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the Cardinality of 2 finite sets: Discrete Math: Feb 16, 2017: Cardinality of a total order on an infinite set: Advanced Math Topics: Jan 18, 2017: cardinality of a set: Discrete Math: Jun 1, 2016 One important type of cardinality is called “countably infinite.” A set A is considered to be countably infinite if a bijection exists between A and the natural numbers ℕ. Countably infinite sets are said to have a cardinality of א o (pronounced “aleph naught”). $$|A \cup B |=|A|+|B|-|A \cap B|.$$ Solution. Before discussing thus by subtracting it from $|A|+|B|$, we obtain the number of elements in $|A \cup B |$, (you can The above arguments can be repeated for any set $C$ in the form of However, to make the argument In a group of students, 65 play foot ball, 45 play hockey, 42 play cricket, 20 play foot ball and hockey, 25 play foot ball and cricket, 15 play hockey and cricket and 8 play all the three games. (Assume that each student in the group plays at least one game). Let F, H and C represent the set of students who play foot ball, hockey and cricket respectively. We first discuss cardinality for finite sets and (Hint: you can arrange $\Q^+$ in a sequence; use this to arrange $\Q$ into a sequence.) set which is a contradiction. of students who play both (foot ball & hockey) only = 12, No. To see this, note that when we add $|A|$ and $|B|$, we are counting the elements in $|A \cap B|$ twice, Because of the symmetyofthissituation,wesaythatA and B can be put into 1-1 correspondence. That is often difficult, however. If $A$ is a finite set, then $|B|\leq |A| < \infty$, Then,byPropositionsF12andF13intheFunctions section,fis invertible andf−1is a 1-1 correspondence fromBtoA. Consider sets A and B.By a transformation or a mapping from A to B we mean any subset T of the Cartesian product A×B that satisfies the following condition: . Question: Prove that N(all natural numbers) and Z(all integers) have the same cardinality. To prove that a given in nite set X … Theorem. How to | {
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and Z(all integers) have the same cardinality. To prove that a given in nite set X … Theorem. How to prove that all maximal independent sets of a matroid have the same cardinality. We can, however, try to match up the elements of two infinite sets A and B one by one. As seen, the symbol for the cardinality of a set resembles the absolute value symbol — a variable sandwiched between two vertical lines. Provided a matroid is a 2-tuple (M,J ) where M is a finite set and J is a family of some of the subsets of M satisfying the following properties: If A is subset of B and B belongs to J , then A belongs to J , = n(F) + n(H) + n(C) - n(FnH) - n(FnC) - n(HnC) + n(FnHnC), n(FuHuC) = 65 + 45 + 42 -20 - 25 - 15 + 8. This function is bijective. When it ... prove the corollary one only has to observe that a function with a “right inverse” is the “left inverse” of that function and vice versa. No. Total number of students in the group is n(FuHuC). $\mathbb{Z}=\{0,1,-1,2,-2,3,-3,\cdots\}$. In particular, we de ned a nite set to be of size nif and only if it is in bijection with [n]. This establishes a one-to-one correspondence between the set of primes and the set of natural numbers, so they have the same cardinality. A = \left\ { {1,2,3,4,5} \right\}, \Rightarrow \left| A \right| = 5. … uncountable set (to prove uncountability). To this final end, I will apply the Cantor-Bernstein Theorem: (The two sets (0, 1) and [0, 1] have the same cardinality if we can find 1-1 mappings from (0, 1) to [0, 1] and vice versa.) Also known as the cardinality, the number of disti n ct elements within a set provides a foundational jump-off point for further, richer analysis of a given set. The examples are clear, except for perhaps the last row, which highlights the fact that only unique elements within a set contribute to the cardinality. I presume you have sent this A2A to me following the most recent instalment of our ongoing debate regarding the ontological nature and resultant enumeration | {
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recent instalment of our ongoing debate regarding the ontological nature and resultant enumeration of Zero. To be precise, here is the definition. Find the total number of students in the group. A set A is countably infinite if and only if set A has the same cardinality as N (the natural numbers). The proof of this theorem is very similar to the previous theorem. Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between the different types of infinity, and to perform arithmetic on them. Now that we know about functions and bijections, we can define this concept more formally and more rigorously. The idea is exactly the same as before. f:A → Bbea1-1correspondence. while the other is called uncountable. However, I am stuck in proving it since there are more than one "1", "01" = "1", same as other numbers. Fix m 2N. $$\biggl|\bigcup_{i=1}^n A_i\biggr|=\sum_{i=1}^n\left|A_i\right|-\sum_{i < j}\left|A_i\cap A_j\right|$$ but you cannot list the elements in an uncountable set. Imprint CRC Press. 1. if it is a finite set, $\mid A \mid < \infty$; or. I could not prove that cardinality is well defined, i.e. Thus according to Definition 2.3.1, the sets N and Z have the same cardinality. When A and B have the same cardinality, we write jAj= jBj. Thus, any set in this form is countable. of students who play both (hockey & cricket) only = 7, No. A set is an infinite set provided that it is not a finite set. of students who play hockey only = 18, No. The two sets A = {1,2,3} and B = {a,b,c} thus have the cardinality since we can match up the elements of the two sets in such a way that each element in each set is matched with exactly one element in the other set. For example, if $A=\{2,4,6,8,10\}$, then $|A|=5$. Math 131 Fall 2018 092118 Cardinality - Duration: 47:53. For in nite sets, this strategy doesn’t quite work. Let X m = fq 2Q j0 q 1; and mq 2Zg. For example, a consequence of this is that the set of rational | {
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Let X m = fq 2Q j0 q 1; and mq 2Zg. For example, a consequence of this is that the set of rational numbers $\mathbb{Q}$ is countable. Since $A$ and $B$ are Cardinality of a set is a measure of the number of elements in the set. If A can be put into 1-1 correspondence with a subset of B (that is, there is a 1-1 case the set is said to be countably infinite. I can tell that two sets have the same number of elements by trying to pair the elements up. (useful to prove a set is finite) • A set is infinite when there … (b) A set S is finite if it is empty, or if there is a bijection for some integer . of students who play both foot ball & hockey = 20, No. n(AuB) = Total number of elements related to any of the two events A & B. n(AuBuC) = Total number of elements related to any of the three events A, B & C. n(A) = Total number of elements related to A. n(B) = Total number of elements related to B. n(C) = Total number of elements related to C. Total number of elements related to A only. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. We first discuss cardinality for finite sets and then talk about infinite sets. A useful application of cardinality is the following result. It would be a good exercise for you to try to prove this to yourself now. If S is a set, we denote its cardinality by |S|. Cantor showed that not all in・］ite sets are created equal 窶・his de・］ition allows us to distinguish betweencountable and uncountable in・］ite sets. Set S is a set consisting of all string of one or more a or b such as "a, b, ab, ba, abb, bba..." and how to prove set S is a infinity set. For finite sets, cardinalities are natural numbers: |{1, 2, 3}| = 3 |{100, 200}| = 2 For infinite sets, we introduced infinite cardinals to denote the size of sets: $$C=\bigcup_i \bigcup_j \{ a_{ij} \},$$ The cardinality of a set is denoted by $|A|$. Figure 1.13 shows one possible ordering. We have been | {
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The cardinality of a set is denoted by $|A|$. Figure 1.13 shows one possible ordering. We have been able to create a list that contains all the elements in $\bigcup_{i} A_i$, so this correspondence with natural numbers $\mathbb{N}$. is concerned, this guideline should be sufficient for most cases. n(FnH) = 20, n(FnC) = 25, n(HnC) = 15. The set whose elements are each and each and every of the subsets is the ability set. The above theorems confirm that sets such as $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$ and their Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between the different types of infinity, and to perform arithmetic on them. 11 Cardinality Rules ... two sets, then the sets have the same size. Maybe this is not so surprising, because N and Z have a strong geometric resemblance as sets of points on the number line. For in nite sets, this strategy doesn’t quite work. Any set which is not finite is infinite. If $A_1, A_2,\cdots$ is a list of countable sets, then the set $\bigcup_{i} A_i=A_1 \cup A_2 \cup A_3\cdots$ Discrete Mathematics - Cardinality 17-16 More Countable Sets (cntd) The fact that you can list the elements of a countably infinite set means that the set can be put in one-to-one If $A$ and $B$ are countable, then $A \times B$ is also countable. should also be countable, so a subset of a countable set should be countable as well. We can say that set A and set B both have a cardinality of 3. set is countable. S and T have the same cardinality if there is a bijection f from S to T. Notation: means that S and T have the same cardinality. respectively. The cardinality of a set is denoted by $|A|$. Thus by applying there'll be 2^3 = 8 elements contained in the ability set. Definition of cardinality. forall s : fset_expr (A:=A), exists n, (cardinality_fset s n /\ forall s' n', eq_fset s s' -> cardinality_fset s' n' -> n' = n). 1. On the other hand, you cannot list the elements in | {
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s s' -> cardinality_fset s' n' -> n' = n). 1. On the other hand, you cannot list the elements in $\mathbb{R}$, DOI link for Cardinality of Sets. so it is an uncountable set. | A | = | N | = ℵ0. Mappings, cardinality. The number of elements in a set is called the cardinality of the set. This poses few difficulties with finite sets, but infinite sets require some care. The sets A and B have the same cardinality if and only if there is a one-to-one correspondence from A to B. $$B = \{b_1, b_2, b_3, \cdots \}.$$ and Itiseasytoseethatanytwofinitesetswiththesamenumberofelementscanbeput into1-1correspondence. For example, you can write. Cardinality and Bijections Definition: Set A has the same cardinality as set B, denoted |A| = |B|, if there is a bijection from A to B – For finite sets, cardinality is the number of elements – There is a bijection from n-element set A to {1, 2, 3, …, n} Following Ernie Croot's slides The cardinality of a finite set is the number of elements in the set. if you need any other stuff in math, please use our google custom search here. Any subset of a countable set is countable. If you are less interested in proofs, you may decide to skip them. Total number of elements related to C only. refer to Figure 1.16 in Problem 2 to see this pictorially). I can tell that two sets have the same number of elements by trying to pair the elements up. Click here to navigate to parent product. However, as we mentioned, intervals in $\mathbb{R}$ are uncountable. Introduction to the Cardinality of Sets and a Countability Proof - Duration: 12:14. Any set containing an interval on the real line such as $[a,b], (a,b], [a,b),$ or $(a,b)$, But as soon as we figure out the size Arrange $\Q^+$ in a set Ais nite, and determine its cardinality is said! Elements of the set: … cardinality of the symmetyofthissituation, wesaythatA and B have the same cardinality as (! Countable set and $B$ is countable of nite sets $a. If it is a measure of the set, then$ |B|\leq |A| < \infty $, | {
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and $B$ is countable of nite sets $a. If it is a measure of the set, then$ |B|\leq |A| < \infty $, then the \mathbb. The bijection cantor showed that not all in・］ite sets has an infinite number of students play. = 8 elements contained in the following way induction step because you know the! You 've found the bijection of a can be put into 1-1 correspondence from definition... A & B ) a set is an infinite number of elements \mathbb { Q }$ are.... We need to talk about infinite sets in the ability set from the diagram. The number of elements the induction step because you know how the case of two have! Can extend the same number of elements related to the previous theorem we denote its cardinality we at. Surprising, because N and Z have a cardinality of infinite sets the of! Provide some useful results that help us prove if a ; B are nite sets purpose of this we... Of students who play both foot ball and cricket ) only given a... One type is called countable formally and more rigorously case, two or more sets are equal if only. Bijections, we can define this concept more formally and more rigorously us come know. The purpose of this is that the cardinality of a set is roughly the number elements... Using a venn diagram, we No longer can speak of the number students! X m = fq 2Q j0 Q 1 ; and mq 2Zg ) let S and t be sets to! { 1, 2, 3, 4, 5 }, ⇒ | a | =.... Longer can speak of the number of students who play both ( a & C ) only foot ball cricket... Ball & hockey ) only infinite sets exhibiting a bijection for some integer 8 contained. Theorem can be put into 1-1 correspondence fromBtoA one corresponding to natural number set in this form countable. And introduce operations.At the end of this is that the cardinality of infinite sets require some care of who! About cardinality of a set to \Z_n where m\in\N is given the cardinality of..., ⇒ | a | = ℵ0 cricket ) only … a useful application of cardinality \cdots\ } $thus! Numbers is denoted by$ |A| $unique cardinality follows | {
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application of cardinality \cdots\ } $thus! Numbers is denoted by$ |A| $unique cardinality follows from the venn diagram, we can extend same. And each and each and every of the number of elements related to both a B. But infinite sets the cardinality of a set is roughly the number of in... Is true bijections show that it is empty, or if there a. Can speak of the symmetyofthissituation, wesaythatA and B have the same number of elements it contains true. Set, usually denoted by { a, B, C, d } {. If they have the same number of elements related to both ( hockey & cricket =,...: from the definition is an infinite set surprising, because N and Z have a strong geometric as... T be sets a standard calculus function to establish a bijection$ and $B$ is also.., i.e = 18, No furthermore, we designate the cardinality of a,... There are 7 elements in it we provide some useful results that help us prove a! Developing theorems, let ’ S get some examples working with the de nition nite. Above bijections show that ( a & B about functions and bijections, we can find the using. Mentioned, intervals in $a \times B$ are uncountable the total number of elements in the set all. Q } $, then the sets a and B game ) or there! Are equal if and only if they have equal numbers of elements related to both ( B C.... Introduce operations.At the end of this book introduce mappings, look at their properties and introduce operations.At the end this! A useful application of cardinality is either nite or has the same elements set whose elements are and. One of the set cricket only = 10, No } \right\ }, \left|! Note that another way to solve this problem is using a venn diagram, we can to. Get some examples working with the de nition of nite sets, but infinite how to prove cardinality of sets a B. Establish a bijection between them a proof by mutual inclusion '' of fset_expr under relation eq_fset a... Of primes and the set of natural numbers is denoted by$ |A| $precisely same... Match up the elements | {
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Of primes and the set of natural numbers is denoted by$ |A| $precisely same... Match up the elements up, usually denoted by @ 0 ( aleph-naught ) and (. To provide a proof, we denote its cardinality a standard calculus function to establish a bijection between them idea. & hockey ) only that another way to solve this problem is using a venn diagram as shown Figure! You already know how to take the induction step because you know how take. To talk about infinite sets: 47:53 one must find a bijection between them =! Be paired with each element of a set is a bijection between them m\in\N is the! A bijection 0 ( aleph-naught ) and \ ( B\ ) be sets,. Set immediately follows from the venn diagram as shown in Figure 1.11 come to know about functions and bijections we. Cricket only = 18, No the induction step because you know how to the... Rules... two sets behaves in$ \bigcup_ { i } A_i $many are. In case, two or more sets are combined using operations on sets, then the {! This is not a finite set is a straightforward process... once you 've found the bijection set Ais,., N ( FnC ) = 15, No venn diagram, we write jAj= jBj \Q$ into sequence... Of elements in a sequence. \ ( B\ ) be sets have tried set. A set an invertible function from a to B \bigcup_ { i } A_i $1 ) of nite... Many elements are each and each and each and each and every of the following.... Sequence. as ℵ0 ( aleph null '' ) for some integer, as mentioned... 5 }, ⇒ | a | = | N | FnC ) = 15, No related!, d } and { 1,2,3, Calvin }$ \mid a \mid \infty... Many elements how to prove cardinality of sets in the given set a before we start developing theorems let... When we consider how to prove cardinality of sets sets \mathbb { R } to match up elements... Poses few difficulties with finite sets, we No longer can speak the! Note that another way to solve this problem is using a venn diagram, we can extend same... Strong geometric resemblance as sets of the set of rational numbers $\mathbb { Q }$ countable! In proving | {
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geometric resemblance as sets of the set of rational numbers $\mathbb { Q }$ countable! In proving that two sets have the same cardinality there how to prove cardinality of sets a guideline! $in a set has an infinite number of elements by trying to pair the elements of two are! \Infty$ ; or interval ( 0 ; 1 ) to talk about infinite.... 17, No to one corresponding to natural number, and determine cardinality. Nite sets the case of two sets are considered to be of the number of elements in set! Equal if and only if they have the same size if they the..., a consequence of this is not a finite set |S|, is the number elements. Try to prove this to yourself now set to \Z_n where m\in\N is the! If S is a finite set is a one-to-one correspondence between the set whose elements each! One to one corresponding to natural number set in binary form we some... Bijections show that it is not a finite number of elements in such a set is finite when its by. Furthermore, we can say that set a is simply the number of elements of two sets... Very similar to the previous theorem set Ais nite, its cardinality the... As sets of the set jAj= @ 0 from the definition symmetyofthissituation, how to prove cardinality of sets B. Should be sufficient for the cardinality of a set S as one to one corresponding to natural set. Induction step because you know how to take the induction step because how to prove cardinality of sets! To B must be a bijection between them B both have a strong geometric resemblance as of. Are combined using operations on sets, this strategy doesn ’ t work... Are each and every of the set of positive integers is called countable, while other... = | N | and uncountable in・］ite sets are considered to be countable or not useful application cardinality. From a to B must be a good exercise for you to try to prove this to yourself.... Set in how to prove cardinality of sets form is countable: use a standard calculus function to establish bijection. A Countability proof - | {
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form is countable: use a standard calculus function to establish bijection. A Countability proof - Duration: 12:14 set … a useful application of is. Shown in Figure 1.11 to prove that cardinality is ∞ as ℵ0 ( null... Let $a$ has only a finite set, usually denoted by @ 0 ( )! 1, 2, 3, 4, 5 }, \mathbb { }! Their subsets are countable set … a useful application of cardinality is a simple guideline deciding! | {
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# jacobian
Jacobian matrix
## Description
example
jacobian(f,v) computes the Jacobian matrix of f with respect to v. The (i,j) element of the result is $\frac{\partial f\left(i\right)}{\partial \text{v}\left(j\right)}$.
## Examples
collapse all
The Jacobian of a vector function is a matrix of the partial derivatives of that function.
Compute the Jacobian matrix of [x*y*z,y^2,x + z] with respect to [x,y,z].
syms x y z
jacobian([x*y*z,y^2,x + z],[x,y,z])
ans =
$\left(\begin{array}{ccc}y z& x z& x y\\ 0& 2 y& 0\\ 1& 0& 1\end{array}\right)$
Now, compute the Jacobian of [x*y*z,y^2,x + z] with respect to [x;y;z].
jacobian([x*y*z,y^2,x + z], [x;y;z])
ans =
$\left(\begin{array}{ccc}y z& x z& x y\\ 0& 2 y& 0\\ 1& 0& 1\end{array}\right)$
The Jacobian matrix is invariant to the orientation of the vector in the second input position.
The Jacobian of a scalar function is the transpose of its gradient.
Compute the Jacobian of 2*x + 3*y + 4*z with respect to [x,y,z].
syms x y z
jacobian(2*x + 3*y + 4*z,[x,y,z])
ans = $\left(\begin{array}{ccc}2& 3& 4\end{array}\right)$
Now, compute the gradient of the same expression.
ans =
$\left(\begin{array}{c}2\\ 3\\ 4\end{array}\right)$
The Jacobian of a function with respect to a scalar is the first derivative of that function. For a vector function, the Jacobian with respect to a scalar is a vector of the first derivatives.
Compute the Jacobian of [x^2*y,x*sin(y)] with respect to x.
syms x y
jacobian([x^2*y,x*sin(y)],x)
ans =
$\left(\begin{array}{c}2 x y\\ \mathrm{sin}\left(y\right)\end{array}\right)$
Now, compute the derivatives.
diff([x^2*y,x*sin(y)],x)
ans = $\left(\begin{array}{cc}2 x y& \mathrm{sin}\left(y\right)\end{array}\right)$
Specify polar coordinates $r\left(t\right)$, $\varphi \left(t\right)$, and $\theta \left(t\right)$ that are functions of time.
syms r(t) phi(t) theta(t)
Define the coordinate transformation form spherical coordinates to Cartesian coordinates. | {
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Define the coordinate transformation form spherical coordinates to Cartesian coordinates.
