The divisibility relation on the natural numbers is an important example of an antisymmetric relation. Yes. For the number of dinners to be divisible by the number of club members with their two advisers AND the number of club members with their two advisers to be divisible by the number of dinners, those two numbers have to be equal. Solution: The antisymmetric relation on set A = {1,2,3,4} will be; Your email address will not be published. i know what an anti-symmetric relation is. Two types of relations are asymmetric relations and antisymmetric relations, which are defined as follows: Asymmetric: If (a, b) is in R, then (b, a) cannot be in R. Antisymmetric: … A relation becomes an antisymmetric relation for a binary relation R on a set A. Equivalently, R is antisymmetric if and only if whenever R, and a b, R. For relation, R, an ordered pair (x,y) can be found where x and y are whole numbers and x is divisible by y. example of antisymmetric The axioms of a partial ordering demonstrate that every partial ordering is antisymmetric. Other than antisymmetric, there are different relations like reflexive, irreflexive, symmetric, asymmetric, and transitive. (i) R is not antisymmetric here because of (1,2) ∈ R and (2,1) ∈ R, but 1 ≠ 2. (ii) Transitive but neither reflexive nor symmetric. If a relation $$R$$ on $$A$$ is both symmetric and antisymmetric, its off-diagonal entries are all zeros, so it is a subset of the identity relation. For example, if a relation is transitive and irreflexive, 1 it must also be asymmetric. Antisymmetry is different from asymmetry: a relation is asymmetric if, and only if, it is antisymmetric and irreflexive. A relation R on a set a is called on antisymmetric relation if for x, y if for x, y => If (x, y) and (y, x) E R then x = y. Such examples aren't considered in the article - are these in fact examples or is the definition missing something? From the Cambridge English Corpus One of them is the out-of … In Matrix form, if a 12 is present in relation, then a 21 is also present in relation and As we know reflexive relation is part of symmetric relation. Note: If a relation is not symmetric that does not mean it is antisymmetric. Also, i'm curious to know since relations can both be neither symmetric and anti-symmetric, would R = {(1,2),(2,1),(2,3)} be an example of such a relation? 8. Your email address will not be published. (b, a) can not be in relation if (a,b) is in a relationship. An antisymmetric relation satisfies the following property: To prove that a given relation is antisymmetric, we simply assume that (a, b) and (b, a) are in the relation, and then we show that a = b. example of antisymmetric The axioms of a partial ordering demonstrate that every partial ordering is antisymmetric. A relation is antisymmetric if (a,b)\in R and (b,a)\in R only when a=b. (number of members and advisers, number of dinners) 2. Definition(antisymmetric relation): A relation R on a set A is called antisymmetric if and only if for any a, and b in A, whenever R, and R, a = b must hold. Example 2. Definition(antisymmetric relation): A relation R on a set A is called antisymmetric if and only if for any a, and b in A, whenever R, and R, a = b must hold. Question about vacuous antisymmetric relations. Antisymmetric definition: (of a relation ) never holding between a pair of arguments x and y when it holds between... | Meaning, pronunciation, translations and examples Required fields are marked *. A symmetric relation is a type of binary relation.An example is the relation "is equal to", because if a = b is true then b = a is also true. More formally, R is antisymmetric precisely if for all a and b in X, (The definition of antisymmetry says nothing about whether R(a, a) actually holds or not for any a.). Based on the definition, it would seem that any relation for which (,) ∧ (,) never holds would be antisymmetric; an example is the strict ordering < on the real numbers. Hence, it is a … Hence, it is a … You should know that the relation R ‘is less than’ is an asymmetric relation such as 5 < 11 but 11 is not less than 5. Example 6: The relation "being acquainted with" on a set of people is symmetric. In a formal way, relation R is antisymmetric, specifically if for all a and b in A, if R(x, y) with x ≠ y, then R(y, x) must not hold, or, equivalently, if R(x, y) and R(y, x), then x = y. Example 6: The relation "being acquainted with" on a set of people is symmetric. Hence, as per it, whenever (x,y) is in relation R, then (y, x) is not. Symmetric or antisymmetric are special cases, most relations are neither (although a lot of useful/interesting relations are one or the other). (iii) R is not antisymmetric here because of (1,2) ∈ R and (2,1) ∈ R, but 1 ≠ 2 and also (1,4) ∈ R and (4,1) ∈ R but 1 ≠ 4. (ii) R is not antisymmetric here because of (1,3) ∈ R and (3,1) ∈ R, but 1 ≠ 3. Learn how and when to remove this template message, https://en.wikipedia.org/w/index.php?title=Antisymmetric_relation&oldid=996549949, Articles needing additional references from January 2010, All articles needing additional references, Creative Commons Attribution-ShareAlike License, This page was last edited on 27 December 2020, at 07:28. Hence, less than (<), greater than (>) and minus (-) are examples of asymmetric. (i) R = {(1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4)}, (iii) R = {(1,1),(1,2),(1,4),(2,1),(2,2),(3,3),(4,1),(4,4)}. Click hereto get an answer to your question  Given an example of a relation. For example: If R is a relation on set A= (18,9) then (9,18) ∈ R indicates 18>9 but (9,18) R, Since 9 is not greater than 18. A purely antisymmetric response tensor corresponds with a limiting case of an optically active medium, but is not appropriate for a plasma. If 5 is a proper divisor of 15, then 15 cannot be a proper divisor of 5. For example, <, \le, and divisibility are all antisymmetric. The relation “…is a proper divisor of…” in the set of whole numbers is an antisymmetric relation. “Is less than” is an asymmetric, such as 7<15 but 15 is not less than 7. An example of antisymmetric is: for a relation “is divisible by” which is the relation for ordered pairs in the set of integers. 