# Which payoff do you want to go for?

Payoff 1.

Toss 5 coins. You get $1 for each consecutive HT that you get. Payoff 2. Toss 5 coins. You get$1 for each consecutive HH that you get.

For example, if you tossed HTHHH, under payoff 1 you will get $1, but under payoff 2 you will get$2.

Note by Calvin Lin
3 years, 2 months ago

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This is a trick question, both payoffs have same expected value of profit, which is: $1. Consider this: We will count the number of ways HT can occur. We can have: HT_ _ _ : 8 ways _ HT _ _ _ : 8 ways _ _ HT _ : 8 ways _ _ _ HT : 8 ways Total number of ways: 32 Similarly for HH, the total number of ways in which it can occur is 32. What this means is that the sum of the payoffs of HT and HH over the 32 possible outcomes is$32 for both. The payoffs may be distributed in different ways for HT and HH, but their expected value is same.

Mathematically, the expected value is same. But there is one more thing left to do: analyze the probability distribution. We can easily see that payoff 2 might be desirable because it tends to give large profits as compared to payoff 1 for certain events. eg: HHHHH gives $4 in p2 and 0 in p1. HHHHT gives$3 in p2 and $1 in p1. But the thing is that these events are not very common. Using method of reflection and analyzing the 16 possibilities , we see that there are many events for which p2 gives$0 payoff. There are 13 outcomes for which p2 gives $0 payoff, but only 6 where p1 gives$0 payoff.

So If I'm given this choice, I'd take payoff 1, as then I have more chance of leaving with at least a dollar in my pocket.

Since this is in quantitative finance, should I talk about risk and stuff? I do not know.

- 3 years, 2 months ago

Yes, the expected payoff for both scenarios is $1. The easiest way to see this is to look at Indicator Variables, which is essentially what Raghav did (though not phrased in that language). Well, everyone has their own "payoff preference". For example, you stated that you would prefer to "be more likely to leave with at least a dollar in your pocket". As such, this tells me that you are very risk adverse, and that you would prefer the certainty of a positive payoff. If you want to talk about "risk and stuff", what do you need to consider, and why? Staff - 3 years, 2 months ago Log in to reply I am not formally educated in these topics, so I'm just saying all this from a logical standpoint. After a bit of thinking, I think I agree with you on the fact that everyone has their own "payoff preference". One can take a risk to win more, or take less risk to win less. In the scenario mentioned here, the aspects of risk are not conspicuous. The amount to be won is not significant, and there is no penalty for losing. Hence, both the payoffs are inherently attractive offers. When we come to real life situations, I think there are many more things that contribute to the risk: 1. The probability that you will get a payoff. 2. The probability that you stand to lose money. 3. The security of the winnings. Money once won shouldn't be taken away from you. 4. The effects of said decisions on long term/ on your ability to take other decisions. According to me, every decision of ours is based on weighing in these risk factors against the probable prize. The balance may tip either way, and so may our decisions. I don't even know if I'm thinking in the right direction... Help me out here @Calvin Lin - 3 years, 2 months ago Log in to reply Part of this series of posts is for you to figure out what kind of risk appetite you have. It is a personal preference, and there isn't necessarily a "right" answer. Your understanding will help guide you in decisions that you make in future. For the points that you raised, you should answer for yourself if those should matter, and why. Hopefully, you will come up with a consistent, logical risk-reward structure. For example, if you would choose payoff 1 over 2 and payoff 3 over 4, but would prefer a combined 2 and 4 over a combined 1 and 3, then it would be very easy for someone to sell you things in part, and make you overpay for them. Staff - 3 years, 2 months ago Log in to reply I think that there may be some double-counting here. For example, you've counted HTHTT and HTHTH twice each. There are only $$2^{5} = 32$$ possible sequences in total, and there are some, such as TTHHH, that have no HT payouts, so there cannot be $$32$$ HT outcomes. I'm getting an average payout of $$\dfrac{28}{32}$$ dollars in the first game type and $$\dfrac{31}{32}$$ dollars in the second, but I am too tired to double-check. I'll do that tomorrow. :) Edit: Oh. now I see what you've done; very clever. I must have missed out on my count somewhere. I'll still wait to confirm in the morning. And yes, there does seem to be an advantage to choosing p1, (even though the expected winnings are the same), in the sense that you are more likely to at least win some money. - 3 years, 2 months ago Log in to reply Is "more likely to at least win some money" the main consideration that you will use if the expected payoff is equal? For example,what would you choose between: Payoff 1: 50% of$0, 50% of $100 Payoff 2: 90% of$1.01, 1% of $4900 Staff - 3 years, 2 months ago Log in to reply If Payoff 2 was 99% of$40.40, 1% of $1000, (to give it the same expected payoff as Payoff 1), I would go with Payoff 2 since I would be assured of money and would at least have a slim chance of a large amount of money. If Payoff 2 was 99% of$0, 1% of $5000, then it gets interesting, (at least for me). Even a such a slight chance of winning$5000 would be hard to pass up, so I would probably go with Payoff 2, unless I absolutely needed that $100 right away. If Payoff 2 was 99.9% of 0$, 0.1% of$50000 .... hmmmm .... That's a lot of dough, so I'd go with Payoff 2. However, if the options were these last two I've listed, then I might choose the$5000 option.

If Payoff 2 involved a 50% chance of losing $100 and a 50% chance of winning$200, then I would go with Payoff 1, since I'm not comfortable with there being such a good chance of losing a fair bit of money. There is a lot of psychology going on in these choices, and yet the expected winnings is always $50, (which ironically is an amount that would never actually be won in any of these scenarios). - 3 years, 2 months ago Log in to reply (Ooops, apparently I can't do maths. Fixed.) So, there are multiple ideas here. - Is the probability of a positive payoff so important that just because it moves from$0 it would greatly influence your preferences?
- At what point does the potential promise of a huge payoff overweigh the certainty of a small payoff?

Staff - 3 years, 2 months ago

In general I guess it all depends on the specific values, and further, on each person's financial status and willingness to take on risk. For me, the potential, (1% or better), of a large payout would outweigh any guaranteed amount less than $100, so in this case there is no difference to me between$0 and $100. But if the guaranteed amount were, say,$5000, with a 0.01% all-or-nothing chance of winning $1,000,000, I'd probably take the$5000 and run, (although I would be tempted to take the risk, at least for a moment). If the all-or-noting percentage were 1%, though, then I would find it hard not to take the chance; $5000 is a lot of money, but$1,000,000 is a life-changing amount of money, and a 1% chance is realistic in my eyes given the potential reward. The question I would ask myself is: which will I regret more - not taking the guaranteed money or not taking the risk?

- 3 years, 2 months ago

@Calvin Lin sir, Am I right?

- 3 years, 2 months ago

There are 2^5 outcoms of 5 coin tosses.

For payoff 1, there are this many ways to toss a HT. There are {HTHTT,HTHTH,HTTHT,HTHHT,THTHT, HHTHT,HTHHH,HHHTH,HHHHT, HTTTT,THTTT,TTHTT,TTTHT,HTTTH,HHTTT, HHTTH} Expected paayyoff 1 = (6/32$2) + (10/32$1)= $11/16=$22/32

For payoff 2, there are this many ways to toss a HH( or HHH,HHHH and HHHHHH). There are {HHHHH, THHHH,HHHHT, THHHT,HTHHH,HHHTH, HHHTT, TTHHH, HHTHH, HHTTT, THHTT, TTHHT, TTTHH, THTHH, THHTH] Expected payoff 2 =(1/32$4 )+ ( 2/32$3) + ( 5/32$2) + (7/32$1)=$3/4=$27/32

Payoff 2 is better, I guess LOL @Calvin Lin

Total possible outcomes of 32 of which there is one combination [TTTTT] which has no payoff.

- 3 years, 2 months ago

Check your calculations. The expected payoff of both scenarios is 1. Staff - 3 years, 2 months ago Log in to reply Let $$X$$ be an indicator variable which takes on value 1 for each $$HT$$ pair and 0 otherwise, and let $$Y$$ be an indicator variable which takes on value 1 for each $$HH$$ pair and 0 otherwise. There are $$5 - 2 + 1 = 4$$ indicator variables (both $$X$$ and $$Y$$) in 5 coin tosses. The expected value of the payoff is: $\text{E}\left [\sum_{k = 1}^{4} X_{k} \right ], ~ \text{E}\left [\sum_{k = 1}^{4} Y_{k} \right ].$We can apply Linearity of Expectation here, which holds even for dependent events (nice proof of this is given in Brilliant Wiki Page), to get: $\sum_{k = 1}^{4}\text{E}\left [ X_{k} \right ] , ~ \sum_{k = 1}^{4}\text{E}\left [ Y_{k} \right ].$Expected value of both $$X_{k}$$ and $$Y_{k}$$ is: $\text{E}\left [ X_{k} \right ] = \text{E}\left [ Y_{k} \right ] = 1 \cdot \dfrac{1}{4} + 0 \cdot \dfrac{3}{4} = \dfrac{1}{4}.$It follows that expected payoffs in both cases are the same and they equal:$\sum_{k = 1}^{4}\text{E}\left [ X_{k} \right ] = \sum_{k = 1}^{4}\text{E}\left [ Y_{k} \right ] = 4 \cdot \dfrac{1}{4} = 1.$ Bonus: What about the variance of the payoff? It can also influence our decision. Is it also the same? It turns out they are not! Let us first consider payoff 1 and indicator variable $$X$$:\begin{align} \text{Var}[X] &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}\right)^{2}\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}^{2}\right) + \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}^{2}\right)\right] + \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= 1 + \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - 1 \\ &= \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ].\end{align}The same is true for payoff 2 and $$Y$$. Now, let us calculate $$\text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ]$$. We make two groups of $$X_{k}X_{j}$$: • $$\left | k-j \right | = 1$$ representing two consecutive indicator variables which share one common coin. Notice that their product is always 0 since it's impossible to have two $$HT$$s in 3 consecutive coins. • $$\left | k-j \right | > 1$$ representing two non-consecutive indicator variables which share no common coin. Their product is non-zero only when they are both non-zero which happens with probability $$\left(\frac{1}{4}\right)^{2} = \dfrac{1}{16}.$$ There are total of $$4 \cdot 3 = 12$$ $$\left(X_{k}X_{j}\right)$$ pairs, and $$2\cdot 3$$ of them which are consecutive ie. where $$\left | k-j \right | = 1$$. Hence, we have: $\text{Var}[X] = \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] = 0 \cdot 6 + \frac{1}{16} \cdot 6 = 0.375.$ In the same way, we calculate $\text{Var}[Y] = \text{E}\left[ \left(\sum_{k \neq j}Y_{k}Y_{j}\right)\right ] = \frac{1}{8} \cdot 6 + \frac{1}{16} \cdot 6 = 1.125.$ So, variance of the payoff 2 is 3 times greater than variance of the payoff 1. If my reasoning is correct, then payoff 1 is somewhat safer option and it tends to the expected value, while payoff 2 is more riskier but it offers better chances of earning more than 1. Am I right @Calvin Lin?

- 1 month ago

I would go for Payoff 2 , because For payoff 1 ..... one can get maximum number of $1 is 2 which equals$2 i.e for this outcome HTTHT ( MAXIMUM POSSIBLE 'HT' consecutive is 2). For payoff 2 ..... one can get maximum number of $1 is 4 which equals$4 i.e for this outcome HHHHH ( MAXIMUM POSSIBLE 'HH' consecutive is 4).

- 2 months, 2 weeks ago

From a statistical standpoint, I would choose payoff option 1. As the number of H tossed increases, the probability of the consecutively tossing another H decreases. That being said, the HH combination can lead to a higher payoff because HTHTH only results in $2 while HHHHH results in$5. The issue lies in the decreased probability of continuing to toss H. This is a good example related to risk, much like in the decision involved in buying a AAA bond earning 5% vs a B bond earning 13%.

