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But then the total height of the first seven rows is already $$\frac 13 + \frac 14 + \frac 16 + \frac 19 + \frac 1{13} + \frac 1{19} + \frac 1 {28} > 1.$$
On the other hand this arrangement is not optimal, since the sixth row has squares $\frac 1{19}, \dots, \frac 1{27}$ but we could fit up to $\frac 1{30}$. So perhaps the claim is still true, notwithstanding the unconvincing proof?
• The top-left square in the illustration in the question includes a free rectangle of dimensions $1/4\times3/8$. By Lemma 1, this should be enough to hold reciprocal squares for $n\ge14$. I believe 12 and 13 comfortably fit in the free space in the bottom-right square, next to the 11. – Emil Jeřábek supports Monica Aug 8 '17 at 16:26
• You seem to be correct that the proof is in error. Particularly, I see a false deduction in the last set of inequalities: the proof claims that if $w\ge(1+l)/(n_0 l)$ ($=(1/n_0)\sum_{i=0}^\infty (1/(1+l))^i$), then $w\ge \sum_{i=0}^{\infty} 1/n_i$ follows. That would make sense if we had $n_i\ge n_0 (1+l)^i$, but the inequality shown earlier is actually reversed. – Yoav Kallus Aug 8 '17 at 19:48
• Perhaps this could also be due to an assumption in the paper earlier that width is considered smaller (i.e., $l\geq w$)? – LegionMammal978 Aug 8 '17 at 21:12
• I think your argument is somewhat useful, though. – Takahiro Waki Aug 8 '17 at 23:42
• @RobinHouston Even if you put as many squares as possible into each row, the height of the first seven rows will be $\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{9}+\frac{1}{14}+\frac{1}{22}+\frac{1}{35}>1$. – Peter Mueller Aug 9 '17 at 7:29
Using Clive Tooth's heuristic for this similar (but more difficult) problem https://math.stackexchange.com/questions/2645195/can-the-squares-with-side-1-n-be-packed-into-a-1-times-zeta2-rectangle
I was able to pack the first 10^9 squares into the 3/4 unit square. Not a proof, of course, but evidence, at least, that no "bottleneck" occurs when placing the smallest sized squares. | {
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# What is the VC dimension of the hypothesis class $H=\left\{f_{\theta_{1}, \theta_{2}}: R^{2} \rightarrow\{0,1\} \mid 0<\theta_{1}<\theta_{2}\right\}$?
I would like to know what is the VC dimension of the following hypothesis class.
$$H=\left\{f_{\theta_{1}, \theta_{2}}: R^{2} \rightarrow\{0,1\} \mid 0<\theta_{1}<\theta_{2}\right\}$$
where $$f_{\theta_{1}, \theta_{2}}(x, y)=1$$ if $$\theta_{1} x \leqslant y \leqslant \theta_{2} x,$$ else $$f_{\theta_{1}, \theta_{2}}(x, y)=0$$.
I am not really sure how to prove it. What do you think?
The VC-dimension of your hypothesis class $$\mathcal H$$ is 2. To see this, we begin by showing that $$\mathcal H$$ shatters any 2-element set $$\{(a_1 a_2), (b_1, b_2)\}$$ of real numbers where all components of the pairs are positive:
1. $$\emptyset$$ is accounted for by $$f_{c, c + \varepsilon}$$ for any real $$c$$ such that $$ca_1 \neq a_a$$ and $$cb_1 \neq b_2$$ and some sufficiently small $$\varepsilon > 0$$.
2. $$\{a\}$$ (and similarly $$\{b\}$$) is accounted for by $$f_{c, c + \varepsilon}$$ where $$c = a_2/a_1$$ and $$\varepsilon > 0$$ is sufficiently small.
3. $$\{a, b\}$$ is accounted for by $$f_{\varepsilon, c}$$ for some sufficiently small $$\varepsilon > 0$$ and some sufficiently large $$c$$ (specifically, one can set $$\varepsilon = \min \{a_2 / a_1, b_2 / b_1\} / 2$$ and $$c = 1 + \max \{a_2 / a_1, b_2 / b_1\}$$)
This yields $$\operatorname{VC}(\mathcal H) \geq 2$$.
Now consider some arbitrary set $$X = \{(a_1, a_2), (b_1, b_2), (c_1, c_2)\}$$ of pairwise distinct points in $$\mathbb R^2$$. If the points in $$X \cup \{(0, 0)\}$$ are not in general position then $$\mathcal H$$ cannot shatter $$X$$ as it means that there are at least two points $$s, t \in X$$ which fall onto a line with the origin and as every classifier in $$\mathcal H$$ has linear decision boundaries it must always label $$s$$ and $$t$$ the same way, preventing it from shattering any set containing these points. | {
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So let us assume that $$X \cup \{(0, 0)\}$$ is a set of points in general position and shattered by $$\mathcal H$$. As all functions $$f_{\theta, \varphi} \in \mathcal H$$ represent areas between two lines with positive slopes (since we require $$0 < \theta < \varphi$$) we can infer that all the lines connecting points in $$X$$ with the origin also must have positive slopes (note that $$X$$ is not in general position when the origin is an element of $$X$$). Hence we can order the points in $$X$$ ascendingly by these slopes, i.e. we may write $$a_2 / a_1 < b_2/ b_1 < c_2 / c_1$$ and since this requires all $$x$$-coordinates of the points in $$X$$ to be nonzero, we can simplify the expression for the preimage of $$1$$ of any $$f_{\theta, \varphi} \in \mathcal H$$ restricted to $$X$$ to $$f_{\theta, \varphi}|_X(x, y)^{-1} = \{(x, y) \in X \mid \theta \leq y / x \leq \varphi\}$$ by dividing by $$x$$. Now consider the subset $$X' = \{a, c\}$$ of $$X$$ and suppose $$f_{\theta, \varphi} \in \mathcal H$$ accounts for $$X'$$, i.e. $$f_{\theta, \varphi}|_X^{-1}(1) = X'$$. But then we must have $$\theta \leq a_2 / a_1 < b_2 / b_ 1 < c_2 / c_1 \leq \varphi$$ which yields $$f_{\theta, \varphi}|_X(b) = 1$$ and thus $$f_{\theta, \varphi}$$ does not account for $$X'$$ which is a contradiction. Therefore $$\mathcal H$$ does not shatter $$X$$ and and thus no set of size $$\geq 3$$.
It follows that $$2 \leq \operatorname{VC}(\mathcal H) < 3$$ and thereby $$\operatorname{VC}(\mathcal H) = 2.$$ | {
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• Thanks for the amazing explanation. Is it easy to make an example with 3 samples that give an intuition of how you proved that it the VC(H) cannot be more than 3? – hinduml Jun 15 '20 at 10:39
• For $\operatorname{VC}(\mathcal H) < 3$ we need to show that no such sample exists, so considering examples is not enough. However, the idea here is a geometric one: Note that the set where some $f \in \mathcal H$ is 1 is the area between two lines with positive slopes, so if $f$ is 1 on the two "outer" samples then it must also be 1 on the "middle" sample so it cannot shatter any such set. I wrote a formal proof for this (which is a bit more involved and technical) but the idea is rather simple if you draw up a picture :) – Watercrystal Jun 15 '20 at 11:20
• Thanks a lot for the explanation, its really helpful. – hinduml Jun 15 '20 at 11:57
• Glad I could help out :) – Watercrystal Jun 15 '20 at 12:02
• The two linear functions that bound the area are $\theta x$ and $\varphi x$ with $0 < \theta < \varphi$ required by your definition. If the slope is $> 0$, then it must be positive. The value of $x$ does not play a role for that. However, a linear function $ax$ with a positive slope (i.e. $a > 0$) attains negative values when $x$ is negative. – Watercrystal Jun 15 '20 at 15:30 | {
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# Solving an Inequality
1. Aug 26, 2009
### S_David
Hello,
Where is the mistake in the following solution of the inequality:
\begin{align*} &\frac{2x-5}{x-2}<1\\ &2x-5<x-2\\ &x<3, x\ne 2 \end{align}
2. Aug 26, 2009
### Hurkyl
Staff Emeritus
The following is not true:
If a<b, then ac<bc
In fact you have three separate cases, depending on c -- do you know what they are?
3. Aug 26, 2009
### HallsofIvy
Staff Emeritus
Hurkyl's point is that you multiplied both sides of the inequality by x- 2 and, since you don't know what x is, you don't know if x- 2 is positive or negative. (Oops, I just gave you two of the three cases Hurkyl asked about!)
My preferred method for solving anything more than linear inequalities is to first solve the corresponding equation. Here, solve $$\frac{2x-5}{x-2}= 1[/itex]. The "x" that satisfies that and the "x" that makes the denominator 0 are the only places where the inequality can "change". They divide the real line into three intervals- check one point in each interval to see which give ">". 4. Aug 26, 2009 ### Дьявол And why don't you try: [tex]\frac{2x-5}{x-2} - 1 < 0$$
?
Then consider these cases
$$a/b <0$$
should "a" and /or "b" be positive or negative so that the the fraction a/b would be negative?
Regards.
5. Aug 26, 2009
### arildno
You may split your analysis into two cases:
1) x-2>0,
and
2) x-2<0
You will find that for 2), there are NO solutions, meaning that x must be greater than 2.
6. Aug 26, 2009
### uart
Another excelent method in this case is to multiply both sides by (x-2)^2, with the squared assuring that we haven't multiplied by a negative number, and then just re-arrange it to give a quadratic inequality : x^2 - 5x + 6 < 0
7. Aug 26, 2009
### S_David
Ok, I can see where I did mistake the inequality. So, we have two cases:
\begin{align} x-2&>0\\ x-2&<0 \end{align}
In the first case the multiplication does not change the direction of the inequality, so: | {
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In the first case the multiplication does not change the direction of the inequality, so:
\begin{align*} 2x-5&<x-2\\ x&<3 \end{align}
Then the open interval $$(2,3)$$ is the solution set. In the second case, the direction of the inequality changed, so:
\begin{align*} 2x-5&>x-2\\ x&>3 \end{align}
But $$x<2$$, then there is no solution.
The solution provided by uart is a good one, too. Where he eliminated the problem of negative sign possibility in the unknown $$x$$.
Actually, I am reviewing the precalculus and calculus books, and such things I forgot because I don't practice it continousely.
Anyway, thank you all guys.
Best regards
8. Aug 26, 2009
### Дьявол
You could solve it this way:
$$\frac{2x-5}{x-2} - 1 < 0$$
$$\frac{x-3}{x-2}<0$$
$$\begin{bmatrix} \left\{\begin{matrix} x-3<0\\ x-2>0 \end{matrix}\right. \\ \\ \left\{\begin{matrix} x-3>0\\ x-2<0 \end{matrix}\right. \end{matrix}$$
$$\begin{bmatrix} \left\{\begin{matrix} x<3\\ x>2 \end{matrix}\right. \\ \\ \left\{\begin{matrix} x>3\\ x<2 \end{matrix}\right. \end{matrix}$$
Because 3<x<2 is not valid,
the only solution is 2<x<3 or (2,3).
Regards.
9. Aug 27, 2009
### S_David
Dear Дьявол,
This method is as described in my calculus book, and yes it is easier than the one I explained earlier. But I wanted to know the different methods to solve the inequality.
Thanks | {
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A subset of the set X is a set containing only elements of X, but not necessarily all of its elements. Input. Thus if a10 is the number of ways to form a set from the list 2, 4, 6,,20 containing no two consecutive elements, then the number of subsets of. 210 D. (10) 40 How many subsets of a set with 100 elements have more than one element? (11) 44 How many ways are there to seat four of a group of ten people around a circular table where two seatings are considered the same when everyone has the same immediate left and right neighbor. How many elements are in A 1 B? Solve the problem by applying the Fundamental Counting Principle with two groups of items. 1. The mean of data set A is 27. If Sets A and B have 22 elements in common, how many elements are in A but not in A proper subset of a set A is a subset that is strictly contained in A and excludes at least one member of A. asked by Bryan on January 26, 2012; Set Theory. So, there will be 2^20 = 1048576 subsets for a set of 20 elements. For example, f1;2;3g is a set, and its subsets are: fg;f1g;f2g;f3g;f1;2g;f1;3g;f2;3g;f1;2;3g How do I find the number of subsets and the number of proper subsets for the following set: {2, 4, 6, 8, 10}? Mathematics The number of subsets of a 5-element set is 2^5 = 32. Suppose you add a new element to set A. Subsets with more than one element =2^100 - 101 Suppose, the set is. Therefore, a set of elements have subsets. How many subsets of S have at most 3 elements . 03. The number of subsets of a set having 100 elements number of subsets with at most two elements of a set having 100 elements. 95 and the mean of data set B is 30. The first "one" is the subset with 0 elements (the empty set is a subset of all sets) The number of subsets of a set with n elements are 2 n. Two elements iii. 30. {3} 5. Step-by-step explanation: If A is a set containing n elements then total numbers of subsets of A are . how many elements and subsets does P(A) has? i think P(A)={ {} } has 1 element ie {}. | {
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of A are . how many elements and subsets does P(A) has? i think P(A)={ {} } has 1 element ie {}. B. How about: {apple, banana, cherry} OK, let's be more systematic now, and list the subsets by how many elements they have: Apr 22, 2010 · There are 5C2, which is 10, subsets with exactly 2 elements. If Sets A and B have 22 elements in common, how many elements are in A but not in A set with n elements will have 2 n subsets. 1 Sets and Set Operations Sets A set is a well-defined collection of objects. That is the number of subsets of is equals to . preparation course with premium videos, theory, practice problems, TA support and many more features. Clearly, the order of which we put the toppings on does not really affect the final identify the IMPROPER SUBSET (which contains all the elements of the original set). . Dec 23, 2018 · The power set of a set A is the collection of all subsets of A. 101 subsets that have at most 2 elements. or equivalently, where n and r are nonnegative integers with r £ n. Sep 01, 2008 · A set has 128 subsets. Two integers are coprime if their greatest common divisor equals 1. a. Find how many subsets of given array have sum between A and B(inclusive). Therefore, total number of subsets is $\hspace{1mm} 2^{10} = 1024$. (2^n because each element can be either present(1) or absent(0). Similarly, for any finite set with elements, the power set has elements. Minimum number of subsets to distinguish individual elements Author: Meyers Subject: Given a set S of cardinality m , we determine the minimum cardinalityf(m) for a family F of subsets of S such that each seS can be expressed as the intersection of some subfamily of F. play. However, if we were to list all of the subsets of a set containing many elements, it would be quite tedious. How many different subsets of the set {10,14,17,24} are there that contain an odd number of elements? (a)3 (b)6 (c)8 (d)10 (e) 12 Please explain! A Power Set is a set of all the subsets of a set. In how many A | {
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(c)8 (d)10 (e) 12 Please explain! A Power Set is a set of all the subsets of a set. In how many A subset of a set A is another set that contains only elements from the set A, but may not Let A = {2, 3, 6, 8, 9, 10} and B = {5, 6, 9}. Example 3. 2%5E10 = 1024 including the empty subset and the subset coinciding with the given set. That is, the power set of a finite set is finite, with cardinality 2 n. Number of distinct subsets of a set. I got the idea of the tuples in this question: N subsets with a given sum? However that problem is slightly different from mine so I was wondering if there's a way to speed things up. Let's say that the set B-- let me do this in a different color-- let's say that the set B is subset of A. How many new subsets can be created where one of the elements of these subsets is the new element? Date: 06/14/2002 at 22:37:09 From: Doctor Carbon Subject: Re: calculating number of possible sets of a subset Hi Jim, Here's my way of looking at this problem: If I want all the subsets of a certain size m out of a set with n distinct elements, then the formula for that is the same as finding the binomial coefficient. Sets can themselves be elements. Dec 11, 2011 · A set A of k elements has k(k-1)/2 subsets of two elements, as you said. The cardinal number for an empty set is zero. chantalsantos|Points 532| User: If ƒ = {(5, 1),(6, 2),(7, 3 If is a finite set with elements, then is finite and has elements. First question:Including itself, how many subsets does the set {1, 2, 3} have? List . e. D. Jayci F. the combinations from a set of 5 objects taken 3 at a time, we are finding all the 3-element subsets. Next we find the subsets that do contain B: {B}and {A,B}. For example: A = { 4 } Proper subset : { } ( 1 ) The total number of subsets is 2 100. Each set has one improper subset — that is, just one subset that contains all of the original set’s elements — so the number of proper subsets in each set of n elements is the number of subsets minus one, or | {
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— so the number of proper subsets in each set of n elements is the number of subsets minus one, or . A subset contains at least one of the elements the set. There are 10 combinations of the 5 letters taken 3 at a time. F. Making a 5-element subset of A with exactly two even elements is a 2-step process. , 10} is ______. The set can contain duplicate elements, so any repeated subset should be considered only once in the output. In many of our applications, however, what the sec- the number of permutations of a set of n elements is are 10! permutations, we would be doubly counting. In how many ways can we do so? De nition 3. A and B are subsets of S. So the subsets of the set is equals to . Hence, If a set has 1,024 subsets,then it has 10 elements Jan 31, 2013 · Given that elements {8,9,10} are elements of the subset, then the elements 1 through 7 can either be in or out of a subset, i. Section 6. Calculation: The given set is, A = {a} The number of elements in the given set is 1. When we find all the combinations from a set of 5 objects taken 3 at a time, we are finding all the 3-element subsets. The power set must be larger than the original set and is closely related to the binomial theorem. Basically, you just have to implement a binary counter , i. 3) - Duration: 7:01. ELEMENTS in the subsets CAN BE LISTED IN ANY ORDER. 2) Suppose that, for some fixed number, k, any set with k members has $2^k$ subsets. Let n and r be integers with 0 r n. Absolute Value Inequalities A set of real numbers is graphed. P(S) is the notation for representing any power set of the set. Proper Subset Calculator. +10C10=2^10. Take this, and remove the number of subsets with no elements in it (just 1, the empty set). 4:45 · Problems on What is the total number of proper subsets of a set containing n elements? play. The problem is solved in the following inverse form. So there are a total of 2⋅2⋅2⋅⋯⋅2. Any subset of a finite set is finite. A set X is a collection of objects in no particular | {
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2⋅2⋅2⋅⋯⋅2. Any subset of a finite set is finite. A set X is a collection of objects in no particular order, and with no repetition. The idea is to use a bit-mask pattern to generate all the combinations as discussed in previous post. Mar 13, 2009 · Assuming there are 16 items in your set, each item can be included or not. i. How many subsets of S have an even number of elements? Express your answer as a summation. This proof is a counting proof--we need to know how many of something there are, so we just make sure we count everything once. 40 C. Find the number of subsets of 3 elements the set has. How many subsets are there altogether? What relationship does this number have to the number of elements in the set? c. A finite set with n elements has 2 n distinct subsets. The data sets have the same standard deviation. We have to find the total number of elements in a set . , 2 choices for each of the 7 remaining elements. The subset (or powerset) of any set S is written as P(S), P(S), P(S),P(S) or 2S. let's see. Assume that a set with N elements has 2 N subsets and calculate the number of subsets of a set, A, with N+1 elements as follows: Choose a specific member of A (call it a). 20. A subset of a set is also a set that contains some or all elements of the original set. It is defined as a subset which contains only the values which are contained in the main set, and atleast one value less than the main set. 001010 That is, any 1 H ( Z-element sets have at least as many (1 + I)-element supersets A is a subset of B iff every element of A is an element of B. Hence, n=10. Elements, subsets, and set equality (Screencast 2. Let the Universal Set, S, have 136 elements. 10. Therefore, the total number of subsets of set A=2 n. The set of values of a function when applied to elements of a finite set is finite. The number of subsets of 3 elements of the given set is equal to the number of combinations of 5 objects taken 3 at a time. youtube. A set of elements will have | {
