is a convergent of √ d, the Lemma tells us that x2 −dy2 is one of the finitely many integers in the interval (−3 √ d, 3 √ d). Since there are infinitely many convergents, the box principle says that at least one such integer is attained infinitely many times: call this m.
We now have infinitely many pairs of solutions (xi, yi) to an equation x2 −dy2 = m. Modulo m, there are only m2 distinct pairs of integers. A second application of the box principle says that there must be at least two pairs (indeed infinitely many!) which are mutually congruent.
By the above algebraic steps, we can find at least one solution to Pell’s equation in positive integers.
By well-ordering, there is a minimal such: this is the fundamental solution.
The fundamental solution to Pell’s equation can be pinned down more accurately. Recall that the sequence of quotients in the continued fraction for √ d is eventually periodic √ d = [c1; . . . , ck, e1, . . . , el, e1, . . . , el, . . .] It can be shown that the convergent pn qn with n = k + l −1 provides the fundamental solution to either x2 −dy2 = ±1. If the result is −1, then computing (pn + qn √ d)2 in the usual way2 solves x2 −dy2 = 1. The fundamental solution can therefore be very large when √ d has a long period.
Example 2.9.
√ 73 = [8; 1, 1, 5, 5, 1, 1, 16, 1, 1, 5, 5, 1, 1, 16, . . .] has k + l −1 = 1 + 7 −1 = 7. The seventh convergent is 1068 125 , but this produces 10682 −73 · 1252 = −1 The fundamental solution to x2 −73y2 = 1 comes from x + √ 73y = (1068 + 125 √ 73)2 = 2281249 + 267000 √ 73 which corresponds to the 1 + 14 −1 = 14th convergent. Phew!
Exercises 1. Look at the table of convergents of √ 7 in Examples 2.3.
(a) For each of the pairs (pn, qn) in the table which produce p2 n −7q2 n = −3, find their remain-ders modulo 3.
(b) Find the first two pairs in the table which are mutually congruent modulo 3. Labelling these (x, y) and (X, Y), compute x + √ 7y X + √ 7Y where the numerator is the larger of the two. Which solution to Pell’s equation do you obtain?
(c) Now try to be a bit sneakier: by allowing one or more of x, y, X, Y to be negative, see if you can produce the fundamental solution to Pell.
2. Prove Lemma 2.8.
(Hint: recall that every convergent x y satisfies x y − √ d < 1 y2 ) 2This is the n = k + 2l −1th convergent.
8 |
17461 | https://www.sciencedirect.com/science/article/pii/S037843200700379X | Role of Sertoli cell number and function on regulation of spermatogenesis - ScienceDirect
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Animal Reproduction Science
Volume 105, Issues 1–2, April 2008, Pages 23-51
Role of Sertoli cell number and function on regulation of spermatogenesis☆
Author links open overlay panel Larry Johnson a, Donald L.Thompson Jr.b, Dickson D.Varner c
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Abstract
Testicular function is under the control of expression and repression of several genes and gene products, and many of these works through Sertoli cells. The capability of Sertoli cells to regulate spermatogenesis is dependent on Sertoli cell functions and Sertoli cell number. Sertoli cell number has long been thought to be stable in adults with no proliferation of Sertoli cells once adult numbers have been reached. However, adult horses do not have stable Sertoli cell numbers, and new studies indicate that adult Sertoli cells can be made to re-enter mitotic phase under certain experimental conditions. This review discusses roles of Sertoli cells in regulation of spermatogenesis and methods for estimating the number of Sertoli cells, in a testis, that overcome the problems (assumptions) associated with the indented, pear-shaped of Sertoli cell nuclei which make it difficult to estimate the volume of individual nuclei. Using several approaches to overcome the problems associated with any one method, the horse is identified as a species in which Sertoli cell number is not fixed, but it fluctuates with season. In addition to Sertoli cell numbers, the functions of Sertoli cells that are very important in signaling and controlling spermatogenesis are discussed. Recent studies have shown that “post-mitotic terminally differentiated Sertoli cells” from adult animals could, under certain conditions, re-enter the cell division cycle. Can seasonal influences be a natural set of conditions to induce the Sertoli cells of the horse testis to seasonally re-enter the cell division cycle and explain the seasonal differences in Sertoli cell number as summarized in this review? Alternatively, can seasonal differences in Sertoli cell number reflect, in the horse to a greater extent, but in adults of most species, the presence of some mitotic-capable Sertoli cells in adults? In any case, both Sertoli cell number and function are important in regulation of spermatogenesis.
Section snippets
Introduction to Sertoli cell function and morphology
The Sertoli cells play a major role in regulation of spermatogenesis and altering rates of spermatozoa produced. Sertoli cell (Fig. 1) functions include providing structural support and nutrition to developing germ cells, phagocytosis of degenerating germ cells and residual bodies, release of spermatids at spermiation and production of a host of proteins that regulate and/or respond to pituitary hormone release and that influence mitotic activity of spermatogonia (Amann, 1970, Dym and Raj, 1977
Seminiferous tubules
Sertoli cells are one of the two somatic cell types of seminiferous tubules, and together with germ cells, constitute the seminiferous epithelium (Fig. 1). The seminiferous epithelium is surrounded by one or more layers of myoid cells: these somatic cells mark the outer limits of the seminiferous tubules. The peritubular myoid cells (myofibroblasts) border the outer edge of seminiferous epithelium and touch the basal lamina of spermatogonia and Sertoli cells. Myoid cells exhibit contractions
Spermatozoa
Although epididymal spermatozoa possess cytoplasmic droplets attached to them, which migrate down the middle piece to the junction between the principal piece and middle piece of the tail, the normal, mature, ejaculated spermatozoon has no such droplet or looses it soon after ejaculation. Comparing the ultrastructure of equine spermatozoa (Fig. 4; Johnson, 1991b, Johnson et al., 2001) to that first shown in drawings by Leewenhoek (who discovered spermatozoa after inventing the microscope), one
Sertoli cell number relates to magnitude of spermatogenesis
In both humans and horses, the number of Sertoli cells is related to the level of spermatogenesis as measured as daily sperm production per testis (Fig. 5; Johnson and Thompson, 1983, Johnson et al., 1984c, Johnson, 1986b). This relationship is higher for horses (r 2=0.68) than for humans (r 2=0.39). Sertoli cell number is also correlated with testicular weight (r 2=0.68) in the horse. It is not likely that Sertoli cell number in adults is related to the efficiency of spermatogenesis as has been
Sertoli cells and seminiferous tubules in non-conventional views
Using Nomarski optics (that allows interior views of cells) to examine 20 μm Epon sections of equine seminiferous tubules, the shapes and sizes of nuclei and cellular structures can be focused in the optical sections (Fig. 6; Johnson et al., 1990). All eight stages of seminiferous epithelium in the horse are characterized by different types of spermatogonia, spermatocyte phases, or developmental steps of spermatids (Amann, 1970, Swierstra et al., 1974, Johnson et al., 1990). The Sertoli cell is
Spermiogenesis: shaping spermatids into spermatozoa
As spermatids develop in close contact with the Sertoli cells, their nuclei become more spear-shaped and chromatin becomes more condensed (Fig. 8; Johnson, 1991b), the tail forms from the distal centriole and mitochondria move into their position in the middle piece of the spermatozoon. The Golgi apparatus produces the acrosome whose lysosomal contents are not unlike other cellular lysosomes, but they are released by a calcium influx across the plasma membrane not unlike induced secretion (as
Seasonal differences in characteristics of equine Sertoli cells
Histological comparisons of seminiferous epithelium between tissues taken during the breeding season and those taken in the non-breeding season of long-day breeders (North American horse of light-weight breeds), do not yield obvious differences in cell populations between seasons (Fig. 9; Johnson et al., 1991a). However, testicular weight is significantly higher in the natural breeding season of the horse (Table 1). Using homogenates of fixed testicular equine tissue (Fig. 10; which allow
Direct counts to calculate Sertoli cell numbers
In subsequent studies, the number of Sertoli cell nuclei was determined for a large number of horses at different ages during the breeding season and the non-breeding seasons (Johnson and Thompson, 1983, Johnson, 1986a, Johnson and Nguyen, 1986, Johnson and Tatum, 1989, Johnson et al., 1991a). Because of the large number of horses that needed to be assessed, the enumeration of Sertoli cell nuclei in homogenates of fixed testicular tissue (Fig. 10) was used as this provided a rapid evaluation.
Development of new methods to determine Sertoli cell number and function
Given the uniqueness of enumerating Sertoli cell nuclei in fixed testicular homogenates (Fig. 10) and that the finding that seasonal differences in their numbers, in adult horses, was inconsistent with the common dogma that the Sertoli cell population remains unchanged in adults, further methods were developed. These were developed to count the number of Sertoli cells in the testis, to confirm or refute the notion that Sertoli cell numbers are not stable in adult horses, as they appear to
Sertoli cell numbers fluctuate with season in a “dose-dependent” effect
If the number of Sertoli cells is different during the breeding and non-breeding seasons, one would anticipate intermediate values during the transitional periods from one season to the other or a “dose-dependent” effect of season on Sertoli cell number. Using large numbers of adult horses (Fig. 13), it was found that the number of Sertoli cells per horse was greater in the breeding season than the non-breeding season and that transitional periods between seasons yielded intermediate values
Initiation of spermatogenesis in colts
When 1–5 year-old horses (Fig. 14) were evaluated for seasonal differences in Sertoli cell number and for the timing of the initiation of spermatogenesis, it was found that seasonal differences in Sertoli cell number existed and the relationship between Sertoli cell number and level of spermatogenesis was established at an early age of pubertal horses (Fig. 14; Johnson et al., 1991b). As the horse reaches puberty, its small testis with dark shaded parenchyma locally develops a light shade as
Species comparisons in Sertoli cell numbers and function
Compared to other species, the horse has intermediate numbers of Sertoli cells per gram testicular parenchyma and numbers per testis (Fig. 17; Okwun et al., 1996, Johnson et al., 2001). However, the function of individual Sertoli cells, as measured by the ratio of germ cells (spermatocytes and especially round spermatids) per Sertoli cell, was greater in the horse than other species (Fig. 17) even though the horse exhibits considerable germ cell degeneration during spermatogenesis (Fig. 18).
Evidence of Sertoli cell proliferation in adults
Returning to the Wikipedia description of Sertoli cells, it states that the Sertoli cell number could be augmented in adults outside the body by experimentation. Hence, adult Sertoli cell numbers are generally stable, but they can be augmented in adults under certain conditions. This statement is supported by several lines of research indicating that the Sertoli cell population may not be numerically stable. Proliferation of Sertoli cells is influenced by FSH (Orth et al., 1988), thyroid
Implications for other species
Proliferation of Sertoli cells is greatly reduced in the adult compared to the early proliferation rates prior to puberty (Johnson et al., 1996a). The data described herein constitutes compelling evidence that the Sertoli cell population is augmented seasonally in the adult stallion and that Sertoli cell proliferation can be restarted by various treatments in rodents. Hence, proliferation of Sertoli cell number in adults may be a general phenomenon of mammals (albeit a greater extent in the
Acknowledgements
This project was funded in part by the Link Equine Endowment Fund, Texas A&M University, NSF GK12 grant 0338310, and NIH grants R25 ES 10735 and R25 RR022711.
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Local initiation of spermatogenesis in the horse
Biol. Reprod. (1995)
P.S. Cooke et al.
Early hypothyroidism in rats causes increased adult testis and reproductive organ size but does not change testosterone levels
Endocrinology (1991)
P.S. Cooke et al.
Triiodothyronine inhibits proliferation and stimulates differentiation of cultured neonatal Sertoli cells: possible mechanism for increased adult testis weight and sperm production induced by neonatal goitrogen treatment
Biol. Reprod. (1994)
K. De Gendt et al.
A Sertoli cell-selective knockout of the androgen receptor causes spermatogenic arrest in meiosis
Proc. Natl. Acad. Sci. USA (2004)
M. Dym et al.
Response of adult rat Sertoli cells and Leydig cells to depletion of luteinizing hormone and testosterone
Biol. Reprod. (1977)
D. Elliott
Pathways of post-transcriptional gene regulation in mammalian germ cell development
Cytogenet. Genome Res. (2003)
D.W. Fawcett
Ultrastructure and function of the Sertoli cell
L.A. Feig et al.
Sertoli cells contain a mitogenic polypeptide
Proc. Natl. Acad. Sci. USA (1980)
J.A. Hettle et al.
Rat testicular peritubular cells in culture secrete an inhibitor of plasminogen activator activity
Biol. Reprod. (1988)
R.W. Holdcraft et al.
Hormonal regulation of spermatogenesis
Int. J. Androl. (2004)
D.R. Holsberger et al.
Cell-cycle inhibitors p27Kip1 and p21Cip1 regulate murine Sertoli cell proliferation
Biol. Reprod. (2005)
M. Huleihel et al.
Regulation of spermatogenesis by paracrine/autocrine testicular factors
Asian J. Androl. (2004)
L. Johnson
A new approach to quantification of Sertoli cells that avoids problems associated with the irregular nuclear surface
Anat. Rec. (1986)
L. Johnson
Spermatogenesis and aging in the human
J. Androl. (1986)
L. Johnson
Seasonal differences in equine spermatocytogenesis
Biol. Reprod. (1991)
L. Johnson
Images of spermatogenesis in various species
Reproduction Laboratory Handbook
(2007)
L. Johnson et al.
Age-related changes in the Leydig cell population, seminiferous tubules, and sperm production in stallions
Biol. Reprod. (1981)
L. Johnson et al.
Annual cycle of the Sertoli cell population in adult stallions
J. Reprod. Fertil. (1986)
L. Johnson et al.
Temporal appearance of seasonal changes in numbers of Sertoli cells, Leydig cells, and germ cells in stallions
Biol. Reprod. (1989)
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Cited by (178)
Involvement of activating ERK1/2 through G protein coupled receptor 30 and estrogen receptor α/β in low doses of bisphenol A promoting growth of Sertoli TM4 cells
2014, Toxicology Letters Show abstract Sertoli cells play a pivotal role in supporting proliferation of germ cells and differentiation during spermatogenesis in mammals. Nanomolar concentrations of Bisphenol A (BPA) can significantly stimulate the proliferation of mouse immature Sertoli (TM4) cells. However, mechanisms by which BPA caused these effects were still unclear. In the present study, an inverse U-shaped curve was observed when treating TM4 cells with increasing doses of BPA: 1 to 10 nM BPA significantly stimulated the proliferation of TM4 cells and increased the proportion of cells in S phase; >1 μM BPA caused lesser proliferation of cells. Exposure of TM4 cells to G15 or ICI 182,780, which are specific antagonists of GPR30 and estrogen receptor α/β (ERα/β), respectively, abolished BPA-induced proliferation of cells, which suggests that both GPR30 and ERα/β were involved in the observed effects of BPA. Furthermore, exposure to BPA caused rapid (5 min) activation of ERK1/2 via both GPR30 and ERα/β. Blocking the GPR30/EGFR signal transduction pathway by antagonists suppressed both phosphorylation of ERK and BPA-induced cell proliferation. BPA up-regulated mRNA and protein expression of GPR30 in a concentration-dependent manner. In summary, the results reported here indicated that activating ERK1/2 through GPR30 and ERα/β is involved in low doses of BPA that promoted growth of Sertoli TM4 cells. The GPR30/EGFR/ERK signal is the downstream transduction pathway in BPA-induced proliferation of TM4 Sertoli cells.
### The blood-testis barrier: The junctional permeability, the proteins and the lipids
2011, Progress in Histochemistry and Cytochemistry Show abstract The elucidation of how individual components of the Sertoli cell junctional complexes form and are dismantled to allow not only individual cells but whole syncytia of germinal cells to migrate from the basal to the lumenal compartment of the seminiferous epithelium without causing a permeability leak in the blood-testis barrier is amongst the most enigmatic yet, challenging and timely questions in testicular physiology. The intriguing key event in this process is how the barrier modulates its permeability during the periods of formation and dismantling of individual Sertoli cell junctions. The purpose of this review is therefore to first provide a reliable account on the normal formation, maintenance and dismantling process of the Sertoli cells junctions, then to assess the influence of the expression of their individual proteins, of the cytoskeleton associated with the junctions, and of the lipid content in the seminiferous tubules on the regulation of the their permeability barrier function. To help focus on the formation and dismantling of the Sertoli cell junctions, several considerations are based on data gleaned not only from rodents but from seasonal breeders as well because these animal models are characterized by exhaustive periods of junction assembly during development and the onset of the seasonal re-initiation of spermatogenesis as well as by an extensive junction dismantling period at the beginning of testicular regression, something unavailable in normal physiological conditions in continual breeders. Thus, the modulation of the permeability barrier function of the Sertoli cell junctions is analyzed in the physiological context of the blood-epidydimis barrier and in particular of the blood-testis barrier rather than in the context of a detailed account of the molecular composition and signalisation pathways of cell junctions. Moreover, the considerations discussed in this review are based on measurements performed on seminiferous tubule-enriched fractions gleaned at regular time intervals during development and the annual reproductive cycle.
### Autophagy-related proteins are functionally active in human spermatozoa and may be involved in the regulation of cell survival and motility
2016, Scientific Reports
### The Sertoli cell: One hundred fifty years of beauty and plasticity
2016, Andrology
### Sertoli cell is a potential target for perfluorooctane sulfonate-induced reproductive dysfunction in male mice
2013, Toxicological Sciences
### Blood-tissue barriers morphofunctional and immunological aspects of the blood-testis and blood-epididymal barriers
2013, Advances in Experimental Medicine and Biology
View all citing articles on Scopus
☆
This paper is part of the special issue entitled “Understanding and Exploiting Spermatozoa – A Festschrift for Rupert P. Amann” guest edited by George E. Seidel Jr, James K. Graham and D.N. Rao Veeramachaneni.
View full text
Copyright © 2008 Published by Elsevier B.V.
Part of special issue
Special Issue: Understanding and Exploiting Spermatozoa – A Festschrift for Rupert P. Amann
Edited by
George E.Seidel Jr., James K.Graham, D.N.Rao Veeramachaneni
Other articles from this issue
Comparative reliability and sensitivity of different methods for assessing treatment effects on sperm production
April 2008 William E.Berndtson
### Maturing the sperm: Unique mechanisms for modifying integral proteins in the sperm plasma membrane
April 2008 Susan Ruth Marengo
### Variation in male sexual behavior
April 2008 Larry S.Katz
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17462 | https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:rational-exponents-radicals/x2f8bb11595b61c86:radicals/v/introduction-to-square-roots | Intro to square roots (video) | Radicals | Khan Academy
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Skip to lesson content
Algebra 1
Course: Algebra 1>Unit 11
Lesson 2: Radicals
Intro to square roots
Understanding square roots
Square roots
Square root of decimal
Roots of decimals & fractions
Intro to cube roots
Cube roots
5th roots
Higher order roots
Math>
Algebra 1>
Exponents & radicals>
Radicals
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Intro to square roots
CCSS.Math: 8.EE.A.2
Google Classroom
Microsoft Teams
About About this video Transcript
Learn about the square root symbol (the principal root) and what it means to find a square root. Also learn how to solve simple square root equations.
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Chhavie Ghildiyal 7 years ago Posted 7 years ago. Direct link to Chhavie Ghildiyal's post “If we consider square roo...” more If we consider square roots as real numbers then can it be further classified in both rational and irrational numbers? why we need negative root 9 = -3 as we can also write root 9= 3 as well as -3?
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Kim Seidel 7 years ago Posted 7 years ago. Direct link to Kim Seidel's post “Yes, square roots can cre...” more Yes, square roots can create 2 answers -- the positive (principal) root and the negative root. When you are working with square roots in an expression, you need to know which value you are expected to use. The default is the principal root. We only use the negative root when there is a minus in front of the radical.
For example:
8 + sqrt(9) = 11
8 - sqrt(9) = 5
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AbbyTabby 5 years ago Posted 5 years ago. Direct link to AbbyTabby's post “Isn't a negative square r...” more Isn't a negative square root an imaginary number?
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Kim Seidel 5 years ago Posted 5 years ago. Direct link to Kim Seidel's post “Only if the minus sign is...” more Only if the minus sign is inside the square root.
sqrt(-9) creates the complex number 3i
-sqrt(9) just equals -3.
Hope this helps.
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28flatlay 2 years ago Posted 2 years ago. Direct link to 28flatlay's post “When you get the negative...” more When you get the negative square root of 9 and get -3, how does it change to positive 9 when you square both sides?
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David Severin 2 years ago Posted 2 years ago. Direct link to David Severin's post “Are you asking what (-3)^...” more Are you asking what (-3)^2 is? Note that (-3)(-3) has two negatives, and this gives a positive answer.
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ronda.hammond 2 years ago Posted 2 years ago. Direct link to ronda.hammond's post “What if it is not a perfe...” more What if it is not a perfect square root?
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Tanner P 2 years ago Posted 2 years ago. Direct link to Tanner P's post “Then the result will be i...” more Then the result will be irrational. This means that the answer cannot be expressed exactly as a decimal or a fraction (of integers). Pi is another example of an irrational number.
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calebriverk 2 years ago Posted 2 years ago. Direct link to calebriverk's post “So √64 is 8? And √81 is 9...” more So √64 is 8? And √81 is 9?
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Kim Seidel 2 years ago Posted 2 years ago. Direct link to Kim Seidel's post “Yes, those are correct.” more Yes, those are correct.
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Σ-JH.9-Σ 9 months ago Posted 9 months ago. Direct link to Σ-JH.9-Σ's post “Whats the square root of ...” more Whats the square root of zero?
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Kim Seidel 9 months ago Posted 9 months ago. Direct link to Kim Seidel's post “It's 0 because 0^2 = 0” more It's 0 because 0^2 = 0
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Dilan Talasila 2 years ago Posted 2 years ago. Direct link to Dilan Talasila's post “what is the square root o...” more what is the square root of -4?
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Kim Seidel 2 years ago Posted 2 years ago. Direct link to Kim Seidel's post “Sqrt(-4) has no solution ...” more Sqrt(-4) has no solution within the set of real numbers. So, we say "there is no real solution".
The reason is that is you square any real number, you always get a positive number.
In higher level math, usually Algebra 2, you learn about the set of complex numbers. A solution for sqrt(-4) exists in the set of complex numbers.
Hope this helps.
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deekstaples 2 years ago Posted 2 years ago. Direct link to deekstaples's post “...... Is there a way to ...” more ...... Is there a way to bookmark lessons and practices to have quick refrence to?
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red-discipline 2 years ago Posted 2 years ago. Direct link to red-discipline's post “You can save lesson pages...” more You can save lesson pages as bookmarks in your browser, or use the google classroom button above the video to add it to your google class. You can then access them as a quick reference.
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Ishaan 2 years ago Posted 2 years ago. Direct link to Ishaan's post “cant you just do 9/3?” more cant you just do 9/3?
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RiverUnknown 2 years ago Posted 2 years ago. Direct link to RiverUnknown's post “For smaller numbers, like...” more For smaller numbers, like root 9, it’s simple to do that because 33 is an easy problem. However, as the number gets larger, this is impractical. Using trial and error to try to find the root of 544644 is ludicrous. If you happen to try 738 immediately, well, congratulations. You’ve gotten the correct answer. But that’s rather unlikely, so it’s best to use prime factorization.
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Cynthia Goodwyn 7 years ago Posted 7 years ago. Direct link to Cynthia Goodwyn's post “Why do numbers have both ...” more Why do numbers have both a positive and a negative square root?
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THE_ONE_AND_ONLY 5 years ago Posted 5 years ago. Direct link to THE_ONE_AND_ONLY's post “Square roots can be both ...” more Square roots can be both because the factors are the same number and same value, and also because positivepositive = positive squared and negativenegative = negative squared.
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Video transcript
[Voiceover] If you're watching a movie and someone is attempting to do fancy mathematics on a chalkboard, you'll almost always see a symbol that looks like this. This radical symbol. And this is used to show the square root and we'll see other types of roots as well, but your question is, well, what does this thing actually mean? And now that we know a little bit about exponents, we'll see that the square root symbol or the root symbol or the radical is not so hard to understand. So, let's start with an example. So, we know that three to the second power is what? Three squared is what? Well, that's the same thing as three times three and that's going to be equal to nine. But what if we went the other way around? What if we started with the nine, and we said, well, what times itself is equal to nine? We already know that answer is three, but how could we use a symbol that tells us that? So, as you can imagine, that symbol is going to be the radical here. So, we could write the square root of nine, and when you look at this way, you say, okay, what squared is equal to nine? And you would say, well, this is going to be equal to, this is going to be equal to, three. And I want you to really look at these two equations right over here, because this is the essence of the square root symbol. If you say the square root of nine, you're saying what times itself is equal to nine? And, well, that's going to be three. And three squared is equal to nine, I can do that again. I can do that many times. I can write four, four squared, is equal to 16. Well, what's the square root of 16 going to be? Well, it's going to be equal to four. Let me do it again. Actually, let me start with the square root. What is the square root of 25 going to be? Well, this is the number that times itself is going to be equal to 25 or the number, where if I were to square it, I'd get to 25. Well, what number is that, well, that's going to be equal to five. Why, because we know that five squared is equal to, five squared is equal to 25. Now, I know that there's a nagging feeling that some of you might be having, because if I were to take negative three, and square it, and square it I would also get positive nine, and the same thing if I were to take negative four and I were to square the whole thing, I would also get positive 16, or negative five, and if I square that I would also get positive 25. So, why couldn't this thing right over here, why can't this square root be positive three or negative three? Well, depending on who you talk to, that's actually a reasonable thing to think about. But when you see a radical symbol like this, people usually call this the principal root. Principal root. Principal, principal square root. Square root. And another way to think about it, it's the positive, this is going to be the positive square root. If someone wants the negative square root of nine, they might say something like this. They might say the negative, let me scroll up a little bit, they might say something like the negative square root of nine. Well, that's going to be equal to negative three. And what's interesting about this is, well, if you square both sides of this, of this equation, if you were to square both sides of this equation, what do you get? Well negative, anything negative squared becomes a positive. And then the square root of nine squared, well, that's just going to be nine. And on the right-hand side, negative three squared, well, negative three times negative three is positive nine. So, it all works out. Nine is equal, nine is equal to nine. And so this is an interesting thing, actually. Let me write this a little bit more algebraically now. If we were to write, if we were to write the principal root of nine is equal to x. This is, there's only one possible x here that satisfies it, because the standard convention, what most mathematicians have agreed to view this radical symbol as, is that this is a principal square root, this is the positive square root, so there's only one x here. There's only one x that would satisfy this, and that is x is equal to three. Now, if I were to write x squared is equal to nine, now, this is slightly different. X equals three definitely satisfies this. This could be x equals three, but the other thing, the other x that satisfies this is x could also be equal to negative three, 'cause negative three squared is also equal to nine. So, these two things, these two statements, are almost equivalent, although when you're looking at this one, there's two x's that satisfy this one, while there's only one x that satisfies this one, because this is a positive square root. If people wanted to write something equivalent where you would have two x's that could satisfy it, you might see something like this. Plus or minus square root of nine is equal to x, and now x could take on positive three or negative three.
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17463 | https://www.mathbootcamps.com/determine-vector-linear-combination-vectors/ | How to Determine if a Vector is a Linear Combination of Other Vectors
The idea of a linear combination of vectors is very important to the study of linear algebra. We can use linear combinations to understand spanning sets, the column space of a matrix, and a large number of other topics. One of the most useful skills when working with linear combinations is determining when one vector is a linear combination of a given set of vectors.
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Suppose that we have a vector →vv⃗ and we want to know the answer to the question “is →vv⃗ a linear combination of the vectors →a1a⃗ 1, →a2a⃗ 2, and →a3a⃗ 3?”. Using the definition of a linear combination of vectors, this question can be restated to the following:
Are there scalars x1x1, x2x2, and x3x3 such that:
→v=x1→a1+x2→a2+x3→a3v⃗ =x1a⃗ 1+x2a⃗ 2+x3a⃗ 3?
If the vectors are in RnRn for some nn, then this is a question that can be answered using the equivalent augmented matrix:
[→a1→a2→a3→v][a⃗ 1a⃗ 2a⃗ 3v⃗ ]
If this matrix represents a consistent system of equations, then we can say that →vv⃗ is a linear combination of the other vectors.
Example
Determine if the vector is a linear combination of the vectors:
, , , and [−461]
Solution
Remember that this means we want to find constants x1, x2, x3, and x4 such that:
=x1+x2+x3+x4[−461]
This vector equation is equivalent to an augmented matrix. Setting this matrix up and row reducing, we find that:
[218−450416313110]
Is equivalent to:
[100−10329−7429010422913290016293529]
While it isn’t pretty, this matrix does NOT contain a row such as [0000c] where c≠0 which would indicate the underlying system is inconsistent. Therefore the underlying system is consistent (has a solution) which means the vector equation is also consistent.
So, we can say that is a linear combination of the other vectors.
The step-by-step process
In general, if you want to determine if a vector →u is a linear combination of vectors →v1, →v2, … , →vp (for any whole number p>2) you will do the following.
Step 1
Set up the augmented matrix
[→v1→v2⋯→vp→u]
and row reduce it.
Step 2
Use the reduced form of the matrix to determine if the augmented matrix represents a consistent system of equations. If so, then →u is a linear combination of the others. Otherwise, it is not.
In the second step, it is important to remember that a system of equations is consistent if there is one solution OR many solutions. The number of solutions is not important – only that there IS at least one solution. That means there is at least one way to write the given vector as a linear combination of the others.
Writing a Vector as a Linear Combination of Other Vectors
Sometimes you might be asked to write a vector as a linear combination of other vectors. This requires the same work as above with one more step. You need to use a solution to the vector equation to write out how the vectors are combined to make the new vector.
Let’s start with an easier case than the one we did before and then come back to it since it is a bit complicated.
Example
Write the vector →v= as a linear combination of the vectors:
, , and [−200]
Solution
Step 1
We set up our augmented matrix and row reduce it.
[20−2201041002]
is equivalent to
Step 2
We determine if the matrix represents a consistent system of equations.
Based on the reduced matrix, the underlying system is consistent. Again, this is because there are no rows of all zeros in the coefficient part of the matrix and a single nonzero value in the augment. (you could also use the number of pivots to make the argument.)
Unlike before, we don’t only want to verify that we have a linear combination. We want to show the linear combination itself. This means that we need an actual solution. In this case, there is only one:
x1=2, x2=4, x3=1
Using these values, we can write →v as:
→v==(2)+(4)+(1)[−200]
Now let’s go back to our first example (the one with the crazy fractions) but change the instructions a bit.
Example
Write the vector →v= as a linear combination of the vectors:
, , , and [−461]
When we did step 1, we had the following work. This showed that the equivalent vector equation was consistent and verified that →v was a linear combination of the other vectors.
[218−450416313110]
Is equivalent to:
[100−10329−7429010422913290016293529]
What if we wanted to write out the linear combination. This is different from the previous example in that there are infinitely many solutions to the vector equation.
Looking more closely at this augmented matrix, we can see that there is one free variable x4. If we write out the equations, we have:
x1–(10329)x4=−7429
x2+(4229)x4=1329
x3+(629)x4=3529
Since x4 is a free variable, we can let it have any value and find a solution to this system of equations. A really “nice” value would be zero. If x4=0, then:
x1–10329(0)=−7429
x2+4229(0)=1329
x3+629(0)=3529
Using this solution, we can write →v as a linear combination of the other vectors.
→v==(−7229)+(1329)+(3529)+(0)[−461]
This would be one solution, but because x4 is free, there are infinitely many. For each possible value of x4, you have another correct way to write →v as a linear combination of the other vectors. For example, if x4=1:
x1=−7429+10329=2929=1
x2=1329–4229=−2929=−1
x3=3529–629=2929=1
Using this, we can also write:
→v==(1)+(−1)+(1)+(1)[−461]
How nice is that? (note: normally, we wouldn’t write out the 1 in the equation showing the linear combination. I left it there so you could see where each number from the solution ended up).
Again, a problem like this has infinitely many answers. All you have to do is pick a value for the free variables and you will have one particular solution you can use in writing the linear combination.
When the Vector is NOT a Linear Combination of the Others
It is worth seeing one example where a vector is not a linear combination of some given vectors. When this happens, we will end up with an augmented matrix indicating an inconsistent system of equations.
Example
Determine if the vector is a linear combination of the vectors:
, [01−1], and [12−1].
Solution
Step 1
We set up our augmented matrix and row reduce it.
[101111220−1−11]
is equivalent to:
Step 2
We determine if the matrix represents a consistent system of equations.
Given the form of the last row, this matrix represents an inconsistent system of equations. That means there is no way to write this vector as a linear combination of the other vectors. That’s that – nothing else to say! This will be our conclusion any time row reduction results in a row with zeros and a nonzero value on the augment.
Study guide – linear combinations and span
Need more practice with linear combinations and span? This 40-page study guide will help! It includes explanations, examples, practice problems, and full step-by-step solutions.
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Linear Combinations of Vectors – The BasicsIn "Linear Algebra"
Linearly independent vectors with examplesIn "Linear Algebra"
Introduction to linear transformationsIn "Linear Algebra" |
17464 | https://pubmed.ncbi.nlm.nih.gov/15258107/ | Comparison of inhibitory effects of the proton pump-inhibiting drugs omeprazole, esomeprazole, lansoprazole, pantoprazole, and rabeprazole on human cytochrome P450 activities - PubMed
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. 2004 Aug;32(8):821-7.
doi: 10.1124/dmd.32.8.821.
Comparison of inhibitory effects of the proton pump-inhibiting drugs omeprazole, esomeprazole, lansoprazole, pantoprazole, and rabeprazole on human cytochrome P450 activities
Xue-Qing Li1,Tommy B Andersson,Marie Ahlström,Lars Weidolf
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1 DMPK and Bioanalytical Chemistry, AstraZeneca R and D Mölndal, S-431 83 Mölndal, Sweden.
PMID: 15258107
DOI: 10.1124/dmd.32.8.821
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Comparative Study
Comparison of inhibitory effects of the proton pump-inhibiting drugs omeprazole, esomeprazole, lansoprazole, pantoprazole, and rabeprazole on human cytochrome P450 activities
Xue-Qing Li et al. Drug Metab Dispos.2004 Aug.
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. 2004 Aug;32(8):821-7.
doi: 10.1124/dmd.32.8.821.
Authors
Xue-Qing Li1,Tommy B Andersson,Marie Ahlström,Lars Weidolf
Affiliation
1 DMPK and Bioanalytical Chemistry, AstraZeneca R and D Mölndal, S-431 83 Mölndal, Sweden.
PMID: 15258107
DOI: 10.1124/dmd.32.8.821
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The human clearance of proton pump inhibitors (PPIs) of the substituted benzimidazole class is conducted primarily by the hepatic cytochrome P450 (P450) system. To compare the potency and specificity of the currently used PPIs (i.e., omeprazole, esomeprazole, lansoprazole, pantoprazole, and rabeprazole) as inhibitors of four cytochrome P450 enzymes (CYP2C9, 2C19, 2D6, and 3A4), we performed in vitro studies using human liver microsomal preparations and recombinant CYP2C19. Sample analysis was done using selected reaction monitoring liquid chromatography/tandem mass spectometry. With several systems for CYP2C19 activity (two marker reactions, S-mephenytoin 4'-hydroxylation and R-omeprazole 5-hydroxylation, tested in either human liver microsomes or recombinant CYP2C19), the five PPIs showed competitive inhibition of CYP2C19 activity with K(i) of 0.4 to 1.5 microM for lansoprazole, 2 to 6 microM for omeprazole, approximately 8 microM for esomeprazole, 14 to 69 microM for pantoprazole, and 17 to 21 microM for rabeprazole. Pantoprazole was a competitive inhibitor of both CYP2C9-catalyzed diclofenac 4'-hydroxylation and CYP3A4-catalyzed midazolam 1'-hydroxylation (K(i) of 6 and 22 microM, respectively), which were at least 2 times more potent than the other PPIs. All PPIs were poor inhibitors of CYP2D6-mediated bufuralol 1'-hydroxylation with IC(50) > 200 microM. The inhibitory potency of a nonenzymatically formed product of rabeprazole, rabeprazole thioether, was also investigated and showed potent, competitive inhibition with K(i) values of 6 microM for CYP2C9, 2 to 8 microM for CYP2C19, 12 microM for CYP2D6, and 15 microM for CYP3A4. The inhibitory potency of R-omeprazole on the four studied P450 enzymes was also studied and showed higher inhibitory potency than its S-isomer on CYP2C9 and 2C19 activities. Our data suggest that, although the inhibitory profiles of the five studied PPIs were similar, lansoprazole and pantoprazole are the most potent in vitro inhibitors of CYP2C19 and CYP2C9, respectively. Esomeprazole showed less inhibitory potency compared with omeprazole and its R-enantiomer. The inhibitory potency of rabeprazole was relatively lower than the other PPIs, but its thioether analog showed potent inhibition on the P450 enzymes investigated, which may be clinically significant.
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17465 | https://www.quora.com/What-does-the-following-formula-mean-s-2t-3-S-distance-travelled-t-time-Does-it-indicate-acceleration | What does the following formula mean - s=2t^3? S=distance travelled; t=time. Does it indicate acceleration? - Quora
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What does the following formula mean - s=2t^3? S=distance travelled; t=time. Does it indicate acceleration?
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Chris Carter
B.Sc. in Mathematics&Physics, The University of Western Australia (Graduated 1965) · Author has 3K answers and 1.2M answer views
·3y
If s = 2t³, where s is distance travelled and t is time then ds/st is velocity v and d²s/dt² = dv/dt is acceleration a.
We have ds/st = v = 6t²
and dv/dt = a = 12t.
So when t = 0 then s, v and a are all 0.
When t = 1 then s = 2, v = 6 and a = 12.
When t = 2 then s = 16, v = 24 and a = 24.
So the object is momentarily stationary at t = 0 and then it takes off with an ever-increasing rate of acceleration.
———————————————
Addition: After reading other answers to this question I find that the next two time derivatives of position after velocity and acceleration have been called jerk, and jounce or snap. I
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If s = 2t³, where s is distance travelled and t is time then ds/st is velocity v and d²s/dt² = dv/dt is acceleration a.
We have ds/st = v = 6t²
and dv/dt = a = 12t.
So when t = 0 then s, v and a are all 0.
When t = 1 then s = 2, v = 6 and a = 12.
When t = 2 then s = 16, v = 24 and a = 24.
So the object is momentarily stationary at t = 0 and then it takes off with an ever-increasing rate of acceleration.
———————————————
Addition: After reading other answers to this question I find that the next two time derivatives of position after velocity and acceleration have been called jerk, and jounce or snap. I would not have chosen these terms, but you have to call them something if you want to use them. The Wikipedia article on Jerk (physics) has this significant sentence: “In designing the Hubble Space Telescope, NASA set limits on both jerk and jounce.”. If it’s important to NASA then it’s important to me.
An example from real life would be if you were driving at a constant speed on a straight road, and you decide to increase speed. You push down the accelerator. The engine starts to develop more torque and power. The car accelerates for a while, then settles down at a higher speed.
Your position is constantly changing because the car is moving the whole time.
Your velocity is constant until you decide to speed up. Then your velocity increases, quickly at first, and then slower and slower until you reach a new, higher speed, when it remains steady.
Your acceleration is zero until you decide to speed up. Then it increases rapidly to a maximum and then declines slowly back to zero as you get closer and closer to your new maximum speed. When acceleration is high you sink back into your seat but you gradually rebound as the acceleration wears off.
Your jerk is zero while speed is constant but it then becomes positive while you are advancing the throttle and are sinking into your seat. As you reach maximum acceleration your jerk has returned to zero. It then becomes slightly negative as the acceleration dies away as you are reaching the new maximum speed. The seat slowly rebounds and pushes you forward towards your previous position. When speed is constant and acceleration is zero, jerk also becomes zero.
Your jounce is zero most of the time. There is a brief positive blip when you start to advance the throttle and increase the power of the engine. When you stop advancing the throttle there will be a brief negative blip which returns jerk to zero. You are now at maximum acceleration and speed is increasing. As speed increases while power remains constant the relationship power = force × speed means that force is diminishing. The relationship force = mass × acceleration means that acceleration is diminishing. Jounce briefly goes slightly negative to make jerk negative to make acceleration diminish.
As you approach maximum speed, acceleration is still positive but getting smaller and smaller. Jerk is negative but getting closer and closer to zero, i.e. increasing, so jounce must be slightly positive.
Eventually you settle at the higher speed or velocity, which is constant, and acceleration,jerk and jounce all return to zero.
The Wikipedia article has other applications of jerk in Engineering, and links to the higher derivatives. I commend it to you if you are interested.
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Amar Doshi
I experience Physics every day. · Author has 1.7K answers and 1.6M answer views
·Updated 3y
It indicates constant jerk
s=2 t 3 j=d 3 s d t 3=d 3 d t 3(2 t 3)=12 s=2 t 3 j=d 3 s d t 3=d 3 d t 3(2 t 3)=12
Jerk (physics) - Wikipedia Rate of change of acceleration with time Jerk (also known as jolt [ 1 ] ) is the rate of change of an object's acceleration over time. It is a vector quantity (having both magnitude and direction). Jerk is most commonly denoted by the symbol j and expressed in m/s 3 ( SI units ) or standard gravities per second [ 2 ] ( g 0 /s). As a vector, jerk j can be expressed as the first time derivative of acceleration , second time derivative of velocity , and third time derivative of position : j ( t ) = d a ( t ) d t = d 2 v ( t ) d t 2 = d 3 r ( t ) d t 3 {\displaystyle \mathbf {j} (t)={\frac {\mathrm {d} \mathbf {a} (t)}{\mathrm {d} t}}={\frac {\mathrm {d} ^{2}\mathbf {v} (t)}{\mathrm {d} t^{2}}}={\frac {\mathrm {d} ^{3}\mathbf {r} (t)}{\mathrm {d} t^{3}}}} Where: a is acceleration v is velocity r is position t is time. Third-order differential equations of the form J ( x . . . , x ¨ , x ˙ , x ) = 0 {\displaystyle J\left({\overset {\mathbf {...} }{x}},{\ddot {x}},{\dot {x}},x\right)=0} are sometimes called jerk equations . When converted to an equivalent system of three ordinary first-order non-linear differential equations, jerk equations are the minimal setting for solutions showing chaotic behaviour . This condition generates mathematical interest in jerk systems . Systems involving fourth-order derivatives or higher are accordingly called hyperjerk systems . [ 3 ] Physiological effects and human perception [ edit ] Human body position is controlled by balancing the forces of antagonistic muscles . In balancing a given force, such as holding up a weight, the postcentral gyrus establishes a control loop to achieve the desired equilibrium . If the force changes too quickly, the muscles cannot relax or tense fast enough and overshoot in either direction, causing a temporary loss of control. The reaction time for responding to changes in force depends on physiological limitations and the attention level of the brain: an expected change will be stabilized faster than a sudden decrease or increase of load. To avoid vehicle passengers losing control over body motion and getting injured, it is necessary to limit the exposure to both the maximum force (acceleration) and maximum jerk, since time is needed to adjust muscle tension and adapt to even limited stress changes. Sudden changes in acceleration can cause injuries such as whiplash . [ 4 ] Excessive jerk may also result in an uncomfortable ride, even at levels that do not cause injury. Engineers expend considerable design effort minimizing "jerky motion" on elevators , trams , and other conveyances. For example, consider the effects of acceleration and jerk when riding in a car: Skilled and experienced drivers can accelerate smoothly, but beginners often provide a jerky ride. When changing gears in a car with a foot-operated clutch, the accelerating force is limited by engine power, but an inexperienced driver can cause severe jerk because of intermittent force closure over the clutch. The feeling of bei
Image: Thtatithticth, CC0, via Wikimedia Commons
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It indicates constant jerk
s=2 t 3 j=d 3 s d t 3=d 3 d t 3(2 t 3)=12 s=2 t 3 j=d 3 s d t 3=d 3 d t 3(2 t 3)=12
Jerk (physics) - Wikipedia Rate of change of acceleration with time Jerk (also known as jolt [ 1 ] ) is the rate of change of an object's acceleration over time. It is a vector quantity (having both magnitude and direction). Jerk is most commonly denoted by the symbol j and expressed in m/s 3 ( SI units ) or standard gravities per second [ 2 ] ( g 0 /s). As a vector, jerk j can be expressed as the first time derivative of acceleration , second time derivative of velocity , and third time derivative of position : j ( t ) = d a ( t ) d t = d 2 v ( t ) d t 2 = d 3 r ( t ) d t 3 {\displaystyle \mathbf {j} (t)={\frac {\mathrm {d} \mathbf {a} (t)}{\mathrm {d} t}}={\frac {\mathrm {d} ^{2}\mathbf {v} (t)}{\mathrm {d} t^{2}}}={\frac {\mathrm {d} ^{3}\mathbf {r} (t)}{\mathrm {d} t^{3}}}} Where: a is acceleration v is velocity r is position t is time. Third-order differential equations of the form J ( x . . . , x ¨ , x ˙ , x ) = 0 {\displaystyle J\left({\overset {\mathbf {...} }{x}},{\ddot {x}},{\dot {x}},x\right)=0} are sometimes called jerk equations . When converted to an equivalent system of three ordinary first-order non-linear differential equations, jerk equations are the minimal setting for solutions showing chaotic behaviour . This condition generates mathematical interest in jerk systems . Systems involving fourth-order derivatives or higher are accordingly called hyperjerk systems . [ 3 ] Physiological effects and human perception [ edit ] Human body position is controlled by balancing the forces of antagonistic muscles . In balancing a given force, such as holding up a weight, the postcentral gyrus establishes a control loop to achieve the desired equilibrium . If the force changes too quickly, the muscles cannot relax or tense fast enough and overshoot in either direction, causing a temporary loss of control. The reaction time for responding to changes in force depends on physiological limitations and the attention level of the brain: an expected change will be stabilized faster than a sudden decrease or increase of load. To avoid vehicle passengers losing control over body motion and getting injured, it is necessary to limit the exposure to both the maximum force (acceleration) and maximum jerk, since time is needed to adjust muscle tension and adapt to even limited stress changes. Sudden changes in acceleration can cause injuries such as whiplash . [ 4 ] Excessive jerk may also result in an uncomfortable ride, even at levels that do not cause injury. Engineers expend considerable design effort minimizing "jerky motion" on elevators , trams , and other conveyances. For example, consider the effects of acceleration and jerk when riding in a car: Skilled and experienced drivers can accelerate smoothly, but beginners often provide a jerky ride. When changing gears in a car with a foot-operated clutch, the accelerating force is limited by engine power, but an inexperienced driver can cause severe jerk because of intermittent force closure over the clutch. The feeling of bei
Image: Thtatithticth, CC0, via Wikimedia Commons
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Robert Toop
Author has 4.6K answers and 2.9M answer views
·3y
Originally Answered: What does the following formula mean - s=t2^3? S=distance travelled; t=time. Does it indicate acceleration? ·
There are rules for the evaluation of algebraic expressions. This formula has the expression
t2^3
that is the sequence of four separate symbols
t 2 ^ 3
A sequence of digits enclosing space, commas or periods represents a number and it has the highest precedence. For example, 1000.29 and 1,000.23 and 1 000.23 all represent the same number.
So our expression contains two numbers: 2 and 3, surrounding the ^ symbol, which is the operator for exponentiation, and its adjacent elements are its operands. In this case, operands are 2 and 3. Unless overridden by parentheses, exponentiation has higher precede
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There are rules for the evaluation of algebraic expressions. This formula has the expression
t2^3
that is the sequence of four separate symbols
t 2 ^ 3
A sequence of digits enclosing space, commas or periods represents a number and it has the highest precedence. For example, 1000.29 and 1,000.23 and 1 000.23 all represent the same number.
So our expression contains two numbers: 2 and 3, surrounding the ^ symbol, which is the operator for exponentiation, and its adjacent elements are its operands. In this case, operands are 2 and 3. Unless overridden by parentheses, exponentiation has higher precedence than operators like multiplication, division, addition and subtraction, so it must be evaluated next. I have blocked out the stuff outside the exponentiation with …
… 2^3 …
Exponentiation means multiply 2 by itself 3 times;
… 222 … = … 8 …
Now we are left with two symbols:
t 8
This is an implicit multiplication, identical to
t 8
So we evaluate it as the product of symbol t and number 8, normally written with the number first:
8t
So properly evaluated, the formula becomes
s = 8t
where s symbolizes distance, t symbolizes time. From physics, we realize that 8 is the change of distance per change of time: speed.
To answer the question: speed is constant, so there is no acceleration in the formula.
Maybe you think t2 is a symbol for one of a set of different times, like t1, t2, t3 etc. Well, the question should have described it as such, and used parentheses to force evaluation precedence:
s = (t2)^3
but even that is confusing, because (t2) could literally be evaluated as the product (2t), in which case we’d have
s = 8t^3
So a careful person would use a different symbol for t2, like t_2 which does not contain an implicit operator.
The whole purpose of rigor in writing algebraic expressions is clear, unequivocal communication. This question contains other examples of bad communication:
s=t2^3? S=distance travelled; t=time.
(1) No space between symbols and words.
(2) Misplaced question mark. It belongs at the end of the preceding sentence.
(3) Use of lowercase s in formula, uppercase S in description.
(4) No units of measure, especially with constants in the formula.
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Fred Scuttle
BS in Physics&Mathematics, University of California, Davis · Author has 1.2K answers and 550.5K answer views
·3y
Originally Answered: What does the following formula mean - s=t2^3? S=distance travelled; t=time. Does it indicate acceleration? ·
I would answer this as you requested but I don’t understand what it is that you’re asking. If it is the third derivative of position with respect to time, then as noted by Larry Rothstein, it’s “jerk”. Can you edit your question to be clearer if that’s not the case?
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What is S=ut+1/2 at^2 formulae in physics?
What is the time travel formula?
What is acceleration if time is related to distance as t=ax^2 +bx?
What would cause my car to accelerate slowly when I press down on the gas?
What makes electric vehicles like Teslas accelerate so smoothly compared to traditional cars like the BMW 5-series?
Larry Rothstein
B. A. Physics from California State University, Fullerton (Graduated 1971) · Author has 4K answers and 992.5K answer views
·3y
Originally Answered: What does the following formula mean - s=t2^3? S=distance travelled; t=time. Does it indicate acceleration? ·
I am not sure what the 2 stands for, but this equation represents the distance traveled due to what is called Jerk, the time derivative of acceleration
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Dean Rubine
I've watched hundreds of hours of Leonard Susskind's online physics lectures · Author has 10.6K answers and 23.7M answer views
·Feb 20
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From 0.5m/s, a car is accelerated at 15 m/s for 25s. What is the given formula?
Let’s call that v 0=0.5 m/s.v 0=0.5 m/s. Presumably the acceleration is a=15 m/s 2.a=15 m/s 2.
There’s no formula given, and if there was, it would be silly to ask what it was in the same question that gave it.
Presumably we’re talking about constant acceleration. t t is time. Instantaneous velocity v(t)v(t) is a function of time:
Answer: v(t)=v 0+a t v(t)=v 0+a t
That’s a formula; we have to use it:
v(25)=.5+15(25)=375.5 m/s v(25)=.5+15(25)=375.5 m/s
That would be the answer if the question was how fast it was it going. We’d need a different formula (or calculus) if the question was: How far did the car travel over the 25 s
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Let’s call that v 0=0.5 m/s.v 0=0.5 m/s. Presumably the acceleration is a=15 m/s 2.a=15 m/s 2.
There’s no formula given, and if there was, it would be silly to ask what it was in the same question that gave it.
Presumably we’re talking about constant acceleration. t t is time. Instantaneous velocity v(t)v(t) is a function of time:
Answer: v(t)=v 0+a t v(t)=v 0+a t
That’s a formula; we have to use it:
v(25)=.5+15(25)=375.5 m/s v(25)=.5+15(25)=375.5 m/s
That would be the answer if the question was how fast it was it going. We’d need a different formula (or calculus) if the question was: How far did the car travel over the 25 seconds?
Answer: x(t)=x 0+v 0 t+1 2 a t 2 x(t)=x 0+v 0 t+1 2 a t 2
x(t)x(t) is the position at time t t; we have initial position x 0=0 x 0=0 for us. Over the 25 seconds, the car moved:
x(25)=0+.5(25)+(1/2)(15)25 2=4700 m x(25)=0+.5(25)+(1/2)(15)25 2=4700 m
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Gunjan Sharma
Studied at Udaipur, Rajasthan, India
·7y
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What is the acceleration formula?
Acceleration is defined as change in velocity per unit time.
So, acceleration tells us how much our speed is changing in unit time, i.e. meters per second- per second.
And so the S.I. unit of acceleration is m/s^2 or m·s^−2 or m s^−2 (meters per second square).
NOTE: acceleration is a vector quantity (it has both magnitude and direction)
formula for acceleration
There are some other formulas too that involves acceleration, they are as follows:
a=acceleration
v=velocity
r=radius of the circle
THANK YOU :)
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Acceleration is defined as change in velocity per unit time.
So, acceleration tells us how much our speed is changing in unit time, i.e. meters per second- per second.
And so the S.I. unit of acceleration is m/s^2 or m·s^−2 or m s^−2 (meters per second square).
NOTE: acceleration is a vector quantity (it has both magnitude and direction)
formula for acceleration
There are some other formulas too that involves acceleration, they are as follows:
a=acceleration
v=velocity
r=radius of the circle
THANK YOU :)
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Josh Parker
Materials science PhD student. · Author has 106 answers and 363.5K answer views
·8y
Related
The acceleration of a point is given as a=5−t2m/s2. How do you write an equation for the velocity if v=2m/s when t=3s?
We have an initial value problem here that can be solved using basic calculus. The equation in the question is a tad unclear, but I think it is a(t)=5−2 t a(t)=5−2 t, with v(3)=2 v(3)=2.
Acceleration is defined as the instantaneous change in velocity, or the derivative of velocity. That is, a=d v d t a=d v d t. Substituting this gives:
d v d t=5−2 t d v d t=5−2 t
This is a differential equation that shouldn’t be too difficult to solve. Rearranging gives d v=(5−2 t)d t.d v=(5−2 t)d t.
Integrate both sides: ∫d v=∫(5−2 t)d t∫d v=∫(5−2 t)d t
Solving the integrals gives v(t)=5 t−t 2+C v(t)=5 t−t 2+C. The final step is to find what that constant of i
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We have an initial value problem here that can be solved using basic calculus. The equation in the question is a tad unclear, but I think it is a(t)=5−2 t a(t)=5−2 t, with v(3)=2 v(3)=2.
Acceleration is defined as the instantaneous change in velocity, or the derivative of velocity. That is, a=d v d t a=d v d t. Substituting this gives:
d v d t=5−2 t d v d t=5−2 t
This is a differential equation that shouldn’t be too difficult to solve. Rearranging gives d v=(5−2 t)d t.d v=(5−2 t)d t.
Integrate both sides: ∫d v=∫(5−2 t)d t∫d v=∫(5−2 t)d t
Solving the integrals gives v(t)=5 t−t 2+C v(t)=5 t−t 2+C. The final step is to find what that constant of integration C C is, which we can do by substituting in the initial value v(3)=2 v(3)=2.
2=5(3)−3 2+C=6+C 2=5(3)−3 2+C=6+C
Therefore, C=−4 C=−4, and the equation for velocity is v(t)=5 t−t 2−4 v(t)=5 t−t 2−4.
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Simon Bridge
Scientist · Author has 86.2K answers and 38.8M answer views
·2y
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For the following position function: s(t) = t⁴-4t³ find all t when the velocity is increasing. Is it just where acceleration (the second derivative) is positive?
The position function is for magnitude only … so it is describing a distance from some reference position (the origin).
Perhaps we are supposed to infer that the motion is linear … ??
So we can orient the z axis (say) to the motion:
→s=z^k:z(t)=t(1–4 t 2)s→=z k^:z(t)=t(1–4 t 2)
The velocity is the time-derivative of →s s→ !! because velocity is a vector.
You end up with →v=˙z^k v→=z˙k^
What you’ll see here is that the speed |→v|≠˙z|v→|≠z˙ … which is likely what the homework is trying to get you to think about.
We do often write v=˙z v=z˙ to encode the direction of motion by whether it is negative
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The position function is for magnitude only … so it is describing a distance from some reference position (the origin).
Perhaps we are supposed to infer that the motion is linear … ??
So we can orient the z axis (say) to the motion:
→s=z^k:z(t)=t(1–4 t 2)s→=z k^:z(t)=t(1–4 t 2)
The velocity is the time-derivative of →s s→ !! because velocity is a vector.
You end up with →v=˙z^k v→=z˙k^
What you’ll see here is that the speed |→v|≠˙z|v→|≠z˙ … which is likely what the homework is trying to get you to think about.
We do often write v=˙z v=z˙ to encode the direction of motion by whether it is negative or positive… so you can use it for an implied velocity without worrying about the exact reference direction ^k k^
So distance can be increasing in the -z direction and you have a velocity that is less than zero.
In this case we have a different word for the velocity and the magnitude of the velocity… even if we sometimes get lazy and use “velocity” for both.
