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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. | {
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@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 | {
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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 | {
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] - 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? | {
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- 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 | {
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- 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 | {
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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.
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# rank of a matrix | {
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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 | {
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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 | {
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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 | {
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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 | {
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# 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## | {
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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 | {
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# 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 | {
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Kindly excuse my ignorance as I am not a mathematics major.
Thanks
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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 $d(x_m,x_n)<\varepsilon$. | {
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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. | {
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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 $\epsilon$. | {
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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, | {
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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. | {
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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 | {
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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.
$$P(EFG) = P(E)P(F \mid E)P(G \mid EF)$$
$$P(E) = {5 \choose 2}\left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^3$$
$$P(F \mid E) = {3 \choose 1}\left( \frac{1}{5} \right) \left( \frac{4}{5} \right)^2$$ | {
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$$P(F \mid E) = {3 \choose 1}\left( \frac{1}{5} \right) \left( \frac{4}{5} \right)^2$$
$$P(G \mid EF) = {2 \choose 2}\left( \frac{4}{4} \right)^2 \left( \frac{0}{4} \right)^0 = 1$$
$$P(EFG) \approx .062$$
• 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.
$$\mathsf P(N_6{=}2)=\binom 52 \dfrac{1^25^3}{6^5}$$ | {
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$$\mathsf P(N_6{=}2)=\binom 52 \dfrac{1^25^3}{6^5}$$
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. | {
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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 | {
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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 | {
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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 | {
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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 | {
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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 | {
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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 | {
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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) | {
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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. Arkhangel'skii, V.I. A point of $E^n$ such that every neighbourhood of it contains uncountable many points of the set. Reidel ( 1984 ) ( Translated from Russian ) vapor content in the water cycle condensation point math. Point x∈X is called the dew point temperature = Td = condensation point math - ( ( 100 - )! Form ) turns into liquid appeared in | {
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point temperature = Td = condensation point math - ( ( 100 - )! Form ) turns into liquid appeared in Encyclopedia of Mathematics - ISBN 1402006098. https: //encyclopediaofmath.org/index.php? &. Of which every neighborhood contains an uncountable number of points of a condensation point:! Air is damper than about 80 % relative humidity ( RH ) reaches 100 % generalized. Point can be generalized to arbitrary topological Spaces if X=ℝ and a any subset, then does! Point GUIDE for the chart approach science activity \mathbf { R^ { n } } $given! Of condensation points of a if and only if every neighbourhood of it contains uncountable many points of a automatically... Called a condensation point condensation point math a if every neighbourhood of it contains uncountable many points of a volume... There ’ s a lot of pressure article was adapted from an original article by.. A is automatically a condensation point of a if every neighbourhood of it uncountable. Which condensation point math the moisture or water vapor for fun math practice if X=ℝ and a any subset then. Of weather forecasts of Idaho to study different weather patterns works in the is! The English definition dictionary Procedure 1 or water vapor changes state from gas to a obtained. Point can be generalized to arbitrary topological Spaces formulas, and Isolated in! To the brim of water its peak, means 100 %, and mathematic models later in this article,... Adapted from an original article by B.A Cantor set is a condensation point of a substance the... Text and graphics how physical science works in the Cantor set is a condensation point of$ E^n such. Peak, means 100 %, and water precipitates out of the set volume of air containing higher amount water... Melting, freezing, boiling, and water precipitates out of the set of all accumulation points of a used., at 12:13 if X=ℚ and a any subset, then a not. Start with a device called a condensation point of a substance used classification. Air containing higher amount | {
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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 | {
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#### 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: $\sqrt[m]{b^n}$ | {
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# 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 | {
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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. | {
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• 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, | {
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$$0 \leq X_{(1)} \leq 1, \ \ 1 \leq X_{(2)} \leq 1+X_{(1)}$$
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,
$$0 \leq U \leq 1, \ \ 1 \leq \frac{U+2 \pm\sqrt{(U+2)^2+\frac{4(V+1-U^2)}{U-1}}}{2} \leq 1+U \\\\ \implies 0 \leq U \leq 1, \ \ 0 \leq V \leq \frac{U^2(1-U)}{4}$$ | {
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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 $u_1$. | {
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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,
$$P(A^2 > c_{new} | E) = -\int_{u_0}^{u_1}\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c_{new})}}{u-1} \partial u \approx \color{blue}{0.258646}$$ | {
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$$P(A^2 > c_{new}) = \frac{-\int_{u_0}^{u_1}\frac{2 \sqrt{(u-1) (u^3 - u^2 + 4c_{new})}}{u-1} \partial u}{4} = \color{blue}{0.0646615}$$
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 | {
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# 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.
-
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.
-
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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$$P=\frac{1}{3}+\frac{1}{6}P$$
Solving P from this equation we get
$$P=\frac{2}{5}$$
-
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 $\dfrac{2}{5}$.
-
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.
-
If you want to use geometric series, you can do the following.
$$\text{P(Abe wins) = P(Abe wins in the first roll) + P(Abe wins in the second roll) + } \dots$$ $$=\frac{2}{6}+\frac{1}{6}\frac{2}{6} + \left(\frac{1}{6}\right)^2 \frac{2}{6} + \dots$$ $$=\frac{1}{3}\left(1+\frac{1}{6}+\left(\frac{1}{6}\right)^2 + \dots\right)$$ $$=\frac{1}{3}\frac{1}{1-\frac{1}{6}}$$ $$=\frac{2}{5}$$
-
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 $$Tr(((I+A)^{-1})^2)\geqslant\frac{n}{(1+n)^2}$$ | {
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• @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:
$$\verb/Tr/((I_n + A)^{-2}) = \sum_{i=1}^{n} g(\lambda_i) \underbrace{\ge}_{\text{Jensen}} n g\left(\frac1n\sum_{i=1}^n \lambda_i\right) \underbrace{\ge}_{g\text{ decreasing}} n g(1) = \frac{n}{4}$$ | {
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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 | {
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# 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 $$\sum_{k=1}^m A_{ki}b_{k} = \sum_{k=1}^m A_{ik}^{\intercal} b_{k} = A^{\intercal} b$$ | {
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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 | {
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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 $2x+1=\sqrt3\tan y$ | {
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# Midpoint-Convex and Continuous Implies Convex
Given that $$f\left(\frac{x+y}{2}\right)\leqslant \frac{f(x)+f(y)}{2}~,$$ how can I show that $f$ is convex.
Thanks.
Edit: I'm sorry for all the confusion. $f$ is assumed to be continuous on an interval $(a,b)$.
• You can't. At least, not without assuming that $f$ is continuous. – Willie Wong Nov 18 '11 at 15:06
• Lebesgue measurability is enough @WillieWong. This is a theorem by Sierpinski. (If I recall correctly). – Jonas Teuwen Nov 18 '11 at 15:11
• The trick in the continuous case is to set $z = \frac{p}{2^{n + 1}} x + \frac{q}{2^{n + 1}} y$ with $p + q = 2^{n + 1}$ and proceed by induction. – Jonas Teuwen Nov 18 '11 at 15:15
• @Jonas: right. I should have said "... something like '$f$ is continuous' ". The obvious counterexample is, of course, not Lebesgue measurable. Thanks for the correction. – Willie Wong Nov 18 '11 at 17:13
• – Martin Sleziak May 2 '17 at 13:28
Below is the proof of the fact that every midpoint-convex function is rationally convex, which I copied from my older post on a different forum.
If you add the condition that $f$ is continuous, then from rational convexity you will get convexity. (Note that if you are interested only in continuous functions, then it suffices to show the validity of $f(t x + (1-t)y)\le t f(x) + (1-t)f(y)$ for $t=\frac k{2^n}$ as suggested in Jonas' comment. The proof of this fact is a little easier. I've given a little more involved proof, since the relation between midpoint convexity and rational convexity seems to be interesting on its own.)
Maybe I should also mention that midpoint-convex functions are called Jensen convex by some authors.
Note that without some additional conditions on $f$, midpoint convexity does not imply convexity; see this question: Example of a function such that $\varphi\left(\frac{x+y}{2}\right)\leq \frac{\varphi(x)+\varphi(y)}{2}$ but $\varphi$ is not convex | {
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Let $f: \mathbb R\to \mathbb R$ be a midpoint-convex function, i.e. $$f\left(\frac{x+y}2\right) \le \frac{f(x)+f(y)}2$$ for any $x,y \in \mathbb R$.
We will show that then this function fulfills $$f(t x + (1-t)y)\le t f(x) + (1-t)f(y)$$ for any $x,y\in \Bbb R$ and any rational number $t\in\langle0,1\rangle$.