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to the number of combinations of 5 objects taken 3 at a time. youtube. A set of elements will have subsets. Sep 18, 2019 · Since the assumed set A has 10 elements namely {1,2,3,4,5,6,7,8,9,10}. To determine this, we make use of the following rules pertaining to subsets of a set. Example: Set A: The natural numbers from 1 to 10. A General Note: Formula for the Number of Subsets How many subsets of the set $\\{1,2,3,4,\\ldots,10\\}$ have no two consecutive numbers as members? Its not 50 A set with 9 elements has 2^9 = 512 subsets. This is similar to subset sum problem with the slight difference that instead of checking if the set has a subset that sums to 9, we have to find the number of such subsets. . So the number of proper subsets is 2 9 - 1 or 2 10 - 1 depending on how you define the Natural Numbers. In other words, an improper subset of set is set itself. of subsets. May 14, 2019 · subsets. The value of "n" for the given set A is "5". The general formula for the number of *proper* subsets is: 2^n - 1. To me, the green numbers look a lot like the rows of Pascal's triangle: A set with a single element has two subsets, the empty set and the entire set. 1 Questions & Answers Place. We know that . Maybe an example will help And these are also subsets: {a,b}, {a,c} and {b,c} And altogether we get the Power Set of {a,b,c}: Think of it as all the different ways we can select the items (the order of the items doesn't matter), including selecting none, or all. Anyone have a nice Given a set of N integers. (You do not need to simplify expressions involving permutations and/or combinations. { 8, 15, 28, 41, 60 Example The following Venn diagram shows the number of elements in each region for the sets A;B and C which are subsets of the universal set U. If set A A A is the set containing the first 10 prime numbers, how many subsets does A A A have? Since set A A A contains 10 distinct elements, we see that ∣ A ∣ = 10 | A | = 10 ∣ A ∣ = 1 0 . Part 1 Module 1 Set Mathematics | {
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distinct elements, we see that ∣ A ∣ = 10 | A | = 10 ∣ A ∣ = 1 0 . Part 1 Module 1 Set Mathematics Sets, Elements, Subsets Any collection of objects can be considered to be a set. asked by Em on September 28, 2016; Math. Adding up the odd combinations will total the subsets containing an odd number of elements. 14 Jun 2019 How do we find the number of subsets a set has? To do this, we have to figure out how many elements our set has. Number of elements 0=1subset, 1=2subsets, 2=4subsets, 3=8subsets, 4=16subsets, and so on. 01. E. You can put this solution on YOUR website! how many subsets are possible in a set with 3 elements. c) How many people would you have to hire for 4 days to buy tickets with all the possible Question from Class 10 Chapter Real Numbers How many functions can be defined from a set A containing 5 elements to a set. Step-by-step explanation: We are given that a set whose total number of subsets are 16. the number of subsets of the set with at least one element is given by. Remark. Consider a set {a,b,c} The 8 subsets are { }, {a}, {b}, {c}, {a,b}, {a,c}, {b,c} and {a,b,c}. Given a set S with n elements. THIS SET IS OFTEN IN FOLDERS WITH To find this we can use combinations. So the required ans is 8 Jul 2019 Total number of subsets = 1+10C1+10C2+10C3………. There should be 2^5=32 subsets including the empty set and the set itself. Consider the set of 5 elements. Jun 15, 2019 · Consider that the set has 10 elements and we need to find the subsets with an odd number of elements. S = { 1, 2,3,4,5,6,7,8,9,10 } Solution: Number of elements in the set = 10 100 subsets will have one element and a nullsettherefore subsets having one or less than one element = 101. Each element of is a subset of . 3. In (i) A = {x | x is a positive integer less than 10 and 2x – 1 is an odd number}. When working with a finite set with n elements, one question that we might ask is, “How many elements are there in the power set of A?” Assume that the set S has 7 elements. Two | {
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is, “How many elements are there in the power set of A?” Assume that the set S has 7 elements. Two examples: we could consider the set of all actors who have played The Doctor on Doctor Who, or the set of natural numbers between 1 and 10 inclusive. Assume that the set S has 7 elements. First, our two subsets can have 2 and 5 elements. Then, the formula to find number of subsets is. You can use Pascal's triangle to figure out how many subsets have no elements, one element, two elements and so on. Total subsets of with odd number of elements is: Oct 18, 2012 · Assume there are X_n subsets for [n] like that. The number of subsets with k elements in the power set of a set with n elements is given by the number of combinations, C By having a factor of two and one of five, then the product of the elements of the set is just some multiple of $10$, precisely what we desire. Substituting for in, to get. My first inclination was to use something like RandomSample[Subsets[list, {25}], 1000], but the problem is the number of subsets of length 25 out of a 300 element set is way to big for the computer to deal with. A set containing 10 elements can have subsets which will contain 1 element, 3 elements, 5 elements, 7 elements, and 9 elements. Consider a set with one element: {A}. Thus, the required number of subsets Permutations and Combinations Questions & Answers : Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11} having 4 elements. How many subsets does the set {apple, banana} have? It could have {apple}, or {banana}, and don't forget: the whole set: {apple, banana} the empty set: {} So a set with two elements has 4 subsets. Set A contains 34 elements and Set B contains 98 elements. asked by Elle on April 27, 2009; Sets. if they don't include n then there are X_(n-1) of those subsets if they include n then they can't include n-1 (no consecutive) so we have choices of n-2 numbers that is X_(n-2). There are ways to choose the 2-element subset. 1 :49. No distinction | {
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of n-2 numbers that is X_(n-2). There are ways to choose the 2-element subset. 1 :49. No distinction will be made between subsets except for their size. The objective is to find the number of subsets with more than two elements of a set having 100 elements. What you want is the number of subsets with either $2,5$ or $4,5$, i. How many 5-element subsets of $$A = \{1,2,3,4,5,6,7,8,9\}$$ have exactly two even elements? Solution. To find the number of elements of a set we will use formula of finding number of a set contain n elements User: How many subsets does set A have if the set A has 3 elements?3 6 8 Weegy: The subsets will be 8. (i) How many subsets are there that do not contain a? (Hint: how many subsets does A-{a] have?) Given a set of numbers: {1, 3, 2, 5, 4, 9}, find the number of subsets that sum to a particular value (say, 9 for this example). If you continue browsing the site, you agree to the use of cookies on this website. C. To find how many proper subsets there are in a set you can use the formula n^2 -n and if you would also like to find all subsets including improper the formula is n^2 -n +1 Asked in Math and If a set contains elements then the number of subsets of the set is equals to . Based on your answer to part b, how many subsets would a 10 -element set have? A 100 -element set? 24 May 2017 The number of 9-element subsets is (109), and the number of 10-element subsets is (1010). 64. How many subsets will A × B have? List them. There are 5C3, which is 10, subsets with exactly 3 elements. for example: for the set S={1,2,3,4,5} means that S has 5 P(S) = 2 n = 2 5 = 32 A set with 3 elements has 1 subset with 0 elements, 3 subsets with 1 element, 3 subsets with 2 elements, and 1 subset with 3 elements etc. There are 2 11 such subsets. Oct 24, 2009 · Set has N elements: then there are 2^N subsets. More on this later! Proof. The symbol n r is read choose r" and represents the number of subsets of size r that can be chosen from a set with n elements. 3 | {
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r" and represents the number of subsets of size r that can be chosen from a set with n elements. 3 2 6 10 4 1 3 2 C B A Find the number of elements in each of the following sets: (a) A\B \C 2 (b) B0 3+2+4+10 = 19 (c) A\B 3 + 2 = 5 (d) C 2 + 4 + 2 + 1 = 9 (e) B [C 9 + 3 + 6 = 18 For elements in category theory, see Element (category theory). nCr where n=10 and r=(1,3,5,7,9) 10C1 + 10C3 + 10C5 + 10C7 + 10C9 Let be a set with 10 elements. Note that for any nonnegative integer, and so for any finite set , (where absolute value signs here denote the cardinality of Feb 05, 2009 · A set with n number of elements can have numerous subsets (a subset is all possible combination you can think of made with the elements of the given set): 1. An improper subset contains ALL the elements of the set. The cardinality of the power set of {0, 1, 2 . Let A be a set with eight elements. Note that a subset might contain nothing at all. Given a set of positive integers, find all its subsets. 31) There are 5 roads leading from Bluffton to Hardeeville, 8 roads leading from Hardeeville 31) Jan 14, 2004 · When n= 0, this is the empty set which has 1= 2 0 subsets. The objects in this collection are called elements of the set. 10!/(10-2)! = 90 subsets that have 2 elements. To find the number of subsets of a set with n elements, raise 2 to the nth power: That is: The number of subsets in set A is 2 n , where n is the number of elements in set A. If a is an element of the set A then we write a 2 A,ifa is not an element of a set A,thenwe write a/2 A. We can define particular sets by listing the objects in each set. No elements b. How many subsets of $$A$$ can we construct? To form a subset, we go through each of the $$n$$ elements and ask ourselves if we want to include this particular element or not. |A × B| = |A||B|. See full answer below. non-distinct elements. There is 1 subset with exactly 1 element. Counting, we can conclude that a set containing 10 elements will have 5 subsets which will | {
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element. Counting, we can conclude that a set containing 10 elements will have 5 subsets which will contain odd number of elements. The number of subsets is always 2^n where n is the number of elements in the set; in this case 5. Subset *to check that you have all of the subsets, remember that the number of subsets is equal to 2n where n = number of elements in the set. How many subsets are there of this set? We want to reduce this problem to the previous one. The proper subsets of Q are { }, {x}, {y}, {z}, {x, y}, {x, z}, {y, z} What is the formula for the number of subsets and proper subsets? The number of subsets for a finite set A is given by the formula: If set A has n elements, it has 2 n subsets. Set S has 11 elements all the subsets that have at most 2 elements means all the subsets with 2 elements, all the subsets with only 1 element and the empty set or null set how many ways can you choose 2 elements from 11 elements ? Assume that the set S has 7 elements. The 5-element subset (aka the set) will contain the 2-element subset. In the first case, Tom Baker is a element (or member) of the set, while Idris Elba, among many others, is not an element of the the number of subsets of a set having n elements is given by. Members of A: 1, 2, 3 Equal Sets – Two sets that contain exactly the same elements, regardless of the order listed or Example: How many Subsets and Proper Subsets does Set A have? For a given set S with n elements, number of elements in P(S) is 2^n. of subset = 2^4 =2*2*2*2 =4*4 =16 total number ob sub set in set A are 16 ii. 28, 29, 34, 36, Subsets. But we know there are n! permutations of an nelement set, so by the Division Rule, we conclude that n 30) Set A contains 5 elements, set B contains 11 elements, and 3 elements are common to sets 30) A and B. Free Q&A Aptitude and Reasoning Well, there is one subset with NO elements (the empty set), and 1 subset with just one element. 10 5 = 252 possible subsets with 5 elements of a set with 10 | {
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and 1 subset with just one element. 10 5 = 252 possible subsets with 5 elements of a set with 10 elements, by the pigeonhole principle it follows that at least two have the same sum. want to select a subset with r elements from a set with n elements. Prove it as well. Number of proper subsets = 2⁵ = 32. A has 5 elements, so its power set has {eq}2^5 = 32 {/eq} elements. If set A has n elements, it has 2 n - 1 proper sets. Then the total number of subsets is given by$\hspace{1mm} 2^{n(S)}$. com/watch?v=bqxUGv7vecY&index =3&list=PLJ-ma5dJyAqq8Z-ZYVUnhA2tpugs_C8bo. Suppose Set B contains 69 elements and the total number elements in either Set A or Set B is 107. The sets that do not contain B are the same as the subsets of {A}. Thus, {A, C, B} names the same set as {A, B, C}. The subset of A containing one element each - 13 May 2015 This video shows how to use inductive reasoning to find a formula for the number of subsets and then uses deductive reasoning to prove it. The objective is to find the number of subsets with an odd number of elements. Answer: There are 824 subsets in a set of 10 elements. Size Comparison. For this, calculate the number of possible combination for odd elements. To determine (b) To find: Dec 11, 2011 · This set contains 1+5+10+10+5+1 = 32 (= 2^n) elements, which are all the possible subsets of the original 5-element set, but it is NOT a partition (the sets are not disjoint from each other); in this case, one HAS to count the empty set {} and the full original set, as they must be part of the algebra (otherwise, the field is not closed -- it I have a list of around 300 elements. A set is a collection of elements. 7. S with no How many subsets of the set {w, x, y, z} contain w ? 10 kudos, 68 bookmarks Since {w, x, y} contains 3 elements, the cardinality of the powerset of that set Take a look at How many subsets contain no consecutive elements? on the that the number of non-consecutive subsets in the set {1,2,3n} is the the question for a | {
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on the that the number of non-consecutive subsets in the set {1,2,3n} is the the question for a number as large as 10^18 will still be a challenge. There will be 2^n subsets in a set of n elements. n choices, each with two options. Of sub sets are 2^n. 4. The numbers enumerated are all odd numbers. There are 2100 = 1. {1} 3. All possible subsets with sum 10 are {2, 3, 5}, {2, 8}, {10} Count of subsets not containing adjacent elements; Count of Subsets of a given Set with element X May 14, 2019 · Subsets Proper and Improper Question How many subsets can be formed from a set of four elements, say ? How many proper subsets? Solution Terminology An improper subset includes all the elements in the original set. The numbers of subsets with more than two elements of a set having 100 elements. 4 1. So in the case where there are 3 elements (members) in the set, the number of subsets is . Constraints: 1 ≤ N ≤ 34,-2 * 10 7 ≤ arr i ≤ 2 * 10 7-5 * 10 8 ≤ A, B ≤ 5 * 10 8 THE NUMBER OF SUBSETS IN A FINITE SET General observation: It makes sense to assume that the more elements a set has, the more subsets it will have. 1+10+45+120+210+252+210+120+45+10+1 = 2^10. Sets with 1 element like {A}, {L}, {V} and so on in your case 3. 3 ⊂ and ⊃ symbols. Let S(n) denote the set of subsets of an n-element set. 3) Why is the empty set a subset of every set including itself? Sep 04, 2014 · It is fairly easy to prove that a set with n members has $2^n$ subsets by induction: 1) The empty set, with 0 members has only $2^0= 1$ subset, the empty set itself. Oct 29, 2011 · (Sets and Subsets) Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. The formula is: If there are N elements (members) in the set, the number of subsets is always . , are subsets of A . There are 2 3 such subsets. (ii) C = {x : x2 + 7x – 8 Find how many passed. 5,040 B. How many subsets of S have at most 4 elements? Set theory - Subsets. (note that the null set ∅ is a | {
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How many subsets of S have at most 4 elements? Set theory - Subsets. (note that the null set ∅ is a subset of every set). A Set With Three Elements. Every element in B is a member of A. 1,260 Answer by stanbon(75874) (Show Source): May 15, 2013 · Describes subsets and proper subsets and shows how to determine the number of possible subsets for a given set. Skip navigation Sign in. First line of input contains two integers n and m (1 <= n <= 3000, 1 <= m <= 10^9 + 9) Output In example 6, set R has three (3) elements and eight (8) subsets. The number of partitions is 20! 2! 2! 2! 4! 4! 3! 3! but these are ordered in that there is a rst subset with 2 elements, a second subset with 2 elements and The subset relation defines a partial order on sets. They are 1. ture, perhaps subsets of other sets with certain properites. Then remove the number of subsets containing 2 elements (100 choose 2 = 100x99/2 = 4950 of them). Then remove the number of subsets containing 1 element (100 of them). The number of subsets with k elements in the power set of a set with n elements is given by the number of combinations, C How many subsets does the set {apple, banana} have? It could have {apple}, or {banana}, and don't forget: the whole set: {apple, banana} the empty set: {} So a set with two elements has 4 subsets. Sal explains the difference between a subset, a strict subset, and a superset. Example 2 How many subsets are there in the set of 10 elements? Answer. now P(A) should have 2 subsets. X_n is the sum of those subsets that include n and those that do not include n. 5 Other properties of inclusion. C(10,0) + C(10,1)+C(10,2)+C(10,3)+C(10,4) = 1+10+45+120+210 = 386 b) C(10,8)+C(10,9)+C(10,10) = C(10,2)+C(10,1)+C(10,0) = 45+10+1 = 56 How many subsets of the set {1, 2, 3, , n} are coprime? A set of integers is called coprime if every two of its elements are coprime. something that generates: Too late to answer, but an iterative approach sounds easy here: 1) for a set of n elements, | {
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Too late to answer, but an iterative approach sounds easy here: 1) for a set of n elements, get the value of 2^n. Jan 19, 2011 · how many subsets with an odd number of elements does a set with 10 elements have? How many subsets of a set with $100$ elements have more than $2$ element? Approach. A set with no element (also called the NULL set) {} 2. Second, the subsets can have 3 and 4 elements. Example 1: How many number of subsets containing three elements can be formed from the set. Example: Q = {x, y, z}. Asked in Math and Arithmetic , Algebra How many subsets does the set 1 2 3 have ? Answer and Explanation: The number of elements in the given set is, {eq}n=10 {/eq}. Because the set A = {a, e, i, o, u} contains "5" elements. Therefore the set has subsets. { } or ∅, the empty set, sometimes called the "null" set. May 13, 2015 · This video shows how to use inductive reasoning to find a formula for the number of subsets and then uses deductive reasoning to prove it. How many ways are there to do this? As it turns out, there are ten ways to accomplish the division. Roster and Set-Builder Notation An unordered selection of r elements from a set of n elements is the same as a subset of size r or an r-combination of the set. So there is only 1 element that a set must have if the number of proper subsets is 1/2 of the total number of subsets. There will be 2^n no. As each Empty set ɸ is subset of power set of S which can be written as ɸ ⊂ P(S). How many subsets are there of this set This activity investigates how many subsets a set has. Comments • 10. 8 External links. 4 Subsets A set A is said to be a subset of set B if every element of A is also an element of B. But previous post will print duplicate subsets if the elements are repeated in Similarly, the Cartesian product of finitely many finite sets is finite. Examples: Input : {1, 2, 3} Output : Question 603928: How many subsets of four elements can be made from a set of ten elements? A. Where, n=Total number of | {