How about acceleration?
→a=¨z^k a→=z¨k^
Here we have the same thing… |→a|≠¨z|a→|≠z¨
… however, the word “acceleration” is used for both the vector and it’s magnitude.
The (vector) acceleration can be positive for slowing down, but in the -z direction, and can be negative for speeding up in the -z direction. Of course, the magnitude is always positive.
So … is the question asking you for the times when the particle is speeding up or when the velocity is getting more positive?
YOU have to decide.
You can gain more understanding by sketching the v-t graph. So you should do that.
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Allen Ries
Math Major University of Alberta · Author has 25.1K answers and 9.7M answer views
·Jan 9
Related
What is the formula for calculating time (t) using distance (s), displacement (s), and velocity (v)?
What is the formula for calculating time (t) using distance (s), displacement (s), and velocity (v)?
s =vt+1/2at²
s is the displacement in position.
v is the initial velocity
a is the acceleration
t is the time displacement
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Gopal Menon
B Tech in Chemical Engineering, Indian Institute of Technology, Bombay (IITB) (Graduated 1975) · Author has 10.2K answers and 15.2M answer views
·4y
Related
The relation between time t and distance x is t=ax^3+bx where a and b are constants. What is the instantaneous acceleration? Explain step by step showing the formula you used.
The relation between time t and distance x is t=ax^3+bx where a and b are constants. What is the instantaneous acceleration? Explain step by step showing the formula you used.
t=a x 3+b x.t=a x 3+b x.
⇒d t d x=3 a x 2+b.⇒d t d x=3 a x 2+b.
⇒v=d x d t=1 3 a x 2+b.⇒v=d x d t=1 3 a x 2+b.
⇒a=d v d t=d v d x.d x d t⇒a=d v d t=d v d x.d x d t
=−6 a x(3 a x 2+b)2.1 3 a x 2+b=−6 a x(3 a x 2+b)3.=−6 a x(3 a x 2+b)2.1 3 a x 2+b=−6 a x(3 a x 2+b)3.
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9 3
Steve Johnson
BS in Physics, North Dakota State University (Graduated 1966) · Author has 5.7K answers and 2.3M answer views
·4y
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What is the distance travelled in the first 2min, velocity attained after 2min and acceleration after 1sec of an object moving with time t given as S =2t^3 + 3t^2+ 5t + 7 if S is the distance in meters?
Edit: I had occasion this morning to look at this and saw that the acceleration is to be after 1 second. Sorry. I have adjusted my advice below.
Distance: Plug in 2 minutes (in seconds) for each t in the function. S is the position with respect to the reference point.
Velocity: Take the first derivative of the function. Plug in 2 minutes (in seconds) for each t in the result. ds/dt is the velocity.
Acceleration: Take the first derivative of ds/dt from the last part. Plug in 1 second for each t in the result. d2s/dt2 is the acceleration.
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Kip Ingram
PhD in Electrical Engineering, The University of Texas at Austin Cockrell School of Engineering (Graduated 1992) · Author has 20.1K answers and 21.6M answer views
·1y
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The position of a particle is S = (2t² - 8t + 6) m, where t is in seconds. What is the total distance traveled by the particle when t = 3s?
You have to be clear on what you mean by “total distance traveled.” If you just mean finishing position minus starting position, then it’s simple; it’s just S(3) - S(0) = 0 - 6 = -6.
On the other hand, if you mean the total absolute distance swept over during the three seconds, you have to be careful about direction reversals. The velocity of the particle will be given by V(t) = 4t - 8, so the par
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You have to be clear on what you mean by “total distance traveled.” If you just mean finishing position minus starting position, then it’s simple; it’s just S(3) - S(0) = 0 - 6 = -6.
On the other hand, if you mean the total absolute distance swept over during the three seconds, you have to be careful about direction reversals. The velocity of the particle will be given by V(t) = 4t - 8, so the particle travels in the negative direction for the first two seconds and then in the positive direction. So fo...
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Chemistry
Protein Denaturation
Molecular and Cellular Bi...
Acids
Specific Proteins
Denaturants
Biochemistry (enzymology)
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Proteins (biochemistry)
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How do acids cause the denaturation of proteins?
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Joseph Kunkel
Studied Biology at Case Western Reserve University (Graduated 1968) · Author has 3K answers and 3.4M answer views
·5y
Originally Answered: How does acid denature protein? ·
Every protein has an isoelectric point where its positive and negative charges are in balance with its environment and the protein structure will have a relatively stable structure at this pH environment with intra-chain ionic bonds stabilizing its native folded structure. The protein usually has its lowest solubility at this isoelectric point and will precipitate if the salt concentration is lowered. However, the protein is likely not denatured at this isoelectric point precipitation. Now, add protons at this pH balance point and amine sidechains can take on a proton and become positively cha
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Every protein has an isoelectric point where its positive and negative charges are in balance with its environment and the protein structure will have a relatively stable structure at this pH environment with intra-chain ionic bonds stabilizing its native folded structure. The protein usually has its lowest solubility at this isoelectric point and will precipitate if the salt concentration is lowered. However, the protein is likely not denatured at this isoelectric point precipitation. Now, add protons at this pH balance point and amine sidechains can take on a proton and become positively charged and acidic side-chains can take on a proton and become neutral. Adding acid to a protein shifts the charges on its surface and at some point will cause movement of its backbone leading to unfolding and perhaps forces between newly changed charges and refolding in a denatured state. At low enough pH all amines are positive and all carboxyls are neutral which creates a uniformly positive charged protein that is likely unfolded and apt to coagulate with neighboring proteins through hydrophobic bonding creating aggregates of denatured protein. Each protein is different due to the variability of its sequential placement of potentially charged protonate-able sidechains.
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Lamees Mana
Laboratory technician at MLB Clean Chemistry Industires (2017–present) · Author has 80 answers and 375K answer views
·6y
Originally Answered: Why do acids denature proteins if they're made of acids? ·
Greetings
First of all , proteins are made from amino acids , and they interact with each other in acid -base neutralization reaction . The reaction is about transferring H +from acid to amine . More important thing is the salt formation since there many amino acids that contain extra side chains such as carboxyl or amine , so its positive -negative attraction between the carboxyl and amine , therefore to say protein is all acidic that adding acid wont change the nature of protein its wrong . Adding acids may disturb salt bridges which it held by the positive by ionic charges and that leads to
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Greetings
First of all , proteins are made from amino acids , and they interact with each other in acid -base neutralization reaction . The reaction is about transferring H +from acid to amine . More important thing is the salt formation since there many amino acids that contain extra side chains such as carboxyl or amine , so its positive -negative attraction between the carboxyl and amine , therefore to say protein is all acidic that adding acid wont change the nature of protein its wrong . Adding acids may disturb salt bridges which it held by the positive by ionic charges and that leads to denaturation of Protein , we realize that in adding acid to Milk protein .
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Jason Dean
Lives in Toronto, ON (1985–present) · Author has 3.1K answers and 1.2M answer views
·3y
Originally Answered: Why do acids denature proteins if they're made of acids? ·
It is important to understand that proteins are polypeptides; they are formed from subunits that are acids; but the peptide bond that forms with loss of water; is not itself acidic.
There are ,of course,amino acids that have independent side chains that ARE acidic; specifically,Glutamic Acid and Aspartic Acid.
What acid will do is neutralize acid groups,kicking off cations like Na+ and disrupting salt bridges that are important for some tertiary structure as well as catalytic function.
The polypeptide backbone itself is fairly resistant to acid,due to its zwitterionic nature (it carries partial n
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It is important to understand that proteins are polypeptides; they are formed from subunits that are acids; but the peptide bond that forms with loss of water; is not itself acidic.
There are ,of course,amino acids that have independent side chains that ARE acidic; specifically,Glutamic Acid and Aspartic Acid.
What acid will do is neutralize acid groups,kicking off cations like Na+ and disrupting salt bridges that are important for some tertiary structure as well as catalytic function.
The polypeptide backbone itself is fairly resistant to acid,due to its zwitterionic nature (it carries partial negative charges and partial positive charges); but it can be digested by enzymes tailored to the task ,which are activated by acid.
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Syed Moris
Studied at Nazareth Academy, Gaya
·5y
Originally Answered: Why do acids denature proteins if they're made of acids? ·
Proteins have a 3D shape that is formed because of ionic forces and other forces of attraction - A change in shape can lead to denaturation or loss of function.
There is a movement of Protons from -COOH group to -NH3 group, this movement of protons changes the no. of charged regions therefore this has an impact on total ionic forces of attraction that are present between molecules of proteins and this has a direct impact on the 3D STRUCTURE OF PROTEINS. At a stable pH an equilibrium will be attained wherein a fixed no. of COOH lose Protons and a fixed no. Of NH3 gain protons. If the pH changes…
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Proteins have a 3D shape that is formed because of ionic forces and other forces of attraction - A change in shape can lead to denaturation or loss of function.
There is a movement of Protons from -COOH group to -NH3 group, this movement of protons changes the no. of charged regions therefore this has an impact on total ionic forces of attraction that are present between molecules of proteins and this has a direct impact on the 3D STRUCTURE OF PROTEINS. At a stable pH an equilibrium will be attained wherein a fixed no. of COOH lose Protons and a fixed no. Of NH3 gain protons. If the pH changes………..
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More answers below
Why do mineral acids denature proteins?
How can pH denature proteins?
What are two examples of diseases caused by the denaturation of protein?
What can cause the denaturation of enzymes?
What are denaturation proteins?
Sai Janani Ganesan
I study proteins · Upvoted by
David Schramm
, M.Biochem. Biochemistry, University of Oxford (2021) and
Steven Merz
, PhD Candidate in Chemical Engineering · Author has 463 answers and 3.4M answer views
·10y
Related
How does cold denaturation of proteins happen?
Cold denaturation of proteins is a fascinating process, most ordered molecular systems are more stable at low temperatures where the thermal fluctuations are minimum. However, proteins are stable only over a limited range of temperature and pressure.
What does this mean? a protein can be easily unfolded by heating and cooling, apart from different solvents and high/low pressures.
What makes a protein stable? there are multiple elements that contribute to protein stability, hydrogen bonds play a crucial role in driving secondary structures, the hydrophobic effect and other electrostatics contri
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Cold denaturation of proteins is a fascinating process, most ordered molecular systems are more stable at low temperatures where the thermal fluctuations are minimum. However, proteins are stable only over a limited range of temperature and pressure.
What does this mean? a protein can be easily unfolded by heating and cooling, apart from different solvents and high/low pressures.
What makes a protein stable? there are multiple elements that contribute to protein stability, hydrogen bonds play a crucial role in driving secondary structures, the hydrophobic effect and other electrostatics contribute more towards stabilizing the tertiary structure.
What makes a protein unfold? it is the disruption of this balance in the hydrophobic effect and the hydrogen bonding interactions that drive unfolding.
What is the main difference between thermal denaturation and cold denaturation?
Very simply, thermal denaturation is entropically driven, cold denaturation is enthalpically driven.
Let me elaborate:
Thermal and cold denaturation arises from the temperature dependence of a thermodynamic equilibrium. This is described by the Gibbs-Helmholtz equation, which theoretically predicts denaturation with heating and cooling.
Thus changes in system enthalpy, entropy and heat capacity characterize unfolding. Based on the behavior of the Gibbs-Helmholz equation, we know that heat and cold denaturation are driven by different mechanisms.
When proteins are heated, they become more extended. The energy one supplies to the system in the form of heat, breaks enthalpically favorable interactions. As a result, there is an increase in conformational entropy and the polypeptide chain displays typical polymer behavior.
Cold denaturation leads to partial unfolding of the polypeptide chain, and is a result of change in interaction between water and hydrophobic groups.
With decrease in temperature, the free energy cost for hydrophobic effect, or the unfavorable interaction of nonpolar residues with water decreases, thus increasing their hydration.
In most experiments, decrease in temperature leads to loss of tertiary structure, ergo, weakening of the hydrophobic effec t. [Page on buffalo.edu]
In cold denatured systems, the hydrogen bonds involving interfacial water is more favorable than that in the bulk water. And this precisely is what is hypothesized to drive unfolding.
Figure 1: Ordered water molecules around unfolded state in the cold denatured system: Role of hydrophobic hydration in protein stability: a 3D water-expl...
Figure 2: T represents a dimensionless temperature scale. T=0.15 and 0.17 refer to the cold denatured state. T=0.21 is the collapsed state at room temperature and T=0.25 is the thermal denatured state. From : Page on physics.mcgill.ca
To summarize:
In cold water, the protein is surrounded by ordered water molecules in the hydration shell.
Molecules in the shell are highly ordered and have low energy.
Folding of a protein is (largely) driven by the entropic penalty of ordered water at room temperature (called the hydrophobic effect), whereas the low enthalpic contribution of these ordered water molecules is what drives cold denaturation.
It is due to this that during cold denaturation, there is release of heat, whereas heat is absorbed in heat denaturation.
For more:
Cold denaturation of proteins.
Role of hydrophobic hydration in protein stability: a 3D water-expl...
Page on physics.mcgill.ca
Cold denaturation of a protein dimer monitored at atomic resolution
Cold Denaturation and Protein Stability
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Mohana Priya Sivabalaji
Studied Biochemistry (Graduated 2020)
·1y
Acids cause denaturation of protein by disrupting the salt bridges and hydrogen bonds between the amino acid residues.
When acid is added to the protein, the proton concentration increases. As H+ concentration increase, COO- of amino acid side chain is converted to COOH. This disrupts the salt bridges formed between the amino acid residues.
Also this causes the weak hydrogen bonds between amino acid residues to break and eventually the structure of protein is disrupted.
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Abhishek Kumar
Studied at Rajkiya Pratibha Vikas Vidyalaya, Rajniwas Marg
·5y
Originally Answered: How does acid denature protein? ·
Acid affects the hydogen bonds present between various globin chains of protein and thus causes conformational changes in structure of protein. This causes denaturation of protein
Upvote ·
9 2
Sai Janani Ganesan
Postdoctoral researcher @ UCSF · Upvoted by
Robin Corey
, biochemistry postdoc · Author has 463 answers and 3.4M answer views
·10y
Related
How do alcohols denature proteins?
I started answering this question a few weeks back. I had an answer in mind and was merely looking for data to support my answer. Over the past weeks, I came across one great paper after another and have had my mind blown too many times. This is a great question!
Short answer:
Not all alcohols denature proteins, the effect of alcohol is rather complex as different peptides/proteins adopt different secondary structures in different environments. For example, trifluoroethanol (TFE) stabilizes helical conformations at moderate to low concentrations.
Ethanol is often used to denature proteins ,an
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I started answering this question a few weeks back. I had an answer in mind and was merely looking for data to support my answer. Over the past weeks, I came across one great paper after another and have had my mind blown too many times. This is a great question!
Short answer:
Not all alcohols denature proteins, the effect of alcohol is rather complex as different peptides/proteins adopt different secondary structures in different environments. For example, trifluoroethanol (TFE) stabilizes helical conformations at moderate to low concentrations.
Ethanol is often used to denature proteins ,and under the usual laboratory conditions (1atm pressure, 298K) does function as a denaturant by altering hydrophobic interactions/core of the protein. However, the effect of ethanol is not consistent for all proteins, and also has a temperature effect.
Not-so-short answer: This seems like a rather trivial question at first. It is something we all learn in biochemistry 101, alcohols destroy:
hydrophobic core of proteins by preferentially interacting with hydrophobic residues
hydrogen bonding pattern, by forming H-bonds with protein residues.
Ergo, the hydrophobic/hydrophilic interactions breakdown and the protein denatures.
Figure 1: ethanol destroying tertiary H-bonds in prion protein, image from : Denaturation Protein
However, alcohols are a group of molecules. They differ in chain length/branching/hydroxyl content. Which brings me to the following set of questions:
Can we really generalize the effect of alcohols?
Denaturation implies loss of secondary and tertiary structure, can alcohols destroy backbone H-bonds (which are stronger due to dipole alignments)?
If alcohols denature proteins, consequently, they must stabilize the denatured state. That is, the alcohol-protein interaction must win over protein-protein interaction. Is this always the case?
What happens if the alcohol-protein interaction is weak? does it result in higher protein-protein interaction and thus protein aggregation?
Clearly, this question is not trivial.
Can we really generalize the effect of alcohols?
No (this is an obvious answer), TFE/water mixtures have long been known to stabilize helices and sheets (Page on wiley.com)
What is fascinating about TFE/water mixtures is the fact that they can induce secondary structure in an unfolded peptide chain that does not have any secondary structure propensity. (Mechanism of Stabilization of Helical Conformations of Polypeptides by Water Containing Trifluoroethanol)
Why does this happen? TFE is known to strengthen backbone H-bonding that stabilizes proteins by weakening the CO/NH-water H-bonding. This can be shown using the Zimm–Bragg model (Page on stanford.edu).
What about alcohols that do denature proteins?
The "hydrophobic core disruption" hypothesis, was proposed in the 60s.
A 1967 paper on experimental thermodynamics by Brandts and Hunt provide excellent evidence for the above claim (Thermodynamics of protein denaturation. III. Denaturation of ribonuclease in water and in aqueous urea and aqueous ethanol mixtures)
They performed their experiments on RNase A at different ethanol (commonly used alcohol denaturant) concentrations and temperatures.
They compared their results to that of argon, chosen as an analog for non polar side chain residues.
Their results showed that protein stability and argon solubility had extremely similar temperature dependent behavior and were thus able to make a case for the hydrophobic effect.
Ethanol decreases stability of the protein at moderate to high temperatures (>10 degree C), and has the opposite effect at lower temperatures (5-10 degree C).
How can we explain this temperature effect?
The mechanism is closely linked to "clathrate" formation or the cage like ordered structures formed by water molecules around hydrophobic beads, which results in decrease of entropy. These form at low temperatures and can be thought of as a co-operative process.
Image from : Hydrophobic interactions
At low temperatures, it is likely that the exposed hydrophobic groups are accommodated by clathrate structures in aqueous solutions (denoted by "v" in the figure below).
When small amounts of ethanol or nonaqueous solvents are added, the stability of clathrate structures are decreased and hence the activity of water itself.
It is therefore expected that the hydrophobic side chains are preferentially solvated by alcohols, once the ordered water structures have been disrupted.
As more of these solvent molecules are added, it becomes entropically favorable for these molecules to enter the solvation shell of the protein residue. Thus on further increase beyond this stage, denaturation will occur due to preferential solvation.
This "clathrate to preferential" transition occurs only at low temperatures, at higher temperatures both water and alcohol will occupy random positions in the solvation shell as ordered water would have already been destroyed. This will directly lead to denaturation due to preferential solvation.
Image from : Thermodynamics of protein denaturation. III. Denaturation of ribonuclease in water and in aqueous urea and aqueous ethanol mixtures
To summarize, the effect of alcohol(s) on protein denaturation is relatively complex, unlike the effect of urea. This post is limited to exploring ony one aspect of denaturation, that caused by hydrophobic effect. Many proteins also aggregate in neutral alcohols. (Page on jbc.org )
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Ritwik Thota
Mechanical Design Engineer at CADD Centre India (Indian company) (2021–present) · Author has 3.7K answers and 374.7K answer views
·1y
Originally Answered: How does acid denature protein? ·
A protein becomes denatured when its normal shape gets deformed because some of the hydrogen bonds are broken. Weak hydrogen bonds break when too much heat is applied or when they are exposed to an acid
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Sam Mahdi
BS in Chemistry, California State University, Northridge (Graduated 2017) · Author has 243 answers and 1.8M answer views
·7y
Related
How does SDS denature protein?
While the answers below are right, I’d like to answer this on a bit more basic level.
In chemistry, there is a concept called Polarity. Molecules can be polar, or non-polar. Compounds like water are polar. Compounds like Benzene are non-polar. These groups like to stick together. This is observed when you mix water and oil.
As you can see, the water below groups together, and the oil above groups together. This is because oil is non-polar and water is polar. Polar groups are referred to as Hydrophilic, and non-polar groups are considered Hydrophobic. So water is Hydrophilic, but oil is hydrophob
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While the answers below are right, I’d like to answer this on a bit more basic level.
In chemistry, there is a concept called Polarity. Molecules can be polar, or non-polar. Compounds like water are polar. Compounds like Benzene are non-polar. These groups like to stick together. This is observed when you mix water and oil.
As you can see, the water below groups together, and the oil above groups together. This is because oil is non-polar and water is polar. Polar groups are referred to as Hydrophilic, and non-polar groups are considered Hydrophobic. So water is Hydrophilic, but oil is hydrophobic. With that concept in mind now we look at proteins.
Proteins are composed of Amino Acids:
Above is an example of an amino acid lysine. The OH,C=O, and NH groups are polar, but the CH2 groups are not. All amino acid have that OH, C=O, and NH and CH backbone.
That R is the thing that changes. That R can either be hydrophobic or Hydrophilic. Proteins are comprised of a series of amino acids, so they have hydrophobic R groups right next to Hydrophilic R groups. As stated above, these groups like to be next to one another. Since these proteins are usually in water, the hydrophilic R portions of the amino acid like to point out towards the water, and the hydrophobic portions inwards towards one another. This creates a fold. Think of a droplet of oil in water.
It’s important to note, that the structure of the protein is key to it’s function. So change the structure, you change its function.
See each oil droplet forms a little ball. Proteins do the same thing. Now if 2 oil droplets touch, they will fuse into one oil droplet, and the molecules will all re-arrange to support that new structure. So now on to SDS.
SDS (sodium dodecyl sulfate) is a compound that has both Hydrophilic groups and Hydrophobic groups.
As before, the S=O groups are hydrophilic, but the CH2 and CH3 groups are hydrophobic. If this was presented to an oil droplet as the picture above, this would stick to the oil droplet via its hydrophobic tail (the CH3 and CH2), but also interact with the surrounding water too (via the S=O). This is actually how soap works. The dirt and sweat on your hands is hydrophobic, but water is hydrophilic. So it’ll be hard to wash them away with just water since the water will just flow over it (just like the beads of oil suspended in water). But if you add soap, the hydrophobic part of soap will bind to the dirt, and the hydrophilic side will interact with the water, effectively removing the dirt from your hands, and into the water. Now onto the Denaturation.
Denaturation is defined as destroying the characteristic properties of a protein by effects that disrupt its molecular conformation (paraphrasing of course). So based on everything above. If you add SDS to a protein in water, then the hydrophobic group (CH3 and CH2) will interact with the hydrophobic R gropus in the protein, and the hydrophilic gropu (S=O) will interact with the hydrophilic R groups in the protein. This will change the protein structure naturally, and change its function. By definition this is denaturation.
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How does excessive acidity denature proteins?
The 3D, folded structure of proteins is mediated by several bonding interactions including hydrogen bonding, charge-charge interactions, van der Waals interactions, and cystine linkages.
First, excessive acidity can cause H+ to reduce the cystine linkages, so that R-S-S-R' becomes R-SH HS-R'.
Second, excessive acidity can cause basic amino acids like lysine and asparagine to lose their negative charge via protonation, and lose the ability to participate in charge-charge interactions.
When these weak links holding the chains of amino acids together break, the protein denatures.
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Michael Keefer
Former Tutor, Math Through Calculus and Most Sciences at Olympic College, Bremerton (1994–2000) · Author has 2.9K answers and 786K answer views
·6y
Related
What are some examples of things that can denature a protein?
Heat, get a protein too hot and it will denature (hard boiled eggs).
Soap, a good soapy solution given time, will interfere with surface proteins. Washing your hands for at least 20 seconds will disable most viruses.
Acid, a low enough pH will mess with ionic links in proteins. It can even break apart some peptides.
Base, a high enough pH can do the same stuff acid does.
Cold, get temperature low enough to freeze and some proteins will be damaged. Though life forms that are subject to freezing have proteins that are not damaged by the concentrations changed by freezing. Most damage by freezing is
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Heat, get a protein too hot and it will denature (hard boiled eggs).
Soap, a good soapy solution given time, will interfere with surface proteins. Washing your hands for at least 20 seconds will disable most viruses.
Acid, a low enough pH will mess with ionic links in proteins. It can even break apart some peptides.
Base, a high enough pH can do the same stuff acid does.
Cold, get temperature low enough to freeze and some proteins will be damaged. Though life forms that are subject to freezing have proteins that are not damaged by the concentrations changed by freezing. Most damage by freezing is by ice crystals pushing through things (like cell walls).
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9 1
Bishal
Research scholar from University of Nebraska-Lincoln (Graduated 2019) · Author has 63 answers and 1.2M answer views
·Updated 5y
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What is the denaturation of protein?
Denaturation of proteins involves the disruption and possible destruction of both the secondary and tertiary structures. Since denaturation reactions are not strong enough to break the peptide bonds, the primary structure (sequence of amino acids) remains the same after a denaturation process. Denaturation disrupts the normal alpha-helix and beta sheets in a protein and uncoils it into a random shape.
Denaturation occurs because the bonding interactions responsible for the secondary structure (hydrogen bonds to amides) and tertiary structure are disrupted. In tertiary structure there are four t
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Denaturation of proteins involves the disruption and possible destruction of both the secondary and tertiary structures. Since denaturation reactions are not strong enough to break the peptide bonds, the primary structure (sequence of amino acids) remains the same after a denaturation process. Denaturation disrupts the normal alpha-helix and beta sheets in a protein and uncoils it into a random shape.
Denaturation occurs because the bonding interactions responsible for the secondary structure (hydrogen bonds to amides) and tertiary structure are disrupted. In tertiary structure there are four types of bonding interactions between "side chains" including: hydrogen bonding, salt bridges, disulfide bonds, and non-polar hydrophobic interactions. which may be disrupted. Therefore, a variety of reagents and conditions can cause denaturation. The most common observation in the denaturation process is the precipitation or coagulation of the protein.
Heat can be used to disrupt hydrogen bonds and non-polar hydrophobic interactions. This occurs because heat increases the kinetic energy and causes the molecules to vibrate so rapidly and violently that the bonds are disrupted. The proteins in eggs denature and coagulate during cooking. Other foods are cooked to denature the proteins to make it easier for enzymes to digest them. Medical supplies and instruments are sterilized by heating to denature proteins in bacteria and thus destroy the bacteria.
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Related questions
Why do acids denature proteins if they're made of acids?
How does acid denature protein?
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Why do acids denature proteins if they're made of acids?
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12 5 3
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Answer
The answer is 5
Explanation
Evaluate the expression on the right side of the equation
The expression is $$15 \div 3$$15÷3
2. Calculate the division
$$15 \div 3 = 5$$15÷3=5
3. Determine which number from the given options balances the equation
The equation is balanced when the left side equals the right side. Since the right side equals 5, we need to find the number that equals 5.
4. Identify the correct number
The number 5 is among the given options.
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17468 | https://www.nuclear-power.com/nuclear-engineering/thermodynamics/what-is-energy-physics/foot-pound-force-unit-ft-lbf-energy-unit/ | Foot-pound force (unit ft.lbf) - Energy Unit - Nuclear Power
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Foot-pound force (unit ft.lbf) – Energy Unit
Energy is generally defined as the potential to do work or produce heat. This definition causes the SI unit for energy to be the same as the unit of work – thejoule (J). Joule is a derived unit of energy, and it is named in honor of James Prescott Joule and his experiments on the mechanical equivalent of heat. In more fundamental terms, 1 joule is equal to:
1 J = 1 kg.m 2/s 2
Since energy is a fundamental physical quantity and is used in various physical and engineering branches, there are many energy units in physics and engineering.
Joule (unit: J)
Calorie (unit: cal)
Britich Thermal Unit(unit: BTU)
Foot-pound force(unit: ft.lbf)
Kilowatt-hour(unit: kWh)
Megawatt-day(unit: MWd)
Electronvolt(unit: eV)
Foot-pound force (unit: ft.lbf)
Foot-pound force (unit: ft.lbf). Foot-pound force is a derived unit of work and energy. It is equal to the energy transferred to an object when a force of one pound-force (lbf) acts on that object in the direction of its motion through a distance of one foot. The corresponding SI unit is the joule. The foot-pound is often used in ballistics, especially in the United States. Typically muzzle energies of bullets are given in foot-pound force.
1 foot-pound force = 1.356 J
1 foot-pound force = 0.324 cal
1 foot-pound force = 0.00129 BTU
References:
Reactor Physics and Thermal Hydraulics:
1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading,MA (1983).
2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering,Springer; 4th edition, 1994, ISBN:978-0412985317
5. Todreas Neil E., Kazimi Mujid S. Nuclear Systems Volume I: Thermal Hydraulic Fundamentals, Second Edition. CRC Press; 2 edition, 2012, ISBN:978-0415802871
6. Zohuri B., McDaniel P. Thermodynamics in Nuclear Power Plant Systems. Springer; 2015, ISBN:978-3-319-13419-2
7. Moran Michal J., Shapiro Howard N. Fundamentals of Engineering Thermodynamics, Fifth Edition,John Wiley & Sons, 2006, ISBN:978-0-470-03037-0
8. Kleinstreuer C. Modern Fluid Dynamics. Springer, 2010,ISBN 978-1-4020-8670-0.
9. U.S. Department of Energy, THERMODYNAMICS, HEAT TRANSFER, AND FLUID FLOW. DOE Fundamentals Handbook, Volume 1, 2, and 3. June 1992.
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Searching the community
This article is an introduction to searching the AoPS forums. All searching takes place from one of three pages: the main webpage, any of the specific forums, or from the dedicated webpage.
Contents
1 Searching methods
1.1 Searching from the community
1.2 Searching from a forum
1.3 Searching from the main webpage
2 Getting the most out of search
2.1 Searching for a specific post
2.2 General searching tips
2.3 Customizing search queries
3 Searching for LaTeX and Asymptote
3.1 Searching for LaTeX
3.2 Searching for Asymptote
Searching methods
Searching from the community
Searching from the community page
The first of these two options is shown in the image to the right. It is primarily useful for quickly searching, especially when the search is of a general nature and without a specific post in mind. The method is simple: simply type your query into the "Search Community" field, click the magnifying glass (or press the Enter key), and posts containing terms matching your query will be shown to you. The parts of the posts that match your query will be highlighted in yellow. For example, when searching for problems involving a circle, one would simply type "circle" into the indicated field, getting a result similar to the following image:
Result of searching "circle"
Clicking on any of these results will open the full topic in which the post was made, starting from the post you clicked on. For example, if the post you clicked on was the fourth post in its topic, you would originally see that post and could scroll up (or down) to see the rest of the topic.
If these are not the results you were looking for, you can click the "Edit search settings" link in the top right corner to adjust your search parameters.
Searching from a forum
It is possible to initiate a search from a specific forum, using the toolbar on its header. This will limit the search results to posts in that forum.
The header of a forum
To search from a forum, click the magnifying glass on its header (see image to the right). This will redirect you to the main search webpage, with the forum information already filled in.
Searching from the main webpage
Searching from the webpage
This page can be reached in three ways: by clicking on the "Advanced search" button under the quick-search option shown above, through the Community drop-down menu, or directly through the link Compared to the other options, the webpage gives you significantly more ways to customize your search. There are five different fields to be filled in, but any of them can be left blank. For example, to search for all posts by rrusczyk made within the last year, the "search term" and "Posted In forum" fields should be left blank, but the other fields should be filled out with the appropriate information. It is even possible to leave every field blank, in which case the search results will mimic the global feed. The five fields are:
Search fields | Field | Function |
--- |
| Search term | Fairly self-explanatory: enter the term(s) that you are searching for. Note that titles, tags, posts, and sources are all simultaneously searched. To search these separately, click the indicated option. It is possible to search different places for different terms simultaneously; for example, to search for AMC problems using Simon's Favorite Factoring Trick, search for posts containing "SFFT" in topics with source containing "AMC". The indicated option will also allow you the choice of restricting your search to opening posts. This field is blank by default, indicating that all posts are searched regardless of content. Searching for multiple terms will return posts matching at least one of those terms, prioritizing posts "closer" to the query as a whole. See the following section for ways to further customize these queries. Note that it is also possible to search for multiple tags, which will return posts under at least one of those tags. |
| Posted By User | Enter the name(s) of user(s) to restrict your search to. The search results will only contain posts posted by the user(s) you entered. Note that the posters of the original topics may be different from the users you enter here. You can also search by user ID instead of username, if you happen to know the ID of the user you are searching for (you can find this in their profile URL, which is of the form artofproblemsolving.com/community/user/). Users entered must be valid users and will be authenticated prior to searching. This field is blank by default, indicating that posts by all users are searched. |
| Posted In Forum | Enter the name(s) of forum(s) to restrict your search to. The search results will only contain posts posted in the forum(s) you entered. If you reached this webpage through clicking on the magnifying glass in one of the forums, this field will be pre-populated with that forum. Forums entered must be valid forums and will be authenticated prior to searching. This field is blank by default, indicating that all forums are searched. Note: On the previous website, forums were associated with a unique ID that could be used in searching, similar to user ID numbers, but this is no longer the case; type in the name of the forum instead. |
| Dates | Select the timeframe in which posts you're searching for must have been made. The options are "Any" (default), "During the last 24 hours", "During the last week", "During the last month", "During the last year", or any manually entered date range. |
| Sorting | Select whether to sort results by "Relevance" (default), "Newest first", or "Oldest first". If "relevance" is selected, the posts will be sorted according to a scoring algorithm that approximates how well a post correlates to the search query. The other two options, "Newest first" and "Oldest first", are self-explanatory. |
Posts by rrusczyk in the last year
For example, to search for the aforementioned posts made by rrusczyk within the last year, the "Search term" field should be blank, the "Posted by User" field should contain rrusczyk, the "Posted in Forum" field should be blank, the "Dates" field should be set to "During the last year", and the "Sorting" field can be set according to how the searcher wishes the results to be presented. The image to the right shows how this search looks on the webpage.
Topics started by rrusczyk in the last year
To search instead for topics that rrusczyk has started within the last year, the steps are the same, except that the "search only the first post of each topic" option should be selected. To reach that option, click first on the "Click here to search titles, posts, sources, and tags separately" text, then check the box indicated. The image on the left shows how this search looks on the webpage. Of course, in both searches it is possible to further narrow the results by including the forum it was posted in, text that should be present in returned posts, and so on. It is also possible to expand the search to posts by rrusczyk and copeland, or even postsbycopelandaboutrrusczyk.
Getting the most out of search
Searching for a specific post
One of the most common uses of search is to find a specific post that you remember some details about, but cannot find. For example, you might recall having once read a particularly well-written solution, a very nice question, or an important announcement, but you can't remember quite where you saw it before. How frustrating!
Obviously, the more details you remember about the post, the more likely you are to find it. If you remember that, for example, the post was written sometime in the summer of 2013, you can set the date range from (for example) May to September of that year. If you remember the author, that narrows down the possibilities significantly, but even remembering that it was one of several authors is a good way to whittle the possibilities down. Finally, remembering the forum that it was posted in would be excellent, but even if you only vaguely remember the post, you can make some educated guesses. For example, if the post was about a new and exciting technique for the USAMO, it was probably posted in either the Contests & Programs forum or the High School Olympiads forum. If the post was a collection of MATHCOUNTS strategies, it's almost certainly in Middle School Math.
General searching tips
Of course, the above strategies are helpful, but still leave a lot of posts to sift through - unless you remember some of the language used in the post. You might remember, for example, a particularly memorable phrase in the post, or perhaps they used a math problem as reference that you remember a bit about. In these cases,
Include, in your search query, unusual or uncommon words. For example, searching for just the word "circumcenter", along with the tips in the previous section, narrows down the possibilities significantly
Avoid common words such as "a", "the", "of", and so on, even if you remember a word-for-word phrase from the post containing these words. You'll get bogged down in posts matching those words - which there are a lot of!
Avoid searching words that are contained within another common word, as they will get matched to words you certainly didn't intend. For example, search queries containing the word "in" might return results for "logging", "Inequality", and so on. The word "a" is particularly guilty of this, as it often returns words containing the letter "a".
Be sure to avoid searching for words that are special modifiers - see the below section. In particular, avoid searches containing the words "and" and "not". Similarly, do not include quotes ("), asterisks (), question marks (?), plus signs (+), minus signs (-), or parentheses as these all represent special commands.
Search engines in general are not very good at searching for mathematical symbols or numbers, so you are generally better off searching for text instead. For example, searching the text of a problem rather than equations or expressions is likely to produce better results.
For example, if you remember a post contained the phrase "the three perpendicular bisectors of a triangle intersect at the circumcenter", your search query should be something similar to "perpendicular bisectors intersect circumcenter", as these are all specific words that limit the number of posts returned (unlike words like "of", "a", "the", and "at").
Customizing search queries
We've already mentioned that posts are matched to the search query using a scoring algorithm, which approximates how "close" a post is to the given query. However, especially when searching for multiple terms, this algorithm often weights certain factors in different ways than the user intended. As such, there are several ways to customize your search queries.
Customizing search queries Operator Example Result
[[(no adjustment)incenter circumcenter Returns posts containing the word "incenter" and/or the word "circumcenter", with higher weight given to posts containing both.
++incenter circumcenter Returned posts must contain the word "incenter", but may or may not contain "circumcenter". Again, higher weight is given to posts containing both words.
--incenter circumcenter excenter Returned posts must not contain the word "incenter". Returned posts will contain the word "circumcenter" and/or the word "excenter", with higher weight given to posts containing both.
AND incenter AND circumcenter Returned posts must contain both "incenter" and "circumcenter". This is equivalent to the query "+incenter +circumcenter".
NOT incenter NOT circumcenter Returned posts will contain "incenter", but will not contain "circumcenter". Equivalent to the "-" and "!" operators.
"" (quotes)"incenter circumcenter"Returns posts containing the phrase "incentercircumcenter". Punctuation is usually ignored in results, so posts containing "incenter, circumcenter" or "incenter-circumcenter" will also be returned.
? and te?t, inc, ine Wildcard symbols. The? symbol allows any character to replace it, so posts containing the words "test" or "text" will match the query "te?t". The symbol allows any number of characters to replace it, so posts containing the words "incenter", "incircle", "inclusive", etc. will match the query "inc". The symbol can also be used in the middle of a word, so posts containing the words "interface", "incircle", "intermediate", "infinite", etc. will all match the query "ine".
() (parentheses)incenter AND (circumcenter OR excenter)Grouping symbols to allow one boolean command (e.g. AND, NOT, +, -,!) to modify multiple elements. The example returns posts that contain both "incenter" and at least one of "circumcenter" or "excenter". Equivalent to the query (incenter AND circumcenter) OR (incenter AND excenter) (Boolean logic applies to search strings).
\AND An "escape" character that allows you to search for reserved keywords and symbols. For example, the above query will return posts containing the word "and". Without the escape character, the search engine would parse the query "AND" as a command linking two nonexistent terms, and would thus throw an error. Similarly, the search term "incenter \AND circumcenter" would include results containing the word "incenter" and the word "and", but not the word "circumcenter".
Note: in order to search for terms containing backslashes, such as LaTeX commands, the somewhat non-intuitive \ is necessary (the first backslash "escapes" the second one).
Note that, when using search modifiers, the yellow background indicating words that match your query will not generally be entirely accurate; for example, searching for "incenter AND circumcenter" will return posts containing both words as expected, but the word "and" will also be highlighted. Don't get confused by this - the search engine is still searching according to your query (and is not searching for the word "and"), the highlighting is simply slightly misleading.
Searching for LaTeX and Asymptote
Searching for LaTeX or Asymptote (Vector Graphics Language) can be complicated, as they are their own languages, but this can actually be a big advantage! Predicting what code they've used is usually quite easy, and since the keywords are generally quite unique, searching for them will often narrow results down to posts using LaTeX/Asymptote themselves. Below are some common commands that are often used, to help you search for them.
Searching for LaTeX
LaTeX is a programming language for rendering mathematical statements, and is very popular on AoPS (and other mathematical sources). LaTeX commands are enclosed in dollar signs for rendering, but you should not generally include those while searching. If you are searching for "full" commands (e.g. \frac{1}{2}), it is highly recommended that you enclose the entire command with quotation marks (e.g. "\frac{1}{2}"; recall double backslash is needed), as otherwise it may parse as three different search terms (\frac, 1, and 2).
Common LaTeX commands Command Use Example Rendered
\frac Creates fractions\frac{1}{2}
\sqrt Creates square (or, more generally, th) roots\sqrt{3}, \sqrt{3},
\leq, \neq, \geq Less than or equal to, not equal to, greater than or equal to (respectively)a \leq b, a \neq b, a \geq b
\alpha Renders greek letters (most often alpha, beta, epsilon, pi, theta, phi, and omega)\alpha+\beta=\pi-\epsilon
\rightarrow, \implies, \iff Used for implication, algorithms, etc.a>b \implies a^2>b^2, a>b \iff a-b>0, A\rightarrow A+1
\sum, \prod, \int Used for summation, product, and integration symbols\sum_1^5 x=15, \prod_1^5x=120, \int_1^5x=12
See LaTeX:Symbols for a more comprehensive list of LaTeX commands, and use the TeXeR to test how commands look when rendered.
Searching for Asymptote
Asymptote is a programming language for creating diagrams. Unlike LaTeX, dollar signs are not necessary, and Asymptote code is enclosed in [asy] tags. Asymptote commands take the form of most modern programming languages; lines are generally of the form command(param1, param2, ...) (this is different from TeX, where each parameter is enclosed in separate brackets).
"Unfortunately" (for our purposes), most Asymptote commands are simply the word-for-word descriptors of their function; for example, the command that returns the midpoint of a path is simply "midpoint", and the command that returns the circumcenter of a triangle is "circumcenter". This is further complicated by the usage of variables, the names of which are entirely up to the posters (unlike TeX, in which variables are generally not used). Below are commonly used Asymptote commands that are not actual words, so they are less likely to be confused with other posts during searching.
Common Asymptote commands Command Purpose Command Purpose
defaultpen Adjusts the default settings for the pen. Very likely for this command to be in an Asymptote drawing.orthographic Adjust the "camera position" in 3-D drawings. Very likely for this command to be in a 3-D Asymptote drawing.
filldraw Colors a given shape in some manner. Useful for shading in parts of diagrams; used reasonably foten.unitsize Sets the default size of a "unit" to a specified value. Default is very small. Not used particularly often because defaultpen is more general.
intersectionpoint Determines the intersection (if there is exactly one) of two paths. Very useful command, and thus in many drawings.intersectionpoints Determine all the intersections of two paths. Obviously very similar to the previous command, but slightly less used as we generally want to know the intersection point of lines (rather than arbitrary paths).
rightanglemark Marks a given angle as right. Used, obviously, in diagrams containing right angles; there are many of these.anglemark Marks a given angle with a given value. Used surprisingly little, since directly marking angles isn't generally that important, but angle chase solutions will make heavy use of it.
linewidth Sets the (visual -- lines have no real width!) width of a line. Also little used because defaultpen is more general.currentpicture A reference to the current picture. Used as an auxiliary command in other commands. This is the default value, so it is used only in very complicated drawings (or as a formality/copied and pasted sections).
Note that simply searching "asy" will generally turn up Asymptote drawings or discussions of such; especially when combined with a username or date range, simply remembering a post had any Asymptote in it whatsoever is usually a good way to find it. As Asymptote becomes more and more popular, the previous statement will become less true (LaTeX, for example, is in a huge number of posts these days), but for now the numbers are small enough that you can sift through them pretty quickly.
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Expected Cards Left When One Color Is Exhausted
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Q)
We have a shuffled deck of 35 red and 65 black cards. Cards are drawn one by one without replacement until all cards of one color are exhausted. We want the expected number of cards remaining in the deck at that point.
` Thoughts
I have two approaches in mind.
First is going by the "first ace" approach: the 35 35 th red card will appear at 35×101 36=98.19 35×101 36=98.19. So, the expected number of cards left is 100−98.19=1.81 100−98.19=1.81. Since there are 36 36 black cards, the 65 65 th black card will appear at 65×101 66=99.46 65×101 66=99.46, so expected cards left = 100−99.46=0.54 100−99.46=0.54.
With probability 35 100 35 100, the last card is red (i.e., black cards finish earlier), and with probability 65 100 65 100, the last card is black (i.e., red cards finish earlier). So:
E[X]=35 100⋅0.54+65 100⋅1.81≈1.36 E[X]=35 100⋅0.54+65 100⋅1.81≈1.36
Another approach: When the last red card is drawn, there are 35 35 red cards and some number of black cards drawn. The remaining cards are all black. The expected number of remaining black cards is the total number of black cards minus the expected number of black cards drawn when the last red card is drawn.
By symmetry, the expected number of black cards remaining is:
(B−R)⋅R R+1=30⋅35 36=29.17(B−R)⋅R R+1=30⋅35 36=29.17
This was given by the gemini and I don't know which method it used while calculating it but the last number seems to be correct, but which is actually right or both are wrong?
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asked Jul 28 at 7:49
CallahanCallahan
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2 35 66+65 36 35 66+65 36. Add up the probability of each card surviving all cards of the other color.user14111 –user14111 2025-07-28 08:04:08 +00:00 Commented Jul 28 at 8:04
To show how your and @user14111's calculations relate: 100−(35⋅101/36)=(36⋅100−35⋅101)/36=(35⋅100+100−35⋅100−35)/36=(100−35)/36=65/36 100−(35⋅101/36)=(36⋅100−35⋅101)/36=(35⋅100+100−35⋅100−35)/36=(100−35)/36=65/36 and similarly for the other term.Rosie F –Rosie F 2025-07-28 08:19:12 +00:00 Commented Jul 28 at 8:19
Note that your first method fails to include the conditioning last card, so it is off by 1 1.lulu –lulu 2025-07-28 09:13:22 +00:00 Commented Jul 28 at 9:13
@lulu Yes but by that method our answer will be 2.36 and the correct one is 2.336 . In this question it doesn't seem there's much difference but I wonder maybe if we change the parameters a bit and they would vary significantly.Callahan –Callahan 2025-07-29 03:06:19 +00:00 Commented Jul 29 at 3:06
1 Note that in your second approach, if 29 black cards remain, that means 35 red and 36 black cards where drawn. So roughly equal proportions but there are almost twice as many black cards in the deck. Therefore this approach is wrong.quarague –quarague 2025-07-29 09:35:17 +00:00 Commented Jul 29 at 9:35
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Let the random variable X X be the number of cards left when one color is exhausted.
Number the red cards from 1 1 to 35 35. Let R i R i be the indicator variable for the event "the i th i th red card is farther down in the deck than all the black cards." Then E(R i)=P(R i=1)=1 66 E(R i)=P(R i=1)=1 66.
Similarly, number the black cards from 1 1 to 65 65. Let B i B i be the indicator variable for the event "the i th i th black card is farther down in the deck than all the red cards." Then E(B i)=P(B i=1)=1 36 E(B i)=P(B i=1)=1 36.
Since X=∑35 i=1 R i+∑65 i=1 B i X=∑i=1 35 R i+∑i=1 65 B i,
E(X)=∑i=1 35 E(R i)+∑i=1 65 E(B i)=35 66+65 36=925 396≈2.336.E(X)=∑i=1 35 E(R i)+∑i=1 65 E(B i)=35 66+65 36=925 396≈2.336.
More generally, suppose there are k k different colors of cards, and the deck contains n i n i cards of color i i for 1≤i≤k 1≤i≤k; cards are dealt one by one until a color is exhausted. Let X X be the number of cards left in the deck when this happens. Then
E(X)=∑∅≠I⊂k|I|−1∑j∉I n j 1+∑i∈I n i.E(X)=∑∅≠I⊂k|I|−1∑j∉I n j 1+∑i∈I n i.
This is because, if c c is a card of color j j, then the probability that some color is exhausted before c c is dealt is
∑∅≠I⊆[k]∖{j}(−1)|I|−1 1 1+∑i∈I n i∑∅≠I⊆[k]∖{j}(−1)|I|−1 1 1+∑i∈I n i
by the in-and-out formula.
Example. If k=4 k=4 and n 1=n 2=n 3=n 4=13 n 1=n 2=n 3=n 4=13 then
E(X)=∑r=1 3(−1)r−1 13(4−r)1+13 r(4 r)=39 14(4 1)−26 27(4 2)+13 40(4 3)=4199 630≈6.665.E(X)=∑r=1 3(−1)r−1 13(4−r)1+13 r(4 r)=39 14(4 1)−26 27(4 2)+13 40(4 3)=4199 630≈6.665.
Example. If k>0 k>0 and n 1=⋯=n k=1 n 1=⋯=n k=1 then
E(X)=∑r=1 k−1(−1)r−1 k−r r+1(k r)=k+∑r=0 k(−1)r−1 k−r 1+r(k r)=k+∑r=0 k(−1)r r−k r+1(k r)E(X)=∑r=1 k−1(−1)r−1 k−r r+1(k r)=k+∑r=0 k(−1)r−1 k−r 1+r(k r)=k+∑r=0 k(−1)r r−k r+1(k r)
=k+∑r=0 k(−1)r(k r)−(k+1)∑r=0 k(−1)r r+1(k r)=k+0−(k+1)1 k+1=k−1=k+∑r=0 k(−1)r(k r)−(k+1)∑r=0 k(−1)r r+1(k r)=k+0−(k+1)1 k+1=k−1
which is obviously correct since just one card will be dealt. In this computation I used the identities
∑r=0 k(−1)r(k r)=(1−1)k=0 and∑r=0 k(−1)r r+1(k r)=∫1 0(1−x)k d x=1 k+1.∑r=0 k(−1)r(k r)=(1−1)k=0 and∑r=0 k(−1)r r+1(k r)=∫0 1(1−x)k d x=1 k+1.
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edited Aug 1 at 10:59
answered Jul 28 at 8:28
user14111user14111
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Thanks!! great insight.Callahan –Callahan 2025-07-29 03:06:44 +00:00 Commented Jul 29 at 3:06
3 +1 The wonders of linearity of expectation. Related to a similar questionHenry –Henry 2025-07-29 17:42:16 +00:00 Commented Jul 29 at 17:42
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Remembering that expected values are basically averages,
if 35 35 red cards are exhausted first,they had 36 36 gaps where black cards can be there with average size 65 36 65 36 so the average "leftover" blacks are 65 36 65 36
Similarly, if 65 65 black cards are exhausted first, the average "leftover" red cards are 35 66 35 66
Adding up, answer =65 36+35 66=925 396,≈2.336=65 36+35 66=925 396,≈2.336
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edited Jul 28 at 9:32
answered Jul 28 at 9:20
true blue aniltrue blue anil
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Thanks I was looking for this particular approach, If I could accept more than 1 answer I would have accepted Yours and Lourrran also.Callahan –Callahan 2025-07-29 03:14:52 +00:00 Commented Jul 29 at 3:14
Fine, the main thing is that it has been of help ! Glad you liked it ! :):)true blue anil –true blue anil 2025-07-29 03:50:51 +00:00 Commented Jul 29 at 3:50
In the first case the last gap has at least one card or the red cards wouldn't be exhausted first. So the last gap has an above average expected value. Same with the second step. I don't see why the expected remaining cards should be the sum of the two averages. I would say the result should be P(red exausted first)E(red exausted first)+P(black exausted first)E(black exausted first)Etoplay –Etoplay 2025-07-30 14:32:14 +00:00 Commented Jul 30 at 14:32
@Etoplay: If you want to bring in probability, P(red is last card) =35 100=35 100, but now the average group size of remaining reds in the gaps between blacks becomes 34 66 34 66, and 35 100(1+34 66)=35 66 35 100(1+34 66)=35 66, as before, with similar computations for the black cards. You might agree that the way I have solved it is simpler true blue anil –true blue anil 2025-07-30 16:42:08 +00:00 Commented Jul 30 at 16:42
@Etoplay: Of course, the "classical" way is to use indicator variables, as in my answer heretrue blue anil –true blue anil 2025-07-30 18:28:01 +00:00 Commented Jul 30 at 18:28
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Same result as previous answers, but with a different (more detailed ?) explanation.
The last card may be Red (P=35 100 P=35 100) or Black (P=65 100 P=65 100).
If Last card is Red, all Black cards define 66 gaps, and the 34 remaining Red cards are randomly dispatched in those 66 gaps. So the expected number of Red cards in each gap is 34 66 34 66. Especially for the last gap, expected number of cards is 34 66 34 66 or 1+34 66 1+34 66 if we consider the last card
And similarly if the last card is Black.
So E(X)=35 100×(1+34 66)+65 100×(1+64 36)=35 66+65 36=925 396 E(X)=35 100×(1+34 66)+65 100×(1+64 36)=35 66+65 36=925 396
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answered Jul 28 at 11:05
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(1) Simplifying:
Focus only on the scenarios where the very last card is red (until section (4) of this post). It's easy to see the scenarios where the black one is on the last place are similar, and can be solved using the same method.
(2) Equivalence:
For the same deck, notice the equivalence of "Black cards are exhausted right after the n n th card is drawn" and "Cards on positions n+1 n+1 through 100 100 are all red, AND the card on position n n is black".
(3) Computing the distribution:
Given n n, the probability of the latter event in (2) can be computed.
NOTATIONS:
r k r k: k t h k t h last place is red. For instance, r 2 r 2 means that the second last place, which is the 99 t h 99 t h is red.
R k R k: All last k k places are red.
X=k X=k: R k R k and k−1 k−1 th last place is black.
Therefore R k+1=R k∗r k+1 R k+1=R k∗r k+1
P(R k)=P(R k−1)∗P(r(k)|R k−1)P(R k)=P(R k−1)∗P(r(k)|R k−1) (Formula of conditional probability)
The vertical bar means "given that" or "in the presence of..."
Putting in the numbers, obtain
P(R k)=P(R k−1)∗35−k+1 100−k+1 P(R k)=P(R k−1)∗35−k+1 100−k+1 (selecting from 100−k+1 100−k+1 cards, among which 35−k+1 35−k+1 are red)
with initial P(R 1)=P(r 1)=35 100 P(R 1)=P(r 1)=35 100.
P(X=k)=P(R k)∗65 100−k P(X=k)=P(R k)∗65 100−k
P(X=k)P(X=k) can therefore be computed by recursion, the computation process is trivial, and the results should look like ratios of products of factorials.
(4) Solving
Having computed P(X=k)P(X=k), a similar probability distribution for Y=k Y=k : "All last k k cards are black" can be computed likewise. Then add the two distribution functions. This sum is the distribution of number of cards remaining in the deck.
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answered Jul 28 at 10:27
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17471 | https://www.youtube.com/watch?v=m39m8FdyE0g | Converting Parametric Equations to Cartesian Equations
Maths Genie
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A Level Maths revision tutorial video.
For the full list of videos and more revision resources visit www.mathsgenie.co.uk.
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Transcript:
okay now we're going to look at converting from parametric equations to cartesian equations so parametric equations where we have two separate equations one for x and one for y into cartesian equations where we've got x and y linked in one equation so we need to eliminate the parameter eliminate t in the first case so how can i eliminate t well i'm going to make one equation equal to t so if i start with my x equation if i plus 3 to both sides so we start with x equals t minus three if i add three to both sides x plus three equals t so i've made the equation t equals and then t equals x plus 3 and i'm going to substitute it in for the t in the y equation so that will give me y equals 7 minus 2 substituting in x plus 3 for t so now i have eliminated t and we can just simplify this a bit by expanding the bracket 7 minus x minus six and we can write it in the form y equals m x plus c so y equals minus two x plus one so there's our cartesian equation when we have trigonometric equations so here are parameters theta so we're going to eliminate theta we've got x equals sine theta y equals cos theta we use the identity sine squared theta plus cos squared theta equals one so we rearrange this to make it sine squared theta rearrange this equation to make it cos squared theta and substitute them both in to the identity so if we square both sides we get x squared equals sine squared theta if we square both sides here we have y squared equals cos squared theta so we've got sine squared theta so i can substitute sine squared theta for x squared i can substitute cos squared theta for y squared and i would have eliminated theta so sine squared theta is x squared cos squared theta is y squared equals one and there's our equation so with trigonometric equations trigonometric parametric equations we use the identities to eliminate the parameter okay two more questions here so find the cartesian equation of the line given by the parametric equations x equals t plus three and y equals two t plus eleven so for this first question we're gonna make the top equation equal to t so if we take away three from both sides subtract three we get x minus 3 equals t so we can just substitute x minus 3 in for t on the second equation and we would have eliminated t so y equals 2 times x minus 3 plus 11. if we expand the bracket to simplify it we've eliminated t at this stage so we've got a cartesian equation but we're going to simplify it so two x minus six plus eleven which is two x plus five so our cartesian equation is y equals two x plus five the second equation here it's a trigonometric one so we've got our two equations two trigonometric equations and with trigonometric equations we almost always use the identity sine squared t plus cos squared t equals one to eliminate our parameter to eliminate t so we're going to need one of these equations to be sine squared t equals the other one to be cos squared t equals and then we can substitute them in so for the y equation that's easy if we square both sides y squared equals sine squared t so we can grab that y squared substitute it in for sine squared t and we would have eliminated the sine squared t for the x it's a bit harder so we've got x equals tan t how can we make that cos squared t so we're gonna have to change x to get a cos we'll change it to sine over cos so now we've got a cos t if we multiply both sides by cos t so x cos t equals sine t and divide both sides by x cos t equals sine t over x and then square both sides so we've got cos squared d cos squared t is sine squared t over x squared so we can substitute this in for cos squared t so if we substitute them in now we've got y squared plus sine squared t over x squared equals one but we've still got a t um we've still got sine squared t but we know sine squared t is y squared so we can substitute y squared in there y squared equals no y squared plus y squared over x squared equals one so we've eliminated t so we've got our cartesian equation we could leave it there or we could try and simplify it if we were asked for it in the form y equals f of x we'd have to rearrange to make it y equals which we could do in this case um but we'll leave it there we'll leave it as it is so we've eliminated our t so we've got a cartesian equation we weren't asked to put it in a specific form so we'll leave it as is okay two more questions so pause the video give these ones a go and then press play when you're ready to go through okay so the first set of equations we're going to make an equation equal to t so if i half both sides of the top equation x over 2 equals t and then we're just going to substitute that in to our second equation so y equals 4 times x over 2 squared plus six times x over two minus four so x over two squared if i put x so i'm gonna scribble it up here x over two squared so if i that means x over two times x over two which is x squared over four so four times x squared over four is just x squared six times x over two six over two is three so three x plus four so y equals x squared plus three x plus four for the second one again we've got a trigonometric one so we'll use the identity sine squared t plus cos squared t equals one and we'll rearrange one of the equations to be sine squared t one to be cos squared t and then substitute them in so for the x one through plus one to both sides x plus one equals two sine t divide both sides by two so x plus one over two equals sine t and then square both sides so x plus one squared over four equals sine squared t so that's that part um and then for the y same thing so plus two to both sides y plus two equals three cos t divide both sides by three y plus two over three equals cos t and then square both sides so y plus 2 squared over 9 equals cos squared d so we can substitute that one in as well so we have x plus one squared over four plus y plus two squared over nine equals one i think it's best we leave it like that so we've eliminated our t we've eliminated our third variable so that is our cartesian equation |
17472 | https://faculty.uml.edu/dklain/orthogonal.pdf | Lectures notes on orthogonal matrices (with exercises) 92.222 - Linear Algebra II - Spring 2004 by D. Klain 1. Orthogonal matrices and orthonormal sets An n × n real-valued matrix A is said to be an orthogonal matrix if AT A = I, (1) or, equivalently, if AT = A−1.