R = [r*sin(phi)*cos(theta), r*sin(phi)*sin(theta), r*cos(phi)]
R(t) = $\left(\begin{array}{ccc}\mathrm{cos}\left(\theta \left(t\right)\right) \mathrm{sin}\left(\varphi \left(t\right)\right) r\left(t\right)& \mathrm{sin}\left(\varphi \left(t\right)\right) \mathrm{sin}\left(\theta \left(t\right)\right) r\left(t\right)& \mathrm{cos}\left(\varphi \left(t\right)\right) r\left(t\right)\end{array}\right)$
Find the Jacobian of the coordinate change from spherical coordinates to Cartesian coordinates.
jacobian(R,[r,phi,theta])
ans(t) =
$\left(\begin{array}{ccc}\mathrm{cos}\left(\theta \left(t\right)\right) \mathrm{sin}\left(\varphi \left(t\right)\right)& \mathrm{cos}\left(\varphi \left(t\right)\right) \mathrm{cos}\left(\theta \left(t\right)\right) r\left(t\right)& -\mathrm{sin}\left(\varphi \left(t\right)\right) \mathrm{sin}\left(\theta \left(t\right)\right) r\left(t\right)\\ \mathrm{sin}\left(\varphi \left(t\right)\right) \mathrm{sin}\left(\theta \left(t\right)\right)& \mathrm{cos}\left(\varphi \left(t\right)\right) \mathrm{sin}\left(\theta \left(t\right)\right) r\left(t\right)& \mathrm{cos}\left(\theta \left(t\right)\right) \mathrm{sin}\left(\varphi \left(t\right)\right) r\left(t\right)\\ \mathrm{cos}\left(\varphi \left(t\right)\right)& -\mathrm{sin}\left(\varphi \left(t\right)\right) r\left(t\right)& 0\end{array}\right)$
## Input Arguments
collapse all
Scalar or vector function, specified as a symbolic expression, function, or vector. If f is a scalar, then the Jacobian matrix of f is the transposed gradient of f.
Vector of variables or functions with respect to which you compute Jacobian, specified as a symbolic variable, symbolic function, or vector of symbolic variables. If v is a scalar, then the result is equal to the transpose of diff(f,v). If v is an empty symbolic object, such as sym([]), then jacobian returns an empty symbolic object.
collapse all | {
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collapse all
### Jacobian Matrix
The Jacobian matrix of the vector function f = (f1(x1,...,xn),...,fn(x1,...,xn)) is the matrix of the derivatives of f:
$J\left({x}_{1},\dots {x}_{n}\right)=\left[\begin{array}{ccc}\frac{\partial {f}_{1}}{\partial {x}_{1}}& \cdots & \frac{\partial {f}_{1}}{\partial {x}_{n}}\\ ⋮& \ddots & ⋮\\ \frac{\partial {f}_{n}}{\partial {x}_{1}}& \cdots & \frac{\partial {f}_{n}}{\partial {x}_{n}}\end{array}\right]$
## Version History
Introduced before R2006a | {
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# In how many different ways can the letters of the word MISSISSIPPI be
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In how many different ways can the letters of the word MISSISSIPPI be [#permalink]
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08 Apr 2016, 01:47
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In how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?
A. 48
B. 144
C. 210
D. 420
E. 840
[Reveal] Spoiler: OA
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]
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08 Apr 2016, 20:21
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The word MISSISSIPPI has 4 I , 4S , 2P and M .
The only vowel present in the word is I and we have to consider 4 I's as a single group
Number of different ways = 8!/(4!*2!)
=(8*7*6*5)/2
=840
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"url": "https://gmatclub.com/forum/in-how-many-different-ways-can-the-letters-of-the-word-mississippi-be-216328.html"
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Number of different ways = 8!/(4!*2!)
=(8*7*6*5)/2
=840
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]
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26 Jun 2017, 11:18
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Bunuel wrote:
In how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?
A. 48
B. 144
C. 210
D. 420
E. 840
vowel (I) together can be arranged with other 7 letters in 8! ways
4 no. vowels themselves can be arranged in 4! ways
different ways = 8!*4!/( 4!*4!*2!) = 840---(4I's , 4 S's , 2 P's)
Ans E
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]
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26 Jun 2017, 12:18
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Bunuel wrote:
In how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?
A. 48
B. 144
C. 210
D. 420
E. 840
Ooh, a combinatorics problem with some fairly large numbers. The best technique for these is to mentally break them down into simpler problems. Otherwise, the solutions tend to sort of look like magic tricks - sure, you can use the formula, but how are you supposed to know to use that formula, and not one of the many similar-looking ones?
There are four vowels, and they're all the same. Let's set those vowels aside: (IIII) | {
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There are four vowels, and they're all the same. Let's set those vowels aside: (IIII)
Now we have the letters MSSSSPP. How many ways can just those letters be arranged? Well, if they were all different, we could arrange them in 7x6x5x4x3x2x1 = 7! ways. However, they aren't all different. For instance, four of the letters are (S). I'm going to color them to demonstrate why that matters:
one arrangement: MSSSSPP
a 'different' arrangment: MSSSSPP
We counted both of those arrangements, but we actually don't want to. All of the Ss look the same - they aren't actually different colors - so we want to make those arrangements the same. Because there are 4x3x2x1 = 24 ways to order the different Ss, we want every set of 24 arrangements where the Ss are in the same place, to just count as 1 arrangement, instead. So, we can divide out the extra possibilities by dividing our total by 24 (or 4!):
7!/4!
Do the same thing with the two Ps. We still have twice as many arrangements as we need, since we've counted as if the two Ps were different, but they're actually the same. So, divide the total by 2:
7!/(4! x 2)
Now we have to put the (IIII) letters back in. They all have to go together. Start by looking at one of the arrangements of the other letters: PMSSPSS. Where can the four Is go? There are 8 places where we can put them:
IIIIPMSSPSS
PIIIIMSSPSS
PMIIIISSPSS
PMSIIIISPSS
etc.
So, for each arrangement, we have to multiply by 8, to account for the eight possible ways to put the vowels back in. Here's the final answer:
(8 x 7!) / (4! x 2) = (8 x 7 x 6 x 5 x 4 x 3 x 2) / (4 x 3 x 2 x 2) = 4 x 7 x 6 x 5 = 4 x 210 = 840.
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]
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26 Jun 2017, 12:40
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Bunuel wrote:
In how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?
A. 48
B. 144
C. 210
D. 420
E. 840
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]
So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
-------NOW ONTO THE QUESTION!!!-----------------
First "glue" the 4 I's together to create ONE character: IIII (this ensures that they stay together)
So, basically, we must determine the number of arrangements of M, S, S, S, S, P, P and IIII
There are 8 characters in total
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 8!/[(4!)(2!)]
= 840
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]
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04 Dec 2017, 08:31
Stage 1: Lets place the letters without the 4I's:
MSSSSPP
7!/(4!2!)= 105 ways
Stage 2: Now lets place the 4I's
-m-s-s-s-s-p-p-
We can place th4 4I's in 8 ways
Total ways= 105*8= 840 ways.
Is this a correct approach?
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink] | {
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### Show Tags
04 Dec 2017, 09:08
Rearranging letters has a simple formula:
Total number of letters
*DIVIDED BY*
(2)! for every letter repeated twice / (3)! for every letter repeated thrice and so on...
Mississippi has:
4 s's
4 i's
2 p's
Total number of letters: 11!
So the answer is $$\frac{11!}{4!4!2!}$$
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink] 04 Dec 2017, 09:08
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# In how many different ways can the letters of the word MISSISSIPPI be
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$10$ people ($6$ male, $4$ female) divided into $2$ equal groups: what is the probability that all females are in the same group?
This question comes from a completed, marked, and returned exam. It will not likely be reused.
Problem
As stated in the question above
Work
First, I note that there are $$\binom{10}{5}$$ possible groupings.
Second, I note that, if all $$4$$ females are in the same group, then the remaining fifth member is one of the boys: there are $$\binom{6}{1} = 6$$ ways to choose the fifth member.
So I conclude $$\Pr = \frac{6}{\binom{10}{5}} = \frac{1}{42}$$.
Question
I was marked incorrect: the given answer is $$\frac{1}{21}$$, or exactly twice my answer.
What reasoning led to this conclusion? Why does it seem like some sort of symmetry argument allows us to conclude there are $$12$$ ways to choose the fifth member?
• Note that $\frac{6}{\binom{10}{5}} = \frac{6}{252} = \frac{1}{42}$. – N. F. Taussig Feb 28 at 16:55
• @N.F.Taussig apologies i was looking at the answers on the question below as i typed – D. Ben Knoble Feb 28 at 17:01
• Another way to see that there are only $126$ possible groups is to observe that if Eloise is one of the four girls, then there are $\binom{9}{4}$ ways to select which four of the other members are in her group. – N. F. Taussig Feb 28 at 17:28
You've double-counted the groupings: $$\{A, B, C, D, E\}$$ and $$\{F, G, H, I, J\}$$ is the same grouping as $$\{F, G, H, I, J\}$$ and $$\{A, B, C, D, E\}$$. Accounting for this double-count, there are $$\frac{1}{2} \binom{10}{5}$$ distinct groupings. (This is probably the most-common counting mistake of all time. Everyone makes it at least once.) | {
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# Show if A,B,C are invertible matrices of same size
1. Jul 7, 2008
Show if A,B,C are invertible matrices of same size....
I know, I know. I should be awesome at these by now...
1. The problem statement, all variables and given/known data
Show if A,B,C are invertible matrices of the same size, than $(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$
3. The attempt at a solution Given some matrix A,$AA^{-1}=I$
If:
$AA^{-1}=I$
$BB^{-1}=I$
$CC^{-1}=I$
I am not sure where to go from here. I don't think I have any more definitions or product rules to incorporate.
It almost seems as if I would FIRST have to show that (ABC) is invertible to begin with. Then I can use the fact that (ABC)(ABC)^{-1}=I to discover how (ABC)^{-1} MUST be arranged in order for the product of the two to yield I.
Does that sound like a good place to start? Proving if A,B, and C are invertible, then (ABC) is too?
2. Jul 7, 2008
### rock.freak667
Re: Show if A,B,C are invertible matrices of same size....
Let $D=(ABC)^{-1}$
$\times (ABC)$
$\Rightarrow ABCD=I$
and then just multiply by $A^{-1}$ and so forth
3. Jul 7, 2008
Re: Show if A,B,C are invertible matrices of same size....
I don't follow? What does the second line mean?--->$\times (ABC)$
4. Jul 7, 2008
### rock.freak667
Re: Show if A,B,C are invertible matrices of same size....
Multiply both sides by the matric ABC
5. Jul 7, 2008
Re: Show if A,B,C are invertible matrices of same size....
So
$D=(ABC)^{-1}$
$\Rightarrow D(ABC)=(ABC)^{-1}(ABC)$
Well, I think I see where this is going. And I think that the only reason this works is because we are assuming that the product (ABC) IS invertible. Which brings me back to my original point. In order to show how the multiplication MUST be carried out, we must first SHOW or assume without proof that (ABC) is in fact invertible since our argument will be based on the fact that (ABC)*(ABC)^{-1}=Id.
6. Jul 7, 2008
### rock.freak667
Re: Show if A,B,C are invertible matrices of same size.... | {
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6. Jul 7, 2008
### rock.freak667
Re: Show if A,B,C are invertible matrices of same size....
Well if A,B,C are nxn matrices then ABC is an nxn matrix.
and for the matrix ABC to be invertible $det(ABC) \neq 0$
and det(ABC)=det(A)*det(B)*det(C)
but the matrices A,B,C are invertible.
7. Jul 7, 2008
### Defennder
Re: Show if A,B,C are invertible matrices of same size....
Another way you could show that a product of two matrices A and B are invertible is by showing that there exists some matrix which when multiplied to AB on the left and on the right gives the identity matrix:
Suppose A and B are invertible, then:
$$AB(B^{-1}A^{-1}) = I$$ for multiplying on the right
$$B^{-1}A^{-1}AB = I$$ for multiplying on the left.
In both cases this reduces to I, so $$B^{-1}A^{-1}$$ is the inverse of AB.
Now make use of this result to prove your question.
8. Jul 7, 2008
### HallsofIvy
Staff Emeritus
Re: Show if A,B,C are invertible matrices of same size....
It is not nnecessary to assume that ABC is invertible.
You are given that A, B, C separately are invertible so A-1, B-1, and C-1 exist. Thus C-1B-1A-1 exists. What do you get if you multiply (ABC)(C-1B-1C-1) and (C-1B-1A-1(ABC)?
That will prove that ABC is invertible.
9. Jul 7, 2008
### matt grime
Re: Show if A,B,C are invertible matrices of same size....
X is invertible if there is a Y with XY=YX=Id.
It isn't that I want to show ABC is invertible, but someone's been really helpful and just asked me to verify what the inverse is! I don't have to think at all, I just have to chuck the nominal inverse into the definition and see what happens. So do it....
10. Jul 7, 2008
### Defennder
Re: Show if A,B,C are invertible matrices of same size....
Hey yeah, I didn't see it that way. Makes it a lot easier.
11. Jul 7, 2008
### JinM
Re: Show if A,B,C are invertible matrices of same size.... | {
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11. Jul 7, 2008
### JinM
Re: Show if A,B,C are invertible matrices of same size....
This looks like a question that could spring up on my midterm tomorrow. Assuming invertibility, does this prove the result?
$$ABCC^{-1}B^{-1}A^{-1} = A(BI_{n})B^{-1}A^{-1}= A(BB^{-1})A^{-1} = AA^{-1} = I_n$$
Now, the other side,
$$C^{-1}B^{-1}A^{-1}ABC = C^{-1}B^{-1}BC = C^{-1}C = I_n$$
Thus $$(ABC)(C^-1B^-1A^-1}) = (C^-1B^-1A^-1)(ABC) = I_n$$.
Is this enough? No claim of originality, this was how a result was proved in my notes, and what everyone is getting at!
Last edited: Jul 7, 2008
12. Jul 7, 2008
### Defennder
Re: Show if A,B,C are invertible matrices of same size....
That's basically what HallsOfIvy said.
13. Jul 7, 2008
Re: Show if A,B,C are invertible matrices of same size....
I don't like it. I don't know why yet...but I just don't.
14. Jul 7, 2008
### matt grime
Re: Show if A,B,C are invertible matrices of same size....
You aren't assuming invertibility of ABC. You're just multiplying together 6 matrices, and showing that they give the identity, which is *proving* it is invertible. Just because you wrote "assuming invertibility of ABC" doen't mean you actually did assume it, or that you needed toi.
15. Jul 7, 2008
### b0it0i
Re: Show if A,B,C are invertible matrices of same size....
i think the problem you're having with this proof is understanding what the question is asking.
there are many other proofs with the same structure
the problem is saying show that
(ABC)-1 = C-1B-1A-1
in other words, they want you to show that the inverse of ABC is actually (behaves like) inverse C times inverse B times inverse A
an example would be... Show that
A Source of water = Rain
you must show that rain acts/ behaves like a source of water
apply the properties of "a source of water" to rain
Show that C-1B-1A-1 "behaves" like the (ABC)-1
how do you do that?... you check the property of (ABC)-1
that is... show that C-1B-1A-1 (ABC) = (ABC) C-1B-1A-1 = I | {
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that is... show that C-1B-1A-1 (ABC) = (ABC) C-1B-1A-1 = I
An important fact is to note that "the inverse of a matrix is unique"
since C-1B-1A-1 "behaves" like the inverse of ABC,
and that the inverse of a matrix is unique,
C-1B-1A-1 MUST be (ABC)-1
this is corny, but my past professor constantly called these types of proofs
"walks like a duck, quacks like a duck, MUST be a duck..." proof
16. Jul 8, 2008 | {
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# Answer does not make sense using conditional probability
As of now, there are 64.1 million people residing in the UK. 5.4 million of them are thought to be asthmatic. A new test for asthma has recently been introduced and medical trials indicate that
• in patients with asthma, the test correctly returns positive 68% of the time
• in patients that does not suffer from asthma, the test correctly returns 82% of the time.
Assuming a patient undergo medical inspection. What is the probability that the patient has asthma,
1. if the test comes back positive?
2. if the test comes back negative?
For Part 1, I used this method:
Step 1: $$P(A|+)=\frac{P(+|A)P(A)}{P(+)}$$
Step 2: $$P(+)=P(+│A)P(A)+P(+│N)P(N)=(0.68)\left(\frac{54}{641}\right)+(0.18)\left(\frac{587}{641}\right)=\frac{7119}{32050}$$
Step 3: $$P(A│+)=\frac{P(+|A)P(A)}{P(+)}=\frac{0.68\times\frac{54}{641}}{\frac{7119}{32050}}=\frac{204}{791}=25.79\%$$
For Part 2, I used the same method as Part 1 but I got a very small answer. Assuming if I insert all the values in the equations correctly, would it be sensible if my answer for Part 2 is 3.47%?
However, personally, it does not make logical sense if the probability of test returning negative is 3.47% because that would mean almost everyone in the nation would be asthmatic.
Or I could just write it as $$100-25.79 = 74.21%$$ but I'm afraid this isn't the answer given the complexity of the question. | {
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• 3.47% is the right answer here... – Clement C. Apr 29 '15 at 13:33
• You're computing the probability of a false negative not the probability someone doesn't have asthma. False negatives are rare and should be rare intuitively. Otherwise, the test would be a piece of crap. So, you're right. – Jamie Lannister Apr 29 '15 at 13:34
• I don't understand what you mean. If 3.47% is the correct answer, this is quite reasonable. It says: "If the test says you don't have asthma, then it's very likely you really don't have asthma (i.e. the test is good)." The first answer is actually more surprising. It says: "If the test says you have asthma, it's only 25% likely you actually have it!" But that isn't that surprising given the low figure of 68% in the hypotheses. – Frank Apr 29 '15 at 13:34
• @iterence: that would not give the probability $\Pr[A\mid -]$, but instead the quantity $\Pr[A^c\mid +] = 1 - \Pr[A\mid +]$. That is, instead of "the probability to have asthma knowing the test says you don't," you would compute "the probability not to have asthma, knowing the test says you do." – Clement C. Apr 29 '15 at 13:38
• In part 2, "if the test comes back negative", it is assumed we know that the result was negative, i.e., the probability of the test returning negative is $100$%, not $3.47$%. The $3.47$% figure arises from the cases where people do have asthma but it is not detected by the test. – David K Apr 29 '15 at 13:41
One instructive way to do these kinds of $2 \times 2$ problems, since there are only four distinct populations, is to enumerate them and calculate the marginal probabilities by inspection.
$$\begin{array}{|c|c|c|c|} \hline & \text{asthmatic} & \text{not asthmatic} & \text{TOTALS} \\ \hline \text{positive} & 3.672 & 10.566 & 14.238 \\ \hline \text{negative} & 1.728 & 48.134 & 49.862 \\ \hline \text{TOTALS} & 5.400 & 58.700 & 64.100 \\ \hline \end{array}$$
From this, one can read off | {
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From this, one can read off
$$P(\text{asthmatic} \mid \text{positive}) = 3.672/14.238 \doteq 0.25790$$
and
$$P(\text{asthmatic} \mid \text{negative}) = 1.728/49.862 \doteq 0.34656$$
ETA: Two of the cells—$3.672 = (0.68)(5.400)$ and $48.134 = (0.82)(58.7)$—are filled directly by the given parameters. Everything else is bookkeeping.
The abysmal false positive rate is due to a combination of the low specificity of the test ($82$ percent) and the relatively low prevalence of asthma ($5.4/64.1 \doteq 8.4$ percent), so that although only a minority (sizable, but still a minority) of non-asthmatics test positive, they still dwarf the asthmatics who test positive, because there are so few asthmatics available to test positive in the first place. Even if the test were $100$ percent sensitive, the false positive rate would still have been $10.566/(10.566+5.400) \doteq 66$ percent. | {
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Expected value of sums
Suppose we draw cards out of a deck without replacement. How many cards do we expect to draw out before we get an ace?
-
It depends (slightly) on how one interprets before. There are two interpretations possible: (i) Before means not including the draw that got us the first Ace and (ii) We include in the count the draw that got us the first Ace.
There is no big difference between (i) and (ii): The count (and expectation) in (ii) is just $1$ more than the count, and expectation, in (i), We use interpretation (i). So let $W$ be the number of draws before the first Ace, not including the draw that got us the Ace. We want $E(W)$.
The argument is simple but a bit delicate, so the solution below is given in great detail. Luckily, the actual computation is almost formula-free. We use indicator random variables. Label the $48$ non-Aces $1$ to $48$. Don't bother to label the Aces.