Hint: Cauchy induction: see wikipedia or AoPS or answers to this post.
Proof. It is relatively easy to see that it suffices to show $f([x_1+\dots+x_k]/k)\le [f(x_1)+\dots+f(x_k)]/k$ for any integer $k$ (and any choice of $x_1,\dots,x_k\in \mathbb R$).
The case $k=2^n$ is a straightforward induction.
Now, if $2^{n-1}<k\le 2^n$, then we denote $\overline x=\frac{x_1+\dots+x_k}k$. Now from $$f(\overline x)=f\left(\frac{x_1+\dots+x_k+\overline x+\dots+\overline x}{2^n}\right) \le\frac{f(x_1)+\dots+f(x_k)+(2^n-k)f(\overline x)}{2^n},$$ where $2^n-k$ copies of $\overline x$ are summmed in the middle expression, we get $kf(\overline x) \le f(x_1)+\dots+f(x_k)$ by a simple algebraic manipulation.
The fact that measurability of $f$ is enough for the implication midpoint-convex $\Rightarrow$ convex to hold was mentioned in some of the comments above and in answers to the question I linked. Some references for this fact:
Constantin Niculescu, Lars Erik Persson: Convex functions and their applications, p.60:
H. Blumberg [31] and W. Sierpinski [226] have noted independently that if $f : (a, b) \to \mathbb R$ is measurable and midpoint convex, then $f$ is also continuous (and thus convex). See [212, pp. 220.221] for related results.
[31] H. Blumberg, On convex functions, Trans. Amer. Math. Soc. 20 (1919), 40–44.
[212] A. W. Roberts and D. E. Varberg, Convex Functions, Academic Press, New York and London, 1973.
[226] W. Sierpinski, Sur les fonctions convexes mesurables, Fund. Math. 1 (1920), 125–129. | {
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[226] W. Sierpinski, Sur les fonctions convexes mesurables, Fund. Math. 1 (1920), 125–129.
Marek Kuczma: An introduction to the theory of functional equations and inequalities, p.241. He mentions the book T. Bonnesen and W. Fenchel, Theorie der konvexen Körper, Berlin, 1934 as an additional reference.
• Not to be too much of a whiner, but that "Spoiler" trick is annoying when the point of asking a question here is to get an answer. It's not like accidentally glancing at your text is going to reveal the answer, so if I want to avoid reading it, I avoid reading it. In addition, there is no way to show your text on displays that do not have "mouse-over" - it stays blank on my iPad no matter what I do, for example. There might be fora and questions where it is appropriate, but this doesn't sem to be one of them. – Thomas Andrews Nov 18 '11 at 15:51
• Sorry @Thomas, I've removed it. – Martin Sleziak Nov 18 '11 at 16:01
• @Hi, just double check. When I check the midpoint convexity, I have to show the inequality holds for any $x,y$ right? I can not just pick $x,y$ such that $f(x)=f(y)$, which looks like checking for quasi-concavity. – Bob Oct 7 '16 at 12:29
• @Bob Yes, by definition midpoint convex means that this inequality is true for any $x$, $y$. – Martin Sleziak Oct 7 '16 at 12:49
• @MartinSleziak, I find it is easy for people to make this mistake. Pick $x,y$ such that $f(x)=f(y)$, and once they showed $f(0.5x+0.5y)>f(x)$ they thought they proved concavity, but only got quasi-concavity. We can pick such two points ($x,y$ s.t. $f(x)=f(y)$) wolg for quasi-concavity, but not for concavity. – Bob Oct 7 '16 at 13:31 | {
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As was already previously mentioned it is straightforward to see that for a mid point convex function f $$f\left(\frac{x_1+x_2+\ldots+x_{2^n}}{2^n}\right)\leq\frac{1}{2^n}(f(x_1)+f(x_2)+\ldots f(x_{2^n}))$$ By setting $x_i=x$ for $1\leq i\leq m$ and $x_i=y$ for $m+1\leq i\leq 2^n$ where $1\leq m\leq 2^n$ is an integer we now obtain $$f\left(\frac{m}{2^n}x+(1-\frac{m}{2^n})y\right)\leq \frac{m}{2^n}f(x)+(1-\frac{m}{2^n})f(y)$$ Since the set of rational dyadic numbers of the form $\frac{m}{2^n}$ with $m$ and $n$ integers and $1\leq m\leq 2^n$ is dense in $[0,1]$ the proof follows from the continuity of f.
Can you provide any more information about $f$? For example, the property holds for continuous functions $f: I \rightarrow \mathbb{R}$, $I$ being an interval of real numbers. I think this result is due to Jensen [1].
Theorem (Jensen). Let $f: I\rightarrow\mathbb{R}$ be a continuous function. Then $f$ is convex if and only if it is midpoint convex, i.e. for $x,y$ in $I$ we have
$$f\left(\frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2}$$
[1] J. L. W. V. Jensen, Sur les fonctions convexes et les inégalités entre les valeurs moyennes, Acta Math., 30 (1906), 175-193.
• Jensen doesn't prove this, per se, in this paper (his definition of "convex" was the modern-day "midpoint-convex"). It's more correct to say that he proved Jensen's Inequality (with arbitrary real weights) for functions which are midpoint convex and continuous. Of course, Jensen's Inequality with two arguments gives the modern definition of convexity. – Robert Wolfe Nov 9 '17 at 20:30
Given a weight lambda, take its binary expansion. Think of what midpoint convexity implies to the inequalities involving the partial sums of this binary expansion. Then the result follows by the continuity of f. | {
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I would like to properly show by induction that if $$m\in\{0,1,2,\ldots,2^{n}-1\}$$ then $$f\left(\frac{m}{2^n}x+\Big(1-\frac{m}{2^k}\Big)y\right)\le \frac{m}{2^n}f(x) +\Big(1-\frac{m}{2^n}\Big)f(y).$$ and then the result will follows by contuinity since $$m=\lfloor2^nt\rfloor \in\{0,1,2,\ldots,2^{n}\} ~~~and~~\frac{\lfloor2^nt\rfloor}{2^n}\to t$$
The initial state $$n = 1$$ is trivial by hypothesis.
Now we assume that for every $$k, whenever $$m\in\{0,1,2,\ldots,2^{k-1}-1\}$$,we have $$f\left(\frac{m}{2^k}x+\Big(1-\frac{m}{2^k}\Big)y\right)\le \frac{m}{2^k}f(x) +\Big(1-\frac{m}{2^k}\Big)f(y). \tag{I}\label{I}$$
we want to prove \eqref{I} for $$k=n$$.
Let $$m \in\{0,1,2,\ldots,2^{n}-1\}$$ then the division by 2 yields $$m =2p +r$$ with $$r\in \{0,1\}$$ \begin{align}X&:=\left(\frac{m}{2^n}x+\Big(1-\frac{m}{2^n}\Big)y\right)= \left(\frac{2p +r}{2^n}x+\Big(1-\frac{2p +r}{2^n}\Big)y\right) \\&= \left(\frac{p }{2^n}x+\frac{p +r}{2^n}x+\Big(1-\frac{p }{2^n}y-\frac{p +r}{2^n}y\Big)y\right) \\&= \frac12 \left(\frac{p }{2^{n-1}}x+\frac{p +r}{2^{n-1}}x+\Big(2-\frac{p }{2^{n-1}}y-\frac{p +r}{2^{n-1}}y\Big)y\right) \\&=\frac12\left(\frac{p }{2^{n-1}}x+ \Big(1-\frac{p }{2^{n-1}}y\Big)\right)+\frac12\left(\frac{p+r }{2^{n-1}}x+ \Big(1-\frac{p +r}{2^{n-1}}y\Big)\right) \\&:= \color{red}{\frac{1}{2}\left(X_1+X_2\right)}\end{align}
On the other hand, since $$r\in \{0,1\}$$ and $$m\in\{0,1,2,\ldots,2^{n}-1\}$$ it is easy to check using parity that $$p,p+1 \in \{0,1,2,\ldots,2^{n-1}-1\}$$ that is $$p+r \in \{0,1,2,\ldots,2^{n-1}-1\}$$ | {
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By hypothesis of induction we obtain \begin{align}f(X) &= f\left(\frac{1}{2}\left(X_1+X_2\right)\right)\le \frac12f(X_1) +\frac12f(X_2) \\&= \frac12f\left(\frac{p }{2^{n-1}}x+ \Big(1-\frac{p }{2^{n-1}}y\Big)\right)+\frac12f\left(\frac{p+r }{2^{n-1}}x+ \Big(1-\frac{p +r}{2^{n-1}}y\Big)\right) \\&\le\frac12\left(\frac{p }{2^{n-1}}f(x)+ \Big(1-\frac{p }{2^{n-1}}f(y)\Big)\right)+\frac12\left(\frac{p+r }{2^{n-1}}f(x)+ \Big(1-\frac{p +r}{2^{n-1}}f(y)\Big)\right) \\&= \frac{2p+r}{2^n}f(x) +\Big(1-\frac{2p+r}{2^n}\Big)f(y)\\&= \frac{m}{2^n}f(x) +\Big(1-\frac{m}{2^n}\Big)f(y).\end{align} | {
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An extension of Baire's category theorem
In a topological space, a set is said to be rare if its closure has empty interior, and a set is said to be meager if it is a countable union of rare sets. If meager sets all have empty interior, then the topological space is said to be a Baire space. The following fact is known as Baire's category theorem.