If we view the matrix A as a family of column vectors: A = A1 A2 · · · An then AT A = AT 1 AT 2 .
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AT n A1 A2 · · · A3 = AT 1 A1 AT 1 A2 · · · AT 1 An AT 2 A1 AT 2 A2 · · · AT 2 An .
.
.
.
.
.
...
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.
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AT nA1 AT nA2 · · · AT nAn So the condition (1) asserts that A is an orthogonal matrix iff AT i Aj = ( 1 if i = j 0 if i ̸= j that is, iffthe columns of A form an orthonormal set of vectors.
Orthogonal matrices are also characterized by the following theorem.
Theorem 1 Suppose that A is an n×n matrix. The following statements are equivalent: 1. A is an orthogonal matrix.
2. |AX| = |X| for all X ∈Rn.
3. AX · AY = X · Y for all X, Y ∈Rn.
In other words, a matrix A is orthogonal iffA preserves distances and iffA preserves dot products.
Proof: We will prove that 1. ⇒2. ⇒3. ⇒1.
1 1. ⇒2. : Suppose that A is orthogonal, so that AT A = I. For all column vectors X ∈Rn, we have |AX|2 = (AX)T AX = XT AT AX = XT IX = XT X = |X|2, so that |AX| = |X|.
2. ⇒3. : Suppose that A is a square matrix such that |AX| = |X| for all X ∈Rn.
Then, for all X, Y ∈Rn, we have |X + Y |2 = (X + Y )T (X + Y ) = XT X + 2XT Y + Y T Y = |X|2 + 2XT Y + |Y |2 and similarly, |A(X + Y )|2 = |AX + AY |2 = (AX + AY )T (AX + AY ) = |AX|2 + 2(AX)T AY + |AY |2.
Since |AX| = |X| and |AY | = |Y | and |A(X+Y )| = |X+Y |, it follows that (AX)T AY = XT Y . In other words, AX · AY = X · Y .
3. ⇒1. : Suppose that A is a square matrix such that AX ·AY = X ·Y for all X, Y ∈Rn Let ei denote the i-th standard basis vector for Rn, and let Ai denote the i-th column of A, as above. Then AT i Aj = (Aei)T Aej = Aei · Aej = ei · ej = ( 1 if i = j 0 if i ̸= j so that the columns of A are an orthonormal set, and A is an orthogonal matrix.
We conclude this section by observing two useful properties of orthogonal matrices.
Proposition 2 Suppose that A and B are orthogonal matrices.
1. AB is an orthogonal matrix.
2. Either det(A) = 1 or det(A) = −1.
The proof is left to the exercises.
Note: The converse is false. There exist matrices with determinant ±1 that are not orthogonal.
2. The n-Reflections Theorem Recall that if u ∈Rn is a unit vector and W = u⊥then H = I −2uuT is the reflection matrix for the subspace W.
Since reflections preserve distances, it follows from Theorem 1 that H must be an orthogonal matrix. (You can also verify condition (1) directly.) We also showed earlier in the course that H = H−1 = HT and H2 = I.
It turns out that every orthogonal matrix can be expressed as a product of reflection matrices.
2 Theorem 3 (n-Reflections Theorem) Let A be an n × n orthogonal matrix. There exist n × n reflection matrices H1, H2, . . . , Hk such that A = H1H2 · · · Hk, where 0 ≤ k ≤n.
In other words, every n × n orthogonal matrix can be expressed as a product of at most n reflections.
Proof: The theorem is trivial in dimension 1. Assume it holds in dimension n −1.
For the n-dimensional case, let z = Aen, and let H be a reflection of Rn that exchanges z and en. Then HAen = Hz = en, so HA fixes en. Moreover, HA is also an orthogonal matrix by Proposition 2, so HA preserves distances and angles. In particular, if we view Rn−1 as the hyperplane e⊥ n , then HA must map Rn−1 to itself. By the induction assumption, HA must be expressible as a product of at most n −1 reflections on Rn−1, which extend (along the en direction) to reflections of Rn as well. In other words, either HA = I or HA = H2 · · · Hk, where k ≤n.
Setting H1 = H, and keeping in mind that HH = I (since H is a reflection!), we have A = HHA = H1H2 · · · Hk.
Proposition 4 If H is a reflection matrix, then det H = −1.
Proof: See Exercises.
Corollary 5 If A is an orthogonal matrix and A = H1H2 · · · Hk, then det A = (−1)k.
So an orthogonal matrix A has determinant equal to +1 iffA is a product of an even number of reflections.
3. Classifying 2 × 2 Orthogonal Matrices Suppose that A is a 2 × 2 orthogonal matrix. We know from the first section that the columns of A are unit vectors and that the two columns are perpendicular (orthonor-mal!). Any unit vector u in the plane R2 lies on the unit circle centered at the origin, and so can be expressed in the form u = (cos θ, sin θ) for some angle θ. So we can describe the first column of A as follows: A = " cos θ ??
sin θ ??
What are the possibilities for the second column? Since the second column must be a unit vector perpendicular to the first column, there remain only two choices: A = " cos θ −sin θ sin θ cos θ # or A = " cos θ sin θ sin θ −cos θ # 3 The first case is the rotation matrix, which rotates R2 counterclockwise around the origin by the angle θ. The second case is a reflection across the line that makes an angle θ/2 from the x-axis (counterclockwise).
But which is which? You can check that the rotation matrix (on the left) has determinant 1, while the reflection matrix (on the right) has determinant -1. This is consistent with Proposition 4 and Corollary 5, since a rotation of R2 can always be expressed as a product of two reflections (how?).
4. Classifying 3 × 3 Orthogonal Matrices The n-Reflection Theorem 3 leads to a complete description of the 3× orthogonal matri-ces. In particular, a 3 × 3 orthogonal matrix must a product of 0, 1, 2, or 3 reflections.
Theorem 6 Let A be a 3 × 3 orthogonal matrix.
1. If det A = 1 then A is a rotation matrix.
2. If det A = −1 and AT = A, then either A = −I or A is a reflection matrix.
3. If det A = −1 and AT ̸= A, then A is a product of 3 reflections (that is, A is a non-trivial rotation followed by a reflection).
Proof: By Theorem 3, A is a product of 0, 1, 2, or 3 reflections. Note that if A = H1 · · · Hk, then det A = (−1)k.
If det A = 1 then A must be a product of an even number of reflections, either 0 reflections (so that A = I, the trivial rotation), or 2 reflections, so that A is a rotation.
If det A = −1 then A must be a product of an odd number of reflections, either 1 or 3.
If A is a single reflection then A = H for some Householder matrix H. In this case we observed earlier that HT = H so AT = A.
Conversely, if det A = −1 and AT = A then det(−A) = 1 (since A is a 3 × 3 matrix) and −AT = −A = −A−1 as well. It follows that −A is a rotation that squares to the identity. If A ̸= I, then the only time this happens is when we rotate by the angle π (that is, 180o) around some axis. But if −A is a 180o rotation around some axis, then A = −(−A) must be the reflection across the equatorial plane for that axis (draw a picture!). So A is a single reflection.
Finally if det A = −1 and AT ̸= A, then A cannot be a rotation or a pure reflection, so A must be a product of at least 3 reflections.
Corollary 7 Let A be a 3 × 3 orthogonal matrix.
1. If det A = 1 then A is a rotation matrix.
2. If det A = −1 then −A is a rotation matrix.
4 Proof: If det A = 1 then A is a rotation matrix, by Theorem 6. If det A = −1 then det(−A) = (−1)3 det A = 1. Since −A is also orthogonal, −A must be a rotation.
Corollary 8 Suppose that A and B are 3 × 3 rotation matrices. Then AB is also a rotation matrix.
Proof: If A and B are 3 × 3 rotation matrices, then A and B are both orthogonal with determinant +1. It follows that AB is orthogonal, and det AB = det A det B = 1·1 = 1.
Theorem 6 then implies that AB is also a rotation matrix.
Note that the rotations represented by A, B, and AB may each have completely different angles and axes of rotation! Given two rotations A and B around two different axes of rotation, it is far from obvious that AB will also be a rotation (around some mysterious third axis). But this is true, by Corollary 8. Later on we will see how to compute precisely the angle and axis of rotation of a rotation matrix.
Exercises: 1. (a) Suppose that A is an orthogonal matrix.
Prove that either det A = 1 or det A = −1.
(b) Find a 2×2 matrix A such that det A = 1, but also such that A is not an orthogonal matrix.
2. Suppose that A and B are orthogonal matrices. Prove that AB is an orthogonal matrix.
3. Suppose that H = I −2uuT is a reflection matrix. Let v1, . . . , vn−1 be an orthonormal basis for the subspace u⊥. (Here u⊥denotes the orthogonal complement to the line spanned by u.) (a) What is Hu = ?
(b) What is Hvi = ?
(c) Let M be the matrix with columns as follows M = v1 · · · vn−1 u What is HM = ? What are the columns of HM?
(d) By comparing det HM to det M prove that det H = −1.
4. (a) Give a geometric description (and sketch) of the reflection performed by the matrix H1 = " cos θ sin θ sin θ −cos θ # 5 (b) Give a geometric description (and sketch) of the reflection performed by the matrix H2 = " 1 0 0 −1 # (c) Show that a rotation in R2 is a product of two reflections by showing that H1H2 is a rotation matrix. Give a geometric description (and sketch) of the rotation performed by the matrix H1H2.
5. Let u, v, w be an orthonormal basis (of column vectors) for R3, and let A be the matrix given by A = uuT + vvT + wwT .
(a) What is Au =? What is Av =? What is Aw =?
(b) Let P be the matrix P = u v w What is AP =?
(c) Prove that A = I (the identity!).
6. Give geometric descriptions of the what happens when you multiply a vector in R3 by each of the following orthogonal matrices: cos θ −sin θ 0 sin θ cos θ 0 0 0 1 cos θ 0 −sin θ 0 1 0 sin θ 0 cos θ 1 0 0 0 cos θ −sin θ 0 sin θ cos θ Sketch what happens in each case.
6 |
17473 | https://physics30spagtakhan.files.wordpress.com/2019/11/apparent-weight-lab2019.pdf | NEWTON’S LAWS – APPARENT WEIGHT (elevator problems) The “apparent weight” is the reading on the scale in the elevator, and in effect, is equal to the normal force acting on the person. Applying Newton’s Second Law and solving for FN, we can express the apparent weight, FN as: FN = mg + m ( a) The +a refers to an upward acceleration and –a is a downward acceleration. Problems 1. An elevator that weighs 3.0 x 103 N is accelerated upward at 1.0 m/s2. What force does the cable exert to give it this acceleration? 2. A person weighing 490 N stands on a scale in an elevator. a. What does the scale read when the elevator is at rest? b. What is the reading on the scale when the elevator rises at a constant velocity? c. The elevator slows down at -2.2 m/s2 as it reaches the desired floor. What does the scale read? d. The elevator descends, accelerating at -2.7 rn/s2. What does the scale read? e. What does the scale read when the elevator descends at a constant velocity? f. Suppose the cable snapped and the elevator fell freely. What would the scale read? 3. A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 836 N. a. As the elevator moves up, the scale reading increases to 935 N, then decreases back to 836 N. Find the acceleration of the elevator. b. As the elevator approaches the 74th floor, the scale reading drops as low as 782 N. What is the acceleration of the elevator? 2 4. A 65 kg person is standing on a scale in an elevator. What is the person’s apparent weight when a) the elevator accelerates upward at 2.5 m/s2, and b) when it nears the floor and decelerates at 1.5 m/s2? 5. A 95.0 kg person stands on a scale in an elevator. What is the apparent weight when the elevator is a) accelerating upward at 1.80 m/s2, b) moving upward at a constant speed c) accelerating downward at 1.30 m/s2, and d) moving downward at a constant speed? 6. A 55 kg person is riding in a hot air balloon, and a scale shows this person’s weight to be 549 N. The acceleration of gravity at the location is 9.79 m/s2. Determine the magnitude and direction of the vertical component of the balloon’s acceleration? ANSWERS 1. 3300 N 2. a) 490 N b) 490 N c) 380 N d) 355 N e) 490 N f) 0 3. a) 1.16 m/s2 b) –0.633 m/s2 4. a) 800 N b) 540 N 5. a) 1.1 x 103 N b) 931 N c) 808 N d) 931 N 6. 0.20 m/s2, up |
17474 | https://www.thoughtco.com/what-is-colligation-1689763 | Definition and Examples of Colligation in Language
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Humanities›
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Colligation
Glossary of Grammatical and Rhetorical Terms
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English
English Grammar
An Introduction to Punctuation
Writing
By
Richard Nordquist
Richard Nordquist
English and Rhetoric Professor
Ph.D., Rhetoric and English, University of Georgia
M.A., Modern English and American Literature, University of Leicester
B.A., English, State University of New York
Dr. Richard Nordquist is professor emeritus of rhetoric and English at Georgia Southern University and the author of several university-level grammar and composition textbooks.
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Updated on May 12, 2025
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Key Takeaways
Colligation refers to how words are grouped based on their function in sentences.
Colligation is like collocation but focuses on word patterns in syntax, not concrete word combinations.
The term colligation means 'tie together' and was first used by linguist John Rupert Firth.
In English grammar, a colligation is a grouping of words based on the way they function in a syntactic structure--i.e., a syntactic pattern. Verb: colligate.
As linguist Ute Römer has observed, "What collocation is on a lexical level of analysis, colligation is on a syntactic level. The term does not refer to the repeated combination of concrete word forms but to the way in which word classes co-occur or keep habitual company in an utterance" (Progressives, Patterns, Pedagogy).
The word colligation comes from the Latin for "tie together." The term was first used in itslinguisticsense by British linguist John Rupert Firth (1890-1960), who defined colligation as "the interrelation ofgrammatical categoriesin syntactical structure."
Examples and Observations
"According to [John Rupert]Firth (1968:181), colligation refers to the relations between words at the grammatical level, i.e. the relations of 'word and sentence classes or of similar categories' instead of 'between words as such.' But nowadays the term colligation has been used to refer not only to significant co-occurrence of a word with grammatical classes or categories (e.g. Hoey 1997, 2000; Stubbs 2001c:112) but also to significant co-occurrence of a word with grammatical words (e.g. Krishnamurthy 2000). The patterning with grammatical words, of course, can be observed and computed even using a raw corpus."
(Tony McEnery, Richard Xiao, and Yukio Tono, Corpus-Based Language Studies: An Advanced Resource Book. Routledge, 2006)
Kinds of Colligation
"Although based on Firth's concept, the more widespread Sinclairian use of colligation describes the co-occurrence of a class of grammatical items with a specified node. For instance, regarding the node true feelings, [John McH.] Sinclair notes that 'there is a strong colligation with a possessive adjective...' Other kinds of colligation might be a preference for a particular verb tense, negative particles, modal verbs, participles, that-clauses, and so on. The notion that words may prefer (or, indeed, avoid) particular positions in text is picked up by [Michael] Hoey ([Lexical Priming,] 2005) in his more detailed definition of colligation: The basic idea of colligation is that just as a lexical item may be primed to co-occur with another lexical item, so also it may be primed to occur in or with a particular grammatical function. Alternatively, it may be primed to avoid appearance in or co-occurrence with a particular grammatical function.
(Hoey 2005:43) Hoey attributes his use of colligation also to refer to sentential position as derivative from [M.A.K.] Halliday . . .; it can, of course, also be seen as a natural extension of considering punctuation as a grammatical class, because punctuation is one of the most obvious indicators of positioning in text."
(Gill Philip, Colouring Meaning: Collocation and Connotation in Figurative Language. John Benjamins, 2011)
Colligation and Verbs of Perception
"The class of verbs of perception such as hear, notice, see, watch enters into colligation with the sequence of object + either the bare infinitive or the -ing form; e.g. We heard the visitors leave/leaving.
We noticed him walk away/walking away.
We heard Pavarotti sing/singing.
We saw it fall/falling. The term [colligation] is far less general than the contrasting term collocation."
(Sylvia Chalker and Edmund Weiner, Oxford Dictionary of English Grammar. Oxford University Press, 1994)
Collocation and Colligation in Language Instruction
"[C]ontext is not only central to linguistic analysis and description but in language pedagogy too. I strongly believe that it makes sense to pay attention to collocation and colligation in language instruction and to teach lexical items in their typical syntactic and semantic contexts. This belief clearly echoes one of [John] Sinclair's (1997:34) . . . data-centered precepts: '[i]nspect contexts,' in which he 'advocate[s] a much closer inspection of the verbal environment of a word or phrase than is usual in language teaching.'
"A corpus-driven study of progressives, especially when it is in part pedagogically motivated, thus has to closely examine the contexts of the respective items under analysis and investigate which terms are normally selected together by the competent speaker of English."
(Ute Römer, Progressives, Patterns, Pedagogy: A Corpus-driven Approach to English Progressive Forms, Functions, Contexts and Didactics. John Benjamins, 2005)
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Nordquist, Richard. "Colligation." ThoughtCo, May. 12, 2025, thoughtco.com/what-is-colligation-1689763.Nordquist, Richard. (2025, May 12). Colligation. Retrieved from Nordquist, Richard. "Colligation." ThoughtCo. (accessed September 29, 2025).
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17476 | https://www.teacherspayteachers.com/Product/6th-Grade-Differentiated-Unit-Rate-Word-Problems-5007027 | 6th Grade Differentiated Unit Rate Word Problems
Description
Give your students practice with unit rate problems! Use these task cards for math stations, small group intervention, partner activities, or as a classroom scoot. This includes an answer key and options for answer sheets.
This product includes 3 sets of 16 task cards covering unit rate word problems. Set A is on-level word problems, Set B is above-level problems, and Set C is below level problems. All sets require students to practice skills calculating unit rates and unknown numbers in a ratio.
There are two options for an answer sheet. Option one allows all answers to fit on one side, option two can be copied double sided and gives students more work space.
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6th Grade Differentiated Unit Rate Word Problems
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17477 | https://tasks.illustrativemathematics.org/content-standards/tasks/263 | Engage your students with effective distance learning resources. ACCESS RESOURCES>>
Comparing Money Raised
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Alignments to Content Standards: 4.OA.A.2
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Task
Helen raised \$12 for the food bank last year and she raised 6 times as much money this year. How much money did she raise this year?
Sandra raised \$15 for the PTA and Nita raised \$45. How many times as much money did Nita raise as compared to Sandra?
Luis raised \$45 for the animal shelter, which was 3 times as much money as Anthony raised. How much money did Anthony raise?
IM Commentary
The purpose of this task is for students to solve three comparisons problems that are related by their context but are structurally different. Multiplicative comparison is purposefully excluded from third grade (see 3.OA.3 and 3.MD.2), making this task appropriate for fourth but not third grade. In these multiplicative comparison problems, one factor and the product are amounts of money and the other factor represents the number of times bigger one amount is than the other. Sometimes this second factor is called a €œscale factor.€ In part (a), the larger amount (which is the product) is unknown, while in part (b) the scale factor is unknown and in part (c) the smaller amount of money is unknown. Students will study multiplicative comparison problems involving scale factors that are fractions in 5th grade; see 5.NF.B.5. Note that in fourth grade, scale factors must always be bigger than 1, so students often think that €œmultiplying makes bigger€; however in 5th grade they will see that when the scale factor is less than 1, the product will actually be smaller than the initial quantity. Note that the numbers in parts (b) and (c) are related by the fact family $3\times15=45$. This allows for a classroom discussion about the different interpretations of the factors in a multiplicative comparison context. To see an annotated version of this and other Illustrative Mathematics tasks as well as other Common Core aligned resources, visit Achieve the Core.
The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, only one practice connection will be discussed in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.
This particular task helps illustrate Mathematical Practice Standard 1, Make sense of problems and persevere in solving them. Problem solving is based upon students engaging in a task in which a solution pathway is not known in advance. As fourth graders approach these three problems, they will analyze the problems to make sense of what each is asking, working to understand the structures and the unknowns. Through this analysis, students will understand that the numbers in parts (b) and (c) are both factors related by the same fact family but each serves a different role in their respective problems. Students need experience with unknowns that play different roles in multiplicative comparison problems in order to support a deeper understanding of this type of problem situation.
Solutions
Solution: Tape diagram
She raised six times as much money (as shown in the diagram) so she raised $6 \times 12 = 72$.
Helen raised \$72 this year.
2. $? \times 15 = 45$ is equivalent to $45 \div 15 = ?$
Nita raised 3 times as much as Sandra.
3. $3 \times ? = 45$ is equivalent to $45 \div 3 = ?$
Anthony raised \$15.
Solution: Writing multiplication equations for division problems
Helen raised $6 \times \$12$ this year, so she raised \$72 this year.
This is a €œNumber of Groups Unknown€ problem. We can represent the question as $$? \times 15 = 45$$ or $$45 \div 15 = ?$$ So Nita raised 3 times as much money as Sandra.
This is a €œGroup Size Unknown€ problem. We can represent the question as $$3 \times ? = 45$$ or $$45 \div 3 = ?$$ So Anthony raised \$15.
Comparing Money Raised
Helen raised \$12 for the food bank last year and she raised 6 times as much money this year. How much money did she raise this year?
Sandra raised \$15 for the PTA and Nita raised \$45. How many times as much money did Nita raise as compared to Sandra?
Luis raised \$45 for the animal shelter, which was 3 times as much money as Anthony raised. How much money did Anthony raise?
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17478 | https://www.quora.com/How-do-I-differentiate-f-x-lnx-x | How to differentiate: f(x) = (lnx) ^x - Quora
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How do I differentiate: f(x) = (lnx) ^x?
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Divya Dubey
Teacher · Author has 228 answers and 283.2K answer views
·4y
Let ,
f(x)=y=(lnx)^x
➡️ y=(lnx)^x
Now Taking log both side we get,
lny=ln[(lnx)^x]
lny=xln(lnx)
Now diff.w.r.t.x
d/dx(lny)=d/dx{xln(lnx)}
Then,
(1/y).dy/dx=x.d/dx{ln(lnx)}+ln(lnx).d/dx(x)
(1/y)dy/dx=x.(1/lnx).d/dx(lnx)+ln(lnx).(1)
(1/y)dy/dx=x.(1/lnx).(1/x)+ln(lnx)
(1/y)dy/dx=(1/lnx)+ln(lnx)
dy/dx=y[1+lnx.ln(lnx)]/(lnx)
dy/dx=(lnx)^x.(lnx)^-1[1+ln(lnx)^lnx]
dy/dx=(lnx)^(x-1)[1+ln(lnx)^lnx] Ans.👈
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Hilmar Zonneveld
Translator (1985–present) · Author has 58.5K answers and 19.4M answer views
·4y
If there is a power, and “x” appears both in the base and in the exponent, you can conveniently use logarithmic differentiation. That is, you start off by taking the logarithm on both sides:
y = (ln x)^x
ln y = x ln x
Differentiation is now rather easy; note that the first step involves using the chain rule on the left side:
y’/y = x / x + ln x
y’/y = 1 + lnx
y’ = y(1 + lnx)
y’= (ln x)^x (1 + ln x)
From Wikipedia: “The same result can be obtained by rewriting [the function] in terms of exp and applying the chain rule.”
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To differentiate the function f(x)=(ln x)x f(x)=(lnx)x, we can use logarithmic differentiation. Here are the steps:
Take the natural logarithm of both sides:
ln f(x)=ln((ln x)x)lnf(x)=ln((lnx)x)
2. Apply the properties of logarithms:
ln f(x)=x⋅ln(ln x)lnf(x)=x⋅ln(lnx)
3. Differentiate both sides with respect to x x using implicit differentiation. Remember to apply the product rule on the right side:
d d x(ln f(x))=1 f(x)⋅f′(x)d d x(lnf(x))=1 f(x)⋅f′(x)
d d x(x⋅ln(ln x))=ln(ln x)+x⋅1 ln x⋅1 x=ln(ln x)+1 ln x d d x(x⋅ln(lnx))=ln(lnx)+x⋅1 lnx⋅1 x=ln(lnx)+1 lnx
4. Set the derivatives equal:
1 f(x)\cd 1 f(x)\cd
Continue Reading
To differentiate the function f(x)=(ln x)x f(x)=(lnx)x, we can use logarithmic differentiation. Here are the steps:
Take the natural logarithm of both sides:
ln f(x)=ln((ln x)x)lnf(x)=ln((lnx)x)
2. Apply the properties of logarithms:
ln f(x)=x⋅ln(ln x)lnf(x)=x⋅ln(lnx)
3. Differentiate both sides with respect to x x using implicit differentiation. Remember to apply the product rule on the right side:
d d x(ln f(x))=1 f(x)⋅f′(x)d d x(lnf(x))=1 f(x)⋅f′(x)
d d x(x⋅ln(ln x))=ln(ln x)+x⋅1 ln x⋅1 x=ln(ln x)+1 ln x d d x(x⋅ln(lnx))=ln(lnx)+x⋅1 lnx⋅1 x=ln(lnx)+1 lnx
4. Set the derivatives equal:
1 f(x)⋅f′(x)=ln(ln x)+1 ln x 1 f(x)⋅f′(x)=ln(lnx)+1 lnx
5. Solve for f′(x)f′(x):
f′(x)=f(x)(ln(ln x)+1 ln x)f′(x)=f(x)(ln(lnx)+1 lnx)
6. Substitute back f(x)=(ln x)x f(x)=(lnx)x:
f′(x)=(ln x)x(ln(ln x)+1 ln x)f′(x)=(lnx)x(ln(lnx)+1 lnx)
Thus, the derivative of f(x)=(ln x)x f(x)=(lnx)x is:
f′(x)=(ln x)x(ln(ln x)+1 ln x)f′(x)=(lnx)x(ln(lnx)+1 lnx)
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Yohenba Soibam
M.sc from National Institute of Technology, Manipur
·4y
Let y=f(x)=(l n(x))x y=f(x)=(l n(x))x
Taking log on both sides
l n(y)=x.l n(l n(x))l n(y)=x.l n(l n(x))
Differentiating w.r.t x, we get
1 y.y′=x.1 l n(x)(1 x)+l n(l n(x))1 y.y′=x.1 l n(x)(1 x)+l n(l n(x))
⇒y′=(l n(x))x[(1 l n(x))+l n(l n(x))]⇒y′=(l n(x))x[(1 l n(x))+l n(l n(x))]
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How do you differentiate x to the power of 2?
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Donald Hartig
PhD in Mathematics, University of California, Santa Barbara (Graduated 1970) · Author has 7.4K answers and 2.8M answer views
·1y
f(x)=ln(x)x=e ln(ln(x)x)f(x)=ln(x)x=e ln(ln(x)x)
=e x ln(ln(x))=e x ln(ln(x))
=ln(x)x(x 1/x ln(x)+ln(ln(x))=ln(x)x(x 1/x ln(x)+ln(ln(x))
=ln(x)x(1 ln(x)+ln(ln(x))=ln(x)x(1 ln(x)+ln(ln(x))
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Arpan Parpan
Studied General Studies (Graduated 2013) · Author has 434 answers and 438.8K answer views
·4y
f(x) = (lnx)^x
=> ln{f(x)} = xln(lnx)
=> f’(x)/f(x) = ln(lnx) + x{1/lnx(1/x)}
=> f’(x)/f(x) = ln(lnx) + 1/(lnx)
=> f’(x) = (lnx)^x{ln(lnx)+1/(lnx)}
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James Woodland
· Author has 2.4K answers and 1.5M answer views
·Mar 25, 2021
place in brackets , add +1 then deduct pi
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Vasudev Kenkre
4y
Take logarithm of both sides and differentiate. It is straightforward then because the right hand sides logarithm is xln(lnx).
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Subhasish Debroy
Former SDE at Bharat Sanchar Nigam Limited (BSNL) · Author has 6.6K answers and 5.8M answer views
·4y
Let y=(l n x)x y=(l n x)x , taking log to the base on either side,
l n y=x∗l n(l n x)l n y=x∗l n(l n x) differentiating both sides w.r.t. x
(1/y)∗d y/d x(1/y)∗d y/d x = x∗1/l n x∗1/x+l n(l n x)∗1 x∗1/l n x∗1/x+l n(l n x)∗1
d y/d x=y[1/l n x+l n(l n x)]d y/d x=y[1/l n x+l n(l n x)]
d y/d x=(l n x)x[1/l n x+l n(l n x)]d y/d x=(l n x)x[1/l n x+l n(l n x)]
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Calvin L.
2nd year mathematics student · Author has 10K answers and 2.2M answer views
·1y
Let y=f(x).y=f(x).
y=(ln x)x y=(lnx)x
ln y=x ln(ln x)lny=x ln(lnx)
y′y=ln(ln x)+1 ln x y′y=ln(lnx)+1 lnx
∴f′(x)=(ln x)x[ln(ln x)+1 ln x]∴f′(x)=(lnx)x[ln(lnx)+1 lnx]
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·2y
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How do you differentiate f(x) =x^2 at x=2 first by the first principal?
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For a detailed step by step version just click on this video I made…
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Ernest Leung
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How do you differentiate f(x) =2tan³4x?
Suppose the question is: "How do you differentiate f(x) = 2 tan³(4x)?"
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Suppose the question is: "How do you differentiate f(x) = 2 tan³(4x)?"
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Philip Lloyd
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How was this derived: f(x) = (ln x) ^x to f'(x) =lnx^x-1 [(1+lnx) ln(lnx)]?
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SCIENTIFIC EXHIBIT
Pneumomediastinum Revisited
Christopher M. Zylak,
James R. Standen,
George R. Barnes,
Carl J. Zylak
Christopher M. Zylak,
James R. Standen,
George R. Barnes,
Carl J. Zylak
Author Affiliations
1 From the Department of Radiology, University of Arizona, 1718 E Speedway, PMB 108, Tucson, AZ 85719. Presented as a scientific exhibit at the 1998 RSNA scientific assembly. Received April 1, 1999; revision requested May 17 and received July 15; accepted July 15. Address correspondence to C.M.Z. (e-mail: cmzjz@hotmail.com).
Christopher M. Zylak
James R. Standen
George R. Barnes
Carl J. Zylak
Published Online:Jul 1 2000
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Abstract
Pneumomediastinum may result from a variety of causes that may be either intrathoracic (eg, narrowed or plugged airway, straining against a closed glottis, blunt chest trauma, alveolar rupture) or extrathoracic (eg, sinus fracture, iatrogenic manipulation in dental extraction, perforation of a hollow viscus). The radiographic signs of pneumomediastinum depend on the depiction of normal anatomic structures that are outlined by the air as it leaves the mediastinum. These signs include the thymic sail sign, “ring around the artery” sign, tubular artery sign, double bronchial wall sign, continuous diaphragm sign, and extrapleural sign. In distal esophageal rupture, air may migrate from the mediastinum into the pulmonary ligament. Pneumomediastinum may be difficult to differentiate from medial pneumothorax and pneumopericardium. Occasionally, normal anatomic structures (eg, major fissure, anterior junction line) may simulate air within the mediastinum. Iatrogenic entities that may simulate pneumomediastinum include helium in the balloon of an intraaortic assist device. In addition, pneumomediastinum may be simulated by the Mach band effect, which manifests as a region of lucency adjacent to structures with convex borders. The absence of an opaque line, which is typically seen in pneumomediastinum, can aid in differentiation. Computed tomographic (CT) digital radiography and conventional CT can also be helpful in establishing or confirming the diagnosis.
Article History
Published in print: July 2000
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Sheets
How to Use FV in Google Sheets: A Step-by-Step Guide
By Spencer Lanoue
July 11, 2025
Contents
Ever thought about how cool it would be to see into the future? Well, while we can't exactly play fortune teller, Google Sheets has a nifty trick up its sleeve called the FV function that can give you a sneak peek into your financial future. Whether you're planning your savings or trying to figure out how much that monthly investment will grow, FV can be your crystal ball.
In this article, we're rolling up our sleeves and getting into all things FV in Google Sheets. We'll walk through what FV is, how it works, and most importantly, how you can use it to make sense of your own numbers. So grab a coffee (or tea, if that's your thing), and let's get started!
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The Basics of FV
Before we jump into the how-tos, it’s worth taking a moment to understand what FV, or Future Value, is all about. In the simplest terms, FV helps you figure out how much a series of investments will be worth at a future date, given a certain interest rate. It’s a handy tool in finance for anyone who wants to project their savings or investments over time.
Let’s say you put a little bit of money away each month. FV can calculate how much that money will grow based on an interest rate. It’s like watching your piggy bank grow without having to crack it open.
Here's the basic syntax of the FV function in Google Sheets:
=FV(rate, number_of_periods, payment, [present_value], [type])
Rate: The interest rate per period.
Number_of_periods: Total number of payment periods.
Payment: Payment made each period; it remains constant throughout the annuity's life.
Present_value: The initial amount of money (optional).
Type: When payments are due. Use 0 if payments are made at the end of the period, and 1 if at the beginning (also optional).
Now that we've got the basics down, let's dive into each component and see how they fit together.
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Breaking Down the FV Function
Alright, let’s break down the components of the FV function so you can see what each part does. This is where the magic happens!
1. Rate
The interest rate is crucial. It determines how fast your money grows. Typically, this rate is expressed on a per-period basis. For instance, if you have an annual interest rate of 6% and you're making monthly investments, you’d use 0.5% (6% divided by 12) for the rate.
It might sound a bit complex, but once you get the hang of it, it’s straightforward. Just think of it as the growth speed of your money.
2. Number of Periods
This one’s simple. It's the total number of periods your investment will grow. Continuing with the monthly example, if you plan to invest for 5 years, you’d have 60 periods (5 years multiplied by 12 months).
It's essentially the timeline for your money's journey. The longer the timeline, the more time your money has to grow.
3. Payment
This is the amount you’ll be contributing in each period. In our example, it would be how much you put away each month. It’s your regular contribution to the investment or saving plan.
Consistent contributions can significantly impact your future value. It's like watering a plant; the more you water (invest), the more it grows.
4. Present Value
The present value is the amount of money you’re starting with. If you don’t have an initial amount and are starting fresh, you can leave this as zero or omit it altogether.
Think of it as the seed money for your investment. If you’ve already got some savings, this is where you’d include that amount.
5. Type
This optional component is all about timing. It indicates when payments are made. Zero is for end-of-period payments, and one is for those made at the beginning.
While it might seem minor, this can affect your calculations. It’s like choosing whether you want your paycheck at the start or end of the month.
With these components in mind, using the FV function becomes a breeze. It's all about plugging in the right numbers and letting Google Sheets do the heavy lifting.
Setting Up Your Google Sheet
Now that we've got the theory down, it's time to get practical. Let's start by setting up your Google Sheet for FV calculations. No worries if you're new to Google Sheets; we'll walk through it step by step.
1. Open Google Sheets
Head over to Google Sheets. If you’re not familiar, it’s part of the Google Suite and works right in your browser. Just go to sheets.google.com and sign in with your Google account.
2. Create a New Spreadsheet
Once you're in, create a new spreadsheet. You can do this by clicking on the "+" button in the bottom-right corner. A fresh, blank sheet will open up, ready for you to fill in.
3. Organize Your Data
Before jumping into FV, it’s helpful to organize your data. Here’s a simple layout:
A1: “Interest Rate”
B1: “Number of Periods”
C1: “Payment”
D1: “Present Value”
E1: “Type”
F1: “Future Value”
In the rows below, you’ll enter your actual data. This setup makes it easy to see all the components of your FV calculation at a glance.
4. Input Your Data
Next, fill in your data. Let’s use an example:
A2: 0.005 (for a 6% annual interest rate divided by 12 months)
B2: 60 (for a 5-year period with monthly contributions)
C2: -200 (monthly payment)
D2: 0 (no initial investment)
E2: 0 (payments made at the end of the period)
Note: The payment is negative because it represents money going out. It’s a small detail, but important for accurate calculations.
5. Enter the FV Formula
Finally, it’s time to calculate the future value. In cell F2, enter the following formula:
=FV(A2, B2, C2, D2, E2)
Press Enter, and voilà! You’ll see the future value of your investment appear in cell F2. It’s like having a financial crystal ball right in your spreadsheet.
Practical Tips for Using FV
Now that you’ve got the basics down, let’s talk about some practical tips to get the most out of the FV function in Google Sheets. These are the little nuggets of wisdom that can make your experience smoother and more insightful.
1. Experiment with Different Scenarios
Don’t be afraid to play around with different numbers. Adjust the interest rate, the number of periods, or the payment amount to see how they affect your future value. This can help you plan for different financial scenarios.
For instance, what happens if you increase your monthly contribution by $50? Or if you find an investment with a slightly higher interest rate? By testing these scenarios, you can make more informed decisions about your finances.
2. Keep Track of Your Assumptions
When using FV, it’s important to remember that your calculations are based on certain assumptions, like a consistent interest rate or regular payments. Keep a note of these assumptions so you can revisit them if your financial situation changes.
It’s a bit like having a roadmap. If your route changes, you’ll know which direction to adjust.
3. Combine with Other Functions
Don’t hesitate to combine FV with other functions in Google Sheets. For example, you can use IF statements to create more dynamic calculations or SUM to total your contributions.
These combinations can turn Google Sheets into a powerful financial tool, letting you analyze your data in new and creative ways.
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4. Use Conditional Formatting
If you want to make your sheet more visually appealing, consider using conditional formatting. You can highlight cells based on certain criteria, making it easier to spot trends or outliers.
It’s like giving your data a little makeover, making it not only functional but also visually engaging.
With these tips, you’ll be well on your way to making the most of FV in Google Sheets. It’s all about exploring possibilities and finding what works best for your personal financial goals.
Common Mistakes to Avoid
Even the best of us can make mistakes, especially when dealing with numbers and formulas. Let’s look at some common pitfalls with the FV function and how you can steer clear of them.
1. Forgetting to Adjust the Interest Rate
This is a big one. Make sure to adjust the interest rate to match your payment periods. If your rate is annual but payments are monthly, divide by 12. It’s a small step that makes a big difference.
Think of it like baking a cake. If you forget an ingredient, the end result might not be what you expected.
2. Entering the Payment as Positive
Remember, payments should be negative in the FV function because they represent outgoing money. If you enter it as positive, your results will be off.
It’s like writing a check. You’re taking money out, not putting it in.
3. Ignoring the Type Argument
The type argument is optional, but it’s important. It determines when payments are made and can affect your calculations. Make sure to set it to 0 or 1, depending on your scenario.
Skipping this is like forgetting to set your alarm. It might not seem like a big deal until you miss that early meeting.
4. Overlooking Initial Investments
If you already have some savings, don’t forget to include it in the present value. It’s a part of your financial picture and can impact your future value significantly.
It’s like counting all your apples when making a pie. Every piece adds up.
By keeping these common mistakes in mind, you can ensure your FV calculations are accurate and truly reflective of your financial situation.
Advanced FV Applications
Once you’ve mastered the basics, there’s a whole world of advanced applications for the FV function. These can help you take your financial planning to the next level, whether it’s for personal finance or business projects.
1. Planning for Retirement
FV can be an invaluable tool for retirement planning. By setting different contribution levels and interest rates, you can estimate how much you’ll need to save to reach your retirement goals.
You might even consider different scenarios, like retiring earlier or later, to see how they impact your savings. It’s like creating a roadmap for your financial journey.
2. Business Investment Projections
If you’re in business, FV can help project the future value of investments, whether they’re in equipment, real estate, or other ventures. By adjusting the variables, you can evaluate the potential return on investment (ROI).
It’s a powerful way to make informed business decisions based on potential future outcomes.
3. Education Savings
Planning for education expenses? FV can help you calculate how much you need to save for tuition and other costs, taking into account interest rates and contributions.
It’s a way to ensure that you’re financially prepared for those education expenses when the time comes.
These advanced applications show how versatile the FV function can be, making it a valuable tool in various financial scenarios. Whether personal or professional, it can provide insights that guide your financial strategies.
Exploring Alternatives to FV
While FV is a powerful tool, it’s not the only option for financial calculations in Google Sheets. There are other functions and methods that can complement or even replace FV, depending on your needs.
1. NPV (Net Present Value)
If you’re dealing with investments that have irregular cash flows, the NPV function might be more suitable. It calculates the present value of an investment based on a series of cash flows and a discount rate.
It’s like looking at your investment from a different angle, considering the value today rather than in the future.
2. PMT (Payment)
Want to know how much you need to pay each period to reach a future value goal? The PMT function can help you calculate regular payments needed for a loan or investment to reach a specific future value.
It’s about working backward from your goal to see what it takes to get there.
3. Manual Calculations
If you prefer a more hands-on approach, you can always use manual calculations. While more time-consuming, it gives you full control over each step of the process.
It’s akin to building something from scratch, where you see every piece coming together.
These alternatives offer different perspectives and methods for financial calculations, allowing you to choose what works best for your specific situation.
Collaborating with Others
Google Sheets shines when it comes to collaboration, and using the FV function is no exception. Here are some ways to collaborate effectively with others on your financial calculations.
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1. Sharing Your Sheet
Google Sheets makes it easy to share your work with others. Simply click the “Share” button in the top-right corner and enter the email addresses of your collaborators. You can control whether they can view, comment, or edit the sheet.
Sharing is caring, after all. It’s a way to bring others into the fold and work together toward common goals.
2. Commenting and Feedback
Encourage collaborators to leave comments and feedback directly in the sheet. This can be particularly useful for team projects or when seeking financial advice from others.
It’s like having a conversation right in the document, keeping everyone on the same page.
3. Version History
Google Sheets keeps a record of changes made to your document. You can view and restore previous versions, ensuring that you never lose important data.
It’s a safety net that gives you peace of mind, knowing that your work is always backed up.
By leveraging Google Sheets’ collaboration features, you can make the most of the FV function and work more effectively with others.
Using FV for Personal Finance
Beyond business and investment, the FV function has practical applications in personal finance. It’s a tool that can help you plan and manage your money more effectively.
1. Budgeting and Saving
Use FV to set savings goals and see how different saving strategies affect your future value. It’s a way to visualize your financial future and stay motivated to reach your goals.
Think of it as a personal financial coach, guiding you toward a brighter future.
2. Loan Repayment
If you have loans, FV can help you calculate the impact of extra payments or changes in interest rates. It’s a way to strategize your repayment plan and potentially save money on interest.
It’s like having a game plan for tackling debt.
3. Emergency Fund Planning
An emergency fund is essential for financial security. Use FV to calculate how your emergency fund will grow over time and ensure that you’re prepared for unexpected expenses.
It’s about having a financial safety net when life throws you a curveball.
By applying the FV function to personal finance, you can gain a clearer picture of your financial situation and make informed decisions that align with your goals.
Final Thoughts
We’ve covered a lot of ground when it comes to using the FV function in Google Sheets. From understanding the basics to applying it in various scenarios, you now have the tools to project your financial future with confidence. Whether you’re saving for a rainy day, planning a big purchase, or simply curious about your financial growth, the FV function is there to guide you.
If you’re looking to take your spreadsheet game to the next level, consider checking out Bricks. It seamlessly integrates spreadsheets, documents, and presentations, with AI at its core to handle everything from writing formulas to creating charts and graphics. With Bricks, you don’t need to be a spreadsheet expert. It can do it all for you in seconds, making it a fantastic companion for anyone working with data.
Spencer Lanoue
Spencer’s spent 10+ years building products and growing startups like Buffer, UserTesting, and Bump Health. He lives in spreadsheets—crunching data, building dashboards, and creating visuals to drive decisions. At Bricks, he’s focused on figuring out how to use AI to automate the busy work of spreadsheets, letting AI handle the heavy lifting for you.
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17481 | https://www.fpl.fs.usda.gov/documnts/fplgtr/fplgtr113/ch12.pdf | 12–1 Drying and Control of Moisture Content and Dimensional Changes William T. Simpson Chapter 12 Contents Determination of Moisture Content 12–1 Oven-Drying Method 12–2 Electrical Method 12–2 Recommended Moisture Content 12–3 Timbers 12–3 Lumber 12–3 Glued Wood Products 12–3 Drying of Wood 12–5 Air Drying 12–6 Accelerated Air Drying and Pre-Drying 12–6 Kiln Drying 12–6 Drying Mechanism 12–6 Drying Stresses 12–7 Dry Kilns 12–7 Kiln Schedules 12–8 Drying Defects 12–9 Moisture Content of Dried Lumber 12–10 Moisture Control During Transit and Storage 12–14 Plywood and Structural Items 12–14 Finish and Factory Lumber 12–15 Dimensional Changes in Wood 12–15 Estimation Using Dimensional Change Coefficient 12–15 Calculation Based on Green Dimensions 12–17 Design Factors Affecting Dimensional Change 12–18 Framing Lumber in House Construction 12–18 Heavy Timber Construction 12–18 Interior Finish 12–18 Flooring 12–18 Wood Care and Installation During Construction 12–18 Lumber and Trusses 12–18 Exterior Trim and Millwork 12–19 Finished Flooring 12–19 Interior Finish 12–19 Plastering 12–19 References 12–20 n the living tree, wood contains large quantities of water. As green wood dries, most of the water is removed. The moisture remaining in the wood tends to come to equilibrium with the relative humidity of the surrounding air. Correct drying, handling, and storage of wood will minimize moisture content changes that might occur after drying when the wood is in service and such changes are undesirable. If moisture content is controlled within reasonable limits by such methods, major problems from dimensional changes can usually be avoided.
The discussion in this chapter is concerned with moisture content determination, recommended moisture content val-ues, drying methods, methods of calculating dimensional changes, design factors affecting such changes in structures, and moisture content control during transit, storage, and construction. Data on green moisture content, fiber saturation point, shrinkage, and equilibrium moisture content are given with information on other physical properties in Chapter 3.
Wood in service is virtually always undergoing at least slight changes in moisture content. Changes in response to daily humidity changes are small and usually of no conse-quence. Changes that occur as a result of seasonal variation, although gradual, tend to be of more concern. Protective coatings can retard dimensional changes in wood but do not prevent them. In general, no significant dimensional changes will occur if wood is fabricated or installed at a moisture content corresponding to the average atmospheric conditions to which it will be exposed. When incompletely dried mate-rial is used in construction, some minor dimensional changes can be tolerated if the proper design is used.
Determination of Moisture Content The amount of moisture in wood is ordinarily expressed as a percentage of the weight of the wood when ovendry. Four methods of determining moisture content are covered in ASTM D4442. Two of these—the oven-drying and the electrical methods—are described in this chapter. 12–2 The oven-drying method has been the most universally accepted method for determining moisture content, but it is slow and necessitates cutting the wood. In addition, the oven-drying method may give values slightly greater than true moisture content with woods containing volatile extrac-tives. The electrical method is rapid, does not require cutting the wood, and can be used on wood in place in a structure.
However, considerable care must be taken to use and inter-pret the results correctly. Use of the electrical method is generally limited to moisture content values less than 30%.
Oven-Drying Method In the oven-drying method, specimens are taken from repre-sentative boards or pieces of a quantity of lumber. With lumber, the specimens should be obtained at least 500 mm (20 in.) from the end of the pieces. They should be free from knots and other irregularities, such as bark and pitch pockets.
Specimens from lumber should be full cross sections and 25 mm (1 in.) long. Specimens from larger items may be representative sectors of such sections or subdivided incre-ment borer or auger chip samples. Convenient amounts of chips and particles can be selected at random from larger batches, with care taken to ensure that the sample is represen-tative of the batch. Veneer samples should be selected from four or five locations in a sheet to ensure that the sample average will accurately indicate the average of the sheet.
Each specimen should be weighed immediately, before any drying or reabsorption of moisture has taken place. If the specimen cannot be weighed immediately, it should be placed in a plastic bag or tightly wrapped in metal foil to protect it from moisture change until it can be weighed. After weighing, the specimen is placed in an oven heated to 101°C to 105°C (214°F to 221°F) and kept there until no apprecia-ble weight change occurs in 4-h weighing intervals. A lum-ber section 25 mm (1 in.) along the grain will reach a con-stant weight in 12 to 48 h. Smaller specimens will take less time. The constant or ovendry weight and the weight of the specimen when cut are used to determine the percentage of moisture content using the formula Moisture content (%) = − × Weight when cut Ovendry weight Ovendry weight 100 (12–1) Electrical Method The electrical method of determining the moisture content of wood uses the relationships between moisture content and measurable electrical properties of wood, such as conductiv-ity (or its inverse, resistivity), dielectric constant, or power-loss factor. These properties vary in a definite and predictable way with changing moisture content, but correlations are not perfect. Therefore, moisture determinations using electrical methods are always subject to some uncertainty.
Electric moisture meters are available commercially and are based on each of these properties and identified by the property measured. Conductance-type (or resistance) meters measure moisture content in terms of the direct current con-ductance of the specimen. Dielectric-type meters are of two types. Those based principally on dielectric constant are called capacitance or capacitive admittance meters; those based on loss factor are called power-loss meters.
The principal advantages of the electrical method compared with the oven-drying method are speed and convenience.
Only a few seconds are required for the determination, and the piece of wood being tested is not cut or damaged, except for driving electrode needle points into the wood when using conductance-type meters. Thus, the electrical method is adaptable to rapid sorting of lumber on the basis of moisture content, measuring the moisture content of wood installed in a building, or establishing the moisture content of a quantity of lumber or other wood items, when used in accordance with ASTM D4442.
For conductance meters, needle electrodes of various lengths are driven into the wood. There are two general types of electrodes: insulated and uninsulated. Uninsulated electrodes will sense the highest moisture content along their length (highest conductance). Moisture gradients between the surface and the interior can lead to confusion. If the wood is wetter near the center than the surface, which is typical for drying wood, the reading will correspond to the depth of the tip of the insulated electrodes. If a meter reading increases as the electrodes are being driven in, then the moisture gradient is typical. In this case, the pins should be driven in about one-fifth to one-fourth the thickness of the wood to reflect the average moisture content of the entire piece. Dried or par-tially dried wood sometimes regains moisture in the surface fibers, and the surface moisture content is greater than the interior. In this case, the meter with the uninsulated pins will read the higher moisture content surface, possibly caus-ing a significant deviation from the average moisture content.
To guard against this problem, electrodes with insulated shanks have been developed. They measure moisture content of only the wood at the tips of the electrodes.
Dielectric-type meters are fitted with surface contact elec-trodes designed for the type of specimen material being tested. The electric field from these electrodes penetrates well into the specimen, but with a strength that decreases rapidly with depth of penetration. For this reason, the readings of dielectric meters are influenced predominantly by the surface layers of the specimen, and the material near midthickness may not be adequately represented in the meter reading if there is a moisture content gradient.
To obtain accurate moisture content values, each instrument should be used in accordance with its manufacturer’s instruc-tions. The electrodes should be appropriate for the material being tested and properly oriented according to meter manu-facturer’s instructions. The readings should be carefully taken as soon as possible after inserting the electrode. A species correction supplied with the instrument should be applied when appropriate. Temperature corrections should then be made if the temperature of the wood differs considerably from the temperature of calibration used by the manufacturer.
12–3 Approximate corrections for conductance-type (resistance) meters are made by adding or subtracting about 0.5% for each 5.6°C (10°F) the wood temperature differs from the calibration temperature. The correction factors are added to the readings for temperatures less than the calibration tem-perature and subtracted from the readings for temperatures greater than the calibration temperature. Temperature correc-tions for dielectric meters are rather complex and are best made from published charts (James 1988).
Although some meters have scales that go up to 120%, the range of moisture content that can be measured reliably is 4% to about 30% for commercial dielectric meters and about 6% to 30% for resistance meters. The precision of the individual meter readings decreases near the limits of these ranges.
Readings greater than 30% must be considered only qualita-tive. When the meter is properly used on a quantity of lum-ber dried to a reasonably constant moisture content below fiber saturation, the average moisture content from the corrected meter readings should be within 1% of the true average.
Recommended Moisture Content Wood should be installed at moisture content levels as close as possible to the average moisture content it will experience in service. This minimizes the seasonal variation in moisture content and dimension after installation, avoiding problems such as floor buckling or cracks in furniture. The in-service moisture content of exterior wood (siding, wood trim) primarily depends on the outdoor relative humidity and exposure to rain or sun. The in-service moisture content of interior wood primarily depends on indoor relative humidity, which in turn is a complex function of moisture sources, ventilation rate, dehumidification (for example, air conditioning), and outdoor humidity conditions.
The recommended values for interior wood presented in this chapter are based on measurements in well-ventilated build-ings without unusual moisture sources and without air con-ditioning. In air-conditioned buildings, moisture conditions depend to a great extent on the proper sizing of the air-conditioning equipment. Wood installed in basements or over a crawl space may experience a moisture content greater than the range provided, and wood in insulated walls or roofs and attics may experience a moisture content greater or less than the range. Nevertheless, the recommended values for installation provide a useful guideline.
Timbers Ideally, solid timbers should be dried to the average mois-ture content they will reach in service. Although this opti-mum is possible with lumber less than 76 mm (3 in.) thick, it is seldom practical to obtain fully dried timbers, thick joists, and planks. When thick solid members are used, some shrinkage of the assembly should be expected. In the case of built-up assemblies, such as roof trusses, it may be necessary to tighten bolts or other fastenings occasionally to maintain full bearing of the connectors as the members shrink.
Lumber The recommended moisture content of wood should be matched as closely as is practical to the equilibrium moisture content (EMC) conditions in service. Table 12–1 shows the EMC conditions in outdoor exposure in various U.S. cities for each month. The EMC data are based on the average relative humidity and temperature data (30 or more years) available from the National Climatic Data Center of the National Oceanic and Atmospheric Administration. The relative humidity data were the average of the morning and afternoon values, and in most cases would be representative of the EMC attained by the wood. However, in some loca-tions, early morning relative humidity may occasionally reach 100%. Under these conditions, condensation may occur and the surface fibers of wood will exceed the EMC. The moisture content requirements are more exacting for finished lumber and wood products used inside heated and air-conditioned buildings than those for lumber used outdoors or in unheated buildings. For general areas of the United States, the recommended moisture content values for wood used inside heated buildings are shown in Figure 12–1. Values and tolerances for both interior and exterior uses of wood in various forms are given in Table 12–2. If the average mois-ture content is within 1% of that recommended and all pieces fall within the individual limits, the entire lot is probably satisfactory.
General commercial practice is to kiln dry wood for some products, such as flooring and furniture, to a slightly lower moisture content than service conditions demand, anticipat-ing a moderate increase in moisture content during process-ing and construction. This practice is intended to ensure uniform distribution of moisture among the individual pieces. Common grades of softwood lumber and softwood dimension lumber are not normally dried to the moisture content values indicated in Table 12–2. Dry lumber, as defined in the American Softwood Lumber Standard, has a maximum moisture content of 19%. Some industry grading rules provide for an even lower maximum. For example, to be grade marked KD 15, the maximum moisture content permitted is generally 15%.
Glued Wood Products When veneers are bonded with cold-setting adhesives to make plywood, they absorb comparatively large quantities of moisture. To keep the final moisture content low and to minimize redrying of the plywood, the initial moisture content of the veneer should be as low as practical. However, very dry veneer is brittle and difficult to handle without damage, so the minimum practical moisture content is about 4%. Freshly glued plywood intended for interior service should be dried to the moisture content values given in Table 12–2. 12–4 Table 12–1. Equilibrium moisture content of wood, exposed to outdoor atmosphere, in several U.S. locations in 1997 Equilibrium moisture contenta (%) State City Jan.