- 1 year, 11 months ago

To be more specific, payoff 1 would be like the AAA bond and payoff 2 like the B bond. It just depends on what kind of risk is right for you.

- 1 year, 11 months ago

Not quite the same comparison. Note that the expected payoff in both methods are the same.

Whereas in the bond example that you gave, we are trading expected payoff for certainty.

I also strongly disagree with "As the number of H tossed increases, the probability of the consecutively tossing another H decreases". The coin tosses are independent, and that is a common fallacy that "the proportion of realized events must be equal / close to the calculated probability"

Staff - 1 year, 11 months ago

I remember reading about an interesting experiment done by a French mathematician regarding payoffs and expected utility from lotteries. Allais Paradox?

- 3 years, 2 months ago

That's interesting. In experiment 1 here, taking a risk, (even if it is only 1%), means the possibility of losing a guaranteed million dollars; I would deeply regret gambling and ending up losing that money, but if I took the million and played 1B just to see what happened and found that I could have won 5 million, I would have just said "Oh well" and still been perfectly happy with my million. But if there is no guaranteed money, then having a chance at 5 million at the expense of a 1% greater chance of winning a million seems worth the risk.

Experiment 1 makes me think of the old saying, "A bird in the hand is worth two in the bush." If there are no "birds in hand", however, then the risk evaluation process is quite different.

- 3 years, 2 months ago

Let Ii be an indicator rv for an event of getting H at ith toss. Let X be a rv which is how many times HT occured. Similarly, let Y be the same but for HH. Then, X = I1I2 +... +I4I5 and Y = I1(1-I2) +...+I4(1-I5). We need to compare EX and EY. Given EIi = 0.5, we can apply linearity of expectation (even for multiplication since events are independent) and get 1 for both expectations.

- 2 years, 7 months ago

So, if their expectations are the same, does that mean that you are indifferent about which payoff to take? If so, why?

Staff - 2 years, 7 months ago

Yes, because with both payoffs, one can win the same amount of dollars on average.

- 2 years, 7 months ago

So, are you indifferent between these 2 payoffs:

Payoff 3: Always get $0 Payoff 4: 50% chance of -$1,000,000,000 and 50% chance of \$1,000,000,000

Staff - 2 years, 7 months ago

I think we want the choice with the minimum variance involved.

Maximin Principle of Rationality: An epistemically rational person maximizes his minimum payoff

Staff - 2 years, 7 months ago

If I don't have a million dollars at all, I will definitely not take payoff 4. I would take payoff 4 only if I was a millionaire. So, psychologically I am not indifferent.

- 2 years, 7 months ago

Right, so there are other considerations that come into play. For most people, the certainty of a result is preferred to an uncertain result (though with equal expected value). In general, having a low variance is better (though of course, there are exceptions).

Staff - 2 years, 7 months ago

I think this is a combinatorics problem?!

Staff - 3 years, 2 months ago

In the sense that arithmetic is a part of algebra, which is a part of calculus? There are numerous cross-disciplinary ideas. Using combinatorics will only get you one perspective of things.

Notice that I am essentially getting at the risk-reward preference, which is a "finance idea" as opposed to a "combinatorics idea". Yes, combinatorics ideas like expected value is involved, but they don't tell the full story.

Staff - 3 years, 2 months ago


# rank of a matrix

We prove that column rank is equal to row rank. The rank of a matrix is the largest number of linearly independent rows/columns of the matrix. Ask a Question . The rank of A is equal to the dimension of the column space of A. This lesson introduces the concept of matrix rank and explains how the rank of a matrix is revealed by its echelon form.. To define rank, we require the notions of submatrix and minor of a matrix. In mathematics, low-rank approximation is a minimization problem, in which the cost function measures the fit between a given matrix … the matrix in example 1 has rank 2. So maximum rank is m at the most. Guide. The idea is based on conversion to Row echelon form. The rank of a matrix would be zero only if the matrix had no non-zero elements. Each matrix is line equivalent to itself. Recent rank-of-matrix Questions and Answers on Easycalculation Discussion . Or, you could say it's the number of vectors in the basis for the column space of A. Prove that rank(A)=1 if and only if there exist column vectors v∈Rn and w∈Rm such that A=vwt. A rank-one matrix is the product of two vectors. In previous sections, we solved linear systems using Gauss elimination method or the Gauss-Jordan method. I would say that your statement "Column 1 = Column 3 = Column 4" is wrong. by Marco Taboga, PhD. Rank of a Matrix in Python: Here, we are going to learn about the Rank of a Matrix and how to find it using Python code? Rank of a matrix definition is - the order of the nonzero determinant of highest order that may be formed from the elements of a matrix by selecting arbitrarily an equal number of rows and columns from it. How to find Rank? The rank of a matrix m is implemented as MatrixRank… If p < q then rank(p) < rank(q) Rank of a matrix. DEFINITION 2. The rank of a matrix is defined as. Pick the 2nd element in the 2nd column and do the same operations up to the end (pivots may be shifted sometimes). The Rank of a Matrix. 7. tol (…) array_like, float, optional. The rank of a matrix is the dimension of the subspace spanned by its rows. The Rank of a Matrix Francis J. Narcowich Department of Mathematics Texas A&M University January 2005 1 Rank and Solutions to Linear Systems The rank of a matrix A is the number of leading entries in a row reduced form R for A. Introduction to Matrix Rank. 6. What is a low rank matrix? 8. Exercise in Linear Algebra. … If all eigenvalues of a symmetric matrix A are different from each other, it may not be diagonalizable. The non-coincident eigenvectors of a symmetric matrix A are always orthonomal. The system has a nontrivial solution if only if the rank of matrix A is less than n. Matrix Rank. Calculators and Converters. The rank depends on the number of pivot elements the matrix. The column rank of a matrix is the dimension of the linear space spanned by its columns. Rank of a Matrix. This exact calculation is useful for ill-conditioned matrices, such as the Hilbert matrix. Find Rank of a Matrix using “matrix_rank” method of “linalg” module of numpy. The nxn-dimensional reversible matrix A has a reduced equolon form In. The rank of a Matrix is defined as the number of linearly independent columns present in a matrix. Remember that the rank of a matrix is the dimension of the linear space spanned by its columns (or rows). Threshold below which SVD values are considered zero. Input vector or stack of matrices. Rank of unit matrix $I_n$ of order n is n. For example: Let us take an indentity matrix or unit matrix of order 3×3. Coefficient matrix of the homogenous linear system, self-generated. Pick the 1st element in the 1st column and eliminate all elements that are below the current one. Return matrix rank of array using SVD method. This also equals the number of nonrzero rows in R. For any system with A as a coeﬃcient matrix, rank[A] is the number of leading variables. OR "Rank of the matrix refers to the highest number of linearly independent rows in the matrix". The rank of the matrix A is the largest number of columns which are linearly independent, i.e., none of the selected columns can be written as a linear combination of the other selected columns. The rank of a matrix can also be defined as the largest order of any non-zero minor in the matrix. 1) Let the input matrix be mat[][]. the maximum number of linearly independent column vectors in the matrix So often k-rank is one less than the spark, but the k-rank of a matrix with full column rank is the number of columns, while its spark is $\infty$. rank-of-matrix Questions and Answers - Math Discussion Recent Discussions on rank-of-matrix.php . The rank of a matrix or a linear transformation is the dimension of the image of the matrix or the linear transformation, corresponding to the number of linearly independent rows or columns of the matrix, or to the number of nonzero singular values of the map. Some theory. 2010 MSC: 15B99 . linear-algebra matrices vector-spaces matrix-rank transpose. Theorem [thm:rankhomogeneoussolutions] tells us that the solution will have $$n-r = 3-1 = 2$$ parameters. The rank is an integer that represents how large an element is compared to other elements. Rank of Symbolic Matrices Is Exact. 1 INTRODUCTION . Top Calculators. 4. In the examples considered, we have encountered three possibilities, namely existence of a unique solution, existence of an infinite number of solutions, and no solution. We are going to prove that the ranks of and are equal because the spaces generated by their columns coincide. Calculator. Based on the above possibilities, we have the following definition. A matrix obtained by leaving some rows and columns from the matrix A is called a submatrix of A. Finding the rank of a matrix. All Boolean matrices and fuzzy matrices are lattice matrices. Changed in version 1.14: Can now operate on stacks of matrices. Common math exercises on rank of a matrix. Matrix Rank. As we will prove in Chapter 15, the dimension of the column space is equal to the rank. You can say that Columns 1, 2 & 3 are Linearly Dependent Vectors. Firstly the matrix is a short-wide matrix \$(m


# Linear independence

## Homework Statement

Assume vectors ##a,b,c\in V_{\mathbb{R}}## to be linearly independent. Determine whether vectors ##a+b , b+c , a+c## are linearly independent.

## The Attempt at a Solution

We say the vectors are linearly independent when ##k_1a + k_2b +k_3c = 0## only when every ##k_n = 0## - the only solution is a trivial combination.
Does there exist a non-trivial combination such that
##k_1(a+b) + k_2(b+c) + k_3(a+c) = 0##?. Distributing:
##k_1a + k_1b + k_2b + k_2c + k_3a + k_3c = (k_1 + k_3)a + (k_1+k_2)b + (k_2+k_3)c = 0## Since ##a,b,c## are linearly independent, the only way this result can occur is when:
##k_1+k_3 =0\Rightarrow k_1 = -k_3##
##k_1+k_2 =0##
##k_2+k_3 =0##

Substituting eq 1 into eq 2 we arrive at ##k_2 - k_3 = 0## and according to eq 3 ##k_2 + k_3=0##, which means ##k_2 - k_3 = k_2 + k_3##, therefore ##k_3 = 0##, because ##k=-k## only if ##k=0##. The only solution is a trivial combination, therefore the vectors ##a+b, b+c, a+c## are linearly independent.

Last edited:

Mark44
Mentor

## Homework Statement

Assume vectors ##a,b,c\in V_{\mathbb{R}}## to be linearly independent. Determine whether vectors ##a+b , b+c , a+c## are linearly independent.

## The Attempt at a Solution

We say the vectors are linearly independent when ##k_1a + k_2b +k_3c = 0## only when every ##k_n = 0## - the only solution is a trivial combination.
Does there exist a non-trivial combination such that
##k_1(a+b) + k_2(b+c) + k_3(a+c) = 0##?. Distributing:
##k_1a + k_1b + k_2b + k_2c + k_3a + k_3c = (k_1 + k_3)a + (k_1+k_2)b + (k_2+k_3)c = 0## Since ##a,b,c## are linearly independent, the only way this result can occur is when:
##k_1+k_3 =0\Rightarrow k_1 = -k_3##
##k_1+k_2 =0##
##k_2+k_3 =0##

Substituting eq 1 into eq 2 we arrive at ##k_2 - k_3 = 0## and according to eq 3 ##k_2 + k_3=0##, which means ##k_2 - k_3 = k_2 + k_3##, therefore ##k_3 = 0##, because ##k=-k## only if ##k=0##. The only solution is a trivial combination, therefore the vectors ##a+b, b+c, a+c## are linearly independent.
That works for me. (IOW, I agree that the three new vectors are linearly independent.)

Instead of working with the system of equations, you can set up a matrix and row reduce it. If you end up with the identity matrix, what that says is that ##k_1 = k_2 = k_3 = 0##, and that there are no other solutions.

The matrix looks like this, from your system:
$$\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix}$$
After a few row operations, the final matrix is I3.

Ray Vickson
Homework Helper
Dearly Missed
That works for me. (IOW, I agree that the three new vectors are linearly independent.)