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Antisymmetric: The relation is antisymmetric as whenever (a, b) and (b, a) ∈ R, we have a = b. Transitive: The relation is transitive as whenever (a, b) and (b, c) ∈ R, we have (a, c) ∈ R. Example: (4, 2) ∈ R and (2, 1) ∈ R, implies (4, 1) ∈ R. As the relation is reflexive, antisymmetric and transitive. 9. R is antisymmetric x R y and y R x implies that x=y, for all x,y,z∈A Examples: Here are some binary relations over A={0,1}. Return to our math club and their spaghetti-and-meatball dinners. For a finite set A with n elements, the number of possible antisymmetric relations is 2 n  3 n 2-n 2 out of the 2 n 2 total possible relations. (number of dinners, number of members and advisers) Since 3434 members and 22 advisers are in the math club, t… Apart from antisymmetric, there are different types of relations, such as: An example of antisymmetric is: for a relation “is divisible by” which is the relation for ordered pairs in the set of integers. A relation can be antisymmetric and symmetric at the same time. On the other hand the relation R is said to be antisymmetric if (x,y), (y,x)€ R ==> x=y. Other Examples. Consider the ≥ relation. Here x and y are the elements of set A. The “equals” (=) relation is symmetric. Q.2: If A = {1,2,3,4} and R is the relation on set A, then find the antisymmetric relation on set A. The divisibility relation on the natural numbers is an important example of an anti-symmetric relation. That is to say, the following argument is valid. Formally, a binary relation R over a set X is symmetric if: ∀, ∈ (⇔). A relation is a set of ordered pairs, (x, y), such that x is related to y by some property or rule. Antisymmetric : Relation R of a set X becomes antisymmetric if (a, b) ∈ R and (b, a) ∈ R, which means a = b. An example of antisymmetric is: for a relation “is divisible by” which is the relation for ordered pairs in the set of integers. For relation, R, an ordered pair (x,y) can be found where x and y … The relation $$R$$ is said to be symmetric if the relation can go in both directions, that is, if $$x\,R\,y$$ implies $$y\,R\,x$$ for any $$x,y\in A$$. REFLEXIVE RELATION:IRREFLEXIVE RELATION, ANTISYMMETRIC RELATION Elementary Mathematics Formal Sciences Mathematics In that, there is no pair of distinct elements of A, each of which gets related by R to the other. In mathematics, a homogeneous relation R on set X is antisymmetric if there is no pair of distinct elements of X each of which is related by R to the other. Congruence modulo k is symmetric. Note - Asymmetric relation is the opposite of symmetric relation but not considered as equivalent to antisymmetric relation. symmetric, reflexive, and antisymmetric. Antisymmetric Relation Given a relation R on a set A we say that R is antisymmetric if and only if for all (a, b) ∈ R where a ≠ b we must have (b, a) ∉ R. This means the flipped ordered pair i.e. In this short video, we define what an Antisymmetric relation is and provide a number of examples. In this context, antisymmetry means that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m, then m cannot be a factor of n. For example, 12 is divisible by 4, but 4 is not divisible by 12. But, if a ≠ b, then (b, a) ∉ R, it’s like a one-way street. This list of fathers and sons and how they are related on the guest list is actually mathematical! Similarly, the subset order ⊆ on the subsets of any given set is antisymmetric: given two sets A and B, if every element in A also is in B and every element in B is also in A, then A and B must contain all the same elements and therefore be equal: A real-life example of a relation that is typically antisymmetric is "paid the restaurant bill of" (understood as restricted to a given occasion). For a relation R in set A Reflexive Relation is reflexive If (a, a) ∈ R for every a ∈ A Symmetric Relation is symmetric, If (a, b) ∈ R, then (b, a) ∈ R Transitive Relation is transitive, If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ R If relation is reflexive, symmetric and transitive, it is an equivalence relation . In mathematics, a homogeneous relation R on set X is antisymmetric if there is no pair of distinct elements of X each of which is related by R to the other. In mathematics, a relation is a set of ordered pairs, (x, y), such that x is from a set X, and y is from a set Y, where x is related to yby some property or rule. In this context, antisymmetry means that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m, then m cannot be a factor of n. For example, 12 is divisible by 4, but 4 is not divisible by 12. As a simple example, the divisibility order on the natural numbers is an antisymmetric relation. Examples of Relations and Their Properties. Partial and total orders are antisymmetric by definition. Example: { (1, 2) (2, 3), (2, 2) } is antisymmetric relation. That is: the relation ≤ on a set S forces Here's something interesting! If we let F be the set of all f… That is: the