Feb.
Mar.
Apr.
May June July Aug.
Sept.
Oct.
Nov.
Dec.
AK Juneau 16.5 16.0 15.1 13.9 13.6 13.9 15.1 16.5 18.1 18.0 17.7 18.1 AL Mobile 13.8 13.1 13.3 13.3 13.4 13.3 14.2 14.4 13.9 13.0 13.7 14.0 AZ Flagstaff 11.8 11.4 10.8 9.3 8.8 7.5 9.7 11.1 10.3 10.1 10.8 11.8 AZ Phoenix 9.4 8.4 7.9 6.1 5.1 4.6 6.2 6.9 6.9 7.0 8.2 9.5 AR Little Rock 13.8 13.2 12.8 13.1 13.7 13.1 13.3 13.5 13.9 13.1 13.5 13.9 CA Fresno 16.4 14.1 12.6 10.6 9.1 8.2 7.8 8.4 9.2 10.3 13.4 16.6 CA Los Angeles 12.2 13.0 13.8 13.8 14.4 14.8 15.0 15.1 14.5 13.8 12.4 12.1 CO Denver 10.7 10.5 10.2 9.6 10.2 9.6 9.4 9.6 9.5 9.5 11.0 11.0 DC Washington 11.8 11.5 11.3 11.1 11.6 11.7 11.7 12.3 12.6 12.5 12.2 12.2 FL Miami 13.5 13.1 12.8 12.3 12.7 14.0 13.7 14.1 14.5 13.5 13.9 13.4 GA Atlanta 13.3 12.3 12.0 11.8 12.5 13.0 13.8 14.2 13.9 13.0 12.9 13.2 HI Honolulu 13.3 12.8 11.9 11.3 10.8 10.6 10.6 10.7 10.8 11.3 12.1 12.9 ID Boise 15.2 13.5 11.1 10.0 9.7 9.0 7.3 7.3 8.4 10.0 13.3 15.2 IL Chicago 14.2 13.7 13.4 12.5 12.2 12.4 12.8 13.3 13.3 12.9 14.0 14.9 IN Indianapolis 15.1 14.6 13.8 12.8 13.0 12.8 13.9 14.5 14.2 13.7 14.8 15.7 IA Des Moines 14.0 13.9 13.3 12.6 12.4 12.6 13.1 13.4 13.7 12.7 13.9 14.9 KS Wichita 13.8 13.4 12.4 12.4 13.2 12.5 11.5 11.8 12.6 12.4 13.2 13.9 KY Louisville 13.7 13.3 12.6 12.0 12.8 13.0 13.3 13.7 14.1 13.3 13.5 13.9 LA New Orleans 14.9 14.3 14.0 14.2 14.1 14.6 15.2 15.3 14.8 14.0 14.2 15.0 ME Portland 13.1 12.7 12.7 12.1 12.6 13.0 13.0 13.4 13.9 13.8 14.0 13.5 MA Boston 11.8 11.6 11.9 11.7 12.2 12.1 11.9 12.5 13.1 12.8 12.6 12.2 MI Detroit 14.7 14.1 13.5 12.6 12.3 12.3 12.6 13.3 13.7 13.5 14.4 15.1 MN Minneapolis–St.Paul 13.7 13.6 13.3 12.0 11.9 12.3 12.5 13.2 13.8 13.3 14.3 14.6 MS Jackson 15.1 14.4 13.7 13.8 14.1 13.9 14.6 14.6 14.6 14.1 14.3 14.9 MO St. Louis 14.5 14.1 13.2 12.4 12.8 12.6 12.9 13.3 13.7 13.1 14.0 14.9 MT Missoula 16.7 15.1 12.8 11.4 11.6 11.7 10.1 9.8 11.3 12.9 16.2 17.6 NE Omaha 14.0 13.8 13.0 12.1 12.6 12.9 13.3 13.8 14.0 13.0 13.9 14.8 NV Las Vegas 8.5 7.7 7.0 5.5 5.0 4.0 4.5 5.2 5.3 5.9 7.2 8.4 NV Reno 12.3 10.7 9.7 8.8 8.8 8.2 7.7 7.9 8.4 9.4 10.9 12.3 NM Albuquerque 10.4 9.3 8.0 6.9 6.8 6.4 8.0 8.9 8.7 8.6 9.6 10.7 N Y New York 12.2 11.9 11.5 11.0 11.5 11.8 11.8 12.4 12.6 12.3 12.5 12.3 NC Raleigh 12.8 12.1 12.2 11.7 13.1 13.4 13.8 14.5 14.5 13.7 12.9 12.8 ND Fargo 14.2 14.6 15.2 12.9 11.9 12.9 13.2 13.2 13.7 13.5 15.2 15.2 OH Cleveland 14.6 14.2 13.7 12.6 12.7 12.7 12.8 13.7 13.8 13.3 13.8 14.6 OK Oklahoma City 13.2 12.9 12.2 12.1 13.4 13.1 11.7 11.8 12.9 12.3 12.8 13.2 OR Pendleton 15.8 14.0 11.6 10.6 9.9 9.1 7.4 7.7 8.8 11.0 14.6 16.5 OR Portland 16.5 15.3 14.2 13.5 13.1 12.4 11.7 11.9 12.6 15.0 16.8 17.4 PA Philadelphia 12.6 11.9 11.7 11.2 11.8 11.9 12.1 12.4 13.0 13.0 12.7 12.7 SC Charleston 13.3 12.6 12.5 12.4 12.8 13.5 14.1 14.6 14.5 13.7 13.2 13.2 SD Sioux Falls 14.2 14.6 14.2 12.9 12.6 12.8 12.6 13.3 13.6 13.0 14.6 15.3 TN Memphis 13.8 13.1 12.4 12.2 12.7 12.8 13.0 13.1 13.2 12.5 12.9 13.6 T X Dallas–Ft.Worth 13.6 13.1 12.9 13.2 13.9 13.0 11.6 11.7 12.9 12.8 13.1 13.5 T X El Paso 9.6 8.2 7.0 5.8 6.1 6.3 8.3 9.1 9.3 8.8 9.0 9.8 UT Salt Lake City 14.6 13.2 11.1 10.0 9.4 8.2 7.1 7.4 8.5 10.3 12.8 14.9 VA Richmond 13.2 12.5 12.0 11.3 12.1 12.4 13.0 13.7 13.8 13.5 12.8 13.0 WA Seattle–Tacoma 15.6 14.6 15.4 13.7 13.0 12.7 12.2 12.5 13.5 15.3 16.3 16.5 WI Madison 14.5 14.3 14.1 12.8 12.5 12.8 13.4 14.4 14.9 14.1 15.2 15.7 WV Charleston 13.7 13.0 12.1 11.4 12.5 13.3 14.1 14.3 14.0 13.6 13.0 13.5 W Y Cheyenne 10.2 10.4 10.7 10.4 10.8 10.5 9.9 9.9 9.7 9.7 10.6 10.6 aEMC values were determined from the average of 30 or more years of relative humidity and temperature data available from the National Climatic Data Center of the National Oceanic and Atmospheric Administration.
12–5 Hot-pressed plywood and other board products, such as particleboard and hardboard, usually do not have the same moisture content as lumber. The high temperatures used in hot presses cause these products to assume a lower moisture content for a given relative humidity. Because this lower equilibrium moisture content varies widely, depending on the specific type of hot-pressed product, it is recommended that such products be conditioned at 30% to 40% relative humidity for interior use and 65% for exterior use.
Lumber used in the manufacture of large laminated members should be dried to a moisture content slightly less than the moisture content expected in service so that moisture ab-sorbed from the adhesive will not cause the moisture content of the product to exceed the service value. The range of moisture content between laminations assembled into a single member should not exceed 5 percentage points.
Although laminated members are often massive and respond rather slowly to changes in environmental conditions, it is desirable to follow the recommendations in Table 12–2 for moisture content at time of installation.
Drying of Wood Drying is required for wood to be used in most products.
Dried lumber has many advantages over green lumber for producers and consumers. Removal of excess water reduces weight, thus shipping and handling costs. Proper drying confines shrinking and swelling of wood in use to manage-able amounts under all but extreme conditions of relative humidity or flooding. As wood dries, most of its strength properties increase, as well as its electrical and thermal insu-lating properties. Properly dried lumber can be cut to precise dimensions and machined more easily and efficiently; wood parts can be more securely fitted and fastened together with nails, screws, bolts, and adhesives; warping, splitting, checking, and other harmful effects of uncontrolled drying are largely eliminated; and paint, varnish, and other finishes are more effectively applied and maintained. Wood must be relatively dry before it can be glued or treated with decay-preventing and fire-retardant chemicals.
The key to successful and efficient drying is control of the drying process. Timely application of optimum or at least adequate temperature, relative humidity, and air circulation conditions is critical. Uncontrolled drying leads to drying defects that can adversely affect the serviceability and eco-nomics of the product. The usual strategy is to dry as fast as the particular species, thickness, and end-product require-ments allow without damaging the wood. Slower drying can be uneconomical as well as introduce the risk of stain.
Softwood lumber intended for framing in construction is usually targeted for drying to an average moisture content of 15%, not to exceed 19%. Softwood lumber for many other uses is dried to a low moisture content, 10% to 12% for many appearance grades to as low as 7% to 9% for furniture, cabinets, and millwork. Hardwood lumber for framing in construction, although not in common use, should also be dried to an average moisture content of 15%, not to exceed 19%. Hardwood lumber for furniture, cabinets, and millwork is usually dried to 6% to 8% moisture content. 6% average moisture content 8% average moisture content 11% average moisture content Figure 12–1. Recommended average moisture content for interior use of wood products in various areas of the United States.
Table 12–2. Recommended moisture content values for various wood items at time of installation Recommended moisture content (%) in various climatological regions Most areas of the United States Dry southwestern areaa Damp, warm coastal areaa Use of wood Averageb Individual pieces Averageb Individual pieces Averageb Individual pieces Interior: woodwork, flooring, furniture, wood trim 8 6–10 6 4–9 11 8–13 Exterior: siding, wood trim, sheathing, laminated timbers 12 9–14 9 7–12 12 9–14 aMajor areas are indicated in Figure 12–1.
bTo obtain a realistic average, test at least 10% of each item. If the quantity of a given item is small, make several tests. For example, in an ordinary dwelling having about 60 floor joists, at least 10 tests should be made on joists selected at random. 12–6 Lumber drying is usually accomplished by some combina-tion of air drying, accelerated air drying or pre-drying, and kiln drying. Wood species, lumber thickness, economics, and end use are often the main factors in determining the details of the drying process.
Air Drying The main purpose of air drying lumber is to evaporate as much of the water as possible before end use or transfer to a dry kiln. Air drying usually extends until wood moisture content is as low as 20% to 25%, at which time the lumber is transferred to a dry kiln if final drying to a lower moisture content is required. Sometimes, depending on a mill’s scheduling, air drying may be cut short at a higher moisture content before the wood is sent to the dry kiln. Air drying saves energy costs and reduces required dry kiln capacity.
Limitations of air drying are generally associated with uncon-trolled drying. The drying rate is very slow during the cold winter months. At other times, hot, dry winds may increase degrade and volume losses as a result of severe surface check-ing and end splitting. Warm, humid periods with little air movement may encourage the growth of fungal stains, as well as aggravate chemical stains. Another limitation of air drying is the high cost of carrying a large inventory of high value lumber for extended periods. Air drying time to 20% to 25% moisture content varies widely, depending on spe-cies, thickness, location, and the time of year the lumber is stacked. Some examples of extremes for 25-mm- (1-in.-) thick lumber are 15 to 30 days for some of the low density species, such as pine, spruce, red alder, and soft maple, stacked in favorable locations and favorable times of the year, to 200 to 300 days for slow drying species, such as sinker hemlock and pine, oak, and birch, in northern locations and stacked at unfavorable times of the year. Details of important air drying considerations, such as lumber stacking and air drying yard layout, are covered in Air Drying of Lumber: A Guide to Industry Practices (Rietz and Page 1971).
Accelerated Air Drying and Pre-Drying The limitations of air drying have led to increased use of technology that reduces drying time and introduces some control into drying from green to 20% to 25% moisture content. Accelerated air drying involves the use of fans to force air through lumber piles in a shed. This protects the lumber from the elements and improves air circulation com-pared with air drying. Small amounts of heat are sometimes used to reduce relative humidity and slightly increase tem-perature. Pre-dryers take this acceleration and control a step further by providing control of both temperature and relative humidity and providing forced air circulation in a completely enclosed compartment. Typical conditions in a pre-dryer are 27°C to 38°C (80°F to 100°F) and 65% to 85% relative humidity.
Kiln Drying In kiln drying, higher temperatures and faster air circulation are used to increase drying rate considerably. Specific kiln schedules have been developed to control temperature and relative humidity in accordance with the moisture content and stress situation within the wood, thus minimizing shrinkage-caused defects.
Drying Mechanism Water in wood normally moves from high to low zones of moisture content, which means that the surface of the wood must be drier than the interior if moisture is to be removed.
Drying can be broken down into two phases: movement of water from the interior to the surface of the wood and evapo-ration of water from the surface. The surface fibers of most species reach moisture equilibrium with the surrounding air soon after drying begins. This is the beginning of the development of a typical moisture gradient (Fig. 12–2), that is, the difference in moisture content between the inner and outer portions of a board. If air circulation is too slow, a longer time is required for the surfaces of the wood to reach moisture equilibrium. This is one reason why air circulation is so important in kiln drying. If air circulation is too slow, drying is also slower than necessary and mold could develop on the surface of lumber. If drying is too fast, electrical energy in running the fans is wasted, and in certain species, Figure 12–2. Typical moisture gradient in lumber during drying at time increasing from t1 to t3.
12–7 surface checking and other drying defects can develop if relative humidity and air velocity are not coordinated.
Water moves through the interior of wood as a liquid or vapor through various air passageways in the cellular struc-ture of the wood, as well as through the wood cell walls.
Moisture moves in these passageways in all directions, both across and with the grain. In general, lighter species dry faster than heavier species because the structure of lighter wood contains more openings per unit volume and moisture moves through air faster than through wood cell walls. Water moves by two main mechanisms: capillary action (liquid) and diffusion of bound water (vapor). Capillary action causes free water to flow through cell cavities and the small pas-sageways that connect adjacent cell cavities. Diffusion of bound water moves moisture from areas of high concentration to areas of low concentration. Diffusion in the longitudinal direction is about 10 to 15 times faster than radial or tangen-tial diffusion, and radial diffusion is somewhat faster than tangential diffusion. This explains why flatsawn lumber generally dries faster than quartersawn lumber. Although longitudinal diffusion is much faster than diffusion across the grain, it generally is not of practical importance in lumber that is many times longer than it is thick.
Because chemical extractives in heartwood plug up passage-ways, moisture generally moves more freely in sapwood than in heartwood; thus, sapwood generally dries faster than heartwood. However, the heartwood of many species is lower in moisture content than is the sapwood and can reach final moisture content as fast.
The rate at which moisture moves in wood depends on the relative humidity of the surrounding air, the steepness of the moisture gradient, and the temperature of the wood. The lower the relative humidity, the greater the capillary flow.
Low relative humidity also stimulates diffusion by lowering the moisture content at the surface, thereby steepening the moisture gradient and increasing the diffusion rate. The greater the temperature of the wood, the faster moisture will move from the wetter interior to the drier surface. If relative humidity is too low in the early stages of drying, excessive shrinkage may occur, resulting in surface and end checking.
If the temperature is too high, collapse, honeycomb, or strength reduction can occur.
Drying Stresses Drying stresses are the main cause of nonstain-related drying defects. Understanding these stresses provides a means for minimizing and recognizing the damage they can cause.
The cause of drying stresses in the differential shrinkage between the outer part of a board (the shell) and the interior part (the core) can also cause drying defects. Early in drying, the fibers in the shell dry first and begin to shrink. However, the core has not yet begun to dry and shrink; consequently, the core prevents the shell from shrinking. Thus, the shell goes into tension and the core into compression (Fig. 12–3).
If the shell dries too rapidly, it is stressed beyond the elastic limit and dries in a permanently stretched (set) condition without attaining full shrinkage. Sometimes surface cracks, or checks, occur during this initial stage of drying, and they can be a serious defect for many uses. As drying progresses, the core begins to dry and attempts to shrink. However, the shell is set in a permanently expanded condition and pre-vents normal shrinkage of the core. This causes the stresses to reverse; the core goes into tension and the shell into compression. The change in the shell and core stresses and in the moisture content level during drying is shown in Figure 12–4. These internal tension stresses may be severe enough to cause internal cracks (honeycomb).
Differential shrinkage caused by differences in radial, tangen-tial, and longitudinal shrinkage is a major cause of warp.
The distortions shown in Figure 3–3 in Chapter 3 are due to differential shrinkage. When juvenile or reaction wood is present on one edge or face of a board and normal wood is present on the opposite side, the difference in their longitudi-nal shrinkage can also cause warp.
Dry Kilns Most dry kilns are thermally insulated compartments de-signed for a batch process in which the kiln is completely loaded with lumber in one operation and the lumber remains stationary during the entire drying cycle. Temperature and relative humidity are kept as uniform as possible throughout the kiln and can be controlled over a wide range. Tempera-ture and relative humidity are changed as the wood dries based on a schedule that takes into account the moisture content and/or the drying rate of the lumber. All dry kilns Figure 12–3. End view of board showing development of drying stresses (a) early and (b) later in drying. 12–8 use some type of forced-air circulation, with air moving through the lumber perpendicular to the length of the lumber and parallel to the spacers (stickers) that separate each layer of lumber in a stack.
Three general types of kilns are in common use. One is the track-loaded type (Fig. 12–5), where lumber is stacked on kiln trucks that are rolled in and out of the kiln on tracks.
The majority of softwood lumber in the United States is dried in this kiln type. Another major type is the package-loaded kiln (Fig 12–6), where individual stacks of lumber are fork-lifted into place in the kiln. This type of kiln is com-monly used for drying hardwood lumber. These kilns are most commonly heated with steam, although softwood lumber kilns are sometimes directly heated. A third common type of kiln, usually package loaded, is the dehumidification kiln. Instead of venting humid air to remove water, as the other two types of kilns do, water is removed by condensa-tion on cold dehumidifier coils (Fig. 12–7).
Kiln Schedules A kiln schedule is a carefully developed compromise between the need to dry lumber as fast as possible for economic efficiency and the need to avoid severe drying conditions that will lead to drying defects. A kiln schedule is a series of temperatures and relative humidities that are applied at vari-ous stages of drying. In most schedules, the temperature is gradually increased and the relative humidity decreased. The schedule for Southern Pine structural lumber is an exception to this general rule. This is lumber usually dried at a con-stant temperature and relative humidity. Temperatures are chosen to strike this compromise of a satisfactory drying rate and avoidance of objectionable drying defects. The stresses that develop during drying are the limiting factor in deter-mining the kiln schedule. The schedule must be developed so that the drying stresses do not exceed the strength of the wood at any given temperature and moisture content. Other-wise, the wood will crack either on the surface or internally or be crushed by forces that collapse the wood cells. Wood generally becomes stronger as the moisture content decreases, and to a lesser extent, it becomes weaker as temperature increases. The net result is that as wood dries it becomes stronger because of the decreasing moisture content and can tolerate higher drying temperatures and lower relative humid-ities without cracking. This is a fortunate circumstance because as wood dries, its drying rate decreases at any given temperature, and the ability to increase drying temperature helps maintain a reasonably fast drying rate. Thus, rapid drying is achieved in kilns by the use of temperatures as high as possible and relative humidities as low as possible.
Drying schedules vary by species, thickness, grade, and end use of lumber. There are two general types of kiln schedules: moisture content schedules and time-based schedules. Most hardwood lumber is dried by moisture content schedules.
This means that the temperature and relative humidity condi-tions are changed according to various moisture content Figure 12–4. Moisture–stress relationship during six stages of kiln drying 50-mm- (2-in.-) thick red oak.
Figure 12–5. Lineshaft, double-track, compartment kiln with alternately opposing fans. Vents are over fan shaft between fans. Vent on high pressure side of fans becomes fresh air inlet when direction of circulation is reversed.
12–9 levels attained by the lumber during drying. A typical hard-wood schedule might begin at 49°C (120°F) and 80% rela-tive humidity when the lumber is green. By the time the lumber has reached 15% moisture content, the temperature is as high as 82°C (180°F). A typical hardwood drying sched-ule is shown in Table 12–3. Some method of monitoring moisture content during drying is required for schedules based on moisture content. One common method is the use of short kiln samples that are periodically weighed, usually manually but potentially remotely with load cells. Alterna-tively, electrodes are imbedded in sample boards to sense the change in electrical conductivity with moisture content. This system is limited to moisture content values less than 30%.
Softwood kiln schedules generally differ from hardwood schedules in that changes in kiln temperature and relative humidity are made at predetermined times rather than mois-ture content levels. Examples of time-based schedules, both conventional temperature (<100°C (<212°F )) and high temperature (>110°C (>230°F )), are given in Table 12–3.
Drying Defects Most drying defects or problems that develop in wood prod-ucts during drying can be classified as fracture or distortion, warp, or discoloration. Defects in any one of these categories are caused by an interaction of wood properties with process-ing factors. Wood shrinkage is mainly responsible for wood ruptures and distortion of shape. Cell structure and chemical extractives in wood contribute to defects associated with uneven moisture content, undesirable color, and undesirable surface texture. Drying temperature is the most important processing factor because it can be responsible for defects in each category.
Fracture or Distortion Surface checks occur early in drying when the shell of a board is stressed in tension enough to fracture the wood. These checks occur most often on the face of flatsawn boards and are illustrated in Figure 12–8. End checks (Fig. 12–9) are simi-lar to surface checks but appear on the ends of boards. End checks occur because the rapid longitudinal movement of moisture causes the board end to dry very quickly and Figure 12–6. Package-loaded kiln with fans connected directly to motors. 12–10 develop high stresses, therefore fracturing. End coatings, either on the log or freshly sawn lumber, are an effective preventative measure. Collapse is a distortion, flattening, or crushing of wood cells. In severe cases (Fig. 12–10), collapse usually shows up as grooves or corrugations, a washboarding effect. Less severe collapse shows up as excessive thickness shrinkage and may not be a serious problem. Honeycomb (Fig. 12–11) is an internal crack that occurs in the later stages of kiln drying when the core of a board is in tension.
It is caused when the core is still at a relatively high mois-ture content and drying temperatures are too high for too long during this critical drying period. Nondestructive test-ing methods, using speed of sound, have been found to be effective in detecting the presence of these cracks in dried lumber. Knots may loosen during drying because of the unequal shrinkage between the knot and the surrounding wood (Fig. 12–12).
Warp Warp in lumber is any deviation of the face or edge of a board from flatness or any edge that is not at right angles to the adjacent face or edge. Warp can be traced to two causes: (a) differences between radial, tangential, and longitudinal shrinkage in the piece as it dries or (b) growth stresses. Warp is aggravated by irregular or distorted grain and the presence of abnormal types of wood, such as juvenile and reaction wood. The six major types of warp are bow, crook, twist, oval, diamond, and cup (Fig. 12–13).
Discoloration The use of dried wood products can be impaired by discol-oration, particularly when the end use requires a clear, natural finish. Unwanted discoloration can develop in the tree, during storage of logs and green lumber, or during drying.
There are two general types of discoloration: chemical and fungal.
Chemical discoloration is the result of oxidative and enzy-matic reactions with chemical constituents in wood. Discol-orations range from pinkish, bluish, and yellowish hues through gray and reddish brown to dark brown shades.
Brown stain in pines and darkening in many hardwoods is a common problem when drying temperatures are too high (Fig. 12–14). A deep grayish-brown chemical discoloration can occur in many hardwood species if initial drying is too slow (Fig. 12–15).
Fungal stains, often referred to as blue or sap stain, are caused by fungi that grow in the sapwood (Fig. 12–16).
Blue-stain fungi do not cause decay of the sapwood, and fungi generally do not grow in heartwood. Blue stain can develop if initial drying is too slow.
Another common type of stain develops under stickers (Fig. 12–17). This stain results from contact of the sticker with the board. Sticker stains (sometimes called shadow) are imprints of the sticker that are darker or lighter than the wood between the stickers and can be caused by either chemical or fungal action, or both.
Moisture Content of Dried Lumber Although widely used, the trade terms “shipping dry,” “air dry,” and “kiln dry” may not have identical meanings as to moisture content in the different producing regions. Despite the wide variations in the use of these terms, they are some-times used to describe dried lumber. The following state-ments, which are not exact definitions, outline these categories.
Shipping Dry Shipping dry means lumber that has been partially dried to prevent stain or mold during brief periods of transit; ideally the outer 3.2 mm (1/8 in.) is dried to 25% or less moisture content.
Fig. 12–7. A typical dehumidification kiln (top) and dehumidification drying system (bottom).
12–11 Air Dry Air dry means lumber that has been dried by exposure to the air outdoors or in a shed or by forced circulation of air that has not been heated above 49°C (120°F ). Commercial air-dry stock generally has an average moisture content low enough for rapid kiln drying or rough construction use.
Moisture content is generally in the range of 20% to 25% for dense hardwoods and 15% to 20% for softwoods and low-density hardwoods. Extended exposure can bring standard 19- and 38-mm (nominal 1- and 2-in.) lumber within one or two percentage points of the average exterior equilibrium moisture content of the region. For much of the United States, the minimum moisture content of thoroughly air-dried lumber is 12% to 15%.
Table 12–3. Typical dry kiln schedules for lumber Moisture content-based schedule for 25-mm (1-in.) (4/4) black walnut, dried to 7% moisture content Temperature (°C(°F)) Relative Equilibrium Moisture content (%) Dry-bulb Wet-bulb humidity (%) moisture content (%) Above 50 49.0 (120) 45.0 (113) 80 14.4 50 to 40 49.0 (120) 43.5 (110) 72 12.1 40 to 35 49.0 (120) 40.5 (105) 60 9.6 35 to 30 49.0 (120) 35.0 (95) 40 6.5 30 to 25 54.5 (130) 32.0 (90) 22 4.0 25 to 20 60.0 (140) 32.0 (90) 15 2.9 20 to 15 65.5 (150) 37.5 (100) 18 3.2 15 to 7 82.2 (180) 54.5 (130) 26 3.5 Equalize 82.2 (180) 58.3 (137) 30 3.8 Condition 82.2 (180) 76.7 (170) 79 11.1 Time-based schedule for 25- to 50-mm (1- to 2-in.) (4/4 to 8/4) Douglas Fir, upper grades, dried to 12% moisture content Temperature (°C(°F)) Relative Equilibrium Time (h) Dry-bulb Wet-bulb humidity (%) moisture content (%) 0 to 12 76.5 (170) 73.5 (164) 86 14.1 12 to 24 76.5 (170) 71.0 (160) 78 11.4 24 to 48 79.5 (175) 71.0 (160) 69 9.1 48 to 72 82.2 (180) 71.0 (160) 62 7.7 72 to 96 or until dry 82.2 (180) 60.0 (140) 36 4.5 High temperature schedule for 50- by 100-mm to 50- by 250-mm (2- by 4-in. to 2- by 10-in.) Southern Pine, dried to 15% moisture content Temperature (°C(°F)) Relative Equilibrium Time (h) Dry-bulb Wet-bulb humidity (%) moisture content (%) 0 until dry 116 (240) 82.2 (180) 29 2.5 12–12 Figure 12–11. Board machined into millwork shows honeycomb (top). Surface of planed red oak board shows no honeycomb (bottom).
Figure 12–12. Loose knot in Southern Pine.
Kiln Dry Kiln dry means lumber that has been dried in a kiln or by some special drying method to an average moisture content specified or understood to be suitable for a certain use. The average moisture content should have upper and lower toler-ance limits, and all values should fall within these limits.
Kiln-dried softwood dimension lumber generally has an average moisture content of 19% or less; the average mois-ture content for many other softwood uses is 10% to 20%.
Hardwood and softwood lumber for furniture, cabinetry, and millwork usually has a final moisture content of 6% to 8% and can be specified to be free of drying stresses. The impor-tance of suitable moisture content values is recognized, and provisions covering them are now incorporated in some softwood standards as grading rules. Moisture content values in the general grading rules may or may not be suitable for a specific use; if not, a special moisture content specification should be made.
Figure 12–8. Surface checking on Douglas Fir dimension lumber.
Figure 12–9. End checking in oak lumber.
Figure 12–10. Severe collapse in western redcedar.
12–13 Figure 12–15. Gray sapwood stain in southern red oak that was dried green with humid, low temperature conditions and poor air circulation.
Figure 12–16. Sap stain in Southern Pine. Color ranges from bluish gray to black.
Figure 12–13. Various types of warp that can develop in boards during drying.
Figure 12–14. Brown sapwood stain in Southern Pine lumber. 12–14 Moisture Control During Transit and Storage Lumber and other wood items may change in moisture content and dimension while awaiting shipment, during fabrication, in transit, and in storage.
When standard 19-mm (nominal 1-in.) dry softwood lumber is shipped in tightly closed boxcars, shipping containers, or trucks or in packages with complete and intact wrappers, average moisture content changes for a package can generally be held to 0.2% or less per month. In holds or between decks of ships, dry material usually adsorbs about 1.5% moisture during normal shipping periods. If green material is included in the cargo, the moisture regain of the dry lumber may be doubled. On the top deck, if unprotected from the elements, the moisture regain can be as much as 7%.
When standard 19-mm (nominal 1-in.) softwood lumber, kiln dried to 8% or less, is piled solid under a good pile roof in a yard in humid weather, average moisture content of a pile can increase at the rate of about 2% per month during the first 45 days. An absorption rate of about 1% per month can then be sustained throughout a humid season. Comparable initial and sustaining absorption rates are about 1% per month in open (roofed) sheds and 0.3% per month in closed sheds. Stock that was piled for a year in an open shed in a western location increased 2.7% on the inside of solid piles and 3.5% on the outside of the piles. All stock that has been manufactured in any way should be protected from precipita-tion and spray, because water that gets into a solid pile tends to be absorbed by the wood instead of evaporating. The extent to which additional control of the storage environment is required depends upon the use to which the wood will be put and the corresponding moisture content recommenda-tions. The moisture content of all stock should be deter-mined when it is received. If moisture content is not as specified or required, stickered storage in an appropriate condition could ultimately bring the stock within the desired moisture content range. If a large degree of moisture change is required, the stock must be redried.
Plywood and Structural Items Green or partially dried lumber and timbers should be open piled on stickers and protected from sunshine and precipita-tion by a tight roof. Framing lumber and plywood with 20% or less moisture content can be solid piled in a shed that provides good protection against sunshine and direct or wind-driven precipitation. However, a better practice for stock with greater than 12% moisture content is the use of stickered piling to bring moisture content more in line with the moisture content in use. Dry lumber can be piled solid in the open for relatively short periods, but at least a minimum pile cover of waterproofed paper should be used whenever possible. Because it is difficult to keep rain out completely, storing solid-piled lumber in the open for long periods is not recommended. If framing lumber must be stored in the open for a long time, it should be piled on stickers over good supports and the piles should be roofed. Solid-piled material that has become wet again should also be re-piled on stickers.
Figure 12–17. Sticker stain in sapwood of sugar maple after planing.
Table 12–4. Amount by which temperature of storage area must be increased above outside temperature to maintain equilibrium moisture content Outside relative Temperature differential (°C (°F)) for desired equilibrium moisture content humidity (%) 6% 7% 8% 9% 10% 11% 12% 90 18.3 (33) 16.1 (29) 12.8 (23) 10.0 (18) 8.3 (15) 6.1 (11) 5.0 (9) 80 16.7 (30) 13.9 (25) 10.5 (19) 7.8 (14) 6.1 (11) 4.4 (8) 3.3 (6) 70 13.9 (25) 11.1 (20) 8.3 (15) 5.6 (10) 3.9 (7) 2.2 (4) 1.7 (3) 60 11.1 (20) 8.3 (15) 5.0 (9) 3.3 (6) 1.7 (3) — — 50 8.3 (15) 5.6 (10) 2.8 (5) 0.6 (1) — — — 12–15 Finish and Factory Lumber Such kiln-dried items as exterior finish, siding, and exterior millwork should be stored in a closed but unheated shed.
They should be placed on supports raised above the floor, at least 150 mm (6 in.) high if the floor is paved or 300 mm (12 in.) if not paved. Interior trim, flooring, cabinet work, and lumber for processing into furniture should be stored in a room or closed shed where relative humidity is controlled.
Kiln-dried and machined hardwood dimension or softwood cut stock should also be stored under controlled humidity conditions.
Dried and machined hardwood dimension or softwood lumber intended for remanufacture should also be stored under controlled humidity conditions. Under uncontrolled conditions, the ends of such stock may attain a greater mois-ture content than the balance of the length. Then, when the stock is straight-line ripped or jointed before edge gluing, subsequent shrinkage will cause splitting or open glue joints at the ends of panels. The simplest way to reduce relative humidity in storage areas of all sizes is to heat the space to a temperature slightly greater than that of the outside air.
Dehumidifiers can be used in small, well-enclosed spaces.
If the heating method is used, and there is no source of mois-ture except that contained in the air, the equilibrium mois-ture content can be maintained by increasing the temperature of the storage area greater than the outside temperature by the amounts shown in Table 12–4. When a dehumidifier is used, the average temperature in the storage space should be known or controlled. Table 3–4 in Chapter 3 should be used to select the proper relative humidity to give the desired average moisture content. Wood in a factory awaiting or following manufacture can become too dry if the area is heated to 21°C (70°F) or greater when the outdoor tempera-ture is low. This often occurs in the northern United States during the winter. Under such circumstances, exposed ends and surfaces of boards or cut pieces will tend to dry to the low equilibrium moisture content condition, causing shrink-age and warp. In addition, an equilibrium moisture content of 4% or more below the moisture content of the core of freshly crosscut boards can cause end checking. Simple remedies are to cover piles of partially manufactured items with plastic film and lower the shop temperature during nonwork hours. Increased control can be obtained in critical shop and storage areas by humidification. In warm weather, cooling can increase relative humidity and dehumidification may be necessary.
Dimensional Changes in Wood Dry wood undergoes small changes in dimension with nor-mal changes in relative humidity. More humid air will cause slight swelling, and drier air will cause slight shrinkage.
These changes are considerably smaller than those involved with shrinkage from the green condition. Equation (12–2) can be used to approximate dimensional changes caused by shrinking and swelling by using the total shrinkage coeffi-cient from green to ovendry. However, the equation assumes that the shrinkage–moisture content relationship is linear.
Figure 3–4 (Ch. 3) shows that this is not the case, so some error is introduced. The error is in the direction of underes-timating dimensional change, by about 5% of the true change. Many changes of moisture content in use are over the small moisture content range of 6% to 14%, where the shrinkage–moisture content relationship is linear (Ch. 3, Fig. 3–4). Therefore, a set of shrinkage coefficients based on the linear portion of the shrinkage–moisture content curve has been developed (Table 12–5). Approximate changes in dimension can be estimated by a simple formula that in-volves a dimensional change coefficient, from Table 12–5, when moisture content remains within the range of normal use. (Dimensional changes are further discussed in Chs. 3 and 6.) Estimation Using Dimensional Change Coefficient The change in dimension within the moisture content limits of 6% to 14% can be estimated satisfactorily by using a dimensional change coefficient based on the dimension at 10% moisture content: ∆D D C M M = − [ ] I F I T( ) (12–2) where ∆D is change in dimension, DI dimension in units of length at start of change, CT dimensional change coefficient tangential direction (for radial direction, use CR), MF mois-ture content (%) at end of change, and MI moisture content (%) at start of change.
Values for CT and CR, derived from total shrinkage values, are given in Table 12–5. When MF < MI, the quantity (MF − MI) will be negative, indicating a decrease in dimension; when greater, it will be positive, indicating an increase in dimension.
As an example, assuming the width of a flat-grained white fir board is 232 mm (9.15 in.) at 8% moisture content, its change in width at 11% moisture content is estimated as ∆D = 232[0.00245(11 − 8)] = 232(0.00735) = 1.705 mm ∆D = 9.15[0.00245(11 − 8)] = 9.15[0.00735] = 0.06725 or 0.067 in.
Then, dimension at end of change DI + ∆D = 232 + 1.7 (= 9.15 + 0.067) = 233.7 mm (= 9.217 in.) The thickness of the same board at 11% moisture content can be estimated by using the coefficient CR = 0.00112. 12–16 Table 12–5. Coefficients for dimensional change as a result of shrinking or swelling within moisture content limits of 6% to 14% (CT = dimensional change coefficient for tangential direction; CR = radial direction) Dimensional change coefficienta Dimensional change coefficienta Species CR CT Species CR CT Hardwoods Alder, red 0.00151 0.00256 Honeylocust 0.00144 0.00230 Apple 0.00205 0.00376 Locust, black 0.00158 0.00252 Ash, black 0.00172 0.00274 Madrone, Pacific 0.00194 0.00451 Ash, Oregon 0.00141 0.00285 Magnolia, cucumbertree 0.00180 0.00312 Ash, pumpkin 0.00126 0.00219 Magnolia, southern 0.00187 0.00230 Ash, white 0.00169 0.00274 Magnolia, sweetbay 0.00162 0.00293 Ash, green 0.00169 0.00274 Maple, bigleaf 0.00126 0.00248 Aspen, quaking 0.00119 0.00234 Maple, red 0.00137 0.00289 Basswood, American 0.00230 0.00330 Maple, silver 0.00102 0.00252 Beech, American 0.00190 0.00431 Maple, black 0.00165 0.00353 Birch, paper 0.00219 0.00304 Maple, sugar 0.00165 0.00353 Birch, river 0.00162 0.00327 Oak, black 0.00123 0.00230 Birch, yellow 0.00256 0.00338 Red Oak, commercial 0.00158 0.00369 Birch, sweet 0.00256 0.00338 Red oak, California 0.00123 0.00230 Buckeye, yellow 0.00123 0.00285 Red oak: water, laurel, willow 0.00151 0.00350 Butternut 0.00116 0.00223 White Oak, commercial 0.00180 0.00365 Catalpa, northern 0.00085 0.00169 White oak, live 0.00230 0.00338 Cherry, black 0.00126 0.00248 White oak, Oregon white 0.00144 0.00327 Chestnut, American 0.00116 0.00234 White oak, overcup 0.00183 0.00462 Cottonwood, black 0.00123 0.00304 Persimmon, common 0.00278 0.00403 Cottonwood, eastern 0.00133 0.00327 Sassafras 0.00137 0.00216 Elm, American 0.00144 0.00338 Sweet gum 0.00183 0.00365 Elm, rock 0.00165 0.00285 Sycamore, American 0.00172 0.00296 Elm, slippery 0.00169 0.00315 Tanoak 0.00169 0.00423 Elm, winged 0.00183 0.00419 Tupelo, black 0.00176 0.00308 Elm, cedar 0.00183 0.00419 Tupelo, water 0.00144 0.00267 Hackberry 0.00165 0.00315 Walnut, black 0.00190 0.00274 Hickory, pecan 0.00169 0.00315 Willow, black 0.00112 0.00308 Hickory, true 0.00259 0.00411 Willow, Pacific 0.00099 0.00319 Holly, American 0.00165 0.00353 Yellow-poplar 0.00158 0.00289 Softwoods Baldcypress 0.00130 0.00216 Pine, eastern white 0.00071 0.00212 Cedar, yellow 0.00095 0.00208 Pine, jack 0.00126 0.00230 Cedar, Atlantic white 0.00099 0.00187 Pine, loblolly 0.00165 0.00259 Cedar, eastern red 0.00106 0.00162 Pine, pond 0.00165 0.00259 Cedar, Incense 0.00112 0.00180 Pine, lodgepole 0.00148 0.00234 Cedar, Northern whiteb 0.00101 0.00229 Pine, Jeffrey 0.00148 0.00234 Cedar, Port-Orford 0.00158 0.00241 Pine, longleaf 0.00176 0.00263 Cedar, western redb 0.00111 0.00234 Pine, ponderosa 0.00133 0.00216 Douglas-fir, Coast-type 0.00165 0.00267 Pine, red 0.00130 0.00252 Douglas-fir, Interior north 0.00130 0.00241 Pine, shortleaf 0.00158 0.00271 Douglas-fir, Interior west 0.00165 0.00263 Pine, slash 0.00187 0.00267 Fir, balsam 0.00099 0.00241 Pine, sugar 0.00099 0.00194 Fir, California red 0.00155 0.00278 Pine, Virginia 0.00144 0.00252 Fir, noble 0.00148 0.00293 Pine, western white 0.00141 0.00259 12–17 Because commercial lumber is often not perfectly flatsawn or quartersawn, this procedure will probably overestimate width shrinkage and underestimate thickness shrinkage. Note also that if both a size change and the percentage of moisture content are known, Equation (12–2) can be used to calculate the original moisture content.
Calculation Based on Green Dimensions Approximate dimensional changes associated with moisture content changes greater than 6% to 14% , or when one moisture value is outside of those limits, can be calculated by ∆D D M M S M = − ( ) − + I F I T I ( ) / 30 100 30 (12–3) where ST is tangential shrinkage (%) from green to ovendry (Ch. 3, Tables 3–5 and 3–6) (use radial shrinkage SR when appropriate).
Neither MI nor MF should exceed 30%, the assumed moisture content value when shrinkage starts for most species.
Table 12–5. Coefficients for dimensional change as a result of shrinkage or swelling within moisture content limits of 6% to 14% (CT = dimensional change coefficient for tangential direction; CR = radial direction)—con.
Dimensional change coefficienta Dimensional change coefficienta Species CR CT Species CR CT Softwoods—con.
Fir, Pacific silver 0.00151 0.00327 Redwood, old-growthb 0.00120 0.00205 Fir, subalpine 0.00088 0.00259 Redwood, second-growthb 0.00101 0.00229 Fir, grand 0.00112 0.00245 Spruce, black 0.00141 0.00237 Fir, white 0.00112 0.00245 Spruce, Engelmann 0.00130 0.00248 Hemlock, eastern 0.00102 0.00237 Spruce, red 0.00130 0.00274 Hemlock, western 0.00144 0.00274 Spruce, white 0.00130 0.00274 Larch, western 0.00155 0.00323 Spruce, Sitka 0.00148 0.00263 Tamarack 0.00126 0.00259 Imported Woods Andiroba, crabwood 0.00137 0.00274 Light red “Philippine mahogany” 0.00126 0.00241 Angelique 0.00180 0.00312 Limba 0.00151 0.00187 Apitong, keruingb (all Dipterocarpus spp.) 0.00243 0.00527 Mahoganyb Meranti 0.00172 0.00126 0.00238 0.00289 Avodire 0.00126 0.00226 Obeche 0.00106 0.00183 Balsa 0.00102 0.00267 Okoume 0.00194 0.00212 Banak 0.00158 0.00312 Parana, pine 0.00137 0.00278 Cativo 0.00078 0.00183 Paumarfim 0.00158 0.00312 Cuangare 0.00183 0.00342 Primavera 0.00106 0.00180 Greenheartb 0.00390 0.00430 Ramin 0.00133 0.00308 Irokob 0.00153 0.00205 Santa Maria 0.00187 0.00278 Khaya 0.00141 0.00201 Spanish-cedar 0.00141 0.00219 Kokroduab 0.00148 0.00297 Teakb 0.00101 0.00186 Lauans: dark red “Philippine mahogany” 0.00133 0.00267 aPer 1% change in moisture content, based on dimension at 10% moisture content and a straight-line relationship between moisture content at which shrinkage starts and total shrinkage. (Shrinkage assumed to start at 30% for all species except those indicated by footnote b.) bShrinkage assumed to start at 22% moisture content. 12–18 Design Factors Affecting Dimensional Change Framing Lumber in House Construction Ideally, house framing lumber should be dried to the mois-ture content it will reach in use, thus minimizing future dimensional changes as a result of frame shrinkage. This ideal condition is difficult to achieve, but some drying and shrinkage of the frame may take place without being visible or causing serious defects after the house is completed. If, at the time the wall and ceiling finish is applied, the moisture content of the framing lumber is not more than about 5% above that which it will reach in service, there will be little or no evidence of defects caused by shrinkage of the frame. In heated houses in cold climates, joists over heated basements, studs, and ceiling joists may reach a moisture content as low as 6% to 7% (Table 12–2). In mild climates, the minimum moisture content will be greater.
The most common signs of excessive shrinkage are cracks in plastered walls, truss rise, open joints, and nail pops in dry-wall construction; distortion of door openings; uneven floors; and loosening of joints and fastenings. The extent of vertical shrinkage after the house is completed is proportional to the depth of wood used as supports in a horizontal position, such as girders, floor joists, and plates. After all, shrinkage occurs primarily in the width of members, not the length.
Thorough consideration should be given to the type of fram-ing best suited to the whole building structure. Methods should be chosen that will minimize or balance the use of wood across the grain in vertical supports. These involve variations in floor, wall, and ceiling framing. The factors involved and details of construction are covered extensively in Wood-Frame House Construction (Sherwood and Stroh 1991).
Heavy Timber Construction In heavy timber construction, a certain amount of shrinkage is to be expected. A column that bears directly on a wood girder can result in a structure settling as a result of the perpendicular-to-grain shrinkage of the girder. If not provided for in the design, shrinkage may cause weakening of the joints or uneven floors or both. One means of eliminating part of the shrinkage in mill buildings and similar structures is to use metal post caps; the metal in the post cap separates the upper column from the lower column. The same thing is accomplished by bolting wood corbels to the side of the lower column to support the girders.
When joist hangers are installed, the top of the joist should be above the top of the girder; otherwise, when the joist shrinks in the stirrup, the floor over the girder will be higher than that bearing upon the joist. Heavy planking used for flooring should be near 12% moisture content to minimize openings between boards as they approach moisture equilibrium. When standard 38- or 64-mm (nominal 2- or 3-in.) joists are nailed together to provide a laminated floor of greater depth for heavy design loads, the joist material should be somewhat less than 12% moisture content if the building is to be heated.
Interior Finish The normal seasonal changes in the moisture content of interior finish are not enough to cause serious dimensional change if the woodwork is carefully designed. Large mem-bers, such as ornamental beams, cornices, newel posts, stair stringers, and handrails, should be built up from compara-tively small pieces. Wide door and window trim and base should be hollow-backed. Backband trim, if mitered at the corners, should be glued and splined before erection; other-wise butt joints should be used for the wide faces. Large, solid pieces, such as wood paneling, should be designed and installed so that the panels are free to move across the grain.
Narrow widths are preferable.
Flooring Flooring is usually dried to the moisture content expected in service so that shrinking and swelling are minimized and buckling or large gaps between boards do not occur. For basement, large hall, or gymnasium floors, however, enough space should be left around the edges to allow for some expansion.
Wood Care and Installation During Construction Lumber and Trusses Although it should be, lumber is often not protected from the weather at construction sites. Lumber is commonly placed on the ground in open areas near the building site as bulked and strapped packages. Supports under such packages are useful to prevent wetting from mud and ground water and should elevate the packages at least 150 mm (6 in.) off the ground. The packages should also be covered with plastic tarpaulins for protection from rain.
Lumber that is green or nearly green should be piled in stickers under a roof for additional drying before it is built into the structure. The same procedure is required for lumber that has been treated with a waterborne preservative but not fully redried. Prefabricated building parts, such as roof trusses, sometimes lie unprotected on the ground at the building site. In warm, rainy weather, moisture regain can result in fungal staining. Wetting of the lumber also results in swelling, and subsequent shrinkage of the framing may contribute to structural distortions. Extended storage of lumber at moisture contents greater than 20% without drying can allow decay to develop.
If framing lumber has a greater moisture content when in-stalled than that recommended in Table 12–2, shrinkage can 12–19 be expected. Framing lumber, even thoroughly air-dried stock, will generally have a moisture content greater than that recommended when it is delivered to the building site. If carelessly handled in storage at the site, the lumber can take up more moisture. Builders can schedule their work so an appreciable amount of drying can take place during the early stages of construction. This minimizes the effects of addi-tional drying and shrinkage after completion. When the house has been framed, sheathed, and roofed, the framing is so exposed that in time it can dry to a lower moisture con-tent than would ordinarily be expected in yard-dried lumber.
The application of the wall and ceiling finish is delayed while wiring and plumbing are installed. If this delay is about 30 days in warm, dry weather, framing lumber should lose enough moisture so that any additional drying in place will be relatively unimportant. In cool, damp weather, or if wet lumber is used, the period of exposure should be ex-tended. Checking moisture content of door and window headers and floor and ceiling joists at this time with an electric moisture meter is good practice. When these mem-bers approach an average of 12% moisture content, interior finish and trim can normally be installed. Closing the house and using the heating system will hasten the rate of drying.
Before wall finish is applied, the frame should be examined and defects that may have developed during drying, such as warped or distorted studs, shrinkage of lintels over openings, or loosened joints, should be corrected.
Exterior Trim and Millwork Exterior trim, such as cornice and rake mouldings, fascia boards, and soffit material, is normally installed before the shingles are laid. Trim, siding, and window and door frames should be protected on the site by storing in the house or garage until time of installation. Although items such as window frames and sashes are usually treated with some type of water-repellent preservative to resist absorption of water, they should be stored in a protected area if they cannot be installed soon after delivery. Wood siding is often received in packaged form and can ordinarily remain in the package until installation.
Finished Flooring Cracks develop in flooring if it absorbs moisture either before or after it is laid, then shrinks when the building is heated.
Such cracks can be greatly reduced by observing the follow-ing practices: • Specify flooring manufactured according to association rules and sold by dealers that protect it properly during storage and delivery.
• Do not allow flooring to be delivered before masonry and plastering are completed and fully dry, unless a dry storage space is available.
• Install the heating plant before flooring is delivered.
• Break open flooring bundles and expose all sides of flooring to the atmosphere inside the structure.
• Close up the house at night and increase the temperature about 8°C (15°F) greater than the outdoor temperature for about 3 days before laying the floor.
• If the house is not occupied immediately after the floor is laid, keep the house closed at night or during damp weather and supply some heat if necessary.
Better and smoother sanding and finishing can be done when the house is warm and the wood has been kept dry.
Interior Finish In a building under construction, average relative humidity will be greater than that in an occupied house because of the moisture that evaporates from wet concrete, brickwork, plas-ter, and even the structural wood members. The average temperature will be lower because workers prefer a lower temperature than is common in an occupied house. Under such conditions, the finish tends to have greater moisture content during construction than it will have during occupancy.
Before the interior finish is delivered, the outside doors and windows should be hung in place so that they can be kept closed at night. In this way, conditions of the interior can be held as close as possible to the higher temperature and lower humidity that ordinarily prevail during the day. Such protec-tion may be sufficient during dry warm weather, but during damp or cool weather, it is highly desirable that some heat be maintained in the house, particularly at night. Whenever possible, the heating plant should be placed in the house before the interior trim is installed, to be available for sup-plying the necessary heat. Portable heaters can also be used.
The temperature during the night should be maintained about 8°C (15°F) greater than the outside temperature but should not be allowed to drop below about 21°C (70°F) during the summer or 17°C (62°F) when the outside tem-perature is below freezing.
After buildings have thoroughly dried, less heat is needed, but unoccupied houses, new or old, should not be allowed to stand without some heat during the winter. A temperature of about 8°C (15°F) greater than the outside temperature and above freezing at all times will keep the woodwork, finish, and other parts of the house from being affected by dampness or frost.
Plastering During a plastering operation in a moderate-sized, six-room house, approximately 450 kg (1,000 lb) of water are used, all of which must be dissipated before the house is ready for the interior finish. Adequate ventilation to remove the evaporated moisture will keep it from being absorbed by the framework.
In houses plastered in cold weather, the excess moisture can also cause paint to blister on exterior finish and siding. 12–20 During warm, dry weather, with the windows wide open, the moisture will be gone within a week after the final coat of plaster is applied. During damp, cold weather, the heating system or portable heaters are used to prevent freezing of plaster and to hasten its drying. Adequate ventilation should be provided at all times of the year because a large volume of air is required to carry away the amount of water involved.
Even in the coldest weather, the windows on the side of the house away from the prevailing winds should be opened 50 to 75 mm (2 to 3 in.), preferably from the top.
References ASTM. [current edition]. Direct moisture content measure-ment of wood and wood-based materials. ASTM D4442–92.
West Conshohocken, PA: American Society for Testing and Materials.
Forest Products Laboratory. 1961. Wood floors for dwell-ings. Agric. Handb. 204. Washington, DC: U.S. Depart-ment of Agriculture.
Forest Products Laboratory. 1972. Methods of controlling humidity in woodworking plants. Res. Note FPL–RN– 0218. Madison, WI: U.S. Department of Agriculture, Forest Service, Forest Products Laboratory.
James, W.L. 1988. Electric moisture meters for wood. Gen.
Tech. Rep. FPL–GTR–6. Madison, WI: U.S. Department of Agriculture, Forest Service, Forest Products Laboratory.
McMillen, J.M.; Wengert, E.M. 1978. Drying eastern hardwood lumber. Agric. Handb. 528. Washington, DC: U.S. Department of Agriculture.
Rietz, R.C. 1978. Storage of lumber. Agric. Handb. 531.
Washington, DC: U.S. Department of Agriculture.
Rietz, R.C.; Page, R.H. 1971. Air drying of lumber: A guide to industry practices. Agric. Handb. 402. Washington, DC: U.S. Department of Agriculture.
Sherwood, G.E.; Stroh, R.C. 1991. Wood-Frame House Construction. USDA Agric. Handb. 73. Washington, DC: U.S. Department of Agriculture.
Simpson, W.T. 1989. Drying wood: a review. Drying Technology. An International Journal, Pt. 1. 2(2): 235–265, Pt. 2, 2(3): 353–368.
Simpson, W.T., ed. 1991. Dry kiln operator’s manual.
Agric. Handb. 188. Washington, DC: U.S. Department of Agriculture.
USDC. 1970. American softwood lumber standard. NBS Voluntary Prod. Stand. PS 20–70; Washington, DC: U.S.
Department of Commerce.
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17482 | https://www.sciencedirect.com/science/article/abs/pii/S0090825805002428 | Practice patterns of SGO members for stage IIIA endometrial cancer - ScienceDirect
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Gynecologic Oncology
Volume 98, Issue 1, July 2005, Pages 77-83
Practice patterns of SGO members for stage IIIA endometrial cancer
Author links open overlay panel Christine M.Lee a, Brian M.Slomovitz a, Marilyn Greer b, Sheena Sharma b, Mary Ann Gregurich b, Thomas Burke a, Karen H.Lu a, Lois M.Ramondetta a
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Objective.
The purpose of surgical staging is to better determine prognosis and treatment. The International Federation of Gynecology and Obstetrics (FIGO) stage IIIA endometrial cancer is a heterogenous disease, and adjuvant therapy is not well-defined. The aim of this study was to survey the Society of Gynecologic Oncologists (SGO) members and fellows about their approach to the treatment of patients with stage IIIA endometrial cancer.
Methods.
All 850 members of the SGO were mailed surveys that asked how they would manage various case scenarios of stage IIIA endometrial cancer. Data were collected using an Internet survey database. Frequency distributions were determined, and nonparametric tests were performed.
Results.
Fifty-three percent of SGO members and fellows responded. For the treatment of stage IIIA disease with malignant cytology only, adjuvant therapy was recommended 46%, 62%, and 98% of the time for women with grades 1, 2, and UPSC, respectively. Sixty-six percent of respondents would not remove malignant cytology from the current staging criteria. Ninety-nine percent of respondents recommended adjuvant therapy for patients with adnexal or serosal involvement. Eighty-six percent indicated that a hysteroscopy for diagnosis would not alter their treatment recommendations.
Conclusions.
While most gynecologic oncologists in our survey recommend adjuvant therapy for stage IIIA endometrial carcinoma, our results showed that patients with malignant cytology only would receive different treatments than patients with adenxal or serosal involvement. Histology and grade of the tumor are predictors of therapy recommendations over malignant cytology. Most respondents agreed that patients with malignant cytology should remain in stage IIIA.
Introduction
Endometrial cancer is the fourth most common malignancy in women and is the leading gynecologic malignancy in the United States. In 2004, over 40,000 new cases of endometrial cancer will be diagnosed, with approximately 7000 women dying of the disease . Surgical staging for endometrial cancer provides prognostic information and can guide management decisions regarding adjuvant treatment. Stage IIIA endometrial cancer, however, is a heterogeneous disease, where patients can have a broad range of extra-uterine tumor involvement, ranging from minimally invasive uterine disease with malignant cytology to extensive pelvic metastases. Therefore, adjuvant therapy within stage IIIA is tailored towards the extent of extra-uterine spread of disease.
FIGO Stage IIIA, malignant cytology only disease, has been controversial because the long-term survival rates are comparable to patients with stage I , , . Conversely, patients with uterine serosal involvement and/or extra-uterine disease have poorer survival rates, demonstrating the differences in survival for stage IIIA endometrial cancer. Thus, this substage of disease requires several different approaches for adjuvant therapy to avoid both insufficient treatment and over treatment.