Instead of working with the system of equations, you can set up a matrix and row reduce it. If you end up with the identity matrix, what that says is that ##k_1 = k_2 = k_3 = 0##, and that there are no other solutions.

The matrix looks like this, from your system:
$$\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix}$$
After a few row operations, the final matrix is I3.

Alternatively, you can compute the determinant of the matrix to find that it is nonzero. What would that tell you?

Alternatively, you can compute the determinant of the matrix to find that it is nonzero. What would that tell you?
Oh. Cramer's rule.
##k_n = \frac{D_{k_n}}{D}## and since the determinant of the system is non zero, the corresponding determinants for every ##k_n## would be 0 (a full column of 0-s means det = 0) and therefore ##k_1 = k_2 = k_3 = 0##

Ray Vickson


# How can a Cauchy sequence converge to an irrational number?

I am a physics major and would like to clear a confusion regarding complete metric spaces. I am quoting the definition of a Cauchy sequence from wikipedia below

Formally, given a metric space $(X, d)$, a sequence $x_1, x_2, x_3, \ldots$ is Cauchy, if for every positive real number $\epsilon > 0$ there is a positive integer $N$ such that for all positive integers $m, n > N$, the distance

$$d(x_m, x_n) < \epsilon$$

Now, if we have sequence like $x_1=3, x_2=3.1, x_3=3.14, \ldots$ converging to $\pi$, I do not understand how all distances $d(x_m, x_n)$ will be less than all positive real numbers. Since irrational numbers do not terminate and continue forever, how can the distance ever be less than the smallest real number or infinitesimal (hyperreal) as the distance can never become $0$. Does this definition of completeness apply where $\epsilon$ is infinitesimal (hypperreal) ?

Kindly excuse my ignorance as I am not a mathematics major.

Thanks

-
Don't think of it as one particular distance, $d(x_m,x_n)$, is smaller than every positive real number. But rather, any given positive real number, you can go far enough out to guarantee that $d(x_m,x_n)$ is smaller. – I. Cavey Jan 11 at 3:14
There is no "smallest [positive] real number". – Eric Wofsey Jan 11 at 3:15
It is not that the distance is smaller than every positive real number, it is that for any positive real number, there is a point where the distance is eventually less than it. E.g. can you eventually go far enough that the sequence is accurate to the fifth decimal place? (i.e. $\epsilon=10^{-5}$) Can you go far enough that the sequence is accurate to the hundredth decimal place? (i.e. $\epsilon=10^{-100}$) If I were to give you some number of decimal places, could your sequence eventually be that accurate? The answer of course is yes. – JMoravitz Jan 11 at 3:16
I disagree with the vote to close the question. ${}\qquad{}$ – Michael Hardy Jan 11 at 3:17
@AloizioMacedo : Perhaps it is unclear to you, but not to me. ${}\qquad{}$ – Michael Hardy Jan 11 at 4:02

The subject line currently reads “How can a Cauchy Sequence converge to an irrational number?”.

If we construe that literally, then one easy way a Cauchy sequence (lower-case initial "s") can converge to $\pi$ is that every term of the Cauchy sequence is $\pi$. Thus: $x_1=\pi, x_2=\pi, x_3=\pi,\ldots\,{}$. I suspect you meant “How can a Cauchy sequence of rational numbers converge to an irrational number?”.

Consider your sequence $3,\ 3.1,\ 3.14,\ 3.141,\ \ldots\,$.

The definition DOES NOT say that all distances between members of this sequence are less than all positive numbers. That would happen only with a constant sequence like my first example above. It says:

For every positive real number $\varepsilon>0$ there is a positive integer $N$ such that for all positive integers $m,n>N$ we have $d(x_m,x_n)<\varepsilon$.

Notice that $N$ depends on $\varepsilon$. In fact as $\varepsilon$ gets smaller, typically $N$ must get bigger. Suppose $\varepsilon = 0.01$. Then for your example sequence, $N=3$ is big enough: every pair of numbers in the sequence at or after the third place in the sequence differ from each other by less than $\varepsilon=0.01$. Thus $3.14$ and $3.141$ differ by less than $0.01$. But now suppose $\varepsilon=0.00001$. Then you need a bigger value of $N$. If each term of the sequence has one more digit or $\pi$, then $N=5$ would be big enough for that value of $\varepsilon$.

Notice that the definition of convergence to $\pi$ differs from the definition of "Cauchy sequence". It says for every $\varepsilon>0$ there is a positive integer $N$ such that for every positive integer $n\ge N$ we have $|x_n-\pi|<\varepsilon$. Again, $N$ depends on $\varepsilon$. If $\varepsilon=0.00001$, then $N=5$ would be enough: every term at or beyond the $5$th one differs from $\pi$ by less than $\varepsilon=0.00001$.

There is nothing in either of these definitions that says that the distance between two different members of the sequence or the distance between $\pi$ and a member of the sequence is $0$.

You wrote:

Since irrational numbers do not terminate and continue forever

Let's be clear on a definition.

It is certainly not correct that numbers whose decimal expansions do not terminate are necessarily irrational. For example, $1/7 = 0.\ 142857\ 142857\ 142857\ \ldots$ has a non-terminating decimal expansion and is rational.

Nor is it the case that "rational number" is defined as one whose decimal expansion repeats or terminates. Euclid and other ancient Greeks proved some numbers are irrational without ever thinking about decimal expansions. That $\pi$ is irrational means $\pi$ is not a quotient of two integers, like $22/7$. Proving $\pi$ is irrational is so difficult that it was not done until the 18th century. Some numbers are far easier to prove to be irrational. For example, if $\log_2 3 = m/n$ and $m,n$ are positive integers, then $2^m=3^n$, but that can't happen because an even number cannot be equal to an odd number.

The fact that a number is rational if and only if its decimal expansion repeats or terminates takes a bit of work to prove, but it's elementary enough that high-school students will understand it.

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In the last sentence, I think you meant "...a number is rational if and only if its decimal expansion repeats or terminates..." – Nathan Reed Jan 11 at 4:34
I don't understand the section after the quoted statement "Since [the decimal expansions of] irrational numbers do not terminate and continue forever." You appear to be trying to rebut that statement, but the statement is true. The asker never claims that numbers whose decimal expansions don't terminate are rational. They also don't claim that the definition of being rational is that the decimal expansion repeats or terminates. Both of those claims would be false but they're never made so I don't understand why you're rebutting them. – David Richerby Jan 11 at 4:55
Agree with @DavidRicherby; this answer would be improved by taking out the rebuttal that infinite decimals are necessarily irrational, which was never part of the OP's inquiry. – Daniel R. Collins Jan 11 at 5:55
Er, I mean the asker doesn't ever claim that non-terminating decimals are irrational. Doh! – David Richerby Jan 11 at 5:57
@NajibIdrissi : Why would you say that? If someone is learning this subject, they should become aware of things like that. "Cauchy sequence" is not defined as meaning a Cauchy sequence of rational numbers. ${}\qquad{}$ – Michael Hardy Jan 13 at 17:08

It's not the case that "all distances $d(x_m, x_n)$ will be smaller than every real number". That is not even true "eventually", in the sense that for some $N$, it's true for all $m, n > N$. That would imply that all such distances are $0$. Furthermore, of course there is no "smallest [positive] real number". The point is that for any positive real number $\epsilon$, no matter how small, the distances between the terms of the sequence eventually get smaller than $\epsilon$ and stay smaller than $\epsilon$.

Think of it as a challenge: I give you some $\epsilon>0$, and you have to find a position in the sequence (some $N$) such that $d(x_m,x_n)<\epsilon$ for all $m,n>N$. If the sequence is Cauchy, you are guaranteed to win the challenge. Cauchy chose the letter $\epsilon$ to stand for error (or erreur): the finite approximations $x_n$ will have an error of less than $\epsilon$ from some point on.

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In the hyperreal setting mentioned by the OP, the difference can be smaller than every positive real number and still be nonzero. See my answer. – Mikhail Katz Jan 12 at 16:24
The question did not originally mention hyperreals. – BrianO Jan 12 at 16:42
OK, I guess I am coming in late in the game. – Mikhail Katz Jan 12 at 16:42

If $n >m \ge 1$ you have $d(x_n,x_m) < 10^{n-1}$.

Choose $\epsilon>0$ and $N$ such that $10^{N-1} < \epsilon$, then if $n,m \ge N$ we have $d(x_n,x_m) < \epsilon$.

Note that the statement is for any $\epsilon>0$ there is some $N$ such that blah, blah, blah, and not that there is some $N$ such that for all $\epsilon>0$ we have blah, blah, blah.

The latter formulation would imply that the sequence is constant after some $N$.

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I think your confusion comes from "for every positive real number $\epsilon$". In fact you have to fix $\epsilon$ first when finding $N$.

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first of all it is not that all the distances are less than every positive number. given a positive number , one can find a stage after which any two terms are the given number close to each other.

The idea of a sequence converging to some point do not necessarily imply that the distance between two terms of the sequence become zero. What it says is that, as $n$ becomes larger, the terms of the convergent sequence starts coming closer to the previous one in terms of distance,

e.g in a sequence $<1/n>$ the distance between 1 and 1/2 is 1/2, distance between 1/2 and 1/4 is 1/4 and so on.. now if you look at distance between any two terms after 1/2, it is always less 1/2, if you look at the terms after 1/100 the max distance between any to terms will be upto 1/100 and so on. So if the given positive number is 1/100, you can choose here $n_0=101$(rather anything bigger than it). whereas if you want distance to be less than 1/1000 say, then this $n_0$ wont work, so you will need $n_{0}=1001$ atleast. It does not depend on terms being rational or irrational.

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" I do not understand how all distances $d(x_m,x_n)$ will be less than all positive real numbers."

This is the key to your misunderstanding. All distances can't be.

But for any real number no matter who small we can find an infinite number of distances smaller. (So we could say all real numbers are bigger than an infinite number of the distances...)

Let's do an actual example. Consider .9, .99, .999, etc. ($a_n = \sum_{i=1}^n 9/10^i$). This is a cauchy sequence that converges to 1. Okay, 1 is not irrational. But is that really where your misunderstanding lies? Okay, I'll get to that later. But for now.

Cauchy sequence: for any $\epsilon > 0$ ... okay, let's say $\epsilon = 1/1,000,000$ ... we can find a point after which all distances are less than $\epsilon$ ... okay, that'd be simply $n > 7$. $d(.99999\ldots9, .9999\ldots9) < 1/1,000,000$ for any two $.999\ldots99$ with more than 7 digits.

Okay, but let's make $\epsilon$ really small. Let's make it 1/googol. ($10^{-100}$) Well, now if n > 100 we still have $(.999\ldots9, .9999\ldots) < 1/\text{googol}$ if those terms have more that 100 nines. How about a googleplex? Then the terms need a googol nines to be that close. But we can find them with more than a googol nines. No matter how small we can find a point where the following difference of terms is smaller.

That's a cauchy sequence.

Okay, that was a cauchy sequence of rational numbers converging to a rational number. You wanted to know about a cauchy sequence of rational numbers converging to an irrational number and you suggested the decimal expansion of $\pi$. Well, same thing. $\pi$ expanded to a million decimal places will be off but it'll be of by a very small amount. For any real number no matter how small, we can find a point after which all decimal expansions are closer that that to $\pi$. These expansions are still finite but they are long enough to be closer to $\pi$ than the small real number we chose. (And if they aren't, we simply choose the finite decimal expansions that are further out. There'll always be finite expansions further out. That's the point.)

It's true we never get there. And we can't really choose rational values that can get there (that's why the number is irrational). but we've trapped and honed it in with cauchy sequences.

