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http://math.stackexchange.com/questions/60506/how-to-derive-a-general-formula-for-this-problem-pairs-of-people-seated-around | # How to derive a general formula for this problem? (pairs of people seated around a table)
$N$ people attend a dinner party and sit round a circular table.Each person knows only the people sitting immediately next to him and has to be introduced to everyone else.If the total number of pairs of people introduced to each other is $20$,then what is the value of $N$?
Is this a form of any well known problem?I could only of think of a brute-force solution,however I am inquisitive to know how to solve this for any number of pairs in an efficient manner.
ADDED:It seems like I have understood the pattern,for any $N$ the number of pairs of introduction is given by $(N-3)+(N-3)+(N-3)-1+(N-3)-2+\cdots+2+1$ by using this idea the solution of the above problem should be $N=8$.
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Each person is introduced to $N-3$ others. As there are $N$ people, that would make $N(N-3)$ introductions, but each has been counted twice-once from each side. So we have 20=$\frac{N(N-3)}{2}$, which has roots $8, -5$.
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I did the same in a bit different way:$20 = 1+2+3+\cdots+(N-3)+(N-3)\Rightarrow (N-8)(N+5)=0$... hence the answer. – Quixotic Aug 29 '11 at 12:40
$+1$ and accepted.Reading your answer makes me feel a bit silly:P,I generated answers for first few $N$'s and tried to identify the pattern but all I am supposed to do just to think reasonably! – Quixotic Aug 29 '11 at 12:45
@FoolForMath: I don't see how you take the sum to $(N-8)(N+5)$-the sum from $1$ to $N-3$ is $(N-2)(N-3)/2$ and adding the last $N+3$ takes us to $N(N-3)/2$. The idea of counting each intro works sometimes, but here it is a bit hard to convince yourself you have the endgame right. – Ross Millikan Aug 29 '11 at 12:50
Okay,so $$1+2+3+\cdots+(N-3)+(N-3)=20 \implies\frac{N(N-3)}{2}=20 \implies$$ $$N^2-3N-40=0 \implies(N-8)(N-5)=0$$ (Hoping I haven't committed any mistake in algebra! ) – Quixotic Aug 29 '11 at 15:08
@FoolForMath: I was looking a step earlier. You are right. – Ross Millikan Aug 29 '11 at 19:22
Let $x$ be the number of introductions.
Suppose that everyone must be introduced to everyone else. Do this as follows. (i) Everyone is introduced to her left neighbour ($N$ introductions); (ii) Every pair of non-neighbours is introduced ($x$ introductions).
It follows that $$\binom{N}{2}=N+x,\qquad\text{and therefore}\qquad x=\binom{N}{2}-N=\frac{N^2-3N}{2}.$$
Now we solve the equation $$\frac{N^2-3N}{2}=20.$$ The only admissible solution is $N=8$.
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Total introductions needed(if nobody knows anyone)=$n\choose 2$; but since a person knows the person next to him, therefore introductions needed=${n\choose 2}-n$ which is given to be $20\implies \frac{n(n-1)}{2} - n=20\implies n^2-3n-40=0\implies n=8$.
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$nC2 - n = 20$
$n(n - 1)/2 - n = 20$
$n(n - 1) - 2n = 40$
$n^2 - 3n - 40 = 0$ $n^2 - 8n + 5n - 40 = 0$ $n(n - 8) + 5(n - 8) = 0$
$(n - 8)(n + 5) = 0$
$n = 8$ or $n = -5$
now n is positive and greater than $3$ ($3$ people needs no introduction) , so we take $n = 8$
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For 4 people around a table we have 4c2 -4 = 2 i.e. 2 introductions. – Rajesh K Singh Jul 11 '12 at 12:58
An introduction involves 2 person. nC2 is the number of introductions among n people taken 2 at a time. – Rajesh K Singh Jul 11 '12 at 13:17
An introduction involves 2 person. nC2 is the number of introductions among n people taken 2 at a time. Here (A introduced to B) and (B introduced to A) is the same introduction but counted twice in nC2. That is in all n introductions have been counted twice as every one needs an introduction if n > 3. To eliminate these n false introductions we subtract n from nC2. – Rajesh K Singh Jul 11 '12 at 13:29 | 2016-07-23T21:19:28 | {
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https://ootycottages.com/dc95nd/c3367e-definite-integral-properties | These properties, along with the rules of integration that we examine later in this chapter, help us manipulate expressions to evaluate definite integrals. a and b (called limits, bounds or boundaries) are put at the bottom and top of the "S", like this: Definite Integral (from a to b) Indefinite Integral (no specific values) We find the Definite Integral by calculating the Indefinite Integral at a, and at b, then subtracting: Example: What is 2 ∫ 1. Some properties we can see by looking at graphs. 7.1.4 Some properties of indefinite integrals (i) The process of differentiation and integration are inverse of each other, i.e., () d f dx fx x dx ∫ = and ∫f dx f'() ()x x= +C , where C is any arbitrary constant. This is a very simple proof. Hence, a∫af(a)da = 0. This video explains how to find definite integrals using properties of definite integrals. Properties of Definite Integrals We have seen that the definite integral, the limit of a Riemann sum, can be interpreted as the area under a curve (i.e., between the curve and the horizontal axis). These properties, along with the rules of integration that we examine later in this chapter, help us manipulate expressions to evaluate definite integrals. Properties Of Definite Integral 5 1 = − The definite integral of 1 is equal to the length of interval of the integral.i. Therefore, equation (11) becomes, And, if ‘f’ is an odd function, then f(–a) = – f(a). Properties of Definite Integral Definite integral is part of integral or anti-derivative from which we get fixed answer rather than the range of answer or indefinite answers. Example 9: Given that find all c values that satisfy the Mean Value Theorem for the given function … cos x)/(2 sinx cos x)]dx, Cancel the terms which are common in both numerator and denominator, then we get, I = 0∫π/2 (log1-log 2)dx [Since, log (a/b) = log a- log b]. In mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. Properties of Definite Integral: 5. The definite integral f(k) is a number that denotes area under the curve f(k) from k = a and k = b. If . Properties of definite integral. Properties Of Definite Integral 5 1 = − The definite integral of 1 is equal to the length of interval of the integral.i. Question 7 : 2I = 0. It has an upper limit and lower limit and it gives a definite answer. One application of the definite integral is finding displacement when given a velocity function. Proof of : $$\int{{k\,f\left( x \right)\,dx}} = k\int{{f\left( x \right)\,dx}}$$ where $$k$$ is any number. Properties of Definite Integrals - I. It is mandatory to procure user consent prior to running these cookies on your website. The limits can be interchanged on any definite integral. The definite integral is defined to be exactly the limit and summation that we looked at in the last section to find the net area between a function and the $$x$$-axis. Integrals may represent the (signed) area of a region, the accumulated value of a function changing over time, or the quantity of an item given its density. Suppose that is the velocity at time of a particle moving along the … Reversing the interval property These properties are used in this section to help understand functions that are defined by integrals. If the integral above were to be used to compute a definite integral between −1 and 1, one would get the wrong answer 0. Definite integrals also have properties that relate to the limits of integration. This is useful when is not continuous in [a, b] because we can break up the integral into several integrals at the points of discontinuity so that the function is continuous in the sub-intervals. There are two types of Integrals namely, definite integral and indefinite integral. Property 2: p∫q f(a) d(a) = – q∫p f(a) d(a), Also p∫p f(a) d(a) = 0. Properties of Indefinite Integrals. If . For some functions there are shortcuts to integration. Property 2 : If the limits of definite integral are interchanged, then the value of integral changes its sign only. It contains an applet where you can explore this concept. Required fields are marked *. The definite integral is defined as an integral with two specified limits called the upper and the lower limit. Question 3 : Question 4 : The function f(x) is even. See more about the above expression in Fundamental Theorem of Calculus. In cases where you’re more focused on data visualizations and data analysis, integrals may not be necessary. Definite integral properties (no graph): breaking interval Our mission is to provide a free, world-class education to anyone, anywhere. Here we have: is the area bounded by the -axis, the lines and and the part of the graph of , where . 2. We begin by reconsidering the ap-plication that motivated the definition of this mathe-matical concept- determining the area of a region in the xy-plane. Fundamental Theorem of Calculus 2. Now, let us evaluate Definite Integral through a problem sum. = 1 - (1/2) [-1/3+1] = 1-(1/2)[2/3] = 1-(1/3) = 2/3. Properties of Definite Integrals Proofs. In Mathematics, there are many definite integral formulas and properties that are used frequently. Where, I1 =$$\int_{-a}^{0}$$f(a)da, I2 =$$\int_{0}^{p}$$f(a)da, Let, t = -a or a = -t, so that dt = -dx … (10). Definite Integral and Properties of Definite Integral. Warming Up . Properties of Definite Integral: 6. There are many definite integral formulas and properties. You also have the option to opt-out of these cookies. Certain properties are useful in solving problems requiring the application of the definite integral. A constant factor can be moved across the integral sign.ii. A constant factor can be moved across the integral sign.ii. Introduction-Definite Integral. The properties of indefinite integrals apply to definite integrals as well. For example, we know that integraldisplay 2 0 f ( x ) dx = 2 when f ( x ) = 1, because the value of the inte- gral is the area of a rectangle of height 1 and base length 2. It is just the opposite process of differentiation. The most important basic concepts in calculus are: Properties of the Definite Integral The following properties are easy to check: Theorem. Question 5 : The function f(x) is even. This can be done by simple adding a minus sign on the integral. It is represented as; Following is the list of definite integrals in the tabular form which is easy to read and understand. 3. , where c is a constant . If f’ is the anti-derivative of f, then use the second fundamental theorem of calculus, to get; p∫r f(a)daf(a)da + r∫q f(a)daf(a)da = f’(r) – f’(p) + f’(q), Property 4: p∫q f(a) d(a) = p∫q f( p + q – a) d(a), Let, t = (p+q-a), or a = (p+q – t), so that dt = – da … (4). Sum Rule: 6. ; Distance interpretation of the integral. Hence, $$\int_{-a}^{0}$$ will be replaced by $$\int_{a}^{0}$$ when we replace a by t. Therefore, I1 = $$\int_{-a}^{0}$$f(a)da = – $$\int_{a}^{0}$$f(-a)da … from equation (10). ; is the area bounded by the -axis, the lines and and the part of the graph where . It’s based on the limit of a Riemann sum of right rectangles. . () = . () Definite integral is independent of variable od integration.iii. Property 8: $$\int_{-p}^{p}$$f(a)da = 2$$\int_{0}^{p}$$f(a)da … if f(-a) =f(a) or it is an even function and $$\int_{-a}^{a}$$f(a)da = 0, … if f(-a) = -f(a) or it is an odd function. The definite integral of a non-negative function is always greater than or equal to zero: The definite integral of a non-positive function is always less than or equal to zero. The definite integral of a function on the interval [a, b] is defined as the difference of antiderivative of the given function, which is calculated for the upper bound of integration minus lower bound of integration. A definite integral retains both lower limit and the upper limit on the integrals and it is known as a definite integral because, at the completion of the problem, we get a number which is a definite answer. Properties of Definite Integrals - II. Section 7-5 : Proof of Various Integral Properties. If f (x) is defined and continuous on [a, b], then we have (i) Zero Integral property If the upper and lower limits of a definite integral are the same, the integral is zero. Use this property, to get, Property 5: $$\int_{0}^{p}$$f(a)da = $$\int_{0}^{p}$$f(p-a)da, Let, t = (p-a) or a = (p – t), so that dt = – da …(5). A function f(x) is called odd function if f (-x) = -f(x). It encompasses data visualization, data analysis, data engineering, data modeling, and more. Moving along the … properties of integrals for ease in using the definite,. This whole section, assume that f ( x ) is even defined an... Of solving definite integral and many more Mathematical topics of indefinite integrals apply to definite integrals ; Why Should. -P, t =0 of variables provided the limits of definite integrals have! Is differentiated? ” ) has a summation at its heart a b! Calculus are: function limits integral Derivatives simple adding a minus sign on the integral at specified. 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http://kmle.rallystoryeventi.it/simplify-the-following-boolean-expression.html | Or you could also use set operations on the given expression and reduce it. It works as a portable calculator to simplify the Boolean expression on the. NOT is also written as A’ and A • Using the above notation we can write Boolean expressions for functions F(A, B, C) = (A * B) + (~A * C) • We can evaluate the Boolean expression with all. Browse other questions tagged boolean boolean-logic boolean-expression boolean-operations demorgans-law or ask your own question. Simplify the following Boolean expressions to a minimum number of literals (a) xy+ xy’ (b) (x+y)(x+y’) (c) xyz + x’y+ xyz’ (d) (A+B)’(A’+B’)’ 2. Convert the following to Decimal and then to Octal 11. not A => ~A (Tilde) A and B => AB A or B => A+B A xor B => A^B (circumflex). F = x’y’z + x’y z + x y’ •First we can see the use of the distributive property. In Studies in Logic and the Foundations of Mathematics, 2000. 2: Simplify the following Boolean functions, using three-variable maps: (a) F(x,y,z) = sum(0,1,5,7) (b) F(x,y,z) = sum(1,2,3,6,7) (c) F(x,y,z) = sum(0,1,6,7) (d. of literals ((ABC)I + (AB)I. d) Obtain the truth table of the function from the simplified expression and show that it is the same as the one in part (a) e) Draw the logic diagram from the simplified expression and compare the total number of gates with the diagram of part (b). a∧[]b∨(a∧b)∧[]a∨(~a∧b) 11. Boolean Algebra Calculator is an online expression solver and creates truth table from it. a’bc + abc’ + abc + a’bc’ 5. A boolean expression evaluates to either true or false. It is also used in Physics for the simplification of Boolean expressions and digital circuits. Is there any way to simplify a combination of XOR and XNOR gates in the following expression? I have tried multiple boolean theorems and I have not been able to simplify this any further: The simp. As with a lot of things in Boolean Algebra, the laws are logical. (A ∪ B)' = A' ∩ B' (A ∩ B)' = A' ∪ B' De Morgan's Theorem can be used to simplify expressions involving set operations. Generating Schematic Diagrams from Boolean Expressions. Use de Morgan's laws and the fact that !!p is equivalent to p to simplify each expression. With assumptions of the form ForAll [vars, axioms], FullSimplify can simplify expressions and equations involving symbolic functions. For instance, the Boolean expression ABC + 1 also reduces to 1 by means of the “A + 1 = 1” identity. Look at the first circuit. Boolean Algebra Simplification of (x'y'+z)'+z+xy+wz. View EGR265 Homework 4. 5 Boolean functions In the same way as algebraic functions describe the relationship between the domain, (a set of inputs) and the range (a set of outputs), a Boolean function can be described by a Boolean expression. Mark each term of the SOP expression in the correct cell of the k-map. Example 1 F = A. This simplifier can simplify any boolean algebra. Every Boolean algebra is isomorphic to an algebra of sets. Introduction We have defined De Morgan's laws in a previous section. Also note that the term Y is included in the term X · Y, so that, by inclusión, the term X · Y can be eliminated from the expression. More complex boolean expressions can be built out of simpler expressions, using the following boolean operators:. Simplify the following Boolean expression: AB(A + B)(C + C) Design the combinatorial circuit for: (p' *r) + q Design the combinatorial circuit for: Complete the truth table of the following Boolean expression: Prove or disprove that the following 2 expressions are equivalent. The other two terms also simplify:. You have to be thorough with the SOP & POS. These theorems can be used to simplify expressions as shown in the examples below. The tabular method reduces the function to a set of prime implicants. Parenthesized forms¶. By using Boolean laws and theorems, we can simplify the Boolean functions of digital circuits. Simplify the following Boolean expression. Simplify The Following Expressions Using Boolean Algebra - Free download as Word Doc (. Browse other questions tagged boolean boolean-logic boolean-expression boolean-operations demorgans-law or ask your own question. Simplify the following Boolean expression AB +A'C +BC 5. Simplify the following Boolean expression ABC+ABC'+AB'C+AB'C' 4. How can I simplify the following Boolean expression? A'C+A'. (CLO 3—Boolean/Comb. The following shows an example of using algebraic techniques to simplify a boolean expression. Obtain the truth table for F. The form of a mathematical expression that is simplest for one problem turns out to be complicated or even unsuitable for another problem. I have added simplification by Bollean algebra. That is, any given boolean expression can be completely represented by using the a functionally complete boolean operator. 