relation ≤ on a set S forces The standard example for an antisymmetric relation is the relation less than or equal to on the real number system. The divisibility relation on the natural numbers is an important example of an antisymmetric relation. A relation ℛ on A is antisymmetric iff ∀ x, y ∈ A, (x ℛ y ∧ y ℛ x) → (x = y). Another example of an antisymmetric relation would be the ≤ or the ≥ relation on the real numbers. Examples. Which is (i) Symmetric but neither reflexive nor transitive. (iv) Reflexive and transitive but … Antisymmetric Relation. In a set A, if one element less than the other, satisfies one relation, then the other element is not less than the first one. In set theory, the relation R is said to be antisymmetric on a set A, if xRy and yRx hold when x = y. i don't believe you do. In this article, we have focused on Symmetric and Antisymmetric Relations. The divisibility relation on the natural numbers is an important example of an antisymmetric relation. That means that unless x=y, both (x,y) and (y,x) cannot be elements of R simultaneously. (iii) Reflexive and symmetric but not transitive. A relation can be both symmetric and antisymmetric (in this case, it must be coreflexive), and there are relations which are neither symmetric nor antisymmetric (e.g., the "preys on" relation on biological species). So from total n 2 pairs, only n(n+1)/2 pairs will be chosen for symmetric relation. The usual order relation ≤ on the real numbers is antisymmetric: if for two real numbers x and y both inequalities x ≤ y and y ≤ x hold then x and y must be equal. Asymmetric Relation In discrete Maths, an asymmetric relation is just opposite to symmetric relation. the truth holds vacuously. so neither (2,1) nor (2,2) is in R, but we cannot conclude just from "non-membership" in R that the second coordinate isn't equal to the first. Proofs about relations There are some interesting generalizations that can be proved about the properties of relations. Typically some people pay their own bills, while others pay for their spouses or friends. Suppose that Riverview Elementary is having a father son picnic, where the fathers and sons sign a guest book when they arrive. And what antisymmetry means here is that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m , then m cannot be a factor of n . The Antisymmetric Property of Relations The antisymmetric property is defined by a conditional statement. It is … In this context, anti-symmetry means that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m , then m cannot be a factor of n . The relation $$R$$ is said to be antisymmetric if given any two distinct elements $$x$$ and $$y$$, either (i) $$x$$ and $$y$$ are not related in any way, or (ii) if $$x$$ and $$y$$ are related, they can only be related in one direction. A relation R is not antisymmetric if there exist x,y∈A such that (x,y) ∈ R and (y,x) ∈ R but x ≠ y. Antisymmetric: The relation is antisymmetric as whenever (a, b) and (b, a) ∈ R, we have a = b. Transitive: The relation is transitive as whenever (a, b) and (b, c) ∈ R, we have (a, c) ∈ R. Example: (4, 2) ∈ R and (2, 1) ∈ R, implies (4, 1) ∈ R. As the relation is reflexive, antisymmetric and transitive. Call it G. both can happen. This is called Antisymmetric Relation. Or it can be defined as, relation R is antisymmetric if either (x,y)∉R or (y,x)∉R whenever x ≠ y. In other words, the intersection of R and of its inverse relation R^ (-1), must be It is not necessary that if a relation is antisymmetric then it holds R(x,x) for any value of x, which is the property of reflexive relation. Another example of an antisymmetric relation would be the ≤ or the ≥ relation on the real numbers. (ii) Let R be a relation on the set N of natural numbers defined by 2006, S. C. Sharma, Metric Space, Discovery Publishing House, page 73, (i) The identity relation on a set A is an antisymmetric relation. Thus, it will be never the case that the other pair you're looking for is in $\sim$, and the relation will be antisymmetric because it can't not be antisymmetric, i.e. Equivalently, R is antisymmetric if and only if whenever R, and a b, R. In this context, antisymmetry means that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m, then m cannot be a factor of n. In discrete Maths, a relation is said to be antisymmetric relation for a binary relation R on a set A, if there is no pair of distinct or dissimilar elements of A, each of which is related by R to the other. Example 6: The relation "being acquainted with" on a set of people is symmetric. The definitions of the two given types of binary relations (irreflexive relation and antisymmetric relation), and the definition of the square of a binary relation, are reviewed. Both ordered pairs are in relation RR: 1. As long as no two people pay each other's bills, the relation is antisymmetric. In Matrix form, if a 12 is present in relation, then a 21 is also present in relation and As we know reflexive relation is part of symmetric relation. for example the relation R on the integers defined by aRb if a < b is anti-symmetric, but not reflexive. A relation that is antisymmetric is not the same as not symmetric. So from total n 2 pairs, only n(n+1)/2 pairs will be chosen for symmetric relation. If a relation is antisymmetric is not symmetric with '' on a set a {! Than ( < ), ( 2, 2 ) } is antisymmetric and irreflexive acquainted with '' on set! A guest book when they arrive as not symmetric divisor of… ” in set. 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# Calculating probability for three consecutive successes