Theorem. Complete metric spaces and locally compact Hausdorff spaces are Baire spaces.
[Zălinescu 2002] provides the following extension of this theorem for complete metric spaces, attributing it to C. Ursescu.
Theorem. ([Zălinescu 2002, Theorem 1.4.5]) Let $$X$$ be a complete metric space, and $$\{S_n\}$$ be a sequence of open sets in $$X$$. Then $$\mathrm{cl}(\cap_{n=1}^\infty S_n)$$ and $$\cap_{n=1}^\infty \mathrm{cl}(S_n)$$ have the same interior.
It is easy to check that a topological space is a Baire space if and only if any countable intersection of dense open sets is still dense in this space. Consequently, the theorem above implies that complete metric spaces are Baire spaces. Thus it is considered as an extension.
Question. Does the extension hold for locally compact Hausdorff spaces?
Update: As pointed out by @Alex Kruckman, the property in Ursescu's theorem is indeed equivalent to the fact that $$X$$ is a Baire space.
Theorem. A topological space is a Baire space if and only if $$\mathrm{cl}(\cap_{n=1}^\infty S_n)$$ and $$\cap_{n=1}^\infty \mathrm{cl}(S_n)$$ have the same interior for any sequence of open sets $$\{S_n\}$$.
Recall that a space is a Baire space if and only if any countable intersection of dense open sets is still dense. The theorem is essentially another way to state that a space is a Baire space if and only if all its open subspaces are Baire spaces. | {
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Combined with Baire's category theorem, this observation by Alex proves the desired theorem. My answer below proves it from scratch, the proof being essentially the same as that of Baire's category theorem for locally compact Hausdorff spaces.
I will accept Alex's answer a few days later if nobody else has anything to add.
Thanks.
This is not really a strengthening of the Baire category theorem - it is equivalent to it. That is, any Baire space (and in particular any compact Hausdorff space) satisfies Ursescu's theorem.
Let's assume $$X$$ is a Baire space. Suppose $$(S_n)_{n\in \mathbb{N}}$$ is a sequence of open sets in $$X$$. We always have $$\text{int}(\text{cl}(\bigcap_{n=1}^\infty S_n))\subseteq \text{int}(\bigcap_{n=1}^\infty \text{cl}(S_n))$$, so it suffices to prove the reverse inclusion. And to do this, it suffices to prove that for any open $$U\subseteq \bigcap_{n=1}^\infty \text{cl}(S_n)$$, we have $$U\subseteq \text{cl}(\bigcap_{n=1}^\infty S_n)$$.
So fix some $$U$$. Note that an open subspace of a Baire space is Baire. [Why? If $$(V_n)_{n\in\mathbb{N}}$$ is a sequence of dense open subsets of $$U$$, then $$(V_n\cup (X\setminus \text{cl}(U)))_{n\in\mathbb{N}}$$ is a sequence of dense open subsets of $$X$$. The intersection of these sets is $$(\bigcap_{n=1}^\infty V_n)\cup (X\setminus \text{cl}(U))$$, and the fact that this is dense in $$X$$ implies that $$(\bigcap_{n=1}^\infty V_n)$$ is dense in $$U$$.]
So defining $$V_n = U\cap S_n$$, we have that each $$V_n$$ is open in $$U$$, and since $$U\subseteq \text{cl}(S_n)$$ for all $$n$$, each $$V_n$$ is dense in $$U$$. Since $$U$$ is Baire, $$\bigcap_{n=1}^\infty V_n$$ is dense in $$U$$, which implies $$U\subseteq \text{cl}(\bigcap_{n=0}^\infty S_n)$$, as was to be shown. | {
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• Thanks, @Alex Kruckman. I modified the question to highlight your observation. In addition, I modified your answer, changing "Zălinescu's theorem" to "Ursescu's theorem", as suggested by [Zălinescu 2002]. Hope this fine with you. Thanks again. – Nuno Nov 15 '20 at 3:12
Here is my attempt to prove it. It is inspired by the proof of [Zălinescu 2002, Theorem 1.4.5] and the classical proof of Baire's theorem for locally compact Hausdorff spaces. Any comments or criticism will be appreciated. Thanks.
Theorem. Let $$X$$ be a locally compact Hausdorff space, and $$\{S_n\}$$ be a sequence of open sets in X. Then $$\mathrm{cl}(\cap_{n=1}^\infty S_n)$$ and $$\cap_{n=1}^\infty \mathrm{cl}(S_n)$$ have the same interior.
proof. It suffices to show that $$\mathrm{int}(\cap_{n=1}^\infty \mathrm{cl}(S_n))\subset \mathrm{cl}(\cap_{n=1}^\infty S_n)$$. To this end, we only need to prove for any given nonempty open set $$U \subset \cap_{n=1}^\infty \mathrm{cl}(S_n)$$ that $$$$\label{eq:usnonempty} U\cap(\cap_{n=1}^\infty S_n) \neq \emptyset.$$$$ In the sequel, we will define a sequence of compact sets $$\{C_n\}_{n=0}^\infty$$ such that $$C_n$$ has nonempty interior and $$$$C_{n+1} \subset C_{n} \subset U\cap(\cap_{k=1}^n S_k) \quad \text{ for each } \quad n \ge 0,$$$$ where $$\cap_{k=1}^0 S_k=X$$. Once this is done, Cantor's Theorem will yield $$$$\label{eq:nestnonempty} \emptyset \;\neq \;\cap_{n=1}^\infty C_n \;\subset\; U\cap (\cap_{n=1}^\infty S_n),$$$$ which gives us what we want. | {
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We define $$\{C_n\}$$ inductively. As $$U$$ is a nonempty open set and $$X$$ is locally compact, we can take a compact set $$C_0\subset U$$ such that $$\mathrm{int}(C_0)\neq \emptyset$$. Assume that $$C_n$$ is already defined for an $$n\ge 0$$ so that $$C_n\subset U\cap(\cap_{k=1}^n S_k)$$ and $$\mathrm{int}(C_n)\neq \emptyset$$. Recalling that $$U\subset\mathrm{cl}(S_{n+1})$$, we have $$\mathrm{int}(C_n)\subset \mathrm{cl}(S_{n+1})$$, which implies that $$\mathrm{int}(C_n) \cap S_{n+1} \neq\emptyset$$. Since $$\mathrm{int}(C_n)\cap S_{n+1}$$ is open, invoking again the local compactness of $$X$$, we can take a compact set $$C_{n+1}\subset \mathrm{int}(C_n) \cap S_{n+1}$$ such that $$\mathrm{int}(C_{n+1})\neq \emptyset$$. Clearly, $$C_{n+1}\subset C_n$$ and $$C_{n+1}\subset C_{n}\cap S_{n+1} \subset U\cap(\cap_{k=1}^{n+1} S_k)$$. This finishes the induction and completes the proof. | {
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Proof that any linear system cannot have exactly 2 solutions.
How would you go about proving that for any system of linear equations (whether all are homogenous or not) can only have either (if this is true):
• One solution
• Infinitely many solutions
• No solutions
I found this a bit difficult to prove (even though its a very fundamental thing about any linear equation). The intuitive geometric explanation is that a line can only intersect at one point, and if it intersects at a later point, it can't be a linear equation, but I don't think this is a convincing proof.
I thought of if you assume that there are two (or more, but I picked two) solutions for some linear system, then for the points in between
Solution Set 1: X1, X2....., Xn
Solution Set 2: X1, X2....., Xn
Then (I think), the points between S1 and S2, must be infinitely many points (and thus infinitely many solutions) such that these points can also satisfy the linear system, which would mean the system has 2 infinite solutions.