6 and proceeding dramatically. Two boolean expressions are logically equivalent if and only if they have the same truth table. Complement of a variable is represented by an overbar. Schematic Diagram of Two Level Logic Karnaugh Map: A 2-dimensional truth table. 29) Implement the following four Boolean expressions with three half adders a) D = A ⊕ B ⊕ C b) E = A′BC + AB′C c) F = ABC′ + (A′ +B′)C d) G = ABC. For example: (XY) + (YX) + (AB) + (BA) + (EE) COMMUTATIVE (XY) + (XY) + (AB) + (AB) + (EE) IDEMPOTENT (XY) + (AB) + (E) ASSOCIATIVE XY+AB+E Simplify the following Boolean expressions by using the required laws. A karnaugh map provides an organized way for simplifying boolean expressions and helps in constructing the simplest minimized SOP and POS expression. State the principle of duality in Boolean algebra and give the dual of the Boolean expression : (X + Y). The most practical law is DeMorgan's law: one form explains how to simplify the negation of a conjunction ( and ) and the other form explains how to simplify the negaion of a disjunction ( or ). A lambda expression (ramda) describes a block of code (an anonymous function can be passed to a construct or method for subsequent execution). Online minimization of boolean functions. Verilo Module 6. Here are some examples of Boolean algebra simplifications. Karnaugh Map (truth table in two dimensional space) 4. Alyazji 2 2. Simplify the following Boolean expression AB +A'C +BC 5. Is there any way to simplify a combination of XOR and XNOR gates in the following expression? I have tried multiple boolean theorems and I have not been able to simplify this any further: The simp. Simplification. Simplify the following Boolean expression. Simplifying Variable Expressions Date_____ Period____ Simplify each expression. name: nkemdirim chimezirim miracle iti1100c assignment student number: 8869343 demonstrate the validity of the following identities means of truth tables. One way to analyze, and perhaps simplify, boolean expressions is to create truth tables. Simplifying logic circuits Obtain the expression of the circuit’s function, then try to simplify We will look at two methods: Algebraic and Karnaugh maps E1. A boolean expression evaluates to either true or false. • Some standardized forms are required for Boolean expressions to simplify communication of the expressions. simplify 2. Thus the circuit computes P ⊕ Q except possibly when P = Q = 0. In all of the following examples, we refer to the theorems above by number for simplicity. (A + B) = Q. A Boolean expression is an expression that evaluates to a value of the Boolean Data Type: True or False. F1 F2 = ∑ mi ∑mj where mi mj = 0 if i ≠ j and mi mj = 1 if i = j. Let us first define some basic formulas of boolean algebra that we're gonna need for the solution 1. - 17528657. It has been fundamental in the development of digital electronics and is provided for in all modern programming languages. To do this, carry. 2: Simplify the following Boolean functions, using three-variable maps: (a) F(x,y,z) = sum(0,1,5,7) (b) F(x,y,z) = sum(1,2,3,6,7) (c) F(x,y,z) = sum(0,1,6,7) (d. For example, NAND gates can be used to implement the NOT gate, the OR gate and the AND gate. Simplify complex Boolean algebra expressions using the 14 Boolean rules and apply DeMorgan's Theorem. It returns true when the expression on its left, the expression on its right, or both are true. Learning how to simplify algebraic expressions is a key part of mastering basic algebra and an extremely valuable tool for all mathematicians to have under their belt. 6 Simplify the following Boolean expressions, using four-variable maps: (b)* x' z + w'xy' + w(x'y + xy') 3. Mapping truth tables to logic gates Given a truth table: Write the Boolean expression Minimize the Boolean expression Draw as gates Map to available gates Determine number of packages and their connections Winter 2010 CSE370 - IV - Canonical Forms 3 4 C F B A 7 nets (wires). If the delay of a NAND gate is 15ns and that of a NOR gate is 12ns, which implementation is faster. Simplifying an expression often means removing a pair of parentheses; factoring an expression often means applying them. Find minimal SOP expressions for the following: 1. It Solves logical equations containing AND, OR, NOT, XOR. ) Write the Boolean expression equivalent to the following logic circuit. A Boolean expression is composed of a combination of the Boolean constants (True or False), Boolean variables and logical connectives. A Boolean expression over Boolean algebra B is defined as. By using Boolean laws and theorems, we can simplify the Boolean functions of digital circuits. Heavy example. • Values and variables can indicate some of the following binary pairs of values:. The following are the logic gates −. 4 Laws of Boolean Algebra The manipulation of algebraic expressions is based on fundamental laws. • Boolean expressions allow us to write programs that. For checking the same you can create a truth table for your simplified expression and match the outputs with those of the initial expression. Minimization of Boolean expressions •The minimization will result in reduction of the number of gates (resulting from less number of terms) and the number of inputs per gate (resulting from less number of variables per term) •The minimization will reduce cost, efficiency and power consumption. Boolean Satisfiability There are many alternative ways to define a Boolean expression, but for our discussion, we must fix one of them. WX(Z YZ) X(W WYZ). The study of boolean functions is known as Boolean logic. Homework 4 1. Look at the first circuit. F (A, B) = (A. Simplify the following Boolean expression AB +A'C +BC 5. Viewed 109 times 2 \$\begingroup\$ Simplify the following expression using Boolean Algebra: $$x = \bar{A} \bar{B} \bar{C} + \bar{A}BC + ABC + A \bar{B} \bar{C} + A \bar{B} C$$ Answer: \begin. Boolean Expression Simplification using AND, OR, ABSORPTION and DEMORGANs THEOREM. Simplify the following Boolean expression using a k-map of size 2x2. Dansereau; v. Simplify the following Boolean expression. Each line gives a form of the expression, and the rule or rules used to derive it from the previous one. Draw the logic circuit for the DeMorgan equivalent Boolean equation you found in 4(c). " Simplify means writing an equivalent expression using the fewest number of operators. Every element of B is a Boolean expression. 10Draw the logic diagrams for the following Boolean expressions: (a) Y =. DIGITAL LOGIC DESIGN ECOM 2012 ENG. As with a lot of things in Boolean Algebra, the laws are logical. Boolean Expressions: true or false. Lorem ipsum dolor sit amet, consectetur adipiscing elit. So, to simplify the Boolean equations and expression, there are some laws and theorems proposed. We can use algebraic manipulation to produce canonical forms even though the canonical forms are rarely optimal. But it is also possible to implement the. NAND: x · y = x + y NOR: x + y = x · y Redundancy laws. Simplify the following Boolean expression AB +A'C +BC 5. pdf), Text File (. 27 / 28 Homework 2-15 Given the Boolean function F = xy'z + x'y'z + w'xy + wx'y + wxy (a) Obtain the truth table of the function. a∨(~ a∧b) 2. October 9, 2011 Performance up! Reduce time out errors. boolean algebra simplification questions and answers pdf Simplify, complement, multiply out and factor an expression. SOP with K-Ma NAND/NOR Lo ic 9. Also, an increase in the number of variables results in an increase of complexity. Every element of B is a Boolean expression. Boolean algebra is a branch of algebra wherein the variables are denoted by Boolean values. •Example: • Two methods for simplifying - Algebraic method (use Boolean algebra theorems) - Karnaugh mapping method (systematic, step-by-step approach) E1. 1 Karnaugh Maps • Applications of Boolean logic to circuit design - The basic Boolean operations are AND, OR and NOT - These operations can be combined to form complex expressions, which can also be directly translated into a hardware circuit - Boolean algebra helps us simplify expressions and circuits • Karnaugh Map: A graphical technique for simplifying an expression into a. Simplify the Boolean Expressions (x+y). Simplify the following Boolean expression: (solution should be one term) XY+XY 2. (c) Simplify the function to a minimum number of literals using Boolean algebra. A boolean expression is an expression that results in a boolean value, that is, in a value of either true or false. Mapping truth tables to logic gates Given a truth table: Write the Boolean expression Minimize the Boolean expression Draw as gates Map to available gates Determine number of packages and their connections Winter 2010 CSE370 - IV - Canonical Forms 3 4 C F B A 7 nets (wires). Every variable name is a Boolean expression. Orienting Questions. Or you could also use set operations on the given expression and reduce it. implementation of Boolean expressions much more systematic and easier. The most practical law is DeMorgan's law: one form explains how to simplify the negation of a conjunction ( and ) and the other form explains how to simplify the negaion of a disjunction ( or ). The first extracts comments from code for documentation purposes. The Boolean expressions A + 1 and A + ~A are also considered tautologies. Using these laws and theorems, it becomes very easy to simplify or reduce the logical complexities of any Boolean expression or function. Boolean allows you to simplify expressions built from these operators, and to test properties like equivalence, subset property etc. Boolean logic allows us to understand if a statement is true or false. 3* Simplify the following Boolean expressions, using three-variable maps: (a)* F(x,y,z) = xy + x'y'z' + x'yz'. The SOP and POS, both forms are used for representing the expressions and also holds equal importance. 1) −3 p + 6p 2) b − 3 + 6 − 2b 3) 7x − x 4) 7p − 10 p 5) −10 v + 6v 6) −9r + 10 r 7) 9 + 5r − 9r 8) 1 − 3v + 10 9) 5n + 9n 10) 4b + 6 − 4 11) 35 n − 1 + 46 12) −33 v − 49 v 13) 30 n + 8n 14) 7x + 31 x. Boolean algebra is the mathematics we use to analyses digital gates and circuits. Simplifying Logic Circuits • First obtain one expression for the circuit, then try to simplify. If a non-logic expression is specified, a syntax. a=f(1+e)+fg. • They are a visual representation of a truth table. In order to analyze and troubleshoot digital circuits, it is necessary to be able to write Boolean expressions from logic circuits OR to draw circuits given only Boolean expressions. Simplify the logic diagram below. Boolean algebra, so this section will: Define Boolean algebras and derive those properties most useful for the design of gate networks. The final simplified expression is: Z = Bear in mind that any unticked terms in any list must be included in the final expression (none occured here except from the last list). Boolean algebra finds its most practical use in the simplification of logic circuits. The following options can be given:. To illustrate how a circuit can be constructed (or implemented) using the Boolean expression, consider the expression for the X-OR gate of Figure 2. Simplify the following Boolean expression: (solution should be one term) (X+Y)(X+Y)(X'+Z”) 3. Use Boolean Algebra to simplify Boolean functions to produce simpler circuits Example: Simplify to a minimum number of literals F = x + x’ y ( 3 Literals) = x + ( x’ y ) = ( x + x’ ) ( x + y ) = ( 1 ) ( x + y ) = x + y ( 2 Literals) Distributive law (+ over •) 10. The key to understanding the different ways you can use De Morgan's laws and Boolean algebra is to do as many examples as you can. Also, if you're learning this as a student, often you will be required to remember them for an exam. Simplify the expression Double invert it If Boolean function has only one term then implement by observation. Even Wolfram Alpha can simplify Boolean expressions, but you need to use a different syntax. TOPIC 7: Converting Truth Tables to Boolean Expressions. We can simplify expressions using the nine key laws of. There are a number of boolean expression patterns that can easily be rewritten to make them simpler. Simplify the following Boolean expression: (solution should be one term) XY+XY 2. Karnaugh Maps are useful for finding minimal implementations of Boolean expressions with only a few variables. (a) a= s*t+v*w+r*s*t. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Boolean Expressions Simplifier; How to simplify / minify a boolean expression? What are boolean algebra simplifications methods? How to show/demonstrate that 2 boolean expression are equal? What is De Morgan's law? What is Disjunctive or Conjuctive Normal Form? How to show step by step calculations?. Find the complement of the following Boolean expression using DeMorgan's law: F = (A + BC'). Appendix Adiscussesthe relation of those trees and other well-known constructs— for example, to Boolean expressions in. Invert all of the input variables; also exchange any 0s for 1s, and vice versa. Again, the simpler the Boolean expression the simpler the resultingthe Boolean expression, the simpler the resulting logic. If A is a linear ordering, then we form the corresponding interval algebra I(A). Here you can check the properties of your boolean function: You can enter your boolean function in either its truth table, or its algebraic normal form(ANF) or its trace representation. Simplify the following expressions using the laws and theorems of Boolean Algebra: (a) S(A,B,C) = A' B' C + A' B C' + A B' C' + A B C = A' (B'C + BC') + A (B'C' + BC). Simplifying Logic Circuits with Karnaugh Maps • The circuit at the top right is the logic equivalent of the Boolean expression: f abc abc abc • Now, as we have seen, this expression can be simplified (reduced to fewer terms) from its original form, using the Boolean identities as shown at right. They can be used in if and while statements. Every variable name is a Boolean expression. NOT is also written as A’ and A • Using the above notation we can write Boolean expressions for functions F(A, B, C) = (A * B) + (~A * C) • We can evaluate the Boolean expression with all. Use de Morgan's laws and the fact that !!p is equivalent to p to simplify each expression. Problems 1-4: Simplify each Boolean expression to one of the following ten expressions: 0, 1, A, B, AB, A+B, , +, A, B. Simplify the following Boolean expression ABC+ABC'+AB'C+AB'C' 4. Any lowercase letter may be used as a variable. Use of “Laws of Boolean” to both reduce and simplify a complex Boolean expression in an attempt to reduce the number of logic gates required. Rules 1 through 9 will be viewed in terms of their application to logic gates. Though, in general, the number of 1 s per product term varies with the number of variables in the product term compared to the size of the K-map. This case is easily checked. C How many gates do you save = A. Here are some examples of Boolean algebra simplifications. October 9, 2011 Performance up! Reduce time out errors. A variable is a symbol in Boolean algebra used to represent (a) data (b) a condition (c) an action (d) answers (a), (b), and (c) 2. Generating Schematic Diagrams from Boolean Expressions. Interchanging the 0 and 1 elements of the expression. The dualof a Boolean expression is obtained by interchanging Boolean sums and Boolean products and interchanging 0s and 1s. Instead, they are stored as integers: true becomes the integer 1, and false becomes the integer 0. Use the rules of Boolean Algebra to change the expression to a SOP expression. This method of proving the equality of two expressions is known as the (1). Return an expression that evaluates to $$1$$ if and only if the input parity is odd. A Boolean expression over Boolean algebra B is defined as. The book's answer is: x = BC+ ˉB(ˉC+A). Karnaugh maps are effective for. The expression can be simplified to A•C + D•( ( A + B’ )' + C ) 3. Determining Standard Expressions from a Truth Table −Given a truth table, the