Original Question:

On a TV news channel, the evening news starts at same time every day. The probability that Mr Li gets home from work in time to watch the news is $0.3$

In a particular week of five working days, what is the probability that Mr Li gets home in time to watch the news on three consecutive days?

Attempt:

Possible arrangements with three consecutive successes in 5 trials:

SSS..
FSSS.
FFSSS
SFSSS

I calculated the respective probabilities for these arrangements as follows:

$0.3^3 + 0.3^3 0.7 + 0.3^3 0.7^2+0.3^4 0.7 = 0.0648$

Actual answer (at the back of the book) = $0.05913$

My question is what I am doing wrong here (or is the book wrong?)

Also, I would really appreciate if someone could tell me a general way to solve the problems of this type where probability of $k$ consecutive success in $n$ trials is asked. In this question I was able to manually find the possible arrangements with consecutive successes but if the number of trials is high for example 200, how would I approach this problem then.

Thanks very much.

For a more general approach, consider the probability of not having 3 successes in a row out of $n$ trials; call this probability $P(n)$, and let's say the probability of a single success is $p$, with $q=1-p$. If we have $n$ trials, condition the probability on the number of successes at the end of the $n$ trials: the $n$ trials must end in $...F$, $...FS$, or $...FSS$, so we have the recursion

$$P(n) = q \; P(n-1) + pq \; P(n-2) + p^2q \; P(n-3) \qquad \text{for } n \ge 3$$ with $P(0) = P(1) = P(2) = 1$.

It's not hard to see how to extend this approach to longer strings of successes.

Let $$X_i=\begin{cases} S, & \text{with prob 0.3 } \\ F, & \text{with prob 0.7 } \\ \end{cases}$$ for $$i=1,2,3,4,5$$ be IID for the outcome of the i-th day. The probability you are asked is the following (if we are talking for exactly 3 days succes in a row)

P(Mr Li get home in time 3 days in a row)= $$P(X_1=X_2=X_3=S)+P(X_1=F,X_2=X_3=X_4=S)+P(X_1=X_2=F,X_3=X_4=X_5=S)= 0.3^3+0.3^30.7+0.3^30.7^2=0.05913$$

(which is the correct answer in the book)

• This is actually incorrect because it does not take into account $P(X_1=X_3=X_4=X_5=S, X_2=F)$ – Couchy311 May 31 at 2:47
• I agree with you technically, that's why I said "if we are talking for exactly 3 days S in a row"...it seems like the problem should have clarified that or give another answer in the back.. – sakas May 31 at 3:08
• I see, thanks for clarifying :) – Couchy311 May 31 at 3:11


# Is there a fast way to get the characteristic polynomial of this symmetric matrix?

I have to find the characteristic polynomial equation of this matrix

$$A= \begin{bmatrix}2 &1 &1&1 \\1&2&1&1\\1&1&2&1\\1&1&1&2 \end{bmatrix}$$

Is there another way than the rather long $\det(A-\lambda I)$ method ?

Maybe the fact that $A$ is symmetric ($A =A^t$) may be helpful ?

• Could I do something like $A-I\lambda= \begin{bmatrix}2-\lambda &1 &1&1 \\1&2-\lambda&1&1\\1&1&2-\lambda&1\\1&1&1&2-\lambda \end{bmatrix}$ then simplify and calculate the $det(...)$ , or would it change the value of$\lambda$? – HowCanIHelpYou Jan 20 '17 at 18:35
• Subtract the identity matrix and then it becomes pretty easy to show that for $A-I$ the eigenvalues will be 4 and zero (with multiplicity 3) – Sergei Golovan Jan 20 '17 at 18:35
• Interestingly, it seems that the characteristic polynomial of the $n\times n$ matrix with $2$ for each diagonal entry and $1$ for all other entries is $(n+1-\lambda)(1-\lambda)^{n-1}$. This suggests an inductive approach to me. – MPW Jan 20 '17 at 19:00
• Clearly, the eigenvalues of $A$ and $A-I$ differ. Exactly by a unity each. – Sergei Golovan Jan 20 '17 at 19:05
• And it seems that the characteristic polynomial of the $n\times n$ matrix with $a$ for each diagonal entry and $b$ for all other entries is $(a+(n-1)b - \lambda)(a-b-\lambda)^{n-1}$. – MPW Jan 20 '17 at 19:15

There's a formula for determinants of block matrices of the form $\begin{bmatrix} A&B\\B&A\end{bmatrix}$, where $A$ and $B$ are square matrices of the same size: $$\det\begin{bmatrix} A&B\\B&A\end{bmatrix}=\det(A-B)\det(A+B).$$ Applying this formula , we obtain \begin{align}\det(A-\lambda I)&=\begin{vmatrix}2-\lambda&1&1&1\\1&2-\lambda&1&1\\1&1&2-\lambda&1\\1&1&1&2-\lambda \end{vmatrix}=\begin{vmatrix}1-\lambda&0\\0&1-\lambda \end{vmatrix}\cdot\begin{vmatrix}3-\lambda&2\\2&3-\lambda \end{vmatrix}\\ &=(1-\lambda)^2\Bigl[(3-\lambda)^2-4\Bigr]=(\lambda-1)^3(\lambda-5).\end{align}