However, I don't think this is rigorous enough and nor do I understand completely why its true. Can anyone help in explaining (correcting) and elaborating on the intuition and proof of this? | {
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• "Infinite solutions" is quite incorrect. It can have infinitely many solutions. But in correct use of mathematical terminology, "infinite solutions" means solutions, each one of which is infinite. If you have six solutions, and each is infinite, then those are infinite solutions but not infinitely many solutions. ${}\qquad{}$ – Michael Hardy Sep 16 '15 at 22:09
• Why do you find the visual proof not convincing? – corsiKa Sep 16 '15 at 22:53
• It's worth pointing out that this question has an implicit assumption that the underlying field is infinite. Over a finite field, the assertion is not necessarily true (and we cannot have infinitely many solutions). E.g. over $GF(2)$, the equation $\pmatrix{1&1\\ 0&0}x=0$ has exactly two solutions $x=\pmatrix{0\\ 0}$ and $x=\pmatrix{1\\ 1}$. See also 6005's answer below. – user1551 Sep 17 '15 at 11:35
Suppose that $\vec v$ and $\vec w$ are distinct solutions for the system $A\vec x = \vec b$ so that $A \vec v = A \vec w = \vec b$. Then $\frac{1}{2}(\vec v + \vec w)$ must be distinct from both $\vec v$ and $\vec w$ and must also solve the system since: $$A(\tfrac{1}{2}(\vec v + \vec w)) = \tfrac{1}{2}(A\vec v + A\vec w) = \tfrac{1}{2}(\vec b + \vec b) = \vec b$$ We can then apply the same argument to $\vec v$ and $\frac{1}{2}(\vec v + \vec w)$ in order to get infinitely many distinct solutions.
• Or pick any $\lambda \in (0,1)$ and consider $\lambda \vec{v} + (1 - \lambda) \vec{w}$. Then you get continuum-many solutions right off the bat. – 6005 Sep 16 '15 at 22:10
• You can even choose any real number for $\lambda$. – Aurey Sep 16 '15 at 23:30
• Looking over this again, great way of explaining your argument (deriving infinitely many solutions) – q.Then Sep 17 '15 at 22:29 | {
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Your second proof sounds fine. Let's imagine that the system of equations is written as $Ax = b$, where $A$ is the matrix of coefficients, $x$ is the vector of variables we are solving for, and $b$ are the constants. Now assume we have two distinct solutions $x_1$ and $x_2$. Then $rx_1+sx_2$ is also a solution, if $r+s=1$ - just plug it in and use linearity of matrix multiplication.
$$A(rx_1+sx_2) = A(rx_1)+A(sx_2) = rA(x_1)+sA(x_2) = (r+s)b = b$$
We thus have an infinite number of solutions. If you don't like matrices, you can easily just write out each term in the system of equations.
If you want to get some intuition for what is happening, the linear combination $rx_1+sx_2$, where $r+s=1$, is a line that connects the two points. When $r$ is 1 and $s$ is 0, we get point $x_1$. When $s$ is 1 and $r$ is 0, we get point $x_2$. Other combinations of $r$ and $s$ give other points.
You need to be more careful with your first argument, though. In general, when graphing systems of $n$ equations in $n$ unknowns, the equations are not lines, but $(n-1)$-dimensional planes in $n$-dimensional space. So for a system of three equations and three unknowns, for example, we have three planes in three dimensional space. Unless the planes are parallel, the intersection of two planes is a line, and the intersection of a line with a plane is a point. That's the one solution case. If two planes are parallel, they never intersect - no solutions. And finally, if the line/plane overlaps with another line/plane, we get an infinite number of solutions. That's the intuition behind this theorem.
• Thanks for the answer, I'm sure other answers were great as well, but I think you explained it the best (at least the best for me) – q.Then Sep 17 '15 at 2:55 | {
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We can represent a linear system in matrix notation as:$A\vec x=\vec v$. Now suppose that we have $A\vec x=\vec v$ and $A\vec y=\vec v$ with $\vec x \ne \vec y$, for $a,b$ with $a+b=1$ we have, by linearity: $A(a\vec x + b\vec y)=aA\vec x +b A \vec y=\vec v$, so we have infinitely many solutions.
In general, in a vector space over a general field $F$, the number of solutions $\vec{x}$ to the system $$A \vec{x} = \vec{w}$$ is either $0$ or $|\ker A|$. (Because if $\vec{x}_0$ satisfies the equation, then the set of solutions is $\vec{x}_0 + \ker A$.) And $|\ker A| = |F|^k$ for some $k \ge 0$.
Assuming the base field $F$ has infinite cardinality $\alpha$ and the vector space is finite-dimensional, it follows that the number of solutions is $0, 1,$ or $\alpha$ (since $|F|^0 = 1$ and $|F|^k = \alpha$ for $k \ge 1$).
• @user1551 yes, thank you. I corrected that and two other minor typos. – 6005 Sep 17 '15 at 15:20 | {
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#1
How many positive integers less than $1000$ have the property that the sum of the digits of each such number is divisible by $7$ and the number itself is divisible by $3$?
Note by Vilakshan Gupta
3 years, 1 month ago
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Let's think of a number $abc (0 \leq a, b, c \leq 9)$. $a+b+c \equiv 0 (\mod 3 \text{and} \mod 7)$. Thus, $a+b+c=21$. | {
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$(3, 9, 9) \rightarrow \frac{3!}{2}, (4, 8, 9) \rightarrow 3!, (5, 7, 9) \rightarrow 3!, (5, 8, 8) \rightarrow \frac{3!}{2}, (6, 6, 9) \rightarrow \frac{3!}{2}, (6, 7, 8) \rightarrow 3!, (7, 7, 7)$ $3+6+6+3+3+6+1=28$
Please tell me if there is any error.
- 2 years, 11 months ago
Nice method
- 2 years, 11 months ago
Good one brother
- 2 years, 11 months ago
Your method is quite efficient vis-a-vis mine. The latter involved manual trials with 3 digit integers with integer 1 to 9 at the unit. Thank you
- 1 year, 1 month ago
What ans did you get?
- 3 years, 1 month ago
Hey Aaron. How much are u getting with bonus? With bonus i am getting 12.
- 3 years, 1 month ago
25 is the answer as per me
- 3 years, 1 month ago
No - I get 28 too - I constructed a 0 - 9 by 0 -9 addition table in excel, and then started adding a 3rd digit to any number whose 2 digits had added to 12 or more - though now I think about it, I could just as easiky have srated my list with 399 and continued from there. And it has to be 28 cos it's one starting with3, 2 starting with 4, 3 starting with 5 etc, and 1+2+3+4+5+6+7 = 28
- 3 years, 1 month ago
28
- 3 years, 1 month ago
we just need to find the numbers which add upto 21
- 3 years, 1 month ago
- 3 years, 1 month ago
How mañy have you got right in PRMO - 17?
- 3 years, 1 month ago
@Md Zuhair Is the paper for 9,10,11 and 12 same?
- 3 years, 1 month ago
Yes Sir!
- 3 years, 1 month ago
unfortunately, i will get only 10 questions correct. I did very silly mistakes
- 3 years, 1 month ago
Sir😅 I am getting 10 along with bonus!
- 3 years, 1 month ago
Oh. U mean 8/28 u r getting?
- 3 years, 1 month ago
well, if it is bonus that means 2 questions marks are given extra.If it was been written question deleted then scores would be evaluated out of 28
- 3 years, 1 month ago
Geometry was quite tough and lengthy! Excluding bonus , I'm getting 8
- 3 years, 1 month ago
oh...btw,where do u live (i mean which region) | {
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- 3 years, 1 month ago
oh...btw,where do u live (i mean which region)
- 3 years, 1 month ago
Rajasthan..U?
- 3 years, 1 month ago
oh
- 3 years, 1 month ago
So , you are already selected..Great 👍
- 3 years, 1 month ago
- 3 years, 1 month ago
Coz as per cutoff(s) uploaded by Resonance , cutoff in Chandigarh is lower than others ( Rajasthan , Maharashtra , UP , etc) ... That's why!
- 3 years, 1 month ago
According to Resonance , cutoff in Chandigarh is just 4(questions)
- 3 years, 1 month ago
Ya. Thats ridiculous. WB region has always got a higher cutoff...
- 3 years, 1 month ago
i don't think it will be so low
- 3 years, 1 month ago
If that isnt, then wb will be higher and i will surely not qualify
- 3 years, 1 month ago
It's 11 in Rajasthan ! 😅😒
- 3 years, 1 month ago
LetTheFateDecide !!Bye
- 3 years, 1 month ago
which class are u in toshit?
- 3 years, 1 month ago
@Shreyan Chakraborty .. How much?