SOP expression is developed from where the output is 1 and the POS expression is developed from where the output is 0 −e. Solution - The following is a 4 variable K-Map of the given expression. Boolean algebra simplification calculator is an advanced calculator that immediately gives the result in the form of a math expression by performing the operations, such as multiplication, addition, etc. Once the Boolean expression for the action of a circuit has been found, the circuit can be sketched out - but finding the Boolean expression may be a problem. The "x" symbol is seldom used. (3 marks). Just use the laws of boolean algebra. A truth table shows the evaluation of a Boolean expression for all the combinations of possible truth values that the variables of the expression can have. Thus if B = 0 then B=1 and B = 1 then B= 0. With assumptions of the form ForAll [vars, axioms], FullSimplify can simplify expressions and equations involving symbolic functions. A boolean expression is an expression that results in a boolean value, that is, in a value of either true or false. ), if the final column of the truth table has more than one true value, apply an OR(+) We need to get a combination of A, B that gives the result shown above. • Values and variables can indicate some of the following binary pairs of values:. Variable used can have only two values. Use MathJax to format equations. js takes an expression split over one operator (e. Karnaugh Map : 1. Some of these laws extend to the manipulation of boolean expressions. Active 3 years, 9 months ago. • Expression are most commonly expressed in sum of products form. The notation boolExp stands for any expression that evaluates to a boolean. The following options can be given:. Karnaugh map gallery. Start studying CIS116: QUIZ 4. For example, NAND gates can be used to implement the NOT gate, the OR gate and the AND gate. , when an oven is on, the PLC adjusts temperature. ( b + c ) + a. Practice is the best way to achieve this. The "Karnaugh Map Method", also known as k-map method, is popularly used to simplify Boolean expressions. Look for common factors, just like you would in ordinary algebra. Simplify the following Boolean expression: (solution should be one term) XY+XY 2. When it is off doors can be opened/closed. F1 F2 = ∑ mi ∑mj where mi mj = 0 if i ≠ j and mi mj = 1 if i = j. IIF ( boolean_expression, true_value, false_value ) where, boolean_expression is a logical expression. Simplify the following Boolean expression. The Java Shell provides various commands and features that simplify working with snippets. F = A’B + B’C’ + D’ b. Lorem ipsum dolor sit amet, consectetur adipiscing elit. Using the following K-Maps: i) Find the minimal sum of products expression. Truth tables often makes it easier to understand the Boolean expressions and can be of great help when simplifying expressions. The goal of this lesson is to learn to use Boolean expressions to make comparisons. Sum-of-Products (SOP) Form. Describe the CMOS inverter circuit. (a) WX + WXZ + W Y Z + W XY + WXZ (b) XZ + XYZ + WX Y (a) WX + WXZ + W Y Z + W XY + WXZ F = XY + W + Y Z + X Z. A Boolean function of n-variables is represented by f(x1, x2, x3…. CHAPTER III-2 BOOLEAN VALUES INTRODUCTION BOOLEAN ALGEBRA •BOOLEAN VALUES • Boolean algebra is a form of algebra that deals with single digit binary values and variables. 2 Basic Laws The properties of Boolean algebra are described by the basic laws introduced in this section. Simplify the following Boolean expression: (solution should be one term) XY+XY 2. How would you simplify the following? I'm having a bit of trouble with the first part with negation. A Boolean expression over Boolean algebra B is defined as. Look for common factors, just like you would in ordinary algebra. Number Conversions Chi 1m lementation 5. Reduce using K'Maps 5. Exponents are supported on variables using the ^ (caret) symbol. Simplify the following Boolean expression: AB(A + B)(C + C) Design the combinatorial circuit for: (p' *r) + q; Design the combinatorial circuit for: [(p Λ q') V ( r V q)] Λ s; Complete the truth table of the following Boolean expression: p' Λ (q V r ) Prove or disprove that the following 2 expressions are equivalent. " No OR operations. 27 / 28 Homework 2-15 Given the Boolean function F = xy'z + x'y'z + w'xy + wx'y + wxy (a) Obtain the truth table of the function. If you're having to simplify expressions often it is more convenient if you don't have to look them up constantly. The SOP and POS, both forms are used for representing the expressions and also holds equal importance. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Use de Morgan's laws and the fact that !!p is equivalent to p to simplify each expression. Establish the connection between the two main behavioral models for gate networks, namely logical expressions and. 1 Boolean Logic. 10 LET A% = 0 20 LET B% = NOT (A%). Sum of product form is a form of expression in Boolean algebra in which different product terms of inputs are being summed together. C How many gates do you save = A. Rules 1 through 9 will be viewed in terms of their application to logic gates. Conclusion. In a digital designing problem, a unique logical expression is evolved from the truth table. Reduce the following Boolean expressions to the indicated number of literals Here's my reduction using only the properties of Boolean Algebra. A boolean expression is an expression that results in a boolean value, that is, in a value of either true or false. The following is the resulting boolean expression of each of the gates. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. 2 Use axioms of boolean algebra to simplify expression in terms of boolean connectives used. file 01334 8. Boolean logic allows us to understand if a statement is true or false. They can be used in if and while statements. Here you can check the properties of your boolean function: You can enter your boolean function in either its truth table, or its algebraic normal form(ANF) or its trace representation. Alyazji 2 2. Y = A'B'C' + A'B + ABC' + AC Y = A'(B'C' +B) + A(BC'+C) (Just factor out the terms like algebra) Now work on the sub term. Truth Table or Boolean function 3. Numbers also work in place of boolean expressions following those rules. Generating Schematic Diagrams from Boolean Expressions. Here is another approach. 2 provides the basic Boolean theorems. Boolean expressions evaluate to 0 when false, and a non-zero value (traditional versions of basic use a value of one, although some variants use a value of negative one) when true. Name Boolean laws Dual 1. Interchanging the OR and AND operations of the expression. For instance, the Boolean expression ABC + 1 also reduces to 1 by means of the “A + 1 = 1” identity. NOT is also written as A’ and A • Using the above notation we can write Boolean expressions for functions F(A, B, C) = (A * B) + (~A * C) • We can evaluate the Boolean expression with all. xy + xy’ 2. Instead, they are stored as integers: true becomes the integer 1, and false becomes the integer 0. boolean algebra simplification questions and answers pdf Simplify, complement, multiply out and factor an expression. Sep 24, 2017 One can use a 3 variable Karnaugh map. There are two methods for converting truth tables to boolean expressions. Solution: 2. And here’s a quick fact: you don’t have to capitalize Boolean operators on any of the major job boards and many of the major ATS’s. Ex ressions 2 3. Viewed 5k times -1 $\begingroup$ Closed. : "z > 1" is a Boolean expression, as is "0 < z <= 3" since each expression is either True or False. "A&~B|~A&B" over "|" then "&" to find "A^B") to simplify more complex phrases. (NO DIAGONALS!). C from this simplification? = A + B. In algebra, simplifying and factoring expressions are opposite processes. F1 F2 = ∑ mi ∑mj where mi mj = 0 if i ≠ j and mi mj = 1 if i = j. In Studies in Logic and the Foundations of Mathematics, 2000. (b ) Draw the logic diagram using the original Boolean expression. Search for a tool Search a tool on dCode by keywords:. The most practical law is DeMorgan's law: one form explains how to simplify the negation of a conjunction ( and ) and the other form explains how to simplify the negaion of a disjunction ( or ). Simplify Boolean algebraic expressions using a 4-variable Karnaugh map. (c ) Simplify the function to a minimum number of literals using Boolean algebra. Y = A'B'C' + A'B + ABC' + AC Y = A'(B'C' +B) + A(BC'+C) (Just factor out the terms like algebra) Now work on the sub term. Determining Standard Expressions from a Truth Table −Given a truth table, the SOP expression is developed from where the output is 1 and the POS expression is developed from where the output is 0 −e. Learn vocabulary, terms, and more with flashcards, games, and other study tools. By using this website, you agree to our Cookie Policy. The following shows an example of using algebraic techniques to simplify a boolean expression. Next, write down input. Boolean Algebra Calculator is an online expression solver and creates truth table from it. If the delay of a NAND gate is 15ns and that of a NOR gate is 12ns, which implementation is faster. In addition to de Morgan's laws, there are a number of other laws that can be used to simplify boolean expressions. =XZY+YX′+YZ+XZ+YX′Z+YZ as YY=Y, ZZ=Z. Numbers also work in place of boolean expressions following those rules. Develop the truth table for the following Boolean expression: Simplify the Boolean expression given as: Develop a logic diagram to implement the following Boolean expression: Draw a logic diagram to implement the following Boolean expression: Provide the Sum-of-Product (SOP) form of the Boolean. Boolean expressions involving comparisons with boolean literals, ternary conditionals with a boolean literal as one of the results, double negations, or negated comparisons can all be changed to equivalent and simpler expressions. It is also called as Binary Algebra or logical Algebra. 10 LET A% = 0 20 LET B% = NOT (A%). A secondary operator is a Boolean operator that can be natively represented as a PyEDA expression, but contains more information than the primary operators. Evaluate each expression assuming that the following declarations have been made. 7k points) basics of boolean algebra. Simplify the following Boolean expressions to a minimum number of literals (a) xy+ xy’ (b) (x+y)(x+y’) (c) xyz + x’y+ xyz’ (d) (A+B)’(A’+B’)’ 2. Parsing boolean values with argparse. From some video on Youtube. How to Prove two Boolean expressions are equivalent? Deduction. Any lowercase letter may be used as a variable. Z=DDF+DEF+DFF+DEF+EEF+FEF. Show that the functions can be implemented with logic diagrams that have only OR gates and inverters. DO NOT use a Karnaugh map except possibly to check your work. Z = A * ~B * C + A * B * ~C + A * B * C. In Studies in Logic and the Foundations of Mathematics, 2000. Summary In this chapter you’ve covered: • Truth Tables for the following functions/gates: • -AND & NAND Gates • -OR & NOR Gates • -NOT & XOR Gates • Logic Gate Diagrams • De Morgan’s Laws to simplify any Boolean Expression. a=f(1+e)+fg. Simplify the following Boolean expressions. Recall our first definition of bigger3. For example, NAND gates can be used to implement the NOT gate, the OR gate and the AND gate. Viewed 5k times -1 $\begingroup$ Closed. Answer to Use properties of Boolean algebra to simplify the following Boolean expression (showing all the steps): [x' + (yz)'][x +z']'. A truth table shows the evaluation of a Boolean expression for all the combinations of possible truth values that the variables of the expression can have. When P is true and Q is true the combined expression (P Or Q) is also true. boolean_factor() in boolean_factor. F (A, B) = (A. Solution: (Figure below) Write the Boolean expression for the original logic diagram as shown below Transfer the product terms to the Karnaugh map Form groups of cells as in previous examples Write Boolean expression for groups as in previous examples Draw simplified logic diagram. Example if we have two variables X and Y then, Following is a canonical expression consisting of minterms XY + X'Y' and Following is a canonical expression consisting of maxterm (X+Y). Simplify the following Boolean expression to a minimum number of literals: F = xyz + x’y +. Boolean Satisfiability There are many alternative ways to define a Boolean expression, but for our discussion, we must fix one of them. Recall our first definition of bigger3. Simplify as much as possible. 8 31 October 2008 Method 1: Minimisation by Boolean Algebra • Make use of rules and theorems of Boolean algebra to simplify the Boolean expression. For all situations described below: A = It is raining upon the British Museum right now (or any other statement that can be true or false) B = I have a cold (or any other statement that can be true or false). Element-by-element boolean expressions can be used wherever comparison expressions can be used. a∧[]b∨(a∧b)∧[]a∨(~a∧b) 11. In this course we will cover Digital Electronics important Topic related to Boolean Algebra and Boolean expression for ugc net computer science and GATE computer science (Hindi) Boolean Algebra and Logic Simplification: NTA-UGC NET. To do this, evaluate the expression, following proper mathematical order of operations (multiplication before addition, operations inside parentheses before anything else), and draw gates for each step. 27 / 28 Homework 2-15 Given the Boolean function F = xy'z + x'y'z + w'xy + wx'y + wxy (a) Obtain the truth table of the function. FullSimplify does transformations on most kinds of special functions. B2SPICE™ version 5. •Example: • Two methods for simplifying - Algebraic method (use Boolean algebra theorems) - Karnaugh mapping method (systematic, step-by-step approach) E1. Each line gives a form of the expression, and the rule or rules used to derive it from the previous one. Draw the membership function for the following Fuzzy expressions a. For example: (XY) + (YX) + (AB) + (BA) + (EE) COMMUTATIVE (XY) + (XY) + (AB) + (AB) + (EE) IDEMPOTENT (XY) + (AB) + (E) ASSOCIATIVE XY+AB+E Simplify the following Boolean expressions by using the required laws. Ex ressions 2 3. Introduction of K-Map (Karnaugh Map) In many digital circuits and practical problems we need to find expression with minimum variables. Write the UN-SIMPLIFIED logic expression for the output. Simplify the following Boolean expressions algebraically. Construct a logic diagram to implement the Boolean function f = (AB)’+AC +BCD using gates. The goal of this lesson is to learn to use Boolean expressions to make comparisons. Boolean Logic George Boole Simplifying a Boolean Expression "It is evident that with the above definitions the following postulates hold. Use the following rules to enter expressions into the calculator. a= st+vw (b) a= t*u*v+x*y+y. Boolean algebra simplification calculator is an advanced calculator that immediately gives the result in the form of a math expression by performing the operations, such as multiplication, addition, etc. A value is considered to be false if it is zero, and true otherwise. 004 Worksheet - 7 of 7 - L06 - Boolean Algebra Problem 6. How would you simplify the following? I'm having a bit of trouble with the first part with negation. We can use these “Laws of Boolean” to both reduce and simplify a complex Boolean expression in an attempt to reduce the number of logic gates required. 3) Simplify the following Boolean expression to a minimum number literals: a) ABC + A′B + ABC′ b) x′yz + xz c) (x+y)′(x′+y′) d) xy + x(wz + wz′) e) (BC′+A′D)(AB′+CD′). Introduction Simplifying logic circuits is a predominant task when designing a digital. The Boolean expression format must be in prefix-form of a LISP list. Consider the problem of simplifying Boolean expressions. Boolean Logic (151)) Use the theorems of Boolean algebra to simplify the following expression: X/ + X/ +X/ + X/ + X/ x-,x3 + X/ x-žx3 Question Max Score 10 15 20 10 10 15 15 15 15 130. Question 4. In a digital designing problem, a unique logical expression is evolved from the truth table. Simplify The Following Boolean Expression: AC' + A'C + A'C' + AC A C AC 1. Good number of problems are asked on EX-OR and EX-NOR gates. docx), PDF File (. 2 Simplify the following Boolean expressions to a minimum number of literals: (a)* xy +xy' (c)* xyz +x'y +xyz' (e) (a+b+c(a' b'+c) Simplify the following Boolean. A variable is a symbol in Boolean algebra used to represent (a) data (b) a condition (c) an action (d) answers (a), (b), and (c) 2. If a 1 and a 2 are Boolean expression, then a 1,'∨ a 2 and a 1 ∧ a 2 are Boolean expressions. The following is the resulting boolean expression of each of the gates. The other two terms also simplify:. Simplify the following Boolean expression into one literal. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. A Normal POS expression is a Product of. Boolean algebra is a system of mathematical logic. by theorem 8. 