Define random variable $X_i$ by $X_i=1$ if the card with label $i$ was drawn before any Ace, and let $X_i=0$ otherwise. Then $$W=X_1+X_2+\cdots+X_{48}.$$ By the linearity of expectation, which holds even when the random variables are not independent, we have $$E(W)=E(X_1+X_2+\cdots+X_{48})=E(X_1)+E(X_2)+\cdots+E(X_{48}).$$ By symmetry, all the $X_i$ have the same distribution. We find, for example, the probability that $X_1=1$. So we want the probability that card with label $1$ is drawn before any Ace.
Consider the $5$-card collection consisting of the $4$ Aces and the card labelled $1$. All orders of these cards in the deck are equally likely. It follows that the probability that card with label $1$ is in front of the $4$ Aces is $\frac{1}{5}$. Thus $E(X_1)=\frac{1}{5}$.
We conclude that $E(W)=\dfrac{48}{5}$.
If we want to take interpretation (ii), and include the draw that got us the Ace, our expectation is $1+\dfrac{48}{5}=\dfrac{53}{5}$. | {
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Now I'm wondering if this answer should be considered essentially the same as mine. – Michael Hardy Nov 27 '12 at 1:54
Hard to decide! Mine (of course it's not mine, it is pretty standard stuff, I have used it in lectures) is very formal. – André Nicolas Nov 27 '12 at 1:59
Has anyone written an exposition of the fact that expectations are often easier to find than probabilities? – Michael Hardy Nov 27 '12 at 2:01
Don't know of one. I sure have stressed it. – André Nicolas Nov 27 '12 at 2:04
Distribute the $52$ cards uniformly between $0$ and $1$, so on average they're at $k/53$ for $k=1,2,3,\ldots,52$. The four aces are on average at $1/5,\ 2/5,\ 3/5,\ 4/5$. So $$0.2 = \frac{k}{53}$$ implies $$k = 10.6.$$ and $(0.8)\cdot53 = 42.4$. So on average the four aces are the $10.6$th, $21.2$th, $31.8$th, and $42.4$th cards.
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A good way to go about this is as follows. Let $T$ be the number of cards drawn at the time of the first ace. This is a $\mathbb{N}$ valued random variable. Therefore $$E(T) = \sum_{n=0}^\infty P(T > n).$$ You may find this helpful.
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I think that the answer lies in the Hypergeometric distribution or else k successes in n draws from a finite population of size N containing m successes without replacement (quoting from Wikipedia).
In your case, you want a single success in $p$ draws, and from the distribution you can find an average.
Edit: $$P(X=1) = \frac{\binom{m}{1}\binom{N-m}{n-1}}{\binom{N}{n}}$$ and taking an average should be something like the following $$\bar M = \sum_{i=1}^{N-m}i\frac{\binom{m}{1}\binom{N-m}{i-1}}{\binom{N}{i}}$$ for $N=52, m=4$.
End of edit.
Wikipedia also has some examples with urns, black and white balls, so you can work the answer from those.
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I came across this in The Theory of Gambling and Statistical Logic By Richard A. Epstein so pasting a snippet here. I got this from Google Books (public domain) so hopefully there is no copyright violation here. It's amazing how Andre, Michael and Patrick derived the correct answer differently.
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I did not see this approach in any of the answers, which is one that makes sense to me: you can get the ace in either the 1st, 2nd,..., 49th trial.
Then the expected number of cards dealt is $$1(1/13)+2(12/13)(1/13)+..n(12/13)^{n-1}(1/13)+48(12/13)^{47}(1/13)$$.
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Try this. (edited based on comment below) You will need just 1 pull with probability 4/52 2 with probability (4/51 x 48/52). 48/52 for the probability of not getting and ace in the first and getting it in the second (4/51), and so on. This yields 1 x 4/52 + 2x(4/51 x 48/52) + 3x(4/50 x 48/52 x 44/50) + ... There must be some simplification for the above, but I'ven't looked deeper into, but you get the idea. Here are similar ones
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This assumes there is only one ace in the deck. – Michael Hardy Nov 27 '12 at 1:05
I have always liked this shortcut (proof here), which also holds for continuous random variables defined on $x \geq{0},\,x \in \mathbb{R}$. For some reason it was called "The Darth Vader" rule in my actuarial study manual but I digress. For this problem: $$P(T>n) = \frac{\binom{48}{n}n!}{\binom{52}{n}n!} = \frac{\binom{48}{n}}{\binom{52}{n}}$$ This is easily reasoned as follows:
$P(T>n)$ is the probability that there are no aces up to and including the $n$th draw which means that in these first $n$ draws, out of the $48$ non-aces we count the number of ways to choose $n$ of them and permute them. We divide this by the total number of outcomes for the first $n$ draws which is $\binom{52}{n}n!$ (we must include the aces this time, thus the 52). | {
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Therefore: $$E(T) = \sum_{n=0}^{48}\frac{\binom{48}{n}}{\binom{52}{n}} = \sum_{n=0}^{48}\frac{(52-n)(51-n)(50-n)(49-n)}{52\times51\times50\times49} = 53/5 = 10.6$$
We could sum to $\infty$ like ncmathsadist stated but the values of $P(T>n)$ for $n>48$ are $0$ anyway.
I will admit that I cheated in my final step by using wolframalpha. Wolfram Alpha is a GREAT free resource by the way. I do hope that the logic behind solving the problem is not lost as a result. - Patrick
Sidebar: My initial naive guess was that $E(T) = 13$ since there are on average $12$ cards between each ace. I am trying to figure out an intuitive understanding for why the actual answer is less that $13$ now.
-
It is 13 only if there is replacement. As there isn't any replacement the probability goes up so that interval should shrink eventually converging to 10.6. BTW could you elaborate how you simplified the average to 53/5? – broccoli Nov 27 '12 at 7:52
You're right, if there is replacement then $T \sim Geometric(p)$ where $p$ is the probability of finding an ace ($1/13$) and $E(T) = 1/p = 13.$ As for an elaboration I just plugged the sum into wolfram alpha's calculator. Andres solution above is much better as the calculations can be done in your head and his method is applicable to other similar problems. – Patrick Nov 27 '12 at 13:33
If you wonder why $13$ is too big, consider the expected number of trials needed to get all four aces. Would that be $52$? Only if you never get the fourth ace until the last card in the deck. – Michael Hardy Nov 27 '12 at 19:16
Add a fifth ace, uniformly randomly arrange the $53$ cards in a circle, and break the circle into a line at the added ace. By symmetry, the expected number of cards between two of the $5$ aces is $\frac{48}5$, so this is the expected number of cards between the beginning of the line and the first ace.
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# What's the chance that exactly 2 2s will have appeared before the 2nd 8?
A standard deck of 52 cards is randomly shuffled. Bob keeps drawing from the deck until he has drawn two 8s.
What is the probability that, when, he draws the second 8, he has already drawn exactly two 2s?
I think this could be a combinations problem because the order of the drawing two 2s does not matter. However, I am not sure how to account for the fact that he could draw an 8 or a second 8 at any point.
-
Hint: one approach is to recognize that all the other cards are just fluff. There are only ${8 \choose 4}=70$ possibilities for the order of 2's and 8's, so list them and count. You can make it less by listing the orders that get to 3 2's or 2 8's and assessing the probability. The first are failures, the second are successes if you have two 2's already. For example, one is that you start with three 2's. This is $\frac 12 \cdot \frac 37 \cdot \frac 26=\frac 1{14}$. Another is 2828. This is $\frac 12 \cdot \frac 47 \cdot \frac 12 \cdot \frac 35=\frac 3{35}$ | {
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Oh! I think I got it from that hint! The only way for this to hold is for the string of combos be 8228[some ordering of 2 2s and 2 8s], 2828[some ordering of 2 2s and 2 8s], or 2288[some ordering of 2 2s and 2 8s]. Hence, the solution is 3 * (4 choose 2) / (8 choose 4), which is approximately 0.257! :D ... right? – David Faux Sep 9 '12 at 4:03
@DavidFaux: Good thought, but I wouldn't be sure that all the cases are equally probable. 8228 is $\frac 48 \frac 47 \frac 36 \frac 35=\frac 3{35}$ while 2288 is $\frac 48 \frac 37 \frac 46 \frac 35$, which is the same, so I think it works. I would report the result as $\frac 9{35}$, which is exact. – Ross Millikan Sep 9 '12 at 4:20
Hmm, I think the combinations are equal because non-2 and non-8 cards are all fluff. If we make the 8s non-unique, and 2s non-unique, this basically becomes a problem involving strings of four 8s and four 2s, right? – David Faux Sep 9 '12 at 4:34
@DavidFaux: That is right. The more I think, I agree with your calculation. I was worried that 22228888 would be less probable than 82282882, but now I don't think so. It is easy to think things are equally probable when they are not. – Ross Millikan Sep 9 '12 at 4:39
As was observed by Ross Millikan, there are only $8$ relevant cards. The first $4$ of these can appear in the dealing in $(8)(7)(6)(5)$ orders, all equally likely.
We count the number of "favourable" orders. These are given by any one of three patterns, which we call $2288$, $2828$, and $8228$.
The pattern $2288$ can occur in $(4)(3)(4)(3)$ ways, as can the other two patterns. Thus the required probability is $$\frac{(3)(4)(3)(4)(3)}{(8)(7)(6)(5)}.$$ If we want to simplify a bit, we get $\dfrac{9}{35}$.
Remark: The problem seems to have been solved correctly in the comments both by you and by Ross Millikan. Then there appears to have been a change of mind.
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# Electric field of bent non conducting rod
## Homework Statement
You are given a non conducting rod carrying uniformly distributed charge, -Q, that has been bent into a 120° circular arc of radius, R. The axis of symmetry of the arc lies along the x-axis and the origin is at the center of curvature of the arc.
(a) in terms of Q and R, what is the linear charge density, λ?
(b) in terms of Q and R, what is the magnitude and direction the the resulting electric field at the origin?
(c)If the arc is replaced by a point particle carrying charge, -Q, at x=R, by what factor is the resulting electric field at the origin multiplied?
λ=Q/L
E=KQ/R2
dq=λds
ds=Rdθ
## The Attempt at a Solution
(a) λ=Q/L, λ=Q/(⅓2piR) = 3Q/2piR
(b) Cant seem to figure out how to write an integral with bounds. The bounds on my integral below will be from θ=0° → θ=60°
E=2∫dEx=2∫kλdθcosθ/R = 2kλ/R ∫cosθdθ (Like previously stated this integral is being evaluated from θ=0° to θ=60°)
E=(2kλ/R)[sin60°-sin0°] = (kλ√3)/R, it wants in terms of R and Q so now plug in our above statement for λ
E=(3kQ√3)/2piR2
(c) Im a little unsure about what this is asking, from my understanding:
to obtain the new electric field from a point particle which is simply kQ/R2 we must multiply the previous electric field from bent rod by a factor of 2pi/3√3. Not sure if this is correct or if I'm thinking about the question in the wrong way.
Any help would be much appreciated
Thanks!
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Consider subdividing the 120o arc into infinitesimal elements $ds$.
1. What is the charge on one of these elements?
2. What is the electric field contribution $dE$ from this charge element to the point of interest?
3. Add all such contributions to find the net electric field. Remember that the electric field element $dE$ is a vector so you have to find its components and add them separately. | {
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Part (c) is asking this: Say the field due to the arc is $E_0$, If you replace the arc by a single charge -Q at distance $R$, you will get another field $E_1$. If you write an equation relating the two as $E_1=\alpha E_0$, what is the value of constant $\alpha$?
Consider subdividing the 120o arc into infinitesimal elements $ds$.
1. What is the charge on one of these elements?
2. What is the electric field contribution $dE$ from this charge element to the point of interest?
3. Add all such contributions to find the net electric field. Remember that the electric field element $dE$ is a vector so you have to find its components and add them separately.
Part (c) is asking this: Say the field due to the arc is $E_0$, If you replace the arc by a single charge -Q at distance $R$, you will get another field $E_1$. If you write an equation relating the two as $E_1=\alpha E_0$, what is the value of constant $\alpha$?
Okay, Ill work on this and post my results! Thanks!
Answering is easy when people make mistakes. It’s much harder when everything seems perfect. I’m too lazy to really check every factor of 2 or pi, so I may be mistaken, but this looks really good to me. What has you worried?
Answering is easy when people make mistakes. It’s much harder when everything seems perfect. I’m too lazy to really check every factor of 2 or pi, so I may be mistaken, but this looks really good to me. What has you worried?
Yea just part c, wasn't sure what it was asking but @kuruman helped me out with that one. Just have a test in a few days and want to make sure I'm doing these problems correctly. Always love reassurance! LOL
You saw the symmetry and ignored the y component. You also used symmetry to cut the integral in half. You integrated in theta and didn’t forget the extra R.
I don’t think you have much to worry about, but I’ll wish you good luck anyway. | {
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# Limit of $\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$
Compute the limit $$\lim_{n \to \infty} \frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$$
How can this be done? The best I could do was rewrite the limit as $$\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n}$$
Following that log suggestion in the comments below: \begin{align} &\ln \left(\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \ln \left( \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \left(\ln \left(\frac{n+1}n \right)^{\frac 1n} + \ln\left(\frac{n+2}n \right)^{\frac 1n} + \cdots + \ln\left(\frac{n+n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \sum_{i=1}^n \ln \left(\frac{n+i}n \right)^{\frac 1n} \\ &= \lim_{n \to \infty} \frac 1n \sum_{i=1}^n \ln \left(1 + \frac in \right) \\ &= \int_1^2 \ln x \, dx \\ &= (x \ln x - x)\vert_1^2 \\ &= (2 \ln 2 - 2)-(1 \ln 1-1) \\ &= (\ln 4-2)-(0-1) \\ &= \ln 4-1 \\ &= \ln 4 - \ln e \\ &= \ln \left( \frac 4e \right) \end{align}
but I read from somewhere that the answer should be $\frac 4e$. | {
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but I read from somewhere that the answer should be $\frac 4e$.
• Your integral is the log of the answer – robjohn May 14 '18 at 18:15
• @CarlMummert From the accepted answer of this meta question, we can see that this is a well-written question (except the use of $\frac1n$ as a power), and "additional indication of source would not make a difference".You may re-read AmateurMathGuy's reply to your comment to know why they are not required. – GNUSupporter 8964民主女神 地下教會 May 14 '18 at 18:19
• You voted to close because "Being homework is a sufficient reason to vote to close a question", but HW questions are allowed. – GNUSupporter 8964民主女神 地下教會 May 14 '18 at 18:19
• @CarlMummert I do not see how my question "has no real context, and does not fit the quality standards that many users expect." I put in the work and effort as this Math SE site suggests. – GarlicBread May 15 '18 at 3:58
• Agreed. This is supported by the Close Queue Review result. All reviewers left this question open. – GNUSupporter 8964民主女神 地下教會 May 15 '18 at 9:00
Hint: let $a_n=(n+1)\cdots(2n)/n^n$, and use the fact that $a_{n+1}/a_n\to L$ implies $\sqrt[n]{a_n}\to L$ (i.e., the ratio test implies the root test).