Here is a matrix in which the columns are eigenvectors of your matrix. Indeed, the columns are pairwise orthogonal.

$$\left( \begin{array}{rrrr} 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 2 & -1 \\ 1 & 0 & 0 & 3 \end{array} \right).$$ It is not an orthogonal matrix, as the columns are of different lengths. However, it can be made orthogonal by dividing each column by its Euclidean length, those being $2, \sqrt 2, \sqrt 6, \sqrt {12}$

The best, and very short way:

First step. Subtract to the rows $2$, $3$ and $4$, the first one.
Second step. Add to the first column the sum of the columns $2$, $3$ and $4.$

We get a triangular determinant.

EDIT: $$\begin{vmatrix}2-\lambda &1 &1&1 \\1&2-\lambda&1&1\\1&1&2-\lambda&1\\1&1&1&2-\lambda \end{vmatrix}\underbrace{=}_{\text{First step}}\begin{vmatrix}2-\lambda &1 &1&1 \\-1+\lambda&1-\lambda&0&0\\-1+\lambda&0&1-\lambda&0\\-1+\lambda&0&0&1-\lambda \end{vmatrix}$$ $$\underbrace{=}_{\text{Second step}}\begin{vmatrix}5-\lambda &1 &1&1 \\0&1-\lambda&0&0\\0&0&1-\lambda&0\\0&0&0&1-\lambda \end{vmatrix}=(5-\lambda)(1-\lambda)^3.$$

There are some simple tricks that you can use. The eigenvalues of $A$ are those values of $\lambda$ such that $A-\lambda I$ is singular, i.e., has rank less than four. Since $$A-1\cdot I=\pmatrix{ 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1}$$ has rank equal to one, it follows that $\lambda=1$ is an eigenvalue of $A$ of multiplicity $4-1=3$. Hence $(\lambda-1)^3$ is a factor in the characteristic polynomial.

In the matrix $$A-5\cdot I=\pmatrix{-3 & 1 & 1 & 1\cr 1 & -3 & 1 & 1 \cr 1 & 1 & -3 & 1 \cr 1 & 1 & 1 & -3},$$ the sum of the elements in each row is zero. Hence $A-5\cdot I$ is singular, and $\lambda-5$ is another factor in the characteristic polynomial of $A$.

It follows that $\det(A-\lambda I = (\lambda-5)(\lambda-1)^3$.


Math Help - Double-check this limit please

Can someone double-check my work here? Thanks, if you can.

$
\lim_{x\to\infty}[x(1-cos(1/x))]
$

Put x to the denominator to use L'Hopital's rule, etc. Applied chain rules, etc.

$
\lim_{x\to\infty}[sin(1/x)]
$

Thus, if x goes to infinity, the above is sin(0) = 0. Final Answer: Zero

2. Yes, it's okay, but you don't need that rule, you can turn that limit into a known one with a simple substitution.

3. Oh, really? What is this substitution?

4. $t=\frac1x.$

5. tx=1

so itequals to lim [(1-cost)/t]
t->0


## Spring 2016, problem 9

The numbers $1, 2, \dots , n$ are written around a circle in some order. What is the smallest and largest possible sum of the absolute differences of adjacent numbers?

1 year ago

The maximum is $\lfloor \frac{n^2}{2} \rfloor$. The minimum is $2 (n - 1)$.

Proof of the maximality part:

For a circular arrangement of $1..n$ call the sum of the absolute differences of neighbours the value of the arrangement. For $a, b \in \left \{1..n \right \}$ write $a \rightarrow b$, if $b$ is the right side neighbour of $a$.

If $n$ is even, call the numbers $\frac{n}{2} .. n$ the large numbers. Call the rest small. If $n$ is odd call the numbers $\frac{n + 3}{2} .. n$ the large numbers. Call $\frac{n + 1}{2}$ the middle number. Call the rest small.

The value $v$ of the arrangement is given by:

$v := \sum_{a \rightarrow b} \left [max(a,b) - min(a,b) \right ] = \sum_{a \rightarrow b} max(a,b) - \sum_{a \rightarrow b} min(a,b)$.

Each of the two sums in the last expression is a sum of $n$ numbers in the range $1..n$, none of which occur more than twice in any of the sums. The value $v$ is clearly no larger than $m$, given by: $m := \left [ n + n + (n -1) + (n-1) + ... \right ] - \left [1 +1 + 2 + 2 + 3 + 3 ... \right ]$, where the first and the second bracket contain $n$ terms each. By a simple calculation $m = \lfloor \frac{n^2}{2} \rfloor$. But the value $m$ is indeed attained:

For $n$ even arrange the numbers like:

large $\rightarrow$ small $\rightarrow$ large $\rightarrow$ small ... $\rightarrow$ large$\rightarrow$ small

For $n$ odd arrange like:

middle $\rightarrow$ large $\rightarrow$ small $\rightarrow$ large $\rightarrow$ small ... $\rightarrow$ large$\rightarrow$ small. Done.