6 Boolean Algebra and Simplification Techniques. Reduced expression using Boolean Algebra 5. The payoff is even greater in boolean algebra. When multiplying and dividing rational expressions what are the rules for canceling numbers, variables etc. Use MathJax to format equations. A) ABC + ABC + ABC + AB`C + ABC (2 term, 5 literals). Sometimes a programmer would like one statement, or group of statements to execute only if certain conditions are true. Construct a logic diagram to implement the Boolean function f = (AB)’+AC +BCD using gates. Simplify the following Boolean expression: AB(A + B)(C + C) Design the combinatorial circuit for: (p' *r) + q Design the combinatorial circuit for: Complete the truth table of the following Boolean expression: Prove or disprove that the following 2 expressions are equivalent. We will now look at some examples that use De Morgan's laws. For instance, the Boolean expression ABC + 1 also reduces to 1 by means of the “A + 1 = 1” identity. a logic probe indicates only a HIGH voltage level. That is, any given boolean expression can be completely represented by using the a functionally complete boolean operator. See {{ ext_info ? 'less' : 'more' }} information Supported operations are AND , OR , NOT , XOR , IMPLIES , PROVIDED and EQUIV. 1 Boolean Expressions 123 • Boolean Expression: Combining the variables and operation yields Boolean expressions. nx is the same as x' (just use a 'n' before the letter instead of ' after it). Draw the Maps and show the loops you use. Algebric method:This method makes use of Boolean postulates, rules and theorems to simplify the expression. Simplify the following Boolean expression AB +A'C +BC 5. Evaluate each expression assuming that the following declarations have been made. Let us first define some basic formulas of boolean algebra that we're gonna need for the solution 1. pdf), Text File (. A karnaugh map provides an organized way for simplifying boolean expressions and helps in constructing the simplest minimized SOP and POS expression. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. a=tuv+y (c) a= f*(e+f+g) a=fe+ff+fg. Application of Boolean Algebra. 2 Basic Laws The properties of Boolean algebra are described by the basic laws introduced in this section. 30 Write the following Boolean expressions in sum of products form: (b + d)(a + b + c)2. K-map Simplification leads to the expression which you have arrived at. Simplification of Boolean functions Using the theorems of Boolean Algebra, the algebraic forms of functions can often be simplified, which leads to simpler (and cheaper) implementations. Boolean theorems and laws are used to simplify the various logical expressions. a∨(~ a∧b) 2. Rules of Boolean Algebra Table 4-1 lists 12 basic rules that are useful in manipulating and simplifying Boolean expressions. 2, Boolean algebra uses binary variables that can have two values, zero and one, which stand in for "false" and "true," respectively. Re: Which one of following Boolean expressions is not logically equivalent to all of Yes, it's c. Online Karnaugh Map solver that makes a kmap, shows you how to group the terms, shows the simplified Boolean equation, and draws the circuit for up to 6 variables. The addition rule for equations tells us that the same quantity can be added to both sides of an equation without changing the solution set of the equation. A + BC = (A + B)(A + C) B. Write the UN-SIMPLIFIED logic expression for the output. It is also called as Binary Algebra or logical Algebra. That is, any given boolean expression can be completely represented by using the a functionally complete boolean operator. Viewed 5k times -1 $\begingroup$ Closed. \$\endgroup\$ - Fizz Feb 11 '15 at 16:14 \$\begingroup\$ In some of the expressions you have a double quote " instead of single quote '. Solution – The following is a 4 variable K-Map of the given expression. 1 Topic – 1 Boolean Operations and Logic Gates1. We can use these “Laws of Boolean” to both reduce and simplify a complex Boolean expression in an attempt to reduce the number of logic gates required. Enter the statement: [Use AND, OR, NOT, XOR, NAND, NOR, and XNOR, IMPLIES and parentheses]. Using these laws and theorems, it becomes very easy to simplify or reduce the logical complexities of any Boolean expression or function. A brief note of different ways of representing a Boolean function is shown below. A Normal SOP expression is a Sum of Products expression with no included product terms. Use single letters for the variable names. Karnaugh Maps are useful for finding minimal implementations of Boolean expressions with only a few variables. A Boolean expression always produces a Boolean value. Boolean algebras are related to linear orderings. AB + A(CD + CD’) b. Boolean Algebra: Simplifying •Using the properties we can simplify Boolean expressions. Why Digital Electronics Boolean Algebra and Logic Simplification? In this section you can learn and practice Digital Electronics Questions based on "Boolean Algebra and Logic Simplification" and improve your skills in order to face the interview, competitive examination and various entrance test (CAT, GATE, GRE, MAT, Bank Exam, Railway Exam etc. Get the free "Boolean Algebra Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. To better understand boolean expressions, it is helpful to construct truth tables. X=X Now let us get to the problem (A+B)(B+C)(A+C) =(AB+AC+BC+BB)(A+C) {Multiplying the first two terms} =(AB+AC. Simplify the following Boolean expression to minimum no. First, I notice that the terms abc' and abc can factor as: abc' + abc = ab(c' + c) = ab(1) = ab. Homework Help: 13: Jan 7, 2016: K: Simplify the following Boolean Expression: Homework Help: 1: Mar 28. How can I simplify the following Boolean expression? A'C+A'. Simplify the expression to a sum of products form. 1 Topic – 1 Boolean Operations and Logic Gates1. Sep 24, 2017 One can use a 3 variable Karnaugh map. Boolean algebra is the mathematics we use to analyze digital gates and circuits. A Boolean function of n-variables is represented by f(x1, x2, x3…. I was wondering if my simplification is complete or can I get a more simplified expression, thanks for the help. Later using this technique Claude Shannon introduced a new type of algebra which is termed as Switching Algebra. Hence, preparing for an interview has become very simpler these days. Find more Computational Sciences widgets in Wolfram|Alpha. A B A B A B A B 2. In programming, you may encounter many different combinations of comparison and logical operators used in an expression. Using Boolean algebra, you can do the following simplifications: Which simplifies the C++ code to just this: bool out = B; Using Boolean algebra to simplify, you'd have to remember (or derive) the identity that , and all the other identities to help you simplify equations. When multiplying and dividing rational expressions what are the rules for canceling numbers, variables etc. 3) Simplify the following Boolean functions, using three-variable maps:. To do this, evaluate the expression, following proper mathematical order of operations (multiplication before addition, operations inside parentheses before anything else), and draw gates for each step. b ) + ( a. An example of a Boolean expression is “y x + y x. Is there any way to simplify a combination of XOR and XNOR gates in the following expression? I have tried multiple boolean theorems and I have not been able to simplify this any further: The simp. If you're having to simplify expressions often it is more convenient if you don't have to look them up constantly. DeMorgan's Theorems are basically two sets of rules or laws developed from the Boolean expressions for AND, OR and NOT using two input variables, A and B. If the delay of a NAND gate is 15ns and that of a NOR gate is 12ns, which implementation is faster. Use properties of Boolean algebra to simplify the following Boolean expression (showing all the steps): [x' + (yz)'][x +z']' Step-by-step answer. There are two methods for converting truth tables to boolean expressions. The following is a list of useful laws (theorems, if you will) of Boolean Algebra. The map method is first proposed by Veitch and then modified by Karnaugh, hence it is also known as "Veitch Diagram". True/False 1. Solutions for Boolean Functions and Computer Arithmetic which you should simplify to (P ∧ ∼Q)∨(∼P ∧Q). We have several Boolean Theorems that helps us to simplify logic expressions and logic circuits. Algebra Assignment Help, Truth table-boolean expressions, 1. Each theorem is described by two parts that are duals of each other. Boolean Algebra: Simplifying •Using the properties we can simplify Boolean expressions. A Boolean expression over Boolean algebra B is defined as. Simplify the following Boolean expression: (solution should be one term) (X+Y)(X+Y)(X'+Z”) 3. Now that you’re practiced simplifying logic expressions, apply your knowledge to simplifying an actual circuit. Any lowercase letter may be used as a variable. Even Wolfram Alpha can simplify Boolean expressions, but you need to use a different syntax. (a + b + c’)(a’b’ + c) 4. Boolean algebra is mathematics of logic. Boolean theorems and laws are used to simplify the various logical expressions. To simplify an expression, enter the expression to cancel and apply the function simplify. Minimize this expression and draw a logic diagram using only a) NAND b) NOR gates. The map method is first proposed by Veitch and then modified by Karnaugh, hence it is also known as "Veitch Diagram". 4 Laws of Boolean Algebra The manipulation of algebraic expressions is based on fundamental laws. A Boolean function is an algebraic form of Boolean expression. A truth table shows the evaluation of a Boolean expression for all the combinations of possible truth values that the variables of the expression can have. As stated in the following steps, a nonstandard SOP expression is converted into standard form using Boolean algebra rule 6 (A + A̅ = 1) Step 1. Solution: 2. Logic) Given the Boolean expression in SOP form, fill in the truth table. A Karnaugh map has zero and one entries at different positions. Simplify the following Boolean expression. This video shows how to simplify a couple of algebraic expressions by combining like terms by adding, subtracting, and using distribution. State if they cannot be simplified A. Also, an increase in the number of variables results in an increase of complexity. Display the output signal to an input (Boolean expression/Logic circuit diagram) through. - 17528657. • Some standardized forms are required for Boolean expressions to simplify communication of the expressions. Parsing boolean values with argparse. C from this simplification? = A + B. Free simplify calculator - simplify algebraic expressions step-by-step This website uses cookies to ensure you get the best experience. Simplify, showing all working: (a) A A A (b) A A A (c) A+0 (d) (A+A) (B +B) (e) A B +A B +B C +A B C 4. Next, write down input. ) with full confidence. xy + xy’ 2. •In digital Logic, we are not using normal mathematics we are using Boolean algebra So, we need to know the laws & rules of Boolean Algebra. boolean_factor_two() splits over two operators (e. Any lowercase letter may be used as a variable. Boolean identities are quick rules that allow you to simplify boolean expressions. WX(Z YZ) X(W WYZ). Simplify Your Life with an Analog Window Detector: AAC Contributors Forum: 0: Jan 24, 2018: A: Simplify the following Boolean equations. com - View the original, and get the already-completed solution here! 1. Evaluate each expression assuming that the following declarations have been made. NOT (B AND C) c. Note that the column marked X · Y matches the one marked X · Y + Y. Minimize this expression and draw a logic diagram using only a) NAND b) NOR gates. ii) Find the minimal product of sums expression. Edwards Columbia University Fall 2012. Homework 4 1. ) Find the complement of a. Simplify: C + BC:. Carry out logic simplification using a 3-variable Karnaugh map. The Boolean expressions A + 1 and A + ~A are also considered tautologies. Do not simplify! 3. Simplify The Following Expressions Using Boolean Algebra. It has been fundamental in the development of digital electronics and is provided for in all modern programming languages. Simplify the following Boolean expressions to the indicated number of literals: (a) AC +ABC +BC (b) (A+B)(A+B) (c) ABC +AC (d) BC +B(AD+CD) (e) (B+C +BC)(BC +AB+AC) 3. Write the following Boolean expressions in their simplest forms. As a Boolean equivalency, this rule may be helpful in simplifying some Boolean expressions. If a 1 and a 2 are Boolean expression, then a 1,'∨ a 2 and a 1 ∧ a 2 are Boolean expressions. Look at the first circuit. Simplify The Following Boolean Expression AB +A’C +BC; Question: Simplify The Following Boolean Expression AB +A’C +BC. A OR (B AND C) Answers to Practice Problems Practice problems - Deriving Boolean Expressions From Truth Tables: What is the Boolean expression for the following truth tables? 1. Carry out logic simplification using a 3-variable Karnaugh map. Examples of use of Boolean algebra theorems and identities to simplify logic expressions. ) with full confidence. Any lowercase letter may be used as a variable. De Morgan's Theorem gives the following equations on set operations. A Veitch diagram is a system of squares on which a given Boolean expression is plotted in order to arrive at the simplest form of the expression. (Note: These expressions can be reduced into a minimal SOP by repeatedly applying the Boolean algebra properties we saw in lecture. Mapping truth tables to logic gates Given a truth table: Write the Boolean expression Minimize the Boolean expression Draw as gates Map to available gates Determine number of packages and their connections Winter 2010 CSE370 - IV - Canonical Forms 3 4 C F B A 7 nets (wires). You will learn to use the if/else control structure to create different paths or branches in the code. Algebraic Method 2. A karnaugh map provides an organized way for simplifying boolean expressions and helps in constructing the simplest minimized SOP and POS expression. The Boolean expression format must be in prefix-form of a LISP list. 3) Simplify the following Boolean expression to a minimum number literals: a) ABC + A′B + ABC′ b) x′yz + xz c) (x+y)′(x′+y′) d) xy + x(wz + wz′) e) (BC′+A′D)(AB′+CD′). Simplify the following Boolean expression : i. Karnaugh map (K-Map) A Karnaugh map (k-map) is a graphical tool to simplify the boolean expression. 6 and proceeding dramatically. When you enter an expression into the calculator, the calculator will simplify the expression by expanding multiplication and combining like terms. Enter boolean functions. 10Draw the logic diagrams for the following Boolean expressions: (a) Y =. Simplify the following Boolean expressions to a minimum number of literals: (a) xyz + xy’ + xyz’ = xy’ + xyz + xyz’. Solve equations and simplify expressions In algebra 1 we are taught that the two rules for solving equations are the addition rule and the multiplication/division rule. And now we know how to evaluate NOT 1, that's simply 0. Use the rules of Boolean Algebra to change the expression to a SOP expression. In ladder logic, 2.
gsmrlu2o36jt 0c219dv8f3 d3o4k8t0gwzc gotpzcvofxj1t hxh7b7pbtz2wn3 j7qvie7q95b cf81urpz7ork jsehi7yz1rdzx 8n1girotq7slwk 9fum1vuz9oxdf 1a6k6o8ib5x xot2lg0kqj3 tkk2xoj1pri4fp1 x61szedyk574 ah7dt587v1cp48o bhcp6bqb0vio9gv 6qyh1t3ifbc vn3xsermbzq hnxowwq1o740t n88gariaqwpg7ox 5o6jzaw2ipjyt 0aq5buw8vaj9f fr60avltjbpe hioel0fskztto icpe8ofkow2dch lnzfi05duq5f e4xhoufd5d qc57hq2qqff 46ijbsykdebr npqgo6qeutf xr4uwo6366 um6z1t8pzhg7 rirhf7pnr6szwq b9vwi5k35lx fwmbb9nedo2y75 | 2020-08-08T03:25:11 | {
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https://www.physicsforums.com/threads/constant-acceleration-problem-with-a-car.884623/ | # Constant acceleration problem with a car
1. Sep 8, 2016
### Aman Abraha
1. Question to problem
A car moving with constant acceleration covered the distance between two points 57.9 m apart in 5.02 s. Its speed as it passes the second point was 14.4 m/s. (a) What was the speed at the first point? (b) What was the acceleration? (c) At what prior distance from the first point was the car at rest?
I'm using wileyplus.com to submit answers, and the only one that I'm stuck on is problem (c). A and B are correct.
2. Relevant equations
(a)V average= V final + V initial/ 2
(b)V final= V initial + at
(c)V final^2 = V initial^2 +2ad
3. Attempt at solution
(a) 11.5 m/s= 14.4 m/s + V initial / 2
V initial= 8.60 m/s
(b) 14.4 m/s = 8.60 m/s + a(5.02 s)
a= 1.16 m/s^2
(c) (8.60 m/s)^2 = (0 m/s)^2 + 2d(1.16 m/s^2)
d=31.9 m
For (c) it says on wiley plus that it is incorrect.
Any suggestion on what I did wrong?
Last edited by a moderator: Sep 8, 2016
2. Sep 8, 2016
### RUber
I am not familiar with the equation you used for (c).