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thanks for your response. My main confusion has been made clear that you have to choose $\epsilon$ first and then you get $N$. But how does this work with infinitesimal (hyperreals)...How does this definition of completeness apply there ? – phd-applicant Jan 11 at 5:55
hyperreals are outside my area of expertise. My understanding though is that cauchy sequences of reals do not nesc converge to hyperreals. Hyperreals are not archimedian (for any x>0 y you can find integer n s.t. (n-1)x <= y < nx) and so have a different definition of completeness. I really don't know. I'm sorry I don't know more. But the key of the reals is that epsilon of choice is not an infintisimal. – fleablood Jan 11 at 6:22
To say that the fact that "we never get there" is "why they are irrational" is nonsense. The sequence $0.3,\ 0.33,\ 0.333,\ 0.3333,\ \ldots$ approaches $1/3$ and never gets there, but $1/3$ is rational. ${}\qquad{}$ – Michael Hardy Jan 11 at 14:29
@fleablood, great question. The answer is, it doesn't. This e is not the smallest element and of course in a field you couldn't have a smallest element. As a first approximation think of a sequence tending to zero as generating an infinitesimal, e.g., $\frac{1}{n})$. Then its square will be so much smaller: $\frac{1}{n^2}$. More details can be found here: math.stackexchange.com/questions/1602977/… – Mikhail Katz Jan 12 at 16:41
@fleablood : This is not about surreal numbers, but about Robinson's hyperreal numbers. An infinitesimal $\varepsilon>0$ is smaller than all reciprocals of finite integers, and hence smaller than all positive real numbers, but it is certainly not a smallest positive hyperreal number, since $\varepsilon/2$ is smaller, and so is $\varepsilon^2$. ${}\qquad{}$ – Michael Hardy Jan 12 at 18:07

I looked over the existing answers and they don't seem to address the concern of the OP as I understood it, so I will try to provide a separate answer. The point is that the $(\epsilon,N)$-type definition of convergence is a first-order property and therefore by the transfer principle is still satisfied over the hyperreals. More specifically, a Cauchy sequence $(x_n)$ will have a natural extension defined even for infinite values of the index $n$. This extended sequence will satisfy the condition you mentioned, namely for every $\epsilon$ there is an $N$ such that, etc. If epsilon is infinitesimal then as you point out $N$ will have to be infinite typically.

Having said this, the definition in question is provably equivalent to the one mentioned by other editors (and that involves fewer quantifiers), namely if both indices are infinite then the difference $x_n-x_m$ will be infinitesimal.

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# Probability that in one rolling of 5 dice we obtain two '6' and one '5'?

First I made a mistake of not taking into account that this event is dependent.

So to get two '$$6$$' the probability would be:

$$P_5(2) = \frac{5!}{2!(5-2)!} \cdot \left(\frac16\right)^2\cdot\left(\frac56\right)^3$$

Where

• $$p = \dfrac16$$
• $$q = \dfrac56$$
• $$N= 5$$ (the number of elements of the system)

Then there are $$3$$ dice left on the table and we want to know the probability that one of them is a '$$5$$'. We rule out the two dice with the '$$6$$' face on. So the number of element in the system is now $$3$$. So the probability would be:

$$P_3(1) = \frac{3!}{1!(3-1)!} \cdot \left(\frac16\right)^1\cdot\left(\frac56\right)^2$$

But then the professor told me that $$p=\dfrac15$$ and not $$p=\dfrac16$$ That's what I don't get.

Of course at the end we multiply those two probabilities to get the final probability that we want.

• The probability is $1/5$ because you CAN"T get another $6.$ Then, you would have more than two $6$s. – StatGuy Oct 12 at 4:50

I am a beginner so see if this makes sense to you. If not let me know. Here are two ways you can solve this.

Method 1

You can solve this using the joint pmf for a multinomial distribution.

$$P(X_6 = 2, X_5 = 1, X_{other} = 2) = \frac{5!}{2!1!2!} \left( \frac{1}{6} \right)^2 \left( \frac{1}{6} \right) \left( \frac{4}{6} \right)^2 \approx .062$$

Method 2

Alternatively you can solve this using the multiplication rule. Here you get the $$1/5$$ that your professor talked about because when you condition you are on a reduced sample space.

Let $$E$$ be the event you get two sixes.

Let $$F$$ be the event you get one five.

Let $$G$$ be the event you get two of the others.

$$P(EFG) = P(E)P(F \mid E)P(G \mid EF)$$

$$P(E) = {5 \choose 2}\left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^3$$

$$P(F \mid E) = {3 \choose 1}\left( \frac{1}{5} \right) \left( \frac{4}{5} \right)^2$$

$$P(G \mid EF) = {2 \choose 2}\left( \frac{4}{4} \right)^2 \left( \frac{0}{4} \right)^0 = 1$$

$$P(EFG) \approx .062$$

• The first method looks like an easy way, can I get the general equation? But still, the $1/5$ doesn't compute to me. How can we rule out a face when it can still happen? To me if we put $p=1/5$ it's would be like considering the 3 other dice have five faces. – Nuz Oct 12 at 5:14
• In regards to method 1 just look up the Multinomial distribution on wiki and check out the pmf. In regards to the $1/5$, personally I don't think about it as a dice with five faces. Personally the way I think about it is you roll the other $3$ dice and if you get any $6$'s you just roll again. So for example, what's the probability you roll a $1$ given that you don't roll a $2,3,4,5,6$? The probability is $1$. Does that mean you have a one faced die? Maybe the way to think of it is if you get something that is not a $1$ you just re roll until you finally do. You are on a reduced sample space. – HJ_beginner Oct 12 at 5:38
• Is there another approach, to calculate the fact that one out of the 3 dice must be a '5' and the 2 others can't be '6's or '5's? Or maybe another analogy or something? – Nuz Oct 12 at 7:11
• Hmmm not sure I understand your question. You can calculate $P(EFG)$ by conditioning on any event first. $P(EFG) = P(G)P(F \mid G)P(E \mid FG)$. There's many many valid approaches. Part of the reason why I find probability challenging/interesting. – HJ_beginner Oct 12 at 7:36

But then the professor told me that $$p=1/5$$ and not $$p=1/6$$ That's what I don't get.

You have correctly evaluated the probability for the event that two from the five dice show 6.

$$\mathsf P(N_6{=}2)=\binom 52 \dfrac{1^25^3}{6^5}$$

Now, of the three dice which don't show 6, each may show faces $$\{1,2,3,4,5\}$$ with equal probability.   Thus the conditional probability for a die showing 5, when given that it does not show 6, is $$1/5$$.   So the conditional probability for obtaining one 5 when given that two 6 have been obtained among the five dice is:

$$\mathsf P(N_5{=}1\mid N_6{=}2)=\binom 31\dfrac{1^14^2}{5^3}$$

Of course at the end we multiply those two probabilities to get the final probability that we want.

And the $$5^3$$ factors will cancel, so:

$$\mathsf P(N_6{=}2,N_5{=}1)=\binom{5}{2}\binom{3}{1}\dfrac{1^21^14^2}{6^5}$$

PS: This is called a multinomial distribution.


Bézout's identity and Diophantine Equation

haki

I am having problems with one exam question.

Does this diophantine equation have a solution(s)

12a+21b+33c=6

as far as I know this is not a linear equation, and what I read online says that Bezout identity only applies for linear diophantine equations.

The solution says gcd(12,21,33) = 3, 6|3 the above equation has infinitely many solutions. Is that correct? Am I correct to assume then that

12a+21b+33c+24d = 6 again has infinite solutions aswell?

17x+6y +3z =73

since gcd(17,6,3) = 1 and 73 | 1 this has also infinitely many solutions?

fresh_42

Mentor
2018 Award
Bézout's Lemma says, that the greatest common divisor $d$ of numbers $a_1,\ldots,a_n$ can always be written as $d= s_1a_1+\ldots +s_na_n$. This immediately applies to all of your examples.

An example where it does not work is $2x+4y+6z = 7$.

WWGD

Gold Member
You can always reduce the case ax+by+cz=d to ax+by=d by letting z=0.

"Bézout's identity and Diophantine Equation"