Proof of the minimality part:

Proof by induction over $n$: The induction start is trivial. So assume, that the minimum value for arrangements of numbers in $1..n$ is given by $m_n := 2 (n -1)$. Consider a minimum-value arrangement $A_{n+1}$ of the numbers $1..(n+1)$. with value $v_{n+1}$. Now take away the number $n+1$ from $A_ {n+1}$. You get an arrangement $A_n$ of the numers $1..n$, with value $v_{n}$. By an easy calculation $v _n \leq v_{n+1} -2$. Thus $v_{n+1}$ cannot be smaller than $2 (n + 1 -1)$ because otherwise $v_n$ would be smaller than $2 (n - 1)$. The minimum can also be no larger than $2 (n + 1 -1)$, since this value is attained for simply placing the numbers in ascending order alongthe circle. Done.

Hello, I get a slightly different answer for the maximum. It must be a whole number.

I get,

maximum = trunc(x^2/2),

i.e., if n is even, maximum = n^2/2

if n is odd, maximum = n^2/2 - 1/2

Jao

cgjoa3 1 year ago

Hi Jao. The notation $\lfloor \cdot \rfloor$ I used is defined by:

$\lfloor x \rfloor := floor(x)$.

For $x> 0$: $floor(x) =trunc(x)$

Nelix 1 year ago
1 year ago

let start with number 1, minimum possible difference for next number is 1 so next number 2 and so on so minimum sum = n

for maximum::

let start with 1 and for max difference next number should be n, number next to n such that max. difference is there should be 2

so the series is 1,n,2,n-1,3,n-2,4,n-3....., and last n/2 or (n-1)/2 for odd and even n

so max = nn/2 or (nn-1)/2

1 year ago

Are these called purdue problems of week or nelix's math proofs of the week?! Jk wish more mathletes would participate !

' half the problem for half the points ' ( lion's share had this been a putnam )Minimality of 2(n-1) Now we will show that $2n-2$ is minimal. To do so, remark that $1$ and $n$ must be on the circle. By the Triangle Inequality, the sum of the positive differences along the minor arc between $1$ and $n$ must be at least $n-1$ and similarly for the otherrh arc


# What are the odds of flipping a coin 100 times and seeing HHHHT? [closed]

What are the odds of flipping a coin 100 times and seeing exactly four consecutive heads? Any more than four heads in a row, such as "HHHHH" would not be considered a string of four consecutive heads. Seeing 10 sets of "HHHHT" would allow a max of 20 consecutive patterns. How would you expect to find the number of times an isolated string of exactly 4 heads in a row in $n$ coin flips?

## closed as off-topic by Did, Michael Albanese, Edward Jiang, Zain Patel, Chill2MachtJul 25 '16 at 0:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Michael Albanese, Edward Jiang, Zain Patel, Chill2Macht
If this question can be reworded to fit the rules in the help center, please edit the question.

• I don't understand, you seem to ask many questions at once. – Jorge Fernández Hidalgo Jul 24 '16 at 17:48
• Is your question the expected number of times you would expect to find $HHHHT$ in a sequence of $100$ flips? – Jorge Fernández Hidalgo Jul 24 '16 at 17:49
• I flip a coin 100 times. How many times should I see HHHH? A HHHHH or anything greater will not be considered as a HHHH. – Triumph Jul 24 '16 at 17:51
• No just HHHHT but THHHHT – Triumph Jul 24 '16 at 17:52
• I am pretty sure the question is "Suppose I flip a coin 100 times. How many subsequences of exactly four heads in a row (no more or less) should I expect to see?" – user326210 Jul 24 '16 at 17:54

In any sequence of $100$ flips, “mark” each spot $HHHH$ occurs (and occurs not within a longer run of $H$’s) between the middle $H$’s. For example, the sequence $$HHTHHHTTTHHHHTHT\dots TTHHHHHTHTHHTTHHHH$$ would be marked $${HHTHHHTTTHH}^{\color{red}|\!}{HHTHT\dots TTHHHHHTHTHHTTHH}^{\color{red}|\!}{HH}^{}.$$

Among the $97$ positions where marks could occur (from after the first two flips to before the final two), a mark does appear in each of the $2^{\rm nd}$ through $96^{\rm th}$ ($95$ in all) of those spots (as $THH^{\color{red}|\!}HHT$) with probability $\frac{1}{2^6}$ (the chance the surrounding sequence of six flips is $THHHHT$). In the first and last positions, the probability is $\frac{1}{2^5}$ (the chance of $HH^{\color{red}|\!}HHT$ for position $1$ and $THH^{\color{red}|\!}HH$ for position $97$).

By the linearity of expectation, the expected number of marks is $95\cdot\frac1{2^6}+2\cdot\frac1{2^5}=\frac{99}{64}$.