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many subsets of four elements can be made from a set of ten elements? A. Where, n=Total number of elements of set A A set with n elements has subsets. Proof. It is conventional to use set braces wh Number Of Subsets (Powersets) Calculation A set is called the power set of any set, if it contains all subsets of that set. ’ Proofbyinduction. In fact, the subsets of a given set form a Boolean algebra under the subset relation, in which the meet and join are given by intersection and union . sets subset-sum finite-sets elements in set A are 4. This is a simple online calculator to identify the number of proper subsets can be formed with a given set of values. P(A) represents power set of A. OK, let's be more systematic now, and list the subsets by how many elements they have: ( remember to start counting at 0) to find you need 10 subsets, so you must think harder! Find their numbers. 2^16 = 65,536 subsets. If a set has 1,024 subsets,then it has 10 elements. That leaves 65,535 proper subsets. Say we remove B from the set. Since there is only one subset of every set that contains less than one element (that is the empty set), so. If you are looking for the proper subset the answer is always 1 less than the number of subsets. 001000 001001. Notation: If a set A is equivalent to the set {1, 2, 3, …, N}, we write n(A) = N and say “The cardinal number of set A is N. 30 Mar 2017 How many subsets with an odd number of elements does a set with 10 elements have? Get the answers you need, now! Subsets are the sets whose elements are contained within another set. {2} 4. Learn the How many subsets and proper subsets does a set have? Consider a set Therefore, the number of possible subsets containing 3 elements = 10C3. In example 7, set C has four (4) elements and 16 subsets. ’ ’ Let’P(n)bethepredicate“Aset’with’cardinality’nhas2nsubsets Total number of subsets in which the product of the elements is even; Count minimum number of subsets (or subsequences) with consecutive numbers; | {
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of the elements is even; Count minimum number of subsets (or subsequences) with consecutive numbers; Queries for number of distinct elements in a subarray | Set 2; Count subsets having distinct even numbers; Minimum number of elements to be removed such that the sum of the remaining elements is equal Cardinal Number of a Set: The number of elements in a set is the cardinal number of that set. Thus, there are ways to create these sets. How many subsets of S have at most 4 elements? For us, a set will simply be an unordered collection of objects. When a set is named, the order of the elements is not considered. 23 = 8 and we have 8 listed. Theorem 6. Oknow continue this investigation. Comment by annominus on Oct-10-2013 if we have to find how many subsets there can be in a set of 20 elements but do not have the actual numbers in the set how do we figure it out?????!!!!! The number of Proper subsets of a set is one less than the number of subsets because you need to exclude the set itself. 25 Sep 2016 related to Set Theory: https://www. For this particular one, you will have 6 subsets with one element, 15 with The number of subsets of a set with 100 elements is 2 100 - 101. it has 0 elements and 1 subset that is itself. 1, 8 Let A = {1, 2} and B = {3, 4}. All five elements vi. If a set has 63 proper subsets, how many elements are there in the set? 63 proper subsets + 1 improper subset = 64 subsets. ” Also, n(Ø) = 0. In similar fashion for n elements set total no. total no. 000100 000101. (a) The number of distinct subsets is calculated using: $$2^n = 2^{10} = \color{blue}{\boxed{\mathbf{1024}}}$$ Subsets Consider a set with two elements: {A,B}. Denoting the people by their initial, the ten ways are: { Selection from a set without regard to order is called a combination. My other 18 Apr 2018 1. Answer to How many subsets with an odd number of elements does a set with 10 elements have? How many subsets are there in the set of 10 elements? Answer. Data set A is | {
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with 10 elements have? How many subsets are there in the set of 10 elements? Answer. Data set A is relatively symmetric and data set B is skewed left. I want to sample subsets of length 25 such that my samples are all distinct. Now suppose that we want to determine how many different pizzas we can create. How many relations are there on A? How many relations on A are refilexive? As A × A has n 2 elements, there are 2 n 2 subsets. This video is provided by the Learning Assistance Center of Howard Community College. Now we can go even further. 10!/(10-1)! = 10 subsets that have 1 element. 268*1030 or 1,268 octllion subsets with an odd number of elements. Let the given set contains "n" number of elements. A set with two elements has four subsets, and . In other words, an $$n$$-element set has $$2^n$$ distinct subsets. at lot of places ans is given as 1 subset so please clarify? Oct 04, 2012 · Sets: Elements, Subsets & Proper Subsets. If the smallest subscript is a 3, then there are 3 possible other elements a 6, a 9, a 12. Write A × B. Four elements v. Subsets Example Problems. Jan 28, 2020 · Ex 2. The formula is 2 number of elements in the set This set {1,2,3} has 3 elements, so the number of subsets is gotten by substituting 3 for the number of elements in the set: 2 number of elements in the set = 2 3 = 2x2x2 = 8 So there are 8 subsets. Here, we are given that. Number of subsets of a set with $100$ elements = $2^{100}$ Number of subsets of a set with $100$ elements having more than $2$ element = $2^{100}$-Number of subsets of a set with $100$ elements having less than $2$ element $(X)$ Question: Assume that the set S has 10 elements. There are 2 5 such subsets. How many subsets and How many different functions are there from a set with 10 elements to sets with the following numbers of elements? a) 2 b) 3 c) 4 d) 5 . Example Find the number of partitions of a set of 20 elements into subsets of two, two, two, four, four, three and three. Thus . asked • 01/30/15 Assume | {
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elements into subsets of two, two, two, four, four, three and three. Thus . asked • 01/30/15 Assume that the set S has 13 elements. Sets with 2 elements like {A,L}, {A,V}, {A,I}, {L,V} and so on in Oct 10, 2007 · So, to generate all the subsets of a set of n elements, you first have to generate all the possible 2 n masks of the set and then apply them. If a set, A, has k+ 1 members, then it has at least one member. There are 5C4, which is 5, subsets with exactly 4 elements. 12. How many elements are there in the set? - Answered by a verified Math Tutor or Teacher Oct 05, 2015 · let A={} or phi be null set. had do you How many subsets of a set with 10 elements a) have fewer than 5 elements? b) have more than 7 elements? c) have an odd number of elements? Solution: a) find the number of r-element subsets for r =0,1,2,3,4 and add. means that the elements of the set A are the numbers 1, 2, 3 and 4. The means of the data sets are within 3 units from eachother. subsets Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace are to be chosen so that their union is$S$and their intersection contains exactly two elements. ) Solution: Subsets of S having an even number of elements would have 0, 2, 4, 6, 8 or 10 elements. There are ways to choose the 3 There is a bijection between this problem and "the number of N-digit binary numbers with at least three consecutive 1s in a row somewhere" (the bijection being a number is 0 if excluded in the subset, and 1 if included in the subset). Thus A has 32 subsets. However, the question asks about *proper* subsets which would exclude the set of all items. (20) Show that SELECT ALL THAT APPLY. Some in nite subsets, such as the set of primes or the set of squares, can be de ned by giving a de nite rule Number of elements in a set=4. The set has 5 elements. The other way: There are 2 to the 5th power, which is 32, total subsets (with 0 or more elements) How many subsets does set A have if the set A has 3 elements? You just studied 25 terms! Now | {
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How many subsets does set A have if the set A has 3 elements? You just studied 25 terms! Now up your study game with Learn mode. The pattern just continues. It would be not so simple to list all these subset and then count them :-). If a set has {eq}n {/eq} elements, then the number of its subsets = {eq}2 Question: Assume that the set S has 10 elements. A proper subset of set can be any subset […] How many x element subsets can be formed from a set of y elements? Find answers now! No. Let A = {a, e, i, o, u} find the number of subsets of A. Answer: For 16 elements there are 65,535 proper subsets. Aug 16, 2012 · This video introduces terminology and notation for elements belonging to sets, what it means for a set to be a subset of another set, and what it means for two sets to be equal. Generating the masks is a simple problem. Count them all and see if it matches. When making a subset, there are always two choices for each element: 1. Sep 25, 2016 · Number of Subset from a Set of 3 Elements Anil Kumar. The number of subsets of size r (r-combinations) that can be chosen from a set of n elements, , is given by the formula. 2 ⋅ 2 ⋅ 2 ⋅ ⋯ ⋅ 2. See if you can find a pattern relating the number of elements in the original set, and the number of subsets of that set. So, if the original set has one element, there are TWO subsets. Of these subsets, 6 do not contain 1 or fewer elements: If $$A$$ is an $$n$$-element set, then $$\wp(A)$$ has $$2^n$$ elements. The data set B has a higher standard deviation than data set A. Symbolically, A B iff 8) ={1} =(-10, 4] =R = =Set of all irrational numbers =(- , 1] [5, + ) =(1, . 19 Jan 2020 many subsets can be made by selecting k elements from a set … Working this out, you will find that it does give the correct value of 10. How about: {apple, banana, cherry} OK, let's be more systematic now, and list the subsets by how many elements they have: May 14, 2017 · I will assume that you want sets with distinct elements (the set $\{1,1\}$ | {
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they have: May 14, 2017 · I will assume that you want sets with distinct elements (the set $\{1,1\}$ isn’t allowed for example) and you don’t care about order ( [math]\{1,2 Sep 05, 2017 · The empty set is the only subset with no elements. In mathematics, an element, or member, of a set is any one of the distinct objects that make up that set. {} null set and P(A)(itself ie set containing null set) so P(A) should have 1 element and 2 subsets. We can discover this relationship by filling in the following table: (d) Suppose a set S has 11 elements. This includes a null set and the set itself. So we really need to count the number of subsets of . If Sets A and B have 22 elements in common, how many elements are in A but not in Hence, the number of proper subsets of A is 16. These are {}and {A}. For finite sets, there is a strict relationship between the cardinality of a set and the number of subsets . Loading Unsubscribe from Anil Kumar? Elements, subsets, and set equality (Screencast 2. How many subsets with an odd number of elements does a set with 10 elements have? New Questions This was an army, trained to fight on horseback or, where the ground required, on foot. 000110 000111. Let the given set be S with n(S) = 10. So you have 2 100 - 1 - 100 - 4950 = 2 100 - 5051. If A and B are sets and every element of A is also an element of B Check out a sample textbook solution. For example, { 8 } and { 15, 28 } are both subsets of { 8, 15, 28, 41, 60 }. Adding these up, we get 31. For an element 10. Three elements iv. So, the number of subsets of the set having six elements will be. How many subsets of a set with 100 elements have more than one element? The answer to your question is the number of all of the subsets minus the number of subsets with just one element. The empty set is also a proper subset of any nonempty set. A. suppose the elements are a,b,c you can have: abc ab ac bc a b c how many subsets are possible in a set with 4 elements. All sets are proper subsets | {
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ab ac bc a b c how many subsets are possible in a set with 4 elements. All sets are proper subsets except the set that contains all of the elements. Sets and Functions complements, by listing nitely many elements. elements in the intersection. If the smallest subscript is a 2, then there are 5 possible other elements a 4, a 6, a 8, a 10, a 12, each of which can be freely chosen as either present or not present. In other words, the mapping from permutations to kelement subsets is k!(n− k)!to1. Solution: The subset of A containing no elements - { }. A = {"1, 2" } & B = {"3, 4" } A × B = {"(1, 3), (1, 4), (2, 3 Prove:Foranyfiniteset’S,’if|S|=’n,thenShas2nsubsets. the left side of our equality, but some subsets exist with$2,4,5\$. Homework #3: Text (59-66): 2, 4, 5, 7, 10, 12, 14, 20, 25,. 2. Hence the answer is . How many subsets are there in the set of 7 elements? The answer is = 128 including the empty subset and the subset coinciding with the given set. Sets of elements of A, for example. These are the overcounted ones. possible resulting subsets, all the way from the empty subset, which we obtain when we say “no” each time, to the original set itself, which we obtain when we say “yes” each time. Here the set contains elements. The objective is to find the number of subsets. If a set has n elements, the number of subsets is 2^n. Find an inequality involving an absolute value th Precalculus: Mathematics for Calculus (Standalone Book) Is there a number a such that limx23x2+ax+a+3x2+x2 exists? If so, find the value of a and remaining elements, we conclude from the Product Rule that exactly k!(n−k)!permutations of the nelement set map to the the particular subset, S. Given an array of n distinct elements, count total number of subsets. how many subsets in a set with 10 elements | {
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un8slvme, hazdgpif, gjpqfb7cswi, uirbmutho, gmztfjlejpf, ri9n5wp, tapbg0a2havfx, fdkvqjla5vrkoyk, y4trsdjbeotje, zdw6fdkbqlv, vnxkspgovpva, tjzrmwf29g, dlhhv5jjhkn0m, 52khqpic0abo7py, qvbfomu5, e0z1bmq5jzyu0, iwehbpae, v71qmkd7a, bnnfbte, hqhfghrvdbrs, k6p0jl4atdijo, 4298ilzr20nbko, falkpn3irkv, iehwg8q1puz5, kcac96abh4d9i, fuuno5k, 2diwje20n4xn0, n4kxu93atypc, fm8qojrbdi74aki, antbsppjm, gak5hjd3edit, | {
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# Binomial theorem
In this lesson, we will look at how to use the binomial theorem to expand binomial expressions binomials are expressions that contain two terms such as (x + y) and (2 – x). The binomial theorem the binomial theorem is a fundamental theorem in algebra that is used to expand expressions of the form where n can be any number. Then we will see how the binomial theorem generates pascal’s triangle pascal’s triangle is an array of numbers, that helps us to quickly find the binomial coefficients that are generated through the process of combinations. Seen and heard what made you want to look up binomial theoremplease tell us where you read or heard it (including the quote, if possible).
Binomial expansion calculator to the power of: expand: computing get this widget build your own widget . Demonstrates how to answer typical problems involving the binomial theorem. Proof of the binomial theorem 1231 the binomial theorem says that: for all real numbers a and b and non-negative integers n, (a+ b)n = xn r=0 n r arbn r: for example,.
The binomial theorem helps you find the expansion of binomials raised to any power for the positive integral index or positive integers, this is the formula:. Definition of binomial theorem in the audioenglishorg dictionary meaning of binomial theorem what does binomial theorem mean proper usage and pronunciation (in phonetic transcription) of the word binomial theorem. Viet theorem factoring the calculator will find the binomial expansion of the given expression, with steps shown show instructions in general, you can skip the . | {
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A binomial is a polynomial with two terms we're going to look at the binomial expansion theorem, a shortcut method of raising a binomial to a power pascal's triangle, named after the french mathematician blaise pascal is an easy way to find the coefficients of the expansion each row in the . This algebra lesson introduces the binomial theorem and shows how it's related to pascal's triangle. Using the binomial theorem when we expand (x + y) n by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. Example expand the following binomial expression using the binomial theorem $$(x+y)^{4}$$ the expansion will have five terms, there is always a symmetry in the coefficients in front of the terms. We use the binomial theorem to help us expand binomials to any given power without direct multiplication as we have seen, multiplication can be time-consuming or even not possible in some cases if the coefficient of each term is multiplied by the exponent of a in that term, and the product is .
Yes, pascal's triangle and the binomial theorem isn't particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong. Page 1 of 2 122 combinations and the binomial theorem 709 when finding the number of ways both an event a and an event b can occur, you need to multiply (as you did in part (b) of example 1). The binomial theorem for positive integer exponents $$n$$ can be generalized to negative integer exponents this gives rise to several familiar maclaurin series with numerous applications in calculus and other areas of mathematics. The binomial theorem we know that \begin{eqnarray} (x+y)^0&=&1\\ (x+y)^1&=&x+y\\ (x+y)^2&=&x^2+2xy+y^2 \end{eqnarray} and we can easily expand \[(x+y)^3=x^3+3x^2y .
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## Binomial theorem
Binomial theorem is a kind of formula that helps us to expand binomials raised to the power of any number using the pascals triangle or using the binomial theorem watch the video to now about the pascal's triangle and the binomial theorem. The binomial theorem is a quick way (okay, it's a less slow way) of expanding (or multiplying out) a binomial expression that has been raised to some (generally inconveniently large) power for instance, the expression (3 x – 2) 10 would be very painful to multiply out by hand. Proof it is not hard to see that the series is the maclaurin series for $(x+1)^r$, and that the series converges when $-1 x 1$ it is rather more difficult to prove that the series is equal to $(x+1)^r$ the proof may be found in many introductory real analysis books.
• Just to show you that binomial theorem allows us to compute the coefficients of all the monomials, not only for the case when we had just a + b where instead we might want to compute the 4th.
• Binomial theorem a binomial is a polynomial with two terms example of a binomial what happens when we multiply a binomial by itself many times .
• The binomial theorem states that the binomial coefficients $$c(n,k)$$ serve as coefficients in the expansion of the powers of the binomial $$1+x$$:.
Binomial theorem, statement that for any positive integer n, the nth power of the sum of two numbers a and b may be expressed as the sum of n + 1 terms of the form in the sequence of terms, the index r takes on the successive values 0, 1, 2,, n the coefficients, called the binomial coefficients . Binomial expression for example, x + a, 2 x – 3y, 3 1 1 4, 7 5 x x x y − − , etc, are all binomial expressions 812 binomial theorem if a and b are real . In this video, i show how to expand the binomial theorem, and do one example using it category education using binomial expansion to expand a binomial to the fourth degree - duration: .
Binomial theorem
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# CLP(R)
CLP(R) is a declarative programming language. It stands for constraint logic programming (Real) where real refers to the real numbers. It can be considered and is generally implemented as a superset or add-on package for a Prolog implementation.
## Example rule
${\displaystyle {\begin{cases}3x+4y-2z=8\\x-5y+z=10\\2x+3y-z=20\end{cases}}}$
are expressed in CLP(R) as:
3*X + 4*Y - 2*Z = 8,
X - 5*Y + Z = 10,
2*X + 3*Y -Z = 20.
and a typical implementation's response would be:
Z = 35.75
Y = 8.25
X = 15.5
Yes
## Example program
CLP(R) allows the definition of predicates using recursive definitions. For example a mortgage relation can be defined as relating the principal P, the number of time periods of the loan T, the repayment each period R, the interest rate per period I and the final balance owing at the end of the loan B.
mg(P, T, R, I, B) :- T = 0, B = R.
mg(P, T, R, I, B) :- T >= 1, P1 = P*(1+I) - R, mg(P1, T - 1, R, I, B).