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# condensation point math

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The dew point is the temperature at which the atmosphere is saturated with water vapor. The Boiling point and the condensation point of water are the same. there is some p2SrCsuch that for any ">0, B(p;") \(SrC) is uncountable. Prove that every uncountable subset S ⊂ Rn contains a condensation point of S. (Hint: use the fact that the countable union of countable sets is countable). In mathematics, a condensation point p of a subset S of a topological space, is any point p such that every open neighborhood of p contains uncountably many points of S.Thus "condensation point" is synonymous with "-accumulation point".Examples. Condensation occurs when the relative humidity (RH) reaches 100%, and water precipitates out of the air. there is some p2SrCsuch that for any ">0, B(p;") \(SrC) is uncountable. It is a condensation point of A if and only if every neighbourhood of it contains uncountably many points of A . The set of condensation points of a set is always closed; if it is non-empty, it is perfect and has the cardinality of the continuum. The term limit point is slightly ambiguous. Every point in the Cantor set is a condensation point. Furthermore, we denote it … This page was last edited on 12 April 2014, at 12:13. During condensation, atoms and molecules form clusters. The set of condensation points of a set is always closed; if it is non-empty, it is perfect and has the cardinality of the continuum. In my above example using artesian heated water at 80 degrees Celsius and using a cooling tank to supply the cold side at about 30 degrees Celsius the tables suggest the example is valid Provided the vacuum is kept above the boiling point of 30 degrees Celsius water. At the dew point, the air is saturated, full to the brim of water. At this temperature, the air becomes saturated with water. (See also [a1].). The temperature at which such condensation would occur is called the DEW POINT TEMPERATURE. Dew Point Calculation You can physically determine the dew point temperature with a device called a hygrometer. The water boils and evaporates into the air forming water vapor. A derivative set is a set of all accumulation points of a set A. It includes unlimited math lessons on number counting, addition, subtraction etc. Yes, JLB. 5. In mathematics, a condensation point p of a subset S of a topological space, is any point p such that every open neighborhood of p contains uncountably many points of S. Thus "condensation point" is synonymous with " {\displaystyle \aleph _ {1}} - accumulation point ". But if X=ℚ and A any subset, then A does not have any condensation points at all. It happens when molecules of water vapor cool and collect together as liquid water. Mold can grow when the air is damper than about 80% relative humidity. We have further classifications of condensation point where the topological space is an ordered field. For example, if X=ℝ and A any subset, then any accumulation point of A is automatically a condensation point. na condensation point of S. Since jjx n pjj<1=nfor all n 1 we see that the sequence (x n) n 1 converges to p. Mental math strategy where a number is added to one and subtracted from other to keep the difference same. Condensation graph, returned as a digraph object.C is a directed acyclic graph (DAG), and is topologically sorted. Editor Daniel F. includes dew point calculations, formulas, and mathematic models later in this article. Take your students through all the steps in the water cycle! unilateral condensation point: x is a condensation point of A and there is a positive ϵ with either (x-ϵ,x)∩A countable or (x,x+ϵ)∩A countable. 3. Consider a Bose gas that follows a linear energy-momentum relation ε = vlp in a space of dimensionality d = 1 and d = 2. Eventually the … Generated on Sat Feb 10 11:12:42 2018 by. Since it’s so full, there’s a lot of pressure. Water tends to evaporate once the temperature increases from the boiling point which is beyond 100 degrees Celsius. Td = dew point temperature; T = temperature observed, expressed in degrees Celsius Condensed definition, reduced in volume, area, length, or scope; shortened: a condensed version of the book. na condensation point of S. Since jjx n pjj<1=nfor all n 1 we see that the sequence (x n) n 1 converges to p. where. It occurs at 212 degrees Fahrenheit or 100 degrees Celsius. an ordinal which is the least among all ordinals of the same cardinality as itself), then a κ-condensation point can be defined similarly. How to use condensate in a sentence. Most material © 2005, 1997, 1991 by Penguin Random House LLC. First let's understand what the dew point is. Condensation is when water vapor changes state from gas to liquid. Jump to: navigation , search. Definition: An open neighborhood of a point $x \in \mathbf{R^{n}}$ is every open set which contains point x. Condensation occurs when a surface is cooler than the dew point temperature of the air next to the surface. If κ is any cardinal (i.e. C = condensation (G) returns a directed graph C whose nodes represent the strongly connected components in G. This reduction provides a simplified view of the connectivity between components. A condensation point of a set S in a metric space X is (by definition) a point p such that, for each ε>0, SBp∩ (,)εis uncountable (uncountably infinite). This fifth-grade worksheet explores condensation and evaporation in a charmingly illustrated science activity. bilateral condensation point: For all ϵ>0, we have both (x-ϵ,x)∩A and (x,x+ϵ)∩A uncountable. Modified entries © 2019 by Penguin Random House LLC and HarperCollins Publishers Ltd. noun. While for the former class the number of components of the system is fixed, for the two other classes it is a fluctuating quantity. Therefore $0$ is a condensation point. The European Mathematical Society. For example, within a cloud, water nucleates around a dust, pollen, or microbial particle. Here is some basic information: Water is constantly being recycled in the water cycle, through evaporation and condensation.You have to understand that water comes in different forms, or states of matter. See more. Condensation is the change of the physical state of matter from the gas phase into the liquid phase, and is the reverse of vaporization.The word most often refers to the water cycle. Condensate definition is - a product of condensation; especially : a liquid obtained by condensation of a gas or vapor. Start with a mini lesson about water and its many forms. The node numbers in C correspond to the bin numbers returned by conncomp.. condensation determines the nodes and edges in C by the components and connectivity in G: Then, by question 3, SrChas a condensation point belonging to SrC, i.e. Then they will answer critical thinking questions about how physical science works in the real world. Water tends to evaporate once the temperature increases from the boiling point which is beyond 100 degrees Celsius. How are the melting, freezing, boiling, and condensation points of a substance used for classification purposes? The temperature when the relative humidity is 100% is called the dew point, and it's at this point that clouds and rain can develop. The value of relative humidity, which represents the moisture or water vapor content in the air at its peak, means 100 %. (a) Let Cdenote the set of condensation points of S. Assume that SrCis uncountable. Dew Point Temperature = Td = T - ((100 - RH)/5.) Condensation and precipitation reduce RH, which is why dehumidification works to … a point of which every neighborhood contains an uncountable number of points of a given set. How to Predict or Calculate a Wall Cavity Dew Point or Condensation Point in Buildings. The boiling and condensation points of a substance are always equal. The Boiling point and the condensation point of water are the same. Both are functions of temperature, pressure, and the water content of … The saturation point is the point at which air reaches its saturation limit, which means air contains the most quantity of water vapor in it. Condensation of water vapor starts when the temperature of air is lowered to its dew point and beyond. Problem 4 [20 points). 2. bilateral condensation point: For all ϵ > 0 , we have both ( x - ϵ , x ) ∩ A and ( x , x + ϵ ) ∩ A uncountable . Furthermore, we denote it by A or A^d.An isolated point is a point of a set A which is not an accumulation point.Note: An accumulation point of a set A doesn't have to be an element of that set. From Encyclopedia of Mathematics. unilateral condensation point: x is a condensation point of A and there is a positive ϵ with either (x-ϵ, x) ∩ A countable or (x, x + ϵ) ∩ A countable. Enrich your vocabulary with the English Definition dictionary www.springer.com Let X be a topological space and A⊂X. Condensation occurs when the relative humidity (RH) reaches 100%, and water precipitates out of the air. SplashLearn is an award winning math learning program used by more than 30 Million kids for fun math practice. That said, let's take a look at two dew point calculation approaches: How to Calculate the Approximate Dew Point - simplified equation. Condensation and Dew Point. Efimov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. https://encyclopediaofmath.org/index.php?title=Condensation_point_of_a_set&oldid=31620, A.V. In other words, what is the Condensation Point of water? Prove that every uncountable subset S ⊂ Rn contains a condensation point of S. (Hint: use the fact that the countable union of countable sets is countable). For each $n$, the subset of the Cantor set contained in the interval $[0,1/3^n]$ is homeomorphic to the entire Cantor set, and in particular it is uncountable. Condensation point of a set. The concept of a condensation point can be generalized to arbitrary topological spaces. A point of $E^n$ such that every neighbourhood of it contains uncountable many points of the set. The generalization to arbitrary spaces is direct: A point $x$ a condensation point (of a set $M$) in a topological space if (the intersection of $M$ with) every neighbourhood of is an uncountable set. Condensation is the process by which water vapor (water in its gas form) turns into liquid. Condensation and Dew Point. This article was adapted from an original article by B.A. The same is true of all perfect subsets of $\mathbb R^n$ Math. Definition: Let $A \subseteq \mathbf{R^{n}}$. A given volume of air containing higher amount of water vapor has a higher dew point than the same volume of drier air. It can also be defined as the change in the state of water vapor to liquid water when in contact with a liquid or solid surface or cloud condensation nuclei within the atmosphere. The water boils and evaporates into the air forming water vapor. Condensation and precipitation reduce RH, which is why dehumidification works to … To answer that question, we first need to define an open neighborhood of a point in $\mathbf{R^{n}}$. in a Euclidean space $E^n$. Condensation Explained. Also learn the facts to easily understand math glossary with fun math worksheet online at SplashLearn. The dew point is an important aspect of weather forecasts. For Problems 6 and 7, let S be a subset of Rn and deﬁne a condensation point of S to be any point x∈ Rn such that every n-ball centered at x contains uncountably many points of S. 6. The temperature at which such condensation would occur is called the DEW POINT TEMPERATURE. A derivative set is a set of all accumulation points of a set A. Dew Point Calculator is a web resource created by the Image Permanence Institute to help express and visualize the relationship between temperature, relative humidity and dew point. Prove that $P$ is perfect. A point x∈X is called a condensation point of A if every open neighbourhood of x contains uncountably many points of A. The difference between the dew point temperature and the actual temperature is … The condensation point of a substance is the temperature at which it changes from a gas to a liquid. Prove: If S is an uncountable set in R (or more generally in Rn) then there exists a condensation point for S. Suggestion: R= , … This is basic equivalent to 100% relative humidity. Adherent, Accumulation, and Isolated Points in Metric Spaces. Procedure 1. Suppose $E\subset \mathbb{R}^k$, $E$ is uncountable, and let $P$ be the set of all condensation points of $E$. The dew point is the point at which liquid water would condense from the atmosphere, and the frost point is the temperature at which ice would form from the atmosphere. That forces the water to condense, creating condensation. They use the Climatology section of the Digital Atlas of Idaho to explore concepts such as relative humidity, dew point, condensation, and cloud formation. We say that a point $x \in \mathbf{R^{n}}$ is an accumulation point of a set A if every open neighborhood of point x contains at least one point from A distinct from x. The difference between the dew point temperature and the actual temperature is … Condensation is when water vapor changes state from gas to liquid. See DEW POINT TABLE - CONDENSATION POINT GUIDE for the chart approach. Assume the Bose gas has spin 1. a) (10 points) In which of these dimensions will the Bose-Einstein condensation occur? Ponomarev, "Fundamentals of general topology: problems and exercises" , Reidel (1984) (Translated from Russian). When condensation occurs, mold and decay are very likely. We address the question of condensation and extremes for three classes of intimately related stochastic processes: (a) random allocation models and zero-range processes, (b) tied-down renewal processes, (c) free renewal processes. If A is a subset of a topological space X and x is a point of X, then x is an accumulation point of A if and only if every neighbourhood of x intersects A ∖ { x }. (a) Let Cdenote the set of condensation points of S. Assume that SrCis uncountable. 5. Students use the digital atlas of Idaho to study different weather patterns. condensation point definition in English dictionary, condensation point meaning, synonyms, see also 'condensation trail',condensation trail',condensational',condemnation'. The dew point gives an indication of the humidity. Namely. For Problems 6 and 7, let S be a subset of Rn and deﬁne a condensation point of S to be any point x∈ Rn such that every n-ball centered at x contains uncountably many points of S. 6. 1) I proved that $P$ is closed set. Then, by question 3, SrChas a condensation point belonging to SrC, i.e. First, students will read through the nonfiction text and graphics. It occurs at 212 degrees Fahrenheit or 100 degrees Celsius. condensation point in American English. Define a point $p$ in a metric space $X$ to be a condensation point of a set $E\subset X$ if every neighborhood of $p$ contains uncountably many points of $E$. Adherent, Accumulation, and Isolated Points in Metric Spaces. 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# find a,b,c,d,e

ab=1
bc=2
cd=3
de=4
ea=5
find a,b,c,d,e

#### anemone

##### MHB POTW Director
Staff member
My solution:

 From $ab=1$ and $bc=2$, we have: $2ab=bc$ $2ab-bc=0$ $b(2a-c)=0$ Since $b \ne 0$, $2a-c=0$ must be true or $c=2a$. From $cd=3$ and $c=2a$, we have: $(2a)d=3$ $2ad=3$ From $de=4$ and $ea=5$ and $2ad=3$, we have: $ade^2=4(5)$ $2ad(e^2)=2(20)$ $3(e^2)=40$ $\therefore e=\pm 2\sqrt{\dfrac{10}{3}}$ \begin{align*}\therefore a&=\dfrac{5}{e}\\&=\pm \dfrac{5}{2}\sqrt{\dfrac{3}{10}}\end{align*} \begin{align*}\therefore d&=\dfrac{3}{2a}\\&=\pm \dfrac{3}{5}\sqrt{\dfrac{10}{3}}\end{align*} \begin{align*}\therefore c&=\dfrac{3}{d}\\&=\pm 5 \sqrt{\dfrac{3}{10}}\end{align*} \begin{align*}\therefore b&=\dfrac{1}{a}\\&=\pm \dfrac{2}{5}\sqrt{\dfrac{10}{3}}\end{align*}

##### Well-known member
As I do not know how to put square root I have put power 1/2

Multiply all 5 to get (abcde)$^2$ = 120
Or abcde = +/-120$^{(1/2)}$
Devide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$

Similarly you can find the rest

##### Well-known member
My solution:

 From $ab=1$ and $bc=2$, we have: $2ab=bc$ $2ab-bc=0$ $b(2a-c)=0$ Since $b \ne 0$, $2a-c=0$ must be true or $c=2a$. From $cd=3$ and $c=2a$, we have: $(2a)d=3$ $2ad=3$ From $de=4$ and $ea=5$ and $2ad=3$, we have: $ade^2=4(5)$ $2ad(e^2)=2(20)$ $3(e^2)=40$ $\therefore e=\pm 2\sqrt{\dfrac{10}{3}}$ \begin{align*}\therefore a&=\dfrac{5}{e}\\&=\pm \dfrac{5}{2}\sqrt{\dfrac{3}{10}}\end{align*} \begin{align*}\therefore d&=\dfrac{3}{2a}\\&=\pm \dfrac{3}{5}\sqrt{\dfrac{10}{3}}\end{align*} \begin{align*}\therefore c&=\dfrac{3}{d}\\&=\pm 5 \sqrt{\dfrac{3}{10}}\end{align*} \begin{align*}\therefore b&=\dfrac{1}{a}\\&=\pm \dfrac{2}{5}\sqrt{\dfrac{10}{3}}\end{align*}

I thought that it may be noted that all are positive or all are -ve. I know you know it but for benefit of others

#### MarkFL

Staff member

I thought that it may be noted that all are positive or all are -ve. I know you know it but for benefit of others
Yes, you can see that all of the values anemone found inherit their sign from $e$.