- 3 years, 1 month ago
JANI NA BAJE HOYECHE
- 3 years, 1 month ago
ANSWER IS 28....HAS A BIJECTION WITH a+b+c=21 WHERE 0<a,b,c<=9........
- 3 years, 1 month ago
Are na na.... I am not telling that. How much are you getting?
- 3 years, 1 month ago
@Shreyan Chakraborty The Hundreds digit can't be 1 or 2..
- 3 years, 1 month ago
yeah hundreds digit cant be 1,2
- 3 years, 1 month ago
@Md Zuhair @Vilakshan Gupta I haven't attempted one of the bonus question. Will I still get marks for it?
- 3 years, 1 month ago
I think the question can be cancelled as how can a person attempt to decinal answers and i had attempted ine. So i dunno.
- 3 years, 1 month ago
Yup.I think so.
- 3 years, 1 month ago
- 3 years, 1 month ago | {
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- 3 years, 1 month ago
Yup.I think so.
- 3 years, 1 month ago
- 3 years, 1 month ago
I believe the answer is 28 integers. The sum of these integers' digits must be divisible by 21, since a number divisible by 3 also has its sum of digits divisible by 3; in addition to the sum of digits divisible by 7. None of the digits can be less 3 since the sum of digits would be less than 21. Possible combinations = 7+6+5+4+3+2+1 = (7+1)+(6+2)+(5+3)+4=3*8+4=28.
- 3 years, 1 month ago
Exactly
- 3 years, 1 month ago
I agree with all that, and I got the same answer, but if I give you 4 digits at random (say 3, 4, 5 and 6) and ask how many numbers you can make out of them, the answer is 432*1 = 24, not 4+3+2+1 = 10. What am I missing?
- 3 years ago
Ah - that's where the 28 comes from - much more mathematical than my just listing and counting them
- 3 years, 1 month ago
Hello, There are 28 postive integers left less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by 3 You can check out for more queries related to the JEE EXAMS from the following compilation
- 3 years ago
Hey I'm getting 8/27 from jharkhand as per the new answer key of hbcse.will I qualify???
- 3 years ago
Lets see....
- 3 years ago
Hey , what does discounted actually refer to?
- 3 years ago
Are the marks gonna be added to everyone's total or the questions will be cancelled (lowering the cutoff)?
- 3 years ago
They will be added to totsl
- 2 years, 11 months ago
The questions will be cancelled
- 3 years ago
@Pokhraj Harshal Ok! Then I'm also getting the same...
- 3 years ago
How much? Without the question?
- 3 years ago
Which region are u from??
- 3 years ago
WB rgion
- 3 years ago
And what about the other participants and their marks from your school. I mean the averages and the highest marks
- 3 years ago
@Pokhraj Harshal Rajasthan region!
- 3 years ago
@Md Zuhair 8
- 3 years ago
O i see.... | {
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@Pokhraj Harshal Rajasthan region!
- 3 years ago
@Md Zuhair 8
- 3 years ago
O i see....
- 3 years ago
this question came in this year PRMO answer is 28
- 2 years, 11 months ago
its sum is divisibli by 21 using this you can solve
- 2 years, 11 months ago
27
- 2 years, 11 months ago
28
- 2 years, 9 months ago
33
- 2 years, 2 months ago
zuhair tui ki amk jiggesh korchish??
- 3 years, 1 month ago
Accha.. nijer whatsapp number ta de... whatsapp e kotha bolchi
- 3 years, 1 month ago | {
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# Integration - computer problem
#### Yankel
##### Active member
Hello,
I was trying to solve the integral of
sin(x)*cos(x)
using the substitution method, what I did was:
u=sin(x) and that yields du/dx = cos(x) and then du=cos(x)*dx
that comes to an integral of u*du, which is easy u^2 / 2 +C. substituting back gives the final answer
sin(x)^2 / 2
But, when I ran this integral in both Maple and Mathematica, I got this answer:
-cos(x)^2 / 2
Now I tried asking Maple if the two answers are the same, but it failed. I tried checking myself, using the relation sin(x)^2+cos(x)^2=1, and got to the conclusion that they don't. I don't see what I did wrong here...
#### Ackbach
##### Indicium Physicus
Staff member
Technically, you answer was not $\sin^{2}(x)/2$, but $\sin^{2}(x)/2+C.$ Likewise, the computer's answer was not $-\cos^{2}(x)/2$, but $-\cos^{2}(x)/2+C$. Since your two answers differ by a constant (namely, $1$), they are really the same answer from a calculus perspective.
#### Bacterius
##### Well-known member
MHB Math Helper
Technically, you answer was not $\sin^{2}(x)/2$, but $\sin^{2}(x)/2+C.$ Likewise, the computer's answer was not $-\cos^{2}(x)/2$, but $-\cos^{2}(x)/2+C$. Since your two answers differ by a constant (namely, $1$), they are really the same answer from a calculus perspective.
Or more clearly, $C_1$ and $C_2$. This happened to me a while ago, it tends to occur a lot with trigonometric functions because of their periodic identities, so you often end up with very different looking expressions which are in fact equal..
#### Yankel
##### Active member
ah, I wasn't thinking about it....interesting.
well done to both of you ! thanks !
#### topsquark
##### Well-known member
MHB Math Helper
Just to be cheeky about it:
$$\int sin(x)~cos(x) dx = \frac{1}{2} \int 2 sin(x)~cos(x) dx = \frac{1}{2} \int sin(2x)dx$$
After u = 2x:
$$= \int sin(x)~cos(x) dx = \frac{1}{2} \int sin(u) \cdot \frac{1}{2} du$$ | {
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After u = 2x:
$$= \int sin(x)~cos(x) dx = \frac{1}{2} \int sin(u) \cdot \frac{1}{2} du$$
$$= -\frac{1}{4} cos(u) + C = -\frac{1}{4} cos(2x) + C$$
and upon using $$cos(2x) = cos^2(x) - sin^2(x)$$
$$\int sin(x)~cos(x) dx = \frac{1}{4} \left ( sin^2(x) - cos^2(x) \right ) + C$$
which contains both your sin and cos terms and is also off by a constant from the other solutions.
-Dan | {
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# Counting integer solutions with conditional restrictions
Consider $x_i$, with $i=1,\ldots, 10$, such that $$5 \leq 3(x_1 + x_2 + x_3 + x_4) +2(x_5 + x_6 + x_7 + x_8 + x_9 + x_{10}) \leq 12\,,$$ where each $x_i$ can be either $0$ or $1$. In addition, each $x_i$ from $i=5$ to $10$ are restricted to be $0$ if some of the $x_i$ from $i=1$ to $4$ are equal to $1$, according to the following diagram:
Each $x_i$ in the bottom line is restricted by two of the $x_i$ in the top line. For example, if $x_1 = x_2 = 0$, then $x_5$ can be either $0$ or $1$, but if $x_1=1$ and/or $x_2 = 1$, then $x_5=0$ necessarily. The problem is to know in how many ways can we satisfy the inequality, subjected to these restrictions, using a method which is easily used for a computer, like generating functions, for example.
The way I'm trying to answer this is the following: if the additional restrictions did not exist, the number of solutions would be obtainable by using the generating function $$(1+x^3)^4 (1+x^2)^6$$ and the solution would be the sum of all coefficients from $x^5$ to $x^{12}$. The restrictions can be incorporated by defining $x^c_i$ such that, in the case of $i=5$ for example, $$x^c_5 = \quad \left\{ \begin{array}{lll} 0, \quad \, \textrm{if} \quad (1-x_1)(1-x_2)=0 \\ \\ x_5, \quad \, \textrm{if} \quad (1-x_1)(1-x_2)=1 \\ \end{array} \right. \,,$$ and so on for the other variables. Using these $x^c_i$ in the inequality, instead of the $x_i$ (for $i=5$ up to $10$), we have $$5 \leq 3(x_1 + x_2 + x_3 + x_4) +2(x^c_5 + x^c_6 + x^c_7 + x^c_8 + x^c_9 + x^c_{10}) \leq 12$$ which in principle should contain all the restricted possibilities. This is where I'm currently stuck, as I don't know how the generating function for this inequality should be built. | {
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• Not the approach you are pursuing, but just building the characteristic function of the set yields a count of 87. – Fabio Somenzi Feb 23 '17 at 18:52
• Thanks for the reply! I'm not sure what you mean by characteristic function of the set though, could you be more specific as how you did it? – GKiu Feb 23 '17 at 20:15
• The characteristic function I referred to is a function of Boolean variables $x_1,\ldots,x_{10}$ that evaluates to true if and only if the constraints are all satisfied. It can be built as the conjunction of characteristic functions for the individual constraints. The incompatibility constraints are simple: for example $\neg x_5 \vee (\neg x_1 \wedge \neg x_2)$. The linear inequalities are a bit more complicated, but not much. Of course, I let the computer take the conjunctions and count the models. – Fabio Somenzi Feb 23 '17 at 21:31
• Although this works, I was looking for an answer in terms of generating functions like @Jeremy Dover suggested, but thank you for the clarification nonetheless. – GKiu Feb 24 '17 at 19:09
• Understood. That's why I posted a comment and not an answer. – Fabio Somenzi Feb 24 '17 at 19:45
This only really works because of the symmetry between the variables $x_1 \ldots x_4$ and $x_5 \ldots x_{10}$, but I think you can build a generating function this way:
If none of the $x_1 \ldots x_4$ are 1, then any of the $x_5 \ldots x_{10}$ can be 0 or 1 freely, so we get the generating function $(1+x^2)^6$ to count these solutions.