The first rule expresses that for a 0 period loan the balance owing at the end is simply the original principal. The second rule expresses that for a loan of at least one time period we can calculate the new owing amount P1 by multiplying the principal by 1 plus the interest rate and subtracting the repayment. The remainder of the loan is treated as another mortgage for the new principal and one less time period.
What can you do with it? You can ask many questions. If I borrow 1000\$ for 10 years at 10% per year repaying 150 per year, how much will I owe at the end?
?- mg(1000, 10, 150, 10/100, B).
The system responds with the answer
B = 203.129.
How much can I borrow with a 10 year loan at 10% repaying 150 each year to owe nothing at the end?
?- mg(P, 10, 150, 10/100, 0).
The system responds with the answer
P = 921.685.
What is the relationship between the principal, repayment and balance on a 10 year loan at 10% interest?
?- mg(P, 10, R, 10/100, B).
The system responds with the answer | {
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?- mg(P, 10, R, 10/100, B).
The system responds with the answer
P = 0.3855*B + 6.1446 * R.
This shows the relationship between the variables, without requiring any to take a particular value.
## References
• Joxan Jaffar, Spiro Michaylov, Peter J. Stuckey, Roland H. C. Yap: The CLP(R) Language and System. ACM Transactions on Programming Languages and Systems 14(3): 339-395 (1992) | {
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# Solve a system of equations for real $x,y,z$
Solve the following system of equations in $x,y,z \in \Bbb R$: $$(x+y)^3 = z$$ $$(x+z)^3 = y$$ $$(y+z)^3 = x$$
I found the solutions $(0,0,0)$, $(\frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}})$ and $(-\frac{1}{2\sqrt{2}}, -\frac{1}{2\sqrt{2}}, -\frac{1}{2\sqrt{2}})$, but I'm not sure if my proof is correct. Also I was wondering if there is a more elegant solution than the one I found (described below).
First, I took the first two equations and solved for $x$, getting $$x = \sqrt[3]{z} - y$$ $$x = \sqrt[3]{y} - z$$
Then substituting $z = c^3$ and $y = b^3$ (since we are working in $\Bbb R$, these values of $b$ and $c$ exist and are unique), we get
$$c - b^3 = b - c^3$$
and thus
$$c + c^3 = b + b^3$$
Since the function $f(x) = x^3 + x$ is one-to-one and onto, this implies $b=c$. By symmetry, this means $a=b=c$. Since $f(x) = x^3$ is one-to-one and onto, this implies $x=y=z$. Plugging this into the original equation gives
$$8x^3 = x$$
which has the solutions $x = 0, \pm \frac{1}{2\sqrt{2}}$. Hence the three solution pairs I found.
Is this solution ok? Is there an alternate, more elegant approach (that perhap more obviously uses the symmetry between $x$, $y$ and $z$?)
• Your third equation is wrong, don't? – MonsieurGalois Dec 8 '16 at 7:58
• @MonsieurGalois Sorry, that was a typo, you're right. Fixed. – aras Dec 8 '16 at 8:01
This problem is very pretty:
Let's suppose that $x\geq y\geq z$ (any order will let you the same, you can check that). Because of this,
$$x+y\geq x+z\geq z+y$$ $$\Rightarrow (x+y)^3\geq (x+z)^3\geq (z+y)^3$$
But that means that:
$$z\geq y\geq x$$
So from that $x=y=z$ then for any equation you will have $8x^3=x$. From this last equation you have $x=0$ is solution or $x=\pm \frac{1}{2\sqrt{2}}$
• This is awesome, thank you! I knew there would be a better argument...will accept when the accept cooldown wears off. – aras Dec 8 '16 at 8:03 | {
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# Uncountability of the Cantor Set
Let $x=\left(0.a_1a_2...\right)_3 \in \mathcal{C}$, where $\mathcal{C}$ denotes the Cantor set and the $a_i$'s are either $0$ or $2$. Let $f:\mathcal{C}\rightarrow \left[0,1\right]$ such that $f(x)=\left(0.(a_1/2)(a_2/2)...\right)_2$.
I ultimately want to show that $\mathcal{C}$ is uncountable. But before I must show that $f$ is continuous and surjective.
This is my attempt for the surjective part:
Let $y\in [0,1]$. Then $y$ has a binary expansion, say, $y=(0.y_1y_2...)$ let $y_i=a_i/2$. Then $f(x)=(0.y_1y_2...)=y$. So $f$ is onto.
Afterwards can I conclude that since $[0,1]$ is uncountable and $f$ is a surjection, $\mathcal{C}$ is uncountable?
My Questions:
1. Is my attempt for the surjective part okay? Also, is my conclusion about the uncountablity of $\mathcal{C}$ okay, or I will have to do more?
2. How do I show that $f$ is continuous?
Thanks. | {
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Thanks.
-
When representing real numbers in base notation, you should also consider the case of an infinite trailing string of 1s (or 2s in in the ternary case), although there are only a countable number of such cases and it won't affect the result. Why do you need to show that f is continuous? I think your argument is pretty convincing already. – Dan Brumleve Sep 14 '11 at 2:56
Maybe this related thread is helpful. – t.b. Sep 14 '11 at 3:10
@Dan: Thanks for your response. Well, I am to show first that $f$ is continuous and surjective and then conclude that $\mathcal{C}$ is uncountable. – Kuku Sep 14 '11 at 3:13
Kuku, it seems to me that showing f is a surjection is enough to establish that the Cantor set is uncountable. – Dan Brumleve Sep 14 '11 at 3:19
"Afterwards can I conclude" has the word order for a question, but the sentence doesn't end in a question mark. Do you mean "Afterwards I can conclude" as a statement, or is the question mark missing? Please clarify this by an edit. (You might also want to tidy up the "My Questions" part while you're at it, though that doesn't lead to ambiguity -- the questions under "1." are both missing question marks, and "I will have" has the word order for a statement.) – joriki Sep 14 '11 at 6:35
You don’t have to show that $f$ is continuous in order to conclude that $\mathcal{C}$ is uncountable: that follows from the fact that $f$ is surjective, assuming that you know that $[0,1]$ is uncountable. Your argument for surjectivity is correct, though it could be stated a bit better, but for clarity you ought to deal with a point raised by Dan Brumleve. Here’s your argument, slightly restated:
Let $y\in [0,1]$ be arbitrary, and let $(0.y_1y_2\dots)$ be a binary representation of $y$. For each positive integer $i$ let $a_i=2y_i$, and let $x$ be the number whose ternary representation is $(0.a_1a_2\dots)$; then $y=f(x)$, so $f$ is surjective. | {
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Since this looks at first sight as if you were actually constructing a function from $[0,1]$ to $\mathcal{C}$, it would help the reader if you were to point out that this isn’t the case. For example, $y=1/2$ has two binary representations, $0.1\bar{0}$ and $0.0\bar{1}$, where the bar indicates a repeating digit, so it’s both $f(0.2_{\text{three}})=$ $f(\frac23)$ and $f(0.\bar{2})=f(\frac13)$.
As I said, continuity of $f$ isn’t needed if all you want is to prove the uncountability of $\mathcal{C}$, but if for some other reason you have to prove the continuity of $f$, here’s one approach. For $x\in\mathbb{R}$ and $r>0$ let $B(x,r)=\{y\in\mathbb{R}:\vert y-x\vert\le r\}$. Suppose that $0.x_1x_2\dots$ is the $1$-less ternary expansion of some $x\in\mathcal{C}$. For $n\in\mathbb{Z}^+$ let $T_n(x)=$ $\{(0.a_1a_2\dots)\in\mathcal{C}:\forall i\le n [a_i=x_i]\}$. Show that $$B(x,3^{-(n+1)})\cap\mathcal{C}\subseteq T_n(x)\subseteq B(x,3^{-n})\cap\mathcal{C}.$$ Then show that for each $r>0$ there is an $n(r)\in\mathbb{Z}^+$ such that $T_{n(r)}(x)\subseteq B(f(x),r)$.
Another approach is to show that $f$ preserves convergent sequences, i.e., that if $\langle x_n:n\in\mathbb{N}\rangle$ is a sequence in $\mathcal{C}$ that converges to $x\in\mathcal{C}$, then $\langle f(x_n):n\in\mathbb{N}\rangle \to f(x)$ in $[0,1]$. A good first step would be to show that if $\langle x_n:n\in\mathbb{N}\rangle$ is a sequence in $\mathcal{C}$ that converges to $x\in\mathcal{C}$, then for each $m\in\mathbb{Z}^+$ there is an $n(m)\in\mathbb{N}$ such that $x_n\in T_m(x)$ whenever $n\ge n(m)$: that is, every term of the sequence from $x_{n(m)}$ on agrees with $x$ to at least $m$ ternary places. Then each term of $\langle f(x_n):n\in\mathbb{N}\rangle$ from $f(x_{n(m)})$ on agrees with $f(x)$ to at least $m$ binary places. From this it’s not hard to conclude that $\langle f(x_n):n\in\mathbb{N}\rangle \to f(x)$. | {
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-
Your first assumption is fine, since for any function $f$, $|dom(f)|\geq|im(f)|$. A simple $\epsilon-\delta$ proof should do for continuity.
Suppose $\mathcal{C}$ is countable, and make a (possibly countably infinite) list of its elements. Then create a new $(0.a_1a_2...)_3$ representation of a number by making $a_i$ anything other than the $i^{th}$ digit of the $i^{th}$ number in your list.
Your constructed number will differ by at least one digit from every number you have listed in your supposedly exhaustive enumeration of the elements of $\mathcal{C}$, yielding a contradiction.
Thus $\mathcal{C}$ is uncountable. For more on the diagonalization argument, see the corresponding Wikipedia page. | {
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Number of ways two knights can be placed such that they don't attack.
What are the number of ways two knights can be placed on a k×k chessboard so that they do not attack each other?
For k from 1 to 8, the answer is given below. How do I find a general formula?
0
6
28
96
252
550
1056
1848
Edit:
Here's my approach after @Peter 's help, I came to a conclusion that number of ways such that they attack is equal to two times the number of possible ways I can put an "L" shape on the board. (2 times because knights can swap positions), am I right? I don't know how do I more forward from here.
I tried finding number of ways to place L by this recursive formula: F[n][n]=4+F[i][i-3]+F[i-2][3]; But it's not working.
• Probably, it is easier to determine the number of ways to place them that they attack each other. This has then only to be subtracted from $\frac{k^2(k^2-1)}{2}$ Jun 18, 2019 at 10:44
• Yes, That helped, So I came to a conclusion that number of ways such that they attack is equal to two times the number of possible ways I can put an "L" shape on the board. (2 times because knights can swap positions), am I right? I don't know how do I more forward from here.
– Het
Jun 18, 2019 at 11:07
• Maybe I can use recursion... @Peter
– Het
Jun 18, 2019 at 11:24 | {
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Note that when we have two knights threatening each other, it actually forms either a $$2\times3$$ or $$3 \times 2$$ board. And for each of $$2 \times 3$$ and $$3 \times 2$$ boards, there are $$2$$ ways of placing two knights so that they threaten each other. So, what we should do is to count how many $$2 \times 3$$ and $$3 \times 2$$ squares on $$n\times n$$ board. For general $$n$$, the answer is $$(n-1)(n-2)+(n-2)(n-1) = 2(n-1)(n-2)$$ And for each $$2\times3$$ and $$3\times2$$ board, there are $$2$$ ways of placing the knights so that they threaten each other. Therefore, in total there are $$2\cdot2(n-1)(n-2)=4(n-1)(n-2)$$ ways of placing two knights so that they threaten each other. So what you are looking for is $$\frac{n^2(n^2-1)}{2}-4(n-1)(n-2)$$ It is also worth mentioning that we are not over-counting because whenever we place two knights so that they threaten each other, either a $$2 \times 3$$ or $$3 \times 2$$ board must contain both of the knights.
• How did you count 2x3 and 3x2 boards?
– Het
Jun 18, 2019 at 12:04
• Since we have a square board, once we count number of $2 \times 3$ boards, number of $3 \times 2$ boards have the same number by symmetry. After that, in order to count number of $3 \times 2$ boards, you can basically start from top left and shift $1$ square to right and find how many times you can shift it to the right. And shifting it to bottom and finding how many times you can shift it to bottom and multiplying them will give you the answer since we are actually finding how many rows of $3$ adjacent squares are there and how many columns of $2$ adjacent squares are there. Jun 18, 2019 at 12:11
• Aren't there 4 ways to place 2 knights on 2x3 or 3x2 board? since both knight can swap positions.
– Het
Jun 18, 2019 at 12:23
• I think knights are to be considered identical in this case. So swapping positions will give the same placement. Jun 18, 2019 at 12:28
• Can you come up with a recursive solution to this problem? Dec 20, 2020 at 6:15 | {
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Finding the minimum value of $a^2+b^2+c^2$ [duplicate]
Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$.
I've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me $a+b+c$, which is unknown. How can I solve this?
marked as duplicate by user354271, kingW3, Frpzzd, Namaste algebra-precalculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 2 '17 at 0:17
Notice that by Inequality of arithmetic and geometric means we know that:
$$2ab \leq a^2+b^2; \\ 2ac \leq a^2+c^2; \\ 2bc \leq b^2+c^2;$$
so we can conclude that: $$2\left(ab+ac+bc\right) \leq 2\left(a^2+b^2+c^2\right) \Longrightarrow \\ \ \ \left(ab+ac+bc\right) \leq \ \ \left(a^2+b^2+c^2\right) \Longrightarrow \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \leq \ \ \left(a^2+b^2+c^2\right) .$$
Note that this in-equlality is sharp for $a=b=c=\sqrt{\dfrac{2}{3}}$;
for which one can see the value of $a^2+b^2+c^2$ is equal to $2$.
Famke's answer is the simplest; however, we can use a variational argument, as well.
For all variations that maintain $ab+bc+ca=2$, we have $$(b+c)\,\delta a+(a+c)\,\delta b+(a+b)\,\delta c=0\tag{1}$$ To minimize $a^2+b^2+c^2$, we must have $$2a\,\delta a+2b\,\delta b+2c\,\delta c=0\tag{2}$$ for all variations that satisfy $(1)$. | {
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Orthogonality says that to have $(2)$ for all variations that satisfy $(1)$, we need $$2a=\lambda(b+c),\quad2b=\lambda(a+c),\quad\text{and}\quad2c=\lambda(a+b)\tag{3}$$ Summing these up, we get that $\lambda=1$, and then solving the equations, we get that $a=b=c$. Finally, to satisfy the constraint for $(2)$, we get that $a=b=c=\sqrt{\frac23}$. Thus, the minimum of $a^2+b^2+c^2$ is $2$.
• My dear robjohn♦ your solution is more creative and interesting. – Davood Sep 1 '17 at 12:46
• Robjohn.Very elegant solution, Peter – Peter Szilas Sep 1 '17 at 14:21
Scalar product:
Let $\vec A : = (a,b,c)$, $\vec B: = (b,c,a)$.
$|\vec A \cdot \vec B| \le |A| |B|$.
$\Rightarrow$ :
$2 \le ab +bc + ca \le a^2 +b^2 + c^2$.
Equality:
$a=b=c = \sqrt{\frac{2}{3}}$.
You can solve this problem by your starting step.
Indeed, by C-S $$3(a^2+b^2+c^2)\geq(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc).$$ Thus, $$3(a^2+b^2+c^2)\geq a^2+b^2+c^2+2(ab+ac+bc)$$ or $$a^2+b^2+c^2\geq ab+ac+bc$$ and since $ab+ac+bc\geq2,$ we obtain: $$a^2+b^2+c^2\geq2.$$ The equality occurs for $(1,1,1)||(a,b,c)$ and $ab+ac+bc=2$,
which says that $2$ is a minimal value.