#### anemone

##### MHB POTW Director
Staff member
As I do not know how to put square root I have put power 1/2
The \sqrt{} command creates a square root surrounding an expression.

Take for example, \sqrt{x} gives $\sqrt{x}$.

Multiply all 5 to get (abcde)$^2$ = 120
Or abcde = +/-120$^{(1/2)}$
Devide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$

Similarly you can find the rest
Well done, kali! And looking more closely, I'd say we're actually approached the problem using quite similar concept!

#### Albert

##### Well-known member
As I do not know how to put square root I have put power 1/2

Multiply all 5 to get (abcde)$^2$ = 120
Or abcde = +/-120$^{(1/2)}$
Devide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$

Similarly you can find the rest
good solution

the use of square root example :type :" \sqrt[m]{b^n} " between two "dollar signals"

you will get: $\sqrt[m]{b^n}$


# Triangle forming probability for area

Say you have a stick which breaks randomly into three pieces (we can choose the points randomly). What is the probability that the area is greater than or equal to $0.4$?

I can see it has something to do with Heron's formula but I just can't put t together.

• what's the length of the stick? – Saketh Malyala Jul 5 '17 at 4:10
• Simulation suggests about $0.26$ for the conditional probability and that the expected value of the area is about $0.0299$, assuming the length of the stick is $1$ – Henry Jul 5 '17 at 7:46
• I get conditional probability = $\displaystyle\;4\int_{a}^{b} \sqrt{x^2(1-x^2)^2 - 0.32^2} dx \approx 0.2586458\;$ where $a \approx 0.37111$, $b \approx 0.76139$ are the two roots of polynomial $x(1-x^2) - 0.32$ in $(0,1)$. – achille hui Jul 5 '17 at 9:51
• @amWhy: it is a duplicate, but the earlier question (from the same person) does not have a response – Henry Jul 5 '17 at 20:46
• When this question has an answer, closing this as a duplicate to a question without answer (even from same user) server no useful purposes. A better choice is ask the OP to delete the old question. – achille hui Jul 5 '17 at 21:11

We will assume the length of the stick is $1$ and we break the stick by picking two points $u, v$ uniformly and independently from $(0,1)$.

Let $a = \min(u,v)$, $b = 1 - \max(u,v)$ and $c = |u-v| = 1 - a - b$. $a, b, c$ forms the sides of a triangle when and only when

$$a \le b+c,\;b \le c+a,\;c \le a+b\;\iff\;a \le \frac12,\;b \le \frac12,\;a+b \ge \frac12$$

Change variable to $x = 1-2a, y = 1-2b$, the lengths $a,b,c$ forms a triangle when $(x,y)$ falls inside another triangle $$\Delta = \big\{ (x,y) : x \ge 0, y \ge 0, x+y \le 1 \big\}$$ with area $\frac12$. It is clear conditional to $a,b,c$ forming a triangle, the probability "density" of picking a particular $(x,y)$ is $2dxdy$.

Let $A$ be the area of a triangle with sides $a,b,c$ and $A_0 = 0.04$. By Heron's formula, we have

\begin{align} A &= \sqrt{s(s-a)(s-b)(s-c)}\\ \iff 16A^2 & = 1(1-2a)(1-2b)(1-2c) = xy(1-x-y)\\ \iff 64A^2 & = ((x+y)^2 - (x-y)^2)(1-x-y) \end{align} Change variable once again to $p = (x+y), q = (x-y)$, the triangle $\Delta$ becomes

$$\Delta' = \big\{ (p,q) : 0 \le |q| \le p \le 1 \big\}$$

Furthermore,

$$A \ge A_0 \quad\iff\quad (p^2 - q^2)(1-p) \ge (8A_0)^2 \quad\iff\quad q^2 \le p^2 - \frac{(8A_0)^2}{1-p}$$ Let $\displaystyle\;f(p) = \sqrt{p^2 - \frac{(8A_0)^2}{1-p}}\;$ and $\lambda_1, \lambda_2$ be the two roots of $f(p)$ in $(0,1)$. The condition above is equivalent to $\lambda_1 \le p \le \lambda_2$ and $|q| \le f(p)$. Since $dpdq = 2dxdy$, the probability we seek equals to

$$\mathbb{P}[A\ge A_0] = \int_{\lambda_1}^{\lambda_2} \int_{-f(p)}^{f(p)} dqdp = 2\int_{\lambda_1}^{\lambda_2} f(p) dp$$

Change variable to $t = \sqrt{1-p}$, this becomes

$$\mathbb{P}[A\ge A_0] = 4\int_{\mu_1}^{\mu_2} \sqrt{t^2(1-t^2)^2 - (8A_0)^2} dt\tag{*1}$$

where $\mu_1, \mu_2$ are now the roots of the polynomial $\;t(1-t^2) - 8A_0\;$ in $(0,1)$.

For the problem at hand where $A_0 = 0.04$, we have

$$\mu_1 \approx 0.3711104191979701,\; \mu_2 \approx 0.7613913530813122$$ and $(*1)$ evaluates numerically to $$\mathbb{P}[A\ge A_0] \approx 0.2586458039398669$$ This is compatible with what another user @Henry obtained through simulation.

• Superb!! (+1). I tried in these lines and confirmed my approach through your answer – Satish Ramanathan Jul 6 '17 at 0:52
• @achillehui What numerical procedure did you follow to compute $(*1)$? My answer is coming out to be $\approx 0.249004$ which is different from our answer. I am trying to understand the reason behind the difference in the error. – Dhruv Kohli - expiTTp1z0 Jul 6 '17 at 5:32
• @expiTTp1z0 I have asked two CAS (maxima and WA) to compute the integral numerically and they return same number (upto last two digits, the difference most likely caused by rounding error on maxima). – achille hui Jul 6 '17 at 5:51
• @achillehui Can you please check this: Let $E$ be the event that three pieces form a triangle, then, $P(E) = 0.25$ and $P(A\geq0) = P(A\geq0|E)P(E) + P(A\geq0|E^C)P(E^C) = P(A\geq0|E)0.25 + 0 = 1 \cdot 0.25 = 0.25$ So how can $P(A\geq 0.04) > 0.25$ be true? Am I doing some mistake. – Dhruv Kohli - expiTTp1z0 Jul 6 '17 at 8:39
• @expiTTp1z0 the probability we are computing is the probability condition to three sticks forming a triangle. It is 4 times of that probability if we include events where triangle inequalities are violated. $\mathbb{P}[A > 0]$, or more precisely $\mathbb{P}[ A > 0 : \text{ forms a triangle } ]$, equals to $1$. There is no a prior reason for $P(A \ge 0.04 ) \le 0.25$. – achille hui Jul 6 '17 at 8:53

Assuming the length of stick is $2$. Let $X_1$ and $X_2$ be randomly sampled from $\text{Uniform}(0,2)$ and let $X_{(1)}, X_{(2)}$ be the order statistic.

The length of the three pieces will be,

$$a=X_{(1)},\ \ b=X_{(2)}-X_{(1)},\ \ c=2-X_{(2)}$$

Using,

• triangle inequality ($a+b\geq c, b+c \geq a, c+a\geq b$) and
• the range of $X_{(1)}$ and $X_{(2)}$ ($0 \leq X_{(1)} \leq X_{(2)} \leq 2$),

one can obtain the following condition for the three pieces to form a triangle,

$$0 \leq X_{(1)} \leq 1, \ \ 1 \leq X_{(2)} \leq 1+X_{(1)}$$

Using Heron's Formula,

$$A^2 = s(s-a)(s-b)(s-c) = 1(1-X_{(1)})(1-X_{(2)}+X_{(1)})(1-2+X_{(2)})$$

$$A^2 = X_{(1)}^2-X_{(2)}^2 +2X_{(2)} +X_{(1)}X_{(2)}^2-X_{(1)}^2X_{(2)}-X_{(1)}X_{(2)}-1$$

Let $E$ be the event that the three pieces form a triangle, then,

$$E = \{0 \leq X_{(1)} \leq 1 \text{ and } 1 \leq X_{(2)} \leq 1+X_{(1)}\}$$

and we need to find,

$$P(A^2 > c) = P(A^2>c|\text{E})P(E) + P(A^2>c|E^C)P(E^C) = P(A^2>c|E)P(E)$$

The joint pdf of $X_{(1)}, X_{(2)}$ is given by,

$$f_{X_{(1)},X_{(2)}}(x_1, x_2) = 2 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}, \ \ x_1, x_2 \in [0,2],\ x_1 \leq x_2$$

And therefore the probability of the event $E$ is,

$$P(E) = \int_{0}^{1}\int_{1}^{1+X_{(1)}}f_{X_{(1)},X_{(2)}}(x_1, x_2) \partial x_2 \partial x_1 = \frac{1}{4}$$

The conditional joint pdf of $X_{(1)}, X_{(2)}$ conditioned on $E$ is,

\begin{align} f_{X_{(1)},X_{(2)}|E}(x_1, x_2) &= \frac{f_{X_{(1)}, X_{(2)}}(x_1\mathbb{I}(x_1 \in [0,1]), x_2\mathbb{I}(x_2 \in [1, 1+x_1]))}{P(E)}\\\\ &= \frac{1/2}{1/4} \\\\ &= 2,\ x_1 \in [0,1], x_2 \in [1,1+x_1]\end{align}

Introduce two random variables $U$ and $V$ as follows,

$$U = X_{(1)}$$

$$V = A^2 = X_{(1)}^2-X_{(2)}^2 +2X_{(2)} +X_{(1)}X_{(2)}^2-X_{(1)}^2X_{(2)}-X_{(1)}X_{(2)}-1$$

Writing $X_{(1)}$ and $X_{(2)}$ in the form of $U$ and $V$,

$$X_{(1)} = U$$

$$X_{(2)}^2(1-U) + X_{(2)}(U^2+U-2)+V+1-U^2=0 \\\\ \implies X_{(2)}^2 - X_{(2)}(U+2)-\frac{V+1-U^2}{U-1}=0 \\\\ \implies X_{(2)} = \frac{U+2 \pm\sqrt{(U+2)^2+\frac{4(V+1-U^2)}{U-1}}}{2}$$

Based on the values $X_{(1)}, X_{(2)}$ can take so as to form a triangle, we obtain the values that $U$ and $V$ can take,

$$0 \leq U \leq 1, \ \ 1 \leq \frac{U+2 \pm\sqrt{(U+2)^2+\frac{4(V+1-U^2)}{U-1}}}{2} \leq 1+U \\\\ \implies 0 \leq U \leq 1, \ \ 0 \leq V \leq \frac{U^2(1-U)}{4}$$

In order to obtain the pdf of $U, V$ conditioned on $E$, we compute the determinant of the Jacobian of $X_{(1)}, X_{(2)}$ with respect to $U, V$ as follows,

$$J = \begin{pmatrix} \frac{\partial X_{(1)}}{\partial U} & \frac{\partial X_{(1)}}{\partial V} \\ \frac{\partial X_{(2)}}{\partial U} & \frac{\partial X_{(2)}}{\partial V} \end{pmatrix} = \begin{pmatrix}1 & 0 \\ \ldots & \frac{\pm 1}{\sqrt{((U+2)(U-1))^2+4(V+1-U^2)(U-1)}}\end{pmatrix} \\ \implies |det J| = \frac{1}{\sqrt{((U+2)(U-1))^2+4(V+1-U^2)(U-1)}}$$

Corresponding to the $\pm$ sign in above matrix, let there be two different matrices $J_+, J_-$. For both matrices, $|det J_{+}| = |det J_{-}| = |det J|$.