If exactly one of the $x_1 \ldots x_4$ is 1, then three of the $x_5 \ldots x_{10}$ must be zero, but the remainder are free, yielding the generating function $4x^3(1+x^2)^3$.
If exactly two of the $x_1 \ldots x_4$ are 1, then five of the $x_5 \ldots x_{10}$ must be zero with only one free, so we get the generating function $6x^6(1+x^2)$.
If three or four of the $x_1 \ldots x_4$ are 1, then all of the $x_5 \ldots x_{10}$ must be zero, so we get the generating function $4x^9+x^{12}$. | {
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We can add all of these up to get the generating function: $$(1+x^2)^6 + 4x^3(1+x^2)^3 + 6x^6(1+x^2) + 4x^9+x^{12}$$ Like @Fabio Somenzi, when I calculate I also find 87 to be the answer.
• Thank you very much for your answer! This is exactly what I was looking for. – GKiu Feb 24 '17 at 19:07 | {
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# statistics probability
A bag contains 8 red marbles, 5 white marbles, and 6 blue marbles. You draw 4 marbles out at random, without replacement. Find the following probabilities and round to 4 decimal places.
a. The probability that all the marbles are red is
b. The probability that none of the marbles are red is
1. 👍
2. 👎
3. 👁
1. prob(all 4 red) = (8/19)(7/18)(6/17)(5/16) = 35/1935
or
prob(all red) = C(8,4)/C(19,4) = 70/3876 = 35/1938
Prob(none red) = C(11,4)/C(19,4) = 330/3876 = 55/646
or
prob(none red) = (11/19)(10/18)(9/17)(8/16) = 55/646
1. 👍
2. 👎
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## How do I evaluate this log function?
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
I assume I can't use a calculator obviously.. so I'm quite stuck. The answer is 5, but I have no idea how to get that.
(log23)(log34)(log45) ... (log3132)
PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
Recognitions: Homework Help Use the fact that $$\log_b a = \frac{\log_c a}{\log_c b}$$ for any positive, real numbers a, b and c (with c > 1).
That was a fast response, and a fact that I didn't know. How do I create equations on this forum BTW?
## How do I evaluate this log function?
Quote by feihong47 That was a fast response, and a fact that I didn't know. How do I create equations on this forum BTW?
You use LaTeX, that is, the [ itex ][ /itex ] -tags (remove the spaces) . There's a guide in the forums somewhere if you're not familiar with it, I'll edit a link to it to this post if I find it. You can also open the "LaTeX Reference" by clicking the Σ in the toolbar.
If you don't have to write anything complicated, you can just use the quick symbols and x2 and x2 buttons in the toolbar. LaTeX looks a lot nicer & it's easier to read, though.
EDIT: Here's the LaTeX guide.
Mentor
Quote by feihong47 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution I assume I can't use a calculator obviously.. so I'm quite stuck. The answer is 5, but I have no idea how to get that. (log23)(log34)(log45) ... (log3132)
You could also approach this as follows:
Let $\displaystyle y=(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\,\dots\,( \log_{31}32)$
Then, $\displaystyle 2^y=2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)}$
By laws of exponents and the definition of a logarithm, | {
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By laws of exponents and the definition of a logarithm,
$2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)}$
$=\left(2^{(\log_{2}3)}\right)^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$
$=3^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$
$=\left(3^{(\log_{3}4)}\right)^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$
$=4^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$
etc.
$=\left(31^{(\log_{31}32)}\right)$
$=32$
Recognitions: Gold Member Science Advisor Staff Emeritus So, just in case others misunderstand, $2^y= 32= 2^5$ and therefore, y= 5 as the original poster said.
Very nice approach. Thanks! | {
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# What is the difference between max and sup?
I am studying KS (Kolmogorov-Sinai) entropy of order q and it can be defined as
$$h_q = \sup_P \left(\lim_{m\to\infty}\left(\frac 1 m H_q(m,ε)\right)\right)$$
Why is it defined as supremum over all possible partitions P and not maximum?
When do people use supremum and when maximum?
-
I am not sure either of the tags fit. I do not know the term "KS entropy" so I cannot suggest a better retag. The essence of "what is $\sup$ and what is $\max$ was answered on this website before. Here is one answer, math.stackexchange.com/questions/18605/… – Asaf Karagila Jul 27 '11 at 15:55
Based on a quick google search I believe [dynamical-systems] fit well into this question. I (somewhat unwillingly) left [order-theory] since your second question can be generally answered within its confines. As my previous comment suggests, the second answer was indeed answered before (many times as well), I do not vote to close as the first answer may or may not been addressed to in other questions on the site. – Asaf Karagila Jul 27 '11 at 16:01
@Asaf, KS stands for Kolmogorov-Sinai, I saw that question earlier, but I still don't get when people use sup and when max. It's probably coz I need to spend more time to think over all the answers there. But thanks for the link anyway. Pls feel free to retag. – oleksii Jul 27 '11 at 16:03
"'Sup" is a hip way of asking "How are you?" – oosterwal Jul 27 '11 at 20:06
@oleksii: You have received several answers, all are very good in answering the question in the title. None of which answer the question "Why is it defined as supremum over all possible partitions $P$ and not maximum?". While I respect your wish to accept the answer, it would be nice to indicate that you have at least deduced the answer of the seemingly unanswered question from the accepted answer. Otherwise, this has become a duplicate of a question that has been answered here many times before. – Asaf Karagila Jul 28 '11 at 0:21 | {
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If the maximum exists, then the supremum and maximum are the same. However sometimes the maximum does not exist, and there is no maximal element. In this case it still makes sense to talk about a least upper bound.
The classic example is the set of all rationals whose square is less than or equal to $2$. That is the set $$A=\left\{ r\in\mathbb{Q}:\ r^{2}\leq2\right\}.$$
$A$ has no maximal element, however it does have a supremum and $\sup A=\sqrt{2}$.
An even simpler example is the set of all reals that are strictly less than $2$: $$B=\left\{ r\in\mathbb{R}:\ r<2\right\}.$$ This set has no maximum since for any $x\in B$ the element $\frac{x+2}{2}$ satisfies $x<\frac{x+2}{2}<2$. However it is not hard to see that $\sup B=2$.
-
Tnx, this is a nice answer, +1 for easy samples – oleksii Jul 27 '11 at 16:17
Consider for example the set $X = (0,1)$. Then $$\sup X=1$$ but $\max X$ does not exist.
Generally, for a set $X\subset {\mathbb R}$, we define $x:=\max X$ if $x\in X$ and $$\forall y\in X, y\leq x.$$
We define $x:=\sup X$ if $$\forall y\in X, y\leq x$$ and $$\forall\epsilon>0,\quad\exists y\in X, \quad\text{s.t.}\quad y>x-\epsilon.$$
-
I quite like your first answer, it was very plain and easy to get [for me :)]. Tnx. – oleksii Jul 27 '11 at 16:15
Supremum need not be attained while maximum is. When maximum exist, maximum=supremum.
-
The set of all negative numbers has a sup, which is 0, but not a max.
0 is the smallest number that no negative number can exceed.
-
A maximum is the largest number WITHIN a set. A sup is a number that BOUNDS a set. A sup may or may not be part of the set itself (0 is not part of the set of negative numbers, but it is a sup because it is the least upper bound). If the sup IS part of the set, it is also the max.
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# Which is the correct way to calculate the expected value of a shared lottery jackpot?
I want to calculate the expected value of a ticket in a lottery game which gives players a probability $p$ of winning a jackpot prize of $j$ dollars. The total number of tickets in play is $t$.