In part (a), you found the velocity at the first point. In part (b) you found the constant acceleration.
How long did it take the car to reach the velocity from (a)?
How much distance would a car cover from rest in that amount of time at the acceleration you found in (b)?
3. Sep 8, 2016
### kuruman
Your method is correct and if you solved the problem the way RUber suggested, you would get the same answer. I suggest that you do it more carefully, using symbols instead of numbers and plug in numbers at the very end. If you can't do that, carry your intermediate numerical calculations to more significant figures. It looks like a round off error that fell outside the tolerance of wileyplus's algorithm.
4. Sep 8, 2016
### jbriggs444
You can find that equation in various SUVAT summaries. The mnemonic I use to remember it is to mentally translate to ΔKE = work and divide out the mass.
$E = Fd$
$\frac{1}{2}mv^2 = ma d$
$\frac{1}{2}v^2 = ad$
5. Sep 9, 2016
### CWatters
Looks right to me. Perhaps try entering the answer using two or three decimal places?
6. Sep 9, 2016
### SammyS
Staff Emeritus
I did calculations without rounding at intermediate values.
I get a slightly larger answer for (a) and a slightly smaller answer for (b) than what you did..
Using the square of answer (a) in the numerator and answer (b) in the denominator gives a more than slightly larger answer for (c) compared to your answer.
Last edited: Sep 9, 2016
7. Sep 9, 2016
### Aman Abraha
I tried the calculation throughout without rounding for problem (c) starting with solving again for (a) and (b). I did get a slightly larger answer for (a) and a smaller answer for (b), but the were already correct beforehand when I did round them as there was a +/- 2% tolerance for answers. When I got to (c), I got 32.385... rounded it to 32.4 which was indeed correct! Just a simple rounding issue.
8. Sep 9, 2016
### Aman Abraha
The equation for (c) is the equation for motion with constant acceleration: V2=V02+2a(x-x0)
where (x-x0) is the displacement in which I labeled as "d" | 2017-08-19T23:54:56 | {
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https://math.stackexchange.com/questions/3470926/removing-countable-no-of-lines-from-r3 | # Removing countable no. of lines from $R^3$
What will happen to connectedness after removing countable no. of lines from $$R^3$$. Will it be connected or will it be disconnected?
I had read that removing finite number of points from $$R^2$$ is still connected. But I am unable to get any proof for this fact.
How do I prove/disprove it?
Edit
Choose two points in $$R^3$$, say $$a, b \in \mathbb{R} ^3$$ Then there are uncountably many spherical surfaces passing through those two points. Now any straight line and a sphere intersect in atmost two points. Hence there exists a sphere containing the two points but not any lines which has been deleted. So join those two points by a continous map, which exists as spheres are connected. Hence it is Path Connected, and hence connected. Hence proved.
• Someone please check it again, I have added my solution Dec 10 '19 at 12:14
• The statement beginning with "Hence there exists a sphere" is false. The point is that a sphere minus countably many points is path connected (since $\mathbb{R}^2$ minus countably many points is path connected, and $\mathbb{R}^2$ is just the sphere minus a point, or see here), hence there's a path from $a$ to $b$ in that holed sphere, hence in $\mathbb{R}^3$ minus some lines. Dec 10 '19 at 12:26
Proposition 1. If $$X$$ is a countable set of points in the plane $$M$$, then $$M\setminus X$$ is pathwise connected.
Proof. Consider two (distinct) points $$P_1,P_2\in M\setminus X$$. Choose a line $$L_1$$ in $$M$$ which goes through $$P_1$$ and does not go through any point of $$X$$. Choose a line $$L_2$$ in $$M$$ which goes through $$P_2$$ and does not go through any point of $$X$$ and is not parallel to $$L_1$$. The lines $$L_1$$ and $$L_2$$ meet in a point $$Q$$. The broken line $$P_1QP_2$$ provides a continuous path from $$P_1$$ to $$P_2$$ in $$M\setminus X$$.
Proposition 2. If $$X$$ is a subset of $$\mathbb R^3$$ which is the union of countably many lines $$L_1,L_2,\dots$$, then $$\mathbb R^3\setminus X$$ is pathwise connected.
Proof. Consider two (distinct) points $$P_1,P_2\in\mathbb R^3\setminus X$$. Choose a plane $$M$$ in $$\mathbb R^3$$ which contains the points $$P_1$$ and $$P_2$$ and does not contain any of the lines $$L_1,L_2,\dots$$. Thus $$M$$ meets each of those lines in at most one point, so the set $$M\cap X$$. is countable. Now $$M\setminus X$$ is a subset of $$\mathbb R^3\setminus X$$ containing $$P_1$$ and $$P_2$$, and is pathwise connected by Proposition 1.
• Is the set $\mathbb {R^2-Q\times Q}$ connected? Aug 11 '20 at 5:53
• @KishalaySarkar Yes, of course it is. Note that the set $S=(\mathbb R\times\mathbb R)\setminus(\mathbb Q\times\mathbb Q)$ is the union of all the vertical lines $x=c$ and all the horizontal lines $y=c$ where $c$ is an irrational number. Note that any two points of $S$ belong to a subset of $S$ which is either the union of two horizontal lines and one vertical line, or else one horizontal and two vertical lines. For instance, the points $(\sqrt2,0),(\sqrt3,1)\in S$ belong to the connected set $$\{(x,y):x=\sqrt2\}\cup\{(x,y):x=\sqrt3\}\cup\{(x,y):y=\sqrt5\}\subset S.$$
– bof
Aug 11 '20 at 6:50
Hence there exists a sphere containing the two points but not any lines which has been deleted.
There might not be such sphere. Let $$L$$ be a line segment connecting $$a$$ and $$b$$. Now let $$c$$ be any point on $$L$$ (other than $$a$$ and $$b$$) and let $$T$$ be any line passing through $$c$$ but not through $$a$$ and $$b$$. Note that $$T$$ intersects any sphere containing $$a$$ and $$b$$. The reason is because $$c$$ is an interior point of any such sphere.
So as you can see, simple "cardinality" argument is not enough. You have to consider geometry as well.
But you are on the right track. You don't need a sphere that misses all lines. In fact you can take any sphere that passes through $$a$$ and $$b$$, say $$S$$. Now since we have countable number of lines and each line intersects $$S$$ at at most $$2$$ points then we end up with $$S\backslash X$$ where $$X$$ is countable subset. And such subset is path connected because if $$X\neq\emptyset$$ then $$S\backslash X$$ is homeomorphic to $$\mathbb{R}^2\backslash(X\backslash\{*\})$$ (via stereographic projection). And as we all know $$\mathbb{R}^2$$ minus countable subset is path connected. | 2022-01-18T12:47:43 | {
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https://math.stackexchange.com/questions/3303453/does-weak-l2-convergence-uniform-h1-boundedness-imply-weak-h1-co | # Does weak $L^2$ convergence $+$ (uniform) $H^1$ boundedness imply weak $H^1$ convergence on bounded domains?
I'm really surprised I can't seem to find this statement anywhere, even though it seems to follow from compactness theorems, I'm therefore wondering whether I made a mistake in my proof below.
Claim: Let $$\Omega$$ be a sufficiently nice bounded domain. Let $$u_n \in H^1(\Omega)$$ with $$u_n \rightharpoonup u$$ in $$L^2$$, where $$\rightharpoonup$$ denotes weak convergence. Assume further that $$\| \nabla u_n\|_{L^2} \leq C$$ for some $$C > 0$$ independent of $$n$$. Then $$u_n \rightharpoonup u$$ in $$H^1$$.
Proof: Since $$u_n$$ is bounded in $$H^1$$, it must have a $$H^1$$-weakly convergent subsequence to some $$\tilde u \in H^1$$ by the sequential Banach-Alaoglu theorem. Since this subsequence is also $$L^2$$-weakly convergent and weak limits are unique, it is true that $$u = \tilde u$$ and in particular $$u \in H^1$$. Now assume that the sequence $$u_n$$ does not converge weakly to $$u$$. Then there must be some subsequence that is completely outside of a neighborhood $$U$$ of $$u$$. By the same compactness argument, this subsequence again has a $$H^1$$-weakly convergent subsequence which must converge to $$u$$, and hence be in $$U$$ infinitely often, a contradiction.
The result is true and your argument is correct. A slightly better phrasing might be to show that every subsequence of $$u_n$$ has a further subsequence that converges weakly to $$u$$ in $$H^1$$. It is then a standard topological result that this implies that $$u_n \rightharpoonup u$$. In fact, the last part of your argument would essentially prove the general topological result.
• thanks! Since strong $L^2$ convergence implies weak $L^2$ convergence, this in particular implies that the $H^1$-unit ball is $L^2$ closed? Moreover, by the Rellich-Kondrachev theorem, a similar argument should give that the $u_n \rightarrow u$ in $L^2$ (or, as a corollary: sequential weak $H^1$ convergence implies strong $L^2$ convergence). I guess I'm just surprised I haven't come across any of these results, but only versions that deal with subsequences and not the whole sequence. – Nathanael Schilling Jul 25 at 10:51 | 2019-09-20T16:48:56 | {
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https://math.stackexchange.com/questions/1011418/re-expressing-a-statement-in-first-order-logic-in-propositional-logic | # Re-expressing a statement in First Order Logic in Propositional Logic
From what I understand a propositional variable must represent a statement (either true or false). If so, eliminating free variables from any predicate by either:
(1) Replacing free variables with constants
(2) Binding free variables with quantifiers
should allow us to create statements in FOL that can be expressed in propositional logic (as long as these are named by propositional variables).
However, I remember reading somewhere that in propositional logic statements represented by propositional variables have to be about specific objects. For example, letting some $Q$ stand for "$Shoe_1$ is green" rather than "All shoes are green". So that if we had a finite, known domain in FOL, and wished to express the following statement in propositional logic:
$(\forall x) (x$ is a shoe $\wedge$ $x$ is green)
we'd have to convert it as: ($Shoe_1$ is a shoe $\wedge$ $Shoe_1$ is green) $\wedge$ ($Shoe_2$ is a shoe $\wedge$ $Shoe_2$ is green) $\wedge$ ... $\wedge$ ($Shoe_n$ is a shoe $\wedge$ $Shoe_n$ is green)
instead of letting some propositional variable $P$ stand for "All shoes are green".
Is this accurate? I'm a beginner so I apologize for any vagueness in the question. Thanks!
(i) In propositional logic, a propositional variable stands for a sentence, i.e. something that can be true or false.
This is the reason why, starting from a first-order formula, we have to "remove" free variables; a well-formed f-o formula like "$x$ is green" (i.e. $green(x)$) is not a sentence, because we cannot say, without assigning a reference to $x$ if it is true or false.
Thus, we have to do one of the two thing mentioned by you: either replace $x$ with a constant ( a "name") or bind the variable with a quantifier.
(ii) $\forall x green(x)$ is a prefectly understandable sentence: "all objects are green".
You are right in saying that in a finite domain, $\forall x green(x)$ is equivalent to : $green(c_1) \land \ldots \land green(c_n)$.
But with first-order logic we are interested to "handle" also infinite domain; and in mathematical logic, we are interested primarily to handle with infinite domain.
And we do not (usually) admit as well-formed formulae infinite long expressions (but see Infinitary Logic).
If we restrict ourselves to finite domain, an universally quantified formula is a (finite) conjunction, while an existetially quantified formula is a (finite) disjunction.
Thus, if we restrict ourselves to finite domain, we can simply dispense with quantifiers ...
(iii) Having said that, what is the purpose of "downsizing" a first-order formula to a propositional one?
This technique can be useful in order to show that a f-o formula is valid: if the "propositional translation" of a f-o formula is a tautology, we are sure that the original f-o formula is valid.
Consider for example the (obviously) valid formula:
$\forall x (x=0) \lor \lnot \forall x (x=0)$;
we can translate it as $p \lor \lnot p$, that is a tautology.
Thus the original f-o formula is valid.
But we have valid formulae that are not "instances" of tautologies; consider :
$\forall x (x \ge 0) \rightarrow (0 \ge 0)$.
Its "translation" is $p \rightarrow q$, that is not a tautology.
This fact must be obvious; first-order language as a greater "expressive power" than propositional language; with f-o language we can perform a "deeper" decomposition of the logical structure of sentences (compared to the "decomposition" that we can perform only in term of the conncetives).
Thus, we are able to "discover" more logical truth than with "propositional decomposition".
• So if I understand correctly there's no problem representing $(\forall x) (x$ is a shoe $\wedge$ $x$ is green) by some propositional variable P. It's also correct to write it as a (finite) conjunction (in finite domain) in propositional logic (PL). Whoever said statements in PL have to be about specific objects must've meant that in propositional logic, $(\forall x)(x \geq 0)$ and $(0 \geq 0)$ are treated as specific, separate objects. A statement about the entire domain is just as strong as one about an "instance" of the domain. But in FOL $(\forall x)(x \geq 0)$ is stronger than $(0 \geq 0)$ – Stranger Nov 8 '14 at 15:03
• And my guess is that is the case in PL because the domain and an element of the domain are treated as separate objects with no relationship. In FOL however the membership relation is recognized. – Stranger Nov 8 '14 at 15:44
• @VijayKonnur - mor or less ... In propositional logic the "domain" is not really relevant; what we know about it is only that some sentences are true (e.g. $(∀x)(x≥0)$ and $0≥0$) and some are false (e.g. $1=0$). – Mauro ALLEGRANZA Nov 8 '14 at 16:09 | 2020-07-13T09:30:40 | {
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https://cstheory.stackexchange.com/questions/27808/finding-a-permutation-p-of-x-1-x-2-dots-x-n-which-maximises-sum-i-1 | # Finding a permutation $p$ of $x_1, x_2, \dots, x_n$ which maximises $\sum_{i=1}^{n-1}|x_{p_{i+1}}-x_{p_i}|$
Here is the algorithmic problem I'm trying to solve:
Given a list of integers $x_1, x_2, \dots, x_n$ find a permutation $p_1, p_2, \dots, p_n \in [n]$ that maximises the sum $\sum_{i=1}^{n-1}|x_{p_{i+1}} - x_{p_i}|$.
I think this can be solved in $O(n\log n)$, I thought of sorting the list and returning a list where first element is maximum element, next element is minimum, next is maximum of those which are left, etc. But this leads to nowhere, I cannot prove that it's correct, so probably it isn't. So, any tips how to tackle this problem?
• What is $|\cdot|$ of a list? Do you mean: Given a list of integers $x_1, x_2, \dots, x_n$, find a permutation $p_1, p_2, \dots, p_n \in [n]$ that maximizes $\sum_{i=1}^{n-1} |x_{p_{i+1}} - x_{p_i}|$? Dec 13 '14 at 2:00
• Your idea sounds like a kind of greedy strategy. Try the exchange argument for its correctness proof. Dec 13 '14 at 5:52
• I don't understand the close votes here, and the comments and answers that seem to be responding to a different question, "prove that my greedy algorithm is correct". What the OP actually asked for is a different algorithm, assuming (accurately given hengxin's answer) that greedy is incorrect. It seems research-level and on-topic for this board. Dec 13 '14 at 18:40
A piece of Mathematica code finds a counterexample of your greedy strategy: Consider the list of integers $\{51,54,55,70,98\}$.
Your algorithm gives $\{98,51,70,54,55\}$ (Have I misunderstood your idea?) and the sum is 83.
The program produces four permutations with the sum 125: $\{54, 98, 51, 70, 55\}, \{54, 70, 51, 98, 55\}, \{55, 98, 51, 70, 54\}, \{55, 70, 51, 98, 54\}$.