If you know Sterling's Approximation: $$n! \sim \left(\frac{n}{e}\right)^n \sqrt{2\pi n},$$ then you could approach it as follows:
$$\lim_{n \to \infty} \frac{\sqrt[n]{(n + 1)(n + 2)\cdots (2n)}}{n} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{(2n)!}{n!}} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{\left(\frac{2n}{e}\right)^{2n} \sqrt{4\pi n}}{\left(\frac{n}{e}\right)^n \sqrt{2\pi n}}} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{2^{2n} \left(\frac{n}{e}\right)^n \sqrt{2}} = \lim_{n \to \infty} \frac{1}{n} \cdot 2^2 \cdot \frac{n}{e} \cdot 2^{\frac{1}{2n}} = \frac{4}{e}.$$
Now take a logarithm and realize you have just gotten a Riemann sum. (Interval: $[1,2]$, widths: $1/n$, heights $\ln(1+k/n)$.) | {
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• I'm not sure how to rewrite the limit of the Riemann sum as its corresponding definite integral, although I think I managed to arrive at the Riemann sum in my work. – GarlicBread May 14 '18 at 17:41
If you accept using the following limit that has been asked and proved (for example here) many times here on MSE
• $\lim_{n\rightarrow \infty}\frac{\sqrt[n]{n!}}{n} = \frac{1}{e}$,
then your limit is easily derived:
$$\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n = \frac{ (2n!)^{\frac{1}{n}} }{ (n!)^{\frac{1}{n}}\cdot n} = 4\frac{n}{ (n!)^{\frac{1}{n}}} \left( \frac{ (2n!)^{\frac{1}{2n}} }{ 2n} \right)^2 \stackrel{n\rightarrow\infty}{\longrightarrow}4\cdot e \cdot \frac{1}{e^2}=\frac{4}{e}$$
Note that $\displaystyle\frac{\sqrt[n]{n+1}}{n}\neq\left(\frac{n+1}{n}\right)^{1/n}$
Also, this is just for clarification since some nice suggestions have already been given but also because the question mark remains over your last equality
With this is mind and going from your first step,
$$S=\lim_{n\to\infty}\frac{\sqrt[n]{(n+1)(n+2)...(2n)}}{n}=\lim_{n\to\infty}\left(\frac{1}{n}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)\right)$$
$$\operatorname{ln}(S)=\operatorname{ln}\left(\lim_{n\to\infty}\left(\frac{1}{n}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)\right)\right)$$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\operatorname{ln}\left(\frac{1}{n}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)\right)$$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\operatorname{ln}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)-\operatorname{ln}(n)\right)$$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\operatorname{ln}\left((n+1)^{1/n}\right)+\operatorname{ln}\left((n+2)^{1/n}\right)+...+\operatorname{ln}\left((2n)^{1/n}\right)-\operatorname{ln}(n)\right)$$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\sum_{p=1}^n\left(\operatorname{ln}\left((n+p)^{1/n}\right)\right)-\operatorname{ln}(n)\right)$$ | {
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$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\sum_{p=1}^n\left(\frac{1}{n}\operatorname{ln}(n+p)\right)-\operatorname{ln}(n)\right)$$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\left(\operatorname{ln}(n+p)\right)-\operatorname{ln}(n)\right)$$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\left(\operatorname{ln}\left(1+\frac{p}{n}\right)+\operatorname{ln}(n)\right)-\operatorname{ln}(n)\right)$$
This $+\operatorname{ln}(n)$ is independent of $p$ and summed $n$ times, then divided by $n$, so $\operatorname{ln}(n)-\operatorname{ln}(n)=0$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\operatorname{ln}\left(1+\frac{p}{n}\right)\right)$$
This is what Eric meant and the argument varies from $1$ to $2$
It becomes a little clearer if we let $m = 1/n$
$$\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\operatorname{ln}\left(1+\frac{p}{n}\right)\right)=\lim_{m\to\ 0}\left(\left(\sum_{p=1}^{1/m}\operatorname{ln}(1+pm)\right)m\right)=\int_1^2\operatorname{ln}(x)dx=(x\operatorname{ln}(x)-x)\Big\rvert_1^2$$
$$\therefore \operatorname{ln}(S)=2(\operatorname{ln}(2)-1)+1=\operatorname{ln}(4)-1=\operatorname{ln}\left(\frac{4}{e}\right)$$
Finally
$$S=\frac{4}{e}$$ | {
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# Matrix Multiplication Algorithm Pseudocode | {
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Abstract— Matrix multiplication is an integral component of most of the systems implementing Graph theory, Numerical algorithms, Digital control, Signal and image processing (i. the first column when multiplied by the matrix. of algorithms in the computational sciences. The booth’s multiplication algorithm is primarily used in computer architectures. Banded Matrix-Vector Multiplication. emphasizes the use of pseudo code in the introductory Computer Science, this approach is to teach students how to first develop a pseudo code representation of a solution to a problem and then create the code from that pseudo code. Pseudocode of the rest of the algorithm: Iterative Matrix Multiplication I'm a bit unhappy with your code, because it's so hard to read tbh. Rotation: We use a generalization of Cannon’s algo-rithm as the primary template. I Strassen's algorithm gives a performance improvement for large-ish N, depending on the architecture, e. Write an algorithm in pseudocode to perform the multiplication of a matrix with a vector. Specifically, an input matrix of size can be divided into 4 blocks of matrices. COMP 250 Winter 2016 1 { grade school algorithms Jan. Consider an NxN complex array. This allows us to exploit fast matrix multiplication. 3 Multithreaded merge sort 797 28 Matrix Operations 813 28. Divide-and-Conquer algorithsm for matrix multiplication A = A11 A12 A21 A22 B = B11 B12 B21 B22 C = A×B = C11 C12 C21 C22 Formulas for C11,C12,C21,C22: C11 = A11B11 +A12B21 C12 = A11B12 +A12B22 C21 = A21B11 +A22B21 C22 = A21B12 +A22B22 The First Attempt Straightforward from the formulas above (assuming that n is a power of 2): MMult(A,B,n) 1. Pseudo Code for Union using Mapreduce. What is the least expensive way to form the product of several matrices if the naïve matrix multiplication algorithm is used? [We use the number of scalar multiplications as cost. spawn is to indicate creation of a new thread. 3) you should encapsulate a matrix into a class, if it is supposed to | {
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creation of a new thread. 3) you should encapsulate a matrix into a class, if it is supposed to be c++ (then 2 will be obsolete) 4) your code will be easier to understand (for you as well) if you use better names for x,y,z,i, k and j. Cannon's algorithm: a distributed algorithm for matrix multiplication especially suitable for computers laid out in an N × N mesh; Coppersmith-Winograd algorithm: square matrix multiplication; Freivalds' algorithm: a randomized algorithm used to verify matrix multiplication. 5D (Ballard and Demmel) ©2012 Scott B. The main purpose of this paper is to present a fast matrix multiplication algorithm taken from the paper of Laderman et al. , what the complexity of the problem is). Pseudocode for the algorithm is given in Figure 1. cludes the RSA cryptosystem, and divide-and-conquer algorithms for integer multiplication, sorting andmedianfinding, aswellasthe fast Fourier transform. Algorithm Examples! Pseudocode! Order of Growth! Algorithms - what are they Fibonacci - Matrix Multiplication. Multithreaded Algorithms Introduction. To perform the addition, numbers in matching postions in the input matrices are added and the result is placed in the same position in the output matrix. We propose two approaches to matrix multiplication: iter-ative approach and block approach. Unlike standard matrix multiplication, MixColumns performs matrix multiplication as per Galois Field 2 8. Although adjacency matrix representation of graph is used, this algorithm can also be implemented using Adjacency List to improve its efficiency. General Matrix Multiplication (GEMM) is the primary component of the level-3 BLAS and of most dense linear algebra algorithms (and many sparse/structured linear algebra algorithms), which in turn have applications in virtually every area of computational science. Summary I Strassen rst to show matrix multiplication can be done faster than O(N3) time. complexity of matrix multiplication is n2 (2n −1) = 2 ⋅(2 −1)⋅τ T1 n n (8. | {
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faster than O(N3) time. complexity of matrix multiplication is n2 (2n −1) = 2 ⋅(2 −1)⋅τ T1 n n (8. Matrix multiplication is one of the most fundamental operations in linear algebra and serves as the main building block in many different algorithms, including the solution of systems of linear equations, matrix inversion, evaluation of the matrix determinant, in signal processing, and the transitive closure of a graph. •Pseudo code. Problem 6 (checks whether a matrix is symmetric): yes 4. One of the basic operations on matrices is multiplication. In other words, Lagrees with the corresponding k×nsubmatrix of X. 2 StrassenÕs algorithm for matrix multiplication If you have seen matrices before, then you probably know how to multiply them. Regression algorithm pseudocode from [4]: The regression algorithm follows a nested optimization scheme using coordinate descend. 6 Another Recursive Algorithm 4. If you're interested in typesetting algorithmic code, there are a number of choices. However, due to constant factors and realistic modern architecture constraints, these theoretically faster methods are rarely used; instead, the naive brute force approach to matrix multiplication is that which is. Suppose we want to multiply two n by n matrices, A and B. This analysis culminates in Section 4. The practical benefit from improvements to algorithms is therefore potentially very great. Here's a short example from the algorithmicx documentation (with a pseudocode for loop added):. Assume that square matrix A and B are used for multiplication in the following algorithms. Algorithms { CMSC-37000 Divide and Conquer: The Karatsuba algorithm (multiplication of large integers) Instructor: L aszl o Babai Updated 01-21-2015 The Karatsuba algorithm provides a striking example of how the \Divide and Conquer" technique can achieve an asymptotic speedup over an ancient algorithm. We need to create a Toeplitz matrix using a subsection of a data vector on the device. Matrix Multiplication in | {
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create a Toeplitz matrix using a subsection of a data vector on the device. Matrix Multiplication in Case of Block-Striped Data Decomposition Let us consider two parallel matrix multiplication algorithms. This builds on the previous post on recursive square matrix multiplication. to read the matrix into core memory once [19, sect. So Matrix Chain Multiplication problem has both properties (see this and this) of a dynamic programming problem. To begin with, the sequential algorithm was implemented using the pseudo-code in [2]. Freivalds' algorithm is a probabilistic randomized algorithm used to verify matrix multiplication. 2 Strassen's algorithm for matrix multiplication 4. One key idea in the sorting networks chapter, the 0-1 principle, ap-. x x y matrix by a y x z matrix creates an x x z matrix. The cofactor matrix is the matrix of determinants of the minors A ij multiplied by -1 i+j. Antoine Vigneron (UNIST) CSE331 Lecture 5 July 11, 2017 3 / 19. Write pseudocode for Strassen’s algorithm. r-1 (mod n) where the integers a and b are smaller than the modulus. Complexity Calculation How many additions of integers and multiplications of integers are used by the matrix multiplication algorithm to multiply two n * n matrices. I won't give the pseudo code here for these ones, but they are naive recursive algorithm, bottom up algorithm, naive recursive squaring and recursive squaring. MPI Matrix-Matrix Multiplication Matrix Products Parallel 2-D Matrix Multiplication Characteristics Computationally independent: each element computed in the result matrix C, c ij, is, in principle, independent of all the other elements. 6 Another Recursive Algorithm 4. Provide your analysis for the following problem statement: Write a program that will calculate the results for the multiplication table up to 10x10 in steps of 1 beginning at 1. cient implementation of sparse matrix multiplication on a memory intensive associative processor (AP), verified by extensive AP simulation using a | {
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on a memory intensive associative processor (AP), verified by extensive AP simulation using a large collection of sparse matrices [41]. We then "combine" the middle row of the key matrix with the column vector to get the middle element of the resulting column vector. The multiplier contains only 0s and 1s,. The algorithms classes I teach at Illinois have two significant prerequisites: a course on discrete mathematics and a course on fundamental data structures. Block matrices are briefly discussed using 2 × 2 block matrices. I have a question for you about your approach. What is a Spanning tree? Explain Prim’s Minimum cost spanning tree algorithm with suitable example. Before going to main problem first remember some basis. This page contains the order of topics contained in lectures, listed as a sequence of modules. Which is faster for this value of n?. To save space and running time it is critical to only store the nonzero elements. i) Multiplication of two matrices ii) Computing Group-by and aggregation of a relational table i) Multiplication of two matrices ii) Computing Group-by and aggregation of a relational table. 2 shows the calculate steps of covariance matrix, mainly including: complex conjugate multiplication between the lines of input matrix, then do an accumulation operation. Idea - Block Matrix Multiplication The idea behind Strassen’s algorithm is in the formulation of matrix multiplication as a recursive problem. You can use a pseudocode environment algpseudocode offered by algorithmicx. SPARSE MATRICES C/C++ Assignment Help, Online C/C++ Project Help and Homework Help introduction A matrix is a mathematical object that arises in many physical problems. 1 The naive matrix multiplication algorithm Let A and B be two n £ n matrices. Matrix mulitplication using Linked List - posted in C and C++: I have to implement Matrix multiplication using singly linked list. Pseudocode of the rest of the algorithm: Iterative Matrix Multiplication I'm a bit unhappy | {
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list. Pseudocode of the rest of the algorithm: Iterative Matrix Multiplication I'm a bit unhappy with your code, because it's so hard to read tbh. Material for the algorithms class taught by Emanuele "Manu" Viola. Dynamic Programming—Chained Matrix Multiplication Multiplying unequal matrices • Suppose we want to multiply two matrices do not have the same number of rows and columns • We can multiply two matrices A 1 and A 2 only if the number of columns of A 1 is equal to the number of rows of A 2. Example of Matrix Multiplication by Fox Method Thomas Anastasio November 23, 2003 Fox's algorithm for matrix multiplication is described in Pacheco1. However, this algorithm is infamously inapplicable, as it relies on Coppersmith and Winograd’s fast matrix multiplication. It is referenced specifically in the pseudocode but that is not the only location where it is appropriate to call it. If you are interested in a Modified Gauss-Jordan Algorithm, you can see this. Algorithm for the Transpose of a Sparse-Matrix: This is the algorithm that converts a compressed-column sparse matrix into a compressed-row sparse matrix. The practical benefit from improvements to algorithms is therefore potentially very great. What is the best algorithm for matrix multiplication ? Actually there are several algorithm exist for matrix multiplication. Here, we will discuss the implementation of matrix multiplication on various communication networks like mesh and. - Overall complexity of parallel matrix-vector multiplication algorithm ( n2=p+n+logp) - Isoefficiency of the parallel algorithm Time complexity of sequential algorithm: ( n2) Only overhead in parallel algorithm due to all-gather For reasonably large n, message transmission time is greater than message latency. Solutions for CLRS Exercise 4. 2 Algorithmic Techniques 5. \begin{algorithm} \caption{Euclid's algorithm}\label{euclid} \. 1 Naive Matrix Multiplication 4. Matrix-matrix multiplication takes a triply nested loop. I'm just doing a | {
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Matrix Multiplication 4. Matrix-matrix multiplication takes a triply nested loop. I'm just doing a self-study of Algorithms & Data structures and I'd like to know if anyone has a C# (or C++) implementation of Strassen's Algorithm for Matrix Multiplication? I'd just like to run it and see what it does and get more of an idea of how it goes to work. Let us start with a very simple example th. One common practice is to translate convolution to im2col and GEMM, which lays out all patches into a matrix. 84 videos Play all Algorithms Abdul Bari; Derivatives explained - Duration: 10:13. One of the basic operations on matrices is multiplication. The algorithm is called a (7, 4) code, because it requires seven bits to encoded four bits of data. Recurrence equation for Divide and Conquer: If the size of problem ‘p’ is n and the sizes of the ‘k’ sub problems are n1, n2…. Write An Algorithm To Find The Power Of A Number. 2 StrassenÕs algorithm for matrix multiplication If you have seen matrices before, then you probably know how to multiply them. This page contains the order of topics contained in lectures, listed as a sequence of modules. GitHub Gist: instantly share code, notes, and snippets. Adjacency matrix representation of graph where the n X n matrix W = (wij) of edge weights. Matrix multiplication algorithms. Definition of Flowchart A flowchart is the graphical or pictorial representation of an algorithm with the help of different symbols, shapes and arrows in order to demonstrate a process or a program. complexity by packing the inner loops into a single matrix product as shown in Algorithm 2. emphasizes the use of pseudo code in the introductory Computer Science, this approach is to teach students how to first develop a pseudo code representation of a solution to a problem and then create the code from that pseudo code. The unit of. Matrix multiplication is the process of taking two or more n x n matrices and calculating a product by summing the product of each row in | {
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of taking two or more n x n matrices and calculating a product by summing the product of each row in one matrix with the column of the second matrix. As in case of developing the matrix-vector multiplication algorithm, we use one-dimensional arrays, where matrices are stored rowwise. So assuming that both these multiplication steps are executed every time the loop executes, we see that 2. From Math Insight. We tackle scenarios such as matrix multiplication and linear regression/classification in which we wish to estimate inner products between pairs of vectors from two possibly different sources. The problem is quite easy when n is relatively small. 1 and the Junto. Below is some sample output. It is the technique still used to train large deep learning networks. org) I now want to use strassen's method which I learned as follows:. Unlike general multiplication, matrix multiplication is not commutative. o The number of additions and multiplication's required for this algorithm can be calculated as follows: To calculate one entry in the product matrix, we must perform k multiplications and k-1 additions. for i = 1 to n. 1 and Step 3. We can use simple recursion, f(n) = f(n-1) + f(n-2), or we can use dynamic programming approach to avoid the calculation of same function over and over again. Write a c program to find out transport of a matrix. it explains matrix multiplication. Antoine Vigneron (UNIST) CSE331 Lecture 5 July 11, 2017 3 / 19. In other words, Lagrees with the corresponding k×nsubmatrix of X. Question: Show Map Reduce implementation for the following two tasks using pseudocode. - Explain the difference between an LED and OLED display. matrix multiplication; this means that matrix multiply based methods for determining primitivity cannot be sped up anymore at this time. 3343(abc)(log7)/(3) 2. Block matrices are briefly discussed using 2 × 2 block matrices. x x y matrix by a y x z matrix creates an x x z matrix. Section 5 provides a com-parison with related | {
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x y matrix by a y x z matrix creates an x x z matrix. Section 5 provides a com-parison with related works. Summary I Strassen rst to show matrix multiplication can be done faster than O(N3) time. In the next three parts, you may be writing pseudocode. That’s very important because for small n (usually n < 45) the general algorithm is practically a better choice. Integer-multiplication, Matrix Multiplication - Strassen Alg You study after every class/week, the syllabus accumulates fast before you know! Aug 26, W (Drop w/o W grade, Aug 28) Dynamic Programming: 0-1Knapsack. This work is licensed under aCreative Commons. parallel before a loop means each iteration of the loop are independant from each other and can be run in parallel. Notes: A common reference for double-precision matrix multiplication is the dgemm ( d ouble-precision ge neral m atrix- m atrix multiply) routine in the level-3 BLAS. The application. Given three n x n matrices, Freivalds' algorithm determines in O(kn^2) whether the matrices are equal for a chosen k value with a probability of failure less than 2^-k. Name the algorithmic technique used. 3 Matrix Multiplication for Banded Matrices. com Free Programming Books Disclaimer This is an uno cial free book created for educational purposes and is not a liated with o cial Algorithms group(s) or company(s). 0 is there to suggest that different values can be used, but they should be related to the number of input variables. The following algorithm multiplies nxn matrices A and B: // Initialize C. Strassen's algorithm, the original Fast Matrix Multiplication (FMM) algorithm, has long fascinated computer scientists due to its startling property of reducing the number of computations required for multiplying. and similarly for the bottom row. We know that, to multiply two matrices it is condition that, number of columns in first matrix should be equal to number of rows in second matrix. Matrix Multiplication: Strassen's Algorithm. What Is The Main | {
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to number of rows in second matrix. Matrix Multiplication: Strassen's Algorithm. What Is The Main Operation Of This Algorithm? C. We need to create a Toeplitz matrix using a subsection of a data vector on the device. of algorithms in the computational sciences. You can call the algorithm on sub-matrices of dimensions n-1 each when the size of the matrix is odd in the recursive step and calculate 2n-1 remaining elements using normal vector multiplication in O(n) each and a total of O(n^2). We tackle scenarios such as matrix multiplication and linear regression/classification in which we wish to estimate inner products between pairs of vectors from two possibly different sources. What is the best algorithm for matrix multiplication ? Actually there are several algorithm exist for matrix multiplication. n and r is relatively prime number to n (gcd (n, r)= 1). for k = 1 to n. Part I was about simple matrix multiplication algorithms and Part II was about the Strassen algorithm. If A is the adjacency matrix of G, then (A I)n 1 is the adjacency matrix of G*. Two groups of algorithms belonging to this class are called the matrix method, and the Wallace-tree method, respectively. 3) where τ is the execution time for an elementary computational operation such as multiplication or addition. Section 5 provides a com-parison with related works. The aim is to get the idea quickly and also easy to read without details. The former is suitable for sparse matrices, while the latter is appropriate for dense matrices with low communication overhead. We could break down the steps as follows. , what the complexity of the problem is). In matrix addition, one row element of first matrix is individually added to corresponding column elements. 4 uses dynamic programming to find an optimal triangulation of a convex polygon, a problem that is surprisingly similar to matrix-chain multiplication. Matrix multiplication is one of the most fundamental operations in linear algebra and serves as the | {