Again from Cauchy-Schwarz from a different perspective
$\sqrt{a^2c^2}+\sqrt{b^2c^2}+\sqrt{a^2c^2} \leq \sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}$
$a^2+b^2+c^2 \ge ab+bc+ac \ge 2$
$a^2+b^2+c^2 \ge 2$
İnequality holds for $a=b=c=\sqrt{\dfrac{2}{3}}$ | {
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Similarly, we can find the area of a parallelogram using vector product. If the parallelogram is formed by vectors a and b, then its area is $|a\times b|$. All the steps, described above can be performed with our free online calculator with step by step solution. Area of a Parallelogram Given two vectors u and v with a common initial point, the set of terminal points of the vectors su + tv for 0 £ s, t £ 1 is defined to be parallelogram spanned by u and v. We can explore the parallelogram spanned by two vectors in a 2-dimensional coordinate system. Solution Begin a geometric proof by labeling important points You may have to extend segment AB as you draw the height from C. Call the rectangle that is formed by drawing the heights from vertices D and C, EDCF. Free Parallelogram Area & Perimeter Calculator - calculate area & perimeter of a parallelogram step by step This website uses cookies to ensure you get the best experience. Dot product of two vectors on plane, Exercises. 5.5. c. -6.0. d. none of above. So the area of your parallelogram squared is equal to the determinant of the matrix whose column vectors construct that parallelogram. The area of a parallelogram farmed from the vector A = i - 2 j + 3k and vector B = 3 i - 2 j + k as adjacent side is asked Oct 8, 2019 in Vector algebra by KumarManish ( 57.6k points) vector algebra The area of triangle as cross product of vectors representing the adjacent sides. Cross product of two vectors (vector product), Online calculator. Dot product of two vectors in space, Exercises. Length of a vector, magnitude of a vector in space. Contacts: support@mathforyou.net. Addition and subtraction of two vectors in space, Exercises. The other multiplication is the dot product, which we discuss on another page. If you want to contact me, probably have some question write me email on support@onlinemschool.com, Online calculator. Or if you take the square root of both sides, you get the area is equal to the absolute value of the | {
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Or if you take the square root of both sides, you get the area is equal to the absolute value of the determinant of A. The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides. $\vec {u}, \vec {v} \in \mathbb {R}^3$. Download 1,300+ Royalty Free Parallelogram Vector Images. Scalar-vector multiplication, Online calculator. This is true in both $R^2\,\,\mathrm{and}\,\,R^3$. Find its area. More in-depth information read at these rules. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how find area of parallelogram formed by vectors. Vector Cross Product Magnitude: equal to the area of the parallelogram defined by the two vectors Direction: determined by the Right-Hand-Rule A × B = A B sin θ = A B sin θ = A sin θ B (0 ≤ θ ≤ π) equals to the area of the parallelogram, build on corresponding vectors: Therefore, to calculate the The formula is actually the same as that for a rectangle, since it the area of a parallelogram is basically the area of a rectangle which has for sides the parallelogram's base and height. Vector magnitude calculator, Online calculator. More in-depth information read at these rules. ABDC is a parallelogram with a side of length 11 units, and its diagonal lengths are 24 units and 20 units. Finding the area of a parallelogram using the cross product.Followup: see http://youtu.be/9o9yx95rFAo for details on how to draw the parallelogram Prove the parallelogram law: The sum of the squares of the lengths of both diagonals of a parallelogram equals the sum of the squares of the lengths of all four sides. Suppose, we are given a triangle with sides given in vector form. The area of a parallelogram is twice the area of a triangle created by one of its diagonals. The below figure illustrates how, using trigonometry, we can calculate that the area of the parallelogram spanned by a and b is a bsinθ, where θ is the | {
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we can calculate that the area of the parallelogram spanned by a and b is a bsinθ, where θ is the angle between a and b. If two sides of a parallelogram are represented by two vectors A and B, then the magnitude of their cross product will be equal to the area of parallelogram i.e. This is true in both $R^2\,\,\mathrm{and}\,\,R^3$. The area of the parallelogram formed by vectors a=(−1,3,1) and b=(1,2,0), rounded to one decimal, is: Select one: a. Press the button "Find parallelogram area" and you will have a detailed step-by-step solution. PROOF Consider a parallelogram OABC whose two sides are represented by two vectors A and B as shown. Direction cosines of a vector, Online calculator. You can input only integer numbers, decimals or fractions in this online calculator (-2.4, 5/7, ...). The area of a polygon is the number of square units inside the polygon. I consider this as revision I have looked at several examples but most are complex and so i want to be helped on this one. Then, draw heights from vertices D and C to segment AB. If the area of parallelogram whose diagonals coincide with the following pair of vectors is ,then vectors are 0:26 000+ LIKES. In Geometry, a parallelogram is a two-dimensional figure with four sides. So the area of this parallelogram is the absolute value of the determinant of . Component form of a vector with initial point and terminal point, Online calculator. Finding the area of the parallelogram spanned by vectors <-1,0,2> and <-2,-2,2> I have not tried anything since I have no idea. The End Opposite sides are equal in length and opposite angles are equal in measure. The figure shows t… And the area of parallelogram using vector product can be defined using cross product. The area of parallelogram as cross product of vectors representing adjacent sides. Area of triangle formed by vectors, Online calculator. Is equal to the determinant of your matrix squared. Magnitude of the Find the area of the parallelogram having diagonals: | {
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of your matrix squared. Magnitude of the Find the area of the parallelogram having diagonals: A=3i+j-2k B=i-3j-4k also Find the volume of the parallelepiped whose edges are represented by: A=2i-3j+4k B=i+2j-k C=3i-j=2k Thus we can give the area of a triangle with the following formula: (5) Area of Parallelogram Formula. area of the parallelogram, build on vectors, one need to find the vector which is the vector product of the initial vectors, then find the magnitude of this vector. Math can be an intimidating subject. Proof Area of Parallelogram Forluma Suppose two vectors and in two dimensional space are given which do not lie on the same line. Addition and subtraction of two vectors on plane, Exercises. Start your proof of the area of a parallelogram by drawing a parallelogram ABCD. Now construct a parallelogram OACB by assuming a scale (say 1cm=50 gwt) corresponding to the weights P and Q. Chapter 4: Area of a Parallelogram, Determinants, Volume and Hypervolume, the Vector Product Introduction. The leaning rectangular box is a perfect example of the parallelogram. Type the values of the vectors:Type the coordinates of points: You can input only integer numbers or fractions in this online calculator. Guide - Area of parallelogram formed by vectors calculator To find area of parallelogram formed by vectors: Select how the parallelogram is defined; Type the data; Press the button "Find parallelogram area" and you will have a detailed step-by-step solution. Statement of Parallelogram Law . To find the area of the parallelogram, multiply the base of the perpendicular by its height. It is a special case of the quadrilateral. A parallelogram has two pairs of parallel sides with equal measures. Chapter 4: Area of a Parallelogram, Determinants, Volume and Hypervolume, the Vector Product Introduction. More in-depth information read at these rules. The formula is actually the same as that for a rectangle, since it the area of a parallelogram is basically the area of a | {
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the same as that for a rectangle, since it the area of a parallelogram is basically the area of a rectangle which has for sides the parallelogram's base and height. 5.4. b. the parallelogram whose adjacent sides are the vectors $\vc{a}$ and $\vc{b}$, as shown in below figure). Addition and subtraction of two vectors in space, Exercises. Therefore, to calculate the area of the parallelogram, build on vectors, one need to find the vector which is the vector product of the initial vectors, then find the magnitude of this vector. Latest Blog Post. The area of a parallelogram is equal to the magnitude of cross-vector products for two adjacent sides. We note that the area of a triangle defined by two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$ will be half of the area defined by the resulting parallelogram of those vectors. A parallelogram is a 4-sided shape formed by two pairs of parallel lines. Area of a parallelogram. The diagonal of the parallelogram OC will give the resultant vector. We consider area of a parallelogram and volume of a parallelepiped and the notion of determinant in two and three dimensions, whose magnitudes are these for figures with their column vectors as edges. 300+ VIEWS. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is asked Oct 9, 2019 in Vector algebra by KumarManish ( 57.6k points) vector algebra The best selection of Royalty Free Parallelogram Vector Art, Graphics and Stock Illustrations. I designed this web site and wrote all the mathematical theory, online exercises, formulas and calculators. The sum of the interior angles in a quadrilateral is 360 degrees. vector product of the vectors Addition and subtraction of two vectors, Online calculator. Length of a vector, magnitude of a vector on plane, Exercises. Find the area of the parallelogram whose one side and a diagonal are represented by conitial vectors i ̂ - j ̂ + k ̂ and 4i ̂ + 5k ̂ Next: Question 27→ Class 12 | {
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are represented by conitial vectors i ̂ - j ̂ + k ̂ and 4i ̂ + 5k ̂ Next: Question 27→ Class 12 Decomposition of the vector in the basis, Exercises. We consider area of a parallelogram and volume of a parallelepiped and the notion of determinant in two and three dimensions, whose magnitudes are these for figures with their column vectors as edges. The magnitude (or length) of the vector $\vc{a} \times \vc{b}$, written as $\|\vc{a} \times \vc{b}\|$, is the area of the parallelogram spanned by $\vc{a}$ and $\vc{b}$ (i.e. There are two ways to take the product of a pair of vectors. Area of a parallelogram is a region covered by a parallelogram in a two-dimensional plane. Area of parallelogram formed by vectors, Online calculator. The Relationship of the Area of a Parallelogram to the Cross Product. Next: solution Up: Area of a parallelogram Previous: Area of a parallelogram Example 1 a) Find the area of the triangle having vertices and . Volume of pyramid formed by vectors, Online calculator. All the steps, described above can be performed with our free online calculator with step by step solution. Let’s see some problems to find area of triangle and parallelogram. Dot product of two vectors, Online calculator. The cross product is anticommutative (i.e., a × b = − b × a) and is distributive over addition (i.e., a × (b + c) = a × b + a × c). One of these methods of multiplication is the cross product, which is the subject of this page. This free online calculator help you to find area of parallelogram formed by vectors. The area of a parallelogram can easily be computed from the direction vectors: Simply treat the vectors as a matrix and take the absolute value of the determinant: Compare with Area: A Parallelogram … Length of a vector, magnitude of a vector on plane, Exercises. You can navigate between the input fields by pressing the keys "left" and "right" on the keyboard. . It should be noted that the base and the height of the parallelogram are perpendicular to | {
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. It should be noted that the base and the height of the parallelogram are perpendicular to each other, whereas the lateral side of the parallelogram is … Any line through the midpoint of a parallelogram bisects the area. By using this website, you agree to our Cookie Policy. $\vec {u} \times \vec {v}$. © Mathforyou 2021 Component form of a vector with initial point and terminal point on plane, Exercises. 300+ SHARES. Vector Cross Product Magnitude: equal to the area of the parallelogram defined by the two vectors Direction: determined by the Right-Hand-Rule A × B = A B sin θ = A B sin θ = A sin θ B (0 ≤ θ ≤ π) If two vectors acting simultaneously at a point can be represented both in magnitude and direction by the adjacent sides of a parallelogram drawn from a point, then the resultant vector is represented both in magnitude and direction by the diagonal of the parallelogram passing through that point. To find the area of a parallelogram, multiply the base by the height. Decomposition of the vector in the basis, Exercises. The Area of a Triangle in 3-Space. The weight of … This is possible to create the area of a parallelogram by using any of its diagonals. Welcome to OnlineMSchool. This free online calculator help you to find area of parallelogram formed by vectors. These two vectors form two sides of a parallelogram. It can be shown that the area of this parallelogram ( which is the product of base and altitude ) is equal to the length of the cross product of these two vectors. More generally, the magnitude of the product equals the area of a parallelogram with the vectors for sides; in particular, the magnitude of the product of two perpendicular vectors is the product of their lengths. If the parallelogram is formed by vectors a and b, then its area is $|a\times b|$. b) Find the area of the parallelogram constructed by vectors and , with and . Area is 2-dimensional like a carpet or an area rug. This web site owner is mathematician Dovzhyk Mykhailo. Component | {
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# Two players alternate flipping a coin until the result is head. How to derive that the probability for the first player to win is $2/3$? [duplicate]
Two players, $$A$$ and $$B$$, alternately and independently flip a coin and the first player to obtain a head wins. Player $$A$$ flips first. What is the probability that $$A$$ wins?
Official answer: $$2/3$$, but I cannot arrive at it.
Thought process: Find the probability that a head comes on the $$n$$th trial. That's easy to do, it's just a Geometric random variable. Then find the probability that $$n$$th turn is player's A turn. Finally, multiply both probabilities.
When I came up with each probability, both of them depended on the amount of trials $$n$$, so my answer was a non-constant function of $$n$$.
However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.
## marked as duplicate by Key Flex, Theo Bendit, José Carlos Santos, Kemono Chen, mrtaurhoFeb 10 at 10:46
• What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that. – Pakk Feb 8 at 21:41
• I have understood the flaw in my method by understanding the correct answer, but thank you anyways – Victor S. Feb 8 at 21:56
So $$A$$ wins if head comes first time in odd number of tosses. Say $$P_k$$ is probability that head comes first time in $$k$$-th toss then $$P_k = q^{k-1}p$$ and
$$\begin{eqnarray} P &=& P_1+P_3+P_5+... \\ \\ &=&p+q^2p+q^4p+q^6p+....\\ \\ &=& {p\over 1-q^2}\\ \\& =& {1\over 1+q} \end{eqnarray}$$
where $$p$$ is probability that head comes in one toss and $$q=1-p$$.
So if the coin is fair, then $$p=1/2=q$$, so $$P= {2\over 3}$$ | {
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So if the coin is fair, then $$p=1/2=q$$, so $$P= {2\over 3}$$
Let $$p$$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=\frac12+\frac12(1-p)\implies p=\frac23$$
• Is this way doable even if the coin is not fair? – Aqua Feb 9 at 7:43
• @greedoid Sure. If the coin lands heads with probability $h,$ then $p=h+(1-h)(1-p)$ – saulspatz Feb 9 at 14:19
Here's another approach.
Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.
In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).
But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.
This gives A a $$\frac23$$ chance of winning to $$\frac13$$ for B.
Let $$p$$ denote the probability that $$A$$ wins. There is a $$.5$$ probability $$A$$ wins on the first flip. If he flips tails, then in order to win $$B$$ must then flip tails. This situation occurs with probability $$.25$$ ($$1/2 * 1/2$$ for both tails) and in this case we are back to where we started. So $$p = .5+ .25p$$ which yields $$p=2/3$$.
P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)
A loses first round, B loses second round, A wins the third round (0.5)^3
And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2
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# Number of digits in a series of numbers
I have a list of the numbers from 1 to 1000.
How can I find the number of 0's, 1's, 2's, and 9's that are used?
The answers are 192, 301, 300, 300 respectively, but I'm interested in the process itself.
Just an explanation for one of these digits will suffice, I'll then be able to solve for the other 3 cases.
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I'm guessing this is from 1 to 1000 inclusively? – Nicolas Villanueva Jun 23 '11 at 17:24
Yes, it includes both. – Chandler Jun 23 '11 at 17:28
Taking 2's. There are two approaches. First, you can count how many are in each place. There are 1000 numbers in the list. How many 2's are in the ones place? How many in the 10's place? How many in the 100's place.
Second, you can count from the front. How many one digit numbers contain a 2? How many 2 digit numbers start with 2? How many other 2's are there among the 2 digit numbers?
Either works. Probably the first is easier, but in either case you would have to work a bit harder if you were counting up to 2357.
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In the 1's, I would have it as 2, 12, 22, ..., 992. = 10*10 = 100 In the 10's, 20, 120, ..., 920. = 10*10 = 100 In the 100's, 200, 201, ..., 299. = 100 100*3=300 But for 0's I get 102, not 192. I'm getting 10,20,...,990 for 1's, which is 90+1 0's. Then 100,200,...,1000, which is 100 's, and then the 1 0 from 1000 in the 100's place. What am I counting wrong? – Chandler Jun 23 '11 at 17:41
You say you got 90 in the 1's place, but you showed 99 numbers and missed 1000. So there are 100 there. For the 10's place, you get 10 in every block of a hundred, 90, plus the one in 1000=91. Then the one in the hundreds place of 1000=1. Total is 100+91+1=192. – Ross Millikan Jun 23 '11 at 17:53
The problem has been solved already, so I will solve a different problem. | {
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The problem has been solved already, so I will solve a different problem.
Note first that in the original problem the $1000$ kind of sticks out. So I would have preferred dealing with $1$ to $999$. (We can always deal with $1000$ at the end. And to tell the truth I really prefer $0$ to $999$.)
Let's bump the problem up a bit, by going from $0$ to $9999$.
For the sake of symmetry, "pad" the little numbers with initial $0$'s, so that everybody has $4$ digits.
The reason I am doing this is to make stuff as symmetrical as possible: symmetry is our friend.
So the numbers are
$0000$
$0001$
$0002$
and so on, ending with
$9998$
$9999$
Think of any digit, like $7$, or even $0$.
How many $7$'s in total in the units place? Once we have put a $7$ there, we can fill out the rest of the spots in $7^3$ ways.
How many $7$'s in the next place? Again, once you have placed the $7$, there are $10^3$ ways to fill out the rest. Continue. We find that the total number of $7$'s is $$4 \times 10^3.$$
This is also the total number of $0$'s! But some of the zeros have been obtained from the "padding" and should now be removed.
The numbers $0$ to $9$ have a padding of $0$ in the tens place, for a total of $10^1$. The ones from $0$ to $99$ have padding in the hundreds place, for a total of $10^2$. And the numbers $0$ to $999$ have padding in the thousands place, a total of $10^3$. So the number of genuine legitimate $0$'s is $$4 \times 10^3 -(10^1+10^2+10^3).$$
In the case of your problem, the corresponding expression would be $3\times 10^2 -(10^1+10^2)$. This is $190$. But your list started at $1$, which eliminates one $0$, and ended at $1000$, which adds $3$, for a total of $192$.
Generalization: It is easy now to see what the counts would be if we listed, say, all numbers from $0$ to $10^k-1$.
-
Count the units digits which cycle 1,2 ... 0 an integral number of times. So you know the number of zeros etc in the units position. | {
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Then look at the tens digits, which have a similar cycle, but in batches of ten, and without anything for the integers 1-9 (no leading zeros).
Then the hundreds digits.
Then the single thousands digit.
If you understand the pattern, it should be easy to extend to higher numbers.
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# Some formulae for a periodic sequence $-1,-1,1,1…$?
Some formulae for a periodic sequence?
when $T = 2$, we have $-1,1,-1,1,-1,1,\text{...}$, the formula is
\begin{align*}(-1)^n\end{align*}
when $T = 4$, we have $-1,-1,1,1,-1,-1,1,1\text{...}$, the formula is?
And how about the case $T=k$?
-
It could be useful in the future to keep in mind that it is rather, "a formula ..." than "the formula ...". – ABC Sep 3 '13 at 2:17
Are you familiar with the floor function? – Blue Sep 3 '13 at 2:19
I think you can do this pretty easily with a combination of trig functions. – Daniel Rust Sep 3 '13 at 2:24
@DanielRust My first thought is to seek something just some combination of $n$, Andre gave one trig answer. – Sequence Sep 3 '13 at 6:46
@Blue I've thought of that without deep thinking, not so familiar with that in doing math, just played in softwares. Dan gave one answer with floor. – Sequence Sep 3 '13 at 6:48
The clearest is $a_n=-1$ if $n$ leaves remainder $1$ or $2$ on division by $4$, and $a_n=1$ otherwise.
There are also conventionally "closed form" formulas, such as $$a_n=\sqrt{2}\cos\left(\frac{(2n+1)\pi}{4}\right).$$
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This is good, is it have relations with the form in Dan's answer? – Sequence Sep 3 '13 at 6:44
Well, they give the same sequence, that's about it. – André Nicolas Sep 3 '13 at 6:47
Yes, that's a wonderful thing in math. – Sequence Sep 3 '13 at 6:49
Hi, how about the case $k=3$?, it's not $\sqrt{3} \cos \left(\frac{1}{6} \pi (3 n+1)\right)$ – Sequence Sep 4 '13 at 2:51
When $n=3$, the formula in my answer yields $\sqrt{2}\cos(7\pi/4)$, which is $1$, exactly as your sequence asks for. The $\sqrt{3}\cos(\frac{1}{6}\pi(3n+1))$ has no connection with the formula that I gave. – André Nicolas Sep 4 '13 at 2:56 | {
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It is a general fact that every periodic sequence $(x_n)$ of period $T$ is in the linear span of the sequences $\{e^T_k\,;\,1\leqslant k\leqslant T\}$, where, for every $n$, $$(e_k^T)_n=\mathrm e^{2\mathrm i \pi nk/T}.$$ That is, there exists $(a_k)_{1\leqslant k\leqslant T}$ such that, for every $n$, $$x_n=\sum_{k=1}^Ta_k(e_k^T)_n=\sum_{k=1}^Ta_k\mathrm e^{2\mathrm i \pi nk/T}.$$ To find $(a_k)_{1\leqslant k\leqslant T}$, one considers the equations above over one period, say for $1\leqslant n\leqslant T$, as a Cràmer system with unknowns $(a_k)_{1\leqslant k\leqslant T}$.