Because of various modalities used to treat stage IIIA endometrial cancer, we surveyed SGO members and fellows to determine current practice patterns for stage IIIA endometrial cancer. Our hypothesis was that stage IIIA patients with malignant cytology only are treated differently in the adjuvant setting than patients with adnexal or uterine serosal involvement. Furthermore, we postulated that the histology or grade of the tumor was more likely to influence recommendations regarding adjuvant therapy rather than the presence or absence of malignant cytology.
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Methods
After receiving approval from the Institutional Review Board at The University of Texas M.D. Anderson Cancer Center and permission from the SGO to obtain mailing addresses of its members, surveys were mailed to the 850 members and fellows enrolled in SGO as of September 2003. Questions were designed after reviewing historical treatment recommendations for stage IIIA endometrial cancer. The survey questions are provided in Table 1.
The initial survey was mailed in September 2003. Each survey was
Demographics of respondents
Table 2 lists the demographics of the respondents to the survey. Forty-three (5%) of the 850 surveys that were mailed were returned because of incorrect addresses. Of the 807 surveys that were not returned, 425 recipients responded (53%).
Scenario 1
This patient underwent surgical staging for endometrial cancer and was found to have stage IIIA, grade 1 endometrioid adenocarcinoma of the uterus on the basis of malignant cytology alone. The tumor showed less than 50% myometrial invasion. Table 3 lists the
Discussion
We observed a difference in the current practice patterns followed for the treatment of stage IIIA endometrial cancer in patients with malignant cytology only as opposed to uterine serosal and/or adnexal involvement. Almost all respondents replied that patients with uterine serosal or adnexal involvement would be offered adjuvant therapy with the majority offering either radiation therapy or chemotherapy. Interestingly, the grade or histology of the tumor influenced recommendations for adjuvant
Recommended articles
References (10)
M.J. Zerbe et al.
Retrograde seeding of malignant cells during hysteroscopy in presumed early endometrial cancer
Gynecol. Oncol. (2000)
C. Egarter et al.
Abdominal dissemination of malignant cells with hysteroscopy
Gynecol. Oncol. (1996)
K.W. Lo et al.
Hysteroscopic dissemination of endometrial carcinoma using carbon dioxide and normal saline: a retrospective study
Gynecol. Oncol. (2002)
G. Arikan et al.
Are endometrial carcinoma cells disseminated at hysteroscopy functionally viable?
Gynecol. Oncol. (2001)
A. Jemal et al.
Cancer statistics, 2003
CA Cancer J. for Clin. (2003)
There are more references available in the full text version of this article.
Cited by (24)
Prognostic significance of peritoneal cytology in low-intermediate risk endometrial cancer
2017, Gynecologic Oncology Citation Excerpt :
However, after controlling for other risk factors, there was no longer a significant association between cytology and outcome. The utility and clinical significance of malignant cells in peritoneal washings in endometrial cancer has long been debated and the gynecologic oncology community remains divided . In a survey of gynecologic oncologists, 66% felt that cytology should remain in the current staging system, however recommendations regarding adjuvant therapy were variable in the setting of isolated positive peritoneal cytology. Show abstract There is uncertainty surrounding the prognostic value and clinical utility of peritoneal cytology in endometrial cancer. Our primary objective was to determine if positive cytology is associated with disease-free and overall survival in women treated surgically for endometrial cancer, specifically those with low or intermediate risk disease. This was a retrospective population-based cohort study of British Columbia Cancer Registry patients who underwent surgery with peritoneal washings for endometrioid-type endometrial cancer from 2003 to 2009. Low risk was defined as Stage IA grade 1 or 2, and intermediate risk defined as Stage IA grade 3, or Stage IB grade 1 or 2 tumours. Five-year overall and disease free-survival were assessed using Kaplan-Meier estimation. Potential covariates including peritoneal cytology, grade, depth of myometrial invasion, LVSI, age, and adjuvant therapy were evaluated in a multivariable Cox proportional hazards model. There were 849 patients, of whom 370 (43.6%) and 298 (35.1%) had low- and intermediate-risk disease, respectively. Overall, forty-nine (5.8%) patients had positive cytology, including 6 and 9 with low- and intermediate-risk respectively (2.2% within low and intermediate risk combined). Positive peritoneal cytology was not significantly associated with disease-free (HR 3.17, 95% CI 0.91–11.03) or overall survival (HR 1.33, 95% CI 0.47–3.76) in low and intermediate risk patients. Only age and extensive LVSI were associated with lower overall survival (HR 1.10, 95% CI 1.08–1.13, and HR 2.39, 95% CI 1.02–5.61, respectively). Positive peritoneal cytology was not associated with disease-free and overall survival in women with low and intermediate risk endometrial cancer.
### Evaluation of the quality of the management of cancer of the corpus uteri-Selection of relevant quality indicators and implementation in Belgium
2013, Gynecologic Oncology Citation Excerpt :
These examples underline the importance of adequate initial surgery, complete staging and histopathology and evidence-based decision regarding the choice of adjuvant treatment. Literature shows a high variability in practices at all steps of the management of uterine cancer [16–22]. This leads to variation in the Quality of Care in comparison with guidelines as demonstrated by a few single-center or regional studies assessing the Quality of Care in comparison with guidelines [23–26]. Show abstract Describe the methodology and selection of quality indicators (QI) to be implemented in the EFFECT (EFFectiveness of Endometrial Cancer Treatment) project. EFFECT aims to monitor the variability in Quality of Care (QoC) of uterine cancer in Belgium, to compare the effectiveness of different treatment strategies to improve the QoC and to check the internal validity of the QI to validate the impact of process indicators on outcome. A QI list was retrieved from literature, recent guidelines and QI databases. The Belgian Healthcare Knowledge Center methodology was used for the selection process and involved an expert's panel rating the QI on 4 criteria. The resulting scores and further discussion resulted in a final QI list. An online EFFECT module was developed by the Belgian Cancer Registry including the list of variables required for measuring the QI. Three test phases were performed to evaluate the relevance, feasibility and understanding of the variables and to test the compatibility of the dataset. 138 QI were considered for further discussion and 82 QI were eligible for rating. Based on the rating scores and consensus among the expert's panel, 41 QI were considered measurable and relevant. Testing of the data collection enabled optimization of the content and the user-friendliness of the dataset and online module. This first Belgian initiative for monitoring the QoC of uterine cancer indicates that the previously used QI selection methodology is reproducible for uterine cancer. The QI list could be applied by other research groups for comparison.
### Positive peritoneal cytology is an independent risk-factor in early stage endometrial cancer
2013, Gynecologic Oncology Citation Excerpt :
These results support the placement of patients with stage I/II endometrial cancer and PPC in a more advanced stage group, as opposed to the current FIGO staging criteria for accurate risk-stratification. In fact, in a survey study of the members of the society of gynecologic oncologists, a majority of respondents supported the retention of PPC in the current FIGO stage IIIA classification . Unfortunately, we are not able to comment on whether or not the presence of PPC should influence postoperative adjuvant therapy recommendations in patients with stage I–II endometrial cancer. Show abstract In light of the recent changes in the International Federation of Gynecology and Obstetrics (FIGO) staging system, the objective of this study was to determine the prognostic significance of positive peritoneal cytology (PPC) among patients with early stage endometrial cancer. Data were extracted from the Surveillance, Epidemiology, and End Results database between 1988 and 2005. Only those patients with stage I/II endometrial cancer who had undergone a complete staging procedure (lymph-node removal) were included. Statistical analyses used Chi-square test, Kaplan–Meier log rank, and Cox proportional hazards models. A total of 14,704 patients were identified: 14,219 with negative peritoneal cytology (NPC) and 485 with positive peritoneal cytology. More patients with PPC compared to those with NPC were diagnosed with high-risk factors such grade III disease (40.2% vs. 23.8%, p<0.0001), and unfavorable histologic types such as clear cell/serous carcinoma (17.5% vs. 7.5%, p=<0.0001) and carcinosarcoma (9.3% vs. 5.6%, p<0.0001). When compared to patients with negative peritoneal cytology, survival was significantly worse among patients with positive peritoneal cytology (p<0.0001): 5-year disease specific survival 95.1% vs. 80.8% in endometrioid adenocarcinoma; 78.0% vs. 50.4% in clear cell/serous cancer; and 64.7% vs. 32.3% in carcinosarcoma. After adjusting for other contributing factors in the multivariable model, PPC remained an independent predictor of poor survival (p<0.0001) in all histologic types examined. PPC is an independent risk factor in patients with early stage endometrial cancer. Although, no longer a part of the current FIGO staging criteria, peritoneal cytology status should still be considered for accurate risk-stratification of these patients.
### Prognostic significance and treatment implications of positive peritoneal cytology in endometrial adenocarcinoma: Unraveling a mystery
2009, Gynecologic Oncology Citation Excerpt :
In contrast, 98% of members would recommend chemotherapy or radiation for patients who have stage IIIA2 disease. Interestingly, one-third of respondents believed positive cytology should remain a component of the staging criteria . Recent discussion as to the impact of positive cytology in changing treatment protocols has raised the question of whether peritoneal cytology should be included in the staging system. Show abstract Review the literature on positive peritoneal cytology in endometrioid endometrial adenocarcinoma, its prognostic value, proposed treatment strategies, and future avenues of investigation. PubMed search of articles pertaining to stage IIIA endometrioid endometrial adenocarcinoma identified over 50 articles that were reviewed. Low-risk stage IIIA1 was defined as those with grade 1 or 2 disease, no evidence of cervical involvement, myometrial invasion less than 50% and no lymph vascular space invasion (LVSI). Analysis of available data was completed with weighted averages, Student's-t-test and Chi square statistical analyses. Incidence of positive washings is approximately 11%. In low-risk stage IIIA1 endometrial carcinoma patients, the rate of recurrence is 4.1%. In contrast, in high-risk stage IIIA1 endometrial carcinoma patients the rate of recurrence is 32%, a statistically significant difference (p<0.001). Positive peritoneal cytology is associated with an increased incidence of extrauterine disease but is not consistently linked to other high-risk factors such as positive lymph node status, extent of myometrial invasion or LVSI. To date there is no definitive consensus on the prognostic significance of positive peritoneal cytology alone. However, even in the low-risk cohort with stage IIIA1 some patients will recur. Adjuvant therapy for low-risk stage IIIA diseased may or may not be of benefit. High-risk disease should be treated with chemotherapy, radiation or a combination thereof. A prospective, multicenter trial of comprehensively surgically staged patients with stage IIIA endometrial cancer is indicated in order to clearly define prognosis and treatment for these patients.
### Stage IIIA endometrial carcinoma
2009, Obstetrics and Gynecology
### Fertility-sparing therapy for young women with endometrial cancer
2006, Expert Review of Anticancer Therapy
View all citing articles on Scopus
View full text
Copyright © 2005 Elsevier Inc. All rights reserved.
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17483 | https://www.mathsisfun.com/algebra/rationalize-denominator.html | Rationalize the Denominator
"Rationalizing the denominator" is when we move a root (like a square root or cube root) from the bottom of a fraction to the top.
Oh No! An Irrational Denominator!
The bottom of a fraction is called the denominator. Numbers like 2 and 3 are rational. But many roots, such as √2 and √3, are irrational.
Example: 1 √2 has an Irrational Denominator
To be in "simplest form" the denominator should not be irrational!
Fixing it (by making the denominator rational)is called "Rationalizing the Denominator"
Note: an irrational denominator is not wrong, it still works. But it is not "simplest form" and so can cost you marks.
Also equations can be easier to solve and calculations can be easier without an irrational denominator, so you should learn how.
So ... how?
1. Multiply Both Top and Bottom by a Root
Sometimes we can just multiply both top and bottom by a root:
Example: 1 √2 has an Irrational Denominator. Let's fix it.
Multiply top and bottom by the square root of 2, because: √2 × √2 = 2:
1 √2 × √2 √2 = √2 2
Now the denominator is a rational number (2). Done!
Having an irrational number in the top (numerator) of a fraction is preferred.
See below for a comparison between calculating 1 √2 and √2 2
2. Multiply Both Top and Bottom by the Conjugate
There is another special way to move a square root from the bottom of a fraction to the top ... we multiply both top and bottom by the conjugate of the denominator.
The conjugate is where we change the sign in the middle of two terms:
Example Expression:
x2 − 3
Its Conjugate:
x2 + 3
Another Example:
a − b3
Its Conjugate:
a + b3
The method we will see works because when we multiply something by its conjugate we get squares like this:
(a+b)(a−b) = a2 − b2
Here's how we can do it:
Example: here is a fraction with an "irrational denominator":
1 3−√2
How can we move the square root of 2 to the top?
We can multiply both top and bottom by 3+√2 (the conjugate of 3−√2), which won't change the value of the fraction:
1 3−√2 × 3+√2 3+√2 = 3+√2 32−(√2)2 = 3+√2 7
(Did you see that we used (a+b)(a−b) = a2 − b2 in the denominator?)
Use your calculator to work out the value before and after ... is it the same?
There is another example on the page Evaluating Limits (advanced topic) where I move a square root from the top to the bottom.
Useful
So try to remember these little tricks, it may help you solve an equation one day!
For fun let's try calculating both 1 √2 and √2 2
They should be the same. Let's use 1.414 as a rough estimate for √2
For 1 √2 ≈ 1 1.414 we need long division:
0.70... 1.414 )1.000 0 1.0000 0.9898 0.0102 etc... give up, too hard!
The answer is 0.70... something.
Now let's try √2 2 ≈ 1.414 2:
1.414 2 = 0.707 easy!
Which form do you prefer?
355, 2248, 2249, 2250, 2251, 3196, 3197, 3198, 3199, 97
Rational Number Algebra Index
Copyright © 2025 Rod Pierce |
17484 | https://www.machiningdoctor.com/es/charts/metric-thread-charts/ | Ir al contenido
Tu experto en mecanizado
Roscas Métricas: Tablas y Fórmulas
Índice
Lista de Tablas
Acceso a todos los gráficos de datos que aparecen en esta página
Normas ISO
Lista de Símbolos
Ejemplos de Designación
Posibles Combinaciones de Diámetro/Paso.
Dimensiones Básicas
Clases Recomendadas
Ajuste/Desviaciones (es/EL)
Tolerancias (Td/TD)
Longitud Normal de Conexión
Visión General
La gama de roscas métricas ISO es serie de roscas con forma de angulo de 60° para uso general. Es la norma más utilizada en todo el mundo. Fue uno de los primeros temas estandarizados por el comité ISO ya en 1947 y todavía se utiliza.
ISO Estándares
Varias normas ISO definen juntas el sistema de rosca métrica ISO:
| Norma ISO | Tema | Qué incluye? |
---
| ISO 261 | Posibles Combinaciones | Lista en la tabla todas las posibles combinaciones de Diámetro/Paso. Define las series de Paso Grueso y Paso Fino . |
| ISO 68 | Perfil Básico | Define el perfil básico de la rosca, incluido el Paso, Diámetro del Paso, Alturas, etc. La imagen con los parámetros de la rosca. Formulas para los parámetros de arriba |
| ISO 965-1 | Tolerancias | El documento principal para roscas métricas ISO: Lista y definición de símbolos. Definición de clases de tolerancia. Reglas de Designaciones Tabla de Ajustes: por clase y paso. Longitud de Conexión Tabla – Por diámetro y paso. Sector de Tolerancias Tabla – Por diámetro y paso.. Radio de la Raíz Tablas y fórmulas. Clases de tolerancia recomendadas: Tablas por calidad de tolerancia y longitud de la conexión. Fórmulas que se utilizaron para construir los gráficos.. |
| ISO 965-2 | Límites | Tablas para el rango permitido de paso, diámetros mayor y menor para las clases 6H y 6g. Los valores se calculan a partir de las definiciones de ISO 965-1. Tienen un uso limitado ya que proporcionan datos sólo para las clases predeterminadas. |
Designaciones
La descripción comienza siempre con M mayúscula, seguida con el diámetro nominal de la rosca en milímetros. Por ejemplo, M8 es una rosca con un diámetro nominal de 8 mm (0,315″). Si no hay ningún carácter después de la M y el diámetro, significa que todos los demás parámetros están de acuerdo con los valores definidos en ISO 965-1:
Valores predeterminados de Rosca ISO Métrica:
Paso– Según la serie de paso Grueso.
Número de Entradas: Una.
Clase: 6g para roscas externas y 6H para roscas internas. (Válido por encima de 1.6 mm / 0.063 «)
Dirección – Rosca derecha.
Designación de Rosca Métrica Ejemplos
Celdas con fondo dorado – Los valores se derivan de los valores predeterminados anteriores.
Celdas con Fondo Blanco – Los valores se derivan de la descripción.
| Designación | Ø | Paso | Interna / Externa | Avance | Número de Entradas | Diámetro del Paso Clase | Mayor/Menor Clase de Diámetro | Dirección de Roscado |
--- --- --- ---
| M8 | 8 | 1.25 | Ambos | 1.25 | 1 | 6H/6g | 6H/6g | Rosca Derecha |
| M8 X 0.75 | 8 | 0.75 | Ambos | 0.75 | 1 | 6H/6g | 6H/6g | Rosca Derecha |
| M8 X Ph2.25P0.75 | 8 | 0.75 | Ambos | 2.25 | 3 | 6H/6g | 6H/6g | Rosca Derecha |
| M8 - 4g8e | 8 | 1.25 | Externa | 1.25 | 1 | 4g | 8e | Rosca Derecha |
| M8 X 0.75 - 5G | 8 | 0.75 | Interna | 1.25 | 1 | 5G | 5G | Rosca Derecha |
| M8 - RH | 8 | 1.25 | Ambos | 0.75 | 1 | 6H/6g | 6H/6g | Rosca Derecha |
| M8 Ph2.25P0.75 - 4g8e - LH | 8 | 0.75 | Externa | 2.25 | 3 | 4g | 8e | Rosca Izquierda |
Explicación:
Paso: Diseñado por «X P«. Por ejemplo, M8X 0,75 significa una rosca de 8 mm (0,315″) con un paso de 0,75 mm (0,03″ o 34 Hilos por pulgada). Si se omite «X P», el paso está definido por la serie de Paso Grueso según a ISO-261.
Número de Entradas: Diseñado por «Ph». Por ejemplo, M8 X Ph2.25P0.75. En este caso, 2.25 mm es el avance y 0.75 mm es el paso. Ph debe ser un múltiplo de P ya que el número de entradas es igual a Ph/P. Para este ejemplo, el número de entradas es 3. Caso se omite Ph, la rosca tiene una sola entrada.
Class: The class appears after the pitch. It is preceded by a «-» and consists of two or four characters. For example, M8 X 0.75– 5g6g.
Las letras minúsculas representan roscas externas, mientras que las letras mayúsculas representan roscas internas.
Dos cartas de clase larga significan que la clase es válida tanto para el paso como para los diámetros mayor/menor.
Cuatro cartas largas significan una clase separada para el diámetro de paso (los dos primeros caracteres) y los diámetros mayor/menor (los dos primeros caracteres).
Para explicaciones detalladas sobre las clases, vea abajo la sección de clases.
Si se omite la clase, se utiliza la clase predeterminada. (Vea la tabla arriba)
Dirección del Roscado: La dirección aparece después de la clase y está precedida por «-«. Si se omite, la dirección es derecha. Si aparece «- LH», designa una rosca a la izquierda. Por ejemplo, M8 X 0,75 – 5g6g – LH es una designación de rosca a la izquierda.
Estándar ISO serie métrica
El Sistema de Rosca Métrica ISO consta de dos series:
| Serie de Roscado | Desde el Diámetro | Hasta al Diámetro | Combinaciones de pasos por diámetro | Número de Combinaciones |
--- ---
| Paso Grueso | 1 mm (0.04") | 68 mm (2.7") | 1 | 50 |
| Paso Fino | 1 mm (0.04") | 300 mm (11.8") | 1-5 | 118 |
Combinaciones Posibles (Diámetro/Paso)
Las siguientes tablas muestran todas las posibles combinaciones de diámetros/paso de rosca como se define en ISO-261.
MC – Serie de Rosca Métrica Gruesa
MF – Serie de Rosca Métrica Fina
Haga clic para obtener la hoja de datos del roscado para la combinación específica.
Posibles combinaciones para M1 – M24
| Diámetro | P=0.2 | 0.25 | 0.3 | 0.35 | 0.4 | 0.45 | 0.5 | 0.6 | 0.7 | 0.75 | 0.8 | 1.0 | 1.25 | 1.5 | 1.75 | 2.0 | 2.5 | 3.0 |
--- --- --- --- --- --- --- --- ---
| 1 | MF | MC | | | | | | | | | | | | | | | | |
| 1.1 | MF | MC | | | | | | | | | | | | | | | | |
| 1.2 | MF | MC | | | | | | | | | | | | | | | | |
| 1.4 | MF | | MC | | | | | | | | | | | | | | | |
| 1.6 | MF | | | MC | | | | | | | | | | | | | | |
| 1.8 | MF | | | MC | | | | | | | | | | | | | | |
| 2 | | MF | | | MC | | | | | | | | | | | | | |
| 2.2 | | MF | | | | MC | | | | | | | | | | | | |
| 2.5 | | | | MF | | MC | | | | | | | | | | | | |
| 3 | | | | MF | | | MC | | | | | | | | | | | |
FirstPrevious123NextLast
Posibles combinaciones para M27 – M300
| Diámetro | P=1 | 1.5 | 2..0 | 3.0 | 3.5 | 4.0 | 4.5 | 5.0 | 5.5 | 6.0 | 8.0 |
--- --- --- --- --- --- |
| 27 | MF | MF | MF | MC | | | | | | | |
| 30 | MF | MF | MF | MF | MC | | | | | | |
| 33 | | MF | MF | MF | MC | | | | | | |
| 36 | | MF | MF | MF | | MC | | | | | |
| 39 | | MF | MF | MF | | MC | | | | | |
| 42 | | MF | MF | MF | | MF | MC | | | | |
| 45 | | MF | MF | MF | | MF | MC | | | | |
| 48 | | MF | MF | MF | | MF | | MC | | | |
| 52 | | MF | MF | MF | | MF | | MC | | | |
| 56 | | MF | MF | MF | | MF | | | MC | | |
FirstPrevious123456NextLast
Dimensiones Básicas de Rosca
Las dimensiones básicas son dimensiones nominales de un perfil de rosca métrica sin márgenes y tolerancias (La clase de rosca lo define). Se basan de la norma ISO 68-1. Las dimensiones básicas se pueden utilizar para el diseño del perfil. Sin embargo, para la fabricación y el mecanizado, necesita el rango permitido de cada dimensión. Estos datos se pueden encontrar en ISO 951-1 o en la siguiente sección de Límites y Dimensiones . Todos parámetros se derivan de fórmulas simples basadas del diámetro nominal y el paso de la rosca.
Símbolos
Lista de símbolos utilizados en las tablas y fórmulas de Roscas ISO Métricas.
| Símbolo | Explicación |
--- |
| Parámetros Básicos - Diámetros y Paso | |
| D | Diámetro mayor (básico) de la rosca interna |
| D1 | Diámetro menor de rosca interna |
| D2 | Diámetro de paso de la rosca interna |
| d | Diámetro mayor (básico) de la rosca externa |
| d1 | Diámetro menor de rosca externa |
| d2 | Diámetro de paso de la rosca externa |
| d3 | Diámetro menor para límites de raíz de rosca externa |
| P | Paso |
| Ph | Avance |
| Parámetros de altura | |
| H | Altura del triángulo fundamental |
| hs | Altura real del hilo - Rosca Externa |
| has | Adendo de la rosca - Rosca Externa |
| hn | Altura real del hilo - Rosca Interna |
| hdn | Dedende de la rosca - Rosca Interna |
| Parámetros de Longitud | |
| LE | Longitud de conexión |
| S / N / L | Designación del grupo de roscado 'Corto' / 'Medio' / 'Largo' |
| Ajuste / Tolerancia | |
| TD1 | Tolerancia - D1 |
| TD2 | Tolerancia - D2 |
| Td | Tolerancia - d |
| Td2 | Tolerancia - d2 |
| EI | Ajuste - internal thread |
| es | Ajuste - Rosca Externa |
| Raiz / Cresta | |
| R | Radio de la Raíz - Rosca Externa |
| C | Truncamiento de la Raíz - Rosca Externa |
| Fcs | Ancho de la Cresta - Rosca Externa |
| Frs | Ancho de la Raíz - Rosca Externa |
| Fcs | Ancho de la Cresta - Rosca Interna |
| Fcs | Ancho de la Raíz - Rosca Interna |
El siguiente vídeo proporciona una explicación de los términos de roscado.
Fórmulas para dimensiones básicas.
Rosca Externa
Rosca Interna
Las fórmulas de arriba proporcionan los resulta dos exactos. Puedes usarlos directamente, utilize nuestro[Thread Calculator , o navegar por lassiguientes tablas.
Dimensiones de rosca Métrica Tablas
Dimensiones Básicas de Rosca: Serie de Rosca Métrica de Paso Grueso
Haga clic en el enlace del Roscado para obtener losdatos completos de la Rosca para todas Clases
Todas las dimensiones están en mm. Para convertir en pulgadas, haga clic en el enlace del Roscado
| Roscado | Diámetro Nominal | Paso | Diámetro del Paso | Diámetro Interior | Altura real del hilo | Cresta / Raiz | Diámetro del Taladro | Angulo de Helice |
--- --- --- ---
| M1 | 1 | 0.25 | 0.838 | 0.729 | 0.135 | 0.063 | 0.77 | 5.43 |
| M1.1 | 1.1 | 0.25 | 0.938 | 0.829 | 0.135 | 0.063 | 0.87 | 4.85 |
| M1.2 | 1.2 | 0.25 | 1.038 | 0.929 | 0.135 | 0.063 | 0.97 | 4.39 |
| M1.4 | 1.4 | 0.3 | 1.205 | 1.075 | 0.162 | 0.075 | 1.13 | 4.53 |
| M1.6 | 1.6 | 0.35 | 1.373 | 1.221 | 0.189 | 0.088 | 1.28 | 4.64 |
| M1.8 | 1.8 | 0.35 | 1.573 | 1.421 | 0.189 | 0.088 | 1.48 | 4.05 |
| M2 | 2 | 0.4 | 1.74 | 1.567 | 0.217 | 0.1 | 1.64 | 4.18 |
| M2.2 | 2.2 | 0.45 | 1.908 | 1.713 | 0.244 | 0.113 | 1.79 | 4.29 |
| M2.5 | 2.5 | 0.45 | 2.208 | 2.013 | 0.244 | 0.113 | 2.09 | 3.71 |
| M3 | 3 | 0.5 | 2.675 | 2.459 | 0.271 | 0.125 | 2.55 | 3.4 |
| M3.5 | 3.5 | 0.6 | 3.11 | 2.85 | 0.325 | 0.15 | 2.95 | 3.51 |
| M4 | 4 | 0.7 | 3.545 | 3.242 | 0.379 | 0.175 | 3.36 | 3.6 |
| M4.5 | 4.5 | 0.75 | 4.013 | 3.688 | 0.406 | 0.188 | 3.82 | 3.4 |
| M5 | 5 | 0.8 | 4.48 | 4.134 | 0.433 | 0.2 | 4.27 | 3.25 |
| M6 | 6 | 1 | 5.35 | 4.917 | 0.541 | 0.25 | 5.09 | 3.4 |
| M7 | 7 | 1 | 6.35 | 5.917 | 0.541 | 0.25 | 6.09 | 2.87 |
| M8 | 8 | 1.25 | 7.188 | 6.647 | 0.677 | 0.313 | 6.86 | 3.17 |
| M9 | 9 | 1.25 | 8.188 | 7.647 | 0.677 | 0.313 | 7.86 | 2.78 |
| M10 | 10 | 1.5 | 9.026 | 8.376 | 0.812 | 0.375 | 8.64 | 3.03 |
| M11 | 11 | 1.5 | 10.026 | 9.376 | 0.812 | 0.375 | 9.64 | 2.73 |
| M12 | 12 | 1.75 | 10.863 | 10.106 | 0.947 | 0.438 | 10.41 | 2.94 |
| M14 | 14 | 2 | 12.701 | 11.835 | 1.083 | 0.5 | 12.18 | 2.87 |
| M16 | 16 | 2 | 14.701 | 13.835 | 1.083 | 0.5 | 14.18 | 2.48 |
| M18 | 18 | 2.5 | 16.376 | 15.294 | 1.353 | 0.625 | 15.73 | 2.78 |
| M20 | 20 | 2.5 | 18.376 | 17.294 | 1.353 | 0.625 | 17.73 | 2.48 |
| M22 | 22 | 2.5 | 20.376 | 19.294 | 1.353 | 0.625 | 19.73 | 2.24 |
| M24 | 24 | 3 | 22.051 | 20.752 | 1.624 | 0.75 | 21.27 | 2.48 |
| M27 | 27 | 3 | 25.051 | 23.752 | 1.624 | 0.75 2.18 |
| M30 | 30 | 3.5 | 27.727 | 26.211 | 1.894 | 0.875 2.3 |
| M33 | 33 | 3.5 | 30.727 | 29.211 | 1.894 | 0.875 2.08 |
| M36 | 36 | 4 | 33.402 | 31.67 | 2.165 | 1 2.18 |
| M39 | 39 | 4 | 36.402 | 34.67 | 2.165 | 1 2 |
| M42 | 42 | 4.5 | 39.077 | 37.129 | 2.436 | 1.125 2.1 |
| M45 | 45 | 4.5 | 42.077 | 40.129 | 2.436 | 1.125 1.95 |
| M48 | 48 | 5 | 44.752 | 42.587 | 2.706 | 1.25 2.04 |
| M52 | 52 | 5 | 48.752 | 46.587 | 2.706 | 1.25 1.87 |
| M56 | 56 | 5.5 | 52.428 | 50.046 | 2.977 | 1.375 1.91 |
| M60 | 60 | 5.5 | 56.428 | 54.046 | 2.977 | 1.375 1.78 |
| M64 | 64 | 6 | 60.103 | 57.505 | 3.248 | 1.5 1.82 |
| M68 | 68 | 6 | 64.103 | 61.505 | 3.248 | 1.5 1.71 |
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Dimensiones Básicas de Rosca: Serie de Rosca Métrica de Paso Fino
Todas las dimensiones están en mm. Para convertir en pulgadas, haga clic en el enlace del Roscado
Haga clic en el enlace del Roscado para obtener los datos completos de la Rosca para todas Clases
Todas las dimensiones están en mm. Para convertir en pulgadas, haga clic en el enlace del Roscado
| Roscado | Diámetro Nominal | Paso | Diámetro del Paso | Diámetro Interior | Altura real del hilo | Cresta / Raiz | Diámetro del Taladro | Angulo de Helice |
--- --- --- ---
| M1 X 0.2 | 1 | 0.2 | 0.87 | 0.783 | 0.108 | 0.05 | 0.82 | 4.18 |
| M1.1 X 0.2 | 1.1 | 0.2 | 0.97 | 0.883 | 0.108 | 0.05 | 0.92 | 3.75 |
| M1.2 X 0.2 | 1.2 | 0.2 | 1.07 | 0.983 | 0.108 | 0.05 | 1.02 | 3.4 |
| M1.4 X 0.2 | 1.4 | 0.2 | 1.27 | 1.183 | 0.108 | 0.05 | 1.22 | 2.87 |
| M1.6 X 0.2 | 1.6 | 0.2 | 1.47 | 1.383 | 0.108 | 0.05 | 1.42 | 2.48 |
| M1.8 X 0.2 | 1.8 | 0.2 | 1.67 | 1.583 | 0.108 | 0.05 | 1.62 | 2.18 |
| M2 X 0.25 | 2 | 0.25 | 1.838 | 1.729 | 0.135 | 0.063 | 1.77 | 2.48 |
| M2.2 X 0.25 | 2.2 | 0.25 | 2.038 | 1.929 | 0.135 | 0.063 | 1.97 | 2.24 |
| M2.5 X 0.35 | 2.5 | 0.35 | 2.273 | 2.121 | 0.189 | 0.088 | 2.18 | 2.81 |
| M3 X 0.35 | 3 | 0.35 | 2.773 | 2.621 | 0.189 | 0.088 | 2.68 | 2.3 |
| M3.5 X 0.35 | 3.5 | 0.35 | 3.273 | 3.121 | 0.189 | 0.088 | 3.18 | 1.95 |
| M4 X 0.5 | 4 | 0.5 | 3.675 | 3.459 | 0.271 | 0.125 | 3.55 | 2.48 |
| M4.5 X 0.5 | 4.5 | 0.5 | 4.175 | 3.959 | 0.271 | 0.125 | 4.05 | 2.18 |
| M5 X 0.5 | 5 | 0.5 | 4.675 | 4.459 | 0.271 | 0.125 | 4.55 | 1.95 |
| M6 X 0.75 | 6 | 0.75 | 5.513 | 5.188 | 0.406 | 0.188 | 5.32 | 2.48 |
| M7 X 0.75 | 7 | 0.75 | 6.513 | 6.188 | 0.406 | 0.188 | 6.32 | 2.1 |
| M8 X 0.75 | 8 | 0.75 | 7.513 | 7.188 | 0.406 | 0.188 | 7.32 | 1.82 |
| M8 X 1 | 8 | 1 | 7.35 | 6.917 | 0.541 | 0.25 | 7.09 | 2.48 |
| M9 X 0.75 | 9 | 0.75 | 8.513 | 8.188 | 0.406 | 0.188 | 8.32 | 1.61 |
| M9 X 1 | 9 | 1 | 8.35 | 7.917 | 0.541 | 0.25 | 8.09 | 2.18 |
| M10 X 0.75 | 10 | 0.75 | 9.513 | 9.188 | 0.406 | 0.188 | 9.32 | 1.44 |
| M10 X 1 | 10 | 1 | 9.35 | 8.917 | 0.541 | 0.25 | 9.09 | 1.95 |
| M10 X 1.25 | 10 | 1.25 | 9.188 | 8.647 | 0.677 | 0.313 | 8.86 | 2.48 |
| M11 X 0.75 | 11 | 0.75 | 10.513 | 10.188 | 0.406 | 0.188 | 10.32 | 1.3 |
| M11 X 1 | 11 | 1 | 10.35 | 9.917 | 0.541 | 0.25 | 10.09 | 1.76 |
| M12 X 1 | 12 | 1 | 11.35 | 10.917 | 0.541 | 0.25 | 11.09 | 1.61 |
| M12 X 1.25 | 12 | 1.25 | 11.188 | 10.647 | 0.677 | 0.313 | 10.86 | 2.04 |
| M12 X 1.5 | 12 | 1.5 | 11.026 | 10.376 | 0.812 | 0.375 | 10.64 | 2.48 |
| M14 X 1 | 14 | 1 | 13.35 | 12.917 | 0.541 | 0.25 | 13.09 | 1.37 |
| M14 X 1.25 | 14 | 1.25 | 13.188 | 12.647 | 0.677 | 0.313 | 12.86 | 1.73 |
| M14 X 1.5 | 14 | 1.5 | 13.026 | 12.376 | 0.812 | 0.375 | 12.64 | 2.1 |
| M16 X 1 | 16 | 1 | 15.35 | 14.917 | 0.541 | 0.25 | 15.09 | 1.19 |
| M16 X 1.5 | 16 | 1.5 | 15.026 | 14.376 | 0.812 | 0.375 | 14.64 | 1.82 |
| M18 X 1 | 18 | 1 | 17.35 | 16.917 | 0.541 | 0.25 | 17.09 | 1.05 |
| M18 X 1.5 | 18 | 1.5 | 17.026 | 16.376 | 0.812 | 0.375 | 16.64 | 1.61 |
| M18 X 2 | 18 | 2 | 16.701 | 15.835 | 1.083 | 0.5 | 16.18 | 2.18 |
| M20 X 1 | 20 | 1 | 19.35 | 18.917 | 0.541 | 0.25 | 19.09 | 0.94 |
| M20 X 1.5 | 20 | 1.5 | 19.026 | 18.376 | 0.812 | 0.375 | 18.64 | 1.44 |
| M20 X 2 | 20 | 2 | 18.701 | 17.835 | 1.083 | 0.5 | 18.18 | 1.95 |
| M22 X 1 | 22 | 1 | 21.35 | 20.917 | 0.541 | 0.25 | 21.09 | 0.85 |
| M22 X 1.5 | 22 | 1.5 | 21.026 | 20.376 | 0.812 | 0.375 | 20.64 | 1.3 |
| M22 X 2 | 22 | 2 | 20.701 | 19.835 | 1.083 | 0.5 | 20.18 | 1.76 |
| M24 X 1 | 24 | 1 | 23.35 | 22.917 | 0.541 | 0.25 | 23.09 | 0.78 |
| M24 X 1.5 | 24 | 1.5 | 23.026 | 22.376 | 0.812 | 0.375 | 22.64 | 1.19 |
| M24 X 2 | 24 | 2 | 22.701 | 21.835 | 1.083 | 0.5 | 22.18 | 1.61 |
| M27 X 1 | 27 | 1 | 26.35 | 25.917 | 0.541 | 0.25 0.69 |
| M27 X 1.5 | 27 | 1.5 | 26.026 | 25.376 | 0.812 | 0.375 1.05 |
| M27 X 2 | 27 | 2 | 25.701 | 24.835 | 1.083 | 0.5 1.42 |
| M30 X 1 | 30 | 1 | 29.35 | 28.917 | 0.541 | 0.25 0.62 |
| M30 X 1.5 | 30 | 1.5 | 29.026 | 28.376 | 0.812 | 0.375 0.94 |
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Límites de Dimensiones de la Rosca
Para fabricar o medir una rosca es necesario conocer los valores máximos y mínimos permitidos de las dimensiones básicas. Estos valores se calculan según la clase de la rosca (vea abajo). Para comprender las clases, primero debe comprender los términos de Ajuste (a veces denominado a desviación) y Tolerancias. Los siguientes datos y explicaciones se basan en ISO 965-1.
Definición de términos:
Permisión (Desviación): La distancia mínima permitida entre el perfil básico y el perfil actual.
Tolerancia: El ancho de tolerancia de un diámetro en el perfil actual de la rosca. (Paso, Mayor & Menor diámetros)
Un pequeño limite significa que la montaje de una rosca macho y hembra será más difícil, pero después del montaje habrá menos libertad de movimiento.
Un grande limite significa que la montaje de una rosca macho y hembra será más fácil, pero después del montaje habrá más libertad de movimiento.
El tamaño del limite no influye la dificultad de producción ni en el precio de la rosca.
Una tolerancia amplia es más fácil y económica de producir, pero produce una mayor dispersión entre las roscas
Una tolerancia estrecha es difícil de producir y más cara, pero produce una dispersión menor entre las roscas.
Clases de Roscas Métricas
La clase se escribe después del paso. Está precedido por un “–” y consta de dos o cuatro caracteres. Por ejemplo, para «M8 X 0.75 – 5g6g«, la clase es 5g6g. La clase define la desviación (ajuste) y la tolerancia de la rosca.
Las letras minúsculas representan roscas externas, las letras mayúsculas representan roscas internas.
Dos cartas de clase larga significan que la clase es válida tanto para el paso como para los diámetros mayor/menor.
Cuatro cartas largas significan una clase separada para el diámetro de paso (Los dos primeros caracteres) y los diámetros mayor/menor (Los dos últimos caracteres).
Si se omite la clase, se utiliza la clase por defecto .
Rosca Métrica Clase Sintaxis:
Valores permitidos para Clase de Roscado:
Roscas Externas:
D: e, f, g, h
T (Diámetro Mayor): 4, 6, 8
T (Diámetro del Paso):3, 4, 5, 6, 7, 8, 9
Símbolos: es, Td, Td2
Roscas Internas:
D: G, H
T (Diámetro Menor): 4, 5, 6, 7, 8
T (Diámetro del Paso):4, 5, 6, 7, 8
Símbolos: EL, TD1, TD2
Ejemplos de Clases de Roscas Métricas
| | | | | | | | |
--- --- --- --- |
| Clase | Perfil de Rosca | Diámetro del Paso | | Diámetro Exterior | | Diámetro Interior | |
| T | D | T | D | T | D |
| No especificado | External (Valores Predeterminados) | 6 | g | 6 | g | | |
| No especificado | Internal (Valores Predeterminados) | 6 | H | | | 6 | H |
| 4e | Externa | 4 | e | 4 | e | | |
| 3g8f | Externa | 3 | g | 8 | f | | |
| 5G | Interna | 5 | G | | | 5 | G |
| 5G7H | Interna | 5 | G | | | 7 | H |
Selección de Clase
El ingeniero puede seleccionar cualquier clase para las roscas macho y hembra para lograr sus objetivos de diseño. No obstante, la norma ISO 965-1 recomienda limitar la elección según una tabla de «Clases recomendadas».
La elección de la clase recomendada se hace en función de dos parámetros:
Calidad de Tolerancia:
F – Fino: para roscas de precisión, cuando se necesita una variación mínima de ajuste.
M – Médium: para uso general.
C – Coarse: para casos en los que pueden surgir dificultades de fabricación, por ejemplo al roscar barras laminadas en caliente y agujeros ciegos largos.
Longitud de Conexión:
N – Normal: Definido para cada rosca en la tabla siguiente.
S – Short: Más corto que la longitud «Normal».
L – Long: Más larga que la «longitud normal».
Elige «N» Cuando se desconoce la longitud de la conexion .
Longitud normal de la conexión para roscas métricas
Escriba la descripción de su rosca en el cuadro de búsqueda.
| Roscado | Longitud de compromiso normal (milímetros) | Longitud de compromiso normal (pulgadas) |
---
| M1 | 0.6 -1.7 | 0.024 -0.067 |
| M1 X 0.2 | 0.5 -1.4 | 0.020 -0.055 |
| M1.1 | 0.6 -1.7 | 0.024 -0.067 |
| M1.1 X 0.2 | 0.5 -1.4 | 0.020 -0.055 |
| M1.2 | 0.6 -1.7 | 0.024 -0.067 |
| M1.2 X 0.2 | 0.5 -1.4 | 0.020 -0.055 |
| M1.4 | 0.7 -2.0 | 0.028 -0.079 |
| M1.4 X 0.2 | 0.5 -1.4 | 0.020 -0.055 |
| M1.6 | 0.8 -2.6 | 0.031 -0.102 |
| M1.6 X 0.2 | 0.5 -1.5 | 0.020 -0.059 |
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Selección de clase recomendada para roscas métricas
Fondo azul – Clases por defecto
Fondo verde – Clases de primera elección.
Fondo amarillo – Clases de segunda opción.
| | | | | | | | | | | |
--- --- --- --- ---
| | | Longitud de conexión | | | | | | | | |
| Tolerancia | Perfil de Rosca | S (Corto) | | M (Medio) | | | | L (Long) | | |
| Paso Fino | Interna | 4H | | 5H | | | | 6H | | |
| Externa | 3h4h | | 4h | 4g | | | 5h4h | 5g4g | |
| Paso Medio | Interna | 5H | 5G | 6H | 6G | | | 7H | 7G | |
| Externa | 5h6h | 5g6g | 6g | 6f | 6e | 6h | 7h6h | 7g6g | 7e6e |
| Paso Grueso | Interna | | | 7H | 7G | | | 8H | 8G | |
| Externa | | | 8g | 8e | | | 9g8g | 9e8e | |
Limites de Rosca Métrica (desviaciones)
Lo limite EL para roscas internas y es para roscas externas se deriva de la clase. Lo limite aproximado se puede calcular de acuerdo con las siguientes fórmulas:
ELH = esh = 0
ELG = esg ≈15 + 11 P
ese ≈50 + 11 P
esf ≈30 + 11 P
P es el paso expresado en micras
El resultado es/EL está en micras.
Lamentablemente, las fórmulas de abajo son aproximadas.Los valores formales deben obtenerse de la norma ISO 965-1 o de la tabla siguiente
Permisiones/Tolerancias de Rosca Métrica
Escriba la descripción de su rosca en el cuadro de búsqueda.
Los valores están en Micras.
Para milímetros, divida por 1,000. Para Pulgadas,dividir por 25,4000.
| Roscado | e | f | g/G | h/H |
--- ---
| M1 X 0.2 | | | 17 | 0 |
| M1 | | | 18 | 0 |
| M1.1 X 0.2 | | | 17 | 0 |
| M1.1 | | | 18 | 0 |
| M1.2 X 0.2 | | | 17 | 0 |
| M1.2 | | | 18 | 0 |
| M1.4 X 0.2 | | | 17 | 0 |
| M1.4 | | | 18 | 0 |
| M1.6 X 0.2 | | | 17 | 0 |
| M1.6 | | 34 | 19 | 0 |
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Rosca Métrica Tolerancias
Las tolerancias TD para roscas internas y Tdroscas externas son suministradas por la clase. Las tolerancias aproximadas se pueden calcular según las fórmulas siguientes.
Td – Tolerancia del Diámetro Mayor Externo
Td2 – Tolerancia del Diámetro del Paso Externo
TD1 – Tolerancia del Diámetro Menor Interno
TD2 – Tolerancia del Diámetro del Paso Interno
P & D – En mm
T– Los resultados de Tolerancias están en micras. (para mm dividir por 1,000 y para pulgadas por 254,000).
T(i) – Tolerancia para el diámetro externo grado i. Por ejemplo, Td(6) es la clase de tolerancia g6, y Td(8) es para la clase g8.
Fórmulas para tolerancias de Roscas Métricas.
Externa
Interna
Las fórmulas anteriores son aproximadas. Para los valores normales, utilice la siguiente tabla o la norma ISO 965-1.
Rosca Métrica Tabla de Tolerancias
Escriba la descripción de su rosca en el cuadro de búsqueda.
Los valores están en Micras.
Para milímetros, divida por 1,000. Para Pulgadas,dividir por 25,4000.
Los números (3 -9) en la fila superior representan la calificación de la clase. (e.g., h6, g8, etc.)
| Símbolo | Diámetro | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
--- --- --- ---
| M1 X 0.2 | | | | | | | | |
| Td | External Major Diameter | | 36 | | 56 | | | |
| Td2 | External Pitch Diameter | 24 | 30 | 38 | 48 | | | |
| TD1 | Internal Minor Diameter | | 38 | | | | | |
| TD2 | Internal Pitch Diameter | | 40 | | | | | |
| M1 | | | | | | | | |
| Td | External Major Diameter | | 42 | | 67 | | | |
| Td2 | External Pitch Diameter | 26 | 34 | 42 | 53 | | | |
| TD1 | Internal Minor Diameter | | 45 | 56 | | | | |
| TD2 | Internal Pitch Diameter | | 45 | 56 | | | | |
| M1.1 X 0.2 | | | | | | | | |
| Td | External Major Diameter | | 36 | | 56 | | | |
| Td2 | External Pitch Diameter | 24 | 30 | 38 | 48 | | | |
| TD1 | Internal Minor Diameter | | 38 | | | | | |
| TD2 | Internal Pitch Diameter | | 40 | | | | | |
| M1.1 | | | | | | | | |
| Td | External Major Diameter | | 42 | | 67 | | | |
| Td2 | External Pitch Diameter | 26 | 34 | 42 | 53 | | | |
| TD1 | Internal Minor Diameter | | 45 | 56 | | | | |
| TD2 | Internal Pitch Diameter | | 45 | 56 | | | | |
| M1.2 X 0.2 | | | | | | | | |
| Td | External Major Diameter | | 36 | | 56 | | | |
| Td2 | External Pitch Diameter | 24 | 30 | 38 | 48 | | | |
| TD1 | Internal Minor Diameter | | 38 | | | | | |
| TD2 | Internal Pitch Diameter | | 40 | | | | | |
| M1.2 | | | | | | | | |
| Td | External Major Diameter | | 42 | | 67 | | | |
| Td2 | External Pitch Diameter | 26 | 34 | 42 | 53 | | | |
| TD1 | Internal Minor Diameter | | 45 | 56 | | | | |
| TD2 | Internal Pitch Diameter | | 45 | 56 | | | | |
| M1.4 X 0.2 | | | | | | | | |
| Td | External Major Diameter | | 36 | | 56 | | | |
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Cálculos de Límites de Rosca Métrica
Una vez que los valores Td, TD, es & ELson conocidos, los límites finales pueden se calcular de acuerdo con las siguientes fórmulas.
La norma de rosca Métrica ISO 965-2 proporciona los límites sólo para las clases 6H & 6g. En todos los otros casos, los cálculos deben hacerse manualmente de acuerdo las siguientes fórmulas o por nuestro Calculador de Roscado
Para roscas superiores de 1.4 mm y clases 5H y 6h para roscas hasta y incluyendo 1.4 mm
Formulas de los límites de Rosca Unificada Externa
Formulas de los límites de Rosca Unificada Interna
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Para otros usos de este término, véase tensión.
| Tensión(σ) |
| Tensiones residuales en el interior de un plástico reveladas por fotoelasticidad mediante luz polarizada |
| Magnitud | Tensión(σ) |
| Tipo | Magnitudvectorial |
| Unidad SI | Pascal |
| Otras unidades | Megapascal (MPa) Bar (unidad de presión) |
| [editar datos en Wikidata] |
Vector tensión en una superficie S con vector unitario normal n{\displaystyle \mathbf {n} \,!}. Dependiendo de la orientación del plano en cuestión, el vector tensión puede no ser necesariamente perpendicular a ese plano, es decir, paralelo a n{\displaystyle \mathbf {n} \,!}, y puede descomponerse en dos vectores: un componente normal al plano, llamado tensión normal σ n{\displaystyle \sigma {\mathrm {n} }\,!}, y otro componente paralelo al plano, denominado _tensión cortante τ{\displaystyle \tau \,!}.
Componentes del tensor tensión en un punto P de un sólido deformable.
En física e ingeniería, se denomina tensión mecánica a la magnitud física que representa la fuerza por unidad de área en el entorno de un punto material sobre una superficie real o imaginaria de un medio continuo. Es decir posee unidades físicas de presión. La definición anterior se aplica tanto a fuerzas localizadas como fuerzas distribuidas, uniformemente o no, que actúan sobre una superficie. Con el objeto de explicar cómo se transmiten a través de los sólidos las fuerzas externas aplicadas, es necesario introducir el concepto de tensión, siendo este el concepto físico más relevante de la mecánica de los medios continuos, y de la teoría de la elasticidad en particular.
El estrés expresa la fuerzas interna que partículas vecinas de un material continuo ejercen entre sí, mientras que "strain" es la medida de la deformación del material. Por ejemplo, cuando una barra vertical sólida soporta un peso elevado, cada partícula de la barra empuja a las partículas inmediatamente inferiores. Cuando un líquido está en un recipiente cerrado bajo presión, cada partícula es empujada contra todas las partículas circundantes. Las paredes del recipiente y la superficie que induce la presión (como un pistón) empujan contra ellas en reacción (newtoniana). Estas fuerzas macroscópicas son en realidad el resultado neto de un gran número de fuerzas intermoleculares y colisiones entre las partículas de esas moléculas. La tensión se suele representar con la letra griega minúscula sigma (σ).
La deformación dentro de un material puede surgir por varios mecanismos, como la tensión aplicada por fuerzas externas al material (como la gravedad) o a su superficie (como fuerzas de contacto, presión externa o fricción). Cualquier esfuerzo (deformación) de un material sólido genera una tensión elástica interna, análoga a la fuerza de reacción de un resorte, que tiende a restaurar el material a su estado original no deformado. En los líquidos y gases, sólo las deformaciones que cambian el volumen generan tensiones elásticas persistentes. Si la deformación cambia gradualmente con el tiempo, incluso en los fluidos suele haber cierta tensión viscosa, que se opone a ese cambio. Las tensiones elásticas y viscosas suelen combinarse bajo el nombre de tensión mecánica.
Esfuerzo mecánico
Puede existir una tensión significativa incluso cuando la deformación es insignificante o inexistente (una suposición común cuando se modela el flujo de agua). Puede existir tensión en ausencia de fuerzas externas; esta tensión incorporada es importante, por ejemplo, en el hormigón pretensado y el vidrio templado. La tensión también puede imponerse en un material sin la aplicación de las fuerzas netas, por ejemplo por cambios de temperatura o química, o por campo electromagnético externo (como en los materiales piezoeléctricos y magnetostrictivos).
La relación entre la tensión mecánica, la deformación y el tasa de cambio de deformación puede ser bastante complicada, aunque una aproximación lineal puede ser adecuada en la práctica si las cantidades son suficientemente pequeñas. Una tensión que supere ciertos límites de resistencia del material dará lugar a una deformación permanente (como flujo plástico, fractura, cavitación) o incluso cambiará su estructura cristalina y composición química.
Historia
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Puente de la época del Romano en Suiza. Los arcos de piedra del puente están sometidos a esfuerzos de compresión.
Puente inca sobre el río Apurímac. La cuerda del puente está sometida a esfuerzos de tracción.
El ser humano conoce las tensiones en el interior de los materiales desde la antigüedad. Hasta el siglo XVII, este conocimiento era en gran medida intuitivo y empírico, aunque esto no impidió el desarrollo de tecnologías relativamente avanzadas como el arco compuesto y el soplado de vidrio.
A lo largo de varios milenios, los arquitectos y constructores en particular aprendieron a unir vigas de madera y bloques de piedra cuidadosamente moldeados para soportar, transmitir y distribuir la tensión de la manera más eficaz, con ingeniosos dispositivos como los capiteles, arcos, cúpulas, cerchas y los arbotantes de las catedrales góticas.
Los arquitectos antiguos y medievales desarrollaron algunos métodos geométricos y fórmulas sencillas para calcular el tamaño adecuado de pilares y vigas, pero la comprensión científica de los esfuerzos sólo fue posible tras la invención de las herramientas necesarias en los siglos XVII y XVIII: El riguroso método experimental de Galileo Galilei, las coordenadas y la geometría analítica de René Descartes, y las leyes del movimiento y el equilibrio y el cálculo de infinitesimales de Newton. Con esas herramientas, Augustin-Louis Cauchy fue capaz de dar el primer modelo matemático riguroso y general de un cuerpo elástico deformado introduciendo las nociones de tensión y deformación. Cauchy observó que la fuerza a través de una superficie imaginaria era una función lineal de su vector normal; y, además, que debía ser una función simétrica (con momento total nulo). La comprensión de la tensión en los líquidos comenzó con Newton, quien proporcionó una fórmula diferencial para las fuerzas de fricción (tensión cortante) en flujo laminar paralelo.
Definición
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La tensión se define como la fuerza a través de una pequeña frontera por unidad de área de esa frontera, para todas las orientaciones de la frontera. Derivada de una cantidad física fundamental (fuerza) y de una cantidad puramente geométrica (área), la tensión es también una cantidad fundamental, como la velocidad, el par o la energía, que puede cuantificarse y analizarse sin consideración explícita de la naturaleza del material o de sus causas físicas.
La tensión a través de un elemento superficial (disco amarillo) es la fuerza que el material de un lado (bola superior) ejerce sobre el material del otro lado (bola inferior), dividida por el área de la superficie
Siguiendo las premisas básicas de la mecánica del continuo, la tensión es un concepto macroscópico. Es decir, las partículas consideradas en su definición y análisis deben ser lo suficientemente pequeñas como para ser tratadas como homogéneas en composición y estado, pero lo suficientemente grandes como para ignorar los efectos de la cuántica y los movimientos detallados de las moléculas. Así, la fuerza entre dos partículas es en realidad el promedio de un gran número de fuerzas atómicas entre sus moléculas; y las cantidades físicas como la masa, la velocidad y las fuerzas que actúan a través de la masa de los cuerpos tridimensionales, como la gravedad, se supone que se distribuyen suavemente sobre ellos.: p.90-106 Dependiendo del contexto, también se puede suponer que las partículas son lo suficientemente grandes como para permitir el promediado de otras características microscópicas, como los granos de una barra de metal o las fibras de un trozo de madera.
Cuantitativamente, la tensión se expresa mediante el vector de tracción de Cauchy T definido como la fuerza de tracción F entre partes adyacentes del material a través de una superficie de separación imaginaria S, dividida por el área de S.: p.41-50 En un fluido en reposo la fuerza es perpendicular a la superficie, y es la conocida presión. En un sólido, o en una flujo de líquido viscoso, la fuerza F puede no ser perpendicular a S; por lo tanto, la tensión a través de una superficie debe considerarse una cantidad vectorial, no escalar. Además, la dirección y la magnitud dependen generalmente de la orientación de S. Así, el estado de tensión del material debe ser descrito por un tensor, llamado tensor de tensiones; que es un función lineal que relaciona la vector normaln de una superficie S con el vector de tracción T a través de S. Con respecto a cualquier coordenadas elegido, el tensor de tensiones de Cauchy puede representarse como una matriz simétrica de 3×3 números reales. Incluso dentro de un cuerpo homogéneo, el tensor de tensiones puede variar de un lugar a otro, y puede cambiar con el tiempo; por lo tanto, la tensión dentro de un material es, en general, un campo tensorial que varía con el tiempo.