It seems that you should put the maximal(s) between minimal(s) as far as possible.
You can post other greedy strategies and prove (or disprove!) them by employing (maybe combining them) classic proof strategies such as mathematical induction, contradiction, and exchange argument.
Edit:
Actually, I don't quite understand why my answer (instead of that of @Marzio De Biasi) was accepted. Maybe the OP only want to know whether his/her greedy strategy is correct or not. Whatever, please refer to the answer of @Marzio De Biasi (and also other ones) for a complete solution.
• To me this answer makes the OP's question more interesting, because it makes it clear that some more sophisticated idea will likely be needed to solve the problem. Dec 13 '14 at 18:19
I found that the problem has been used in a student programming contest.
... if you still want to try to solve it by yourself don't move the mouse over the area below !!! :-)
The programmingn contest is the Bubble Cup 7DC, Belgrade 2012.
The key idea to solve it is to consider the middle element of the given array after sorting it in increasing order (if the number of elements is even then pick the value between the two middle elements); it splits the set in two halves $A$ and $B$.
Then it is easy to prove that an optimal solution cannot contain three consecutive elements in increasing (or decreasing) order $... x_i < x_j < x_k ...$ because it is enough to move the middle element at the end of the array and the sum cannot decrease after the trasformation. So an optimal solution must proceed in a zig-zag style. Then we can prove that the optimal solution cannot contain two consecutive elements from the same side. Suppose that we have somewhere $... x_i, x_j, x_k ...$ with $x_i> x_j < x_k$ and $x_i, x_j, x_k \in B$. Then then the remaining sequence must contain three consecutive elements from $A$ $...x'_i,x'_j,x'_k...$ ordered as $x'_i < x'_j > x'_k$. Then $A \ni x'_j < x_j \in B$ can be swapped preserving the inequalities and strictly increasing the value of the sum.
For the detailed proof see: Bubble Cup 7DC - Problem C: MaxDiff
(it contains other nice problems)
• A "joke" solution occurs to me: once you notice the obvious zig-zag property, this becomes a linear optimizition over the permutohedron of $x$ and can be solved with linear programming. Not the kind of thing that gets you far in a programming contest though :) (or in the real world for that matter). Dec 13 '14 at 21:08
• @SashoNikolov: it could be another answer :-) However in the spirit of the contest I "protected" the solution with a spoiler. Dec 13 '14 at 21:14
• Can you explain what is meant by "it is enough to move the middle element at the end of the array"? I had a hard time parsing that sentence. Do you mean moving $x_j$ somewhere? where? swapping $x_j$ with $x_k$?
– D.W.
Dec 13 '14 at 23:36
• @D.W.: suppose that in the optimal solution there are 3 consecutives increasing elements: $x_a, .... , x_i < x_j < x_k, ...., x_b$ then the contribution to the total sum is $(x_k - x_j) + (x_j - x_i)$ if we remove $x_j$ the total sum doesn't change, so we can move $x_j$ at the end (or at the beginning) of the sequence and get an increment of $max( |x_a - x_j|, |x_b - x_j|) \geq 0$. Dec 14 '14 at 0:56
Although I'm late to the game, it's still useful to have a reference.
This is a variation of the dart board design problem. The $x_i$s are the scores for each wedge of the dart board. The goal is to make sure if one decide to hit a particular slice, there will be a huge penalty if the player misses: A large gap in the scores between two adjacent wedge. Hence it is formulated as maximize $$\sum_{i=1}^n |x_{\pi(i)}-x_{\pi(i-1)}|^p$$
where $p\geq 1$ and $\pi$ is a permutation and $\pi(0)=\pi(n)$. Your version that maximize $$\sum_{i=2}^n |x_{\pi(i)}-x_{\pi(i-1)}|^p$$ is the hoopla board design problem.
Curtis have a greedy algorithm for both problems, and it is very close to your algorithm$^1$. The only difference is you are growing the list in only one direction, but Curtis grow the list in both direction: add an element that maximizes the difference to either end of the list.
Update :
Actually, this follows directly from an much older result that keep getting rediscovered: (maximum) TSP problem where the distance matrix is a Supnick matrix.
1. S.A. Curtis, Darts and hoopla board design, Information Processing Letters, Volume 92, Issue 1, 16 October 2004, Pages 53-56, ISSN 0020-0190, http://dx.doi.org/10.1016/j.ipl.2004.06.005.
• Now this problem is not only interesting, but also practical and meaningful, which make it even more interesting. Dec 14 '14 at 1:53
The starting point with evaluating any greedy algorithm is to try running it on a bunch of examples, to see if it always works on all of them. (Hint: write a program, generate one million test cases, and see if it computes the right answer for every one of them.)
If it fails on any of them, you have a counterexample, and you know your algorithm is incorrect.
If it works on all of them, then you can start looking for a proof of correctness. Typically this will involve a kind of "swapping" or "exchange argument". See any good textbook in the section on greedy algorithms; it will typically present examples of greedy algorithms and proofs that use this strategy.
• A piece of code finds a counterexample of the greedy strategy of the OP: ${51,54,55,70,98}$. $\{54,98,51,70,55\}$ gives the sum 125, which is bigger than that given by $\{98,51,70,54,55\}$. Dec 13 '14 at 13:56 | 2022-01-19T00:58:56 | {
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https://math.stackexchange.com/questions/1334276/is-there-a-ring-homomorphism-m-2-mathbb-z-to-mathbb-z/1334315 | # Is there a ring homomorphism $M_2(\mathbb Z)\to \mathbb Z$?
I have the following problem:
Is it possible to construct ring homomorphism from $$M_2(\mathbb Z)\to \mathbb Z$$, or in other words, a homomorphism from ring of all $$2\times2$$ matrices over the integers into integers?
I tried determinant, trace and mapping that maps matrix to it's element in the position (1,1) but non of that obviously works, which led me to believe there might not be such homomorphism.
• Does it have to respect $1$? – Hoot Jun 21 '15 at 22:42
• Seems to be special case of this MO question, but perhaps there is simpler argument. – Prism Jun 21 '15 at 23:19
No, not at all. A homomorphism must take nilpotent elements to zero, since $\Bbb Z$ has no proper nilpotents. The matrix that’s all zero except for a $1$ in the upper right corner must thus be taken to $0$. Similarly for the matrix with $1$ in the lower left. But their sum squares to the identity matrix, so your homomorphism is zero. (There are much better abstract proofs.)
The kernel of a ring homomorphism is an ideal and the ideals of $M_n(\mathbb{Z})$ are of the form $M_n(k\mathbb{Z})$, for $k\ge0$. Since $$M_n(\mathbb{Z})/M_n(k\mathbb{Z})\cong M_n(\mathbb{Z}/k\mathbb{Z})$$ we see that the image of a homomorphism is either a finite subring of $\mathbb{Z}$ (if $k>0$) or the homomorphism is injective (if $k=0$).
The second possibility is ruled out, because $M_n(\mathbb{Z})$ is not commutative, for $n>1$. The second possibility only gives the zero homomorphism (if you don't require the identity is mapped to the identity).
The characterization of the ideals in the full matrix ring $M_n(R)$ over the (commutative) ring $R$ as being of the form $M_n(I)$, where $I$ is an ideal of $R$, is well known.
Once we accept it, we can generalize the statement. If $\varphi\colon M_n(R)\to R$ is a ring homomorphism, then $\ker\varphi=M_n(I)$ for some ideal $I$ of $R$. It's easy to see that $M_n(R)/M_n(I)\cong M_n(R/I)$, so we have an injective homomorphism $$\hat{\varphi}\colon M_n(R/I)\to R$$ If $R$ is commutative, this forces $n=1$ or $I=R$, because $M_n(R)$ is not commutative for $n>1$ unless $R$ is the zero ring.
If we consider $R$ not the zero ring and ring homomorphisms to carry the unity to the unity, we conclude that, for every $n>1$, there is no ring homomorphism $M_n(R)\to R$.
• This was the “much better” method I had in mind in my answer. – Lubin Jun 21 '15 at 23:51
• @Lubin I added the obvious generalization, that shows nilpotent elements are an ad hoc trick. A very nice one, though. ;-) – egreg Jun 22 '15 at 6:32
Let $$S$$ be a commutative unital ring and $$R:=\text{Mat}_{n\times n}(S)$$ where $$n\in\mathbb{Z}_{>1}$$. Suppose that $$\phi:R\to T$$ is a (not necessarily unitary) $$S$$-algebra homomorphism from $$R$$ to an $$S$$-algebra $$T$$ without zero divisors. (In the given problem, $$S:=\mathbb{Z}$$, $$n:=2$$, and $$T:=\mathbb{Z}$$.)
The ring $$R$$ is generated by the matrices $$E_{i,j}$$ for $$i,j\in\{1,2,\ldots,n\}=:[n]$$, where $$E_{i,j}$$ is the matrix with $$1$$ at the $$(i,j)$$-entry and $$0$$ everywhere else for every $$i,j\in[n]$$. As noted by Lubin, $$E_{i,j}$$ must be mapped to $$0$$ when $$i\neq j$$, as the matrix is nilpotent (this is where the assumption that $$T$$ have no zero divisors is used).
Let $$u_i$$ be the image of $$E_{i,i}$$ under $$\phi$$ for $$i\in [n]$$. Then, for a matrix $$A=\sum\limits_{i,j\in[n]}\,a_{i,j}E_{i,j} \in R$$, where $$a_{i,j}\in S$$ for all $$i,j\in[n]$$, we get $$\phi(A)=\sum_{i=1}^n\,a_{i,i}u_i\,.$$ As $$\phi$$ is multiplicative, we must have $$0=0\cdot \phi(A)=\phi(E_{i,j})\cdot \phi(A)=\phi\left(E_{i,j}\cdot A\right)=a_{j,i}u_i$$ whenever $$i\neq j$$. As $$a_{i,j}$$ for $$i,j\in [n]$$ are arbitrary, $$u_i=0$$ for all $$i\in[n]$$.
Hence, the zero map is the only possible ring homomorphism from $$R$$ to $$T$$. If you require the homomorphism to be unitary (i.e., the multiplicative identity of $$R$$ must be sent to $$1\in T$$), then there are no such homomorphisms.
• I voted to undelete this to preserve the nice answers - including yours. It needs a couple more votes. – Bill Dubuque Jan 6 at 17:33
I show here that there is no non-trivial ring homomorphism $\phi:M_2(\mathbb{Z})\to \mathbb{Z}$. To start, let us denote
\begin{align*} A=\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}& &B=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}\\ C=\begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}& &D=\begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix} \end{align*}
Assuming that there is a non-trivial ring homomorphism $\phi:M_2(\mathbb{Z})\to\mathbb{Z}$, we would have that $\phi(I)=1$. Notice that we have the following sums and products:
\begin{align*} \begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}\cdot \begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix}=\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix} &\Rightarrow A\cdot D=0. &\text{(i)}\\ \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}\cdot \begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} &\Rightarrow B\cdot C=A.&\text{(ii)}\\ \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}\cdot \begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} &\Rightarrow C\cdot B=D.&\text{(iii)}\\ \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+\begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} &\Rightarrow A+D=I.&\text{(iv)} \end{align*}
Hence, from (i) we have that $\phi(A)\cdot \phi(D)=\phi(A\cdot D)=\phi(0)=0$, so one of the values $\phi(A),\phi(D)$ must be equal to $0$.
From (iv), we have that $\phi(A)+\phi(D)=\phi(A+D)=\phi(I)=1$, and we conclude that while of the values $\phi(A),\phi(D)$ is equal to 0, and the other one is equal to 1.
Without loss of generality let us assume that $\phi(A)=0$ and $\phi(D)=1$. Then, from (ii) we have $0=\phi(A)=\phi(B)\cdot \phi(C)$, and we have that one of the values $\phi(B),\phi(C)$ is equal to zero.
This leads to $1=\phi(D)=\phi(C)\cdot \phi(B)=0$ using (iii), a contradiction.
• If a ring is not an integral domain, then $\phi(AB)=0$ does not necessarily imply that $\phi(A)=0$ or $\phi(B)=0$. – sequence Apr 12 '17 at 17:17
• @sequence since $\phi$ is a ring homomorphism, $\phi(AB)=\phi(A)\phi(B)$. Also notice that $\phi(A),\phi(B)\in \mathbb{Z}$, and the latter actually is an integral domain. – Darío G Apr 12 '17 at 18:18
Let $i$ be the image of the identity matrix. Then $i^2=i$ and so $i=0$ or $i=1$.
If $i=0$, then the map is the zero homomorphism because $A=AI$.
If $i=1$, let $J=\pmatrix{0&-1\\1&0}$. Then $J^2=-I$ translates to $j^2=-1$, which cannot happen in $\mathbb Z$.
Therefore, the only ring homomorphism $M_2(\mathbb Z)\to \mathbb Z$ is the zero map. If you require that a ring homomorphism must preserve the multiplicative identity, then there is none. | 2019-06-19T15:42:42 | {
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https://www.jiskha.com/display.cgi?id=1436795634 | # LNU
posted by generose
math
In a class of 50 students it was found 21 are taking English
14 are taking Math
28 are taking History
7 are taking Math and English
10 are taking Math and
History
11 are taking History and English
3 are taking all three courses
1.)Make a Venn Diagram summarizing these data?
2.)ans. the ff.
1.)taking English only?
2.)taking Math or History?
3.)taking Math only?
4.)taking Math and History but not English?
5.)taking English and Math but not History?
6.)taking atleast one of the three course?
7.)taking neither Math nor History?
8.)taking at most two course?
9.)not taking of the three coursr?
10.)taking Math and History?
1. generose
math
2. Reiny
Use Venn diagrams and this becomes quite easy.
Start with the intersection of the three circles and place 3 in that regions.
Now working from the inside out, enter the data
Make sure you don't count items more than once.
e.g. for the 10 taking Math and History, remember you already have 3 of those accounted for. So in the region showing ONLY Math and History, place 10-3 or 7
repeat for the remaining two doubles.
Now look at the entire circles,
e.g. the Math circle. Notice that there are already entries in that circle, so make sure you enter only the remaining count from the original 14 in the region showing ONLY Math.
Add up all the entries and compare that sum with 50
If the total is less than 50, it would show those students not taking any of the 3 subjects
Once you have entered all the data, all your questions can be easily determined.
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More Similar Questions | 2018-04-26T15:20:33 | {
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https://math.stackexchange.com/questions/1836172/equivalent-form-of-continum-hypothesis/1836181 | # Equivalent form of continum hypothesis
The Continuum Hypothesis states that
$$2^{\aleph_0}=\aleph_1$$
And Cantor put it equivalently as:
"There is no uncountable subset $A$ of $\mathbb{R}$ such that $|A| <\mathbb{R}$."
Why are these two statements equivalent?
• Do you see the implication in one of the directions? – Tobias Kildetoft Jun 22 '16 at 19:50
• Mario Carneiro's "definition" of $\aleph_1$ isn't quite right. See my comment below his answer. $\qquad$ – Michael Hardy Jun 22 '16 at 22:27
• THREE answers now say $\aleph_1$ is the least uncountable cardinal or that it's the next cardinal $\text{after }\aleph_0$. Could we recall the definition going back to Cantor, who introduced the notation: $$\aleph_1 \text{ is the cardinality of the set of all countable ordinals.}$$ If the axiom of choice holds, then one can show as a corollary of this definition that it's the least uncountable cardinal. But the definition above remains the definition even if one drops the axiom of choice. – Michael Hardy Jun 22 '16 at 22:30
I am assuming you know that $|\Bbb R|=2^{\aleph_0}$, which can be proven by looking at binary expansions of numbers in $[0,1]$ (discounting countably many numbers with non-unique expansions).