Example: Consider some sequence $x=(x_1,x_2,x_3,x_1,x_2,x_3,\ldots)$, then $T=3$, $$e_1^3=(j,j^2,1,j,j^2,1,\ldots),\quad e^3_2=(j^2,j,1,j^2,j,1,\ldots),\qquad e^3_3=(1,1,1,1,1,1,\ldots),$$ with $j=\mathrm e^{2\mathrm i\pi/3}$, and one looks for $(a_1,a_2,a_3)$ such that $x=a_1e^3_1+a_2e^3_2+a_3e^3_3$, that is, $$x_1=a_1j+a_2j^2+a_3,\quad x_2=a_1j^2+a_2j+a_3,\quad x_3=a_1+a_2+a_3.$$ Thus, $$3a_1=j^2x_1+jx_2+x_3,\quad 3a_2=jx_1+j^2x_2+x_3,\quad 3a_3=x_1+x_2+x_3,$$ which yields $x_n$ as a linear combination of $j^n$, $j^{2n}$ and $1$, namely, for every $n$, $$x_n=a_1j^n+a_2j^{2n}+a_3.$$
-
You can write it as $(-1)^{\large \lfloor \frac{2 \cdot n+T-2}{T} \rfloor}$.
If you defined things a bit differently (start counting from zero, parameterize the half period instead of the full period) you could just write $(-1)^{\large \lfloor \frac{n}{T} \rfloor}$. | {
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-
But why? ${}{}{}{}$ – Pedro Tamaroff Sep 3 '13 at 2:26
hi, I would also like to know a little about why and how to obtain the formula, I've thought something about Mod, Ceiling and Floor. This is good, is it have some relations with the form in Andre's answer? – Sequence Sep 3 '13 at 6:44
One intuition is that you can associate raising $-1$ to a power with "alternating" and associate the floor function with "doubling", "tripling", etc. You can find the relationship between the answers in the Taylor series for cosine. – Dan Brumleve Sep 3 '13 at 6:49 | {
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# Sequence of Rationals Converging to a Limit
I’m trying to show that for every real number $r$, there exists a sequence of rational numbers $\{q_{n}\}$ such that $q_{n} \rightarrow r$.
Could I get some comments on my proof?
I know that between 2 reals $r, b$ there exists a rational number $m$ such that $r < m < b$.
So I can write
$r < q_{1} < b$ ; Now check if $q_{1} = r$ or not. If it does I’m done, and if not, I consider the interval $(a, q_{1})$.
$r < q_{2} < q_{1}$ check if $q_{2} = r$ or not. If it does I’m done, and if not, I consider the interval $(a, q_{3})$
If I continue in this manner, I see that $|r – q_{n+1}| < |r – q_{n}|$. So whether $r$ is rational or irrational, I’m making my the size of the interval $(r, q_{n})$ closer to 0 and as $n \rightarrow \infty$. And so given any $\epsilon > 0$, I know that $|q_{n} – r| < \epsilon$.
Revision
Let $\{q_{n} \}$ be a sequence of rational numbers and $q_{n} \rightarrow r$ where $r \in \mathbb{R}$.
If $r$ is rational, then let every element of $q_{n} = r$.
But if $r$ is irrational, then consider the interval $(r, b)$ where $b \in \mathbb{R}$. Since we can always find a rational number between two reals, consider
$r < q_{1} < b$. Now pick $q_{2}$ such that $q_{2}$ is the midpoint of $r$ and $q_{1}$. So we get that $r < q_{2} < q_{1}$. Then repeat the process so that $r < q_{n} < q_{n-1}$. Note that $|r – q_{n}| = \frac{1}{2}|r – q_{n-1}|$. As we take more values for $q_{n}$, it is clear that $|q_{n} – r| \rightarrow 0$. So given any $\epsilon > 0$, $|q_{n} – r| < \epsilon$.
#### Solutions Collecting From Web of "Sequence of Rationals Converging to a Limit"
You have the right idea but you are handwaving. You are not telling exactly how you will pick the rationals, and how their limit is $r$.
Also $|r – q_{n+1}| \lt |r – q_n|$ does mean $q_n \to r$. For instance, $\left|-1 – \frac{1}{n+1}\right| \lt \left|-1 – \frac{1}{n}\right|$, but $\frac{1}{n} \to 0$, and not $-1$. | {
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Also, you don’t have to check if $q_i = r$, as you pick $r \lt q_i$.
(Also, the above writeup contains a possible hint, try to find it :-))
Since you are almost there (based on your Revision) (but note: you still are handwaving and don’t have a “proper” proof yet..)
Here is the hint I was referring to:
Since there is at least one rational in $(r, r+\frac{1}{n})$. Pick one and call it $q_n$. What is the limit of $q_n$?
A different proof:
Consider the set $S = \{q: q \ge r, q \in \mathbb{Q}\}$. Show that $\inf S = r$ and show that that implies there is a sequence $q_n \in S$ whose limit is $r$.
Here are some critiques to your proof.
1. When you write “So I can write $r<q_1<b$ “, what is $b$?
2. If $q_1 = r$, how are you done? (I agree that you are, but what would $q_2, q_3,…$ be in this case?)
3. “So whether $r$ is rational or irrational, I’m making my the (sic) size of the interval $(r,q_n)$ closer to $0$ and as $n\rightarrow\infty$.” As $n\rightarrow \infty$, what happens?
4. As noted in the comments, the condition $|r-q_{n+1}|<|r-q_n|$ does not guarantee that $|q_n-r|\rightarrow 0$.
Your idea for the proof is great, you just need a slight adjustment to force $|q_n-r|\rightarrow 0$. You could, for example, pick $q_n$ to be between $r$ and the midpoint of $(r,q_{n+1})$. This would force $|q_n-r|$ to be less than half of $|q_{n-1}-r|$, which will force $|q_n-r|\rightarrow 0$. Fix this, and fix up the general presentation a bit and you’ll have a great proof. | {
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Your question set me thinking, and I came up with an argument inspired by it, but a bit different from your solution attempt. I do not imagine that the idea is new, but I do not recall seeing it done in this way (this may only mean that my education is incomplete- no pun intended). The idea is to take an irrational real number $r,$ and to recursively construct two sequences $(x_n)$ and $(y_n)$ of rational numbers which approach $r$ respectively from below and above (and are respectively non-decreasing and non-increasing). We start with a rational number $x_1 < r$ and a rational number $y_1 >r.$ Having found rational $x_n <r$ and rational $y_n >r,$ we set $z_n = \frac{x_n +y_n}{2}.$ Then $x_n < z_n < y_n.$ If $z_n <r,$ we set $x_{n+1} = z_n$ and $y_{n+1} = y_n,$ while if $z_n > r,$ we set $x_{n+1} = x_n$ and $y_{n+1} = z_n.$
(Note that $z_n \neq r$ as $z_n$ is rational, but $r$ is irrational). In the first case, we have $y_{n+1}-x_{n+1} = \frac{y_n – x_n}{2}$ and in the second case, we also have $y_{n+1} -x_{n+1} = \frac{y_n – x_n}{2}.$ Hence for each $n > 1,$ we have $y_n -x_n = \frac{y_1 – x_1}{2^{n-1}}.$ For large enough $n,$ we can make this less than any chosen real positive quantity $\varepsilon$. Then for large enough $n,$ we have both $0 < y_n -r < \varepsilon$ and $0 < r – x_n < \varepsilon.$ Hence the sequences $(x_n)$ and $(y_n)$ both have limit $r.$ | {
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# Is there an equation for the sum of alternating cubes?
The following sequences for sum of alternating cubes:
Odd cubes: [1, 28, 153, 496, 1225, 2556, 4753, 8128, 13041, 19900, 29161, 41328, 56953, 76636, 101025, 130816, 166753, 209628, 260281, 319600]
Even cubes: [8, 72, 288, 800, 1800, 3528, 6272, 10368, 16200, 24200, 34848, 48672, 66248, 88200, 115200, 147968, 187272, 233928, 288800]
Nothing on OEIS. Trying to find equations for generating such values.
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You may also want to look at: en.wikipedia.org/wiki/Cube_(algebra) in addition to the nice answer by Ross M. – Amzoti Nov 26 '12 at 5:02
Not sure why there are three answers and yet none of them seem to answer the question directly -- figured it out: Odd sum: $a(n) = n^2(2n^2-1)^2$ Even sum: $a(n) = 2n^2(n+1)^2$ – KaliMa Nov 26 '12 at 5:14
Are you sure the odd sum is correct? – Amzoti Nov 26 '12 at 5:30
@Amzoti Typo'd it, shouldn't be squared at the end – KaliMa Nov 26 '12 at 5:33
Still looks like there is a problem, shouldn't it be: type "Sum[(2*i+1)^3,{i,0,n}]" at woframalpha and see if it matches. You can remove the +1 for the even result. – Amzoti Nov 26 '12 at 5:39
$$\sum\limits_{i=1}^n(2i)^3=8\sum\limits_{i=1}^ni^3$$ $$\sum\limits_{i=1}^n(2i-1)^3=\sum\limits_{i=1}^{2n}i^3-\sum\limits_{i=1}^n(2i)^3$$
Of course, the sum of cubes is (for some strange reason...) the square of the sum of consecutive integers.
$$\sum\limits_{i=1}^ni^3=\left(\sum\limits_{i=1}^ni\right)^2=\frac{n^2(n+1)^2}4$$
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Here is a question regarding the strange reason. – robjohn Feb 3 '13 at 13:39
There is a very general way for producing formulae for functions $F:\mathbb{N} \rightarrow \mathbb{Z}.$ Let me explain. | {
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Consider the set of functions $F:\mathbb{N} \rightarrow \mathbb{Z}.$ On such functions there is a discrete analog of differentiation called the shift operator, denoted $\Delta$, which given a function $F$ produces a function $\Delta F$ via the formula $\Delta F(x) := F(x+1) - F(x).$ Now classically, if one is given an infinitely differentiable function $f:\mathbb{R} \rightarrow \mathbb{R}$ one can consider the Taylor series $\displaystyle\sum_{i=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^k$ and ask if this series converges and is equal to $f$ near $0.$ In the discrete case, note that iterated powers of the shift operator are always well defined. So it is natural to ask if there is a discrete analog of Taylor series. These analogs are called Mahler series. Given a function $F:\mathbb{N} \rightarrow \mathbb{Z},$ we define it's Mahler series via the formula
$$\sum_{i=0}^{\infty} \Delta F(0){{x}\choose{k}}.$$
The next question to ask is when is $F$ equal to it's Mahler expansion. And this is where the discrete case is nicer than the classical, for the answer is always. That is
$$F(x) = \sum_{i=0}^{\infty} \Delta F(0){{x}\choose{k}}$$
for all $x \in \mathbb{N}$
(Note if $k>x,$ we have ${x\choose k} = 0$ so there is no issue of convergence in the above series).
So if I have such a function $F:\mathbb{N} \rightarrow \mathbb{Z}$ which I want to express via a nice formula, a good place to begin is to try to write down it's Mahler expansion.
$$F(x) = \sum_{i=0}^{x} (2i)^3.$$
Then
$$\Delta F(x) = (2(x + 1))^3$$
$$\Delta^2 F(x) = (2(x + 2))^3 - (2(x + 1))^3 = 24x^2 +72x + 56$$
$$\Delta^3 F(x) = \Delta^2 F(x+1) - \Delta^2 F(x) = 192x + 208$$
$$\Delta^4 F(x) = \Delta^3 F(x+1) - \Delta^3 F(x) = 192$$
and
$$\Delta^k F(x) = 0 \text{ if } k\ge5.$$
It follows
$$F(x) = {{x}\choose{1}} + 56{{x}\choose{2}} + 208{{x}\choose{3}} + 192{{x}\choose{4}}.$$ | {
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$$F(x) = {{x}\choose{1}} + 56{{x}\choose{2}} + 208{{x}\choose{3}} + 192{{x}\choose{4}}.$$
Following the same mechanical procedure we can also produce a formula for the some of the cubes of the odd natural numbers and many other functions of this sort.
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For the odd cubes, you are asking about $\sum_{i=0}^n (2i+1)^3=\sum_{i=0}^n 8i^3+12i^2+6i+1$. Now feed each term to Faulhaber's formula
For the even cubes, you want $\sum_{i=1}^n (2i)^3=8\sum_{i=1}^n i^3$
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I think I figured out the odd cubes via n^2*(2n^2 - 1). Having trouble with evens. – KaliMa Nov 26 '12 at 5:02
@KaliMa: For the evens, the sum is just $\frac 16 n(n+1)(2n+1)$ – Ross Millikan Nov 26 '12 at 5:03
I thought that equation is for the sum of first n general cubes (even and odd together) – KaliMa Nov 26 '12 at 5:05
@KaliMa: True, but the upper index is half the last number that is cubed. The idea is if you want to do something with all the evens, just do it with all the numbers and double them. The factor 8 accounts for the doubling. – Ross Millikan Nov 26 '12 at 5:07
@KaliMa: No, $40=8(1+4)=8(1+2^2), 8+64=8(1+2^3)=8(1+8)=72$ – Ross Millikan Nov 26 '12 at 5:11 | {
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# Why does applying a small angle approximation to equivalent forms of $\frac{\cos^2 \theta}{\sin \theta\tan \theta}$ yield different results?
In a question about small angle approximations I had answered, I simplified the trigonometric expression using identities and then applied the small and approximation.
I hadn't made any arithmetic mistakes, but the answer I got was different to the mark scheme, where they applied the small angle approximation straight away with no simplification.
My question is: Why do these different approaches yield different answers?
Edit: The question is:
$$\frac{\cos^2 \theta}{\sin \theta\tan \theta}$$
$$\theta^{-2}-{\frac 12}$$
The mark scheme says:
$${\frac 14}{\theta^2}+\theta^{-2}-1$$
• It might help if you could post both approaches, including the simplifications you made. Also, why are you sure that you have no errors? – Dirk Apr 24 at 8:48
• @Dirk. My teacher and I went through it many times and we found nothing arithmetically wrong. I will add the question, although it may not look pretty – Adam Cummings Apr 24 at 8:50
• Your LaTeX looked reasonably pretty (congratulations!), but I think I made it a little prettier. :) Be that as it may ... You should also include your simplification, so that everyone can be as certain as you and the teacher that you made no mistake there. – Blue Apr 24 at 9:20
• We cannot judge your work if you don't explain it. – Yves Daoust Apr 24 at 9:22
The correct approximation using Taylor expansion is $$\theta^{-2}-\dfrac76+O(\theta^2)$$. However, since your approximation is not taking higher order terms into account, it undoubtedly depends on the expression you start with, even if they are equivalent.
For instance, you could write
$$\frac{\cos^2 \theta}{\sin\theta\tan\theta}\simeq\frac{(1-\frac12\theta^2)^2}{\theta\cdot\theta}=\theta^{-2}-1+\frac{\theta^2}4$$
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Or
$$\frac{\cos^2 \theta}{\sin\theta\tan\theta}=\frac{\cos^3\theta}{\sin^2\theta}\simeq\frac{(1-\frac12\theta^2)^3}{\theta^2}=\theta^{-2}-\frac{3}{2}+\frac{3\theta^2}4-\frac{\theta^4}{8}$$
Note that only the leading term $$\theta^{-2}$$ is correct.
Here is a plot to compare the variants. The difference with $$\frac{\cos^2\theta}{\sin \theta\tan \theta}$$ is plotted.
From the top:
• orange: $$\theta^{-2}-\frac12$$
• green: $$\theta^{-2}-1+\frac{\theta^2}4$$
• blue: $$\theta^{-2}-\dfrac76$$
• purple: $$\dfrac{(1-\frac12\theta^2)^3}{1-(1-\frac12\theta^2)^2}$$, obtained from $$\dfrac{\cos^3\theta}{1-\cos^2\theta}$$, yet another equivalent expression
• red: $$\theta^{-2}-\frac{3}{2}+\frac{3\theta^2}4-\frac{\theta^4}{8}$$
Note that all these approximations are local: they are valid near $$0$$, and the farther from $$0$$, the worst the approximation. With Taylor expansion, you can add more terms to get a better approximation on a larger (but still finite) interval. Notice the behavior of the purple solution on a larger interval is better: it's because it's not a polynomial approximation but a rational function.
Note also that the difference, which is plotted, is (locally) much lower than the leading term $$\theta^{-2}$$.