Introducción
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Si se considera un cuerpo sometido a un sistema de fuerzas y momentos de fuerza, se puede observar la acción de las tensiones mecánicas si se imagina un corte mediante un plano imaginario π que divida el cuerpo en dos partes. Para que cada parte estuviera en equilibrio mecánico, sobre la superficie de corte de cada una de las partes debería restablecerse la interacción que ejercía la otra parte del cuerpo. Así, sobre cada elemento de la superficie (dS), debe actuar una fuerza elemental (dF), a partir de la cual se define un vector tensión (tπ) como el resultado de dividir dicha fuerza elemental entre la superficie del elemento.
t π=d F d A{\displaystyle \mathbf {t} _{\pi }={\frac {d\mathbf {F} }{dA}}}
Este vector tensión depende del estado tensional interno del cuerpo, de las coordenadas del punto escogido y del vector unitario normal al plano π (n π). Se puede probar que t π y n π están relacionados por una aplicación lineal T o campo tensorial llamado tensor tensión:
t π=T(n π){\displaystyle \mathbf {t} {\pi }=\mathbf {T} \left(\mathbf {n} {\pi }\right)}
La tensión mecánica se expresa en unidades de presión, es decir, fuerza dividida entre área. En el Sistema Internacional, la unidad de la tensión mecánica es el pascal (1 Pa = 1 N/m²). No obstante, en ingeniería también es usual expresar otras unidades como kg/cm² o kg/mm², donde «kg» se refiere a kilopondio o kilogramo-fuerza, no a la unidad de masa kilogramo.
Principio de Cauchy
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Artículo principal:Postulados de Cauchy
Sea B{\displaystyle B\,} un medio continuo deformado, entonces en cada subdominio V⊂B{\displaystyle V\subset B\,} existe un campo vectorialt{\displaystyle t\,}, llamado campo de tensiones, tal que las fuerzas de volumen f∈R 3{\displaystyle f\in \mathbb {R} ^{3}} y el campo de tensiones t∈R 3{\displaystyle t\in \mathbb {R} ^{3}} satisfacen las siguientes ecuaciones de equilibrio
∫V f(x)d V+∫∂V t(x,n)d A=0{\displaystyle \int {V}f(\mathbf {x} )dV+\int {\partial V}t(\mathbf {x} ,n)dA=0}∫V x×f(x)d V+∫∂V x×t(x,n)d A=0{\displaystyle \int {V}\mathbf {x} \times f(\mathbf {x} )dV+\int {\partial V}\mathbf {x} \times t(\mathbf {x} ,n)dA=0}
Este principio fue enunciado por Augustin Louis Cauchy en su forma más general, aunque previamente Leonhard Euler había hecho una formulación menos general. De este principio puede demostrarse el teorema debido a Cauchy para el tensor tensión que postula que el principio de Cauchy equivale a la existencia de una aplicación lineal, llamada tensor tensiónT∈C 1(B,R 3){\displaystyle T\in C^{1}(B,\mathbb {R} ^{3})} con las siguientes propiedades:
t(x,n)=T(x),{\displaystyle t(\mathbf {x} ,n)=T(\mathbf {x} ),}
div T(x)+f(x)=0,{\displaystyle \operatorname {div} \,T(\mathbf {x} )+f(\mathbf {x} )=0,}
T(x)=T T(x){\displaystyle T(\mathbf {x} )=T^{T}(\mathbf {x} )}
Con el principio, enunció también los dos postulados que definen la actuación de los vectores sobre una superficie
Tensión normal y tensión tangencial
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Si consideramos un punto concreto de un sólido deformable sometido a tensión y se escoge un corte mediante un plano imaginario π que lo divida al sólido en dos, queda definido un vector tensión tπ que depende del estado tensional interno del cuerpo, de las coordenadas del punto escogido y del vector unitario normal n π al plano π definida mediante el tensor tensión:
t π=T(n π){\displaystyle {\mathbf {t} {\pi }}={T(\mathbf {n} {\pi })}\,}
Usualmente ese vector puede descomponerse en dos componentes que físicamente producen efectos diferentes según el material sea más dúctil o más frágil. Esas dos componentes se llaman componentes intrínsecas del vector tensión respecto al plano π y se llaman tensión normal o perpendicular al plano y tensión tangencial o rasante al plano, estas componentes vienen dadas por:
{σ π=t π⋅n π τ π=‖t π×n π‖⇒‖t π‖2=σ π 2+τ π 2{\displaystyle {\begin{cases}\sigma {\pi }=\mathbf {t} {\pi }\cdot \mathbf {n} {\pi }\\tau {\pi }=\|\mathbf {t} {\pi }\times \mathbf {n} {\pi }\|\end{cases}}\Rightarrow \qquad \|\mathbf {t} {\pi }\|^{2}=\sigma {\pi }^{2}+\tau _{\pi }^{2}}
Análogamente cuando existen dos sólidos en contacto y se examinan las tensiones entre dos puntos de los dos sólidos, se puede hacer la descomposición anterior de la tensión de contacto según el plano tangente a las superficies de ambos sólidos, en ese caso la tensión normal tiene que ver con la presión perpendicular a la superficie y la tensión tangencial tiene que ver con las fuerzas de fricción entre ambos.
Un caso particular: tensión uniaxial (problema unidimensional)
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Un caso particular es el de tensión uniaxial, que se define en una situación en que se aplica fuerzaF uniformemente distribuida sobre un área A. En ese caso la tensión mecánica uniaxial se representa por un escalar designado con la letra griega σ (sigma) y viene dada por:
σ=F A{\displaystyle \sigma ={\frac {F}{A}}}
El concepto de esfuerzo longitudinal parte en dos observaciones simples sobre el comportamiento de cables sometidos a tensión:
Cuando un cable con elasticidad lineal se estira bajo la acción de una fuerza F, se observa que el alargamiento unitario Δ L/L es proporcional a la carga F dividida por el área de la sección transversal A del cable, esto es, al esfuerzo, de modo que podemos escribir
σ=E Δ L L{\displaystyle \sigma =E{\frac {\Delta L}{L}}\,}
donde E es una característica del material del cable llamado módulo de Young.2. El fallo resistente o ruptura del cable ocurre cuando la carga F superaba un cierto valor F rupt que depende del material del cable y del área de su sección transversal. De este modo queda definido el esfuerzo de ruptura
σ rupt=F rupt A{\displaystyle \sigma {\text{rupt}}={\frac {F{\text{rupt}}}{A}}\,}
Estas observaciones ponen de manifiesto que la característica fundamental que afecta a la deformación y al fallo resistente de los materiales es la magnitud σ, llamada esfuerzo o tensión mecánica. Medidas más precisas ponen de manifiesto que la proporcionalidad entre el esfuerzo y el alargamiento no es exacta porque durante el estiramiento del cable la sección transversal del mismo experimenta un estrechamiento, por lo que A disminuye ligeramente. Sin embargo, si se define la tensión real σ = F/A' donde A' representa ahora el área verdadera bajo carga, entonces se observa una proporcionalidad correcta para valores pequeños de F.
El coeficiente de Poisson se introdujo para dar cuenta de la relación entre el área inicial A y el área deformada A'. La introducción del coeficiente de Poisson en los cálculos estimaba correctamente la tensión al tener en cuenta que la fuerza F se distribuía en un área algo más pequeña que la sección inicial, lo cual hace que σ >s.
Véase también
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diferenciarla de: Tension (physics) / Tensión (mecánica), hoy, coinciden el número de sus (30) iw, no sus catª, cfr.; e ítem/wikidata (Q206175)
Tensión cortante
Tensor tensión
Tensor deformación
Deformación
Factor de carga
Referencias
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↑org/what-is-shear-stress-materials-definition/ «Qué es el estrés de cizallamiento - Materiales - Definición». Propiedades de los materiales. 31 de julio de 2020. Consultado el 2 de noviembre de 2022.
↑Gordon, J.E. (2003). Estructuras, o, Por qué las cosas no se caen (2. Da Capo Press edición). Cambridge, MA: Da Capo Press. ISBN0306812835.
↑Jacob Lubliner (2008). "Plasticity Theory"Archivado el 31 de marzo de 2010 en Wayback Machine. (revised edition). Dover Publications. ISBN0-486-46290-0
↑ pp.107-110
↑Wai-Fah Chen and Da-Jian Han (2007), "Plasticity for Structural Engineers". J. Ross Publishing ISBN1-932159-75-4
↑Peter Chadwick (1999), "Continuum Mechanics: Concise Theory and Problems". Dover Publications, series "Books on Physics". ISBN0-486-40180-4. pages
↑I-Shih Liu (2002), "Continuum Mechanics". Springer ISBN3-540-43019-9
Bibliografía
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Luis Ortiz Berrocal (2007). Resistencia de materiales, Madrid: Ed. McGraw-Hill. ISBN 978-84-481-5633-6
Enlaces externos
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Artículos sobre tensión en inglés por Andrés Melo y Geraint Wiggins, formato.PDF
| Control de autoridades | Proyectos Wikimedia Datos:Q206175 Multimedia:Stress (mechanics) / Q206175 Identificadores GND:4134428-5 NDL:00568941 AAT:300056434 Diccionarios y enciclopedias Britannica:url Identificadores médicos MeSH:D013314 UMLS:C0038442 |
Datos:Q206175
Multimedia:Stress (mechanics) / Q206175
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17486 | https://www.youtube.com/watch?v=bfK983FRxmI | 35. Conjugate Harmonic Functions | Complex Variables and Applications | Brown and Churchill
Mathematics Helpline
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Conjugate Harmonic Functions
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Course Contents:
The Concept of Analytic Functions
• Complex numbers, complex planes, complex functions
• Analytic functions
• Entire functions
• Harmonic functions
• Elementary functions: complex exponential, logarithmic and hyperbolic functions
Infinite Series
• Power series, derived series, radius of convergence
• Taylor series and Laurent series
Conformal Representation
• Transformation, conformal transformation
• Linear transformation
• Möbius transformations
Complex Integration
• Complex integrals
• Cauchy-Goursat theorem
• Cauchy’s integral formula and their consequences
• Liouville’s theorem
• Morera’s theorem
• Derivative of an analytic function
Recommended Books:
1. D. G. Zill and P. D. Shanahan, Complex Analysis, (Jones and Bartlett Publishers,
2003)
2. H. S. Kasana, Complex Variables: Theory and Applications, (Prentice Hall, 2005)
3. J. W. Brown and R. V. Churchill, Complex Variables and Applications, 7th edition,
(McGraw Hill Company, 2004)
4. M. R. Spiegel, Complex Variables, (McGraw Hill Book Company, 1974)
5. Louis L. Pennisi, Elements of Complex Variables, (Holt, Linehart and Winston, 1976)
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में विराजमान ही में स्टूडेंट हम डिस्कस करते हैं और मोनिका डुब्लीकेट हमलेट कर लेते हैं यह हमारे पास यह फंक्शन है लैटर फॉर्म्स एंड इज इक्वल टू यू X1 अधिसमय यह ऑप्शन है तो अब हम देखेंगे कि यह जो यह शर्मा निकुंज बेस्ट आफ घी जो है वह करेंगे इसे अब हम इसको करेंगे उसके लिए क्या है कि इस फंक्शन एंड फंक्शंस द फंक्शंस हाउ आर यू [संगीत] के लिए बेड विद और शर्मा निक पिक कि इन उम्र एनडी कर दो कि एंड एक तो यह है कि फंक्शन जो है वह क्या हूं दोनों हर मानिक हूं हम्म और माने कानूनविद की बात करने जा रहे हैं तो दोनों फंक्शंस है वह अरमान एक उपहार मानिक होने का मतलब क्या है यह यू कब हार मानिक होगा इसका मतलब यह है कि यूरो एक्स एक्स प्लस न्यू व्हाई व्हाई इज इक्वल टू जीरो विक्रम वर्मा ने होगा व्यक्ति सभी वाइज है वह किसके हो जाए तो सबसे पहले रिक्वायरमेंट है यह हार्मोन कार गिफ्ट की प्यूरी और विजय है वह शर्मा ने दूसरा यह है कि जो फर्स्ट ऑर्डर शोल्ड बे थे फर्स्ट प्रेसिडेंट ऑफ कि एंड फर्स्ट ऑर्डर थे फर्स्ट ऑर्डर पार्शल डेरिवेटिव्स सेटिस्फाई अधिकाश घृणित वेंस सेटिस्फाइड रघु दीक्षित के रिमांड पर कि सिक्वेंस ये कोशिश आजमानी सीक्वेंस कार्टिज इंफोर्मेशन स्टूडेंट्स क्या होती हैं इसमें यू एक्स इज इक्वल टू बी वायलेंस एंड यू वेयर इज इक्वल टू माइनस बी एक्स यह जो है वह सेंस करती है वह यह है यह दूसरा यह है कि वह कश्मीर को है वह सब्सक्राइब करें तो हम कि युद्ध शर्मा ने गेट आफ यू इन थिस 220 वीएस एप्स हार मानी कन्ज्यूगेट कर दो मैं ऑफिस यू यह याद रखिए यह देखिए भी है तो बिल्कुल हम कहेंगे उसका हार्मोन यानि ग्रेट है लेकिन हम यह नहीं कह सकते कि वह बीच का फरमान एक प्लेट है ठीक है हम लेंगे वीएस डीडी शर्मा निकुंज बेस्ट ऑफ़ यू कब जब यह दोनों ओर मानिक मानिक हूं और अच्छी तो है वह क्या होती है तो इसके बाद हमारे पास है उसको इंप्रूव करता है कि यह है कि फंक्शन सब्सक्राइब सब्सक्राइब के इस विडियो को जरूर सब्सक्राइब करें और माने तो फिर फंक्शन घृत हो तो माफ कर दो थैंक्यू होता है अब यह जितनी हमने बेसिकली पहले बातें की हैं उनका करने के लिए ही यह नैचुरल है हम यह बात की के लैटविया इजहार माने कन्ज्यूगेट ऑफ यू एन डी वी क्या है हर माह निकुंज बेस्ट तो हर माने का हृदय की डेफिनेशन क्या है हर माह निकाल लेंगे इसमें हम यह क्या है कि यू और वी जो दोनों है यह क्या होंगे हार मानी केहुके डी के अंदर और का शिरोमणि क्वेश्चंस है वह क्या होंगी सेटिस्फाया होंगी अगर का निर्माण इवेंट्स बाय होती है तो इसका मतलब यह क्या है अनांइट इंडियन है रेफर सेट किया है अनलिमिटेड है कन्वर्सली गैस का लेते हैं कि एप्स है वह अनांइट एक है अगर ऐप चैनल टेक है तो फिर मना लिट्टे के लिए अभी हमने जुर्म पड़ा कि इसके जो कंपोनेंट्स है वह और मानिक होते हैं जो कंप्लीट है वह क्या होते हैं और माने और का शिरोमणि क्वेश्चंस पेज है वह क्या होती है सेटिस्फाई होती हैं तो इस बुनियाद पर हम कह सकते हैं विजय हारमोनिका ब्रिगेड आफ यू तो यह वह जो पहले हम बातें करके आ चुके हैं म्यूजिकली हमने किस फॉर में लिख लिया है एक सीरम की फॉर्म में जो है वह ली है अब हम जो है वह एग्जांपल के जरिए यह चीज स्टूडेंट्स प्रूफ करेंगे कि अगर विजय हारमोनिक पेट हारमोनिक ऑफ यू तो यूज है वह लाजमी नहीं है बीच का जो है वह शर्मा निकुंज ब्रिगेड को ठीक है इश्क कब होगा वह हमारा इस चैप्टर का जो है उसको नहीं है वहां के चैप्टर में हम जो है वह इसके ऊपर कंडीशन स्कोर इंपोज करेंगे अभी हम यह प्रूफ करेंगे कि अगर जो है वह भी यू का हारमोनिका डुब्लीकेट है तू लाज़मी नहीं है युवी का भी है वह शर्मा निकुंज ब्रिगेड हो [संगीत] कि अब स्टूडेंट्स यह हमारे पास एक एग्जांपल है कि अफसर डीजी को ड्रोस एंड स्किमिंग इसमें यह जो अ फॉर किड्स केयर यह इंटर फंक्शन है और अगर यह एंटर फंक्शन है रूबी कर लेते हैं यह डर फंक्शन है और उसके बाद जो है वह इसके जो कंपोनेंट्स है वह में वी शर्मा ने काजू एट होगा यूके और हम प्रूफ करेंगे यू हार मानने कन्ज्यूगेट नहीं है बी के कैसे अब सबसे पहले तो यह है कि स्टूडेंट्स f-18 जो है वह में दिया गया है और इस घृणित प्ले करो subscribe to subscribe इसको हम कि युद्ध सब्सक्राइब प्लीज डू है इस वजह है वह लिखा जा सकता है अयस्क स्टेपल होता है स्टूडेंट्स माइनस वन के एक होता है तो यह क्या बन जाएगा यह बनेगा एक्स क्लियर - वॉइस केयर प्लस अयोध्या टू एक्स वाय यह हमारे पास हफ्ते गया इसको हम टू कर लेते हैं यह हमारे पास वह क्या है यह अब यहां पर अगर हम इसका कंपीटीशन अधिक तो इसमें अब यह देखिए कि जूते यह किसके गया शेयर - शेयर करें और सब्सक्राइब यह तो एक स्वाभाविक है अच्छा अब यहां पर इसके पास डेरिवेटिव्स अगर हम फोल्ड करें तो यह एक्स्ट्रा है वह क्या आएगा 2X yuva21 के गुल आएगा - टू लाइक इसी तरह subscribe To My के और व्यक्ति का या 21 अब यहां से सब्सक्राइब वह कैसे subscribe To My है और इसी तरह जो है वह व्वे वीडियोस के और यह हमारे पास डेरिवेटिव्स यह पॉलिनॉमियल्स है और घ्र डेढ-डेढ सब्सक्राइब सब्सक्राइब टो कि फेज़ इज इक्वल टू जेड तक केयर इन विंटर फंक्शन कि हेंस को रेफर साइड इस चैनल टेक ऑफ मेडिकल रिसर्च के रिज़ल्ट एक जय हिंद होल कंपलेक्स ए प्लेन अच्छा ठीक है तो इसका मतलब क्या हुआ कि जो B है हमारे पास चीफ वाक्य ग्रॉफिक है तो बी एक्स वाय जो की मेक्सिको यह हमारे पास क्या है और मोनिका शर्मा निकुंज बेस्ट टर्न ऑफ यू एक्स वाय ठीक है और वह किसके इक्वल है एक हेयर - वास क्वाइट है तो अब हम क्या करते हैं आमलेट करते हैं लेट्स सैफरन जो है वह किसके ऊपर है दैट इज इक्वल टू अब हम इन्हें वापस मीटर चेंज कर देते हैं एक्स केयर यह जो तो एक्स्ट्रा है इसे पहले लेते हैं लास्ट आयो टाटा एक्स क्लियर - वॉइस केयर को बाद में लिख लेते हैं ठीक है जी हां हैं इनके अब हम चेक करते हैं कि क्या यह जो है वह यूविका कन्ज्यूगेट है क्यों कि अब हमने बेसिकली क्या नेट किया है कि फाइबर हमारे पास किसके पीपल हैव ई एक्स वाय इधर कर दिया है और यह एक्सरसाइज है वह हमने इधर करके अब चेक करने लगे हैं कि क्या यूज है वह इसका चाहिए अब यहां पर हमने जो है वह क्या करेंगे इसी तरह ही किसकी है अब स्कूल लेते हैं तो यहां पर जूते यह घृणित और व्यक्ति से घृणा के अभियुक्त और Y2 इसके वी एक्स किसके बुलाएगा तो ऐज के और विवाह किसके जाएगा - 21806 देखें जूते इससे नॉट डिफिकल्ट विवाह हैंड यू वेयर इज नॉट डिफिकल्ट टो माइनस बी एक्स इसका मतलब क्वेश्चन होल्ड नहीं होती नहीं होती तो यह ऑप्शनल है अब फंक्शनल नहीं है तो इसका मतलब यह जो हमारे पास नंबर है 11 - के इसका यह यह हमारे पास है वह इस हार मानने वाले नहीं है तो इसका मतलब यह है कि अगर जो है वह माने तो लाजमी है कि जो भी होगा तो यह हमने इसमें जो है वह उसकी इसके बाद जो है वह नेक्स्ट एग्जांपल अगर हम एक डुबो दिया गया हो कन्ज्यूगेट दिया गया हो तो दूसरे एलिमेंट को हम जो है वह कैसे करेंगे इस ठीक है यह माने कि न माने फंक्शन फंक्शन दिया गया है तो उसका जो है वह हम कैसे करेंगे इसको हम डिस्कस करते हैं झाल |
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The Instance and Object are from Object Oriented Programming.
For some programming languages like Java, C++, and Smalltalk, it is important to describe and understand code. In other languages that used in Structured Programming, this concept doesn't exist.
This is a view from Structural Programming. There's no real significant difference that should consume too much of your time. There might be some fancy language that some people might take up a lot of spaces to write about, but at the end of the day, as far as a coder, developer, programmer, architect, is concerned, an instance of a class and an object mean the same thing and can often be used interchangeably. I have never met anyone in my career that would be picky and spend a half-hour trying to point out the differences because there's really none. Time can be better spent on other development efforts.
UPDATE With regards to Swift, this is what Apple who invented Swift prefers :
An instance of a class is traditionally known as an object. However,
Swift classes and structures are much closer in functionality than in
other languages, and much of this chapter describes functionality that
can apply to instances of either a class or a structure type. Because
of this, the more general term instance is used.
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edited Oct 23, 2020 at 2:41
Smart Manoj
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answered May 21, 2010 at 20:52
Kevin Le - KhnleKevin Le - Khnle
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What about languages that do not have classes but do have structs? Could I say "an instance of a struct" or would that be intrinsically wrong? How should I call it in that case? Thanks!
sagitta
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sagitta
06/21/2021 13:50:33
Commented
Jun 21, 2021 at 13:50
It wouldn't necessarily be wrong, but it would sound awkward, like using British English terms like "lorry", "lift", and "chips" in the US instead of "truck", "elevator", and "fries".
chepner
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chepner
07/29/2022 12:52:30
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Jul 29, 2022 at 12:52
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Excellent question.
I'll explain it in the simplest way possible:
Say you have 5 apples in your basket. Each of those apples is an object of type Apple, which has some characteristics (i.e. big, round, grows on trees).
In programming terms, you can have a class called Apple, which has variables size:big, shape:round, habitat:grows on trees. To have 5 apples in your basket, you need to instantiate 5 apples. Apple apple1, Apple apple2, Apple apple3 etc....
Alternatively: Objects are the definitions of something, instances are the physical things.
Does this make sense?
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answered May 21, 2010 at 20:51
Yuval KarmiYuval Karmi
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It doesn't. :) "Objects are the definitions of something" What you call objects here are classes.
T-Gergely
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T-Gergely
02/10/2016 09:50:35
Commented
Feb 10, 2016 at 9:50
The analogy was fine up to the "Alternative" line. (Although I'd argue it's the tree that instantiates Apple to produce individual apples, not the person with the basket.)
chepner
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chepner
07/29/2022 12:54:31
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Instance: instance means just creating a reference(copy).
object: means when memory location is associated with the object (is a run-time entity of the class) by using the new operator.
In simple words, Instance refers to the copy of the object at a particular time whereas object refers to the memory address of the class.
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edited Dec 16, 2013 at 16:52
community wiki
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saying that an instance is a reference to an object really cleared things up for me.
John C
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John C
05/31/2017 22:36:21
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May 31, 2017 at 22:36
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@JohnC It shouldn't, because it isn't correct. An object is an instance of a class. That's all there is to it.
user207421
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user207421
12/04/2019 07:16:16
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Dec 4, 2019 at 7:16
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Object:
It is a generice term basically it is a Software bundle that has state(variables) and behaviour(methods)
Class:
A blue print(template) for an object
instance-it's a unique object thing for example you create a object two times what does that mean is yo have created two instances
Let me give an example
```
Class student()
{
private string firstName;
public student(string fname)
{
firstName=fname;
}
Public string GetFirstName()
{
return firstName;
}
}
```
Object example:
Student s1=new student("Martin");
Student s2=new student("Kumar");
The s1,s2 are having object of class Student
Instance:
s1 and s2 are instances of object student
the two are unique.
it can be called as reference also.
basically the s1 and s2 are variables that are assigned an object
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edited Aug 14, 2014 at 13:29
Shajo
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answered Mar 22, 2013 at 15:33
Durai Amuthan.HDurai Amuthan.H
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An object is an instance of a class.
user207421
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user207421
12/04/2019 07:17:15
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Objects and instances are mostly same; but there is a very small difference.
If Car is a class, 3 Cars are 3 different objects. All of these objects are instances. So these 3 cars are objects from instances of the Car class.
But the word "instance" can mean "structure instance" also. But object is only for classes.
All of the objects are instances.
Not all of the instances must be objects. Instances may be "structure instances" or "objects".
I hope this makes the difference clear to you.
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edited May 29, 2020 at 4:56
PG328
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answered Mar 1, 2016 at 10:00
HarshaHarsha
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Let's say you're building some chairs.
The diagram that shows how to build a chair and put it together corresponds to a software class.
Let's say you build five chairs according to the pattern in that diagram. Likewise, you could construct five software objects according to the pattern in a class.
Each chair has a unique number burned into the bottom of the seat to identify each specific chair. Chair 3 is one instance of a chair pattern. Likewise, memory location 3 can contain one instance of a software pattern.
So, an instance (chair 3) is a single unique, specific manifestation of a chair pattern.
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edited Dec 5, 2019 at 21:52
answered May 21, 2010 at 21:03
Brian ShowalterBrian Showalter
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An object is an instance of a class. An instance isn't an instance of an object. Answer is meaningless.
user207421
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user207421
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Wow, after nine years someone finally pointed out the flaw in my explanation. Thanks! Better now?
Brian Showalter
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Brian Showalter
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Quick and Simple Answer
Class : a specification, blueprint for an object...
Object : physical presence of the class in memory...
Instance : a unique copy of the object (same structure, different data)...
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edited Oct 17, 2022 at 6:34
Gaai
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Syed Ekram UddinSyed Ekram Uddin
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An object is a construct, something static that has certain features and traits, such as properties and methods, it can be anything (a string, a usercontrol, etc)
An instance is a unique copy of that object that you can use and do things with.
Imagine a product like a computer.
THE xw6400 workstation is an object
YOUR xw6400 workstation, (or YOUR WIFE's xw6400 workstation) is an instance of the xw6400 workstation object
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edited Nov 6, 2016 at 11:40
Luís Cruz
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Greg OlmsteadGreg Olmstead
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Your first and second paragraphs describe classes, not objects.
user207421
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user207421
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Java is an object-oriented programming language (OOP). This means, that everything in Java, except of the primitive types is an object.
Now, Java objects are similar to real-world objects. For example we can create a car object in Java, which will have properties like current speed and color; and behavior like: accelerate and park.
That's Object.
Instance, on the other side, is a uniquely initialized copy of that object that looks like Car car = new Car().
Check it out to learn more about Java classes and object
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answered Apr 14, 2019 at 15:49
JohnnyJohnny
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I can't believe, except for one guy no one has used the code to explain this, let me give it a shot too!
```
// Design Class
class HumanClass {
var name:String
init(name:String) {
self.name = name
}
}
var humanClassObject1 = HumanClass(name: "Rehan")
```
Now the left side i.e: "humanClassObject1" is the object and the right side i.e: HumanClass(name: "Rehan") is the instance of this object.
```
var humanClassObject2 = HumanClass(name: "Ahmad") // again object on left and it's instance on the right.
```
So basically, instance contains the specific values for that object and objects contains the memory location (at run-time).
Remember the famous statement "object reference not set to an instance of an object", this means that non-initialised objects don't have any instance.
In some programming languages like swift the compiler will not allow you to even design a class that don't have any way to initialise all it's members (variable eg: name, age e.t.c), but in some language you are allowed to do this:
```
// Design Class
class HumanClass {
var name:String // See we don't have any way to initialise name property.
}
```
And the error will only be shown at run time when you try to do something like this:
```
var myClass = HumanClass()
print(myClass.name) // will give, object reference not set to an instance of the object.
```
This error indicates that, the specific values (for variables\property) is the "INSTANCE" as i tried to explain this above!
And the object i.e: "myClass" contains the memory location (at run-time).
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edited Nov 30, 2019 at 22:36
answered Nov 30, 2019 at 22:12
Rehan Ali KhanRehan Ali Khan
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This answer may be seen as trite, but worrying about the differences between an instance and object is already trite city.
I think its best depicted in javascript:
```
let obj= {"poo":1}
// "obj" is an object
```
verses
```
Class Trash {
constructor(){this.poo = 1;}
}
let i = new Trash();
// "i" is an instance
```
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answered Dec 12, 2019 at 18:01
heugheug
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When a variable is declared of a custom type (class), only a reference is created, which is called an object. At this stage, no memory is allocated to this object. It acts just as a pointer (to the location where the object will be stored in future). This process is called 'Declaration'.
```
Employee e; // e is an object
```
On the other hand, when a variable of custom type is declared using the new operator, which allocates memory in heap to this object and returns the reference to the allocated memory. This object which is now termed as instance. This process is called 'Instantiation'.
```
Employee e = new Employee(); // e is an instance
```
However, in some languages such as Java, an object is equivalent to an instance, as evident from the line written in Oracle's documentation on Java:
Note: The phrase "instantiating a class" means the same thing as "creating an object." When you create an object, you are creating an "instance" of a class, therefore "instantiating" a class.
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edited May 8, 2021 at 11:08
answered Mar 21, 2020 at 13:47
Srishti GuptaSrishti Gupta
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This is just an example that looks intuitive, but wrong. In Employee e, e points to nowhere and does not exist. While in theory object is some real entity that exists out of blue print (Class).
Manish Kumar
–
Manish Kumar
05/08/2021 09:54:47
Commented
May 8, 2021 at 9:54
@ManishSoni What you are speaking of is an 'instance' unless you are speaking in context of languages which consider an 'object' and an 'instance' as synonyms! That's what the entire answer just explains.
Srishti Gupta
–
Srishti Gupta
05/08/2021 11:08:07
Commented
May 8, 2021 at 11:08
The example here is misleading, it is synonymous with Java syntax where this does not make sense. However, your answer conveys the same meaning (conceptually right) this is what I told 'it looks intuitive but wrong'.
Manish Kumar
–
Manish Kumar
05/08/2021 14:09:36
Commented
May 8, 2021 at 14:09
@ManishSoni The example is not synonymous to only Java. If you look at other languages (C++, C#, JavaScript, etc.), many of them use the new operator and classes. Don't look at the example keeping Java or any language's syntax in mind. The example is language-agnostic.
Srishti Gupta
–
Srishti Gupta
05/09/2021 08:04:41
Commented
May 9, 2021 at 8:04
"When a variable is declared of a custom type (class), only a reference is created, which is called an object." :- Do you think this statement is right? reference is called object? Your elaboration for the answer made this debatable. There is one concise straight answer. why complicate and confuse this?
Manish Kumar
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Manish Kumar
05/10/2021 04:36:22
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May 10, 2021 at 4:36
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There are 3 things you need to understand : Class , Object and Instance.
Class : Class is the blueprint code from which you will create an Object(s)
Object : When memory is allocated to the data entity (created from blueprint class) , that data entity or reference to it is called Object
Instance : When data is filled in an Object , it becomes an instance of that Object. It can also be called a state of that Object.
Example : In context with C# (objects are reference type here)
Lets say we have a class like this (This is your blueprint code)
```
public class Animal
{
//some fields and methods
}
```
We create an object like this
```
Animal a = new Animal();
Animal b = a;
Animal c = a;
Animal d = b;
```
So here is the question : How many objects and instances are here ?
Answer : There is only 1 object but 4 instances.
Why ?
In first line (Animal a = new Animal();),we created an Object from class Animal with new Operator. That Object is somewhere on your RAM. And the reference to that Object is in "a".
We have 1 object and 1 instance at this time.
Now in next line, we assign b with a. Here Object is not copied but the reference of object from "a" is stored in "b" too. Thus , we have 2 instances , "a and b".
This goes on and we only copy reference of same object located at some memory.
Finally , we have 4 instances "a,b,c,d" of a single object that was created with new Operator.
(Read how reference type works in C# for more. I hope you understand my language)
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edited May 14, 2021 at 15:19
answered May 14, 2021 at 14:38
machinemanmachineman
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Once you instantiate a class (using new), that instantiated thing becomes an object. An object is something that can adhere to encapsulation, polymorphism, and abstraction principles of object oriented programming and the real thing a program interacts with to consume the instance members defined in the class. The object contains instance members (non-static members).
Thus, an instance of a class is an object. The word ‘instance’ is used when you are referring to the origin from where it was born, it's clearer if you say ‘instance of a class’ compared to ‘object of a class’ (although the latter can be used too).
Can also read the 'Inner classes' section of this java document on nested classes -
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edited May 31, 2023 at 4:54
Ahmad Adibzad
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Praveen TiwariPraveen Tiwari
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An instance is a specific representation of an object. An object is a generic thing while an instance is a single object that has been created in memory. Usually an instance will have values assigned to it's properties that differentiates it from other instances of the type of object.
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answered May 21, 2010 at 20:42
TLiebeTLiebe
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Object - An instance of a class that has its own state and access to all of the behaviour defined by its class.
Instance - Reference to an memory area for that particular class.
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answered Oct 20, 2014 at 11:10
prathipati sprathipati s
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Class : A class is a blue print.
Object : It is the copy of the class.
Instance : Its a variable which is used to hold memory address of the object.
A very basic analytical example
Class House --> Blueprint of the house. But you can't live in the blue print. You need a physical House which is the instance of the class to live in. i.e., actual address of the object is instance. Instances represent objects.
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answered Jan 18, 2017 at 19:41
SindhuSindhu
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Instance refers to the object, not to a variable.
user207421
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user207421
12/04/2019 07:19:17
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Dec 4, 2019 at 7:19
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If we see the Definition of Object and Instance object -
Memory allocated for the member of the class at run time is called object or the object is the instance of the Class.
Let us see the Definition of instance -
Memory allocated For Any at run time is called an instance variable.
Now understand the meaning of any runtime memory allocation happen in C also through Malloc, Calloc, Realloc such:
```
struct p
{
}
p t1
t1=(p) malloc(sizeof(p))
```
So here also we are allocating runtime memory allocation, but here we call as an instance so t1 is an instance here we can not say t1 as an object, so Every object is the instance of Class, but every Instance is not an Object.
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edited Jun 4, 2023 at 9:05
Ahmad Adibzad
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answered Sep 10, 2014 at 10:37
Vivek KumarVivek Kumar
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There's a correct answer lurking in here somewhere but the language needs cleaning up and simplifying.
user207421
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user207421
12/04/2019 07:18:38
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Dec 4, 2019 at 7:18
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I think if we consider other approaches than OOP (mainly by assuming the term Class hasn't always been used, as it's the case for many C projects, which still applied the concept of Objects), following definitions would make the most sense:
A Class defines an interface that objects adhere to.
An Object is an aggregate of different fields. (It doesn't have to "physically" exist, but it can).
All Objects of the same Class can be used in the same way, defined by the Class.
An Instance is a unique realization of an Object.
As many OOP languages use static typing, the Object description is usually part of the Class already. As such, when talking about an Object in C/C++, what usually is meant is the Instance of an Object.
In languages that do not have static typing (such as JavaScript), Objects can have different fields, while still sharing the same Class.
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answered Dec 15, 2021 at 12:35
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each object said to be an instance of its class, but each instance of the class has its own value for each attribute
instances share the attribute name and operation with their instances of class but an object contains an implicit reference to its own class
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edited May 31, 2023 at 2:20
Ahmad Adibzad
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I can't believe this could be hard to explain, but it is actually easier than all the answers I read. It is just simple like this.
Firstly, you need to understand the definition:
Instance is a unique copy-product of an Object.
unique - have different characteristics but share the same class compare to object
Object is a name that has been used to keep the Class information (i.e
method)
Let's say, there is a toy_1 as an object.
There is also toy_2 as an object ----> which is ALSO an INSTANCE to toy_1.
At the same time, toy_1 is also an INSTANCE to toy_2. (remember again INSTANCE is a COPY-PRODUCT)
That is why most of the answers I found said it is INTERCHANGABLE. Thank you.
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edited Jun 1, 2023 at 5:49
Ahmad Adibzad
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Regarding the difference between an object and an instance, I do not think there is any consensus.
It looks to me like people change it pretty much interchangeably, in papers, blog posts, books or conversations.
As for me, the way I see it is, an object is a generic and alive entity in the memory, specified by the language it is used in. Just like the Object class in Java. We do not much care its type, or anything else associated with it, whether it is managed by a container or not.
An instance is an object but associated with a type, as in this method accepts Foo instances, or you can not put Animal instances in an instance of
a List of Vehicles.
objects for example have locks associated with them, not instances, whereas instances have methods. objects are garbage collected, not instances.
But as I said, this is only how I see it, and I do not think there is any organisation we can refer to for a standard definition between them and everyone will pretty much have their slightly different understanding / definitions (of course within limits).
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answered Jan 26, 2019 at 15:47
Koray TugayKoray Tugay
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An object can be a class, say you have a class called basketball.
but you want to have multiple basketballs so in your code you create more than 1 basketball
say basketball1 and basketball2.
Then you run your application.
You now have 2 instances of the object basketball.
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answered May 21, 2010 at 20:42
HexBlitHexBlit
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That would mean a Object is a Instance of a Class Constructor ?
streetparade
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streetparade
05/21/2010 20:47:35
Commented
May 21, 2010 at 20:47
think of the object like a blueprint. say we have a blueprint for the basketball, that is the class. when the basketball is created and made it now exists, so that would mean we have 1 instance of the object basketball. if we built another basketball from the object(blueprint). we now have 2 instances of the basketball. There is always just 1 object, but we can make many instances of that object.
HexBlit
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HexBlit
05/21/2010 21:08:53
Commented
May 21, 2010 at 21:08
@Darxval then what is a class
Chyanit Singh
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Chyanit Singh
12/30/2019 18:36:00
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An object is a generic thing, for example, take a linear function in maths
ax+b is an object, While 3x+2 is an instance of that object
Object<<< Instance
General<<< Specific
There is nothing more to this
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edited May 29, 2019 at 20:33
answered May 29, 2019 at 15:23
Chyanit SinghChyanit Singh
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17488 | https://arxiv.org/pdf/2002.01988 | arXiv:2002.01988v2 [math.CO] 12 Feb 2020
Lozenge Tiling Function Ratios for Hexagons with Dents on Two Sides
Daniel Condon February 13, 2020
Abstract
We give a formula for the number of lozenge tilings of a hexagon on the triangular lattice with unit triangles removed from arbitrary positions along two non-adjacent, non-opposite sides. Our formula implies that for certain families of such regions, the ratios of their numbers of tilings are given by simple product formulas.
1 Introduction
The triangular lattice is a tiling of the plane by unit equilateral triangles, and a region on the triangular lattice is a connected union of finitely many of those triangles 1. We say triangles that share an edge are adjacent . A lozenge on the triangular lattice is the union of two adjacent triangles, and a lozenge tiling or simply tiling of a region is a set of lozenges within that region, which cover the region, and which do not overlap. A region with at least one tiling is called
tileable . A hexagonal region is said to be semiregular if its opposite sides are the same length. We identify congruent regions and let Ha,b,c denote the semiregular hexagon(al region) with opposite sides of lengths a, b, and c.A formula for the number of lozenge tilings of a semiregular hexagon was first given by MacMahon in the context of plane partitions. A bijection between plane partitions and lozenge tilings was later given by David and Tomei . Letting M( R) denote the number of lozenge tilings of a region R, MacMahon’s formula can be expressed M( Ha,b,c ) =
c
∏
i=1
(a + i)b
(i)b
=: P (a, b, c ) (1) where ( x)y is the pochhammer symbol, ( x)y = ∏y−1
i=0
(x+ i). When Rx denotes a family of regions defined by parameters x (as with {Ha,b,c : a, b, c ∈ N}) we say
1
When we say ‘triangle’ in this paper, we always mean a unit triangle on the lattice. (a) (b) Figure 1: (a) H3,4,2; (b) a tiling of H3,4,2
that M( Rx) denotes the tiling function of Rx with parameters x. So P (a, b, c )is the tiling function for semiregular hexagons. Hexagons in the triangular lattice need not be semiregular, but the lengths of opposite sides necessarily differ by the same amount, say t. When t > 0 we say the longer of each opposite pair is a long side and the rest are short sides. We can therefore define hexagonal regions by the lengths of their short sides
a, b, c , and the difference t, using the notation Ha,b,c,t . The sides of the hexagon alternate around the perimeter between short and long, so without a loss of generality we will assume that the side lengths of the hexagon appear in the clockwise order a, b + t, c, a + t, b, c + t starting from the north side. The hexagon in Figure 2 (a) is H4,3,2,4. Semiregular hexagons have t = 0, so Ha,b,c = Ha,b,c, 0.
The region H4,3,2,4 has no lozenge tilings at all. A lozenge covers exactly one up-pointing triangle and one down-pointing triangle, so that a region is tileable if it contains the same number of triangles with each orientation: we call such a region balanced . The hexagon Ha,b,c,t contains an excess of t triangles with one orientation, so semiregular hexagons are the only hexagons which are balanced. The orientation in excess is that along the long sides of the hexagon (by our conventions, these are up-pointing triangles). In this paper we study regions we call dented hexagons , obtained by removing unit triangles along 2 two long sides of a hexagon. We suppose m triangles are removed from the northeast side, and n triangles are removed from the northwest side. For balanced regions, this implies t = n+m. Figure 2 (b) gives one example of such a region. The black triangles have been removed from the region; we call these dents . The grey areas are part of the dented hexagon but are covered by forced lozenges , which are described at the end of Section 2. Given non-negative integers a, b, c, t suppose ~u = ( ui)mi=1 and ~v = ( vj )nj=1
are vectors of integers with 1 ≤ ui < u i+1 ≤ b + t for 1 ≤ i < m , and 1 ≤
2When we say a unit triangle is along a side of the the hexagon, we mean it shares an edge with the border of that side. When we say we a triangle is removed from a side, we mean that the removed triangle is along that side.
2a
c+tb+t
c
b
a+t
(a)
u1
u2
v1
v2
(b) Figure 2: (a) H4,3,2,4; (b) H4,3,2,4,(1 ,4) ,(3 ,4) with forced lozenges shaded grey. For this region u1 = 5 , u 2 = 3, v1 = v2 = 2
vj < v j+1 ≤ c + t for 1 ≤ j < n , so that a > 0 or u1 > 1 or v1 > 1. We use Ha,b,c,t,~ u,~ v to denote the dented hexagon formed by removing dents from
Ha,b,c,t at locations indexed by ~u and ~v. Specifically, we remove each uith unit triangle from the northeast side of Ha,b,c,t and each vj th unit triangle from the northwest side of Ha,b,c,t . The indexing in both cases starts at the northmost triangle along each side. It is straightforward to check that Ha,b,c,t,~ u,~ v with parameters as described is well-defined and unique. Other natural parameters of dented hexagons we will use are ui := b + n +
i − ui and vj := c + m + j − vj . The parameter ui counts the number of triangles along the northeast side which are below the ith dent along that side but have not been removed from the region: these are the up-pointing triangles immediately southeast of that dent, as shown in Figure 2 (b). The parameter vj
analogously counts up-pointing triangles directly southwest of the dent indexed by vj .The main result of this paper is a remarkably simple product formula for the ratio between the tiling functions of Ha,b,c,t,~ u,~ v and H0,b,c,t,~ u,~ v so long as the latter region is well defined. It follows that the first family of regions has a ‘nice’ tiling function when the latter family does, and we identify some instances of this. This is not the first paper to identify families of regions with nice ratios of tiling functions; similar results were recently found by Lai, Ciucu, Rohatgi, and Byun , , , , , . One specific subfamily of dented hexagons was studied by Lai, who found a lovely tiling function . A general formula for the tiling function of a hexagonal region with dents in arbitrary positions along its border has been discovered by Ciucu and Fischer , but that formula is given only as the determinant of a matrix. 32 Further Background
A region on the triangular lattice can be identified with a graph, so that each unit triangle corresponds to a vertex and adjacent triangles correspond to adja-cent vertices. This graph is bipartite, with its vertex bipartition defined by the orientations of the triangles, since triangles are only adjacent to triangles of the opposite orientation. A lozenge tiling of a region naturally partitions the trian-gles of that region into adjacent pairs, and so a lozenge tiling can be identified with a perfect matching on the graph of that region. We will therefore borrow two graph theoretic techniques. The following is a direct application of the Graph Splitting Lemma, which appears as Lemma 3.6 in a 2014 paper by Lai , and is implicit in earlier work by Ciucu .
Lemma 1 (Region Splitting Lemma) . Let R be a balanced region of the trian-gular lattice with a partition into regions P and Q such that unit triangles in P
which are adjacent to unit triangles in Q are all of the same orientation, and that this orientation is not in excess within P . Then M( R) = M( P ) · M( Q), and in particular M( R) = 0 if either P or Q is not balanced.
We will also use Kuo’s graphical condensation method. The following is a special case of his Theorem 5.4 from , expressed in the language of this paper.
Lemma 2. Let R be a simply connected region of the triangular lattice, and let
α, β, γ, δ be unit triangles R of the same orientation which touch the boundary of R at a corner or edge, so that the places they touch the boundary appear in the cyclic order α, β, γ, δ . Then
M( R − α − β) · M( R − γ − δ)= M( R − α − δ) · M( R − β − γ) − M( R − α − γ) · M( R − β − δ) (2)
The lozenge tilings of a region on the triangular lattice often have a natural interpretation as families of nonintersecting lattice paths with fixed start and end points; this relationship is shown in Figure 4. We will make use of the follow-ing classical result originally due to Lindstr¨ om , and independently to Gessel and Viennot . The formulation we use could be considered a restatement of Lemma 14 from .
Proposition 3. Let S = {(si, t i) : i ∈ [n]} and E = {(pj , q j ) : j ∈ [n]} be sets of coordinates on the square lattice Z2, such that every Up-Right lattice path from
(si, t i) to (pj , q j ) intersects any Up-Right lattice path from (sj , t j ) to (pi, q i) for
i 6 = j. The number of families of n non-intersecting Up-Right lattice paths which each start at a point in S and end at a point in E is given by det(( ai,j )ni,j =1 ),where ai,j = (pj +qj −si −ti
qj−ti
) is the number of Up-Right lattice paths from (si, t i)
to (pj , q j ).
4When a lozenge is in all possible tilings of a region, we say that lozenge is
forced . Often, a region contains a unit triangle which is adjacent to only one other triangle - the lozenge covering those two triangles is forced. One can remove the triangles covered by a forced lozenge to produce a new region, the tilings of which are in a natural bijection with the tilings of the original region; the bijection is defined by removing or replacing the forced lozenge. The new region may itself have a triangle only adjacent to one other triangle, which means it too is covered by a lozenge that is forced in the new region, and therefore in the original region. Removing triangles covered by forced lozenges may thus yield a sequence of regions that all have the same number of tilings. Figure 2 (b) serves as an example. The forced lozenges are shown and colored grey.
3 Main Results
Our first remark about dented hexagons is a characterization of when they have any tilings at all.
Proposition 4. Let a balanced dented hexagon H be given. Let LN be the Nth horizontal lattice line south of H’s northern side 3, and μN be the number of dents north of LN . Then H has a tiling iff μN ≤ N for all N ∈ N.Proof. Let ~u and ~v and a balanced dented hexagon H = Ha,b,c,t,~ u,~ v be given. Suppose for some N , μN > N ; assume that N is minimal such that this is the case. It is straightforward to check the region above LN is unbalanced. By Lemma 1, applied to the regions above and below LN , the dented hexagon has no tilings. If μN ≤ N for each N , we can exhibit a partition of the overall region into tileable regions, proving the existence of a tiling for the entire region. We induct on the number of dents. If the region has no dents, it is a balanced semiregular hexagon and has at least one tiling. Suppose Proposition 4 holds for dented hexagons with fewer than t dents. Suppose WLOG that um ≥ vn; in other words the southmost dent is on the eastern side of the region. We sketch the semiregular hexagon
H1,u m,c directly underneath the southmost dent, so that its northern side is the border of that dent. Figure 3 (a) depicts this hexagon (dark). We also sketch a parallelogram consisting of all triangles directly west of the southwest edge of the dark hexagon. This parallelogram is also depicted (light) in Figure 3 (a). Both the hexagon and parallelogram are tileable, and the rest of the region is tileable by the induction hypothesis; so the entire region is tileable. The fact that μt−1 ≤ t− 1 implies that um ≥ t, and um ≤ b. This guarantees the northeast and southwest sides of the dark hexagon are of length ≤ b, so that the light region west of the dark hexagon is actually a parallelogram. Had it been the case that vn > u m, we would simply flip the picture.
3LNis then
√3
2Nunits south of H’s northern side.
5(a) (b) Figure 3: H5,5,3,4,(5 ,6) ,(2 ,5) (a) partitioned into three tileable regions; (b) parti-tioned into many regions. Our application of the inductive hypothesis in the proof implies that one could continue subdividing the entire region into parallelograms and semiregular hexagons. Such a subdivision is depicted in Figure 3 (b). Proposition 4 shows that untileable dented hexagons are characterized by having too many dents too far north, which yields the following consequence.
Corollary 5. Let H = Ha,b,c,n +m, (ui)m
1,(vj)n
1
and H′ = Ha′,b ′ ,c ′,n +m, (u′
i)m
1,(v′
j)n
1
be dented hexagons with ui ≤ u′
i
and vj ≤ v′
j
for each i, j . If H has a tiling, so does H′.
Our main result is that families of dented hexagons with fixed parameters,
b, c, t, ~ u, ~ v have a tiling function given by the following rational function in a (in fact, a polynomial in a):
Theorem 6. Where Ha,b,c,t, (ui)m
1,(vj)n
1
is a tileable dented hexagon (implying
t = n + m),
M( Ha,b,c,t,~ u,~ v) = M( H0,b,c,t,~ u,~ v)
m
∏
i=1
(ui)ui
n
∏
j=1
(vj )vj
× P (a, b + n, c + m)
∏mi=1 (a + ui)ui
∏nj=1 (a + vj )vj
.
When H0,b,c,t,~ u,~ v has a nice tiling function, Theorem 6 shows that Ha,b,c,t,~ u,~ v
also has a nice tiling function. In particular, this explains a result found by Lai (Theorem 3.1 from , with q = 1), that when the dents along each side of the region are adjacent and vn = um the tiling function for the region is relatively simple. We discuss this and related results in section 6. 6(a) (b) Figure 4: Tilings of H3,4,2,5,(3 ,6) ,(2 ,5,6) correspond to lattice paths which start on edges along the southwest borders of the region (including along the southwest side of the hexagon and also along the western dents) and end at edges along the northeast borders, the steps of which are directly east or northeast. The correspondence is as follows: given a tiling, place dots in the middle of these edges as exhibited in (a). Then, connect each dot by a path to the middle of the opposite edge of the lozenge containing it, and then to the middle of the opposite edge of the next lozenge, and so on until the dots are connected by paths. Warping the picture as in (b) transforms the paths into non-intersecting Up-Right lattice paths. This process is reversible.
4 Groundwork
In this section we lay the technical foundation for the arguments that follow.
Lemma 7. If H0,b,c,t,~ u,~ v is a tileable dented hexagon, then M( Ha,b,c,t,~ u,~ v) is a polynomial in a when fixing b, c, t and the vectors ~u and ~v.Proof. The tilings of H0,b,c,t, (ui)m
1,(vj)n
1
are in bijection with families of ( b + n)non-intersecting lattice paths which start along the southwest boundaries of dented hexagon and end at the northeast boundaries. This bijection is explained briefly in Figure 4. Using this type of bijection is a standard technique when enumerating lozenge tilings, and similar examples of the technique explained in greater detail can be found in . The “start points” along the southwest side of the region can be interpreted as having coordinates {(−i, i ) : i ∈ [b]}, and the start points from the dents would then have coordinates {(−b, b + c + t + 1 − vi) : i ∈ [n]}. The “end points” then have coordinates {(a− b− 1 + j, b +c+ t+1 −j) : j ∈ [b+ t]− { ui : i ∈ [m]}} .It is clear these satisfy the path-intersection condition of Proposition 3 when each set is labeled with indices increasing from north to south. Applying Proposition 3, the number of lattice paths with these start and 7end points is given by the determinant of a matrix, the entries of which are of the form (a−1+ vi
vi−j
) and ( a+c+tb+c+t+1 −j−i
). Since each entry in the matrix is a polynomial on a, so is the determinant.
The arguments that follow are largely about the proportionality of polyno-mials. For p, q polynomials in a with we will use the non-standard notation
p(a) ≡
a
q(a)to mean there is a nonzero constant c with c · p(a) = q(a). We will sometimes use this notation where p, q are functions over many parameters which are poly-nomials over a when all other parameters are fixed, in which case c may depend on parameters other than a. It is easy to check ≡
a
is an equivalence relation and satisfies the following properties, where p1, p 2, and q are polynomials in a,and q is not identically 0.
p1(a) ≡
a
p2(a) ⇒ p1(a) + p2(a) ≡
a
p1(a) ≡
a
p2(a) (3)
p1(a) ≡
a
p2(a) ⇐⇒ p1(a)q(a) ≡
a
p2(a)q(a) (4) We will make use of the following technical lemma.
Lemma 8. For d ∈ Z and y ∈ Z+ and z ∈ N,
P (a + d, y − 1, z + 1) ≡
a
P (a + d, y, z ) (a + d + z + 1) y−1
(a + d + y)z
. (5)
Proof. It is straightforward to show from the definition of P (x, y, z ) that P (x,y,z +1)
P(x,y,z )
=
(x+z+1) y
(z+1) y
; in turn it is straightforward to show
P (x, y − 1, z + 1) = P (x, y, z ) (y)z (x + z + 1) y−1
(x + y)z (z + 1) y−1
. (6) Equation (5) follows from equation (6) where x = a + d.
5 Proof of Theorem 6
For b, c ∈ N and ~u = ( ui)m
1
, ~ v = ( vj )n
1
vectors over N so that there exists a well-defined region H0,b,c,m +n,~ u,~ v, define
fb,c,~ u,~ v(a) := P (a, b + n, c + m)
∏mi=1 (a + ui)ui
∏nj=1 (a + vj )vj
.
The main method of our proof is to show that M( Ha,b,c,m +n,~ u,~ v) interpreted as a function in a is proportional to f . We will do this by showing f is a polynomial 8(a) (b) Figure 5: H4,3,2,3,(1 ,6) ,(3 ,4) (a) the top row is forced since u1 = 1; (b) the south-east side is forced since um = 0. in a which can be defined recursively and obeys the same recursion as the tiling function. Observe that the equation in Theorem 6 can be expressed M( Ha,b,c,m +n,~ u,~ v) ≡
a
fb,c,~ u,~ v(a).
It will sometimes be convenient for us to assume that u1 > 1 or v1 > 1. A dented hexagon with u1 = v1 = 1 has no tilings by Proposition 4, in which case Theorem 6 holds by vacuously. In the case that u1 = 1, and v1 > 1 or n = 0, the top row of triangles of the region Ha,b,c,t,~ u,~ v are covered by forced lozenges, and may be removed from the region. Figure 5 (a) depicts an example of this. Removing the forced lozenges gives the region Ha+1 ,b,c,t −1,(ui−1) mi=2 ,(vj −1) nj=1 ,which thus has the same tiling function as Ha,b,c,t,~ u,~ v. We should like to show that our definition of fb,c,~ u,~ v respects this and the analogous case with v1 = 1 .
If um = 0 (or vn = 0), then the southeast (or southwest) side of the region is covered by forced lozenges. If um = 0, then Ha,b,c,t,~ u,~ v thus has the same tiling function as Ha,b,c +1 ,t −1,(ui)m−1
i=1 ,~ v
, which is the region that is obtained by removing the forced tiles. This is depicted in Figure 5 (b). We wish to show our definition of fb,c,~ u,~ v respects this fact, and the analogous fact when vn = 0.
Lemma 9. If Ha,b,c,t, (ui)m
1,(vj)n
1
is tileable then a. If u1 = 1 , and v1 > 1 or n = 0 , then fb,c,~ u,~ v(a) ≡
a
fb,c, (ui−1) m
2,(vj−1) n
1
(a+1) .
b. If v1 = 1 , and u1 > 1 or m = 0 , then fb,c,~ u,~ v(a) ≡
a
fb,c, (ui−1) m
1,(vj−1) n
2
(a+1) .
c. If um = 0 then fb,c,~ u,~ v(a) = fb,c +1 ,(ui)m−1
i=1 ,(vj)n
1
(a).
d. If vn = 0 then fb,c,~ u,~ v(a) = fb+1 ,c, (ui)mi=1 ,(vj )n−11
(a).
Proof. These identities are straightforward to check when written explicitly.