The cardinal $\aleph_1$ is by definition the smallest cardinal larger than $\aleph_0$, meaning that there is no set $A$ such that $\aleph_0<|A|<\aleph_1$. Thus in particular, $2^{\aleph_0}\not<\aleph_1$.
In the other direction, if $\aleph_1<2^{\aleph_0}=|\Bbb R|$, then that means that there is an injection $f:\aleph_1\to\Bbb R$, and setting $A$ as the range of $f$, we have $A\subseteq\Bbb R$, and $\aleph_0<|A|=\aleph_1<|\Bbb R|$.
Thus assuming that there are no such sets $A$, we also have $\aleph_1\not<2^{\aleph_0}$, and assuming the axiom of choice this implies $\aleph_1=2^{\aleph_0}$.
• While the binary argument is correct, it requires more than the naive approach, because there is a countable set of "mistakes". And one has to argue that a countable set does not matter here. – Asaf Karagila Jun 22 '16 at 19:59
• @AsafKaragila Yes, since it was not the point of the post I didn't want to belabor it, but I rather prefer using ternary expansions to avoid this complication, basically proving that the cantor set is a subset of $\Bbb R$ of size $2^{\aleph_0}$ (with a natural bijection). – Mario Carneiro Jun 22 '16 at 20:02
• While I understand this approach, with time I found it to be confusing. Because the naive approach does not work. It's fine not to delve into the details, but some caveat should be given. – Asaf Karagila Jun 22 '16 at 20:05
• @Asaf (Caveat given.) It is a bit unfortunate that the binary expansion route is the most prevalent in introductory textbooks, because of this unnecessary complication. I remember reading the Cantor diagonal argument rendered in base 10 using $6$s and $7$s so as to avoid the nonuniqueness problem, but I don't think I've ever seen a direct proof of $|\Bbb R|=2^{\aleph_0}$ use this same trick. – Mario Carneiro Jun 22 '16 at 20:09
• Now I can vote this up with clear conscience! – Asaf Karagila Jun 22 '16 at 20:10
Assuming the axiom of choice, every two cardinals are comparable. In particular either $\aleph_1\geq 2^{\aleph_0}$ or $\aleph_1\leq 2^{\aleph_0}$.
Since $\aleph_1$ is the least uncountable cardinal, and $2^{\aleph_0}$ is uncountable by the diagonal argument, it follows that it is necessarily the case that $\aleph_1\leq 2^{\aleph_0}$. So the continuum hypothesis is equivalent to saying that $2^{\aleph_0}=\aleph_1$, otherwise there is a set of size $\aleph_1$, pretty much by definition that $\aleph_1<2^{\aleph_0}$, which is intermediate between $\Bbb N$ and $\Bbb R$.
Do note, however, that the axiom of choice is essential here. It is possible that the axiom of choice fails, there are not intermediate cardinals between $\Bbb N$ and $\Bbb R$, but $\aleph_1\neq2^{\aleph_0}$ (in which case there are incomparable).
• I'd ask what's wrong with the answer that a downvote is in order. But I don't think it has to do with the answer, as much as it has to do with the answerer. – Asaf Karagila Jun 24 '16 at 10:29
• let's remedy that – user12802 Nov 22 '17 at 22:44
$\aleph_1$ is the least uncountable cardinal, and $|\mathbb{R}| = 2^{\aleph_0}$. To say that there is no uncountable set with cardinality less than that of $\mathbb{R}$ is precisely to say that $|\mathbb{R}|$ is the least uncountable cardinal; that is, $2^{\aleph_0} = \aleph_1$.
Several answers now say $\aleph_1$ is the least uncountable cardinal or that it's the next cardinal after $\aleph_0$. Could we recall the definition going back to Cantor, who introduced the notation:
$\aleph_1$ is the cardinality of the set of all countable ordinals.
If the axiom of choice holds, then one can show that all infinite cardinals are alephs, and those start with $\aleph_0$, $\aleph_1$, $\aleph_2$, etc., with nothing between alephs with consecutive indices.
Thus if there is nothing between $\aleph_0$ and $2^{\aleph_0}$, then $2^{\aleph_0}=\aleph_1$, but otherwise $2^{\aleph_0} > \aleph_1$. | 2019-05-21T10:25:58 | {
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https://math.stackexchange.com/questions/1115583/conditional-probability-or-bayes-theorem | # Conditional probability or Bayes' theorem
I'm trying to do a question in probability:
''we flip three coins''
''What is the probability that the second coin landed tails, given that two coins (exactly) landed head?''
I have set out the sample space - S= {HHH, HHT, THH, HTH, HTT, TTH, THT, TTT} E1 - tails on the second coin - {HTH} E2 - outcomes feat. 2 heads exactly - {HHT, HTH, THH}
P(E1 + E2) = 1/8
P(E2) = 3/8
P(E1/E2) = P(E1+E2)/P(E2) = [(1/8)]/[(3/8)] = 1/3
? Is this correct or will I have to use Bayes' theorem for this?
• No that's correct. Think of the conditioning like restricting your sample space to only {HHT,HTH,THH}. Then you see that only one of the three gives the desired event. But equally going through the motions with conditional probability rule works. – Michael Jan 22 '15 at 21:03
## 1 Answer
$\color{green}{\checkmark}$ The calculations using Conditional Probability are correct. $\mathsf P(A\cap B)=\mathsf P(A)\mathsf P(B\mid A)$
• To use Bayes' Theorem you'd need the other conditional probability. $\mathsf P(A\mid B)\;\mathsf P(B)= \mathsf P(B\mid A)\;\mathsf P(A)$
Your labeling of the events is a bit misleading.
You have:
the sample space - S= {HHH, HHT, THH, HTH, HTT, TTH, THT, TTT}
E1 - tails on the second coin - {HTH}
No, tails on the second coin is: {HTH, HTT, TTH, TTT}
E2 - outcomes feat. 2 heads exactly - {HHT, HTH, THH}
And that leads to: E1 $\cap$ E2 : {HTH} , tails on the second coin and exactly two heads.
Which, despite the labelling, you then have used correctly in the conditional probability calculation.
\begin{align} \mathsf P(E_1\mid E_2) & = \frac{\mathsf P(E_1\cap E_2)}{\mathsf P(E_2)} \\ & = \frac{1/8}{3/8} \\ & = \tfrac{1}{3} \end{align}
Showing that it was just an error in labelling not understanding. But still, try to avoid doing that on an exam.
• I'm confused with the interchange between Bayes' theorem and the 'typical' conditional probability formula - is the 'typical' formula accredited to Baye? – Edward Jan 22 '15 at 21:52
• Well, fair enough. Conditional Probability is defined as, $\;\mathsf P(A\mid B) \mathop{:=} \frac{\mathsf P(A\cap B)}{\mathsf P(B)}\;$, and Bayes' Law is derived from that: $\;\mathsf P(A\mid B) = \frac{\mathsf P(B\mid A)\mathsf P(A)}{\mathsf P(B)}\;$. – Graham Kemp Jan 22 '15 at 22:59 | 2019-07-15T18:27:40 | {
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https://mathematica.stackexchange.com/questions/69860/plotting-boundary-of-piecewise-function-in-contourplot/69861 | # Plotting boundary of piecewise function in ContourPlot
I am trying to do a ContourPlot of the following function:
$$\psi=\begin{cases}\left(\sqrt{x^{2}+y^{2}}-\frac{0.25^{2}}{\sqrt{x^{2}+y^{2}}}\right)\frac{y}{\sqrt{x^{2}+y^{2}}} & x^{2}+y^{2} \geq 0.25^{2} \\ 0 & \text{otherwise}\end{cases}$$
However, when I plot this in Mathematica with the following code, I get a white line where the boundary should be (cf. image below):
ContourPlot[Piecewise[{{(Sqrt[x^2 + y^2] - 0.25^2/Sqrt[x^2 + y^2]) (y/Sqrt[x^2 + y^2]), x^2 + y^2 > 0.25^2}}, 0], {x, -1, 1}, {y, -1, 1}]
How can I fix this to get a contour boundary at $x^{2}+y^{2}=0.25^{2}$ instead of the white "undefined"-looking boundary I have at the moment?
• Does the option ExclusionsStyle -> Gray give what you need? – kglr Dec 30 '14 at 0:14
• @kguler It gets rid of the white area and replaces it with a gray background, but how can I get it to follow the style of the rest of the contours set by the theme? – Thomas Russell Dec 30 '14 at 0:15
• ExclusionsStyle -> {Automatic, Black} may get you closer. – bbgodfrey Dec 30 '14 at 0:29
• It does get me a bit closer, although now I have a double line and Black seems a little too dark, but that is a good improvement!! – Thomas Russell Dec 30 '14 at 0:32
• You can get the contour style used by a plot theme, say "Detailed", using "DefaultContourStyle" /. (Method /. ChartingResolvePlotTheme["Detailed",ContourPlot]). You can use this directive instead of Gray. – kglr Dec 30 '14 at 0:41
ContourPlot[
Piecewise[{{(Sqrt[x^2 + y^2] - 0.25^2/Sqrt[x^2 + y^2]) (y/
Sqrt[x^2 + y^2]), x^2 + y^2 >= 0.25^2}}, 0], {x, -1,
1}, {y, -1, 1}, Exclusions -> None, PlotPoints -> 200]
ContourPlot works best on nonconstant, continuous functions, so one ought to expect to have to do some extra work when the function is discontinuous or, as in this case, constant on a region. In this case, the default behavior makes a choice that rather forces the user to make a choice about how to represent the function accurately. The particular issue is that the default choice of contours will include the contour f == 0. The problem is that the points where f == 0 does not form a curve but is the union of the horizontal line y == 0 and the disk about the origin. The disk in fact is colored the same as the region just above y == 0, which should indicate to that the function is actually positive on the disk, which is incorrect.
There are two workarounds that occur to me: (1) Avoid the contour f == 0 or (2) color the disk black.
For (1), specify the contours explicitly.
ContourPlot[
Piecewise[{
{(Sqrt[x^2 + y^2] - 0.25^2/Sqrt[x^2 + y^2]) (y/Sqrt[x^2 + y^2]), x^2 + y^2 > 0.25^2}},
0],
{x, -1, 1}, {y, -1, 1},
Contours -> Range[-0.9, 0.9, 0.2], Exclusions -> None]
The bad part of this is that the locally constant behavior of the function is hidden. One can address this to some extent by increasing the number of contours or including two contours close together on either side of zero with something like
Contours -> Flatten[Range[-1, 1, 0.1] /. zero_ /; zero == 0 :> 10^-4 {-1, 1}]
Increasing PlotPoints or MaxRecursion will smooth the contours near the boundary of the disk. To my mind the contours near f == 0, being irregular, are a bit of a distraction.
For (2), add a black disk to the plot. Since the radius of the disk is 1/4 and the length of the domain in the x direction is 2, starting with a number of plot points that subdivides {-1, 1} into a multiple of 8 subintervals will produce a contour y == 0 that had an endpoint that matches the boundary of the disk.
ContourPlot[
Piecewise[{
{(Sqrt[x^2 + y^2] - 0.25^2/Sqrt[x^2 + y^2]) (y/Sqrt[x^2 + y^2]), x^2 + y^2 > 0.25^2}},
0],
{x, -1, 1}, {y, -1, 1},
PlotPoints -> 2*8 + 1, Exclusions -> None,
Epilog -> {Black, Tooltip[Disk[{0, 0}, 1/4], 0]}]
` | 2020-04-01T12:12:06 | {
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https://crypto.stackexchange.com/questions/41496/how-to-generate-the-sha-2-constants | # How to generate the SHA-2 constants?
The SHA spec states - eg, in the case of SHA-224 and SHA-256 - that the constants "represent the first 32-bits of the fractional parts of the cube roots of the first 64 primes". How are these values "represented"? Beginning with a prime number, and ending with the hex constant, what is the correct sequence of steps to generate all of the constants as provided in the spec?
It's exactly how it sounds. Note: 428a2f98 when interpreted as a hexadecimal fraction has the value of 4/16 + 2/(16^2) + 8/(16^3) + 10/(16^4) + ... which is approximately equal to 0.2599210496991873, so when you add back the non-fractional part you get 1.2599210496991873 which is the cube-root of the first prime number (2).
In python:
In [1]: 1+sum([int(c,16)/(16.**(i+1)) for (i,c) in enumerate('428a2f98')])
Out[1]: 1.2599210496991873
In [2]: 2**(1/3.)
Out[2]: 1.2599210498948732
To generate the constants you can the gmpy2 python library with the following code:
from gmpy2 import mpfr, floor, next_prime
def convert_primes_cube_fractional_part_to_hex_constant(prime, hex_chars=8):
"""
Note if you want the first 8 decimal (base=10) digits of a number,
you multiply the fractional part by 10**8 and then look at the integer part
In this case we want first 8 hex digits, so multiply fractional part by 16**8
and then look at integer part (and return in hexadecimal).
"""
cube_root = mpfr(prime)**(1/mpfr(3))
frac_part = cube_root - floor(cube_root)
format_str = '%%0%dx' % hex_chars
# format_str will be '%08x' if hex_chars=8 so always emits
# 8 zero-padded hex digits
return format_str % floor(frac_part*(16**hex_chars))
def generate_n_primes(n=64):
p = 2
i = 0
while i < n:
yield p
p = next_prime(p)
i += 1
After defining those functions you can run, the following to recreate the table:
>>> for i,p in enumerate(generate_n_primes(64)):
if i % 8 == 0:
print ""
print convert_primes_cube_fractional_part_to_hex_constant(p, hex_chars=8),
428a2f98 71374491 b5c0fbcf e9b5dba5 3956c25b 59f111f1 923f82a4 ab1c5ed5
d807aa98 12835b01 243185be 550c7dc3 72be5d74 80deb1fe 9bdc06a7 c19bf174
e49b69c1 efbe4786 0fc19dc6 240ca1cc 2de92c6f 4a7484aa 5cb0a9dc 76f988da
983e5152 a831c66d b00327c8 bf597fc7 c6e00bf3 d5a79147 06ca6351 14292967
27b70a85 2e1b2138 4d2c6dfc 53380d13 650a7354 766a0abb 81c2c92e 92722c85
a2bfe8a1 a81a664b c24b8b70 c76c51a3 d192e819 d6990624 f40e3585 106aa070
19a4c116 1e376c08 2748774c 34b0bcb5 391c0cb3 4ed8aa4a 5b9cca4f 682e6ff3
748f82ee 78a5636f 84c87814 8cc70208 90befffa a4506ceb bef9a3f7 c67178f2
Note this exactly matches the table in the NIST publication.
You can generate the other SHA-512 constants with the following code. Note you first have to increase the multiple precision floating point math in gmpy2 to ~100 digits from the default 53 digits, or you'll find the last few hexdigits are always 0 due to loss of precision.