If the goal is to get a better approximation on a bounded interval, there are other methods, such as uniform approximation by a polynomial (on a bounded interval) using Chebyshev's equioscillation theorem, Padé approximants, splines... | {
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• I don't want to add a new answer that would be almost identical to this one, but I would try to emphasize the following point: $$\frac{\cos\theta}{\tan^2\theta}=\frac{\cos^2\theta}{\sin\theta\tan\theta}=\frac{\cos^3\theta}{\sin^2\theta}.$$ While the denominator is always approximated by $\theta^2$ when ignoring higher order terms, the numerator is obviously different in each case. – Ennar Apr 24 at 9:30
• @Ennar And the first is certainly the simplifiation the OP did. – Jean-Claude Arbaut Apr 24 at 9:32
• Yes. Hopefully looking at these expressions side by side will help them. – Ennar Apr 24 at 9:33
• Thanks you, this is very helpful. Apologies for the barrage of questions, but which of these result is "more correct"? (If that makes any sense). If there are multiple ways of obtaining the approximation, why has the mark scheme only specified one as correct? – Adam Cummings Apr 24 at 10:28
• @AdamCummings None of these approximations is very good (you can expect one correct term, and even that may fail), but if you want the "best", it's the one with the constant term closest to $-7/6$ (you could also try $\frac{\cos^3 x}{1-\cos^2 x}$, slightly better). If you want several terms, the good way is the Taylor formula. Now, regarding the mark scheme: as dumb as it seems, you had only to apply the approximation without simplification. – Jean-Claude Arbaut Apr 24 at 10:36 | {
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$P(X=x)=k$ for $x=4,5,6,7,8$, where $k$ is constant. inverse f ( x) = cos ( 2x + 5) $inverse\:f\left (x\right)=\sin\left (3x\right)$. The input to the floor function is any real number x and its output is the greatest integer less than or equal to x. \end{aligned} $$, Now, Variance of discrete uniform distribution X is,$$ \begin{aligned} V(X) &= E(X^2)-[E(X)]^2\\ &=100.67-[10]^2\\ &=100.67-100\\ &=0.67. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. To analyze our traffic, we use basic Google Analytics implementation with anonymized data. The probability mass function (pmf) of $X$ is, \begin{aligned} P(X=x) &=\frac{1}{5-0+1} \\ &= \frac{1}{6}; x=0,1,2,3,4,5. Given a discrete random variable, $$X$$, its probability distribution function, $$f(x)$$, is a function that allows us to calculate the probability that $$X=x$$. The probability mass function (pmf) of X is, \begin{aligned} P(X=x) &=\frac{1}{11-9+1} \\ &= \frac{1}{3}; x=9,10,11. If the function is one-to-one, there will be a unique inverse. f ( x) = x3. Step 4 - Click on âCalculateâ for discrete uniform distribution, Step 6 - Calculate cumulative probabilities. Find more Mathematics widgets in Wolfram|Alpha. Convolution calculator online. 1. Free online calculators for exponents, math, fractions, factoring, plane geometry, solid geometry, algebra, finance and trigonometry Calculator,Discrete Uniform distribution, Discrete uniform distribution examples, Discrete uniform distribution calculator, uniform distribution definition,mean,variance If the function is one-to-one, there will be a unique inverse. \end{aligned} $$, Mean of discrete uniform distribution X is,$$ \begin{aligned} E(X) &=\sum_{x=9}^{11}x \times P(X=x)\\ &= \sum_{x=9}^{11}x \times\frac{1}{3}\\ &=9\times \frac{1}{3}+10\times \frac{1}{3}+11\times \frac{1}{3}\\ &= \frac{9+10+11}{3}\\ &=\frac{30}{3}\\ &=10. Step 1: Enter the expression you want to evaluate. To configure the | {
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&=\frac{30}{3}\\ &=10. Step 1: Enter the expression you want to evaluate. To configure the filter for discrete time, set the Sample time property to a positive, nonzero value, or to -1 to inherit the sample time from an upstream block. y = x2 + x + 1 x. \end{aligned} $$, And variance of discrete uniform distribution Y is,$$ \begin{aligned} V(Y) &=V(20X)\\ &=20^2\times V(X)\\ &=20^2 \times 2.92\\ &=1168. \end{aligned} . 3. It is a tool used to convert the finite sequence of equally-spaced samples of any function into an equivalent-length sequence. Comparing the efficiently of different algorithms that solve the same problem. There is an extremely powerful tool in discrete mathematics used to manipulate sequences called the generating function. \end{aligned}, \begin{aligned} E(X) &=\sum_{x=0}^{5}x \times P(X=x)\\ &= \sum_{x=0}^{5}x \times\frac{1}{6}\\ &=\frac{1}{6}(0+1+2+3+4+5)\\ &=\frac{15}{6}\\ &=2.5. The identity function f(z)=z in the complex plane is illustrated above. \end{aligned}, The variance of discrete uniform distribution $X$ is, \begin{aligned} V(X) &=\frac{(8-4+1)^2-1}{12}\\ &=\frac{25-1}{12}\\ &= 2 \end{aligned}, c. The probability that $X$ is less than or equal to 6 is, \begin{aligned} P(X \leq 6) &=P(X=4) + P(X=5) + P(X=6)\\ &=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\\ &= \frac{3}{5}\\ &= 0.6 \end{aligned}. c. The mean of discrete uniform distribution $X$ is, \begin{aligned} E(X) &=\frac{1+6}{2}\\ &=\frac{7}{2}\\ &= 3.5 \end{aligned} Hope you like article on Discrete Uniform Distribution. Derivative numerical and analytical calculator A4:A11 in Figure 1) and R2 is the range consisting of the frequency values f(x) corresponding to the x values in R1 (e.g. Below are the few solved examples on Discrete Uniform Distribution with step by step guide on how to find probability and mean or variance of discrete uniform distribution. By millions of students & professionals equivalent-length sequence → 4: Status → about ID may look like: BD92F455... ( pmf ) of | {
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# The ideals $\langle y-x-1\rangle$ and $\langle x-2,y-3\rangle$ in $\mathbb C[x,y]$ are prime
The following is a quote from Wolfram MathWorld article about prime ideals.
A maximal ideal is always a prime ideal, but some prime ideals are not maximal. In the integers, $\{0\}$ is a prime ideal, as it is in any integral domain. Note that this is the exception to the statement that all prime ideals in the integers are generated by prime numbers. While this might seem silly to allow this case, in some rings the structure of the prime ideals, the Zariski topology, is more interesting. For instance, in polynomials in two variables with complex coefficients $\mathbb C[x,y]$, the ideals $$\langle 0\rangle \subset \langle y-x-1\rangle \subset \langle x-2,y-3\rangle$$ are all prime.
How can we prove that $\langle y-x-1\rangle$ and $\langle x-2,y-3\rangle$ are prime ideals in $\mathbb C[x,y]$?
Perhaps showing that quotient rings are $\mathbb C[x,y]/\langle y-x-1\rangle\cong \mathbb C[x]$ and $\mathbb C[x,y]/\langle x-2,y-3\rangle\cong\mathbb C$ would be way to go? (This would mean the quotient ring is an integral domain and therefore the ideal is prime.) | {
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• Yes to your questions at the very end. They are kernels of evaluation maps. – anon Apr 24 '16 at 5:22
• @user26857 Thanks for the edit. As you can see, I quoted the text from the link, so I written in in the same way. (As it was an exact quote.) What somewhat surprise me was the removal of (commutative-algebra) tag. According to the tag-excerpt this tag is for: "Questions about commutative rings, their ideals, and their modules." (But you are certainly more active in questions from this area as I am. So you probably know the norms of this math.SE subcommunity better.) – Martin Sleziak Apr 24 '16 at 6:42
• To see that $<y - x- 1>$ is a prime ideal, you can also note that, like every other degree $1$ polynomial, it is irreducible, and that irreducible elements are prime in, for example, UFDs (polynomial rings over $\mathbb{C}$ are certainly unique factorization domains). – Badam Baplan Aug 11 '18 at 15:50
• @BadamBaplan Maybe you could expand your comment a bit and post it as an answer. – Martin Sleziak Aug 11 '18 at 17:09
Let me try to expand anon's comment to an answer.
Let $R$ be a commutative ring. If $f(x)\in R[x]$ and $a\in x$ then there exist uniquely determined $q(x)\in R[x]$ and $b\in R$ such that $$f(x)=q(x)(x-a)+b.$$ Moreover, we have $b=f(a)$.
This follows from the fact that we can divide by monic polynomials in $R[x]$ for any commutative ring $R$. Using this observation we can also show that the map $f(x)\mapsto f(a)$ is a homomorphism from $R[x]$ to $R$ and the kernel is precisely the ideal $\langle x-a\rangle$.
Since we have $\mathbb C[x,y]=\mathbb C[x][y]$ we can use the above to get a homomorphism $$\varphi \colon \mathbb C[x,y] \to \mathbb C[x] \qquad \varphi \colon f(x,y) \mapsto f(x,x+1).$$ Kernel of this homomorphism is $\langle y-x-1 \rangle$. This shows that $$\mathbb C[x,y]/\langle y-x-1 \rangle \cong C[y].$$ So the quotient ring is an integrity domain and the ideal is a prime ideal. | {
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Similarly, we can get, using the above observation, the homomorphisms $$\psi_1 \colon \mathbb C[x][y] \to \mathbb C[x] \qquad \psi_1 \colon f(x,y) \mapsto f(x,3)$$ and $$\psi_2 \colon \mathbb C[x] \to \mathbb C \qquad \psi_2 \colon g(x) \mapsto g(2).$$ By composing these two homomorphisms we get $\psi=\psi_2\circ\psi_1$ from $\mathbb C[x,y]$ to $\mathbb C$ given by $f(x,y)=f(2,3)$.
Again we have that $\operatorname{Ker} \psi_2 = \langle x-2 \rangle$.
The polynomials in $\operatorname{Ker} \psi$ are precisely the polynomials such that $\psi_1(f)\in \operatorname{Ker} \psi_2$. So $f\in \operatorname{Ker} \psi$ if and only if $$f(x,y) = q(x,y)(y-3) + h(x)(x-2).$$ Every such polynomial belongs to $\langle x-2,y-3 \rangle$. And since both $x-2$ and $y-3$ belong to the kernel of $\psi$, we get $$\operatorname{Ker} \psi = \langle x-2,y-3 \rangle.$$
From this we get $\mathbb C[x,y]/\langle x-2,y-3 \rangle\cong\mathbb C$. Again, since the quotient ring is an integrity domain, we get that the ideal is prime. | {
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# Math nomenclature
I kinda feel ashamed about asking this, but could someone explain me what this means? $$\binom{k}{i}$$
That should indicate the number of nodes at a certain depth in a binomial heap..but I can't remember what that actually means and I wasn't successful with Google!
Thanks
==================================
so if i had $$\binom{0}{0}$$ would that be $$\dfrac{0!}{0!(0-0)!}$$ ...what would that be...zero?
-
Search for binomial coefficient and use \binom{k}{i} to get MathJax to format it properly for display as $\binom{k}{i}$. – Dilip Sarwate Feb 9 '13 at 15:47
No need to feel ashamed, Seb! That's what we're here for...ask away! Your question was good, and showed context, was clear, and it's hard to google for something when you don't know what to call it! – amWhy Feb 9 '13 at 16:07
thanks for the edits and for the patient :)! – TriRook Feb 9 '13 at 16:07
@Seb Once you know that $0! = 1$, you can answer your own question about $\binom00$. In general $\binom{k}{i}$ is only zero when $i < 0$ or when $i > k$. – Erick Wong Feb 9 '13 at 19:26
I assume you mean $$\dbinom{n}k$$ $\dbinom{n}k$ is a short hand for the number of ways in which you can choose $k$ objects from $n$ distinguishable objects. These number are called the binomial coefficients. Some textbooks and articles also denote this as $C(n,k)$. | {
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As an example, if we have three colored balls, $\color{red}{\text{red}}$, $\color{blue}{\text{blue}}$ and $\color{brown}{\text{brown}}$, then there are three ways of choosing two balls. \begin{matrix} \color{red}{\text{red}} & \color{blue}{\text{blue}}\\ \color{blue}{\text{blue}} & \color{brown}{\text{brown}}\\ \color{brown}{\text{brown}} & \color{red}{\text{red}} \end{matrix} Note that when we say we choose, we are not interested in the order in which these are picked i.e. $\color{red}{\text{red}} \,\, \color{blue}{\text{blue}}$ and $\color{blue}{\text{blue}} \,\, \color{red}{\text{red}}$ refer to the same choice of two balls. Hence, we have $\dbinom{3}2 = 3$.
Similarly, if we have $5$ colored balls, say $\color{red}{\text{red}}$, $\color{blue}{\text{blue}}$, $\color{brown}{\text{brown}}$, $\color{orange}{\text{orange}}$, and $\color{lightgreen}{\text{green}}$, there are $10$ ways of choosing $3$ balls. \begin{matrix} \color{red}{\text{red}} & \color{blue}{\text{blue}} & \color{brown}{\text{brown}}\\ \color{red}{\text{red}} & \color{blue}{\text{blue}} & \color{orange}{\text{orange}}\\ \color{red}{\text{red}} & \color{blue}{\text{blue}} & \color{lightgreen}{\text{green}}\\ \color{red}{\text{red}} & \color{orange}{\text{orange}} & \color{brown}{\text{brown}}\\ \color{red}{\text{red}} & \color{orange}{\text{orange}} & \color{lightgreen}{\text{green}}\\ \color{red}{\text{red}} & \color{brown}{\text{brown}} & \color{lightgreen}{\text{green}}\\ \color{blue}{\text{blue}} & \color{brown}{\text{brown}} & \color{orange}{\text{orange}}\\ \color{blue}{\text{blue}} & \color{brown}{\text{brown}} & \color{lightgreen}{\text{green}}\\ \color{blue}{\text{blue}} & \color{orange}{\text{orange}} & \color{lightgreen}{\text{green}}\\ \color{orange}{\text{orange}} & \color{brown}{\text{brown}} & \color{lightgreen}{\text{green}} \end{matrix}
In general, $$\dbinom{n}r = \dfrac{n!}{r!(n-r)!}$$ where $k! = k \times (k-1) \times (k-2) \times \cdots \times 2 \times 1$. | {
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The name binomial coefficient arises from binomial theorem. When we expand $(x+y)^n$, the coefficient of $x^k y^{n-k}$ is given by $\dbinom{n}k$ i.e. $$(x+y)^n = \sum_{k=0}^n \dbinom{n}k x^k y^{n-k}$$ There are a lot of wonderful properties these binomial coefficients satisfy and I highly recommend you to go through the wiki-page for these properties.
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Nice answer! Your examples would probably look better as matrixes than as aligns. – Rahul Feb 9 '13 at 16:36
@ℝⁿ. Thanks. Yes. These indeed look better. I have updated them. – user17762 Feb 9 '13 at 16:39
awesome answer thanks for taking time to go through it so well!! I have a question but i'll post it below cause i'm trying to put a formula in.. – TriRook Feb 9 '13 at 17:02
$$\binom{k}{i}\; \text{ is called a \color{blue}{\bf{binomial\;coefficient}}, which is read as "k choose i"}$$
"k choose i" comes from the fact that if gives you the number of ways to choose $i$ elements from a set of $k$ elements. The term "binomial coefficient" makes explicit its relationship to the binomial theorem. When we expand $(x+y)^n$, the coefficient of $x^k y^{n-k}$ is given by $\large\binom{n}k$ i.e. $$(x+y)^n = \sum_{k=0}^n \dbinom{n}k x^k y^{n-k}$$
You can compute your binomial coeffient by noting that $$\displaystyle \;\binom{k}{i} = \frac{k!}{i!(k-i)!} = \frac{k\cdot (k-1)\cdots (k - i + 1)}{i\cdot (i-1) \cdots 2 \cdot 1}$$
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I've not in recent times ever heard $\binom{k}{i}$ read as anything other than "$k$ choose $i$". (But maybe I don't get out enough ...) – Peter Smith Feb 9 '13 at 16:12
@PeterSmith Good point; "often" should have been "usually", which should probably be omitted altogether! Edited accordingly. – amWhy Feb 9 '13 at 16:16 | {
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$k \choose i$ is a binomial coefficient. It is the coefficient of $x^i$ in expanding $(1+x)^k$. It is also the number of $i$-element subsets of a $k$-element set. It can be computed via $\frac{k!}{i!(k-i)!}$ or with several recursive formulas. Values are often listed with help of Pascal's triangle.
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For the question in the edit: We have $0! = 1$, so $\binom{0}{0} = 1$ which is the first entry in Pascal's triangle.
A previous question on this site asked about the definition of zero factorial; there are several excellent answers there. As Zhen Lin points out in a comment, the best explanation of why $0! = 1$ depends on how you define the factorial.
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# Logical functions with no arguments
I am trying to understand how logical functions are evaluated if no arguments are passed to them.
{Or[], Nor[], And[], Nand[], Xor[], Xnor[]}
{False, True, True, False, False, True}
I tried Trace but it didn't offer many clues. Mathematica (12.2.0 on Win7-x64) evaluates these expressions without any messages.
• If you have a look in the Properties and relations section you will see that it zero-argument evaluates to False. reference.wolfram.com/language/ref/Or.html
– kcr
Feb 28, 2022 at 8:39
• Similar situation in Java developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…
– kcr
Feb 28, 2022 at 8:39
• Perhaps this is the way they are supposed to be constructed based on theoretical coding principles? Sorry for the triple comment...I am horrible
– kcr
Feb 28, 2022 at 8:40
• These are simply the trivial cases for these functions. And, for example, is True unless one of it's arguments is False. The zero-argument case is just an extension of that. Feb 28, 2022 at 11:05
• One sees similar behavior with Plus and Times (where it might be more obviously plausible). Symbols with the OneIdentity attribute will, when given an empty argument list, evaluate to their respective identity elements. Feb 28, 2022 at 15:57
I am pretty sure that Syed figured out the question by now, but for future visitors I thought it a good idea to write something.
1. Why does Or[] return False?
From the docs we read
It evaluates its arguments in order, giving True immediately if any of them are True, and False if they are all False.
which is another way of saying that it returns True if any or all of its arguments is true and False otherwise.
1. Why does Nor[] return True?
The easiest way to realize this -I think- is again to have a look at the documentation where we find that it is equivalent to Not[Or[]]. So, Not[False] is True.
1. Why does And[] return True? | {
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1. Why does And[] return True?
Well, again this can be seen directly from the docs and also from the comment by @Sjoerd Smit.
It evaluates to False if any of its arguments are false, and to True otherwise.
1. Why does Nand[] return False?
Similar to the explanation of Nor in this case. Nand[] is equivalent to Not[And[]] and hence Not[True] which is False.
1. Why does Xor[] return False?
Well, it has to evaluate all of its argument and yields True if an odd number of them is True and False otherwise. Since, we do not have an odd number of True statements, it returns False
1. Why does Xnor[] return True?
Xnor[] is equivalent to Not[Xor[]]
• Thanks for the write up.
– Syed
Jan 14 at 12:56
• @Syed if you want to add more to it, let me know and I will turn it into a wiki :-)
– bmf
Jan 14 at 12:58
• It seems that in your description of Nor True and False should be reversed. Jan 14 at 19:26
• @yarchik yes, of course you are right. thanks for spotting this :-)
– bmf
Jan 15 at 1:38
The answer of @bmf is perfect, but I would like to bring some arguments that allow us to better remember the results. It also shows a connection to other MA functions.
The Xor function can be seen as Sum modulo 2. In fact
Sum[1, {i, 0}]
(* 0 *)
which explains why Xor[] == False.
The And function can be regarded as the Product modulo 2. In fact
Product[1, {i, 0}]
(* 1 *)
which explains why And[] == True.
See also a comment of @DanielLichtblau above, which argues in terms of Plus and Times.
• (+1) that's a very neat description!!!
– bmf
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# Is the following generalization of Strong Law of Large Numbers valid?
According to SLLN, if $$X_1, X_2, \ldots$$ is an infinite sequence of i.i.d. random variables with expected value $$\mu$$ and $$S_n := \sum_{i=1}^n X_i/n$$ then $$S_n \to \mu$$ almost surely.
If the sequence is instead $$X_{1,1}, X_{2,1}, X_{2,2}, \ldots, X_{n,1}, \ldots,X_{n,n}, X_{n+1,1}, \ldots$$ whose terms are i.i.d. random variables with expected value $$\mu$$ and $$S_n := \sum_{i=1}^n X_{n,i}/n$$, can we still say $$S_n \to \mu$$ almost surely? If not are there any extra conditions that make this true?