Let ~∅ denote the empty vector. We will now verify that Theorem 6 holds when ~u = ~∅. The case where ~v = ~∅ follows by symmetry. 9Lemma 10. The dented hexagon Ha,b,c,n, ~∅,(vj )n
1
is tileable and
M( Ha,b,c,n, ~∅,(vj )n
1
) ≡
a
fb,c, ~∅,~ v(a) := P (a, b + n, c )
∏nj=1 (a + vj )vj
. (7) This is equivalent to a result by Cohn, Larsen, and Propp , but the equiva-lence is not obvious so we will prove this result directly using Kuo Condensation.
Proof. Dented hexagons with dents on just one side are tileable by Proposition 4. We will induct on n + c, with base cases at n = 0 and c = 0. Note that if
c = 0 then the region has a unique tiling 4 and fb, 0,~∅,~ v(a) := P (a,b +n, 0) ∏nj=1 (a+vj )0 = 1 . In the case n = 0, equation (7) reduces to MacMahon’s formula (1). For our inductive hypothesis, assume that (7) holds for dented hexagons
Ha,b ′,c ′,n ′,~∅,(v′
j)n−11
with c′ +n′ < c +n. Consider dented hexagons Ha,b,c,n, ~∅,(vj )n
1
.In the case v1 = 1, Ha,b,c,n, ~∅,(vj )n
1
has forced lozenges along its northern side so has the same tiling function as Ha+1 ,b,c,n −1,~∅,(vj −1) n
2
, so M
(
Ha,b,c,n, ~∅,(vj )n
1
)
≡
a
M
(
Ha+1 ,b,c,n −1,~∅,(vj −1) n
2
)
(by IH) ≡
a
fb,c, ~∅,(vj −1) n
2
(a + 1) (by Lemma 9b) ≡
a
fb,c, ~∅,(vj )n
1
(a).
In the case vn = 0, Ha,b,c,n, ~∅,(vj )n
1
has forced lozenges along its southwest side so has the same tiling function as Ha,b +1 ,c,n −1,~∅,(vj )n−11
, so M
(
Ha,b,c,n, ~∅,(vj )n
1
)
≡
a
M
(
Ha,b +1 ,c,n −1,~∅,(vj )n−11
)
(by IH) ≡
a
fb,c +1 ,~∅,(vj )n−11
(a)(by Lemma 9d) ≡
a
fb,c, ~∅,(vj )n
1
(a).
We therefore assume that n > 0, v1 > 1, and vn > 0. Regard Ha,b,c,n, ~∅,~ v
as a subregion of the (unbalanced) region Ra = Ha,b,c −1,n +1 ,~∅,(vj +1) n−1
j=1
. Within each region Ra, let α be the unit triangle indexed by vn, let β be the northmost triangle on the northwest side, let γ be the southmost triangle along the north-east side, and let δ be the southmost triangle along the northwest side. These placements are depicted in Figure 6. We will apply Kuo Condensation (Lemma 2) to Ra, α, β, γ, δ . Figure 7 de-picts each region referenced in the condensation formula, in some cases with forced lozenges shaded (we regard these as not being part of the region). Note they are all balanced dented hexagons. Since these regions can be regarded as dented hexagons with dents on only one side, each is tileable by Proposition 4. The parameter c + n is strictly minimal on Ra − α − γ, so we may apply the inductive hypothesis to each of the other regions.
4
The entire northwest side is comprised of dents, and the unique tiling extends to the unique tiling of the a × b parallelogram.
10 αβγδ
(a) (b) Figure 6: (a) H3,5,4,3,~∅,(2 ,3,5) ; (b) R3 with α, β, γ, δ labeled, and the border of
H3,5,4,3,~∅,(2 ,3,5) depicted with a thick line. Lemma 2 can be expressed
M (R−α−γ)M (R−β−δ) = M (R−α−δ)M (R−β−γ)−M (R−α−β)M (R−γ−δ).
Removing forced lozenges as depicted in Figure 7, we can rewrite this:
M
(
Ha,b,c,n, ~∅,(vj )n
1
)
M
(
Ha+1 ,b +1 ,c −1,n −1,~∅,(vj −1) n−11
)
= M
(
Ha,b +1 ,c −1,n, ~∅,(vj )n
1
)
M
(
Ha+1 ,b,c,n −1,~∅,(vj −1) n−11
)
− M
(
Ha+1 ,b,c −1,n, ~∅,(vj −1) n
1
)
M
(
Ha,b +1 ,c,n −1,~∅,(vj )n−11
)
(8) Applying the inductive hypothesis to each region except Ha,b,c,n, ~∅,(vj )n
1
, this implies
M
(
Ha,b,c,n, ~∅,(vj )n
1
) P (a + 1 , b + n, c − 1)
∏n−1
j=1
(a + vj )vj
≡
a
P (a, b + n + 1 , c − 1)
∏nj=1 (a + vj )vj −1
· P (a + 1 , b + n − 1, c )
∏n−1
j=1
(a + vj )vj +1
− P (a + 1 , b + n, c − 1)
∏nj=1 (a + vj )vj
· P (a, b + n, c )
∏n−1
j=1
(a + vj )vj
(9) We shall show that when the term M (Ha,b,c,n, ~∅,(vj )n
1
) is replaced with fb,c, ~∅,(vj )n
1
(a), the products on each line of (9) are ≡
a
-equivalent, so that
M (Ha,b,c,n, ~∅,(vj )n
1
) ≡
a
fb,c, ~∅,(vj )n
1
(a)11 Ra − α − γαβγδ
α
γ
Ha,b,c,n, ~∅,(vj )n
1
Ra − β − δαβγδ
β
δ
Ha+1 ,b +1 ,c −1,n −1,~∅,(vj −1) n−11
Ra − α − δαβγδ
α
δ
Ha,b +1 ,c −1,n, ~∅,(vj )n
1
Ra − β − γαβγδ
β
γ
Ha+1 ,b,c,n −1,~∅,(vj −1) n−11
Ra − α − βαβγδ
α
β
Ha+1 ,b,c −1,n, ~∅,(vj −1) n
1
Ra − γ − δαβγδ
γ
δ
Ha,b +1 ,c,n −1,~∅,(vj )n−11
Figure 7: Ra with two of α, β, γ, δ removed and forced lozenges shaded. 12 by relations (3) and (4). It is clear that the product on the first line would be equal to the product on the third line when fb,c, ~∅,(vj )n
1
(a) is written out explicitly, so it remains to show the products on the last two lines are ≡
a
-equivalent. We will manipulate the terms on the second line to show this. We rewrite
n
∏
j=1
(a + vj )vj −1 =
n
∏
j=1
(a + vj )vj /(a + c)n (10)
n−1
∏
j=1
(a + vj )vj +1 = ( a + c + 1) n−1
n−1
∏
j=1
(a + vj )vj . (11) We can therefore rewrite:
P (a, b + n + 1 , c − 1)
∏nj=1 (a + vj )vj −1
· P (a + 1 , b + n − 1, c )
∏n−1
j=1
(a + vj )vj +1
(by eqns. (10), (11)) = P (a, b + n + 1 , c − 1)
∏nj=1 (a + vj )vj
· P (a + 1 , b + n − 1, c )
∏n−1
j=1
(a + vj )vj
· (a + c)n
(a + c + 1) n−1
(by eqn. (5)) ≡
a
P (a, b + n, c )
∏nj=1 (a + vj )vj
· (a + b + n + 1) c−1
(a + c)b+n
× P (a + 1 , b + n, c − 1)
∏n−1
j=1
(a + vj )vj
· (a + 1 + c)b+n−1
(a + b + n + 1) c−1
× (a + c)n
(a + c + 1) n−1
= P (a + 1 , b + n, c − 1)
∏nj=1 (a + vj )vj
· P (a, b + n + 1 , c )
∏n−1
j=1
(a + vj )vj
,
which is exactly the product on the third line of equation (9). This completes the proof.
Note that H0,b,c,n, ~∅,(vj )n
1
has forced lozenges at its northern tip that when removed leave a region of the form H1,b,c +1 −v1,n −1,~∅,(vj −v1 )n
2
, which is also a dented hexagon with fewer dents which are still all along the northwest side. Through repeated application of Lemma 10, one can obtain a complete product formula for M( Ha,b,c,n, ~∅,~ v); in particular, when b = c = 0, this may be regarded as an independent proof for the number of tilings of a trapezoid with an arbi-trary number of dents along the long base, which Cohn, Larsen, and Propp 13 attribute to Gelfand and Tsetlin . Our main theorem can be interpreted as a generalization of that result. We are now ready to prove Theorem 6. The proof is similar to that of Lemma 10, and uses that lemma as a base case.
Proof of Theorem 6. We shall show by induction that if Ha,b,c,t,~ u,~ v is tileable then M( Ha,b,c,t,~ u,~ v) ≡
a
fb,c,~ u,~ v(a) := P (a, b + n, c + m)
∏mi=1 (a + ui)ui
∏nj=1 (a + vj )vj
. (12) We will induct on the number of dents m + n, using m = 0 and n = 0 as base cases. In these cases, equation (12) follows immediately from Lemma 10. For our inductive hypothesis, suppose equation (12) holds for dented hexagons with fewer than m + n dents. Consider a dented hexagon with dents indexed by ( ui)m
1
, (vj )n
1
.In the case where um = 0, Ha,b,c,t, (ui)m
1,(vj)n
1
has forced lozenges along its southeast side and therefore has the same tiling function as Ha,b,c +1 ,t −1,(ui)m−11 ,(vj )n
1
,so M( Ha,b,c,t, (ui)m
1,(vj)n
1
) ≡
a
M( Ha,b,c +1 ,t −1,(ui)m−11 ,(vj )n
1
)(IH) ≡
a
fb,c +1 ,(ui)m−11 ,(vj )n
1
(a)(Lemma 9c) ≡
a
fb,c, (ui)m
1,(vj)n
1
(a).
Equation (12) holds when vn = 0 by symmetric reasoning. Assume therefore that m, n > 0, um > 0 and vn > 0. Regard Ha,b,c,t, (ui)m
1,(vj)n
1
as a subregion of the (unbalanced) region Ra := Ha,b,c,t, (ui)m−11 ,(vj )n−11
. Within
Ra let α denote the unit triangle indexed by vn, let β denote the triangle in-dexed by um, let γ denote the southmost triangle along the northeast side of Ra,and let δ denote the southmost triangle along the northwest side of Ra. Since
m, n > 0 and um > 0 and vn > 0 these locations are clearly defined within Ra
and are distinct, as depicted in Figure 8. We will apply Kuo Condensation (Lemma 2) to Ra, α, β, γ, δ . Figure 9 de-picts each region referenced in the condensation formula, in some cases with forced lozenges shaded (we regard these as not being part of the region). Note they are all families of balanced dented hexagons. Since Ha,b,c,t,~ u,~ v is tileable so are each of these regions by Corollary 5. The number of dents is strictly minimal on Ra −α−β, so we may apply the inductive hypothesis to each of the other regions. Recall lemma 2 states
M (R−α−β)M (R−β−δ) = M (R−α−δ)M (R−β−γ)−M (R−α−γ)M (R−β−δ).
Removing forced lozenges as depicted in Figure 9, we can rewrite this:
M (Ha,b,c,t, (ui)m
1,(vj)n
1
) M
(
Ha,b +1 ,c +1 ,t −2,(ui)m−11 ,(vj )n−11
)
= M
(
Ha,b +1 ,c,t −1,(ui)m−11 ,(vj )n
1
)
M
(
Ha,b,c +1 ,t −1,(ui)m
1,(vj)n−11
)
− M
(
Ha,b,c +1 ,t −1,(ui)m−11 ,(vj )n
1
)
M
(
Ha,b +1 ,c,t −1,(ui)m
1,(vj)n−11
)
(13) 14 α β
γδ
(a)
α βγδ
(b) Figure 8: (a) H3,4,2,5,(3 ,6) ,(2 ,5,6) ; (b) R3 with α, β, γ, δ labeled. Applying the inductive hypothesis to each region except Ha,b,c,t, (ui)m
1,(vj)n
1
, this implies
M (Ha,b,c,t, (ui)m
1,(vj)n
1
) P (a, b + n, c + m)
∏m−1
i=1
(a + ui)ui
∏n−1
j=1
(a + vj )vj
≡
a
P (a, b + n + 1 , c + m − 1)
∏m−1
i=1
(a + ui)ui+1
∏nj=1 (a + vj )vj −1
P (a, b + n − 1, c + m + 1)
∏mi=1 (a + ui)ui −1
∏n−1
j=1
(a + vj )vj +1
− P (a, b + n, c + m)
∏m−1
i=1
(a + ui)ui
∏nj=1 (a + vj )vj
P (a, b + n, c + m)
∏mi=1 (a + ui)ui
∏n−1
j=1
(a + vj )vj
(14) We shall show that when the term M (Ha,b,c,t, (ui)m
1,(vj)n
1
) is replaced with
fb,c, (ui)m
1,(vj)n
1
(a), the products on each line of (14) are ≡
a
-equivalent, so that
M (Ha,b,c,n, (ui)m
1,(vj)n
1
) ≡
a
fb,c, (ui)m
1,(vj)n
1
(a)by relations (3) and (4). It is clear that the product on the first line would be equal to the product on the third line when fb,c, ~∅,(vj )n
1
(a) is written out explicitly, so it remains to show the products on the last two lines are ≡
a
-equivalent. We will manipulate the terms on the second line to show this, employing the following analogues to equations (10) and (11):
k
∏
j=1
(a + vj )vj −1 =
k
∏
j=1
(a + vj )vj /(a + c + m)k (15)
k
∏
i=1
(a + ui)ui−1 =
k
∏
i=1
(a + uj )ui /(a + b + n)k (16) 15 Ra − α − β
α β
γδ
Ha,b,c,t, (ui)m
1
,(vj )n
1
Ra − γ − δ
α β
γ
δ
Ha,b +1 ,c +1 ,t −2,(ui)m−11 ,(vj )n−11
Ra − α − δ
α βγ
δ
Ha,b +1 ,c,t −1,(ui)m−11 ,(vj )n
1
Ra − β − γ
α β
γ
δ
Ha,b,c +1 ,t −1,(ui)m
1
,(vj )n−11
Ra − α − γ
α β
γ
δ
Ha,b,c +1 ,t −1,(ui)m−11 ,(vj )n
1
Ra − β − δ
α β
γ
δ
Ha,b +1 ,c,t −1,(ui)m
1
,(vj )n−11
Figure 9: Ra with two of α, β, γ, δ removed and forced lozenges shaded. 16 k∏
j=1
(a + vj )vj +1 = ( a + c + m + 1) kk∏
j=1
(a + vj )vj (17)
k
∏
i=1
(a + ui)ui−1 = ( a + b + n + 1) kk∏
i=1
(a + uj )ui . (18) We can therefore rewrite:
P (a, b + n + 1 , c + m − 1)
∏m−1
i=1
(a + ui)ui +1
∏nj=1 (a + vj )vj −1
· P (a, b + n − 1, c + m + 1)
∏mi=1 (a + ui)ui−1
∏n−1
j=1
(a + vj )vj +1
(by eqns. (15) - (18)) = P (a, b + n + 1 , c + m − 1)
∏m−1
i=1
(a + ui)ui
∏nj=1 (a + vj )vj
· (a + c + m)n
(a + b + n + 1) m−1
× P (a, b + n − 1, c + m + 1)
∏mi=1 (a + ui)ui
∏n−1
j=1
(a + vj )vj
· (a + b + n)m
(a + c + m + 1) n−1
(by Lemma 8) ≡
a
P (a, b + n, c + m)
∏m−1
i=1
(a + ui)ui
∏nj=1 (a + vj )vj
· (a + b + n + 1) c+m−1
(a + c + m)b+n
· (a + c + m)n
(a + b + n + 1) m−1
× P (a, b + n, c + m)
∏mi=1 (a + ui)ui
∏n−1
j=1
(a + vj )vj
· (a + c + m + 1) b+n−1
(a + b + n)c+m
· (a + b + n)m
(a + c + m + 1) n−1
= P (a, b + n, c + m)
∏m−1
i=1
(a + ui)ui
∏nj=1 (a + vj )vj
· P (a, b + n, c + m)
∏mi=1 (a + ui)ui
∏n−1
j=1
(a + vj )vj
,
which is exactly the product from third line from (14). This is the last case in the inductive step, so equation (12) holds in general.
6 Hexagons with Two Large Dents
When the dents along each side of a dented hexagon are all adjacent, as in
Ha,b,c,t, (u+i)mi=1 ,(v+j)nj=1 , the region has forced lozenges that form large triangular dents. Figure 10 depicts this forcing, and the region that results if the forced lozenges and omitted dents are removed entirely. The original goal of this paper was finding a general tiling function for hexagons with two large dents, and indeed Theorem 6 simplifies to a ratio of tilings of semiregular hexagons when the dents along each side are all adjacent:
Corollary 11. A dented hexagon Ha,b,c,t, (u+i)mi=1 ,(v+j)nj=1 has tilings exactly if
17 u
m
b+n−u
v
n
c+m−v
(a) (b) Figure 10: (a) H5,4,2,5,(4+ i)21,(3+ j)31 ; (b) the region with forced lozenges removed
u ≥ n or v ≥ m, in which case
M
(
Ha,b,c,t, (u+i)mi=1 ,(v+j)nj=1
)
= M
(
H0,b,c,t, (u+i)mi=1 ,(v+j)nj=1
)
× P (a, b + n, c + m)P (u, b + n − u, m )P (v, c + m − v, n )
P (a + u, b + n − u, m )P (a + v, c + m − v, n ) .
This generalizes a specific case of a result by Lai who studied the problem when the southern borders of the dents are level (see Theorem 3.1 from , with q = 1). We give an expression of that result below in the language of this paper; it follows from Corollary 11.
Corollary 12. Given a dented hexagon Ha,b,c,t, (u+i)mi=1 ,(v+j)nj=1 with u + m =
v + n, let D := u − n. If D < 0 the region has no tilings. Otherwise,
M
(
Ha,b,c,t, (u+i)mi=1 ,(v+j)nj=1
)
= P (a, b + n, c + m)P (u, b − D, m )P (v, c − D, n )
P (a + u, b − D, m )P (a + v, c − D, n )
× P (c − D, n + m, b )P (D, n, m )
P (c − D + n, m, D ) · P (D, m, b − D).
Proof of Corollary 11 . It is straightforward to check for arbitrary values that
m > v, n > u ⇐⇒ v < u + m ≤ v + n, and m + ( u + m − v) > u + m
OR u < v + n ≤ u + m, and n + ( v + n − u) > v + n.
(19) We shall show the second set of expressions hold exactly when the region
Ha,b,c,t, (u+i)m
1,(v+j)n
1
has no tilings. Suppose that u + m ≤ v + n (meaning the southern border of the eastern dent is weakly north of the southern border of the western dent) and the region has no tilings. Then ( μN − N ) is maximized at N = u + m: so μu+m > u + m.If v > u + m then μu+m = m ≤ u + m, giving a contradiction. So it must be 18 n
c−D
m
b−D
c
n+m
b
(a)
c−D
n
D
m
b−D
c
n+m
b
(b) Figure 11: (a) H0,6,4,5,(4+ i)31,(5+ j)21 with forced lozenges; (b) the region inter-preted as a hexagon with one dent ( HD,m,b −D shaded red). that v ≤ u + m. This implies μu+m = m + ( u + m − v) > u + m, so that the first line on the left side of equation (19) holds. The argument works in reverse: if v < u + m ≤ v + n and ( u + m) − v > u then μu+m > u + m; so these three inequalities are equivalent to the eastern dent being weakly north of the western dent and the region having no tilings. The inequalities u < v + n ≤ u + m and n + ( v + n − u) > v + n are equivalent to the western dent being weakly north of the eastern dent and the region having no tilings. The formula given follows immediately from Theorem 6.
Proof of Corollary 12 . The region H0,b,c,t, (u+i)mi=1 ,(v+j)nj=1 has forced lozenges that when removed give a region congruent to Hc−D,n +m,b −D,D, (n+i)Di=1 ,~∅, as depicted in figure 11. The region H0,n +m,b −D,D, (n+i)Di=1 ,~∅ also has forced lozenges, that when re-moved give a semiregular hexagon HD,m,b −D. The result then follows by apply-ing Corollary 11 to both Ha,b,c,t, (u+i)mi=1 ,(v+j)nj=1 and Hc−D,n +m,b −D,D, (n+i)Di=1 ,~∅.
7 Final Remarks
The method of proof used for Corollary 12 is applicable whenever the tiling func-tion of H0,b,c,t,~ u,~ v is simple to express. For example, let H = H0,b,c,t, (u+i)m
1,(v+j)n
1
be a region with vn = 1, as depicted in Figure 12 (a). 19 n
1
m
b+n−u
c
m+n
b
(a) (b) Figure 12: (a) H0,b,c,~ u,~ v with vn = 1 and a split-line; (b) R3 with forced lozenges shaded grey. Observe that the blue split-line in the figure partitions the region into two unbalanced hexagons Hn,b, 0,1 and Hm−1,b +n−u,c −1,1. It can be seen by modify-ing the proof of the Region Splitting Lemma that if R is a balanced region with a partition into regions P , Q, so that unit triangles in P which are adjacent to unit triangles in Q are all of the same orientation, and this orientation is in excess within P by some amount, say d, then all tilings of R include exactly d
lozenges covering one triangle from P and one triangle from Q.It follows that all tilings of the entire region must include exactly one lozenge which crosses the split-line. Let Si be the tilings of the region so that the split-line-crossing lozenge’s northern border is i units north of H’s southern side. Let
Ri be the region obtained by removing that lozenge from H, and observe that
Si has a natural bijection with the tilings of Ri. Furthermore, Lemma 1 applies to each region Ri with respect to the blue split-line, and partitions Ri into two regions with known tiling functions: M( Ri) = M( H1,n,b +1 −i) M( Hm−1,b +n−u,c −1,1,(i),~∅)= P (m − 1, b + n − u, c )( b + n − u)!
n!( c − 1)!( b + m + n − u − 1)!
× (b + 2 − i)n(b + n + 2 − u − i)c−1(i)m−1
M( Ha,b,c,t, (u+i)m
1,(v+j)n
1
) = P (a, b + n, c + m)P (u, b + n − u, m )( v + n)!( a + v)!
P (a + u, b + n − u, m )P (a + v, 1, n )( v)!( a + v + n)!
×
b+n+1 −u
∑
i=1
M( Ri)20 1
c+m−v
m
b+ 1 −u
c
1 + m
b
(a) (b) Figure 13: (a) H0,b,c,m +1 ,~ u,~ v with n = 1 and a split-line; (b) R3 with forced lozenges shaded grey. A similar calculation could be made when 1 ≤ vn ≤ m, indexing over the positions of |vn| distinct lozenges which cross the split-line. We can apply this method to a different family of dented hexagons, with v
arbitrary, u < b + 1, and n = 1, employing a split-line which cuts southwest from the eastern dent, as depicted in Figure 13. Again, each tiling of this region has exactly one lozenge which crosses the split-line. If Ri is obtained by removing the lozenge in the ith position from the southwest end of the split-line, then tilings of H0,b,c,m +1 ,(u+i)m
1,(v+1)
are in bijection with the union of the tilings of {Ri}c+m−v+1
i=1
:M( Ri) = M( H1,u −1,c +m+1 −v−i) M( Hb−u,c,m, 1,(i),~∅)= P (b − u, c, m + 1) c!
(u − 1)!( b − u + c)! m!
× (c + m + 2 − v − i)u−1(c − i + 2) m(i)b−u
Ha,b,c,m +1 ,(u+i)m
1,(v+1)
= P (a, b + 1 , c + m)P (u, b + 1 − u, m )v!( c + m + a)! c!
P (a + u, b + 1 − u, m )( c + m)!( a + v)!
×
c+m−v+1
∑
i=1
M( Ri).
We note some dead ends. We have shown that the tiling function for the dented hexagon Ha,b,c,t,~ u,~ v may be given as a polynomial of entirely linear factors (of a when the other parameters are fixed). The same is not true for obvious other parameters of the region, such as c or b. Similarly when dents were placed along more than two sides of the hexagon, or along short sides of the hexagon, the tiling function of the region could not be interpreted as a polynomial of 21 linear factors over any obvious single parameter of the region.
Acknowledgments. The author would like to thank M. Ciucu for intro-ducing me to this subject, for his insights into this problem, and for his feedback on this paper.
References
S. H. Byun, Identities involving Schur functions and their applications to a shuffling theorem (2019), submitted, available at arXiv:1906.04533[math.CO] . M. Ciucu and I. Fischer, Lozenge Tilings of Hexagons with Arbitrary Dents , Adv. Appl. Math. 73 C (2016), 1-22. M. Ciucu, Enumeration of Perfect Matchings in Graphs with Reflective Symmetry , J. Combin. Theory Ser. A 77 (1997), no. 1, 67-97. M. Ciucu, T. Eisenk¨ obl, C. Krattenthaler, and D. Zare, Enumeration of Lozenge Tilings of Hexagons with a Central Triangular Hole , J. Combin. Theory Ser. A 95 (2001), 251-334. M. Ciucu, T. Lai, and R. Rohatgi, Tilings of hexagons with a removed triad of bowties
(2020), available at arXiv:1909.04070[math.CO] . M. Ciucu and T. Lai, Lozenge tilings of doubly-intruded hexagons , J. Combin. Theory Ser. A 167 (2019), 294-339. H. Cohn, M. Larsen, and J. Propp, The shape of a typical boxed plane partition , New York J. Math. 4 (1998), 137-165. G. David and C. Tomei, The Problem of the Calissons , Amer. Math. Monthly 96 (935)
(1989), 429-431. I. Gessel and X. Viennot, Binomial determinants, paths, and hook length formulae. , Adv. in Math. 58 (1985), no. 3, 300-321. E. Kuo, Applications of Graphical Condensation for Enumerating Matchings and Tilings ,Theoret. Comput. Sci. 319 (2004), no. 1-3, 29-57. T. Lai, Enumeration of Hybrid Domino-Lozenge Tilings , J. Combin. Theory Ser. A 122
(2014), 53–81. , A q-enumeration of lozenge tilings of a hexagon with three dents , Adv. in Appl. Math. 82 (2017), 23-57. , A Shuffling Theorem for Reflectively Symmetric Tilings (2019), submitted, avail-able at arXiv:1905.09268[math.CO] . , A Shuffling Theorem for Centrally Symmetric Tilings (2019), submitted, avail-able at arXiv:1906.03759[math.CO] . T. Lai and R. Rohatgi, A shuffling theorem for lozenge tilings of doubly-dented hexagons
(2019), submitted, available at arXiv:1905.08311[math.CO] . B. Lindstr¨ om, On the vector representations of induced matroids , Bull. Lond. Math. Soc.
5 (1973), 85-90. P. A. MacMahon, Combinatory Analysis , Vol. 2, Cambridge University Press, 1916; reprinted in P. A. MacMahon, Combinatory Analysis , Vol. 1-2, Chelsea, New York, 1960. M. Gelfand and M. L. Tsetlin, Finite-dimensional representations of the group of uni-modular matrices , Dokl. Akad. Nauk. 71 (1950), 825-828 (Russian); English transl., Izrail M. Gelfand: Collected Papers 2 (1988), 653–656.
22 |
17489 | https://bio.libretexts.org/Bookshelves/Introductory_and_General_Biology/General_Biology_(Boundless)/11%3A_Meiosis_and_Sexual_Reproduction/11.02%3A_The_Process_of_Meiosis_-_Meiosis_I | Skip to main content
11.2: The Process of Meiosis - Meiosis I
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: Nov 22, 2024
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11.1: The Process of Meiosis - Introduction to Meiosis
11.3: The Process of Meiosis - Meiosis II
Page ID
: 13249
Boundless
Boundless
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
Describe the stages and results of meiosis I
Meiosis I
Meiosis is preceded by an interphase consisting of three stages. The G1 phase (also called the first gap phase) initiates this stage and is focused on cell growth. The S phase is next, during which the DNA of the chromosomes is replicated. This replication produces two identical copies, called sister chromatids, that are held together at the centromere by cohesin proteins. The centrosomes, which are the structures that organize the microtubules of the meiotic spindle, also replicate. Finally, during the G2 phase (also called the second gap phase), the cell undergoes the final preparations for meiosis.
Prophase I
During prophase I, chromosomes condense and become visible inside the nucleus. As the nuclear envelope begins to break down, homologous chromosomes move closer together. The synaptonemal complex, a lattice of proteins between the homologous chromosomes, forms at specific locations, spreading to cover the entire length of the chromosomes. The tight pairing of the homologous chromosomes is called synapsis. In synapsis, the genes on the chromatids of the homologous chromosomes are aligned with each other. The synaptonemal complex also supports the exchange of chromosomal segments between non-sister homologous chromatids in a process called crossing over. The crossover events are the first source of genetic variation produced by meiosis. A single crossover event between homologous non-sister chromatids leads to an exchange of DNA between chromosomes. Following crossover, the synaptonemal complex breaks down and the cohesin connection between homologous pairs is also removed. At the end of prophase I, the pairs are held together only at the chiasmata; they are called tetrads because the four sister chromatids of each pair of homologous chromosomes are now visible.
Prometaphase I
The key event in prometaphase I is the formation of the spindle fiber apparatus where spindle fiber microtubules attach to the kinetochore proteins at the centromeres. Microtubules grow from centrosomes placed at opposite poles of the cell. The microtubules move toward the middle of the cell and attach to one of the two fused homologous chromosomes at the kinetochores. At the end of prometaphase I, each tetrad is attached to microtubules from both poles, with one homologous chromosome facing each pole. In addition, the nuclear membrane has broken down entirely.
Metaphase I
During metaphase I, the tetrads move to the metaphase plate with kinetochores facing opposite poles. The homologous pairs orient themselves randomly at the equator. This event is the second mechanism that introduces variation into the gametes or spores. In each cell that undergoes meiosis, the arrangement of the tetrads is different. The number of variations is dependent on the number of chromosomes making up a set. There are two possibilities for orientation at the metaphase plate. The possible number of alignments, therefore, equals 2n, where n is the number of chromosomes per set. Given these two mechanisms, it is highly unlikely that any two haploid cells resulting from meiosis will have the same genetic composition.
Anaphase I
In anaphase I, the microtubules pull the attached chromosomes apart. The sister chromatids remain tightly bound together at the centromere. The chiasmata are broken in anaphase I as the microtubules attached to the fused kinetochores pull the homologous chromosomes apart.
Telophase I and Cytokinesis
In telophase I, the separated chromosomes arrive at opposite poles. In some organisms, the chromosomes decondense and nuclear envelopes form around the chromatids in telophase I. Then cytokinesis, the physical separation of the cytoplasmic components into two daughter cells, occurs without reformation of the nuclei. In nearly all species of animals and some fungi, cytokinesis separates the cell contents via a cleavage furrow (constriction of the actin ring that leads to cytoplasmic division). In plants, a cell plate is formed during cell cytokinesis by Golgi vesicles fusing at the metaphase plate. This cell plate will ultimately lead to the formation of cell walls that separate the two daughter cells.
Two haploid cells are the end result of the first meiotic division. The cells are haploid because at each pole there is just one of each pair of the homologous chromosomes. Therefore, only one full set of the chromosomes is present. Although there is only one chromosome set, each homolog still consists of two sister chromatids.
Key Points
Meiosis is preceded by interphase which consists of the G1 phase (growth), the S phase ( DNA replication), and the G2 phase.
During prophase I, the homologous chromosomes condense and become visible as the x shape we know, pair up to form a tetrad, and exchange genetic material by crossing over.
During prometaphase I, microtubules attach at the chromosomes’ kinetochores and the nuclear envelope breaks down.
In metaphase I, the tetrads line themselves up at the metaphase plate and homologous pairs orient themselves randomly.
In anaphase I, centromeres break down and homologous chromosomes separate.
In telophase I, chromosomes move to opposite poles; during cytokinesis the cell separates into two haploid cells.
Key Terms
crossing over: the exchange of genetic material between homologous chromosomes that results in recombinant chromosomes
tetrad: two pairs of sister chromatids (a dyad pair) aligned in a certain way and often on the equatorial plane during the meiosis process
chromatid: either of the two strands of a chromosome that separate during meiosis
11.1: The Process of Meiosis - Introduction to Meiosis
11.3: The Process of Meiosis - Meiosis II |
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Principle of Inclusion-Exclusion
The Principle of Inclusion-Exclusion (abbreviated PIE) provides an organized method/formula to find the number of elements in the union of a given group of sets, the size of each set, and the size of all possible intersections among the sets.
Contents
[hide]
1 Important Note(!)
2 Application
2.1 Two Set Example
2.2 Three Set Example
2.3 Four Set Example
2.3.1 Problem
2.3.2 Solution
2.4 Five Set Example
2.4.1 Problem
2.4.2 Solution
3 Statement
4 Proof
5 Remarks
6 Examples
7 See also
Important Note(!)
When using PIE, one should understand how to strategically overcount and undercount, in the end making sure every element is counted once and only once. In particular, memorizing a formula for PIE is a bad idea for problem solving.
Application
Here, we will illustrate how PIE is applied with various numbers of sets.
Two Set Example
Assume we are given the sizes of two sets, and , and the size of their intersection, . We wish to find the size of their union, .
To find the union, we can add and . In doing so, we know we have counted everything in at least once. However, some things were counted twice. The elements that were counted twice are precisely those in . Thus, we have that:
.
Three Set Example
Assume we are given the sizes of three sets, and , the size of their pairwise intersections, , and , and the size their overall intersection, . We wish to find the size of their union, .
Just like in the Two Set Example, we start with the sum of the sizes of the individual sets . We have counted the elements which are in exactly one of the original three sets once, but we've obviously counted other things twice, and even other things thrice! To account for the elements that are in two of the three sets, we first subtract out . Now we have correctly accounted for them since we counted them twice originally, and just subtracted them out once. However, the elements that are in all three sets were originally counted three times and then subtracted out three times. We have to add back in . Putting this all together gives:
.
Four Set Example
Problem
Six people of different heights are getting in line to buy donuts. Compute the number of ways they can arrange themselves in line such that no three consecutive people are in increasing order of height, from front to back. (2015 ARML I10)
Solution
Let be the event that the first, second, and third people are in ordered height, be the event that the second, third, and fourth people are in ordered height, be the event that the third, fourth, and fifth people are in ordered height, and be the event that the fourth, fifth and sixth people are in ordered height. By a combination of complementary counting and PIE, we have that our answer will be . Now for the daunting task of evaluating all of this.
For , we just choose people and there is only one way to put them in order, then ways to order the other three guys for . Same goes for , , and .
Now, for , that's just putting four guys in order. By the same logic as above, this is . Again, would be putting five guys in order, so . is just choosing guys out of , then guys out of for . Now, is just the same as , so , is so , and is so .
Moving on to the next set: is the same as which is , is ordering everybody so , is again ordering everybody which is , and is the same as so .
Finally, is ordering everybody so .
Now, lets substitute everything back in. We get a massive expression of .
Five Set Example
Problem
There are five courses at my school. Students take the classes as follows: 243 take algebra. 323 take language arts. 143 take social studies. 241 take biology. 300 take history. 213 take algebra and language arts. 264 take algebra and social studies. 144 take algebra and biology. 121 take algebra and history. 111 take language arts and social studies. 90 take language arts and biology. 80 take language arts and history. 60 take social studies and biology. 70 take social studies and history. 60 take biology and history. 50 take algebra, language arts, and social studies. 50 take algebra, language arts, and biology. 50 take algebra, language arts, and history. 50 take algebra, social studies, and biology. 50 take algebra, social studies, and history. 50 take algebra, biology, and history. 50 take language arts, social studies, and biology. 50 take language arts, social studies, and history. 50 take language arts, biology, and history. 50 take social studies, biology, and history. 20 take algebra, language arts, social studies, and biology. 15 take algebra, language arts, social studies, and history. 15 take algebra, language arts, biology, and history. 10 take algebra, social studies, biology, and history. 10 take language arts, social studies, biology, and history. 5 take all five. None take none.
How many people are in my school?
Solution
Let A be the subset of students who take Algebra, L-languages, S-Social Studies, B-biology, H-history, M-the set of all students. We have:
Thus, there are people in my school.
Statement
If are finite sets, then:
.
Proof
We prove that each element is counted once.
Say that some element is in sets. Without loss of generality, these sets are
We proceed by induction. This is obvious for
If this is true for we prove this is true for For every set of sets not containing with size there is a set of sets containing with size In PIE, the sum of how many times these sets are counted is There is also one additional set of sets so is counted exactly once.
Remarks
Sometimes it is also useful to know that, if you take into account only the first sums on the right, then you will get an overestimate if is odd and an underestimate if is even. So,
,
,
,
and so on.
Examples
2011 AMC 8 Problems/Problem 6
2017 AMC 10B Problems/Problem 13
2005 AMC 12A Problems/Problem 18
2001 AIME II Problems/Problem 9
2002 AIME I Problems/Problem 1
2020 AIME II Problems/Problem 9
2001 AIME II Problems/Problem 2
2017 AIME II Problems/Problem 1
See also
Combinatorics
Overcounting
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17491 | https://www.itl.nist.gov/div898/handbook/eda/section3/eda358.htm | | | | |
---
| 1. Exploratory Data Analysis 1.3. EDA Techniques 1.3.5. Quantitative Techniques | | | --- | | 1.3.5.8. | Chi-Square Test for the Variance | |
| Purpose: Test if the variance is equal to a specified value | A chi-square test ( Snedecor and Cochran, 1983) can be used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value. The one-sided version only tests in one direction. The choice of a two-sided or one-sided test is determined by the problem. For example, if we are testing a new process, we may only be concerned if its variability is greater than the variability of the current process. |
| Definition | The chi-square hypothesis test is defined as: | | | --- | | H0: | ( \sigma^2 = \sigma_0^2 ) | | Ha: | | | | | --- | ( \sigma^2 < \sigma_0^2 ) | | for a lower one-tailed test | | ( \sigma^2 > \sigma_0^2 ) | | for an upper one-tailed test | | ( \sigma^2 \ne \sigma_0^2 ) | | for a two-tailed test | | | Test Statistic: | ( T = (N-1)(s/\sigma_0)^2 ) where N is the sample size and s is the sample standard deviation. The key element of this formula is the ratio s/σ0 which compares the ratio of the sample standard deviation to the target standard deviation. The more this ratio deviates from 1, the more likely we are to reject the null hypothesis. | | Significance Level: | α. | | Critical Region: | Reject the null hypothesis that the variance is a specified value, σ02, if | | | | --- | ( T > \chi^2_{1-\alpha, \, N-1} ) | | for an upper one-tailed alternative | | ( T < \chi^2_{\alpha, \, N-1} ) | | for a lower one-tailed alternative | | ( T < \chi^2_{\alpha/2, \, N-1} ) or ( T > \chi^2_{1-\alpha/2, \, N-1} ) | | for a two-tailed alternative | where ( \chi^2_{., \, N-1} ) is the critical value of the chi-square distribution with N - 1 degrees of freedom. | The formula for the hypothesis test can easily be converted to form an interval estimate for the variance: [ \sqrt{\frac{(N-1)s^2}{\chi^2_{1-\alpha/2, \, N-1}}} \le \sigma \le \sqrt{\frac{(N-1)s^2}{\chi^2_{\alpha/2, \, N-1}}} ] A confidence interval for the standard deviation is computed by taking the square root of the upper and lower limits of the confidence interval for the variance. |
| Chi-Square Test Example | A chi-square test was performed for the GEAR.DAT data set. The observed variance for the 100 measurements of gear diameter is 0.00003969 (the standard deviation is 0.0063). We will test the null hypothesis that the true variance is equal to 0.01. H0: σ2 = 0.01 Ha: σ2 ≠ 0.01 Test statistic: T = 0.3903 Degrees of freedom: N - 1 = 99 Significance level: α = 0.05 Critical values: Χ 2α/2,N-1 = 73.361 Χ 21-α/2,N-1 = 128.422 Critical region: Reject H0 if T < 73.361 or T > 128.422 The test statistic value of 0.3903 is much smaller than the lower critical value, so we reject the null hypothesis and conclude that the variance is not equal to 0.01. |
| Questions | The chi-square test can be used to answer the following questions: 1. Is the variance equal to some pre-determined threshold value? 2. Is the variance greater than some pre-determined threshold value? 3. Is the variance less than some pre-determined threshold value? |
| Related Techniques | F Test Bartlett Test Levene Test |
| Software | The chi-square test for the variance is available in many general purpose statistical software programs. Both Dataplot code and R code can be used to generate the analyses in this section. These scripts use the GEAR.DAT data file. | |
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Three Dimensions>How to Find Intersection in GeoGebra
How to Find Intersection in GeoGebra
You can quickly find intersections between lines, planes and spheres in GeoGebra using the following sets of instructions.
GeoGebra Instruction 1
1.
: Open 3D Graphics and Algebra View under View in Menu.
2.
: Add the two objects that you want to intersect by entering them into Algebra View. If one or both of your objects are lines, then they must be added by using one of the Line commands.
3.
: Enter
Intersect(
and fill in the names of the objects as they appear in `Algebra View`.
4.
: The intersection points or lines are now shown in Graphics View and Algebra View.
Example 1
Enter the equation of the sphere
| |
| |
in Algebra View. Create a line with the command Line(, ). Choose as the point on the line, and as the direction vector of the line. Then, use the command
Intersect(, )
to find where the line intersects the sphere.
Previous entry
How to Make a Sphere in GeoGebra
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17493 | https://oeis.org/A001481/internal | The OEIS is supported by the many generous donors to the OEIS Foundation.
Numbers that are the sum of 2 squares. (Formerly M0968 N0361)
%I M0968 N0361 #256 Aug 04 2025 10:13:57
%S 0,1,2,4,5,8,9,10,13,16,17,18,20,25,26,29,32,34,36,37,40,41,45,49,50,
%T 52,53,58,61,64,65,68,72,73,74,80,81,82,85,89,90,97,98,100,101,104,
%U 106,109,113,116,117,121,122,125,128,130,136,137,144,145,146,148,149,153,157,160
%N Numbers that are the sum of 2 squares.
%C Numbers n such that n = x^2 + y^2 has a solution in nonnegative integers x, y.
%C Closed under multiplication. - _David W. Wilson_, Dec 20 2004
%C Also, numbers whose cubes are the sum of 2 squares. - _Artur Jasinski_, Nov 21 2006 (Cf. A125110.)
%C Terms are the squares of smallest radii of circles covering (on a square grid) a number of points equal to the terms of A057961. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Apr 16 2007. [Comment corrected by _T. D. Noe_, Mar 28 2008]
%C Numbers with more 4k+1 divisors than 4k+3 divisors. If a(n) is a member of this sequence, then so too is any power of a(n). - _Ant King_, Oct 05 2010
%C A000161(a(n)) > 0; A070176(a(n)) = 0. - _Reinhard Zumkeller_, Feb 04 2012, Aug 16 2011
%C Numbers that are the norms of Gaussian integers. This sequence has unique factorization; the primitive elements are A055025. - _Franklin T. Adams-Watters_, Nov 25 2011
%C These are numbers n such that all of n's odd prime factors congruent to 3 modulo 4 occur to an even exponent (Fermat's two-squares theorem). - _Jean-Christophe Hervé_, May 01 2013
%C Let's say that an integer n divides a lattice if there exists a sublattice of index n. Example: 2, 4, 5 divide the square lattice. The present sequence without 0 is the sequence of divisors of the square lattice. Say that n is a "prime divisor" if the index-n sublattice is not contained in any other sublattice except the original lattice itself. Then A055025 (norms of Gaussian primes) gives the "prime divisors" of the square lattice. - _Jean-Christophe Hervé_, May 01 2013
%C For any i,j > 0 a(i)a(j) is a member of this sequence, since (a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2. - _Boris Putievskiy_, May 05 2013
%C The sequence is closed under multiplication. Primitive elements are in A055025. The sequence can be split into 3 multiplicatively closed subsequences: {0}, A004431 and A125853. - _Jean-Christophe Hervé_, Nov 17 2013
%C Generalizing Jasinski's comment, same as numbers whose odd powers are the sum of 2 squares, by Fermat's two-squares theorem. - _Jonathan Sondow_, Jan 24 2014
%C By the 4 squares theorem, every nonnegative integer can be expressed as the sum of two elements of this sequence. - _Franklin T. Adams-Watters_, Mar 28 2015
%C There are never more than 3 consecutive terms. Runs of 3 terms start at 0, 8, 16, 72, ... (A082982). - _Ivan Neretin_, Nov 09 2015
%C Conjecture: barring the 0+2, 0+4, 0+8, 0+16, ... sequence, the sum of 2 distinct terms in this sequence is never a power of 2. - _J. Lowell_, Jan 14 2022
%C All the areas of squares whose vertices have integer coordinates. - _Neeme Vaino_, Jun 14 2023
%C Numbers represented by the definite binary quadratic forms x^2 + 2nxy + (n^2+1)y^2 for any integer n. This sequence contains the even powers of any integer. An odd power of a number appears only if the number itself belongs to the sequence. The equation given in the comment by Boris Putievskiy 2013 is Brahmagupta's identity with n = 1. It proves that any set of numbers of the form a^2 + nb^2 is closed under multiplication. - _Klaus Purath_, Sep 06 2023
%D J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag, p. 106.
%D David A. Cox, "Primes of the Form x^2 + n y^2", Wiley, 1989.
%D L. Euler, (E388) Vollständige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 417.
%D S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 98-104.
%D G. H. Hardy, Ramanujan, pp. 60-63.
%D P. Moree and J. Cazaran, On a claim of Ramanujan in his first letter to Hardy, Expos. Math. 17 (1999), pp. 289-312.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H T. D. Noe, Table of n, a(n) for n = 1..10000
%H Michael Baake, Uwe Grimm, Dieter Joseph and Przemyslaw Repetowicz, Averaged shelling for quasicrystals, arXiv:math/9907156 [math.MG], 1999.
%H Chris Bispels, Matthew Cohen, Joshua Harrington, Joshua Lowrance, Kaelyn Pontes, Leif Schaumann, and Tony W. H. Wong, A further investigation on covering systems with odd moduli, arXiv:2507.16135 [math.NT], 2025. See p. 3.
%H Henry Bottomley, Illustration of initial terms
%H Richard T. Bumby, Sums of four squares, in Number theory (New York, 1991-1995), 1-8, Springer, New York, 1996.
%H John Butcher, Quadratic residues and sums of two squares
%H John Butcher, Sums of two squares revisited
%H Leonhard Euler, Vollständige Anleitung zur Algebra, Zweiter Teil; see also Deutsches Textarchiv
%H Steven R. Finch, Landau-Ramanujan Constant [broken link]
%H Steven R. Finch, Landau-Ramanujan Constant [From the Wayback Machine]
%H Steven R. Finch, On a Generalized Fermat-Wiles Equation [broken link]
%H Steven R. Finch, On a Generalized Fermat-Wiles Equation [From the Wayback Machine]
%H J. W. L. Glaisher, On the function which denotes the difference between the number of (4m+1)-divisors and the number of (4m+3)-divisors of a number, Proc. London Math. Soc., 15 (1884), 104-122. [Annotated scanned copy of pages 104-107 only]
%H Leonor Godinho, Nicholas Lindsay, and Silvia Sabatini, On a symplectic generalization of a Hirzebruch problem, arXiv:2403.00949 [math.SG], 2024. See p. 17.
%H Darij Grinberg, UMN Spring 2019 Math 4281 notes, University of Minnesota, College of Science & Engineering, 2019. [Wayback Machine copy]
%H Shuo Li, The characteristic sequence of the integers that are the sum of two squares is not morphic, arXiv:2404.08822 [math.NT], 2024.
%H Thomas Nickson and Igor Potapov, Broadcasting Automata and Patterns on Z^2, arXiv preprint arXiv:1410.0573 [cs.FL], 2014.
%H Michael Penn, Sums of Squares, Youtube playlist, 2019, 2020.
%H Peter Shiu, Counting Sums of Two Squares: The Meissel-Lehmer Method, Mathematics of Computation 47 (1986), 351-360.
%H N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)
%H William A. Stein, Quadratic Forms:Sums of Two Squares
%H Eric Weisstein's World of Mathematics, Square Number
%H Eric Weisstein's World of Mathematics, Generalized Fermat Equation
%H Eric Weisstein's World of Mathematics, Landau-Ramanujan Constant
%H Eric Weisstein's World of Mathematics, Gaussian Integer
%H A. van Wijngaarden, A table of partitions into two squares with an application to rational triangles, Proceedings of the Koninklijke Nederlandse Akademie van Wetenschappen, Series A, 53 (1950), 869-875.
%H Gang Xiao, Two squares
%H Index entries for sequences related to sums of squares
%H Index entries for "core" sequences
%F n = square 2^{0 or 1} {product of distinct primes == 1 (mod 4)}.
%F The number of integers less than N that are sums of two squares is asymptotic to constantN/sqrt(log(N)), hence lim_{n->infinity} a(n)/n = infinity.
%F Nonzero terms in expansion of Dirichlet series Product_p (1 - (Kronecker(m, p) + 1)p^(-s) + Kronecker(m, p)p^(-2s))^(-1) for m = -1.
%F a(n) ~ knsqrt(log n), where k = 1.3085... = 1/A064533. - _Charles R Greathouse IV_, Apr 16 2012
%F There are B(x) = x/sqrt(log x) (K + B2/log x + O(1/log^2 x)) terms of this sequence up to x, where K = A064533 and B2 = A227158. - _Charles R Greathouse IV_, Nov 18 2022
%p readlib(issqr): for n from 0 to 160 do for k from 0 to floor(sqrt(n)) do if issqr(n-k^2) then printf(%d,,n); break fi: od: od:
%t upTo = 160; With[{max = Ceiling[Sqrt[upTo]]}, Select[Union[Total /@ (Tuples[Range[0, max], {2}]^2)], # <= upTo &]] ( _Harvey P. Dale_, Apr 22 2011 )
%t Select[Range[0, 160], SquaresR[2, #] != 0 &] ( _Jean-François Alcover_, Jan 04 2013 )
%o (PARI) isA001481(n)=local(x,r);x=0;r=0;while(x<=sqrt(n) && r==0,if(issquare(n-x^2),r=1);x++);r \ _Michael B. Porter_, Oct 31 2009
%o (PARI) is(n)=my(f=factor(n));for(i=1,#f[,1],if(f[i,2]%2 && f[i,1]%4==3, return(0))); 1 \ _Charles R Greathouse IV_, Aug 24 2012
%o (PARI) B=bnfinit('z^2+1,1);
%o is(n)=#bnfisintnorm(B,n) \ _Ralf Stephan_, Oct 18 2013, edited by _M. F. Hasler_, Nov 21 2017
%o (PARI) list(lim)=my(v=List(),t); for(m=0,sqrtint(lim\=1), t=m^2; for(n=0, min(sqrtint(lim-t),m), listput(v,t+n^2))); Set(v) \ _Charles R Greathouse IV_, Jan 05 2016
%o (PARI) is_A001481(n)=!for(i=2-bittest(n,0),#n=factor(n)~, bittest(n[1,i],1)&&bittest(n[2,i],0)&&return) \ _M. F. Hasler_, Nov 20 2017
%o (Haskell)
%o a001481 n = a001481_list !! (n-1)
%o a001481_list = [x | x <- [0..], a000161 x > 0]
%o -- _Reinhard Zumkeller_, Feb 14 2012, Aug 16 2011
%o (Magma) [n: n in [0..160] | NormEquation(1, n) eq true]; // _Arkadiusz Wesolowski_, May 11 2016
%o (Python)
%o from itertools import count, islice
%o from sympy import factorint
%o def A001481_gen(): # generator of terms
%o return filter(lambda n:(lambda m:all(d & 3 != 3 or m[d] & 1 == 0 for d in m))(factorint(n)),count(0))
%o A001481_list = list(islice(A001481_gen(),30)) # _Chai Wah Wu_, Jun 27 2022
%Y Disjoint union of A000290 and A000415.
%Y Complement of A022544.
%Y Cf. A004018, A000161, A002654, A064533, A055025, A002828, A000378, A025284-A025320, A125110, A118882, A125022.
%Y A000404 gives another version. Subsequence of A091072, supersequence of A046711.
%Y Cf. A057961, A232499, A077773, A363763.
%Y Column k=2 of A336820.
%K nonn,nice,easy,core
%O 1,3
%A _N. J. A. Sloane_
%E Deleted an incorrect comment. - _N. J. A. Sloane_, Oct 03 2023
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17495 | https://www.youtube.com/watch?v=JPe0AyfUAaQ | Percentage of amounts using bar model
Mr Mahmud
37400 subscribers
117 likes
Description
2932 views
Posted: 26 Feb 2020
Percentage of amounts using bar model. In maths, a bar model can be defined as a pictorial representation of a number in the form of bars used to solve number problems such as percentage. Bar models help us understand how a problem needs to be solved and calculated. Due to it being pictorial, it helps students solve percentage of amounts problems without following any set steps.
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Transcript:
today we want to do percentage and I'm going to teach you one method for three different types of percentage questions so you can do percentage over males with it it can do percentage increase you can do percentage decrease with it and also you can do first percentage with it [Music] hello everyone this is mr. moody helping you do battery math by giving you tips and tricks on how to devise and going through a massive vision on YouTube so if you have been subscribed yet please do consider subscribing and don't forget to press the bell icon so that you get notified when I upload my next video now exams are coming up now I don't want to remind you of how many days are left till the exams start because it could be quite intimidating and also it could be quite daunting for some people who haven't started revising yet so if you're one of them if you know one of them I'm not going to tell you the exam days you probably know in the back of your head those dates okay it's around about May June let's keep you up to here I'm not going to make it any worse for you well anyway today we want to do percentage and I'm going to teach you one method for three different types of percentage questions so you can do percentage of emails with it it can be percentage increase you can do percentage decrease with it and also you can do first percentage with it so for different things pretty much okay so let's get my whiteboard and let's get started now with this method you probably already know about this method for those people maybe Corinne in primary or corner seven eight you seven eight you've probably seen this method choir regular so we call it a power method so you're going to use bars you know to solve all four different types of sentences so let's start off with a very simple question let's say we want to find out what 20 percent of 500 is so let's write down 20 percent of 500 now usually what would happen is you'll find ten percent by dividing by ten and then multiply the answer by 2 that will give you 20 percent or five hundred simple I know where our aim is to be able to answer more difficult questions on percentages and this one method will answer well kind of all four types of percentage questions so to start off with what we need to do is draw a bar okay now let's start off with zero now this is zero percent and this is zero let's say you call it pounds so that's zero pound so the here is of pounds over here we go percentages okay now 20 percent now what's the hundred percent now if you write wonder percent here now the hundred percent we know it is the full amount so we're going to write down five hundredths okay now think about some percentages that you can find on this bar so I'm going to point my whiteboard somewhere on the bar and I want to think about what would that centage be and what would the amount possibly be so let's say well here so I'll draw a line here so yes halfway try to be as lucky as possible so percentage-wise yes fifty percent exactly so that's 50 percent and half of 50 percent well half of the ml to 50 percent is 250 okay now I want to find twenty percent now I know you can probably do in this in your head you're thinking this is an easy question well the thing is it's not about the easy question it's about knowing the method that's why I chose an easy question so that you can relate I can know you know the answer then you will see how the thing actually works the method okay so that's 50 percent what are the percentage can you wear count or what the percentage do we need to find in order to get to say 20 percent let's say we each find we can find say 20 percent no 20 percent 10 percent let's say this is 10 percent so write down the 10 percent here so what do I need to do to 100 to get to 10 percent think about it we need to divide it by 10 yes so we divided 200 by 10 to get 10% therefore we need to divide the 500 by 10 as well so 500 divided by 10 is 50 okay now can you think of a number what do we need to do 200 percent to get to 20 percent so let's say 20 percent is our next percentage you need 2 divided by 5 yeah 5 so we need to divide this by 5 so 500 divided by 5 is 100 let's write it here it's gonna be a bit mixed opss with 50 were here for this 100 pounds okay or well you can do it can you double the 10% answer find out the 20% answer which is 100 okay now this was a really simple example I using this method you can find out what yes--we we said 20 percent let's say 75 percent okay now how would you find 75 percent now I want to change this to 75 percent now so we found 20 percent let's say 75 percent try make it a bit more difficult so half of 100 percent is 50 percent which is 250 we have 50 percent we get 25 percent so let's say that's 25 percent try to be as accurate as possible 3 good to have this again so half of 250 is 125 now we've got 25% and we go 50 percent everything KAC on the diagram the power you got 50 and 25 makes 75 cents we say we're gonna find 75 percent so now 75 percent and that will give us 250 out of 225 if you're not sure did it working pretty sure you can do this in your head 375 so answer is 375 pounds if you want to find what 80 percent is what do we do so let's say 80 percent round over here 80 percent and this will give us so we know what 75 percent is profound out what 5% is yes let's go here that's about five percent smooth this so five percent half of 10 percent is 5 percent 10 percent is 50 so how five percent is going to be 25 so 75 is 375 I add it to 5 percent to 75 percent which gives us 80 percents we need to add these two numbers so 7 3 and so 2 5 and 25 gives us 400 so using one method you can find us so many different percentages of just one amount okay now this video was about percentage of the mouse and the next video I'm going to show you how to do percentage increase and decrease using the same method and also the following video I'm going to do on verse percentage and that's the interesting part so make sure you keep following the percentage series videos and don't forget to subscribe and if you got any questions let me know in the comments now that's it for today good luck with revisions and see you in the next video mr. Mahmud helping you do better in maths what's happening everybody is missing my mood here helping you do better in math by giving you tips and tricks on how to devise and going through massive Asia and you too let's try again let's try a cup chair so that's what's happening everybody is missing I'm really helping you be a mess on YouTube no let's try this one more time this was was cool okay what's up everybody its mr. Mahmud helping you do better in math by Kim you tips and tricks on how to advise and going through a massive a vision on a YouTube helping you do better in maths by giving you tips and tricks on how to divine eyes on YouTube no I don't like this lesson hello I'm mr. Mahmud and this is my board pen and this is my white board today we are going to learn maths how to find percentage of amounts using bars alright let's try this again what's up and everybody is missing I'm in here helping you do but didn't mass by Kimmy you take some takes another advise and going through a massive June on YouTube did you see it move I think I was written was moving my table was moving let's try again what's up and everybody is mr. moody a helping you do between math might give you tips and rigs and revise and going through math vision I think I need to change my intro I think I'm a bit bored of it anyway let's do it the normal way okay let's get on with it now ever tried this thing my head spinning now |
17496 | https://www.statlect.com/mathematical-tools/k-permutations | StatLect
Index > Mathematical tools
k-permutations
by Marco Taboga, PhD
This lecture introduces -permutations, a basic concept in combinatorial analysis. Before reading this lecture, you should read the lecture on permutations.