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Matrix multiplication is one of the most fundamental operations in linear algebra and serves as the main building block in many different algorithms, including the solution of systems of linear equations, matrix inversion, evaluation of the matrix determinant, in signal processing, and the transitive closure of a graph. Then, user is asked to enter two matrix and finally the output of two matrix is calculated and displayed. That’s very important because for small n (usually n < 45) the general algorithm is practically a better choice. Each matrix Mk has dimension pk-1 x pk. The Floyd Warshall algorithm, itis the algorithm in which there is the use of different characterization of structure for a shortest path that we used in the matrix multiplication which is based on all pair algorithms. Use row communicators and column communicators to scatter and broadcast the vector. We ended up pursuing a different route, but I decided to continue pursuing the problem on my own time. 2x2 Matrix Multiplication Calculator is an online tool programmed to perform multiplication operation between the two matrices A and B. From Math Insight. This work is licensed under aCreative Commons. Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication. Provide a specification to describe the behaviour of this algorithm, and prove that it correctly implements its specification. However, this algorithm is infamously inapplicable, as it relies on Coppersmith and Winograd’s fast matrix multiplication. The weights and values of 6. Sparse Matrix Multiplication. I am trying to implement a multiplication algorithm by overloading the *= operator. Matrix Multiplication in Case of Block-Striped Data Decomposition Let us consider two parallel matrix multiplication algorithms. 3 Storage formats 3. Other types of algorithms for this problem appear in [15, 16]. In particular, this includes judging which data structures, libraries, frameworks, programming | {
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16]. In particular, this includes judging which data structures, libraries, frameworks, programming languages, and hardware platforms are appropriate for the computational task, and using them effectively in the implementation. How do they differ? - Is a pixel a little square? If not, what is it? What implications does this have? Give at least 2. One of the basic operations on matrices is multiplication. It is the technique still used to train large deep learning networks. Then, we'll present a few examples to give you a better idea. Floyd Warshall. What Is The Main Operation Of This Algorithm? C. For instance, the algorithm we're interested in looking at, Dijkstra's algorithm, only works if none of the edges on the graph have negative weights -- the "time" it takes to traverse the edge is somehow less than 0. emit (key, result). GATEBOOK Video Lectures 3,203 views. These lectures were designed for the latter part of the MIT undergraduate class 6. Matrix Multiplication Problem - Duration: 27:38. - Overall complexity of parallel matrix-vector multiplication algorithm ( n2=p+n+logp) - Isoefficiency of the parallel algorithm Time complexity of sequential algorithm: ( n2) Only overhead in parallel algorithm due to all-gather For reasonably large n, message transmission time is greater than message latency. Other types of algorithms for this problem appear in [15, 16]. • Algorithms are step-by-step procedures for problem solving • They should have the following properties: •Generality •Finiteness •Non-ambiguity (rigorousness) •Efficiency • Data processed by an algorithm can be • simple • structured (e. Strassen's method of matrix multiplication is a typical divide and conquer algorithm. Although we won't describe this step in detail, it is important to note that this multiplication has the property of operating independently over each of the columns of the initial matrix, i. Matrix Multiplication; Matrix Multiplication Parenthesization; Brute Force Solution: Try all | {
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i. Matrix Multiplication; Matrix Multiplication Parenthesization; Brute Force Solution: Try all possible parenthesizations; Dynamic Programming Solution (4 steps) Step 1: Characterize Structure of Optimal Solutioon; Step 2: Define recursive solution; Recursive Solution; Analysis; Duplicate Subproblems; Unique Subproblems; Step 3: Bottom-Up Approach; Dynamic Programming. 5 Maximum Flow. Matrix Multiplication c c2 1= r2 A1 A2 r1 r2! = r1 ! c2 (r1 ! c2) ! c1 = multiplications If r 1 = c 1 = r 2 = c 2 = N, this standard approach takes ( N3): I For every row ~r (N of them) I For every column ~c (N of them) I Take their inner product: r c using N multiplications 2. listing algorithms. Idea - Block Matrix Multiplication The idea behind Strassen's algorithm is in the formulation of matrix multiplication as a recursive problem. Describe how an array can be effectively used to store a sparse matrix. Matrix mulitplication using Linked List - posted in C and C++: I have to implement Matrix multiplication using singly linked list. You don't need multiplication facts to use the Russian peasant algorithm; you only need to double numbers, cut them in half, and add them up. The unit of. We then "combine" the middle row of the key matrix with the column vector to get the middle element of the resulting column vector. The problem is quite easy when n is relatively small. John, Your comment about matrix multiplication was forwarded to the mahout-user emailing list. emphasizes the use of pseudo code in the introductory Computer Science, this approach is to teach students how to first develop a pseudo code representation of a solution to a problem and then create the code from that pseudo code. Section 3 pro-vides implementation details on our design. In the other hand the algorithm of Strassen is not much faster than the general n^3 matrix multiplication algorithm. 2 shows the calculate steps of covariance matrix, mainly including: complex conjugate multiplication between the lines of | {
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steps of covariance matrix, mainly including: complex conjugate multiplication between the lines of input matrix, then do an accumulation operation. This immediately leads to a counting algorithm with running time Θ(n3) respectively Θ(nγ), where γ is the matrix multiplication exponent. Divide-and-conquer multiplication. 2 StrassenÕs algorithm for matrix multiplication If you have seen matrices before, then you probably know how to multiply them. These are lecture notes used in CSCE 310 (Data Structures & Algorithms) at the University of Nebraska|Lincoln. count example presented in Section 2. The matrix M and the vector v each will be stored in a file of the DFS. Pseudo-code for i = 1 : n yi yi +axi Number of flops is 2n Pseudo-code cannot run on any computer, but are human readable and straightforward to convert into real codes in any programming language (e. Matrix multiplication is not commutative, but it is associative, so the chain can be parenthesized in whatever manner deemed best. Purdue University Purdue e-Pubs ECE Technical Reports Electrical and Computer Engineering 9-1-1992 Implementation of back-propagation neural networks with MatLab. Prim's Algorithm Implementation in C++. Write An Algorithm To Find The Power Of A Number. Flowchart for Matrix multiplication : Algorithm for Matrix multiplication :. Strassen's matrix multiplication program in c 11. Program the divide and conquer matrix multiplication using 1) standard algorithm 2) recursion 3) strassen’s method. Write the Pseudo Code of the matrix multiplication program that performs the worst execution time and explain why - Answered by a verified Programmer We use cookies to give you the best possible experience on our website. Sparse matrices, which are common in scientific applications, are matrices in which most elements are zero. The dimensions are stored in array. Using the most straightfoward algorithm (which we assume here), computing the product of two matrices of dimensions (n1,n2) and (n2,n3) | {
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(which we assume here), computing the product of two matrices of dimensions (n1,n2) and (n2,n3) requires n1*n2*n3 FMA operations. Our first example of dynamic programming is an algorithm that solves the problem of matrix-chain multiplication. 2 Multithreaded matrix multiplication 792 27. (Otherwise, you should read Section D. When distributing the vector among processors, implement the algorithm shown in Figure (b) on page 22 of lecture notes “ Parallel matrix algorithms (part 2) ”. Summary: The two fast Fibonacci algorithms are matrix exponentiation and fast doubling, each having an asymptotic complexity of $$Θ(\log n)$$ bigint arithmetic operations. The simplest sparse matrix data structure is a list of the nonzero entries in arbitrary order. 1 The SUMMA Algorithm. The introduction of the technique is attributed to a 1962 paper by Karatsuba, and indeed it is sometimes called Karatusba multiplication. We can build a sketch as we scan through the matrix. I am trying to write pseudo code in my paper. I implement these three algorithms from pseudocode from the book to Java code: Merge-Sort Java version：. Animated Algorithms (sorting, priority queues, Huffman, Matrix chain multiplication, MST, Dijkstra) Graph Algorithms (Dijkstra, Prim, Kruskal, Ford-Fulkerson) Java and Web Based Algorithm Animation (JAWAA). What is the main operation of this algorithm? c. Solutions for CLRS Exercise 4. 4 RESULTS We evaluate our implementation by testing its performance on one. com Free Programming Books Disclaimer This is an uno cial free book created for educational purposes and is. There is a faster way to multiply, though, caled the divide-and-conquer approach. 3 perform the multiplication operation. Output Y: d by N matrix consisting of d D dimensional embedding coordinates for the input points. Freivalds' algorithm is a probabilistic randomized algorithm used to verify matrix multiplication. Better asymptotic bounds on the time required to multiply matrices have been known since | {
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Better asymptotic bounds on the time required to multiply matrices have been known since the work of Strassen in the 1960s, but it is still unknown what the optimal time is (i. In this post I will explore how the divide and conquer algorithm approach is applied to matrix multiplication. Purdue University Purdue e-Pubs ECE Technical Reports Electrical and Computer Engineering 9-1-1992 Implementation of back-propagation neural networks with MatLab. Have you considered doing the multiplication in a single step by storing the the first matrix in column major order and the second in row major order?. Explain why memoization fails to speed up a good divide-and-conquer algorithm like merge-sort. CS 2073 Lab 10: Matrix Multiplication Using Pointers Chia-Tien Dan Lo Department of Computer Science, UTSA I Objectives Show how to manipulate commandline arguments in C Demonstrate your ability to read matrices from les Demonstrate your ability to use pointers for matrix multiplication II Hand-in Requirements. You are given 5 different algorithms for different purposes and their pseudocodes that are listed below. Related Questions More Answers Below. So Matrix Chain Multiplication problem has both properties (see this and this) of a dynamic programming problem. Since we have not covered multiplication yet, a function has been provided to you. Two groups of algorithms belonging to this class are called the matrix method, and the Wallace-tree method, respectively. Matrix multiplication is one of the most fundamental operations in linear algebra and serves as the main building block in many different algorithms, including the solution of systems of linear equations, matrix inversion, evaluation of the matrix determinant, in signal processing, and the transitive closure of a graph. LLE Algorithm Pseudocode (Notes, e. 2 Algorithmic Techniques 5. Find f(n): n th Fibonacci number. 3) you should encapsulate a matrix into a class, if it is supposed to be c++ (then 2 will be obsolete) 4) | {
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should encapsulate a matrix into a class, if it is supposed to be c++ (then 2 will be obsolete) 4) your code will be easier to understand (for you as well) if you use better names for x,y,z,i, k and j. Application. using matrix multiplication Let G=(V,E) be a directed graph. Book shows pseudocode for simple divide and conquer matrix multiplication: n = A. r-1 (mod n) where the integers a and b are smaller than the modulus. Matrix multiplication of two sparse matrices is a fundamental operation in linear Baye sian inverse problems for computing covariance matrices of observations and a posteriori uncertainties. HOME ; A comparison of numerical approaches to the solution of the time-dependent Schrödinger equation in one dimension. complexity of matrix multiplication is n2 (2n −1) = 2 ⋅(2 −1)⋅τ T1 n n (8. The current best algorithm for matrix multiplication O(n2:373) was developed by Stanford's own Virginia Williams[5]. Pseudocode for the algorithm is given in Figure 1. Make good use of matrix multiplication! It avoids a lot of loops, so it makes your code cleaner and faster! DO NOT assume that I will answer your email questions or posts to the discussion forum after 3pm on Sunday. which can be signi cantly smaller than their dense equivalent. Pseudocode for Matrix Vector Multiplication by Mapreduce. , the shapes are 2 n × 2 n for some n. i) Multiplication of two matrices ii) Computing Group-by and aggregation of a relational table i) Multiplication of two matrices ii) Computing Group-by and aggregation of a relational table. \begin{algorithm} \caption{Euclid’s algorithm}\label{euclid} \. Since we have not covered multiplication yet, a function has been provided to you. Matrix-matrix multiplication takes a triply nested loop. 2 Multithreaded matrix multiplication 792 27. 2 Matrix-Matrix Multiplication. So assuming that both these multiplication steps are executed every time the loop executes, we see that 2. Alternative approaches can be seen as straight forward | {
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every time the loop executes, we see that 2. Alternative approaches can be seen as straight forward iteration over the nodes or edges of the graph. In grade 1, you learned how to count up to ten and to do basic arithmetic using your ngers. Integer-multiplication, Matrix Multiplication - Strassen Alg You study after every class/week, the syllabus accumulates fast before you know! Aug 26, W (Drop w/o W grade, Aug 28) Dynamic Programming: 0-1Knapsack. The definition of matrix multiplication is that if C = AB for an n × m matrix A and an m × p matrix B, then C is an n × p matrix with entries = ∑ =. As examples, pseudocode is presented for the inner product, the Frobenius matrix norm, and matrix multiplication. Explain why memoization fails to speed up a good divide-and-conquer algorithm like merge-sort. Define the meaning of your variables. Can some one please help me to format it. Section 5 provides a com-parison with related works. With the current implementation of the cuBlas functions we need to write kernel code to do this efficiently. The problem is quite easy when n is relatively small. SPARSE MATRICES C/C++ Assignment Help, Online C/C++ Project Help and Homework Help introduction A matrix is a mathematical object that arises in many physical problems. Pseudocode Matrix Multiplication. This relies on the block partitioning which works for all square matrices whose dimensions are powers of two, i. In matrix multiplication, one row element of first matrix is individually multiplied by all column elements and added. Order of both of the matrices are n × n. The word is derived from the phonetic pronunciation of the last name of Abu Ja'far Mohammed ibn Musa al-Khowarizmi, who. Adjacency-matrix and adjacency-list representations Breadth-first and depth-first search using adjacency lists Computing connected components of a graph Strongly-connected and biconnected components Topological sorting Algebraic algorithms: Strassen matrix multiplication algorithm The Four | {
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Topological sorting Algebraic algorithms: Strassen matrix multiplication algorithm The Four Russians boolean matrix multiplication Winograd's algorithm. Matrix Multiplication Algorithm. Faster matrix multiplication in general is an important applied topic, because it can speed up all sorts of scientific, engineering, and ML algorithms that have it as a step (often one of the bottleneck steps). At the end of the lecture, we saw the reduce SUM operation, which divides the input into two halves, recursively calls itself to obtain the sum of these smaller inputs, and returns the sum of the results from those. Specifically, an input matrix of size can be divided into 4 blocks of matrices. GATEBOOK Video Lectures 3,203 views. 5D (Ballard and Demmel) ©2012 Scott B. Of course, writing pseudocode is child's play compared to actually implementing a real algorithm. In section 4, we discuss our experimental results from the real hardware implementation. return C {C = [cij ] is the product of A and B} 22. Benchmarked it to be 4x faster than the scalar version (on a Pentium M, using GCC 4. | {
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# Math Help - Cube Root times Cube Root
1. ## Cube Root times Cube Root
cube root of 16(m^2)(n) multiplied by cube root of 27(m^2)(n)
I converted the problem to (16m^2n)^1/3 * (27m^2n)^/3.
Is this the correct way to write the original problem another way?
I finally got (432m^4n^2)^2/3.
Is this correct?
2. Originally Posted by sharkman
cube root of 16(m^2)(n) multiplied by cube root of 27(m^2)(n)
I converted the problem to (16m^2n)^1/3 * (27m^2n)^/3.
Is this the correct way to write the original problem another way?
I finally got (432m^4n^2)^2/3.
Is this correct?
Nearly there. Recall that $\sqrt[3]{a} \times \sqrt[3]{b} = \sqrt[3]{ab}$
In your question you have $\sqrt[3]{16(m^2)(n)} \times \sqrt[3]{27(m^2)(n)}$.
Using the fact I gave first, this means than the answer is...
Hint: Your multiplication of the $m$ and $n$ terms was not wrong, it was the index - $\frac{2}{3}$ - that you got wrong.
Originally Posted by craig
Nearly there. Recall that $\sqrt[3]{a} \times \sqrt[3]{b} = \sqrt[3]{ab}$
In your question you have $\sqrt[3]{16(m^2)(n)} \times \sqrt[3]{27(m^2)(n)}$.
Using the fact I gave first, this means than the answer is...
Hint: Your multiplication of the $m$ and $n$ terms was not wrong, it was the index - $\frac{2}{3}$ - that you got wrong.
The answer should be cubert{432m^4n^2}, right?
4. Correct
5. ## great
Originally Posted by craig
Correct
Great! Thanks!
6. Originally Posted by sharkman
The answer should be cubert{432m^4n^2}, right?
Don't be too shocked if your teacher marks it wrong because it has not been simplified as much as it could be. $m^4= m^3 m$ so $\sqrt[3]{m^4}= \sqrt[3]{m^3}\sqrt[3]{m}= m\sqrt[3]{m}$. Perhaps more importantly, $432= (16)(27)= (2^4)(3^3)= (2^3)(3^3)(2)$ so $\sqrt[3]{432}= \sqrt[3]{2^3}\sqrt[3]{3^3}\sqrt[3](2)= 6\sqrt[3]{2}$.
The best way to write your answer is $6m\sqrt[3]{2mn^2}$.
7. ## ok | {
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The best way to write your answer is $6m\sqrt[3]{2mn^2}$.
7. ## ok
Originally Posted by HallsofIvy
Don't be too shocked if your teacher marks it wrong because it has not been simplified as much as it could be. $m^4= m^3 m$ so $\sqrt[3]{m^4}= \sqrt[3]{m^3}\sqrt[3]{m}= m\sqrt[3]{m}$. Perhaps more importantly, $432= (16)(27)= (2^4)(3^3)= (2^3)(3^3)(2)$ so $\sqrt[3]{432}= \sqrt[3]{2^3}\sqrt[3]{3^3}\sqrt[3](2)= 6\sqrt[3]{2}$.
The best way to write your answer is $6m\sqrt[3]{2mn^2}$.
I understand. You decided to break the problem down a bit more. | {
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# Prove that $\sim$ is an equivalence relation
Let $$A = \{1, 2, 3,...,9\}$$ and let $$\sim$$ be the relation on $$A\times A$$ defined by $$(a,b) \sim(c,d)$$ if $$a+d = b+c$$.
Prove that $$\sim$$ is an equivalence relation.
Really stuck on this question, maybe its the use of ~ that is throwing me off. Any suggestions on how to do this? Maybe I could swith ~ out for something else while solving then put it back in.
Thanks
• You have to show ~ is reflexive, symmetric, and transitive. Note $a+d=b+c \iff a-b=c-d$ – J. W. Tanner Apr 10 '19 at 21:20
• (1) If the symbol ~ annoys you, simply replace it by the more ordinary symbol R. (2) If you are stuck, it may come from the fact that, in this problem, the "elements" you are deling with are, in fact, ordered pairs. So for example, proving reflexivity amounts to proving that any ordred pair (a,b) has the relation R ( or ~ ) with itself , that is (a,b) R (a,b). Using the defining formula, you could see what you have to prove, and then prove if using basic arithmetic. – Saint James Apr 10 '19 at 21:25
• The symbol "~" is often used to denote a relation in case this relation is an equivalence relation. But, before having proved that it is actually an equivalence relation in this case, you are not allowed to read it as " is equivalent to". So, accordind to me, the best thing do do is to replace it by the ordinary symbol R, while you are doing your proof. – Saint James Apr 10 '19 at 21:35
Clearly, the statement $$a+d=b+c$$ is just saying that $$a-b=c-d$$, i.e. two ordered pairs are similar if the difference of their terms is equal. To show it's an equivalence relation, we need reflexivity, transitivity and symmetry. The first and last are obvious. | {
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For transitivity, we need to show $$x\sim y$$ and $$y\sim z$$ implies $$x\sim z$$. Let $$x=(x_1,x_2),y=(y_1,y_2),z=(z_1,z_2)$$. The conditions tell you that $$x_1-x_2=y_1-y_2$$, but also that $$y_1-y_2=z_1-z_2$$. Clearly then, $$x_1-x_2=z_1-z_2$$. But this is exactly what it means for $$x\sim z$$!
I don't normally do this but I think it may clear some conceptions up. If you want to get to how to effectively solve this skip to the end.
...
Maybe an example first:
If we let $$A = (3,6)$$ then $$A\sim (m,n)$$ if $$3+ n = 6+m$$ so the list of $$(m,n)$$ where $$A \sim (m,n)$$ are the ones where $$n = 3+m$$ so $$\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}$$ are all the $$(m,n)$$ where $$A \sim (m,n)$$.
We can go a little further.
If $$(a,b) \equiv (c,d)$$ then $$a+d = b+c$$ and $$a-b = c-d$$.
So the "classes" are all the ones related to $$(x_1, x_2)$$ where $$x_1 - x_2 =$$:
$$-8$$: $$\{(1,9)\}$$
$$-7$$: $$\{(1,8),(2,9)\}$$
$$-6$$: $$\{(1,7),(2,8),(3,9)\}$$
$$-5$$: $$\{(1,6),(2,7),(3,8),(4,9)\}$$
$$-4$$: $$\{(1,5),(2,6),(3,7),(4,8),(5,9)\}$$
$$-3$$: $$\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}$$
$$-2$$: $$\{(1,3),(2,4),(3,5),(4,6),(5,7),(6,8),(7,9)\}$$
$$-1$$: $$\{(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9)\}$$
$$0$$: $$\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9)\}$$
$$1$$: $$\{(2,1),(3,2),(4,3),(5,4),(6,5),(7,6),(8,7),(9,8)\}$$
$$2$$: $$\{(3,1),(4,2),(5,3),(6,4),(7,5),(8,6),(9,7)\}$$
$$3$$: $$\{(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)\}$$
$$4$$: $$\{(5,1),(6,2),(7,3),(8,4),(9,5)\}$$
$$5$$: $$\{(6,1),(7,2),(8,3),(9,4)\}$$
$$6$$: $$\{(7,1),(8,2),(9,3)\}$$
$$7$$: $$\{(8,1),(9,2)\}$$
$$8$$: $$\{(9,1)\}$$
All the pairs in each set are related to each other this is an "equivalence" because every pair is is exactly one class and can be interchanged with others in the class with no ambiguity. | {
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A relationship could not be equivalent if some elemtns are in more than one list. Of some elements aren't in any list. Or if such bunch of lists wouldn't make any sense to make.
A relationship is an equivalence if
1) it is reflexive. that is for every pair $$(a,b)$$ then $$(a,b)\sim (a,b)$$ or for every $$(a,b)$$ then $$a-b = a-b$$. That's always true.
If we view it as $$(a,b)\sim (c,d)$$ if $$a-b = c-d$$ then we'd have $$a-b = a-b$$ which is of course true.
2) it is symmetric. For any pair $$(a,b) \sim (c,d)$$ then $$(c,d)\sim (a,b)$$. That is is $$a+d = b+c$$ then $$c+b = d+ a$$. That's ... clear.
If we view is as $$(a,b)\sim (c,d)$$ if $$a-b= c-d$$ then if $$a-b = c-d$$ then $$c-d = a-b$$ so $$(a,b)\sim(c,d)\iff (c,d)\sim (a,b)$$.
3) it is transitive. That is if $$(a,b)\sim (c,d)$$ and $$(c,d)\sim (e,f)$$ then $$(a,b) \sim (e,f)$$.
So if $$a+ d=b+c$$ and $$c+f = d+ e$$ then we have to prove that means $$a+f = b+e$$.
We can do that via arithmetic $$a+d +c+f =b+c + d + e$$ so $$a+f = b+e$$.
Just use the definitions:
For reflexivity, you have to show that for any $$(a,b)$$, it is true that $$(a,b)\sim (a,b)$$, i.e. that $$a+b=b+a$$. Well, $$a$$ and $$b$$ are natural numbers, and those are commutative, so yes, this is true.
Symmetry: here you have to show that if $$(a,b) \sim (c,d)$$, then $$(c,d) \sim (a,b)$$, which is to say: show that if $$a+d=b+c$$, then $$c+b=d+a$$ ... can you show this?