Now, we compute the pdf of $U,V$ conditioned on $E$,

\begin{align} f_{U,V|E}(u,v) &= f_{X_{(1)}, X_{(2)}|E}\left(u, \frac{u+2 +\sqrt{(u+2)^2+\frac{4(v+1-u^2)}{u-1}}}{2}\right) |det J_{+}| \\\\ & \ \ \ + f_{X_{(1)}, X_{(2)}|E}\left(u, \frac{u+2 -\sqrt{(u+2)^2+\frac{4(v+1-u^2)}{u-1}}}{2}\right) |det J_{-}| \\\\ &= 2 \cdot 2 \cdot |det J| \\\\ &= \frac{4}{\sqrt{((u+2)(u-1))^2+4(v+1-u^2)(u-1)}}\end{align}

It is easy to $\color{red}{\text{verify}}$ that,

$$\int_{0}^{1}\int_{0}^{\frac{u^2(1-u)}{4}}\frac{4}{\sqrt{((u+2)(u-1))^2+4(v+1-u^2)(u-1)}} \partial v \partial u = \int_{0}^{1} 2u \partial u = 1$$

Integrate over $U$ to obtain the marginal distribution of $V$ conditioned on $E$,

$$f_{V|E}(v) = \int_{u:\ 0 \leq u \leq 1,\ \ 0 \leq v \leq \frac{u^2(1-u)}{4}} f_{U,V|E}(u,v) \partial u$$

The cdf of $V$ conditioned on $E$ will then be,

$$P(V \leq c | E) = \int_{0}^{c}\int_{u:\ 0 \leq u \leq 1,\ \ 0 \leq v \leq \frac{u^2(1-u)}{4}} f_{U,V|E}(u,v)\partial u \partial v$$

Consider the equation $v = f(u) = \frac{u^2(1-u)}{4}, \ u \in [0,1]$. For a particular value of $v = c > 0$, there are two values of $u$ satisfying the equation. Let those values be $u_0$ and $u_1$.

Note that $u_0 \in [0,f^{-1}(v_{max})] \equiv [0,\frac{2}{3}]$ and $u_1 \in [f^{-1}(v_{max}), 1] \equiv [\frac{2}{3},1]$. With this argument, the above integral can be written as,

\begin{align}P(V \leq c | E) &= \int_{0}^{u_0}\int_{0}^{\frac{u^2(1-u)}{4}} f_{U,V|E}(u,v)\partial v \partial u + \int_{u_0}^{u_1}\int_{0}^{c} f_{U,V|E}(u,v) \partial v \partial u \\\\ & \ \ \ \ + \int_{u_1}^{1}\int_{0}^{\frac{u^2(1-u)}{4}} f_{U,V|E}(u,v)\partial v \partial u\\\\ &= \int_{0}^{u_0}2udu + \int_{u_0}^{u_1}\left(\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c)}}{u-1} + 2u \right) \partial u + \int_{u_1}^{1}2u \partial u\\\\ &= u_0^2 + \int_{u_0}^{u_1}\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c)}}{u-1} \partial u + u_1^2 - u_0^2 + 1 - u_1^2 \\\\ &= 1 + \int_{u_0}^{u_1}\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c)}}{u-1} \partial u \end{align}

Finally,

\begin{align} P(A^2>c) &= P(A^2 > c|E)P(E) \\\\ &= (1-P(V \leq c|E))P(E) \\\\ &= \left(- \int_{u_0}^{u_1}\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c)}}{u-1} \partial u \right)P(E) \end{align}

Now, the value of $c$ is $0.04^2$ for the case when the length of the stick is $1$. Scaling up this value for the stick of length $2$ so that the probability doesn't change,

$$\frac{c_{new}}{\text{(max area in case of length 2 stick)}^2} = \frac{c_{old}}{\text{(max area in case of length 1 stick)}^2} \\\\ \implies \frac{c_{new}}{\left(\frac{\sqrt{3}}{9}\right)^2} = \frac{c_{old}}{\left(\frac{\sqrt{3}}{36}\right)^2} \\\\ \implies c_{new} = 0.0256$$

For $c > 0$, the values $u_0$ and $u_1$ are the solutions of $c = \frac{u^2(1-u)}{4}$ in the intervals $[0,\frac{2}{3}]$ and $(\frac{2}{3}, 1]$, respectively. Therefore, for $c = c_{new} = 0.0256$, $u_0 = 0.420283$ and $u_1 = 0.862277$. Putting these values in the last equation,

$$P(A^2 > c_{new} | E) = -\int_{u_0}^{u_1}\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c_{new})}}{u-1} \partial u \approx \color{blue}{0.258646}$$

$$P(A^2 > c_{new}) = \frac{-\int_{u_0}^{u_1}\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c_{new})}}{u-1} \partial u}{4} = \color{blue}{0.0646615}$$

Note: Used Wolfram to compute the value of the integral.

• It should be compared to $\sqrt 3/9$ when the stick breaks to 3 equal parts. – Narasimham Jul 5 '17 at 5:45
• I suspect you are supposed to assume the length of the stick is $1$ – Henry Jul 5 '17 at 7:46
• @Henry The main ideas will remain same even if the length of the stick is taken to be $2$. I took the length as $2$ so that the semi perimeter becomes $1$. Spent few hours, still couldn't find the closed form solution. I ll try again later. – Dhruv Kohli - expiTTp1z0 Jul 5 '17 at 10:55
• I voted up for your try. Good try!! – Satish Ramanathan Jul 6 '17 at 0:53
• @satishramanathan Thanks. I have completed my solution now. – Dhruv Kohli - expiTTp1z0 Jul 6 '17 at 5:13


# What's the probability that Abe will win the dice game?

Abe and Bill are playing a game. A die is rolled each turn.

If the die lands 1 or 2, then Abe wins.

If the die lands 3, 4, or 5, then Bill wins.

If the die lands 6, another turn occurs.

What's the probability that Abe will win the game?

I think that the probability is $\frac{2}{5}$ just by counting the number of ways for Abe to win. I'm not sure how to formalize this though in terms of a geometric distribution.

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As a visitor from rpg.stackexchange.com, I can say that it's very common to generate dice which you don't have (such as a d3) by saying "Roll a d4 and reroll 4's". Effectively, any result you reroll from scratch doesn't exist on the die. I wouldn't have been able to state it formally, though, which is why I'm just a visitor here! – Bobson Oct 4 '12 at 4:25
@Bobson: For a d3, wouldn't you normally take a d6 and map 1-2 to 1, 3-4 to 2, and 5-6 to 3? – Joren Oct 4 '12 at 14:03
@Joren - I've done it both ways. The problem with the d6 is sometimes you forget to declare whether you're mapping it that way, or whether you're mapping it 4->1, 5->2, 6->3. Which is also perfectly valid. With the d4, there's no question about it. Admittedly, it may not have been the best choice of examples. – Bobson Oct 4 '12 at 14:11
Could someone with enough rep change this to say die instead of dice. It wont let me save the suggested edits because there aren't enough characters changed. – zzzzBov Oct 4 '12 at 15:26

You are right. You can just ignore rolls of $6$ as they leave you back in the same situation. To formalize this, the chance Abe wins on turn $n$ is $\frac 13 \left(\frac 16 \right)^{n-1}$ and the chance that Bill wins on turn $n$ is $\frac 12 \left(\frac 16 \right)^{n-1}$. You can sum these if you want.

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Let $P$ be the chance that Abe wins, then we have

$$P=\frac{1}{3}+\frac{1}{6}P$$

Solving P from this equation we get

$$P=\frac{2}{5}$$

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I agree with you that $\dfrac{2}{5}$ is obvious. End of story. But if you really want to sum a series, abbreviate by $A$ the event "$1$ or $2$" and by $S$ the event "$6$." Then Abe can win in various ways. These are $A$ (wins immediately), $SA$ (get a $6$, then win), $SSA$, $SSSA$, and so on.

These have probabilities $\dfrac{2}{6}$, $\:\dfrac{1}{6}\cdot\dfrac{2}{6}$, $\:\dfrac{1}{6}\cdot\dfrac{1}{6}\cdot\dfrac{2}{6}$, and so on. So we want to sum the series $$a+ar+ar^2+ar^3+\cdots,$$ where $a=\dfrac{2}{6}$ and $r=\dfrac{1}{6}$.

By the usual formula for the sum of an infinite geometric series, this is $\dfrac{a}{1-r}$, which simplifies to $\dfrac{2}{5}$.

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The key point to realise is that if a 6 is rolled, then the probability of each player winning after that roll is the same as it was before. So if $p_A$ and $p_B$ denote the probability of each player winning, then we have $$p_A = \frac 26 + \frac 16 p_A\\ p_B = \frac 36 + \frac 16 p_B$$ which is easy to solve.

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If you want to use geometric series, you can do the following.

$$\text{P(Abe wins) = P(Abe wins in the first roll) + P(Abe wins in the second roll) + } \dots$$ $$=\frac{2}{6}+\frac{1}{6}\frac{2}{6} + \left(\frac{1}{6}\right)^2 \frac{2}{6} + \dots$$ $$=\frac{1}{3}\left(1+\frac{1}{6}+\left(\frac{1}{6}\right)^2 + \dots\right)$$ $$=\frac{1}{3}\frac{1}{1-\frac{1}{6}}$$ $$=\frac{2}{5}$$

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The chance Abe win is 2/5

You can forget about the reroll. At each reroll, the chance that Abe win is 2/5. So no matter how many rerolls out there, the chance that Abe will win is the weighted average of chance that Abe will win multiplied by probability distribution of the rerolls number.

That average is easy to compute because it's always 2/5.

All answers are the same but I try to do this as intuitive as possible because I can't type equation :)

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# Computing $\operatorname{Tr} \bigl( \bigl( (A+I )^{-1} \bigr)^2\bigr)$

Suppose that $A \in \mathbb{R}^{n \times n}$ is a symmetric positive semi-definite matrix such that $\operatorname{Tr}(A)\le n$. I want a lower bound on the following quantity $$\operatorname{Tr} \bigl( \bigl( (A+I )^{-1} \bigr)^2\bigr)$$ where $I$ denotes the identity matrix.

My intuition tells me it should be

$$\operatorname{Tr} \bigl( \bigl( (A+I )^{-1} \bigr)^2\bigr) \ge 1/4$$

However, I'm not sure how to show it. Also my intuition might be wrong?