[Added] It may be easier to focus on the $H$’s and $T$’s instead of the spaces between them, so another way to mark the runs-of-$4$ is to color in red the last $H$ of each run, like $$HHTHHHTTTHHH\color{red}HTHT\dots TTHHHHHTHTHHTTHHH\color{red}H.$$ The first three $H$’s are colored red with probability $0$, the fourth with probability $\frac{1}{2^5}$, the next $95$ with probability $\frac{1}{2^6}$, and the last with probability $\frac{1}{2^5}$, giving the same result.

• Is it possible for you to let me know what the answer would be for 8 out of 100? – Triumph Jul 27 '16 at 1:48
• Try to modify my answer for runs of $8$. There are no new situations that come up, so it shouldn’t be very hard. – Steve Kass Jul 27 '16 at 18:32

This seems to fall easily to linearity of expectation. Let $a_1,a_2\dots a_n$ be your sequence of outcomes.

Let $X_1$ be the indicator variable that $a_1,a_2,a_3,a_4=H$ and $a_5=T$

Let $X_{97}$ be the indicator variable that $a_{96}=T$ and $a_{97},a_{98},a_{99},a_{100}=H$

Finally, for $2\leq n \leq 96$ let $X_n$ be the indicator that $a_{n-1}=T,a_n,a_{n+1},a_{n+2},a_{n+3}=H,a_{n+4}=T$

The random variable you want is $\sum_{i=1}^{97} X_n$.

By linearity of expectation its expectation is $E[X_1]+95E[X_2]+E[X_{97}]=\frac{1}{32}+\frac{95}{64}+\frac{1}{32}=\frac{99}{64}$

• What I answered is the expected number of times you can find a substring of the form $HHHH$ that does not have another $H$ to either side. Assuming the flips are independent, and each outcome has probability $\frac{1}{2}$. – Jorge Fernández Hidalgo Jul 24 '16 at 18:16
• When they are four $H$ it is $\frac{99}{64}$ and when there are three It is $\frac{1}{16}+\frac{96}{32}+\frac{1}{16}$. – Jorge Fernández Hidalgo Jul 24 '16 at 18:45
• then it is $\frac{1}{2^{11}}+\frac{89}{2^{12}}+\frac{1}{2^{11}}$ – Jorge Fernández Hidalgo Jul 24 '16 at 19:31
• if you throw $10000$ times then you should get $10$ consecutive (and not more) an expected number of $\frac{1}{2^{11}}+\frac{9989}{2^{12}}+\frac{1}{2^{11}}$ – Jorge Fernández Hidalgo Jul 24 '16 at 19:52
• I told you, $\frac{1}{2^{11}}+\frac{9989}{2^{12}}+\frac{1}{2^{11}}$, which is approximately $2.43$ – Jorge Fernández Hidalgo Jul 24 '16 at 21:10

For $i=1$ to $97$, let random variable $X_i$ be equal to $1$ if there is a string of HHHH that begins at $i$ and does not extend, and let $X_i=0$ otherwise. We want the expectation of $X_1+\cdots+X_{97}$. By the linearity of expectation, this is $E(X_1)+\cdots+E(X_{97})$.

So we need only calculate the $\Pr(X_i)=1$.

These are not all equal. If $i=1$ or $i=97$, we have $\Pr(X_i=1)=\frac{1}{32}$. For all the other $i$, we have $\Pr(X_i=1)=\frac{1}{64}$. That is because in all but "end" cases, the string of $4$ H must be flanked by T on both sides.

The required expectation is therefore $2\cdot \frac{1}{32}+95\cdot\frac{1}{64}$.

• Would this factor in the possibility of 10 heads in a row that would change you 100 coin flips into 90. From the start you should automatically drop the 100 flips down to 50 with the assumption that the other 50 flips would be tails correct? What are the odds of 5 heads or 6 heads each one of those will take more flips away from what you need to achieve 4 heads. – Triumph Jul 24 '16 at 18:14
• The expectation calculation (indirectly) takes care of $10$ heads in a row by not counting them, because then $X_i=0$ for $i=1$ to at least $11$. – André Nicolas Jul 24 '16 at 18:17


# Simple Probability Question: Perfect quiz score

a) Students have two quizzes that they must take in their undergraduate studies unit. The probability of getting a perfect score on the first quiz is 68%, the probability of getting a perfect score on the second quiz is 74%, and the probability of falling short of a perfect score on both quizzes is 19%. What is the probability of scoring a perfect score on both quizzes?

My attempt: Pr(Scoring a perfect score on both quizes) $= 0.68 \cdot 0.74 = 0.5032$

b) Refer to the probabilities described in question a. If a student has achieved a perfect score on the first quiz, what is the conditional probability of getting a perfect score on the second quiz?