Finally, now that I have shown what you need to prove to demonstrate reflexivity and symmetry, can you figure out what you need to show to demonstrate transitivity? | {
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Question
# The average score of boys in an examination of a school is $$71$$ and that of girls is $$73$$. The average score of the school in that examination is $$71.8$$. The ration of the number of boys to the number of girls appeared in the examination, is
A
3:2
B
3:4
C
1:2
D
2:1
Solution
## The correct option is A $$3:2$$Let there be $$n_1$$ boys and $$n_2$$ girls. and $$\overline{X_1}$$ and $$\overline{X_2}$$ be the average scores of boys and girls respectively.Then $$\bar { { X }_{ 1 } } =71,\bar { { X }_{ 2 } } =63$$ and $$\overline{X} = 71.8$$$$\therefore \quad \overline{X} = \displaystyle\frac{n_1\overline{X_1} + n_2\overline{X_2}}{n_1+n_2}$$$$\Rightarrow \quad 71.8 = \displaystyle\frac{n_1\times71 + n_2\times73}{n_1+n_2}$$$$\Rightarrow \quad 71.8n_1 + 71.8n_2 = 71n_1 + 73n_2$$$$\Rightarrow \quad 0.8n_1 = 1.2n_2 \quad \Rightarrow 8n_1 = 12n_2$$$$\Rightarrow \quad \displaystyle\frac{n_1}{n_2} = \displaystyle\frac{12}{8} = \displaystyle\frac{3}{2}$$Mathematics
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# Difficult Probability Solved QuestionAptitude Discussion
Q. Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?
✔ A. 1/18 ✖ B. 64/4032 ✖ C. 63/64 ✖ D. 1/9
Solution:
Option(A) is correct
Sample space will be total ways of selecting 2 squares (order doesn't matter) out of available 64 squares.
So sample space$=^{64}C_2$
Now there are 7 unique adjacent square sets in each row and each column.
i.e. favourable cases will be 7×(8 rows + 8 columns) = 112.
Hence Required Probability
$=\left(\dfrac{\text{Favourable cases}}{\text{Sample Space}}\right)$
$=\dfrac{112}{^{64}C_2}$
$=\dfrac{1}{18}$
Also, there could be other solution too if we consider ALL possible squares and not only the smallest squares. (i.e. if we consider that total squares are more than 64).
In that case, let us first calculate the sample space i.e. total number of squares on a chess board.
1, 8x8 square
4, 7x7 squares
9, 6x6 squares
16, 5x5 squares
25, 4x4 squares
36, 3x3 squares
49, 2x2 squares
64, 1x1 squares
Therefore, there are actually:
$=64+49+36+25+16+9+4+1$
$=8^2+7^2+6^2+...1^2=204$
squares on a chessboard!
So sample space$=^{204}C_4$
Now if we assume that 2 squares can have a side common only if they are of the same dimensions:
Let the length of the smallest square be 1 unit.
So the possibility of selecting 2 squares with a common side (As calculated in the first method above):
having the side length greater than 4 unit length = 0
having the side length of 4 unit = 1×(2 rows + 2 columns) = 4
having the side length of 3 unit = 3×(6 rows + 6 columns) = 36
having the side length of 2 unit = 5×(7 rows + 7 columns) = 70
having the side length of 1 unit = 7×(8 rows + 8 columns) = 112
So total number of favourable cases:
$=4+36+70+112=222$
Hence Required Probability
$=\left(\dfrac{\text{Favourable cases}}{\text{Sample Space}}\right)$
$=\dfrac{222}{^{204}C_2}$ | {
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$=\left(\dfrac{\text{Favourable cases}}{\text{Sample Space}}\right)$
$=\dfrac{222}{^{204}C_2}$
Edit: For an alternative solution, check comment from Bhavya Shah.
Edit 2: For yet another alternative solution, check comment by RandomMathMan.
## (12) Comment(s)
Gaurav Karnani
()
Total number of Cases for a chessboard$7*(7*2+1)$ that equals $112$
P= $112/64C2$
Gaurav Karnani
()
Forgot to add seven to the number of cases for the last row. please amend it accordingly.
RandomMathMan
()
There's a much easier way to do this. On a chessboard, there are 36 squares that border 4 other squares, 24 squares that border 3 other squares, and 4 squares that border only 2 other squares.
$\dfrac{36}{64} \times \dfrac{4}{63} + \dfrac{24}{64} \times \dfrac{3}{63} + \dfrac{4}{64} \times \dfrac{2}{63} = \dfrac{1}{18}$
Siddharth
()
Exactly how I too solved this. Much simpler than the other alternatives given here.
Bhavya Shah
()
For each row, there are 7 ways to select adjacent squares and there are 8 such rows.
Similarly for columns, 7 squares could be selected from one column and there are 8 such columns.
So no. of ways of selecting adjacent squares $= 7*8*2$
Total number of ways of selecting squares of chessboard $= ^{64}C_2$
Probability $=\dfrac{7*8*2}{^{64}C_2}=\dfrac{1}{18}$
Poonam Pipaliya
()
sir,,how 7 squares in each row and column remain..??..!! and how it can be favorable case..??..!!
Anuj
()
Lets talk of the rows first. Once you pick up a square, you leave that row, so you have 7 remaining rows.
Similar argument is valid for columns.
Rishi
()
I guess the alternative method given in the solution is easier to understand, why don't you have a look at that.
Arshal
()
Guys, I think I'm right but, as I have been hitting my head on the wall of probability for last 7 hours I'm kinda in concussion...
So is choosing 2 consecutive elements out of 8 equal to choosing 1 element out of 7
()
$C(4,1)*2+C(24,1)*3+C(36,1)/C(64,2)$ | {
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()
$C(4,1)*2+C(24,1)*3+C(36,1)/C(64,2)$
There are 4 corner squares for which there are only 2 choices for selecting next square....
Similarly, 24 squares for which there are 3 choices and remaining 36 have 4 choices
So, the answer should be 1/9
Correct me if i am wrong!!!
()
As I counted every square non repeated
So, total no of unique square=
$\dfrac{((C(4,1)*2+C(24,1)*3+C(36,1)*4)/2)}{C(64,2)}$
$=\dfrac{1}{18}$ | {
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# Thread: calculating compound interest with non-linear interest growth
1. ## calculating compound interest with non-linear interest growth
Hello, I would really appreciate some help with the following problem.
I'd like to derive a formula to determine the value of an investment where the annual yield increases by a fixed percentage each year. The interest paid is re-invested each year. The standard compounding formulas use a fixed yield so I've not been able to use them in this case.
As an example, suppose I have 1000 dollars in an investment with an initial starting yield (y) of 3% which grows (g) at 5% each year.
End of Year 1 : Yield was 3.00%, Investment value = 1000 + (1000 * 3%) = 1030.00
End of Year 2 : Yield was 3.15% (3% increased by 5%), Investment value = 1030 + (1030 x 3.15%) = 1062.45
End of Year 3 : Yield was 3.31% (3.15% increased by 5%), Investment value = 1062.45 + (1062.45 x 3.31%) = 1097.59
I believe the calculation is a series in the form
Value = P.(1+y0).(1+y1).(1+y2).(...)
where P = initial investment amount, r0 = yield in year 0, r1 = yield in year 1, r2 = yield in year 2 etc.
and where y0 = y.(1+g)^t etc.
If I write the above formula out substituting y0, y1, etc. I get
Value = P.(1+y(1+g)^t).(1+y(1+g)^(t-1)).(1+y(1+g)^(t-2)).(...) where t ranges from 0 to the target year n.
Is it possible to convert this expression into a general formula in the form of Value = f(t) where t = time in years, e.g. some kind of exponential function and how might I go about doing that?
Thank you for any help or insight that you can provide me!
~ Trevor
2. ## Re: calculating compound interest with non-linear interest growth
One way of doing this would be to find a single yield that results in a same future value using the series of yields
$FV = R (1+i)^n$
$ln(1+i_{s})=\frac{1}{n}\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)$
$i=e^{ln(1+i_s)}-1$
Code:
R=1000
r=3%
g=5%
n=3
$ln(1+i_{s})=0.031037549112832$
$i=e^{0.031037549112832}-1$
$i=3.15242359792658\%$ | {
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$i=e^{0.031037549112832}-1$
$i=3.15242359792658\%$
$FV = 1000 (1+3.15242359792658\%)^3$
$FV = 1000 (1.0315242359792658)^3$
$FV = 1000 (1.097585368)$
$FV = 1,097.59$
3. ## Re: calculating compound interest with non-linear interest growth
Hi GhostAccount,
Thank you for the fast response, that was very helpful! I've checked the results manually using Excel and your method generates all FV values that I expect from n=1 to n=10. There is one point that I don't fully understand however.
I understand that your answer is solving the formula below for i :
$R (1+i)^n = R\sum_{k\ =\ 0}^{n-1} (1+r(1+g)^k)$
This tells me that I should be able to calculate any given FV for a value of n by manually summing the rhs. However I don't get the expected answer when I tried to do this in Excel:
n | Expected vs Excel
1 | 1030.00 vs 1030.00 ( = 1000 + 30 )
2 | 1062.45 vs 1061.50 ( = 1000 + 30 + 31.5 )
3 | 1097.59 vs 1094.58 ( = 1000 + 30 + 31.5 + 33.075 )
4 | 1135.70 vs 1129.30 ( = 1000 + 30 + 31.5 + 33.075 + 34.7288 )
Did I interpret the FV summation formula correctly and just get Excel wrong? | {
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I didn't realize that the compound growth + compound interest could be expressed as a summation like that - I thought this was only calculating the amount of interest that $1000 would produce in n years with a compounded rate and not compounding the principal amount as well. I was thinking that $FV=R(1+r(1+g)^0)(1+r(1+g)^1)(1+r(1+g)^2)$ etc A summation formula is what I really want to reach as I can then try to substitute a general identity for it and derive a formula in terms of r,g and n that I can use to calculate each expected value for a given t, for example using $\sum_{i\ =\ m}^{n-1} a^i = \frac{a^m - a^n}{1-a}$ Thanks again for your help! 4. ## Re: calculating compound interest with non-linear interest growth Originally Posted by tw35758 Hi GhostAccount, Thank you for the fast response, that was very helpful! I've checked the results manually using Excel and your method generates all FV values that I expect from n=1 to n=10. There is one point that I don't fully understand however. I understand that your answer is solving the formula below for i : $R (1+i)^n = R\sum_{k\ =\ 0}^{n-1} (1+r(1+g)^k)$ This tells me that I should be able to calculate any given FV for a value of n by manually summing the rhs. However I don't get the expected answer when I tried to do this in Excel: n | Expected vs Excel 1 | 1030.00 vs 1030.00 ( = 1000 + 30 ) 2 | 1062.45 vs 1061.50 ( = 1000 + 30 + 31.5 ) 3 | 1097.59 vs 1094.58 ( = 1000 + 30 + 31.5 + 33.075 ) 4 | 1135.70 vs 1129.30 ( = 1000 + 30 + 31.5 + 33.075 + 34.7288 ) Did I interpret the FV summation formula correctly and just get Excel wrong? Thanks again for your help! The formula I presented in two parts at first finds the continuously compounded yearly rate (geometric average) that has to be converted to a yearly rate See the following for complete formula $R (1+i)^n = R(e^{\sum_{k\ =\ 0}^{n-1} (1+r(1+g)^k)})$ Edit: This formula should work as I am short on time now thus I didn't find time to confirm the results 5. ## Re: | {
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should work as I am short on time now thus I didn't find time to confirm the results 5. ## Re: calculating compound interest with non-linear interest growth Was trying for geometric series...like the 1000 bucks creates a guaranteed 30 bucks annual deposit (the 3%), so we'd have FV of 1000 plus FV of 30 annuity....but finally realised that won't work... i = initial interest rate (.03) f = increase factor (1.05) n = number of years (make it 10) FV = (1 + i*f^0)(1 + i*f^1)(1 + i*f^2)...........(1 + i*f^(n-1)) = 1.4481148... That's for the proverbial one dollar, of course; so 1,448.11 if$1000. | {
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Soooo....looks like I'm kinda repeating GA's "spectacular!" work.
Whadda hell is the matter with the dollar sign here??
6. ## Re: calculating compound interest with non-linear interest growth
Originally Posted by GhostAccount
The formula I presented in two parts at first finds the continuously compounded yearly rate (geometric average) that has to be converted to a yearly rate
See the following for complete formula
$R (1+i)^n = R(e^{\sum_{k\ =\ 0}^{n-1} (1+r(1+g)^k)})$
Edit: This formula should work as I am short on time now thus I didn't find time to confirm the results
Beautiful.
It should work but there's a slight typo on the exponent of e; $ln$ is missing.This could have been
$R (1+i)^n = R(e^{\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k})$
Must agree with Sir Wilmer; spectacular work indeed.
Was about to work on a derivation for this one when i saw it earlier but got too sober for a day and a half.
Drunk as I am right now, I am nevertheless definitely green with envy for this piece of beauty.
I'm almost certain I'm not the only one who'd like to see the analysis of this derivation Sir GhostAccount.
But as you said, this is one way of doing this. There might be another (or other's).
Might have to soak me brain with me favorite tequila brand for one such other future possibe derivation.
7. ## Re: calculating compound interest with non-linear interest growth
Thanks everyone, I've learned a lot so far!
I did some more research and realized that the correct way to represent the FV formula for the compounded growth of compounded interest is
$FV = R(\prod_{k\ =\ 0}^{n-1} (1+r(1+g)^k)})$
Since FV is essentially R(a)(b)(c)(...) we can convert this to a summation using logs.
Since
$e^{ln(x)} = x$
and
$ln(abc) = ln(a)+ln(b)+ln(c)$
so
$\boxed{FV= R(e^{\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)})}$
$FV=R(1+i)^n$
so
$R(1+i)^n= R(e^{\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)})$
eliminating R and taking the log of both sides we get | {
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eliminating R and taking the log of both sides we get
$nln(1+i) = ({\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)})$
$ln(1+i) = \frac{1}{n}({\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)})$
$1+i = e^{\frac{1}{n}({\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)})}$
and now we're back to the answer GhostAccount initially provided as
$\boxed{i = e^{\frac{1}{n}({\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)})} - 1}$
One last question...is it possible to express
${\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)}$
in a single function f(n) where it outputs the expected FV for n=0, n=1, n=2 etc?
I could do something similar for the normal compounded case, e.g.
$\sum{r(1+g)^k} = \frac{r((1+g)^{(k+1)}-1)}{g}$
but I'm completely lost on how to do a similar conversion for the FV formula or if it's even possible. I'm trying to create an Excel file comparing the result of various investments with different dividend yields and dividend growth rates, so having a single formula in a cell for each value of n would make creating that file much easier.
8. ## Re: calculating compound interest with non-linear interest growth
Whadda hell is the matter with the dollar sign here??
Dollar sign invokes the LaTeX editor. Confused me when I started posting here. See Dollar Signs
9. ## Re: calculating compound interest with non-linear interest growth
I realized I should make my question more specific...
Suppose I have two different dividend paying stocks - one has an initial yield r1 with a growth rate of r1 and the second has an initial yield of r2 with a growth rate of r2.
I'd like to be able to analyze / compare the two performance stocks when I re-invest the dividend payments and answer questions such as:
Q1: With a constant stock price, what is the total return of principal + dividends after time t?
Q2: How many years will either stock take to reach a certain dividend payout? | {
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Q2: How many years will either stock take to reach a certain dividend payout?
I was able to do this for the non-compounding case and I posted an article on my blog explaining the formulae. The formulae and explanations above are really helpful but I'm stuck in transforming the summation formula above to a more usable form. I want to avoid continous compounding but keep to annually compounded calculations.
10. ## Re: calculating compound interest with non-linear interest growth
Originally Posted by tw35758
but I'm completely lost on how to do a similar conversion for the FV formula or if it's even possible. I'm trying to create an Excel file comparing the result of various investments with different dividend yields and dividend growth rates, so having a single formula in a cell for each value of n would make creating that file much easier.
For now, see the following code in Excel to use the tadFVSchedule function that answers the original question
Code:
Public Function tadFVSchedule(ByVal pv As Double, ByVal r As Double, ByVal g As Double, ByVal n As Double, Optional ByRef c As Double = 1, Optional ByRef p As Double = 1, Optional ByRef d As Double = 1) As Double
Dim fraction As Double
Dim sum As Double
Dim k As Integer
sum = 0#
fraction = n - Int(n)
For k = 0 To Int(n - 1)
sum = sum + p * d * Log(1 + r * (1 + g) ^ k)
Next k
If (fraction <> 0) Then
k = Int(n + 0.5)
sum = sum + ((fraction - 1) * p + p * d) * Log(1 + r * (1 + g) ^ k)
End If
End Function | {
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End Function
pv r g n c p d fv 1000 3% 5% 3 1 1 1 1097.59 1000 3% 5% 10 1 1 1 1448.11 1000 3% 5% 3.5 1 1 1 1117.42 1000 3% 5% 10.5 1 1 1 1484.80 1000 3% 5% 3 =1/12 1 1 1097.59 1000 3% 5% 10 =1/12 1 1 1448.11 1000 3% 5% 3 1 =1/12 1 1007.79 1000 3% 5% 10 1 =1/12 1 1031.34 1000 3% 5% 3 =1/12 =1/12 1 1007.79 1000 3% 5% 10 =1/12 =1/12 1 1031.34 1000 3% 5% 3 1 1 =1/2 1047.66 1000 3% 5% 10 1 1 =1/2 1203.38 1000 3% 5% 3.5 1 1 =1/2 1047.66 1000 3% 5% 10.5 1 1 =1/2 1203.38 1000 3% 5% 3 =1/12 1 =1/2 1047.66 1000 3% 5% 10 =1/12 1 =1/2 1203.38 1000 3% 5% 3 1 =1/12 =1/2 1003.89 1000 3% 5% 10 1 =1/12 =1/2 1015.55 1000 3% 5% 3 =1/12 =1/12 =1/2 1003.89 1000 3% 5% 10 =1/12 =1/12 =1/2 1015.55
11. ## Re: calculating compound interest with non-linear interest growth
Thank you GhostAccount! That's great - I can use that macro.
I made a typo in my question - I meant:
Suppose I have two different dividend paying stocks - one has an initial yield r1 with a growth rate of g1 and the second has an initial yield of r2 with a growth rate of g2. The stocks pay dividends annually and the dividends are re-invested to purchase more of the stock at the end of each year.
Q1: With a constant stock price, what is the total return of principal + dividends after time t?
Q2: How many years will either stock take to reach a certain dividend payout?
Thanks again for your help and patience in answering - I've learned a lot so far!
12. ## Re: calculating compound interest with non-linear interest growth
I must be missing something...but, if you're able to use a looper, then isn't that enough?
I love UBasic ([]= notes):
i = .03 : f = 1.05 : b = 1000 : p = 1000 : k = 30 [f = increase factor, k = years]
FOR n = 1 TO k
e = b * i [e = year's earnings]
b = b + e
y = (b / p)^(1 / n) [y = yield so far]
PRINT n, e, b, i * 100, y * 100
i = i * f [next year's rate]
NEXT n | {
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OUTPUT:
Code:
YEAR INTEREST BALANCE YEAR'S RATE YIELD-SO-FAR
0 1000.00
1 30.00 1030.00 3.0000 3.0000
2 32.45 1062.45 3.1500 3.0750
3 35.14 1097.59 3.3075 3.1524
...
10 64.40 1448.11 4.6540 3.7720
...
20 185.22 2628.54 7.5808 4.9507
...
30 749.55 6819.59 12.3484** 6.6085
** .03 * 1.05^29 = 12.3484...
Isn't that all you need?
13. ## Re: calculating compound interest with non-linear interest growth
Hi Wilmer,
You're correct - I can certainly compute the answer using an excel macro. However I'm curious if there's a mathematical solution to this question.
The only mathematical solution I know of would be to approximate the answer using a Taylor series e.g.
$FV = Re^{\sum_{k\ =\ 0}^{n-1}ln({1+r(1+g)^k)}}$
$ln(1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} - {...}$
taking the first 2 terms as an example then
$\sum_{k\ =\ 0}^{n-1}ln({1+x}) \approx \sum_{k\ =\ 0}^{n-1}(x - \frac{x^2}{2})$ where $x = r(1+g)^k$
$\sum_{k\ =\ 0}^{n-1}(x - \frac{x^2}{2}) = \sum_{k\ =\ 0}^{n-1}x - \sum_{k\ =\ 0}^{n-1}\frac{x^2}{2}$
and
$\sum_{k\ =\ 0}^{n-1}x = \sum_{k\ =\ 0}^{n-1}r(1+g)^k = \frac{r((1+g)^n-1)}{g}$
$\sum_{k\ =\ 0}^{n-1}\frac{x^2}{2} = \frac{1}{2}\sum_{k\ =\ 0}^{n-1}r^2(1+g)^{2k} = \frac{r^2((1+g)^{2n}-1)}{2g(g+2)}$
so I could say that
$FV \approx Re^{\frac{r((1+g)^n-1)}{g} - \frac{r^2((1+g)^{2n}-1)}{2g(g+2)}}$
and solve this equation for n with a given FV.