• Could be helpful: math.stackexchange.com/questions/391128/… – Squirtle Jul 31 '15 at 21:10
• The expression is invariant under conjugation, and $A$ is diagonalizable. – anomaly Jul 31 '15 at 21:11
• @Squirtle thank. But the square inside is hard to deal with – Boby Jul 31 '15 at 21:13
• @anomaly Is that a suggestion how to solve it? – Boby Jul 31 '15 at 21:24
• @Boby: Well, yes. – anomaly Jul 31 '15 at 21:27

By spectral theorem, there is invertible matrix $P$ such that $$P^{-1}(I+A)P=\pmatrix{1+\mu_1 \\ & \ddots & \\ & & 1+\mu_n}$$ where $\mu_i$ is eigenvalue of $A$ and $\mu_i\geqslant0$. So $$(I+A)^{-1}=P\pmatrix{\dfrac1{1+\mu_1} \\ & \ddots & \\ & & \dfrac1{1+\mu_n}}P^{-1}$$ And $$((I+A)^{-1})^2=P\pmatrix{\dfrac1{(1+\mu_1)^2} \\ & \ddots & \\ & & \dfrac1{(1+\mu_n)^2}}P^{-1}$$ So $$Tr(((I+A)^{-1})^2)=\sum_{i=1}^n\dfrac1{(1+\mu_i)^2}\leqslant n$$ Then by $\sum_{i=1}^n\mu_i\leqslant n$, we have $\mu_i\leqslant n$ for all $i$. So a lower bound is $$Tr(((I+A)^{-1})^2)\geqslant\frac{n}{(1+n)^2}$$

• @Boby, you should convince yourself (google or prove it yourself) that adding the identity to a matrix will shift the eigenspectrum in the manner of this proof (on line one). – Squirtle Jul 31 '15 at 21:33
• @hermes Sorry, but it is not clear to me how does this help to find the lower bound. Could you be more explicit? – Start wearing purple Jul 31 '15 at 21:48
• @hermes I mean this calculation only implies an obvious lower bound $n\cdot \frac{1}{(1+n)^2}$. But this is very far from the conjectured one and definitely not optimal. – Start wearing purple Jul 31 '15 at 22:10
• @L.G. Not sure but maybe we can use concavity of $\frac{1}{(1+x)^2}$ to show that $\sum _{i=1} \frac{1}{(1+\mu_i)^2} \ge \frac{1}{(1+\sum _{i=1}\mu_i)^2} =\frac{1}{4}$ – Boby Jul 31 '15 at 22:29
• The conjecture may not be true.I mean we can get a lower bound but it may not be $1/4$. The lower bound can be obtained by Lagrangian multiplier. – Math Wizard Jul 31 '15 at 22:35

Since $A$ is real symmetric, there exists a matrix $P$ such that

$$PAP^{-1} = D =\verb/diag/(\lambda_1,\lambda_2,\ldots,\lambda_n)$$ is diagonal with eigenvalues $\lambda_1,\ldots,\lambda_n$. This leads to

$$P (I_n+A)^{-2}P^{-1} = (I_n + PAP^{-1})^{-2} = \verb/diag/\left(\frac{1}{(1+\lambda_1)^2},\ldots,\frac{1}{(1+\lambda_n)^2}\right)\\ \implies \verb/Tr/((I_n + A)^{-2}) = \sum_{i=1}^{n}\frac{1}{(1+\lambda_i)^2}$$

Consider the function $g(x) = \frac{1}{(1+x)^2}$. It is easy to check for $x \in (0,\infty)$, $$g(x) > 0, \quad g'(x) = -\frac{2}{(1+x)^3} < 0 \quad\text{ and }\quad g''(x) = \frac{6}{(1+x)^4} > 0$$ This means $g(x)$ is a positive, monotonic decreasing and convex function over $(0,\infty)$.
By Jensen's inequality, we obtain following lower bound:

$$\verb/Tr/((I_n + A)^{-2}) = \sum_{i=1}^{n} g(\lambda_i) \underbrace{\ge}_{\text{Jensen}} n g\left(\frac1n\sum_{i=1}^n \lambda_i\right) \underbrace{\ge}_{g\text{ decreasing}} n g(1) = \frac{n}{4}$$

Since $\verb/Tr/(I_n) \le n$ and $\verb/Tr/((I_n + I_n)^{-2}) = \frac{n}{4}$, above lower bound $\frac{n}{4}$ is achievable and hence the optimal one.

As a consequence, your intution $\verb/Tr/((I_n + A)^{-1}) \ge \frac14$ is true (but far from the optimal).

• +1 Yet $\frac14\ge\frac{n}{(1+n)^2}$, so the OP's bound is at least better than the one given in the other answer ;D – user1551 Aug 24 '15 at 11:44


# Is there a name for matrix product with reversed indices?

The typical matrix product is as follows: $$(\mathbf{A}\mathbf{B})_{ij} = \sum_{k=1}^m A_{ik}B_{kj}\,.$$

Is there a name or characterization for one such as $$(\mathbf{A}\mathbf{B})_{ij} = \sum_{k=1}^m A_{ki}B_{jk}\,?$$ Furthermore, what can be said about the matrix vector product $$(\mathbf{A}\mathbf{b})_{i} = \sum_{k=1}^m A_{ki}b_{k}\,?$$ Is there any way to express the above matrix-vector product in terms of traditional linear algebra?

• Great questions. The beginner's mind is a precious thing. – orangeskid Jul 11 '15 at 0:24
• Agreed, this is a really good question! It is not trivial but amazing that your formulas turn out to be $A^TB^T$ and $A^TB$. – Eli Rose -- REINSTATE MONICA Jul 11 '15 at 1:00
• @EliRose why is this not trivial? It's pretty much by definition of transposition. – Ruslan Jul 11 '15 at 11:04

Your second formula is just $A^tB^t$ and your third one $A^tb$, where the $(\hskip1ex)^t$ means transposed
• (Also, note that $\;\;\; A^tB^t \: = \: (BA)^t \:\:\:\:$.) $\;\;\;\;\;\;\;\;\;$ – user57159 Jul 11 '15 at 3:13
Given a matrix $\mathbf{A} = (A)_{ij}$ the transpose of a matrix $\mathbf{A}^{\intercal}$ is defined by $\mathbf{A}^{\intercal} = (A)_{ji}$. Intuitively the transpose of a matrix is found by reflecting the matrix across the line through the diagonal coefficients $i = j$. See https://en.wikipedia.org/wiki/Transpose.
With this in mind your first alternate version of multiplication can be written as $\sum_{k=1}^m A_{ki}B_{jk} = \mathbf{A}^\intercal\mathbf{B}^\intercal$.
$$\sum_{k=1}^n A_{ki} B_{jk} = \sum_{k=1}^n B_{jk} A_{ki} = (BA)_{ji} = (BA)^\intercal = A^\intercal B^\intercal$$ and $$\sum_{k=1}^m A_{ki}b_{k} = \sum_{k=1}^m A_{ik}^{\intercal} b_{k} = A^{\intercal} b$$


How can I solve $\int \frac{3x+2}{x^2+x+1}dx$

I want to compute this primitive $$I=\int \frac{3x+2}{x^2+x+1}dx.$$

I split this integral into two part: $$\int \frac{3x+2}{x^2+x+1}dx=\int \frac{2x+1}{x^2+x+1}dx+\int \frac{x+1}{x^2+x+1}dx,$$ For the first part: $$\int \frac{2x+1}{x^2+x+1}dx= \ln(|x^2+x+1|)+c_1$$ For the second part: due a change of variable $u=x+1$, I find $$\int \frac{x+1}{x^2+x+1}dx= \int \frac{x+1}{x(x+1)+1}dx=\int \frac{u}{u(u-1)+1}dx=\dots$$

• $u=x+1\implies x=u-1$, hence $\frac{x+1}{x(x+1)+1}=\frac{u}{(u-1)u+1}\neq\frac{u}{u^2}$. – Workaholic Feb 3 '16 at 15:59
• @Workaholic, I don't have any idea to compute the second integral, can you help me – Achaire Feb 3 '16 at 16:02
• the varible change is no good here $\frac{x+1}{x(x+1)+1}=\frac{x}{x(x+1)+1}+\frac{1}{x(x+1)+1}$ – Achaire Feb 3 '16 at 16:05
• Do you know how to deal with integrals of the form $\displaystyle\int\frac{1}{ax^2+bx+c}\,\mathrm dx$? – Workaholic Feb 3 '16 at 16:07
• It seems that most agree that rewriting $x^2+x+1$ as $x^2+x+\frac{1}{4}+\frac{3}{4}=(x+\frac{1}{2})^2+\frac{3}{4}$ would be helpful. – John Joy Feb 3 '16 at 16:41

Hint For the second:

$$\int \frac{x+1}{x^2+x+1}dx$$

Rewrite as $$\int\bigg( \frac{2x+1}{2(x^2+x+1)}+\frac{1}{2(x^2+x+1)} \bigg)dx$$

$$=\underbrace{\frac 1 2 \int\frac{2x+1}{x^2+x+1}dx}_{:=I}+\underbrace{\frac 1 2 \int \frac{1}{x^2+x+1}dx}_{:=J}$$

For $I$ substitute $t=x^2+x+1$ and $dt=(2x+1)dx$

For $J$ complete the square : $\frac 1 2\displaystyle\int \frac{1}{(x+1/2)^2+3/4}$ and now substitute $\varphi=x+1/2$ and $d\varphi=dx\Longrightarrow \frac 1 2\displaystyle\int \frac{1}{\varphi ^2 +3/4}d \varphi$

factor out $3/4$ from the denominator $\frac 2 3\displaystyle\int \frac{1}{(4\varphi ^2)/(3)+1}d \varphi$ and then substitute $z=\frac{2 \varphi}{\sqrt 3}$ and $dz=\frac{2}{\sqrt 3}d \varphi$ so you will get $\frac {1}{\sqrt 3}\displaystyle\int \frac{1}{z^2+1}dz$ and from here it is easy

HINT: the second integral is given by $$\int\frac{x+1}{x^2+x+1}dx$$ you must write the denominator as $$(x+1/2)^2+3/4)$$

Here is one approach to the second integral:

$$\int \frac{x + 1}{ x^2 + x + 1} dx = \frac12 \left(\int \frac{2x + 1}{ x^2 + x + 1} dx + \int \frac{1}{x^2 + x + 1} dx\right)$$

The first integral is identical to the one you integrated. The second is $\arctan$, which can be seen after you complete the square.

$$\int \frac{1}{x^2 + x + 1} dx = \int \frac{1}{x^2 + x + (1/2)^2 - (1/2)^2 + 1} dx$$ $$\int \frac{1}{(x+1/2)^2 + 3/4} dx = \frac{4}{3} \int \frac{1}{\left( \frac{2x+1}{\sqrt 3}\right) + 1} dx = \frac{2}{\sqrt{3}} \int \frac{1}{u^2 + 1} du = \frac{2}{\sqrt{3}} \arctan \left( \frac{2x+1}{\sqrt{3}}\right) + C$$

Where we used the substitution $u = \frac{2x+1}{\sqrt{3}}$.

Hint:
There's no need to split the integral that way (IMO), we can instead write, \begin{align} \int \frac{3x+2}{x^2+x+1}\,\mathrm dx&=\frac{3}{2}\int\frac{2x+\tfrac{4}{3}}{x^2+x+1}\,\mathrm dx\\ &=\frac32\int\dfrac{2x+1+\tfrac13}{x^2+x+1}\,\mathrm dx\\ &=\dfrac32\left(\int\dfrac{2x+1}{x^2+x+1}\,\mathrm dx+\int\dfrac{{\small 1/3}}{x^2+x+1}\,\mathrm dx\right). \end{align} The first one is obvious. For the second one we should complete the square in the denominator, like this, $$\int\dfrac{1}{x^2+x+1}\,\mathrm dx=\int\dfrac{1}{\color{royalblue}{x^2+x+\tfrac14}+1-\tfrac14}\,\mathrm dx=\int\dfrac1{\color{royalblue}{\left(x+\tfrac1{2}\right)^2}+\tfrac34}\,\mathrm dx.$$ Now try to proceed further by rewriting that last expression so as to exploit the fact that the integral of $\frac{u'(x)}{u(x)^2+1}$ is $\arctan u(x)+\rm C$, you may have to factor something, and do one or more substitutions.

Let $3x+2=A\dfrac{d(x^2+x+1)}{dx}+B$

$\iff3x+2=A(2x+1)+B=2Ax+A+B$

$\implies2A=3,A+B=2$

$$\implies\int\dfrac{3x+2}{x^2+x+1}dx=A\cdot\dfrac{d(x^2+x+1)}{x^2+x+1}+B\dfrac{dx}{x^2+x+1}$$

$$\int\dfrac{dx}{x^2+x+1}=\int\dfrac4{(2x+1)^2+(\sqrt3)^2}dx$$

Set $2x+1=\sqrt3\tan y$