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 {}. 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 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 . 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 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 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 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 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 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 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 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 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 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 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, 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 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; 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 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 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 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 $\{1,1\}$ 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 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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# 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 .

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 .

## 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

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)


# 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


# 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.

-
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.

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 $\langle f(x_n):n\in\mathbb{N}\rangle \to f(x)$.

-
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.


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

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


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 $(1)$.

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 $a=b=c=\sqrt{\dfrac{2}{3}}$


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. 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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. 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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 form of a vector with initial point and terminal point in space, Exercises. As we will soon see, the area of a parallelogram formed from two vectors. can be seen as a geometric representation of the cross product. 2-Dimensional like a carpet or an area rug inside the polygon units inside the.... 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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 $$P= {2\over 3}$$

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

P(A winning) = 0.5/(1-0.25)


# 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.

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 $10^k-1$.

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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.

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 $T=k$?

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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

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 $n$, $$x_n=a_1j^n+a_2j^{2n}+a_3.$$

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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