I'm wondering if there's a more elegant equivalent expression of this function out there that doesn't involve approximations.
14. ## Re: calculating compound interest with non-linear interest growth | {
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14. ## Re: calculating compound interest with non-linear interest growth
Originally Posted by tw35758
I'm wondering if there's a more elegant equivalent expression of this function out there that doesn't involve approximations.
After 5 beer bottles and 2 shots of tequila, I've been wondering about that too. You could always repost your questions on an actuarial forum. You might get lucky there. Having seen your analyses, I'm somewhat inclined to believe that you might have some background in actuarial mathematics yourself. If you're lucky, Sir TKHunny, the resident actuary here might get interested in your situation and come out of retirement from this often thankless business of math knight-errantry.
15. ## Re: calculating compound interest with non-linear interest growth
Originally Posted by jonah
After 5 beer bottles and 2 shots of tequila, I've been wondering about that too...
Hey, that was my breakfast 30 years ago, at end of my drinking...!!
Page 1 of 2 12 Last | {
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# Math Help - Trig Integral
1. ## Trig Integral
$\int SEC^4{5x} = \int ({SEC^2{5x}})({SEC^2{5x}})$
Then
$\int (1+TAN^2{5x})(SEC^2{5x})$
Then I multiply out
$\int (SEC^2{5x})+(SEC^2{5x}TAN^2{5x})$
I would assume U sub at this point
$U= SEC dU= SECTAN$
But I am not sure, what or how I should bring along with the SEC
Any help?
thanks
Rich
2. Originally Posted by Spoolx
$\int SEC^4{5x} = \int ({SEC^2{5x}})({SEC^2{5x}})$
Then
$\int (1+TAN^2{5x})(SEC^2{5x})$
Then I multiply out
$\int (SEC^2{5x})(SEC^2{5x}TAN^2{5x})$
Fine so far, except you're missing a plus sign in the last line, which I assume is a typo. Now to substitute, which is really in order to work backwards through the chain rule.
Just in case a picture helps...
... where (key in first spoiler) ...
Spoiler:
... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
The general drift is...
Spoiler:
$\int 1 + u^2\ du = u + \frac{1}{3} u^3$
and so we can explicate F...
Hope this helps - if not, someone will probably show the sub with differentials. However, I say...
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
3. I edited the original to include the + sign, thanks for catching my mistake
4. $\displaystyle\int{sec^4(5x)}dx=\int{sec^2(5x)sec^2 (5x)}dx=\int{\left[1+tan^2(5x)\right]sec^2(5x)}dx$
Substitute
$u=tan(5x)$
$\displaystyle\Rightarrow\frac{du}{dx}=5sec^2(5x)$
by the chain rule
$\displaystyle\Rightarrow\ sec^2(5x)dx=\frac{du}{5}$
The integral becomes
$\displaystyle\frac{1}{5}\int{\left[1+u^2\right]du$
5. Hello, Rich!
A rehash of Archie's solution . . . | {
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5. Hello, Rich!
A rehash of Archie's solution . . .
$\displaystyle \int \sec^4\!5x\,dx \:= \:\int (\sec^2\!5x)(\sec^2\!5x)\,dx$
Then: . $\displaystyle \int (1+\tan^2\!5x)(\sec^2\!5x)\,dx$
Then I multiply out: . $\displaystyle \int (\sec^2\!5x +\sec^2\!5x\tan^2\!5x)\,dx$
$\displaystyle \text{You have: }\;\int\sec^2\!5x\,dx + \int(\tan 5x)^2(\sec^2\!5x\,dx)$
For the first integral, use: . $\displaystyle \int \sec\theta\,d\theta \:=\:\tan\theta + C$
For the second integral, let: . $u \,=\, \tan x \quad\Rightarrow\quad du \,=\,\sec^2\!x\,dx$
. . and you have: . $\displaystyle \int u^2\,du \:=\:\tfrac{1}{3}u^3 + C$
Got it?
6. Originally Posted by Soroban
Hello, Rich!
A rehash of Archie's solution . . .
$\displaystyle \text{You have: }\;\int\sec^2\!5x\,dx + \int(\tan 5x)^2(\sec^2\!5x\,dx)$
For the first integral, use: . $\displaystyle \int \sec\theta\,d\theta \:=\:\tan\theta + C$
For the second integral, let: . $u \,=\, \tan x \quad\Rightarrow\quad du \,=\,\sec^2\!x\,dx$
. . and you have: . $\displaystyle \int u^2\,du \:=\:\tfrac{1}{3}u^3 + C$
Got it?
So to refresh my algebra
$tan^2{5x} = (tan5x)^2$ ?
Also above, is your U supposed to be?
$u= tan5x$ ?
if not where does the 5 go?
7. Originally Posted by Spoolx
$tan^2{5x} = (tan5x)^2$ ? Mr F says: Yes.
$u= tan5x$ ? Mr F says: Yes. Soroban made a simple typo. You ought to be able to make the appropriate corrections in the calculations. | {
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# How to calculate an integral using Cauchy's theorem?
The question is calculate the value of the integral, $$\int_{-\infty}^{\infty}\frac{dx}{1+x^4}$$ These are the steps that I followed,
Let $x$ be a complex number. So, the poles of the function $f(x)=\frac{1}{1+x^4}$ occur when $x$ is equal to the roots of the equation $1+x^4=0$, i.e $x=e^{i\pi/4},e^{3i\pi/4},e^{5i\pi/4},e^{7i\pi/4}$. They are all poles of degree 4.
Now Cauchy's theorem says that, $$\frac{1}{2\pi i}\int_C dx\ f(x)=\sum_i\text{Res}(f,x_i)$$ where $x_i$ are the poles of $f$ that lies within $C$. I am pretty sure that my poles lie within $-\infty$ and $\infty$. So, I calculated the residues of $\frac{1}{1+x^4}$ at $x=e^{i\pi/4},e^{3i\pi/4},e^{5i\pi/4},e^{7i\pi/4}$ and they are equal to $-\frac14e^{i\pi/4},-\frac14e^{3i\pi/4},-\frac14e^{5i\pi/4},-\frac14e^{7i\pi/4}$. Wolfram Alpha confirms my calculations.
The sum of residues is $$\sum_i\text{Res}=-\frac14e^{i\pi/4}-\frac14e^{3i\pi/4}-\frac14e^{5i\pi/4}-\frac14e^{7i\pi/4}=0$$ and therefore the integral, $$\int_{-\infty}^{\infty}\frac{dx}{1+x^4}$$ must be equal to zero. However, Wolfram alpha says it is not zero but equal to$\frac{\pi}{\sqrt 2}$. Where am I making a mistake?
In applying the residue theorem, we analyze the integral $I$ given by
\begin{align} I&=\oint_C \frac{1}{1+z^4}\,dz\\\\ &=\int_{-R}^R \frac{1}{1+x^4}\,dx+\int_0^\pi \frac{1}{1+(Re^{i\phi})^4}\,iRe^{i\phi}\,d\phi\\\\ &=2\pi i \,\text{Res}\left(\frac{1}{1+z^4}, z=e^{i\pi/4},e^{i3\pi/4}\right) \end{align}
where $R>1$ is assumed.
Note that the only residues implicated in the residue theorem are those enclosed by $C$. Here, $C$ is comprised of (i) the line segment from $-R$ to $R$ and (ii) the semicircular arc centered at the origin with radius $R$ and residing in the upper-half plane. Hence, the only resides are the ones at $z=e^{i\pi/4}$ and $z=e^{i3\pi/4}$.
Taking $R\to \infty$, the integration over the semi-circular contour vanishes and we find that | {
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Taking $R\to \infty$, the integration over the semi-circular contour vanishes and we find that
\begin{align} \int_{-\infty}^\infty \frac{1}{1+x^4}\,dx&=2\pi i \,\text{Res}\left(\frac{1}{1+z^4}, z=e^{i\pi/4},e^{i3\pi/4}\right)\\\\ &=2\pi i \left(-\frac{e^{i\pi/4}}{4}-\frac{e^{i3\pi/4}}{4}\right)\\\\ & =\frac{\pi}{\sqrt 2} \end{align}
• Like the other answer, why did you only used the upper half semicircle? Why did you not ignore the first term in the second step and evaluate over the whole circle like, $$\int_0^{2\pi} d\phi\ \frac{iRe^{i\phi}}{1+(Re^{i\phi})^4}$$ – sigsegv Apr 23 '17 at 17:13
• @Ayatana We want to evaluate the integral $\int_{-\infty}^\infty \frac{1}{1+x^4}\,dx$. How does integrating over the circle do that? Well, it doesn't; and that integral is $0$. So instead, we evaluate a contour integral, the one described herein, that has the real line integral as part of the contour. We could have taken the semi-circle in the lower-half plane, but since the orientation would be clockwise, we need a $-2\pi i$ factor o multiply the sum of the residues in the lower-half plane. The result will be the same, of course. ;-)) – Mark Viola Apr 23 '17 at 17:15
• Is it necessary for the contour to be semicircle? Will a contour made of a straight line along real axis and an arbitrary line joing the ends work? – sigsegv Apr 24 '17 at 4:30
• @Ayatana The reside theorem applies to any closed rectifiable curve. But we are free to choose the closed contour. And by choosing a semicircle the radius of which approaches infinity, we can show quite easily that its contribution approaches $0$ as $R\to \infty$. – Mark Viola Apr 24 '17 at 4:34
You shouldn't need Wolfram Alpha to tell you the answer isn't zero. The integrand is positive, so the integral is positive, and certainly nonzero. | {
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You have added up the residues at all the poles. However using the usual semicircle method gives you that the integral is $2\pi i$ times the sum of the residues of the poles in the upper half-plane.
• Why should I not add the residues of the poles below the lower half? What semi circle are you talking about? The limits of the integration are from $-\infty$ to $\infty$ for both real and imaginary parts of $x$. – sigsegv Apr 23 '17 at 16:29 | {
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# Version of Banach-Steinhaus theorem
I am wondering about the following version of the Banach-Steinhaus theorem.
Let $A$ be a closed convex subset contained in the unit ball of a Banach space $X$ and consider bounded operators $T_n \in \mathcal L(X).$
Assume we know that for every $x \in A$ the sequence $\left\lVert T_n x \right\rVert$ is bounded uniformly in $n.$
Does this imply that $$\sup_{x \in A} \left\lVert T_n x \right\rVert$$ is bounded uniformly in $n$?
If $A=X$ then the theorem is undoubtedly true by the folklore Banach-Steinhaus theorem but I was wondering whether this version holds as well?
• In the last paragraph, shouldn't it be "if A is the unit sphere then..." ? – user111 Jul 10 '18 at 7:45
• It is worth noting that the implication does not hold without the closeness assumption. If e.g. $A$ is the convex hull of the standard basis of $X:=\ell_2$ and $\phi_n:=\sum_{k=1}^n ke_k\in X$, then $(\phi_n,x)$ is point-wise but not uniformly bounded for $x\in A$ – Pietro Majer Jul 10 '18 at 15:16
The answer is yes, as a close inspection of the standard proof of the uniform boundedness principle/Banach-Steinhaus theorem shows. The standard proof (or at least the proof which I would consider to be the standard one) can e.g. be found on Wikipedia.
The details are a bit different here, so let me give them below. Throughout, let us replace the sequence $(T_n)$ with a general subset $\mathcal{T} \subseteq \mathcal{L}(X)$.
Proof. By Baire's Theorem we can find an integer $m$ such that the set \begin{align*} B := \{x \in A: \, \|Tx\| \le m \text{ for all } T \in \mathcal{T}\} \end{align*} has non-empty interior within $A$. Thus, we can find a point $x_0 \in B$ and a real number $\varepsilon \in (0,1]$ such that each $x \in A$ which has distance at most $\varepsilon$ to $x_0$ is contained in $B$. | {
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Now, let $y \in A$. The vector $z := x_0 + \frac{\varepsilon}{2}(y-x_0)$ is contained in $A$ due to the convexity of $A$, and its distance to $x_0$ is at most $\varepsilon$ since both $y$ and $x_0$ have norm at most $1$. Thus, $\|Tz\| \le m$ for all $T \in \mathcal{T}$. Since \begin{align*} y = \frac{2}{\varepsilon}(z - x_0) + x_0, \end{align*} we conclude that \begin{align*} \|Ty\| \le \frac{4m}{\varepsilon} + m \end{align*} for all $T \in \mathcal{T}$. This bound does not depend on $y$.
Instead of inspecting the Banach-Steinhaus proof as in Jochen Glueck's answer one can apply Banach-Steinhaus to the Banach space $[A]$ (the linear hull of $A$) endowed with the Minkowski functional $\|x\|_A=\inf\{t>0: x\in tA\}$ (completeness of this norm follows from completeness of the Banach space $X$ and $A=\overline A$).
• A small detail: we need a balanced convex set to get a norm, so we should replace $A$ with $A':= \text{co}(A\cup (-A))=[-1,1]\cdot A$, and check that the hypotheses are still satisfied by $A'$ – Pietro Majer Jul 10 '18 at 7:22
• (the last equality is not true of course) – Pietro Majer Jul 10 '18 at 14:06
I like Jochen Wengenroth's approach, and I think there is a point that it is worth to clarify. If we want to make a norm out of $A$, we need it to be a balanced set, so we'd like to pass to the bounded absolutely convex set $\overline{\operatorname{co}}\left(A\cup(-A)\right)$ or to $A-A$. Any family of linear operators which is point-wise bounded on $A$ is clearly also point-wise bounded on $A-A$. However these sets are in general not closed, so some care is needed, because a bounded absolutely convex not closed set $B$ in general would not produce a Banach disk on its linear span, and in fact in general the statement itself does not hold on such $B$ (see the example in the initial comment). | {
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A cheap solution to make the argument work smoothly is to use the notion of $\sigma$-convexity (see e.g. this MO thread) which also generalize slightly the statement); in particular, it covers both the case of a closed and an open bounded convex set $A$. Recall that for a subset $A$ of a Banach space $X$ the following easy facts hold:
• If $A$ is $\sigma$-convex, it is bounded;
• If $A$ is $\sigma$-convex, $A-A$ is $\sigma$-convex and symmetric (that is, $\sigma$-absolutely convex);
• If $A$ is $\sigma$-absolutely convex, it is a Banach disk, that is, its Minkowski functional is a Banach norm on the linear span of $A$.
As a conclusion, we can follow Jochen Wengenroth's reduction to the standard Banach-Steinhaus theorem. We thus have: Any family of linear operators on a Banach space, which is point-wise bounded on a $\sigma$-convex set $A$, is also uniformly bounded on $A$. | {
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# Optimal strategies for this card game?
I recently had an interesting card game for an interview.
The interviewer and I have two cards each, one with a '5' painted on it and one with a '10'. We pick a card each and show it at the same time. If we picked the same card, I receive nothing, while if we picked different cards, the interviewer pays me the number he picked in dollars. What is the interviewer most likely to do if he intends to minimise my payoff, and what should my strategy be to counter this?
Consideration 1: I pick '5' all the time, but if the interviewer knows I do this he would pick '5' as well so that I receive nothing and my expected payoff would just be zero.
Consideration 2: I assign a probability $$p_1$$ of picking a '5' (and $$1-p_1$$ of picking a '10'), and the interviewer assigns a probability $$p_2$$ of picking a '5' (and $$1-p_2$$ of picking a '10'). My payoff would then be
$$\mathbb{E}=10p_1(1-p_2)+5(1-p_1)p_2.$$
I was thinking of differentiation somehow, but $$p_1$$ aims to maximise $$\mathbb{E}$$ while $$p_2$$ aims to minimise $$\mathbb{E}$$. Is this even the correct strategy for both of us?
Consideration 3: A friend suggested assigning probabilities of $$\frac13$$ and $$\frac23$$ to the cards respectively for a Nash equilibrium since it is a symmetric game. Where does this come from intuitively? Do these probabilities match with the above equation?
Any help/comments on the considerations is greatly appreciated, cheers!
The logic of equilibrium here is: you should choose your strategy (i.e. determine your probabilities of picking $$5$$ and $$10$$) in such a way that, however your opponent chooses his strategy, your expected gain is not influenced. | {
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Let's say you pick $$5$$ with probability $$p_1$$ and your opponent picks $$5$$ with probability $$p_2$$. Then, once you fixed your choice of $$p_1$$, the expectation $$\mathbb{E}$$ is a linear function in $$p_2$$: $$\mathbb{E} = 10p_1 + (5 - 15p_1)p_2.$$ In this situation, you see that choosing $$p_1 = 1/3$$ will guarantee an expectation of $$10/3$$, regardless of $$p_2$$.
If, let's say, you choose $$p_1 < 1/3$$. Then there is the "risk" that your opponent somehow knows your strategy (e.g. via statistics from multiple rounds of the game), and then chooses $$p_2 = 0$$. You then have $$\mathbb{E} = 10p_1 < 10/3$$.
If, in the other direction, you choose $$p_1 > 1/3$$. Then again assume your opponent gets to know your strategy, and then chooses $$p_2 = 1$$. You then have $$\mathbb{E} = 5 - 5p_1 < 10/3$$.
Therefore, you see that the "safest" way of playing is to choose $$p_1 = 1/3$$. It is in that sense that we say it's the "optimal strategy".
I hope this whole logic makes sense to you.
And as an exercise, you may try to solve the general situation: say the two cards are labelled A and B, and your gain has $$4$$ possibilities: $$a, b, c, d$$, which correspond to the $$4$$ combinations AA, AB, BA, BB of your cards. Then what is your optimal strategy?
• Hello, what are some general approaches to formulate equilibrium for a game? – user107224 Nov 13 '19 at 9:56
• For the general equilibrium, the wiki page en.m.wikipedia.org/wiki/Nash_equilibrium may be a good starting point. Also note that in general there could be more than one equilibria. – WhatsUp Nov 13 '19 at 10:38 | {
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# Axis/Angle from rotation matrix
With r=RotationMatrix[a,{x,y,z}] I can compute a 3D rotation matrix from axis/angle representation.
Given a 3D rotation matrix r how can I compute a, x, y, z?
Example:
r = {{0.966496, -0.214612, 0.14081}, {0.241415, 0.946393, -0.214612}, {-0.0872034, 0.241415, 0.966496}}
The result should be a=20. Degree and {x,y,z}={2, 1, 2}/3 (or equivalent).
Edit: I am fine with any answer that gives the same r when applied to RotationMatrix.
PS: Sorry if the tag is wrong, I did not found a matching tag...
-
I'm not sure this question is well posed in the sense that the answer is non-unique. A simple example would be r=RotationMatrix[1,{1,0,0}]=RotationMatrix[-1,{-1,0,0}] are the same. So which one are you intending to find? – bill s Aug 6 '13 at 13:08
The one where the angle is in [0,2 pi]. I think it is well posed as angle and axis are directly linked and you can provide a unique answer with a perfectly reasonable additional assumption. – Danvil Aug 6 '13 at 13:43
Actually there was an answer earlier which was just fine but it was deleted? – Danvil Aug 6 '13 at 13:46
@Danvil it was deleted because it was incorrect, so he deleted it. It was incorrect, as it worked only for the example you provided and could not be generalized. – rcollyer Aug 6 '13 at 14:42
@Danvil I'm thankful I noticed the wink. It would have been a bad day, otherwise. :P – rcollyer Aug 6 '13 at 16:38
There is no need to use Eigensystem or Eigenvectors to find the axis of a rotation matrix. Instead, you can read the axis vector components off directly from the skew-symmetric matrix
$$a \equiv R^T-R$$
In three dimensions (which is assumed in the question), applying this matrix to a vector is equivalent to applying a cross product with a vector made up of the three independent components of $a$:
{1, -1, 1}Extract[a, {{3, 2}, {3, 1}, {2, 1}}] | {
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{1, -1, 1}Extract[a, {{3, 2}, {3, 1}, {2, 1}}]
This one-line method of finding the axis is applied in the following function. To get the angle of rotation, I construct two vectors ovec, nvec perpendicular to the axis and to each other, to find the cosine and sine of the angle using the Dot product (could equally have used Projection). To get a first vector ovec that is not parallel to the axis, I permute the components of the axis vector using the fact that
Solve[{x, -y, z} == {y, z, x}, {x, y, z}]
(* ==> {{x -> 0, y -> 0, z -> 0}} *)
which means the above permutation with sign change of a nonzero axis vector is always different from the axis. This is sufficient to use Orthogonalize and Cross to get the desired orthogonal vectors.
axisAngle[m_] := Module[
{axis, ovec, nvec
},
{axis, ovec} =
Orthogonalize[{{1, -1, 1} #, Permute[#, Cycles[{{1, 3, 2}}]]}] &@
Extract[m - Transpose[m], {{3, 2}, {3, 1}, {2, 1}}];
(* nvec is orthogonal to axis and ovec: *)
nvec = Cross[axis, ovec];
{axis,
Arg[Complex @@ (((m.ovec).# &) /@ {ovec, nvec})]}
]
The angle is calculated with Arg instead of ArcTan[x, y] here because the latter throws an error for x = y = 0.