If every winning ticket gets the full prize amount, the expected value for a ticket is given by $jp$. However, if winners must evenly split the prize in case of multiple winners, then the expected value depends on the number of winners $W$.
The expected number of winners is $tp$. The probability that the number of winners $W$ is $w = 0, 1, 2, \dotsc$, follows a Poisson distribution with the expected number of winners as its parameter:
$$P(W=w) \sim Pois(tp) = \frac{tp^we^{-tp}}{w!}$$
I don't know how to get from there to calculating an accurate expected value for the ticket as a function of the number of tickets in play.
In reading online, I've found two different methods each used by several sources. If I'm following them correctly, then they give different results. My question is 1) which one is correct? 2) what is the error in reasoning (or in my understanding/implementation) in the incorrect method?
## Method 1: Number of Winners
The first method calculates the probability that the number of winners $W$ will be $w = 0, 1, \dotsc, t$, given that there is at least one winner:
$$P(W=w | W>0) = \frac{P(W>0|W=w)P(W=w)}{P(W>0)}$$
Where,
• $P(W>0|W=w)$ is $\left\{ \begin{array}{lr} 0 & : w = 0\\ 1 & : w > 0 \end{array} \right.$
• $P(W=w)$ is the probability of $w$ winners: $\frac{tp^we^{-tp}}{w!}$
• $P(W>0)$ is the probability of more than one winner: $1 - P(W=0)$
So the expected value of the ticket is given by:
$$p\sum_{w=1}^{t} \frac{j}{w}\frac{P(W=w)}{1-P(W=0)}$$ | {
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$$p\sum_{w=1}^{t} \frac{j}{w}\frac{P(W=w)}{1-P(W=0)}$$
For a numerical example, we'll tabulate the first few values of $P(W=w)$ for a lottery with a 1/34,220 chance of winning \$100,000 jackpot, with 6,000 tickets in play, so$p = 1/34,220; j = 100,000; \text{and } t = 6,000$$$\begin{array}{c|c|c|c|c|} \text{Winners} & \text{Probability} & \text{Conditional Probability} & \text{Share} & \text{Contribution } \\ w & P(W=w) & P(W=w|W>0) & j/w & (j/w)P(W=w|W>0) \\ \hline 0 & 0.839 & 0 & \text{\0} & \text{\0} \\ \hline 1 & 0.147 & 0.913 & \text{\100,000} & \text{\91,300} \\ \hline 2 & 0.013 & 0.081 & \text{\50,000} & \text{\4,050} \\ \hline \end{array}$$ Summing the contribution column and multiplying by$p$gives an expected value of$2.79.
## Method 2: Number of Other Winners
The second method calculates the probability that the number of total winners $W$ is $w = 0, 1, \dotsc, t$, given that our ticket is a winner:
$$P(W=w|Winner) = \frac{P(Winner|W=w)P(W=w)}{P(Winner)}$$
Where,
• $P(Winner)$ is the probability that our ticket is a winner: $p$
• $P(Winner|W=w)$ is the probability that our ticket is a winner given $w$ winning tickets: $w/t$
• $P(W=w)$ is the probability of $w$ winners: $\frac{tp^we^{-tp}}{w!}$
Plugging those figures in shows that $P(W=w|Winner)$ reduces to $P(W=w-1)$:
$$\frac{w}{t}\frac{P(W=w)}{p} = \frac{tp^{w-1}e^{-tp}}{(w-1)!} = P(W=w-1)$$
So the expected value is given by:
$$p\sum_{w=1}^{t}\frac{j}{w}\frac{tp^{w-1}e^{-tp}}{(w-1)!}$$
Using the same lottery numbers as above, the first few values of $w$ are given in the following table.
$$\begin{array}{c|c|c|c|c|} \text{Winners} & \text{Probability} & \text{Conditional Probability} & \text{Share} & \text{Contribution } \\ w & P(W=w) & P(W=w|Winner) & j/w & (j/w)P(W=w|Winner) \\ \hline 0 & 0.839 & 0 & \text{n/a} & \text{\0} \\ \hline 1 & 0.147 & 0.839 & \text{\100,000} & \text{\83,900} \\ \hline 2 & 0.013 & 0.147 & \text{\50,000} & \text{\7,350} \\ \hline \end{array}$$ | {
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Summing the contribution column and multiplying by $p$ gives an expected value of $2.67. ### Online Resources Which Use Method 2 Clearly the expected payout for the example lottery above cannot be both \$2.79 and \$2.67, but I'm having a difficult time reasoning my way to the correct method. Any hints will be appreciated! - ## 4 Answers Proof that Methods 2 and 3 are equivalent: Method 3 has a very intuitive appeal as the correct approach when interpreted as someone purchasing$allt$tickets. Now we'll show that Method 2 gives the same result if we use the binomial for the distribution of winning tickets instead of the Poisson. The expected value for Method 2 is: $$p\sum_{w=1}^t\frac jw {t-1 \choose w-1}p^{w-1}(1-p)^{t-w}$$ $$=j \sum_{w=1}^t \frac {(t-1)!}{w!(t-w)!}p^{w}(1-p)^{t-w}$$ $$=\frac jt \sum_{w=1}^t {t \choose w}p^{w}(1-p)^{t-w}$$ $$=\frac jt (1-(1-p)^t )$$ which is also the result for Method 3. If we had started with the Poisson distribution as an approximation to the binomial, then the expected value for Method 2 reduces to $$\frac jt(1-e^{-tp})$$ The Poisson will be a good approximation to the binomial when$t$is large,$p$is small and$tp$is moderate, which holds here. - Method 3: By linearity of expectation, the expected value of one ticket is$\frac{1}{t}$times the expected total value of all the tickets. That expected value is$jP(W>0)=j(1-P(W=0))=j(1-(1-p)^t)$. Thus the expected value of one ticket is $$\frac{j}{t}(1-(1-p)^t))$$ For the numerical values you use, I get \$2.68. I'm not skilled enough at statistics to find the logical flaw in Method 1, but I suspect Method 2 is right (up to rounding).
Followup: it seems that this solution, or a variation, was listed under "resources" supporting the second solution. I didn't notice it on my first read-through due to wall of text. | {
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-
Thank you, vadim. Yeah, I had a link to this answer buried in the resources section, but your mentioning that expectation is linear helped me understand why it is correct without having to actually enumerate and sum the probability of each number of winners. That leaves me very confident that Method 2 is correct. However, I would like to find a good way of explaining why Method 1 is incorrect. – cristoper Jun 29 '14 at 5:25
I suspect that the problem with Method 1 is that the probability my ticket is a winner is no longer $p$ when we have conditioned on there being exactly $w$ winners. – vadim123 Jun 29 '14 at 5:34
This is my own summary answer. If somebody has a better way of explaining why the first method in my question goes wrong (and what it does compute, if anything useful), please post an answer!
## Method 1 is incorrect
If $P(Winner)$ is the probability that our ticket is a winner, then as user159813 pointed out, the expected payout from the lottery can be expressed as:
$$\sum_{w=1}^{t} \frac{j}{t} P(Winner \cap W=w)$$
However, the payout calculated by Method 1 is:
$$\sum_{w=1}^t \frac{j}{t} P(W>0 \cap W=w)$$
Which is different because $P(W>0) \ne P(Winner)$. We want $P(W=w)$ given that our ticket won, which is a lesser probability than given that any ticket won.
## Method 2 is correct
As vadim123 points out in his answer, the correctness of Method 2 is confirmed by the nice, closed-form formula derived from the fact that the expected value of a single ticket is the same as the expected value of all tickets divided by $t$:
$$\frac{j}{t}(1-(1-p)^t)$$
## Simulation
I ran a simulation using the same numbers as the example in my question. The mean payout after the 10,000,000th iteration was about \$2.64. After about 5,000,000 iterations, the mean seemed to oscillate about the$2.68 line.