My attempt:

$$\frac{\text{Pr(Scoring a perfect score on both quizzes)}}{\text{Pr(Perfect score on second Quiz)}} = 0.68$$

but I feel like there is more to both questions

• what has been tried ? what is the context ( where did you find the question) ? – user451844 Aug 19 '17 at 14:22
• a) Pr(Scoring a perfect score on both quizes) = 0.68 * 0.74 = 0.5032 b) Pr(Scoring a perfect score on both quizes)/Pr(Perfect score on second Quiz) = 0.68 but I feel like there is more to both questions – user473207 Aug 19 '17 at 14:29
• if you fall short 19% of the cases, what happens the other 81% of the cases ? – user451844 Aug 19 '17 at 14:44
• @user473207 Your calculations assume that the two events (obtainoing a perfect score on quiz 1 and obtaining a perfect score on quiz 2) are independent. BUt just conceptually, that is unlikely: SOmeone who gets a perfect score on quiz 1 is probably more likely to get a perfect score on quiz 2 than someone who does not get a perfect score on quiz 1 ... so you are right to suspect there is more to this! – Bram28 Aug 19 '17 at 14:47
• @user473207 Also, the expression 'falling short of a perfect score on both quizzes'' is a bit ambiguous (is it that you don't get a perfect score on quiz 1 and also don't get a perfect score on quiz 2, or is it that it is not true that you do perfect on both quizzes (so you can get a perfect score on one of the two quizzes, just not on both quizzes)). What they undoubtedly mean is the first interpretation. SO, you know that there is a 19% chance of not getting a perfect score on quiz 1 and also not a perfect score on quiz 2. NOw, as Roddy asked: what does the other 81% then represent? – Bram28 Aug 19 '17 at 14:49

Let $A$ be the event that a perfect score is obtained on the first quiz; let $B$ be the event that a perfect score is obtained on the second quiz. We are given \begin{align*} P(A) & = 0.68\\ P(B) & = 0.74\\ P(A^C \cap B^C) & = 0.19 \end{align*} Hence, \begin{align*} P(A^C) & = 1 - P(A) = 0.32\\ P(B^C) & = 1 - P(B) = 0.74 \end{align*} The probability of not obtaining a perfect score on at least one of the quizzes is $$P(A^C \cup B^C) = P(A^C) + P(B^C) - P(A^C \cap B^C) = 0.32 + 0.26 - 0.19 = 0.39$$ Thus, the probability of obtaining a perfect score on both quizzes is $$P(A \cap B) = 1 - P(A^C \cup B^C) = 1 - 0.39 = 0.61$$ so your assumption that the probabilities of obtaining a perfect score on each quiz are independent was false.
The probability that a student who received a perfect mark on the first quiz obtains a perfect mark on the second quiz is $$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.61}{0.68} = \frac{61}{68}$$


# Math Help - Eigenvectors!

1. ## Eigenvectors!

(i) Determine all the eigenvalues of A
(ii) For each eigenvalue of A, find the set of eigenvectors corresponding to

A = ( 1 2 )
( 3 2 )

I found eigenvalues of A to be 4 and -1.

I also found the eigenvectors to be (2/3,1) for =4 and (-1,1) for =-1

BUT the solution in the back of the book says (2,3) for =4 and (1,-1) for =1

I'm soo confusedd! Can someone tell me what's wrongg? Am I calculating the eigenvectors wrong?

Here's how i calculate eigenvector for =4

A-4I = (-3 2)
(3 -2)

then (-3 2 |0)
(3 -2 |0)

and i end up with x1 -(2/3) x2 =0. so x2 = t, and x1 = (2/3)t. eigenvector = (2/3,1) but book says (2,3)

and for =-1

A-(-1)I = (2 3 )
(2 3 )

then (2 3 |0)
(2 3 |0)

and i end up with x1+x2=0. x2= t and x1 = -t. eigenvector = (-1,1) but book says (1,-1)

2. You don't have a problem. Note that your eigenvectors are just multiples of the book's answer. This is fine!

3. Hi
$\texttt{det}(A-I_{2}X)=\texttt{det}\begin{pmatrix}
1-X & 2\\
3 & 2-X
\end{pmatrix}=(1-X)(2-X)-6=X^2-3X-4=(X-4)(X+1)$

and thus your eigenvalues are 4 and -1.
For 4
$A\begin{pmatrix}
x\\
y
\end{pmatrix}=4\begin{pmatrix}
x\\
y
\end{pmatrix}\Leftrightarrow \begin{pmatrix}
x+2y\\
3x+2y
\end{pmatrix}=$
$\begin{pmatrix}
4x\\
4y
\end{pmatrix}\Leftrightarrow \left\{\begin{matrix}
-3x+2y=0\\
3x-2y=0
\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x=\frac{2}{3}t\\
y=t,t\in \mathbb{R}
\end{matrix}\right.$

and thus the eigenvectors are of the form,
$x=x_1(\frac{2}{3},1),x_1\in \mathbb{R}\setminus \left \{ 0 \right \}$

4. The set of all eigenvectors of a matrix A, corresponding to a given eigenvalue, $\lambda$, form a subspace. In particular, any multiple of an eigenvector is also an eigenvector, corresponding to the same eigenvalue: If $Av= \lambda v$ and "r" is any scalar, then $A(rv)= r(Av)= r(\lambda v)= (r\lambda)v= \lambda (rv)$.