I will give an example for why this is not trivial. Consider distribution of variables is Bernoulli with $$p=1/2$$ and the sample $$0, 1, 1, 0, 0, 0, 1, \ldots$$. That is $$(2n-1) \times 0$$ are followed by $$2n \times 1$$ and that is repeated for every $$n > 0$$. For this sequence $$S_n$$ converges to $$1/2$$ in the first case and the limit does not exist in the second. If the probability of all such examples is $$0$$ then convergence will be almost sure, but this is something that can probably be proven on a case by case with union bound inequality. I just wonder whether there is some general result that makes such analysis easier. | {
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First observe that in this setting, the sequence $$\left(S_n\right)_{n\geqslant 1}$$ is independent. For an independent sequence, in view of the Borel-Cantelli lemma, almost sure convergence and complete convergence are equivalent. Hence $$S_n\to \mu$$ almost surely if and only if for all positive $$\varepsilon$$, $$\tag{*} \sum_{n\geqslant 1}\mathbb P\left(\lvert S_n-\mu\rvert \gt\varepsilon\right)<+\infty.$$ Let $$(Y_i)_{i\geqslant 1}$$ be an i.i.d. sequence such that $$Y_1$$ has the same law as $$X_{1,1}$$. Then $$(*)$$ is equivalent to $$\forall \varepsilon>0, \sum_{n\geqslant 1}\mathbb P\left(\left\lvert \sum_{j=1}^n(Y_j-\mu)\right\rvert \gt n\varepsilon\right)<+\infty.$$ By Theorem 3 in this paper by Baum and Katz, this is equivalent to $$\mathbb E[Y_1^2]<\infty$$, hence we do need extra conditions.
Yes, the result holds equally well in both cases. More generally, Let $$(X_i:i=1,2,...)$$ be any infinite sequence of i.i.d. random variables with expected value $$μ$$, and let $$F: \Bbb Z_+\to\mathrm P(\Bbb Z_+)$$ such that $$|F(k)|\in\Bbb Z_+$$ with $$|F(k)|\to\infty$$ as $$k\to\infty$$. Then$$\frac1{|F(n)|}\sum_{i\in F(n)}X_i\;\to_{\text{a.s.}}\;\mu$$as $$n\to\infty$$. The proof is essentially the same as that for the simple average of an initial segment of the original sequence, because the i.i.d. property of any (finite or infinite) subsequence of the original sequence $$(X_i:i=1,2,...)$$ follows from that of the original sequence. | {
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subset of binary space countable?
I know that the set of all binary sequences is uncountable. Now consider the subset of this set, that whenever a digit is $1$, its next digit must be $0$. Is this set countable?
I think it is not because it is like "half" of the set of binary sequences. By that I mean if I just delete the sequences with a $1$ followed by a $1$, then I'm left with the required sequence. But I'm not sure how to show this clearly. Can someone think of a bijection that would prove this? Or am I wrong? Please help.
Thanks
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If you insist to have a bijection, here is one. Your set $S$ is the set of all binary infinite words containing no factor $11$. Any such word can be uniquely written as an infinite word on the alphabet $\{0, 10\}$, that is, $S = \{0, 10\}^\omega$. For instance, $$0100100010010(100)^\omega = \color{blue}{0}\color{red}{10}\color{blue}{0}\color{red}{10}\color{blue}{0}\color{blue}{0}\color{red}{10}0\color{red}{10}(\color{red}{10}\color{blue}{0})^\omega$$
It follows that the map from $\{0,1\}^\omega \to S$ consisting to replace every $1$ by $\color{red}{10}$ is a bijection. No need of the Cantor-Bernstein-Schroeder theorem.
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You can explicitly write an injective function from the set of all binary sequences to the set you are considering: just map every sequence to one with $0$ inserted in every other position.
So no, this is definitely not countable. | {
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So no, this is definitely not countable.
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But how does an injective function suffice here? Don't you need a bijective function to show the same cardinality? – verticese Feb 25 at 8:02
@verticese: it shows that it has at least the same cardinality as the bijective sequences (that is, the continuum), which is all you need to show that it is uncountable. But it is not hard to get a bijection if you want: of course you also have an injection going the other way: the identity. This gives you a bijection by Cantor-Bernstein-Schroeder. In this case, you can probably define a bijection explictly without much difficulty, although there would be no nice and concise formula, I think, but rather a procedure. – tomasz Feb 25 at 8:05
As @tomasz notes: If $A\subseteq B$ then $|A|\le|B|$. If there's an injection $B\to A$, then $|B|\le|A|$, so by Cantor-Schroeder-Bernstein there's a bijection $A\to B$. – BrianO Feb 25 at 8:12
Your set has an uncountable subset: consider the infinite sequences of $a$ and $b$ where $a=10$ and $b=00$, the set of these sequences has the same cardinality of the set of sequences of $0$ and $1$ (just map $a \mapsto 0$ and $b \mapsto 1$), and it is is clearly a subset of your set.
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This is the same as what I suggested in my answer. – tomasz Feb 25 at 9:34
@tomasz I only see that now after reading your comment. I would say this answer is another viewpoint on the same thing you suggested, and sometimes an alternate viewpoint helps (as in how this one helped me understand your answer better). – Todd Wilcox Feb 25 at 16:53
Here's another (possibly) interesting way of proving uncountability, since you tagged this as a topology question. You can inject your set into $[0,1]$ under the evaluation map
$$(x_i) \mapsto \sum_{i = 1}^\infty \frac{x_i}{2^i}$$ | {
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$$(x_i) \mapsto \sum_{i = 1}^\infty \frac{x_i}{2^i}$$
This is injective on your set (but not the set of all binary sequences, since $0.0111\dots = 0.100\dots$). We shall show the image of this map is a perfect set, and hence uncountable (see Rudin, or some other source on metric spaces). Surely no point in the set is isolated, because if we have an expansion $x = 0.a_0 a_1 \dots a_2$, and if this expansion does not eventually end in all zeroes, we may truncate the expansion to find another number in the image which is as close to this number as possible. If the number does end in all zeroes, you may add a one to an arbitrary position to get another number as closed as possible. Now suppose $\lim x_i = x$, where each $x_i$ is in the image of the evaluation. We must have $0 < x < 1$, for if $x = 1$, the $x_i$ must eventually be of the form $0.11\dots$, which is impossible. Write $x = 0.a_0 a_1 \dots$. Suppose $a_k = 1$. If we force $|x_i - x| < 1/2^{k+2}$, then the first $k+2$ digits of $x_i$ must be the same as the first $k+1$ digits of $x$, hence $a_{k+1} = 0$. Thus the image is perfect, and the set is uncountable.
Intuition tells me this set is probably fractal, similar to the Cantor set, but I can't visualize it very well.
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You can inject the set of all sequences of integers into this your set (call it $A$): $$(n_i)_{i<\omega} \mapsto (1 + 0^{n_0+1} + 1 + 0^{n_1+1} + \dotsc)\colon \omega^\omega\to A.$$ where the value is meant as an infinite regular expression. That is, the $i$-th $1$ is followed by $n_i+1$ many $0$s. (The "$+ 1$" guarantees that the value under this function is a member of A.)
Thus, $A$ has cardinality $\lvert A\rvert \ge |\omega^\omega| = \lvert\Bbb R\rvert$. But $A\subseteq 2^\omega$, so $\lvert A\rvert \le \lvert 2^\omega\rvert = \lvert\Bbb R\rvert$. By the Cantor-Schroeder-Bernstein Theorem, $\lvert A\rvert = \lvert\Bbb R\rvert$.
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Notation: $\chi_A (x)=1$ if $x\in A$, and $\chi_A (x)=0$ if $x\not \in A.$ | {
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-
Notation: $\chi_A (x)=1$ if $x\in A$, and $\chi_A (x)=0$ if $x\not \in A.$
Let $E=\{2 n:n\in N\}.$ For $S\subset E \;$ let $S^*=\{n :2 n\in S\}$ and let $f(S)=(\chi_{S^*} (n))_{n\in N}.$
Then $f$ is a bijection from the power-set $P(E)$ to the set $B$ of all binary sequences.
For $S\subset E$ let $g(S)=(\chi_S (n))_{n\in N}.$ Then $g$ is an injection from $P(E)$ into the set $G$ of binary sequences $(b_n)_{n\in N}$ that satisfy $\forall n\in N\;(b_n=1\implies b_{n+1}=0)\}.$
Thus $g f^{-1}$ is an injection from $B$ into $G.$ So $G$ is uncountable because it has the uncountable subset $\{g f^{-1}(x): x\in B\}.$
The Cantor-Schroeder-Bernstein Theorem implies that for any $B,G$ with $G\subset B,$ there is a bijection $h:B\to G$ if there is an injection $i:B\to G.$
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# Variance of number of heads in three flips of a coin selected from a set of fair and unfair coins
We have $10$ coins, $2$ are two-tailed, $2$ are two-headed, the other $6$ are fair ones. We (randomly) pick a coin and we flip it $3$ times. Find the variance of the number of gotten heads.
My attempt:
$X$ - number of heads that we got
$\mathbb{P}\left(X=0\right)=\frac{2}{10}\cdot1\cdot1\cdot1 + \frac{6}{10}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}$ - we picked two-tailed coin or fair one
$\mathbb{P}\left(X=1\right)=\frac{6}{10}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot 3$ - we picked fair one and we have 3 possibilites: HHT, HTH, THH
$\mathbb{P}\left(X=2\right)=\frac{6}{10}\cdot{3\choose2}\cdot\left(\frac{1}{2}\right)^2\cdot\left(\frac{1}{2}\right)^1$ - again we picked fair one and we have to have 2 successes in 3 tries
$\mathbb{P}\left(X=3\right)=\frac{2}{10}\cdot1\cdot1\cdot1+\frac{6}{10}\cdot\left(\frac{1}{2}\right)^3$ - we can pick two-headed or fair one coin
And the rest is simple,
$\text{Var}X=\sum_{i=0}^3i^2\cdot\mathbb{P}\left(X=i\right)-\left(\sum_{i=0}^3i\cdot\mathbb{P}\left(X=i\right)\right)^2$
Is my solution correct?
• The information is symmetric, so the expressions for the probability of 1 head or 2 heads should be the same. – Paul Feb 2 '17 at 18:42
• Yes, your approach is correct. $\checkmark$ – callculus Feb 2 '17 at 18:47
• @Paul well, yea, I should have decided whether I use probability mass function or just standard combinatorics. But the result is the same – SantaXL Feb 2 '17 at 18:50
That's one approach. It is good. Here's another,
We know $X\mid C \sim \mathcal{Bin}(3, C)$ and $\mathsf P(C=c)=\tfrac 15\mathbf 1_{c=0}+\tfrac 35\mathbf 1_{c=1/2}+\tfrac 15\mathbf 1_{c=1}$
So $\mathsf E(C)= \tfrac 1{2}$ and $\mathsf {Var}(C)=\tfrac 1{10}$ and $\mathsf E(C^2)=\frac 7{20}$
By the Law of Total Probability: $\mathsf E(X) \\ = \mathsf E(\mathsf E(X\mid C)) \\ = \mathsf E(3C) \\ = \frac 32$ | {
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Then by the Law of Total Variance: $\quad\mathsf {Var}(X) \\ = \mathsf{Var}(\mathsf E(X\mid C))+\mathsf E(\mathsf{Var}(X\mid C)) \\ = \mathsf {Var}(3C)+\mathsf E(3C(1-C)) \\ = \frac 9{10}+\frac 32-\frac {21}{20} \\ = \frac {27}{20}$
• Hmm, your result is different from mine. In my approach: $\mathbb{P}(X=0)=\mathbb{P}(X=3)=\frac{9}{40}, \ \mathbb{P}(X=1)=\mathbb{P}(X=2)=\frac{11}{40} \\ EX=1\cdot\frac{9}{40}+2\cdot\frac{9}{40}+3\cdot\frac{11}{40}=\frac{9+18+33}{40}=\frac{60}{40}=\frac{3}{2} \\EX^2=1^2\cdot\frac{9}{40}+2^2\cdot\frac{9}{40}+3^2\cdot\frac{11}{40}=\frac{9+36+99}{40}=\frac{144}{40}=\frac{18}{5} \\ VarX=EX^2-E^2X=\frac{18}{5}-\frac{9}{4}=\frac{27}{20}$ – SantaXL Feb 3 '17 at 13:52
• Where is the mistake? – SantaXL Feb 3 '17 at 13:53
• In mine. @SantaXL corrected. – Graham Kemp Feb 3 '17 at 14:49
• ok, thanks for showing me this approach, I wasn't aware such Law exists. – SantaXL Feb 3 '17 at 15:22 | {
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# Proving $24\mid5^{2n}-1$ using modular arithmetic
So, in general I am aware of how to use modular arithmetic to prove a divisibility. But I have the following problem:
Prove that $24\mid5^{2n}-1$ for all $n\in\mathbb Z$.
I know that theoretically, I could show 23 different cases that the expression is congruent to $0\bmod24$, but that seems like it might be excessive to me.
Is there a faster way to show this?
• $5^{2n}=(5^2)^n=25^n$ Now, note that $25^n=(24+1)^n$ and use something that should be familiar to you. Dec 11, 2017 at 3:01
You want to prove what $5^{2n}$ is congruent to in modulo 24
Notice that $5^{2n} = (5^2)^n$
$5^2=25 \equiv 1 \pmod {24}$
So $5^{2n} \equiv 1^n \equiv 1 \pmod {24}$
• Edited, and thanks Dec 11, 2017 at 3:05
• Makes sense, thank you! Dec 11, 2017 at 3:19
You could use $5^{2n}-1=(5^2-1)(5^{2n-2}+5^{2n-4}+\ldots +1)$ where you are summing the geometric series.
Use induction. It is true for $n=1$. Assuming it is true for $n$, show it is true for $n+1$: $$5^{2(n+1)}-1=25(5^{2n}-1)+24 = 0 (mod \ 24).$$
1)
Prove $8|5^{2n} - 1$ and $3|5^{2n} - 1$
a) $5^{2n} - 1 = (5^n -1)(5^n + 1)$. $5^n \pm 1 \equiv 1^n\pm 1 \equiv 0\mod 2$ so $2|5^2 - 1$ and $2|5^2 + 1$ and so $4|5^{2n}-1$. If $5^n - 1 = 2k$ then $5^{2n} + 1= 2k + 2 = 2(k+1)$. Either $k$ is even or $k+1$ is even so either $2k$ is even and $2(k+1)$ is divisible by $4$, or $2k$ is divisible by $4$ and $2(k+1)$ is even. Either way. $2k*2(k+1)= 5^{2n} -1$ is divisible by $8$.
b)$5^{2n}-1\equiv (-1)^{2n}-1 \equiv 1^n - 1 \equiv 0 \mod 3$.
So $3|5^{2n}-1$.
2)
Or notice $5^{2n} -1 = 25^n -1 \equiv 1^n -1 \equiv 0 \mod 24$.
(I will admit, I did not see the obvious right away.) | {
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# Does $i^4$ equal $1?$
I can't seem to find a solution to this for the life of me. My mathematics teacher didn't know either.
Edit: I asked the teacher that usually teaches my course today, and she said it was incredible that the other teacher didn't know.
My logic goes as follows:
any real number: $x$ to the fourth power is equal to $(x^2)^2$. Using this logic, $i^4$ would be equal to $(i^2)^2$. This would result in $(-1)^2$, and $(-1)^2 = 1$.
Obviously, this logic can be applied to any real numbers, but does it also apply to complex numbers?
• Yes, $i^4=1$. But what else could it equal ??
– user65203
Jan 7 '17 at 15:00
• To check if $i^4$ is real or not, why not try conjugating it and see what happens? Jan 7 '17 at 15:11
• Yes, @J.M.isn'tamathematician, but before you proceed to prove something, I'm going to get some paper and work it out for myself Jan 7 '17 at 15:21
• "I'm going to get some paper and work it out for myself" - that was my intention behind asking you those questions. Jan 7 '17 at 15:24
• If you're working within the naive approach to complex numbers often used when they're first introduced the students, "Suppose $i^2=-1$ but that other than that all the usual rules of arithmetic apply", then basically the bottom line is that this is one of the "usual rules" which is simply assumed to apply. Specifically, we assume that complex numbers are associative, that is that for any $a, b, c$, $a(bc)=(ab)c$. You might try showing in detail why this assumption implies $i^4=1$. Jan 12 '17 at 21:03
Yes. The powers of $i$ are cyclic, repeating themselves ever time the exponent increases by 4: $$i^0 = 1$$ $$i^1=i$$ $$i^2 = -1$$ $$i^3 = -i$$ $$i^4 = 1$$ $$i^5 = i$$ $$i^6 = -1$$ $$i^7 = -i$$ $$i^8 = 1$$ etc.
Your reasoning is excellent, and you should feel good about the fact that you figured this out on your own. The fact that your math teacher didn't know this is, in my professional opinion as a mathematics educator, a disgrace. | {
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Edited to add: As Kamil Maciorowski notes in the comments, the pattern persists for negative exponents, as well. Specifically, $$i^{-1}= \frac{1}{i} = -i$$ If $\frac{1}{i}=-i$ seems odd, notice that $i(-i) = -i^2 = -(-1) = 1$, so $i$ and $-i$ are multiplicative inverses; therefore $i^{-1} = -i$. Once you know that, you can extend the pattern: $$i^{-1} = -i$$ $$i^{-2} = -1$$ $$i^{-3} = i$$ $$i^{-4} = 1$$ and so on.
Second update: The OP asks for some additional discussion of the property $\left( x^a \right)^b = x^{ab}$, so here is some background on that:
First, if $a$ and $b$ are natural numbers, then exponentiation is most naturally understood in terms of repeated multiplication. In this context, $x^a$ means $(x\cdot x\cdot \cdots \cdot x)$ (with $a$ factors of $x$ appearing), and $\left( x^a \right)^b$ means $(x\cdot x\cdot \cdots \cdot x)\cdot(x\cdot x\cdot \cdots \cdot x)\cdot \cdots \cdot (x\cdot x\cdot \cdots \cdot x)$, with $b$ sets of parentheses, each containing $a$ factors of $x$. Since multiplication is associative, we can drop the parentheses and recognize this as a product of $ab$ factors of $x$, i.e. $x^{ab}$.
Note that this reasoning works for any $x$, whether it is positive, negative, or complex. It even applies in settings were multiplication is noncommutative, like matrix multiplication or quaternions. All we need is that multiplication is associative, and that $a$ and $b$ be natural numbers.
Once we have established that $\left( x^a \right)^b = x^{ab}$ for natural numbers $a,b$ we can extend the logic to integer exponents. If $a$ is a positive number, and if $x$ has a multiplicative inverse, then we define $x^{-a}$ to mean the same thing as $\left(\frac1x\right)^a$, or (equivalently) as $\frac1{x^a}$. With this convention in place, it is straightforward to verify that for any combination of signs for $a,b$, the formula $\left(x^a\right)^b = x^{ab}$ holds. | {
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