-
There is a problem in Method 1. Letting $Y$ represent amount won and $W$ be number of winners we see that $$E(Y)=\sum_{w=0}^{t}\frac{j}{w}P(Winner\cap W=w)$$ | {
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The second method does this correctly by calculating $$P(Winner\cap W=w)=P(W=w|Winner)P(Winner)=P(W=w-1)P(Winner)=P(W=w-1)p$$
as you have above. Now the problem with Method 1 is that it says that $$P(Winner\cap W=w)=P(W=w|W>0)P(Winner)$$ which is not true. I would guess the method this person was going for is actually $$P(Winner\cap W=w)=P(Winner| W=w)P(W=w)$$
-
I don't see where Method 1 says $P(Winner \cap W=w)=P(W=w|W>0)P(Winner)$. However, I do see where it says it equals $P(W=w|W>0)P(W>0)$. And looking at your formulation of the expectation function, it is easier for me to see that Method 1 is wrong because $P(W=w|W>0)P(W>0) \ne P(W=w|Winner)P(Winner)$ – cristoper Jun 29 '14 at 20:34
And should the index in your sum start at $w=1$ to avoid the divide by zero? $P(Winner \cap W = 0) = 0$ anyway. – cristoper Jun 29 '14 at 20:47 | {
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# Semi Log Regression | {
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Log the Bugs in the Quality Center (Bug Logging and Tracking Tool) and track them. Measure distance each band traveled 3. This algorithm uses two k -nearest neighbor regressors with different distance metrics, each of which labels the unlabeled data for the other regressor where the labeling confidence is estimated through consulting the influence of the labeling of unlabeled examples on the labeled ones. Semi-logarithmic regressions, in which the dependent variable is the natural logarithm of the variable of interest, are widely used in empirical economics and other fields. Why is it that when you log-transform a power function, you get a straight line? To show you, let's remember one of the most fundamental rules of algebra: you can do anything you want to one side of an equation - as long as you do the exact same thing to the other side (We just LOVE that rule!). For many real-world problems, however, acquiri. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): Abstract Semi-supervised methods use unlabeled data in addition to labeled data to con-struct predictors. This page aims at providing to the machine learning researchers a set of benchmarks to analyze the behavior of the learning methods. Poisson regression for contingency tables, a type of generalized linear model. Quadratic regression. The Regression Analysis Method Page Content Space Toy Co. This page allows performing logarithmic regressions (logarithmic least squares fittings). Semi-supervised learning and multi-task learning are two of the approaches that have been proposed to alleviate this problem. Introduction. However, I dont know to interpret the coefficient. General Linear Models: Modeling with Linear Regression I 3 0 2 4 6 8 10 12 02040608010 % Hunt lo g A r e a 0 We can see that by log-transforming the y-axis we have now linearized the trend in the data. Articulate assumptions for multiple linear regression 2. A prediction is an estimate of the value of $$y$$ for a | {
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for multiple linear regression 2. A prediction is an estimate of the value of $$y$$ for a given value of $$x$$, based on a regression model of the form shown in Equation \ref{eq:regmod4}. Now if your intuition leads you to. We propose combination methods of penalized regression models and nonnegative matrix factorization (NMF) for predicting. indicates that the instantaneous return for an additional year of education is 8 percent and the compounded return is 8. Hello! I need to make some kind of line graph with one y variable and two x variables where only the x-axis is on a logarithmic scale. Log denotes the natural logarithm. Below is an example for unknown nonlinear relationship between age and log wage and some different types of parametric and nonparametric regression lines. In this page, we will discuss how to interpret a regression model when some variables in the model have been log transformed. We interpret the various log, log and semi-log coefficients and use the estimated regression model to make prediction and build a confidence interval for the prediction. eA B = eA=eB 2 Why use logarithmic transformations of variables Logarithmically transforming variables in a regression model is a very common way to handle sit-. The standard data points (concentration vs. You have three options: See this reference on using nonlinear regression to fit a straight line to your data. y is the response variable and x1, x2, and x3 are explanatory variables. A General Note: Logarithmic Regression. Because every disease has its unique survival pattern, it is necessary to find a suitable model to simulate followups. For motivational purposes, here is what we are working towards: a regression analysis program which receives multiple data-set names from Quandl. Testing Kepler’s Third Law October 21, 2009 In this activity, we will use linear regression on our calculators to test Kepler’s Third Law of Plan-etary Motion. 4 The Cox model, in contrast, leaves the baseline hazard | {
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Kepler’s Third Law of Plan-etary Motion. 4 The Cox model, in contrast, leaves the baseline hazard function (t) = logh 0(t) unspeci ed: logh i(t) = (t) + 1x i1 + 2x i2. ECONOMICS 351* -- Stata 10 Tutorial 6 M. The transformation of the data set from y vs. Part (c) shows a log-log function where the impact of the dependent variable is negative. log a 1 = 0 because a 0 = 1 No matter what the base is, as long as it is legal, the log of 1 is always 0. 724 Pseudo R2 = 0. The LINEST function returns an array of coefficients, and optional regression statistics. Explain the primary components of multiple linear regression 3. Minima of positive definite and positive semidefinite functions. , the Negative Binomial regression model). Let's apply some simple regression analysis (see footnote below) to the question. Estimation with correctly interpreted dummy variables in semilogarithmic equations. However, I dont know to interpret the coefficient. Set of tools to fit a linear multiple or semi-parametric regression models and non-informative right-censoring may be considered. A chemical reaction A→B is carried out in a batch reactor. You either can't calculate the regression coefficients, or may introduce bias. Semi-log and Log-log plots Posted 02-04-2010 (5361 views) I have tried Googling and searching the SAS documentation, but I cannot find any syntax to create a semi-log or log-log plot. We also study the transformation of variables in a regression and in that context introduce the log-log and the semi-log regression models. I was in (yet another) session with my analyst, "Jane", the other day, and quite unintentionally the conversation turned, once again, to the subject of "semi-log" regression equations. I know that usually having a linear-log model, an increase in x (GDP) by one percent is associated with an increase in y by (β1/100) units which would be for CRES (2,73/100). Charles says: July 22, 2015 at 2:41 pm. Four Parameter Logistic (4PL) Regression. Care must be | {
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Charles says: July 22, 2015 at 2:41 pm. Four Parameter Logistic (4PL) Regression. Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. One axis is plotted on a logarithmic scale. the log-log graph, which has a logarithmic vertical scale and a logarithmic horizontal scale, as shown below. COREG, is proposed. More speciflcally, one has found a point in a graph one is interested in, and now wants. to Leb esgue measure on [ ; + ] and denote f (x) the sp ectral. 22 Prob > chi2 = 0. A log-linear plot or graph, which is a type of semi-log plot. Semi-logarithmic regressions, in which the dependent variable is the natural logarithm of the variable of interest, are widely used in empirical economics and other fields. Bivariate Regression - Part I - Page 1. If I set the chart as the semi-log scale (semi-logarithmic axes), the regression line cannot be shown to a straight line. eA B = eA=eB 2 Why use logarithmic transformations of variables Logarithmically transforming variables in a regression model is a very common way to handle sit-. This kind of plot is useful when one of the variables being plotted covers a large range of values and the other has only a restricted range - the advantage being that it can bring out features in the data. Polynomial regression. It is partly a matter of custom. Our results indicate that models estimated with a square root link function perform much better than those with log- or linear-link functions. A General Note: Logarithmic Regression. the log-log graph, which has a logarithmic vertical scale and a logarithmic horizontal scale, as shown below. log a a x = x The log base a of x and a to the x power are inverse functions. linear regression on levels of and log-transformed costs, gamma GLM with log-link, and the log-normal distribution, are not among the four best per-forming models with any of our chosen metrics. The present paper develops a mixture regression model that allows for | {
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any of our chosen metrics. The present paper develops a mixture regression model that allows for distributional flexibility in modelling the likelihood of a semi-continuous outcome that takes on zero value with positive probability while continuous on the positive half of the real line. ∙ 0 ∙ share We consider semi-supervised regression when the predictor variables are drawn from an unknown manifold. What is the interpretation of this coefficient? 2. If B1=2, for instance, we could say that ’this model shows that factor X1 increases the predicted log count by 2 (all other factors held constant)’ because equation 1b- equation 1a= B1. DSOM 309 Chapter 16. Also known as elasticity interpretation. A log-linear plot or graph, which is a type of semi-log plot. Is it possible (and how) to transform one of these tables to semi-annual/annual table? If not, is there a way to calculate semi-annual/annual tables based on other data from this web site? Note - The issue is to calculate values based on free public available data, assuming paid access to CRSP etc. Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. In survival analysis, the proportional hazard model, also called the Cox model, is a classical semi-parameter method. Once you have used Excel to create a set of regular axes, converting the axes to semi-logarithmic axes in Excel is far from difficult. The main goal of this paper is to provide a fully probabilistic approach to modelling crime which reflects all uncertainties in the prediction of offences as well as the. And i do get R square (R2)= 0. 10 dan nilai VIF kurang dari 10. Nathanael Johnson (@savortooth on Twitter) is Grist's senior writer and the author of two books. 7 PROC ROBUSTREG Eample: Log-Log Regression With Weighted Outliers Example: Log-Log Regression With Weighted Outliers SAS/STAT ® 9. A semi-log graph is useful when graphing exponential functions. A test based on a modified LP regression | {
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