We first deal with -permutations without repetition and then with -permutations with repetition.
Table of contents
k-permutation without repetition
Definition of k-permutation without repetition
Number of k-permutations without repetition
k-permutation with repetition
Definition of k-permutation with repetition
Number of k-permutations with repetition
Solved exercises
Exercise 1
Exercise 2
Exercise 3
k-permutation without repetition
A -permutation without repetition of objects is a way of selecting objects from a list of . The selection rules are:
the order of selection matters (the same objects selected in different orders are regarded as different -permutations);
each object can be selected only once.
A -permutation without repetition is also simply called -permutation.
The following subsections give a slightly more formal definition of -permutation and deal with the problem of counting the number of possible -permutations.
Definition of k-permutation without repetition
Let , ,..., be objects. Let , , ..., be () slots to which of the objects can be assigned. A -permutation (or -permutation without repetition or simple -permutation) of objects from , ,..., is one of the possible ways to choose of the objects and fill each of the slots with one and only one object. Each object can be chosen only once.
Example Consider three objects , and . There are two slots ( and ) to which we can assign two of the three objects. There are six possible -permutations of the three objects (six possible ways to choose two objects and fill the two slots with the two objects):
Number of k-permutations without repetition
Denote by the number of possible -permutations of objects. How much is in general? In other words, how do we count the number of possible -permutations of objects?
We can derive a general formula for by filling the slots in a sequential manner:
First, we assign an object to the first slot. There are objects that can be assigned to the first slot, so there are
Then, we assign an object to the second slot. There were objects, but one has already been assigned to a slot. So, we are left with objects that can be assigned to the second slot. Thus, there areand
Then, we assign an object to the third slot. There were objects, but two have already been assigned to a slot. So, we are left with objects that can be assigned to the third slot. Thus, there areand
An so on, until we are left with objects and only one free slot (the -th).
Finally, when only one free slot remains, we assign one of the remaining objects to it. Thus, there areand
Therefore, by the above sequential argument, the total number of possible -permutations of objects is
can be written as
Remembering the definition of factorial, we can see that the numerator of the above ratio is while the denominator is , so the number of possible -permutations of objects is
The number is usually indicated as follows:
Example The number of possible -permutations of objects is
k-permutation with repetition
A -permutation with repetition of objects is a way of selecting objects from a list of . The selection rules are:
the order of selection matters (the same objects selected in different orders are regarded as different -permutations);
each object can be selected more than once.
Thus, the difference between -permutations without repetition and -permutations with repetition is that objects can be selected more than once in the latter, while they can be selected only once in the former.
The following subsections give a slightly more formal definition of -permutation with repetition and deal with the problem of counting the number of possible -permutations with repetition.
Definition of k-permutation with repetition
Let , ,..., be objects. Let , , ..., be () slots to which of the objects can be assigned. A -permutation with repetition of objects from , ,..., is one of the possible ways to choose of the objects and fill each of the slots with one and only one object. Each object can be chosen more than once.
Example Consider three objects , and and two slots ( and ). There are nine possible -permutations with repetition of the three objects (nine possible ways to choose two objects and fill the two slots with the two objects, being allowed to pick the same object more than once):
Number of k-permutations with repetition
Denote by the number of possible -permutations with repetition of objects. How much is in general? In other words, how do we count the number of possible -permutations with repetition of objects?
We can derive a general formula for by filling the slots in a sequential manner:
First, we assign an object to the first slot. There are objects that can be assigned to the first slot, so there are
Then, we assign an object to the second slot. Even if one object has been assigned to a slot in the previous step, we can still choose among objects, because we are allowed to choose an object more than once. So, there are objects that can be assigned to the second slot andand
Then, we assign an object to the third slot. Even if two objects have been assigned to a slot in the previous two steps, we can still choose among objects, because we are allowed to choose an object more than once. So, there are objects that can be assigned to the second slot andand
An so on, until we are left with only one free slot (the -th).
When only one free slot remains, we assign one of the objects to it. Thus, there are:and
Therefore, by the above sequential argument, the total number of possible -permutations with repetition of objects is
Example The number of possible -permutations of objects is
Solved exercises
Below you can find some exercises with explained solutions.
Exercise 1
There is a basket of fruit containing an apple, a banana and an orange and there are five girls who want to eat one fruit. How many ways are there to give three of the five girls one fruit each and leave two of them without a fruit to eat?
Solution
Giving the 3 fruits to 3 of the 5 girls is a sequential problem. We first give the apple to one of the girls. There are 5 possible ways to do this. Then we give the banana to one of the remaining girls. There are 4 possible ways to do this, because one girl has already been given a fruit. Finally, we give the orange to one of the remaining girls. There are 3 possible ways to do this, because 2 girls have already been given a fruit. Summing up, the number of ways to assign the three fruits is equal to the number of 3-permutations of 5 objects (without repetition). If we denote it by , then
Exercise 2
An hexadecimal number is a number whose digits can take sixteen different values: either one of the ten numbers from 0 to 9, or one of the six letters from A to F. How many different 8-digit hexadecimal numbers are there, if an hexadecimal number is allowed to begin with any number of zeros?
Solution
Choosing the 8 digits of the hexadecimal number is a sequential problem. There are 16 possible ways to choose the first digit and 16 possible ways to choose the second digit. So, there are 16x16 possible ways to choose the first two digits. There are 16 possible ways two choose the third digit and 16x16 possible ways to choose the first two. Thus, there are 16x16x16 possible ways to choose the first three digits. An so on, until we have chosen all digits. Therefore, the number of ways to choose the 8 digits is equal to the number of 8-permutations with repetition of 16 objects:
Exercise 3
An urn contains ten balls, each representing one of the ten numbers from 0 to 9. Three balls are drawn at random from the urn and the corresponding numbers are written down to form a 3-digit number, writing down the digits from left to right in the order in which they have been extracted. When a ball is drawn from the urn it is set aside, so that it cannot be extracted again. If one were to write down all the 3-digit numbers that could possibly be formed, how many would they be?
Solution
The 3 balls are drawn sequentially. At the first draw there are 10 balls, hence 10 possible values for the first digit of our 3-digit number. At the second draw there are 9 balls left, hence 9 possible values for the second digit of our 3-digit number. At the third and last draw there are 8 balls left, hence 8 possible values for the third digit of our 3-digit number. In summary, the number of possible 3-digit numbers is equal to the number of 3-permutations of 10 objects (without repetition). If we denote it by , then
How to cite
Please cite as:
Taboga, Marco (2021). "k-permutations", Lectures on probability theory and mathematical statistics. Kindle Direct Publishing. Online appendix.
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17497 | https://www.connecticutchildrens.org/medical-professionals/clinical-pathways/preseptal-and-orbital-cellulitis | What Are You Looking For?
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Preseptal and Orbital Cellulitis
Pathway Background and Objectives
Orbital cellulitis is a relatively uncommon condition but carries significant morbidity, including orbital abscess, vision loss, cavernous sinus thrombophlebitis, and intracranial abscess. Orbital abscess is a surgical emergency. Though they have distinctly different pathogenesis, differentiating orbital cellulitis from preseptal (periorbital) cellulitis is often difficult, especially in younger children. Preseptal cellulitis is more common than orbital cellulitis and has less risk of serious complications. Preseptal cellulitis arises from a superficial infection of the skin and soft tissues in the periorbital region anterior to the orbital plate. The most common organisms implicated in preseptal cellulitis are Staph aureus, Strep pneumoniae, other Strep species, and anaerobes. In comparison, orbital cellulitis arises from sinusitis, most commonly from the ethmoid sinuses, with spread of infection and inflammation through the thin layers of the lamina papyracea. The most common causative agents of orbital cellulitis are Staph and Strep species, and less commonly anaerobes. Both orbital and preseptal cellulitis can cause periorbital edema, erythema, eye pain, and fever, and therefore can sometimes be difficult to distinguish clinically. Patients with orbital cellulitis are more likely to have pain with extra-ocular movements, restricted extra-ocular movements, diplopia, proptosis, and an absolute neutrophil counts (ANC) >10,000. If there is concern for orbital cellulitis, an urgent CT scan with IV contrast to evaluate for orbital abscess is necessary. If orbital cellulitis/abscess is suspected and/or identified, care coordination between multiple services including, Pediatric Hospital Medicine, ENT, and Ophthalmology is required. The objectives of this pathway are to:
Quickly identify patients with orbital cellulitis who may require surgery
Identify those patients who require a CT Scan
Improve coordination of the multiple subspecialists often involved in care of this group of patients
Standardize antibiotics for these infections
Algorithm Educational Module
Percentage of patients with pathway order set usage
Percentage of patients with ophthalmology consult
Percentage of patients who have a CT scan
Percentage of patients who have a CT scan with documented sxs of orbital cellulitis
Percentage of patients who require surgery
Percentage of patients with appropriate antibiotic choice per pathway recommendation
Percentage of patients with appropriate antibiotic duration per pathway recommendation
ED average length of stay (minutes)
Inpatient average length of stay (days)
Number of returns to ED within 48 hours
Number of readmission within 48 hours
• Botting AM, McIntosh D, Mahadevan M. Paediatric pre- and post-septal peri-orbital infections are different diseases. A retrospective review of 262 cases. Int J Pediatr Otorhinolaryngol. 2008 Mar;72(3):377-83.• Hamilton S, Taylor M, Schneider, JG, et al. Assessing the Diagnostic Performance and Clinical utility of Nasal Methicillin-Resistant Staphylococcus aureus PCR Testing in Pediatric Orbital Cellulitis. Journal of Pediatric Infectious Diseases Society. 2024 Aug; 13(8):430-433. • Nageswaran S, Woods CR, Benjamin DK Jr, Givner LB, Shetty AK. Orbital cellulitis in children. Pediatr Infect Dis J. 2006 Aug;25(8):695-9.• Rudloe TF, Harper MB, Prabhu SP, Rahbar R, Vanderveen D, Kimia AA. Acute periorbital infections: who needs emergent imaging? Pediatrics. 2010 Apr;125(4):e719-726.
• Eric Hoppa, MD• Scott Schoem, MD• Majida Gaffar, MD• Julie Quistorff, APRN
The clinical pathways in the above links have been developed specifically for use at Connecticut Children’s and are made available publicly for informational and/or educational purposes only. The clinical pathways are not intended to be, nor are they, a substitute for individualized professional medical judgment, advice, diagnosis, or treatment. Although Connecticut Children’s makes all efforts to ensure the accuracy of the posted content, Connecticut Children’s makes no warranty of any kind as to the accuracy or completeness of the information or its fitness for use at any particular facility or in any individual case. |
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Beyond the Nozzle: The Dramatic Journey of Stagnation Pressure
Beyond the Nozzle: The Dramatic Journey of Stagnation Pressure
Stagnation pressure, also known as total pressure or pitot pressure, is a concept in fluid mechanics and thermodynamics. It is a measure of the total energy contained in a fluid stream, accounting for both the static pressure and the kinetic energy of the fluid.
When a fluid (liquid or gas) flows, it possesses both static pressure (pressure exerted by the fluid molecules in random motion) and dynamic pressure (pressure due to the fluid's motion). Stagnation pressure is the sum of these two pressures and represents the maximum pressure that the fluid would achieve if it were brought to a complete stop, adiabatically (without any heat transfer) and isentropically (without any entropy change).
The most common way to measure stagnation pressure is by using a Pitot tube, named after Henri Pitot, a French engineer who invented it in the early 18th century. The Pitot tube is a device with two openings, one facing the fluid flow directly to measure the stagnation (total) pressure, and another opening on the side to measure the static pressure. By subtracting the static pressure from the stagnation pressure, we can determine the dynamic pressure of the fluid flow.
Stagnation pressure is essential in various engineering applications, such as aerodynamics, fluid dynamics, and gas dynamics. It is used to calculate important parameters like flow velocity, fluid density, and total energy in the fluid stream. In many practical situations, the conservation of stagnation pressure is applied to analyze the flow behavior through various components like nozzles, diffusers, and turbines.
Mathematically, stagnation pressure (P_0) can be expressed as:
P_0 = P + 0.5 ρ V^2
where: P_0 is the stagnation pressure, P is the static pressure, ρ (rho) is the density of the fluid, and V is the velocity of the fluid flow.
Stagnation pressure is a useful concept for engineers and scientists to understand and analyze fluid flow behavior and energy changes in fluid systems.
Fig 1. Stagnation pressure
What is stagnation pressure?
Stagnation pressure, also known as total pressure or pitot pressure, is a concept in fluid mechanics and thermodynamics. It refers to the total energy contained in a fluid stream, accounting for both the static pressure and the kinetic energy of the fluid.
When a fluid (liquid or gas) flows, it possesses both static pressure (pressure exerted by the fluid molecules in random motion) and dynamic pressure (pressure due to the fluid's motion). Stagnation pressure is the sum of these two pressures and represents the maximum pressure that the fluid would achieve if it were brought to a complete stop, adiabatically (without any heat transfer) and isentropically (without any entropy change).
Stagnation pressure is a fundamental concept used in various engineering applications, such as aerodynamics, fluid dynamics, and gas dynamics. It is measured using devices like a Pitot tube, which can determine both the stagnation pressure and the static pressure of the fluid flow. Stagnation pressure is critical for analyzing flow behavior, energy changes, and pressure distributions in fluid systems and is used to calculate important parameters like flow velocity, fluid density, and total energy in the fluid stream.
Examine our diverse selection of Pressure Sensor to find the best fit for your specific needs.
Can stagnation pressure be negative?
Stagnation pressure, as a physical quantity, cannot be negative in conventional circumstances. Stagnation pressure represents the total energy contained in a fluid stream and is the sum of the static pressure and the dynamic pressure (0.5 ρ V^2). Both static pressure and dynamic pressure are non-negative values, and their sum, the stagnation pressure, will also be non-negative.
However, there are some specific scenarios or conditions in which the stagnation pressure can be mathematically calculated to have a negative value. This usually occurs in theoretical or academic contexts and is not physically meaningful.
One such scenario is in the context of compressible flow through a converging-diverging nozzle. As the flow reaches a point of choking (where the flow velocity reaches the local speed of sound), the pressure downstream of the throat can be calculated to be negative in the theoretical equations. In this case, the negative stagnation pressure represents an extreme situation where the flow becomes supersonic, and certain assumptions used in the equations break down.
It's essential to recognize that such negative values are typically artifacts of the mathematical models or idealizations used in specific theoretical scenarios and do not have a physical interpretation. In real-world applications and practical engineering situations, stagnation pressure will always be a non-negative quantity, representing the total energy in the fluid flow.
What are Stagnation Pressure measuring ways?
Stagnation pressure, also known as total pressure or pitot pressure, can be measured using various methods and instruments. The most common and widely used technique is by employing a Pitot tube. Here are some of the ways stagnation pressure can be measured:
Pitot Tube:A Pitot tube is a simple and effective device used to measure stagnation pressure in fluid flows. It consists of two openings – one facing the fluid flow directly to measure the total pressure (stagnation pressure), and another opening on the side to measure the static pressure. By comparing the stagnation pressure with the static pressure, the dynamic pressure (velocity pressure) can be determined.
Pitot-Static System: In aircraft, a Pitot-static system is used to measure both stagnation pressure (pitot pressure) and static pressure. The Pitot tube measures the total pressure, and a static port measures the static pressure. The difference between these two pressures provides the dynamic pressure, which is used to calculate the aircraft's airspeed.
Pressure Transducers: Modern pressure transducers or pressure sensors can be used to measure stagnation pressure indirectly. By carefully positioning the pressure sensor in the flow, it can measure the total pressure, and the static pressure can be obtained separately using another pressure sensor placed at a point perpendicular to the flow.
Manometers: U-tube or inclined-tube manometers can be used to measure the difference between stagnation pressure and static pressure. The fluid levels in the tubes respond to the pressures applied at their respective ends, allowing for pressure difference measurements.
Pressure Gauges: Specialized pressure gauges, such as differential pressure gauges, can measure the difference between the stagnation pressure and static pressure.
Pitot-Static Tubes: These are similar to Pitot tubes but also include additional openings to measure the static pressure simultaneously. They are commonly used for measuring airspeed in aircraft and wind speed in meteorological applications.
Electronic Pressure Probes: These probes can measure pressure at different points in a flow field, including stagnation pressure, and provide real-time data for analysis and measurements.
It's important to note that the choice of the measuring method depends on the specific application, the accuracy required, the nature of the fluid flow, and the available instrumentation. For most practical purposes, Pitot tubes and Pitot-static systems remain the most commonly used methods for measuring stagnation pressure in various engineering and aerodynamic applications.
Fig 2. Stagnation Pressure
Stagnation Pressure formulas
Stagnation pressure, denoted as P_0, is the sum of the static pressure (P) and the dynamic pressure (q) of a fluid flow. The dynamic pressure is defined as one-half of the fluid's density (ρ) times the flow velocity squared (V^2). The formula for stagnation pressure can be expressed as:
P_0 = P + 0.5 ρ V^2
Where: P_0 = Stagnation pressure (total pressure or pitot pressure) P = Static pressure ρ (rho) = Density of the fluid V = Flow velocity
This formula is applicable to both compressible and incompressible fluid flows, provided that the flow is adiabatic (no heat transfer) and isentropic (no entropy change). Stagnation pressure is an important parameter in fluid mechanics and is used in various engineering applications to analyze fluid flow behavior, energy changes, and pressure distributions.
An example of using these formula
Let's walk through an example of using the stagnation pressure formula to calculate the total pressure at a specific point in a fluid flow.
Example: Suppose we have an air duct with a flow velocity of 20 m/s. The static pressure in the duct is measured to be 100 kPa, and the air density is 1.2 kg/m^3. We want to find the stagnation pressure at this point in the flow.
Given: Flow velocity (V) = 20 m/s Static pressure (P) = 100 kPa = 100,000 Pa Density (ρ) = 1.2 kg/m^3
Using the formula for stagnation pressure:
P_0 = P + 0.5 ρ V^2
Substitute the given values:
P_0 = 100,000 Pa + 0.5 1.2 kg/m^3 (20 m/s)^2
P_0 = 100,000 Pa + 0.5 1.2 kg/m^3 400 m^2/s^2
P_0 = 100,000 Pa + 240 kg/m^3 m^2/s^2
P_0 = 100,000 Pa + 240 N/m^2
P_0 = 100,240 Pa
So, the stagnation pressure at the specified point in the flow is approximately 100,240 Pa, or 100.24 kPa.
In this example, we used the given flow velocity, static pressure, and air density to calculate the stagnation pressure at a particular location in the air duct. This value represents the total pressure at that point, accounting for both the static and dynamic pressures of the flow.
Formula table
Here's a table summarizing the formulas mentioned earlier for stagnation pressure and related parameters:
ParameterSymbolFormula
Stagnation PressureP_0 P + 0.5 ρ V^2
Static PressureP P_0 - 0.5 ρ V^2
Dynamic Pressureq 0.5 ρ V^2
Densityρ (rho)m / V
Flow VelocityV√(2 q / ρ)
Mass Flow Ratem_dot A ρ V
Stagnation TemperatureT_0 T + (V^2 / (2 cp))
Where: P_0 = Stagnation Pressure (total pressure or pitot pressure) P = Static Pressure ρ = Density of the fluid V = Flow Velocity q = Dynamic Pressure (velocity pressure) A = Cross-sectional area of the flow m = Mass of the fluid T = Temperature of the fluid cp = Specific heat at constant pressure
These formulas are used in fluid mechanics, thermodynamics, aerodynamics, and other engineering fields to analyze fluid flow behavior, energy changes, and pressure distributions in fluid systems.
What is the significance of stagnation pressure?
The significance of stagnation pressure lies in its role as a fundamental concept in fluid mechanics and its practical applications in various engineering fields. Some of the key aspects of its significance are:
Total Energy Measurement: Stagnation pressure represents the total energy content of a fluid stream. It includes both the static pressure (pressure due to fluid molecules' random motion) and the dynamic pressure (pressure due to the fluid's motion). This total energy measurement is essential in analyzing fluid flows and energy distribution within a system.
Aerodynamics: In aerodynamics, stagnation pressure is crucial for analyzing aircraft and spacecraft performance. It is used to calculate airspeed, lift, and drag forces, and it provides critical data for optimizing wing design and aerodynamic efficiency.
Gas Dynamics: In gas dynamics and propulsion systems, stagnation pressure plays a vital role in understanding the performance of jet engines, rockets, and nozzles. It helps determine the thrust generated and the overall efficiency of these propulsion systems.
Wind Tunnel Testing: Stagnation pressure measurements are used in wind tunnel testing to evaluate the aerodynamic characteristics of models and prototypes. It allows engineers to simulate real-world conditions and assess how various objects interact with fluid flows.
Fluid Flow Analysis: Stagnation pressure provides valuable insights into the behavior of fluid flow around obstacles, through pipes, or within complex geometries. It helps identify regions of high pressure, low pressure, and areas of separation or recirculation in the flow.
Boundary Layer Analysis: In fluid dynamics, the boundary layer is the region of fluid flow close to a solid surface where viscosity effects are significant. Stagnation pressure measurements are used to study boundary layer thickness and its impact on drag and lift forces.
Compressible Flows: Stagnation pressure is particularly important in compressible flows, where changes in fluid velocity and density significantly affect the flow behavior. It helps determine the critical points such as Mach number and shock waves.
HVAC Systems: In heating, ventilation, and air conditioning (HVAC) systems, stagnation pressure measurements are used to assess airflow rates and pressure differentials in ducts and pipes.
Fluid System Design:Engineers use stagnation pressure calculations to design efficient and safe fluid systems for various applications, such as pipelines, turbines, compressors, and heat exchangers.
Research and Development: Stagnation pressure measurements are essential for researchers and scientists in various fields to study fluid behavior, investigate new technologies, and advance our understanding of fluid mechanics.
Overall, stagnation pressure provides a comprehensive and insightful way to understand fluid flow characteristics, energy changes, and pressure distributions in different engineering applications. Its significance extends across a wide range of industries and scientific disciplines, making it a fundamental concept in fluid dynamics and thermodynamics.
Fig 3. Pressures scale
Stagnation pressure different types
Stagnation pressure can be categorized into different types based on the specific context or application in which it is used. Here are some common types of stagnation pressure:
Stagnation Pressure (Total Pressure): This is the most commonly referred to type of stagnation pressure. It represents the total energy contained in a fluid stream and includes both the static pressure and the dynamic pressure. It is also known as total pressure or pitot pressure.
Stagnation Pressure in Compressible Flows: In compressible fluid flows, the term "stagnation pressure" is often used to refer to the total pressure (including both static and dynamic pressures). This is particularly relevant in supersonic flows where compressibility effects become significant.
Stagnation Pressure in Incompressible Flows: In incompressible fluid flows, stagnation pressure typically refers to the total pressure as well. However, since the density remains nearly constant, the dynamic pressure contribution is usually much smaller than the static pressure contribution.
Stagnation Pressure Rise: Stagnation pressure rise occurs when the stagnation pressure increases along a fluid flow due to various factors, such as compression or adiabatic heating.
Stagnation Pressure Loss: Stagnation pressure loss happens when the stagnation pressure decreases along a fluid flow. This can occur due to factors like friction, shock waves, or expansion.
Stagnation Pressure Recovery: Stagnation pressure recovery refers to the regaining of stagnation pressure after experiencing a pressure loss. For example, in some flow scenarios, the flow may expand after passing through a constriction, leading to a partial recovery of stagnation pressure.
Stagnation Pressure Point: Stagnation pressure point is the specific location in a fluid flow where the flow velocity becomes zero, and the stagnation pressure reaches its maximum value. This is known as the stagnation or stagnation point.
These types of stagnation pressure are used in various engineering applications, such as aerodynamics, gas dynamics, and fluid mechanics, to analyze flow behavior, energy changes, and pressure distributions in fluid systems. The appropriate type of stagnation pressure used in a particular scenario depends on the nature of the flow and the specific parameters being analyzed.
Stagnation Pressure (Total Pressure)
Stagnation pressure, also known as total pressure or pitot pressure, represents the total energy contained in a fluid stream. It accounts for both the static pressure (pressure due to fluid molecules' random motion) and the dynamic pressure (pressure due to the fluid's motion). Stagnation pressure is a fundamental concept in fluid mechanics and is widely used in various engineering applications. It is denoted as P_0 and is expressed as:
P_0 = P + 0.5 ρ V^2
where P is the static pressure, ρ is the density of the fluid, and V is the flow velocity.
Stagnation Pressure in Compressible Flows: In compressible fluid flows, especially in situations where the fluid velocity approaches or exceeds the speed of sound, compressibility effects become significant. Stagnation pressure in such cases still refers to the total pressure, including both static and dynamic pressures. However, compressibility effects cause changes in the density and temperature of the fluid, leading to alterations in stagnation pressure distribution in the flow field.
Stagnation Pressure in Incompressible Flows: In incompressible fluid flows, the density remains nearly constant, and the fluid velocity is much lower than the speed of sound. As a result, the dynamic pressure contribution (0.5 ρ V^2) in the stagnation pressure formula is often much smaller than the static pressure contribution (P). In these flows, the term "stagnation pressure" typically still refers to the total pressure, but the dynamic pressure can sometimes be neglected due to its small magnitude.
Stagnation Pressure Rise
Stagnation pressure rise occurs when the stagnation pressure increases along a fluid flow. This can happen due to various factors, such as compression or adiabatic heating. For example, in a converging nozzle, the fluid is compressed, leading to a rise in the stagnation pressure.
Stagnation Pressure Loss: Stagnation pressure loss, also known as total pressure loss, happens when the stagnation pressure decreases along a fluid flow. This occurs due to factors such as friction, shock waves, or expansion. For instance, when a fluid flows through a diverging nozzle, it expands, leading to a loss in stagnation pressure.
Stagnation Pressure Recovery: Stagnation pressure recovery refers to the phenomenon where stagnation pressure partially recovers after experiencing a pressure loss. This recovery can occur under certain flow conditions, such as after passing through a constriction or a shock wave. For example, in some aerodynamic shapes, the flow downstream of a shock wave may experience partial stagnation pressure recovery due to the pressure wave's interaction with the flow.
Stagnation Pressure Point
Stagnation pressure point is a specific location in a fluid flow where the flow velocity becomes zero, and the stagnation pressure reaches its maximum value. This point is called the stagnation or stagnation point. For example, in aerodynamics, the stagnation point is the point where the fluid directly faces the oncoming flow and experiences maximum stagnation pressure. This occurs at the leading edge of an airfoil or the nose of an aircraft.
These different types of stagnation pressure concepts are essential in understanding fluid flow behavior, energy changes, and pressure distributions in various engineering applications, ranging from aircraft design to gas turbine performance analysis. The choice of the appropriate type depends on the specific flow conditions and parameters being analyzed.
Comparing table
Here's a comparison table summarizing the different types of stagnation pressure:
Type of Stagnation PressureDefinitionFormula
Stagnation PressureTotal pressure in a fluid stream, including both static and dynamic pressures.P_0 = P + 0.5 ρ V^2
Stagnation Pressure in Compressible FlowsTotal pressure in compressible fluid flows, where compressibility effects become significant.P_0 = P + 0.5 ρ V^2
Stagnation Pressure in Incompressible FlowsTotal pressure in incompressible fluid flows, where the dynamic pressure contribution is small.P_0 = P + 0.5 ρ V^2
Stagnation Pressure RiseIncrease in stagnation pressure along a fluid flow, often due to compression or adiabatic heating.Depends on the specific flow scenario and compressibility effects.
Stagnation Pressure LossDecrease in stagnation pressure along a fluid flow, caused by factors like friction or expansion.Depends on the specific flow scenario and compressibility effects.
Stagnation Pressure RecoveryPartial regaining of stagnation pressure after a pressure loss, often seen after passing through a shock wave or constriction.Depends on the specific flow scenario and compressibility effects.
Stagnation Pressure PointSpecific location in a fluid flow where the flow velocity becomes zero, and the stagnation pressure reaches its maximum value.Occurs at the stagnation or stagnation point, often at the leading edge of an airfoil.
The table provides a quick comparison of the different types of stagnation pressure, highlighting their definitions and any unique characteristics. These types are used in various engineering applications to analyze fluid flow behavior, energy changes, and pressure distributions in fluid systems under different flow conditions.
What is stagnation pressure used for?
Stagnation pressure, also known as total pressure or pitot pressure, is used for various purposes in fluid mechanics, thermodynamics, and engineering applications. Some of the key uses of stagnation pressure include:
Airspeed Measurement: Stagnation pressure is crucial for determining the airspeed of an aircraft. By measuring the stagnation pressure using a Pitot tube, pilots can accurately gauge the aircraft's velocity relative to the surrounding air.
Aerodynamic Analysis: Stagnation pressure helps in the analysis of aerodynamic characteristics, lift, and drag forces on objects such as airfoils, wings, and vehicles. It aids in designing efficient and stable flying vehicles.
Wind Tunnel Testing: In wind tunnel experiments, stagnation pressure measurements are used to study and analyze the aerodynamic behavior of scaled-down models. This data provides insights into real-world applications and design improvements.
Gas Turbines and Jet Engines: Stagnation pressure is essential for assessing the performance and efficiency of gas turbines, jet engines, and rocket nozzles. It helps optimize engine design and analyze thrust and exhaust gas behavior.
Flow Behavior Analysis:Stagnation pressure data is used to analyze the behavior of fluid flows around obstacles, through pipes, and within complex geometries. It helps identify regions of high pressure, low pressure, and areas of flow separation or recirculation.
Boundary Layer Studies: Stagnation pressure measurements are used to study boundary layer characteristics near solid surfaces. Understanding boundary layer behavior is vital in optimizing aerodynamic performance and reducing drag.
HVAC Systems: In heating, ventilation, and air conditioning (HVAC) systems, stagnation pressure measurements are used to assess airflow rates and pressure differentials in ducts and pipes, ensuring efficient HVAC operations.
Supersonic and Hypersonic Flows: In supersonic and hypersonic flow regimes, stagnation pressure is critical for analyzing shock waves, expansion waves, and other compressibility effects.
Atmospheric Studies: Stagnation pressure measurements are utilized in meteorology to determine wind speed and direction and analyze weather patterns.
Research and Development: Stagnation pressure data aids researchers and scientists in various fields to study fluid behavior, investigate new technologies, and advance the understanding of fluid mechanics.
In summary, stagnation pressure is a fundamental concept with diverse applications across numerous engineering disciplines. Its role in aerodynamics, gas dynamics, propulsion, and fluid analysis makes it a critical parameter for understanding and optimizing the behavior of fluid systems in both real-world applications and research settings.
How is stagnation pressure different from static pressure?
Stagnation pressure and static pressure are two important concepts in fluid mechanics that represent different aspects of the pressure in a fluid flow. They are related but have distinct characteristics:
Stagnation Pressure: Stagnation pressure, also known as total pressure or pitot pressure, is the total energy contained in a fluid stream. It accounts for both the static pressure and the dynamic pressure of the flow. Stagnation pressure represents the maximum pressure that the fluid would achieve if it were brought to a complete stop, adiabatically (without any heat transfer) and isentropically (without any entropy change). It is the pressure measured when the fluid comes to rest due to compression and is usually measured using a Pitot tube or similar devices.
Formula for Stagnation Pressure (P_0): P_0 = P + 0.5 ρ V^2
where: P_0 = Stagnation Pressure (total pressure) P = Static Pressure ρ = Density of the fluid V = Flow Velocity
Static Pressure: Static pressure is the pressure exerted by fluid molecules in random motion. It represents the pressure at a given point in the fluid flow when the fluid is in motion. It does not consider the kinetic energy of the flow and is the pressure measured when the fluid is not flowing (i.e., at rest). Static pressure is crucial in determining the balance of forces in fluid systems and is used in various engineering applications, such as in designing pressure vessels, calculating loads on surfaces, and determining fluid velocities.
Static Pressure Formula: P = P_0 - 0.5 ρ V^2
where: P = Static Pressure P_0 = Stagnation Pressure (total pressure) ρ = Density of the fluid V = Flow Velocity
Key Differences:
Components of Pressure: Stagnation pressure includes both the static pressure and the dynamic pressure (0.5 ρ V^2), while static pressure considers only the pressure due to fluid molecules' random motion.
Total Energy vs. Local Energy: Stagnation pressure represents the total energy contained in the fluid stream, accounting for the fluid's kinetic energy. Static pressure, on the other hand, represents the local energy at a specific point in the flow.
Flow Conditions: Stagnation pressure is relevant when the fluid flow is in motion, whereas static pressure is relevant whether the fluid is in motion or at rest.
Both stagnation pressure and static pressure are important parameters used in various engineering applications to analyze fluid flow behavior, pressure distributions, and energy changes within fluid systems. Understanding the differences between these pressure types is crucial for accurately assessing fluid dynamics in practical scenarios.
Can stagnation pressure be higher than the static pressure?
Yes, stagnation pressure can be higher than the static pressure in a fluid flow. In fact, this is a common occurrence in various fluid dynamics scenarios.
Stagnation pressure (P_0) is the total pressure in a fluid stream, accounting for both the static pressure (P) and the dynamic pressure (0.5 ρ V^2), where ρ is the density of the fluid, and V is the flow velocity.
The dynamic pressure component (0.5 ρ V^2) is directly related to the flow velocity and represents the pressure due to the fluid's motion. As the flow velocity increases, the dynamic pressure increases proportionally.
If the flow velocity is high enough, the dynamic pressure can significantly contribute to the stagnation pressure, resulting in a higher total pressure (P_0) compared to the static pressure (P).
This situation is common in high-speed flows, such as supersonic or hypersonic flows, where the fluid's velocity approaches or exceeds the speed of sound. In these scenarios, the dynamic pressure can become dominant, leading to a substantial difference between the stagnation pressure and the static pressure.
For example, in a supersonic aircraft flying at high speeds, the stagnation pressure at the aircraft's nose can be significantly higher than the static pressure measured at the surrounding atmospheric conditions.
In summary, stagnation pressure can indeed be higher than static pressure, especially in high-speed flows where the dynamic pressure plays a significant role in the total pressure calculation.
Fig 4. Flow measurement
Is stagnation pressure constant in a flow field?
In an ideal, adiabatic, and isentropic flow, stagnation pressure remains constant along a streamline. This condition is often referred to as "stagnation pressure conservation."
Stagnation pressure conservation is based on the Bernoulli's principle, which states that the total mechanical energy per unit mass of a fluid particle remains constant along a streamline in the absence of external work and heat transfer. In other words, if a fluid particle follows a streamline without any energy exchange with its surroundings, the sum of its static pressure and dynamic pressure (stagnation pressure) will remain constant.
In real-world scenarios, however, there are various factors that can cause the stagnation pressure to change along a flow field. Some of the reasons for stagnation pressure variation include:
Friction: Friction between the fluid and the surfaces it flows over can cause energy losses and lead to a decrease in stagnation pressure.
Heat Transfer: If heat is exchanged between the fluid and its surroundings, adiabatic conditions are not met, and stagnation pressure may be affected.
Compressibility: In compressible flows, such as those with shock waves or regions of high Mach numbers, the conservation of stagnation pressure may not hold.
Expansion or Compression: Changes in flow area can cause changes in the fluid's velocity and, consequently, its stagnation pressure.
Turbulence: Turbulent flows can have significant energy fluctuations, impacting stagnation pressure.
Viscosity: Viscous effects in the fluid can cause energy losses, affecting the stagnation pressure.
In practice, it is essential to consider these factors and the specific flow conditions to determine whether stagnation pressure remains constant or varies along a flow field. While stagnation pressure conservation is an essential concept in fluid mechanics, it is an idealized condition and may not hold in all real-world scenarios.
Why is stagnation pressure important in aerodynamics?
Stagnation pressure is of paramount importance in aerodynamics due to its fundamental role in understanding and analyzing the airflow around objects, such as aircraft, rockets, and other flying vehicles. Here are some key reasons why stagnation pressure is significant in aerodynamics:
Airspeed Measurement: Stagnation pressure is directly related to the airspeed of an aircraft. By measuring the stagnation pressure using a Pitot tube, the airspeed can be determined. This critical information is essential for pilots to ensure safe and optimal flight operations.
Lift and Drag Calculations: Understanding the stagnation pressure distribution around an airfoil or wing helps aerodynamicists calculate lift and drag forces. The difference in stagnation pressure between the upper and lower surfaces of the wing contributes to lift generation, while the stagnation pressure on the aircraft's nose affects drag forces.
Aerodynamic Efficiency:Stagnation pressure analysis enables the assessment of an aircraft's aerodynamic efficiency. By optimizing the pressure distribution on the aircraft's surfaces, designers can improve fuel efficiency and performance.
Flow Separation and Stall Prediction: Stagnation pressure measurements provide insights into flow separation and stall conditions. Studying stagnation pressure patterns can help identify regions of potential flow separation, which is crucial for preventing dangerous stall situations.
Shock Wave Analysis: In supersonic and hypersonic flows, stagnation pressure is essential for analyzing shock waves. Shock waves are regions of abrupt pressure and density changes and have a significant impact on aerodynamic performance.
Boundary Layer Studies: Stagnation pressure measurements are used to study boundary layer characteristics. The boundary layer is the thin layer of fluid near a solid surface, and its behavior significantly affects aerodynamic forces and drag.
Wind Tunnel Testing: In wind tunnel experiments, stagnation pressure is measured to assess airflow behavior around scaled-down models of aircraft and other objects. These tests provide valuable data for design improvements and validating aerodynamic theories.
Performance Optimization: Stagnation pressure analysis allows engineers to optimize aircraft and spacecraft designs, reducing drag, improving lift, and achieving better overall performance.
Mach Number Determination:Stagnation pressure is used to determine the Mach number, a dimensionless parameter that indicates the speed of an aircraft relative to the speed of sound in the surrounding air.
Overall, stagnation pressure plays a central role in aerodynamics by providing critical data for designing and analyzing flying vehicles' performance. The understanding of stagnation pressure and its distribution aids in achieving safe, efficient, and high-performance flight operations.
What is the relationship between stagnation pressure and Bernoulli's principle?
Stagnation pressure and Bernoulli's principle are closely related concepts in fluid mechanics. Bernoulli's principle is a fundamental principle that describes the behavior of an ideal fluid in steady flow. The relationship between stagnation pressure and Bernoulli's principle can be understood through the application of the principle along a streamline.
Bernoulli's principle states that in an ideal, steady, and incompressible flow, the total mechanical energy per unit mass of the fluid remains constant along a streamline. This total mechanical energy includes three components: the fluid's potential energy (due to its elevation), the fluid's kinetic energy (due to its motion), and the fluid's pressure energy.
The equation for Bernoulli's principle along a streamline is as follows:
P + 0.5 ρ V^2 + ρ g h = constant
where: P is the static pressure of the fluid, ρ is the density of the fluid, V is the flow velocity, g is the acceleration due to gravity, h is the height above a reference plane.
Now, consider the case where the fluid is brought to rest adiabatically (without any heat transfer) and isentropically (without any entropy change). At this point, the fluid's kinetic energy becomes zero (V = 0), and the equation simplifies to:
P + ρ g h = constant
This constant value represents the stagnation pressure (P_0) of the fluid. Thus, we can say that in an adiabatic and isentropic flow where the fluid comes to rest, the stagnation pressure is equal to the sum of the static pressure and the hydrostatic pressure (ρ g h).
In summary, the relationship between stagnation pressure and Bernoulli's principle is that the stagnation pressure is the total mechanical energy per unit mass of the fluid when it comes to rest adiabatically and isentropically along a streamline. It is a direct consequence of Bernoulli's principle applied to this specific scenario.
How is stagnation pressure affected by an increase in flow velocity?
Stagnation pressure is directly affected by an increase in flow velocity. As the flow velocity increases, the stagnation pressure also increases.
Stagnation pressure (P_0) is the total pressure in a fluid stream and is given by the sum of the static pressure (P) and the dynamic pressure (0.5 ρ V^2), where ρ is the density of the fluid, and V is the flow velocity.
The dynamic pressure component (0.5 ρ V^2) represents the pressure due to the fluid's motion. It is directly proportional to the square of the flow velocity. Therefore, as the flow velocity increases, the dynamic pressure increases much faster.
The relationship between flow velocity and stagnation pressure is described by the stagnation pressure formula:
P_0 = P + 0.5 ρ V^2
When the flow velocity (V) increases, the dynamic pressure term (0.5 ρ V^2) becomes larger, resulting in a higher stagnation pressure (P_0). Conversely, if the flow velocity decreases, the dynamic pressure decreases, leading to a lower stagnation pressure.
This relationship is important in various engineering applications. For example, in aerodynamics, an increase in flow velocity can result in higher stagnation pressure at specific points on an airfoil or aircraft. Similarly, in gas turbine engines, an increase in airflow velocity can lead to higher stagnation pressure at the engine inlet, affecting the engine's performance.
In summary, an increase in flow velocity results in a higher stagnation pressure, while a decrease in flow velocity leads to a lower stagnation pressure. Understanding this relationship is crucial for analyzing fluid flow behavior and its impact on pressure distributions in various engineering applications.
What are parameters affect on stagnation pressure?
Stagnation pressure is affected by several parameters that influence the fluid flow behavior. These parameters include:
Flow Velocity (V): Stagnation pressure is directly proportional to the square of the flow velocity. As the flow velocity increases, the dynamic pressure component (0.5 ρ V^2) in the stagnation pressure formula becomes larger, resulting in higher stagnation pressure.
Static Pressure (P): Stagnation pressure is the sum of the static pressure and dynamic pressure. An increase in static pressure will lead to a corresponding increase in stagnation pressure.
Density of the Fluid (ρ): Stagnation pressure is directly proportional to the density of the fluid. Higher fluid density will result in higher stagnation pressure for the same flow conditions.
Compressibility Effects: In compressible flows, especially in high-speed flows where the fluid velocity approaches or exceeds the speed of sound, compressibility effects become significant. These effects can alter the stagnation pressure distribution and may lead to changes in the flow behavior.
Flow Area (A): Changes in the cross-sectional area of the flow path can affect the flow velocity and, consequently, the stagnation pressure. For example, a converging section may increase the flow velocity and raise the stagnation pressure.
Temperature (T): Changes in temperature can affect the density of the fluid, which, in turn, influences the stagnation pressure. For ideal gases, the relationship between temperature and density is described by the ideal gas law (ρ = P / (R T)), where R is the gas constant.
Fluid Properties: Different fluids have varying densities and specific heat capacities, affecting the stagnation pressure for the same flow conditions.
Boundary Layer Effects: In fluid flow near solid surfaces, the boundary layer affects the fluid's behavior. This can lead to changes in the stagnation pressure distribution along the surface.
Flow Regime: Stagnation pressure can be influenced by the flow regime, such as laminar or turbulent flow. Turbulent flows may have higher energy losses and lead to variations in stagnation pressure.
Obstacles and Flow Obstructions: Presence of obstacles or flow obstructions in the fluid path can cause pressure changes, influencing the stagnation pressure at specific locations.
Heat Transfer: Stagnation pressure can be affected by heat transfer processes in the fluid, especially if the flow is not adiabatic.
Understanding these parameters and their effects on stagnation pressure is crucial for analyzing fluid flow behavior, optimizing aerodynamic designs, and predicting the performance of various fluid systems.
What happens to stagnation pressure when fluid flows through a nozzle?
When a fluid flows through a nozzle, the stagnation pressure typically decreases. This phenomenon is known as "stagnation pressure loss" or "total pressure loss" and occurs due to the conversion of stagnation pressure into kinetic energy as the flow accelerates through the nozzle.
In a nozzle, the cross-sectional area decreases in the flow direction, leading to an increase in flow velocity according to the principle of continuity (mass conservation). As the fluid accelerates, its kinetic energy increases, and this energy is taken from the stagnation pressure.
The process of pressure decrease in a nozzle can be explained by the conservation of energy. According to Bernoulli's principle, the total mechanical energy per unit mass of the fluid remains constant along a streamline in an ideal, steady, and incompressible flow. In the nozzle, as the flow velocity increases, the kinetic energy component (0.5 ρ V^2) in the stagnation pressure formula becomes larger, resulting in a decrease in stagnation pressure.
The decrease in stagnation pressure is usually more significant in converging nozzles, where the cross-sectional area decreases continuously. This reduction in pressure allows the fluid to maintain energy conservation throughout the flow.
Stagnation pressure loss in a nozzle is a crucial consideration in various engineering applications, such as in gas turbine engines, rocket nozzles, and fluid systems with constricted flow passages. Engineers need to account for the pressure loss to optimize system performance and design efficient nozzles for specific purposes.
How does stagnation pressure relate to the concept of stagnation point in fluid mechanics?
Stagnation pressure and the concept of a stagnation point in fluid mechanics are closely related and interconnected.
The Stagnation Point: A stagnation point is a specific location in a fluid flow where the flow velocity becomes zero (V = 0). At this point, the fluid comes to rest due to an obstacle or flow constriction, and the velocity energy is converted entirely into pressure energy. As a result, the pressure at the stagnation point reaches its maximum value, known as stagnation pressure (P_0).
The Stagnation Pressure: Stagnation pressure (P_0) is the total pressure in a fluid stream and represents the maximum pressure that the fluid would achieve if it were brought to a complete stop adiabatically (without any heat transfer) and isentropically (without any entropy change). Stagnation pressure accounts for both the static pressure (P) and the dynamic pressure (0.5 ρ V^2) of the fluid.
The Relationship: The relationship between stagnation pressure and the concept of a stagnation point is evident in their definitions. At the stagnation point, the fluid's velocity is zero, and all its kinetic energy is converted into pressure energy. Therefore, the pressure at the stagnation point is the maximum in the fluid flow, representing the stagnation pressure (P_0).
Mathematically, at the stagnation point:
V = 0 Dynamic Pressure (0.5 ρ V^2) = 0
As a result, the stagnation pressure formula simplifies to:
P_0 = P
This means that at the stagnation point, the stagnation pressure (P_0) is equal to the static pressure (P). The pressure at any other point in the flow (away from the stagnation point) will be lower than the stagnation pressure due to the presence of flow velocity and dynamic pressure.
In summary, the concept of a stagnation point is related to stagnation pressure in fluid mechanics. The stagnation point represents the location where the fluid velocity is zero, and the pressure reaches its maximum value, which is equal to the stagnation pressure. Understanding the behavior of the flow at the stagnation point and the associated stagnation pressure is crucial for analyzing fluid flow around obstacles, airfoils, and other fluid systems.
What is the impact of friction on stagnation pressure in a fluid flow?
Friction in a fluid flow has a significant impact on the stagnation pressure. As the fluid interacts with surfaces and experiences viscosity, energy losses occur due to fluid-solid interactions, resulting in a decrease in stagnation pressure. This phenomenon is known as "stagnation pressure loss" or "total pressure loss" due to friction.
When a fluid flows over a surface, the fluid molecules in direct contact with the surface experience drag and shear forces. This interaction causes energy dissipation in the form of heat, resulting in a loss of mechanical energy. As a consequence, the fluid's total pressure (stagnation pressure) decreases compared to what it would be in an ideal, frictionless flow.
The extent of stagnation pressure loss due to friction depends on several factors, including the following:
Flow Velocity: Higher flow velocities lead to higher frictional losses, as the fluid interacts more intensely with the surfaces it flows over.
Viscosity of the Fluid: Fluids with higher viscosity experience stronger internal friction, resulting in more significant stagnation pressure losses.
Surface Roughness: Rough surfaces introduce additional frictional effects, causing higher energy losses in the fluid flow.
Flow Path Geometry:The shape and curvature of the flow path affect the magnitude of frictional losses.
Reynolds Number: The Reynolds number, which characterizes the flow regime, influences the level of turbulence and laminar flow. Turbulent flows generally experience higher frictional losses than laminar flows.
Flow Distance:The distance the fluid travels over the surface affects the cumulative effect of friction on the stagnation pressure.
Stagnation pressure loss due to friction is particularly relevant in real-world fluid systems, where ideal conditions are not achieved. In practical engineering applications, it is essential to account for these pressure losses when designing and analyzing fluid flow systems, such as pipelines, ducts, heat exchangers, and other components.
Understanding the impact of friction on stagnation pressure is crucial for accurately predicting the behavior of fluid systems and ensuring efficient and reliable operation in various engineering applications.
How does stagnation pressure change across a shock wave?
Across a shock wave, the stagnation pressure undergoes a significant change. A shock wave is a high-intensity compression wave that occurs when a fluid flows at a supersonic speed and encounters a sudden change in flow conditions, such as a constriction or an object. As the fluid passes through the shock wave, the stagnation pressure experiences a sudden increase, resulting in a phenomenon known as "stagnation pressure rise."
The key changes in stagnation pressure across a shock wave can be understood by considering the conservation of energy and momentum in the flow:
Stagnation Pressure Rise: When a fluid encounters a shock wave, its flow velocity abruptly decreases to subsonic speed. This reduction in velocity causes a sudden increase in the static pressure, resulting in a rise in stagnation pressure. The increase in stagnation pressure is due to the conversion of the fluid's kinetic energy into pressure energy as it slows down across the shock wave.
Flow Deceleration:As the fluid crosses the shock wave, it decelerates from supersonic to subsonic speed. This deceleration is associated with a corresponding increase in the static pressure and, consequently, the stagnation pressure.
Temperature Rise:Shock waves cause an increase in fluid temperature, which is linked to the rise in stagnation pressure. The shock wave compresses the fluid and converts some of the flow's kinetic energy into thermal energy, leading to an increase in temperature and pressure.
It's essential to note that the change in stagnation pressure across a shock wave is not reversible. Once the flow passes through the shock, the pressure rise is permanent, and the fluid remains at a higher stagnation pressure downstream of the shock wave.
Stagnation pressure changes across shock waves play a critical role in aerodynamics, gas dynamics, and supersonic/hypersonic flows. These changes can significantly impact the performance and behavior of various aerodynamic shapes, aircraft, and propulsion systems. Understanding the effects of shock waves on stagnation pressure is essential for designing high-speed vehicles, wind tunnels, and other engineering systems involving compressible fluid flows.
What is the impact of stagnation pressure on boundary layer separation?
Stagnation pressure can have a significant impact on boundary layer separation in fluid flows. Boundary layer separation occurs when the boundary layer, which is the thin layer of fluid adjacent to a solid surface, detaches from the surface and forms a separated flow region. The presence of stagnation pressure gradients can influence the occurrence and extent of boundary layer separation.
Boundary Layer Acceleration: In regions with favorable pressure gradients, where the stagnation pressure increases in the flow direction, the boundary layer tends to accelerate along the surface. This acceleration helps to maintain attachment of the boundary layer to the surface, delaying or preventing separation.
Boundary Layer Deceleration: In regions with adverse pressure gradients, where the stagnation pressure decreases in the flow direction, the boundary layer experiences deceleration. This deceleration can lead to flow separation, where the boundary layer separates from the surface, creating a separated flow region.
Separation Bubble: In some cases, the combination of favorable and adverse pressure gradients can lead to the formation of a separation bubble. A separation bubble occurs when the boundary layer initially separates from the surface due to an adverse pressure gradient but reattaches further downstream due to a favorable pressure gradient. This creates a recirculation zone within the separation bubble.
Flow Reattachment:Stagnation pressure gradients can also influence flow reattachment after separation. In cases where the pressure gradient becomes favorable again, the separated flow region may reattach to the surface, reducing the size and intensity of the separated region.
Impact on Lift and Drag: Boundary layer separation has a direct impact on the aerodynamic forces acting on an object, such as an airfoil or wing. Separation can significantly increase drag and reduce lift, affecting the overall aerodynamic performance of the object.
Flow Control: Understanding the impact of stagnation pressure on boundary layer separation is essential for flow control strategies. By manipulating the pressure distribution, engineers can design aerodynamic shapes and control surfaces to delay separation or promote reattachment, enhancing the efficiency and stability of vehicles.
In summary, stagnation pressure gradients play a crucial role in boundary layer behavior and separation. Proper management of stagnation pressure distributions can be used to control boundary layer separation and improve the aerodynamic performance of various engineering systems, such as aircraft, vehicles, and turbines.
Conclusion
In conclusion, stagnation pressure is a fundamental concept in fluid mechanics that represents the total pressure in a fluid stream, accounting for both the static pressure and the dynamic pressure due to the fluid's motion. It is measured when the fluid comes to rest adiabatically and isentropically, and it plays a crucial role in various engineering applications.
Stagnation pressure is used for a wide range of purposes, including airspeed measurement, aerodynamic analysis, wind tunnel testing, gas turbine and jet engine performance assessment, flow behavior analysis, boundary layer studies, and atmospheric studies.
Several parameters affect stagnation pressure, including flow velocity, static pressure, fluid density, compressibility effects, flow area, temperature, and friction. Understanding these parameters is essential for accurately predicting fluid behavior in different flow scenarios.
Stagnation pressure is related to the concept of a stagnation point in fluid mechanics. At the stagnation point, the fluid velocity is zero, and the pressure reaches its maximum value, which is equal to the stagnation pressure. Stagnation pressure loss due to friction is an important consideration in real-world fluid systems, leading to a decrease in stagnation pressure as the fluid interacts with surfaces and experiences energy losses.
Furthermore, across a shock wave, stagnation pressure undergoes a significant change, experiencing a sudden increase known as "stagnation pressure rise." Stagnation pressure changes can influence boundary layer separation, affecting aerodynamic performance and flow control strategies in various engineering applications.
Overall, understanding stagnation pressure and its effects on fluid flow behavior is essential for optimizing system design, predicting performance, and ensuring safe and efficient operation in numerous engineering and scientific disciplines.
To recap
Q: What is stagnation pressure?
A: Stagnation pressure, also known as total pressure or pitot pressure, is the total energy contained in a fluid stream. It represents the pressure the fluid would have if brought to rest adiabatically and isentropically.
Q: How is stagnation pressure measured?
A: Stagnation pressure is often measured using a device called a Pitot tube, which has a forward-facing opening to measure the total pressure and a side opening to measure the static pressure. The stagnation pressure is then calculated as the sum of the total pressure and the dynamic pressure.
Q: What are the factors affecting stagnation pressure?
A: Stagnation pressure is affected by flow velocity, static pressure, fluid density, compressibility effects, flow area, temperature, friction, boundary layer behavior, and flow regime.
Q: Can stagnation pressure be higher than static pressure?
A: Yes, stagnation pressure can be higher than static pressure, especially in high-speed flows where the dynamic pressure contributes significantly to the total pressure.
Q: What is the relationship between stagnation pressure and Bernoulli's principle?
A: Stagnation pressure is related to Bernoulli's principle, as it represents the total mechanical energy per unit mass of the fluid when it comes to rest along a streamline.
Q: How is stagnation pressure affected by an increase in flow velocity?
A: Stagnation pressure increases with an increase in flow velocity, as the dynamic pressure component in the stagnation pressure formula becomes larger.
Q: What happens to stagnation pressure when fluid flows through a nozzle?
A: Stagnation pressure decreases when fluid flows through a nozzle due to the conversion of stagnation pressure into kinetic energy as the flow accelerates.
Q: How does stagnation pressure change across a shock wave?
A: Across a shock wave, stagnation pressure experiences a sudden increase known as "stagnation pressure rise" due to the abrupt decrease in flow velocity.
Q: What is the impact of stagnation pressure on boundary layer separation?
A: Stagnation pressure gradients can influence boundary layer separation, affecting flow acceleration, deceleration, separation bubbles, and flow reattachment.
Q: What is the significance of stagnation pressure in aerodynamics?
A: Stagnation pressure is crucial in aerodynamics for airspeed measurement, lift and drag calculations, aerodynamic efficiency analysis, flow separation prediction, shock wave analysis, and wind tunnel testing. It plays a central role in designing and optimizing aircraft and other flying vehicles.
Reference
6th Nov 2023
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