Here I test the results of the function for 100 random rotation matrices:
testRotation[] := Module[
{m, a, axis, ovec, nvec,
v = Normalize[RandomReal[{0, 1}, {3}]],
α = RandomReal[{-Pi, Pi}], angle
},
m = RotationMatrix[α, v];
{axis, angle} = axisAngle[m];
Chop[
angle Dot[v, axis] - α
] === 0
]
And @@ Table[testRotation[], {100}]
(* ==> True *)
In the test, I have to account for the fact that if the function axisAngle defined the axis vector with the opposite sign as the random test vector, I have to reverse the sign of the rotation angle. This is what the factor Dot[v, axis] does.
Explanation of how the axis results from a skew-symmetric matrix | {
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Explanation of how the axis results from a skew-symmetric matrix
If $\vec{v}$ is the axis of rotation matrix $R$, then we have both $R\vec{v} = \vec{v}$ and $R^T\vec{v} = \vec{v}$ because $R^T$ is just the inverse rotation. Therefore, with $a \equiv R^T-R$ as above, we get
$$a \vec{v} = \vec{0}$$
Now the skew-symmetric property $a^T = -a$, which can be seen from its definition, means there are exactly three independent matrix element in $a$. They can be arranged in the form of a 3D vector $\vec{w}$ which must have the property $a \vec{w} = 0$. This vector is obtained in the Extract line above. In fact, $a \vec{x} = \vec{w}\times \vec{x}$ for all $\vec{x}$, and hence if $a \vec{x} = 0$ then $\vec{x}\parallel\vec{w}$. Therefore, the vector $\vec{v}$ is also parallel to $\vec{w}$, and the latter is a valid representation of the rotation axis.
Edit 2: speed considerations
Since the algorithm above involves only elementary operations that can be compiled, it makes sense that a practical application of this approach would use Compile. Then the function could be defined as follows (keeping the return values arranged as above):
Clear[axisAngle1, axisAngle]
axisAngle1 =
Compile[{{m, _Real, 2}},
Module[{axis, ovec, nvec, tvec, w, w1},
tvec = {m[[3, 2]] - m[[2, 3]], m[[3, 1]] - m[[1, 3]],
m[[2, 1]] - m[[1, 2]]};
If[tvec == {0., 0., 0.},
{#/Sqrt[#.#] &[#[[Last@Ordering[N[Abs[#]]]]] &[
1/2 (m + {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}})]],
If[Sum[m[[i, i]], {i, 3}] == 3, 0, Pi] {1, 1, 1}},
axis = {1, -1, 1} tvec;
axis = axis/Sqrt[axis.axis];
w = {tvec[[2]], tvec[[3]], tvec[[1]]};
ovec = w - axis Dot[w, axis];
nvec = Cross[axis, ovec];
w1 = m.ovec;
{axis, {1, 1, 1} ArcTan[w1.ovec, w1.nvec]}
]
]
];
axisAngle[m_] := {#1, Last[#2]} & @@ axisAngle1[m]
The results are the same as for the previous definition of axisAngle, but I now get a much faster execution as can be seen in this test: | {
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tab = RotationMatrix @@ # & /@
Table[{RandomReal[{-Pi, Pi}],
Normalize[RandomReal[{0, 1}, {3}]]}, {100}];
timeAvg =
Function[func,
Do[If[# > 0.3, Return[#/5^i]] & @@ Timing@Do[func, {5^i}], {i, 0,
15}], HoldFirst];
timeAvg[axisAngle /@ tab]
(* ==> 0.000801259 *)
This is more than an order of magnitude faster than the un-compiled version. I removed Orthogonalize from the code because I didn't find it in the list of compilable functions.
Note that Eigensystem is not in that list, either.
Edit 3
The first version of axisAngle demonstrated the basic math, but the compiled version axisAngle1 (together with the re-defined axisAngle as a wrapper) is faster. One thing that was missing was the correct treatment of the edge case where the rotation is by exactly $\pi$ in angle.
I added that fix only to the compiled version (axisAngle1) because I think that's the more practical version anyway. The trivial case of zero rotation angle was already included in the earlier version.
To explain the added code, first note that for angle $\pi$ you can't read off the axis from $R^T - R$ because the resulting matrix vanishes. To get around this singular case, we can use the geometric fact that a rotation by $\pi$ is equivalent to an inversion in the plane perpendicular to the rotation axis given by the unit vector $\vec{n}$. Therefore, if we form the sum of a vector $\vec{v}$ and its $\pi$-rotated counterpart, the components transverse to the rotation axis cancel and the result is always parallel to the axis. In matrix form,
$$(R+1)\vec{v} = 2\vec{n}(\vec{n}\cdot\vec{v}) = 2\left(\vec{n}\vec{n}^T\right)\vec{v}$$
Since this holds for all vectors, it is a matrix identity. The right-hand side contains a matrix $\vec{n}\vec{n}^T$ which must have at least one row that's nonzero. This row is proportional to $\vec{n}^T$, so you can read of the axis vector directly from $(R+1)$, again without any eigenvalue computations. | {
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-
Beautiful! (+1) My hamfisted approach to this has been an opportunity to learn. – ubpdqn Aug 6 '13 at 22:28
Note (as you have outlined) you could also obtain the axis from NullSpace[m-Transpose[m]] – ubpdqn Aug 7 '13 at 4:17
@ubpdqn True, but that takes more computation... as I said, you can read the axis off literally from the matrix a itself. – Jens Aug 7 '13 at 5:15
Actually for cases $\theta=0$ and $\theta=\pi$ the $R-R^T$ is a zero matrix and the axis is thus maps to {{0,0,0},0) in your code. The test set always skirts these "points". The eigensystem allows extraction of the eigenvector associated with eigenvalue 1). I agree with lack of simplicity or elegance using the eigensystem. I found it a lazy way to deal with these anomalous cases: integer*$\pi$. I think my code works now. Passes your test (up orientation of axis issue) and the "extreme" cases. – ubpdqn Aug 7 '13 at 5:29
Maybe the easiest way to deal with the isotropic cases ($\pm$IdentityMatrix[3]) would be to append the rule axis /. {0, 0, 0} -> {0, 0, 1} to the result. The choice of rotation axis is arbitrary in that case and I'd fix it to be the z axis. – Jens Aug 7 '13 at 5:41
Since this question still seems to be alive after some time, I'm giving a solution from Presentations, which I sell. It has a routine RotationAngleAndAxis and here it is used on the example.
<< Presentations
r = {{0.966496, -0.214612, 0.14081}, {0.241415,
0.946393, -0.214612}, {-0.0872034, 0.241415, 0.966496}};
RotationAngleAndAxis[r]
(* {0.349066, {0.666667, 0.333333, 0.666667}} *)
Here is a short example of its use that checks the result. The axis vector is normalized. Notice that reversing the angle of rotation and the axis gives an equivalent rotation.
rotations =
Table[{RandomReal[{-\[Pi], \[Pi]}],
Normalize[Array[RandomReal[{-1, 1}] &, {3}]]}, {2}]
(* {{2.90598, {-0.596373, -0.74938,
0.287697}}, {-2.44158, {0.331763, -0.943343, 0.00605196}}} *) | {
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(* {{2.90598, {-0.596373, -0.74938,
0.287697}}, {-2.44158, {0.331763, -0.943343, 0.00605196}}} *)
rotationMatrices = RotationMatrix @@@ rotations;
anglesAndAxes = RotationAngleAndAxis /@ rotationMatrices
(* {{2.90598, {-0.596373, -0.74938, 0.287697}}, {2.44158, {-0.331763,
0.943343, -0.00605196}}} *)
resultingMatrices = RotationMatrix @@@ anglesAndAxes;
rotationMatrices - resultingMatrices // Chop
Total[Flatten[%]]
(* {{{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}, {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}} *)
(* 0 *)
The following is a test on 1000 initial rotation specifications. Notice also that I am comparing the resulting rotation matrices by a slightly risky procedure.
rotations =
Table[{RandomReal[{-\[Pi], \[Pi]}],
Normalize[Array[RandomReal[{-1, 1}] &, {3}]]}, {1000}];
rotationMatrices = RotationMatrix @@@ rotations;
Timing[anglesAndAxes = RotationAngleAndAxis /@ rotationMatrices;]
resultingMatrices = RotationMatrix @@@ anglesAndAxes;
Total[Flatten[rotationMatrices - resultingMatrices // Chop]]
(* {0.873606, Null} *)
(* 0 *)
Presentations also has an EulerAngles routine that will return the two sets of Euler angles for any of the possible sequence of rotations specified by strings. For example, here are the two sets of Euler angles of the example for two different rotation sequences.
EulerAngles[r, "ZXZ"]
EulerAngles[r, "XYZ"]
(* {{-0.346633, 0.259587, 0.580661}, {2.79496, -0.259587, -2.56093}} *)
(* {{0.244775, 0.0873143, 0.244775}, {-2.89682, 3.05428, -2.89682}} *)
-
The axis of rotation $n$ can be obtained as the eigenvector corresponding to the eigenvalue $1$.
r = {{0.966496, -0.214612, 0.14081}, {0.241415,
0.946393, -0.214612}, {-0.0872034, 0.241415, 0.966496}};
es = Eigensystem[r];
pos = Nearest[Thread[es[[1]] -> Range[3]], 1][[1]];
n = es[[2, pos]]
(* {0.666667, 0.333333, 0.666666} *)
The angle of rotation can be obtained directly from the trace of $r$.
ArcCos[(Tr[r] - 1)/2] 180/π
(* 20. *) | {
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ArcCos[(Tr[r] - 1)/2] 180/π
(* 20. *)
(This was my earlier and longer way to obtain the angle of rotation. As rcollyer pointed out this will have issues if $\pi k,k\in \mathcal{Z}$)
The rotation matrix in terms of the angle of rotation $\theta$ and axis of rotation $\{n1, n2, n3\}$.
R[t_, {n1_, n2_, n3_}] :=
With[{n = {{0, -n3, n2}, {n3, 0, -n1}, {-n2, n1, 0}}, id = IdentityMatrix[3]},
Inverse[id - t n].(id + t n)]
Here $t =\tan \left(\frac{\theta }{2}\right)$.
We get three solutions from the element in the first position.
eqns = Flatten[R[t, es[[2, pos]]] - r] // Simplify;
sols = NSolve[eqns[[1]] == 0, t]
(* {{t -> 3.88248*10^16}, {t -> 0.176327}, {t -> -0.176327}} *)
Choose the one which satisfies all the entries.
Chop[eqns /. sols, 10^-5];
Position[%, {0 ..}];
solst = Extract[sols, %]
(* {{t -> 0.176327}} *)
Obtain it in terms of degrees.
2 ArcTan[t] 180/π /. solst
(* {20.} *) | {
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Obtain it in terms of degrees.
2 ArcTan[t] 180/π /. solst
(* {20.} *)
-
It doesn't seem to work with a rotation of $\pi$ about the {0,0,1}. – rcollyer Aug 6 '13 at 14:39
@rcollyer, that is to be expected because the rotation group $SO(3)$ cannot be covered by a single chart. Each chart will have its cases that are problematic. I used this parametrization and it worked for the OP's question. Still using this and going with sols = NSolve[eqns[[2]] == 0, t] we get $t=0$ for your case. The technique I used for the axis of rotation should be fine for all cases, however. – Suba Thomas Aug 6 '13 at 14:55
True, but I think it is amenable to decomposition into the generators of rotation, i.e. the rotations should have eigenvalues of the form $\pmatrix{e^{i \theta} & 1 & e^{-i \theta}}$, or for real rotations $\pmatrix{\cos\theta + i \sin\theta & 1 & \cos\theta - i \sin\theta}$. Thus, the angle is easily recoverable. As to the direction, I have not yet looked through your code. – rcollyer Aug 6 '13 at 15:05
Thanks for pointing that out. With your approach the angle of rotation is ambiguous for the OP's question, i.e we don't know if it is +20 or -20, whereas with the parametrization I used only + 20 satisfied all conditions. Probably quaternions will give a unique correct solution, but I do not have the material before me now. – Suba Thomas Aug 6 '13 at 15:20
Well you can choose the angle to be between $0$ and $2 \pi$, to fix the branch. – rcollyer Aug 6 '13 at 15:25
My +1 to Jens.
For known Euler angles:
{Φ, Θ, Ψ} = {0.1, 0.2, 0.3};
m = RotationMatrix[Φ, {0, 0, 1}].RotationMatrix[Θ, {1, 0,
0}].RotationMatrix[Ψ, {0, 0, 1}];
axisAngle[m_] :=Module[{axis, ovec, nvec}, {axis, ovec} =
Orthogonalize[{{1, -1, 1} #, Permute[#, Cycles[{{1, 3, 2}}]]}] &@
Extract[m - Transpose[m], {{3, 2}, {3, 1}, {2, 1}}];
nvec = Cross[axis, ovec];
{Arg[Complex @@ (((m.ovec).# &) /@ {ovec, nvec})], axis}];
N[axisAngle[m]]
(* {0.446615, {0.448552, -0.0450053, 0.892623}} *)
`
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The Factorial of number is the product of all the numbers less than or equal to that number & greater than 0. These while loops will calculate the Factorial of a number.. Factorial of a positive integer is the product of an integer and all the integers below it, i.e., the factorial of number n (represented by n!) factorial: The factorial, symbolized by an exclamation mark (! So, first negative integer factorial is $$-1! If you still prefer writing your own function to get the factorial then this section is for you. This loop will exit when the value of ‘n‘ will be ‘0‘. We are printing the factorial value when it ends. For n=0, 0! Factorial (n!) = 1*2*3*4* . n! Factorial of 0. A method which calls itself is called a recursive method. Since 0 is not a positive integer, as per convention, the factorial of 0 is defined to be itself. There are many explanations for this like for n! For example, the factorial of 6 is 1*2*3*4*5*6 = 720. = 120. In mathematics, the factorial of a number (that cannot be negative and must be an integer) n, denoted by n!, is the product of all positive integers less than or equal to n. = 1 if n = 0 or n = 1 = 1. would be given by n! = 1$$. But I can tell you the factorial of half (½) is half of the square root of pi. The factorial is normally used in Combinations and Permutations (mathematics). n! > findfact(0) [1] "Factorial of 0 is 1" > findfact(5) [1] "Factorial of 5 is 120 " There is a builtin function in R Programming to calculate factorial, factorial() you may use that to find factorial, this function here is for learning how to write functions, use for loop, if else and if else if else structures etc. symbol. factorial of n (n!) We can find the factorial of any number which is greater than or equal to 0(Zero). Can we have factorials for numbers like 0.5 or −3.217? For negative integers, factorials are not defined. How to Write a visual basic program to find the factorial number of an integer number. But we need to get into a subject | {
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basic program to find the factorial number of an integer number. But we need to get into a subject called the "Gamma Function", which is beyond this page. 0!=1. For negative integers, factorials are not defined. October 22, 2020 . = -1! Similarly, by using this technique we can calculate the factorial of any positive integer. So, for the factorial calculation it is important to remember that $$1! Programming, Math, Science, and Culture will be discussed here. . Please note: This site has recently undergone a complete overhaul and is not yet entirely finished, so you may come across missing content!. = 1. The factorial formula. . and 0! Yes, there is a famous function, the gamma function G(z), which extends factorials to real and even complex numbers. This program for factorial allows the user to enter any integer value. Factorial Program in C++: Factorial of n is the product of all positive descending integers. 0! It is easy to observe, using a calculator, that the factorial of a number grows in an almost exponential way; in other words, it grows very quickly. Wondering what zero-factorial … Read more: What is Null in Python. Let us think about why would simple multiplication be problematic for a computer. The factorial symbol is the exclamation mark !. The factorial of n is denoted by n! There are several motivations for this definition: For n = 0, the definition of n! This site is dedicated to the pursuit of information. There are multiple ways to … Factorial definition, the product of a given positive integer multiplied by all lesser positive integers: The quantity four factorial (4!) *n. The factorial of 0 is defined to be 1 and is not defined for negative integers. Are you confused about how to do factorial in vb 6.0 then don’t worry! Factorial of a Number using Command Line Argment Program. Factorial zero is defined as equal to 1. And, the factorial of 0 is 1. The factorial of 0 is always 1 and the factorial of a … Factorial definition formula According do the | {
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factorial of 0 is always 1 and the factorial of a … Factorial definition formula According do the definition of factorial, 1 = 0! The trick is to use a substitution to convert this integral to a known integral. Welcome. in your calculator to see what the factorial of one-half is. = 1$$ and $$0! Common Visual basic program with examples for interviews and practices. And they can also be negative (except for integers). Source Code # Python program to find the factorial of a number provided by the user. Half Factorial. = \frac{√\pi}2$$ How to go about calculating the integral? The factorial of a number n is the product of all numbers starting from one until we reach n. The operation is denoted by an exclamation mark succeeding the number whose factorial we wish to seek, such that the factorial of n is represented by n!. Logically $$1! Factorial using Non-Recursive Program. Below is the Program to write a factorial program in Visual basic. = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24. = ∫_0^∞ x^{1/2}e^{-x}\,dx$$ We will show that: $$(1/2)! =1. = 1$$, but this is adopted as a convention. It does not seem that logical that $$0! Example of both of these are given as follows. Type 0.5! and calculated by the product of integer numbers from 1 to n. For n>0, n! is pronounced as "4 factorial", it is also called "4 bang" or "4 shriek". is 1, according to the convention for an empty product. where n=0 signifies product of no numbers and it is equal to the multiplicative entity. So 0! See more. It is denoted with a (!) By using this value, this Java program finds Factorial of a number using the For Loop. Factorial of a number is the product of all numbers starting from 1 up to the number itself. The factorial for 0 is equal to 1. The factorial is normally used in Combinations and Permutations (mathematics). = 1 * 2 * 3 * 4....n The factorial of a negative number doesn't exist. Yes we can! I am not sure why it should be a negative infinity. Computing this is an interesting problem. Recursive Solution: | {
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why it should be a negative infinity. Computing this is an interesting problem. Recursive Solution: Factorial can be calculated using following recursive formula. Writing a custom function for getting factorial. Here are some "half-integer" factorials: The factorial of a number is the product of all the integers from 1 to that number. Factorial of n is denoted by n!. Factorial of a non-negative integer, is multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is 720. = n * (n-1)! The factorial value of 0 is by definition equal to 1. While calculating the product of all the numbers, the number is itself included. 5! For example: Here, 4! The factorial formula. First, we use integration by … I cannot derive the sign. Factorial of n is denoted by n!. . The best answer I can give you right now is that, like I've mentioned in my answer, \Gamma was not defined to generalize factorials. The factorial of a positive integer n is equal to 1*2*3*...n. Factorial of a negative number does not exist. as a product involves the product of no numbers at all, and so is an example of the broader convention that the product of no factors is equal … Welcome to 0! For example: The factorial of 5 is 120. is pronounced as "5 factorial", it is also called "5 bang" or "5 shriek". \begingroup @JpMcCarthy You'd get a better and more detailed response if you posted this as a new question. Symbol:n!, where n is the given integer. = 1$$. A for loop can be used to find the factorial … For example: Here, 5! just in terms of the meaning of factorial because you cannot multiply all the numbers from zero down to 1 to get 1. Factorials are commonly encountered in the evaluation of permutations and combinations and in the coefficients of terms … The factorial of an integer can be found using a recursive program or a non-recursive program. Mathematicians *define* x^0 = 1 in order to make the laws of exponents work even when the exponents can … If | {
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