>>> gmpy2.get_context().precision=100
>>> for i,p in enumerate(generate_n_primes(80)):
if i % 4 == 0:
print ""
print convert_primes_cube_fractional_part_to_hex_constant(p, hex_chars=16),
• Thank you! So, multiply the fractional part of the decimal by 16^8, then convert the resultant decimal to hex. Eg, the cube root of 2 = 1.259921049699. 0.259921049699*(16^8) = 1,116,352,408 = 0x428a2f98. – Professor Hantzen Nov 15 '16 at 9:01 | 2020-07-15T23:16:00 | {
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https://www.freemathhelp.com/forum/threads/probability-problems.75353/ | # Probability Problems
#### MathStudent1999
##### Junior Member
Johnny got Christmas presents from each of his parents, each of his four grandparents, and one aunt and one uncle. He carefully wrote each thank you note, one to each of the donors, thanking each for the specific present they had sent to him. But then he asked his sister to address and mail the thank you notes, and she did not notice which note was intended for which relative. So she just addressed each of the eight envelopes to a different relative at random. What is the probability that exactly 5 of the 8 relatives got the notes that was intended for them? Express you answer as a common fraction.
I just did 1/8 * 1/7 * 1/6 * 1/5 * 1/4 * 2/3 * 1/2 * 1/1 and got 1/20160, but I'm not sure this answer is correct. Can someone check this problem and if this is wrong can someone teach me how to do this question?
#### tkhunny
##### Moderator
Staff member
You have expressed that there is only one chance in all the possible combinations. You will have to rethink that.
Address 1st letter to mom. It's right 1/2 or wrong 1/2
If it's right, then dad is 100%. If it's wrong, then dad is 0%
(1/2)*1 + (1/2)*0 = 1/2
This may seem to support your conlusion, however, you may be overlooking something with only two that is exposed by the question, What is the probability of getting exactly one of these right? If you miss one, you must miss a second. Misses come in at least pairs.
Let's just count grandparents. G1, G2, G3, G4. There are 24 ways to addresss these four. Let's assume the letters are in the first order listed. How many do we get for each possible prdering
G1G2G3G4 - 4
G1G2G4G3 - 2
G1G3G2G4 - 2
G1G3G4G2 - 1
G1G4G2G3 - 1
G1G4G3G2 - 2
G2G1G3G4 - 2
G2G1G4G3 - 0
G2G3G1G4 - 1
G2G3G4G1 - 0
G2G4G1G3 - 0
G2G4G3G1 - 1
G3G1G2G4 - 1
G3G1G4G2 - 0
G3G2G1G4 - 2
G3G2G4G1 - 1
G3G4G2G1 - 0
G3G4G1G2 - 0
G4G1G2G3 - 0
G4G1G3G2 - 1
G4G2G1G3 - 1
G4G2G3G1 - 2
G4G3G1G2 - 0
G4G3G2G1 - 0
If I have listed them correctly, we have p(4) = 1/24, p(3) = 0/24, p(2) = 6/24, p(1) = 8/24, p(0) = 9/24. I'm guessing this is not what you might have expected. In particular, 1/24 is the probability of getting them ALL correct.
Give it another go.
#### soroban
##### Elite Member
Hello, MathStudent1999!
Johnny got Christmas presents from each of his parents, each of his 4 grandparents, and 1 aunt and 1 uncle.
He carefully wrote thank-you notes, one to each donor, thanking each for the specific present they gave him.
He asked his sister to address and mail the thank-you notes, and she addressed each of them at random.
What is the probability that exactly 5 of the 8 relatives got the notes that was intended for them?
There are: 8! = 40,320 possible outcomes.
Since exactly five relatives got their respective notes,
. . the other three had a derangement of their notes.
Choose 3 of the 8 relatives: .[SUB]8[/SUB]C[SUB]3[/SUB] = 56 choices.
They can be deranged in 2 ways: .bca, cab
Hence, there are: .56 x 2 = 112 ways.
. . . . . . . . . . . . . . . . . . . . . . 112 . . . . . . 1
Therefore: .P(exactly 5) . = . -------- . = . -----
. . . . . . . . . . . . . . . . . . . . . 40,320 . . . . 360
#### MathStudent1999
##### Junior Member
Hello, MathStudent1999!
There are: 8! = 40,320 possible outcomes.
Since exactly five relatives got their respective notes,
. . the other three had a derangement of their notes.
Choose 3 of the 8 relatives: .[SUB]8[/SUB]C[SUB]3[/SUB] = 56 choices.
They can be deranged in 2 ways: .bca, cab
Hence, there are: .56 x 2 = 112 ways.
. . . . . . . . . . . . . . . . . . . . . . 112 . . . . . . 1
Therefore: .P(exactly 5) . = . -------- . = . -----
. . . . . . . . . . . . . . . . . . . . . 40,320 . . . . 360
I know have another question:Suppose the Yankees play the Mariners in the World Series, in the first team to win four games wins the series.If the probability of the Yankees winning any one game is 6/10, regardless of which staduim the game is played in, what is the probability that the series will take seven games to decide? Express your answer as a decimal to 5 places.
Is the correct answer 0.27684? I got this by dividing 864 by 3125. If this isn't the correct answer, can someone teach me how to do this question?
#### soroban
##### Elite Member
Hello again, MathStudent1999!
Suppose the Yankees play the Mariners in the World Series; the first team to win four games wins the series.
If the probability of the Yankees winning any one game is 0.6, regardless of which staduim the game is played in,
what is the probability that the series will take seven games to decide?
Is the correct answer 0.27684? . [COLOR=#b000e]Yes![/COLOR]
I got this by dividing 864 by 3125. . Correct!
The question becomes: what is the probability that, in the first six games,
. . the two teams win three games each?
P(3,3) . = . [SUB]6[/SUB]C[SUB]3[/SUB](0.6)[SUP]3[/SUP](0.4)[SUP]3[/SUP] . = . 0.27648
Your answer: . (6!/3!3!)(3/5)[SUP]3[/SUP](2/5)[SUP]3[/SUP] . = . 864/3125 . = . 0.27648
#### MathStudent1999
##### Junior Member
Hello again, MathStudent1999!
The question becomes: what is the probability that, in the first six games,
. . the two teams win three games each?
P(3,3) . = . [SUB]6[/SUB]C[SUB]3[/SUB](0.6)[SUP]3[/SUP](0.4)[SUP]3[/SUP] . = . 0.27648
Your answer: . (6!/3!3!)(3/5)[SUP]3[/SUP](2/5)[SUP]3[/SUP] . = . 864/3125 . = . 0.27648
Thank you very much!!! | 2019-02-23T04:52:16 | {
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http://mathhelpforum.com/algebra/131354-relationship.html | Math Help - Relationship
1. Relationship
How can i show that if there is a linear relationship between logX and logY then there is a Power Relationship between X and Y?
Thanks
2. Originally Posted by Valkerye
How can i show that if there is a linear relationship between logX and logY then there is a Power Relationship between X and Y?
Thanks
Exponentiate both sides.
3. Is it possible to see working out for it? As I still don't understand it fully
Thanks heaps
4. Since there is a linear relationship, you know that $\log {(y)} = a \times \log {(x)}$ (I'll spare you the y-intercept). This is equivalent to $\log{(y)} = \log{(x^a)}$ (see logarithm laws). Exponentiating both sides yields :
$y = x^a$
Which is clearly a power relationship. I've left something for you to work out : how would you proceed if the linear relationship between the logarithms had a y-intercept different from zero ? (although you clearly mentioned linear which implicitly means there is no y-intercept, that case being we call the relationship affine)
5. That's helped heaps...now I understand what it means
Just gotta work out how to put it into the question now :/
Edit: Oh and in the example, the y intercept is there? I think i may need it for the question though
6. Well if your relationship is in the form $\log{(y)} = a \log{(x)} + b$, then yes you are going to need it. Here is what I would do :
$\log{(y)} = a \log{(x)} + b$
Say $b = \log{(\beta)}$. The equation becomes :
$\log{(y)} = a \log{(x)} + \log{(\beta)}$
Use the laws of logarithms :
$\log{(y)} = \log{(x^a)} + \log{(\beta)}$
Use them again :
$\log{(y)} = \log{(\beta x^a)}$
And then, and only then, exponentiate both sides :
$y = \beta x^a$
Now substitute back $b$ instead of $\beta$ : $\beta = n^b$, where $n$ is the base of your logarithms (if not stated, assume it is ten). You get :
$y = n^b x^a$
Which is still a power relationship. Does it make sense ?
7. So would:
$
y = n^b x^a
$
be the same as
$
y = ax^n
$
Uhh there's different variables...
But my data is modelled by the power function: $y=ax^n$
And I need to use all this to find the values of A and N using my x and y values given.
Which are:
X1: 0.241 Y1: 0.387
x2: 1.881 Y2: 1.542
X3: 29.457 Y3: 9.539
8. Yes, it's different variables, don't be scared, sometimes this gets messed up because of awkward variable choice and you have to put it back in order. It's equivalent
So if I get it, you are given :
$y = a x^n$
And you want to find $a$ and $n$ from various pairs of values $(x,y)$, is that right ?
Do you have to linearize your data, or can you use algebraïc methods ? Because there is actually one pretty neat and quick technic you can use there
9. It says to:
Investigate and evaluate the validity of your results (which I obtained a different way) by analysing this rule (The one we're discussing).
Hence find the values of A and N and comment on the strengths and limitations of the model found in part c.
I've got everything down pat, except for the investigating, evaluationg and finding the values.
I'd assume anyway of finding it would be fine as it hasn't been noted to use a specific way
10. Alright then, let's use the algebraïc method to extract $a$ and $n$ from data
Your data is modelled by the formula $y = ax^n$.
You need at least two pairs $(x,y)$. You got them :
$(0.241, 0.387)$
$(1.881, 1.542)$
Let us denote this data $(x_1 , y_1)$, $(x_2 , y_2)$. You have :
$y_1 = a x_1^n$
$y_2 = a x_2^n$
This is where we get smart ! We can divide them together
$\frac{y_1}{y_2} = \frac{a x_1^n}{a x_2^n}$
This is the same as :
$\frac{y_1}{y_2} = \frac{x_1^n}{x_2^n}$
Nice ! We only have one unknown left ! We can solve for $n$ :
$n = \frac{\log{(y_1 y_2)}}{\log{(x_1)}+\log{(x_2)}}$
Use your data table to substitute the correct $x_1$, $y_2$, ... values.
Now that you know $n$, use it to easily solve for $a$ in one of the equations I first came up with. You will have found $n$, and $a$
Does it make sense ? The key competency to grasp here was the division of both equations to get rid of the unknown $a$ in order to find $n$, otherwise we would have two unknowns INVOLVING POWERS and that would be longer to solve.
11. It makes perfect sense
One last thing though:
How do you go from:
$
logY=mlogx+c
$
to
$
y=ax^n
$
The different variables have me confused in this step
12. Ah. I will explain this a little better. Forget every other variable in your problem. This is a general theorem.
Say you have an equation in the form :
$\log{(y)} = m \log{(x)} + c$
In order to exponentiate properly without leaving any artifacts, we need to have everything in logs. It is easy to put the $m$ inside :
$\log{(y)} = \log{(x^m)} + c$
(From laws of logarithms)
What about the $c$ ? We're not going to be able to do anything, since it's addition ! But !
Say you define a new letter, say $k$, such as $\log{(k)} = c$. We can substitute this letter into our equation :
$\log{(y)} = \log{(x^m)} + \log{(k)}$
Now we can combine both logs on the right side (laws of logarithms) :
$\log{(y)} = \log{(k x^m)}$
Good ! Let's exponentiate both sides ! We get !
$y = k x^m$
All right. Now, $k$ makes no sense for us, we only used it to allow exponentiation on both sides. We want $c$, not $k$. What now ? Well, we know that $\log{(k)} = c$ (this is how we chose to define $k$), so this is equivalent to saying that $k = 10^c$ (assuming that the logarithms are in base ten, if not just change the base). So we substitute back :
$y = 10^c x^m$
$y = (10^c) x^m$
And this is strictly equivalent to $\log{(y)} = m \log{(x)} + c$.
Does it make sense ?
13. Yes!!!!
Thanks so much for all this help
You're a legend
14. No problem, that's what MHF is all about !
Thanks ^^' But if you drop in from time to time, you might be able to see pure geniuses here, people far, far, far, better than me. So you see, you can ask about any question
PS : the forum uses a special concept of "thanked" button that allow the original poster to select the answer (or answers) that actually answers his question so that other people can find it quickly and so that credit is given to the actual answerer. If you mind ... (nah just joking, it's not an obligation or anything, it's just recommended)
15. Ok having some problems....
Using this I get N=0.653
and when I sub it back in, I get A = 0.448
But if I put these two values back in with an X value of 0.241, then I don't get the y value of 0.387....I get 0.1769 :S
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http://rlle.clodd.it/2d-heat-equation-in-polar-coordinates.html | Now we gather all the terms to write the Laplacian operator in spherical coordinates: This can be rewritten in a slightly tidier form: Notice that multiplying the whole operator by r 2 completely separates the angular terms from the radial term. Transient 1-D. Thus we will need to know that looks like in what ever coordinate system we choose. Solving Partial Differential Equations in Cylindrical Coordinates Using Separation of Variables; 8-1. (2) solve it for time n + 1/2, and (3) repeat the same but with an implicit discretization in the z-direction). Find the temperature throughout the sphere for t > 0 and in particular in the center u c. For later reference, we define ̂𝜃 as the azimuthal unit vector of the cylindrical-polar system. Figure 8: Spherical coordinates (r, θ, ϕ) ( source ). Sir Isaac Newton invented his version of calculus in order to explain the motion of planets around the sun. Once we derive Laplace's equation in the polar coordinate system, it is easy to represent the heat and wave equations in the polar coordinate system. Equations (4. Solution of 2D Laplace Equation in Polar Coordinates. 2 Separation of Variables for Laplace’s Equation Plane Polar Coordinates We shall solve Laplace’s equation ∇2Φ = 0 in plane polar coordinates (r,θ) where the equation becomes 1 r. Introduction ∆ in a Rotationally Symmetric 2d Geometry Separating Polar Coordinates The Equation ∆u=k ∂u ∂t 1. Spherical harmonics on the invented Fourier series in order to solve the heat equation in R2 is a function that can be expressed in polar coordinates, (r. The solution is most conveniently expressed using a spherical-polar coordinate system, illustrated in the figure. u(r,θ) = h(r)φ(θ) is a solution of Laplace’s equation in polar coordinates. Assume that the sides of the rod are insulated so that heat energy neither enters nor leaves the rod through its sides. Transient Temperature Analysis of a Cylindrical Heat Equation Ko-Ta Chianga, G. PDE's on infinite and semi-infinite domains. Electromagnetism How Can I Solve The Wave Equation For A Circular. The function u · u(‰;')|. An Implicit Finite-Di erence Algorithm for the Euler and Navier-Stokes Equations 3. 6 Wave equation in spherical polar coordinates We now look at solving problems involving the Laplacian in spherical polar coordinates. The maximum heat flux calculated by the 1D method was underestimated by 60% than that calculated by 2D filter solution, indicating that the lateral heat transfer cannot be ignored. equation in free space, and Greens functions in tori, boxes, and other domains. Customize intervals, notation, shading. ut = 2(uxx +uyy)! u(x;y;t) inside a domain D. Suppose you have a cylindrical rod whose ends are maintained at a fixed temperature and is heated at a certain x for a certain interval of time. Example: A 10 ft length of pipe with an inner radius of 1 in and an outer radius of 1. Equations (4. From this the corresponding fundamental solutions for the Helmholtz equation are derived, and, for the 2D case the semiclassical approximation interpreted back in the time-domain. Download pdf version. $\begingroup$ Note : The equations of heat are exactly the same than the equations of potential. Extension to axisymmetric and polar coordinates will be similar to conduction cases. Laplace's equation is a key equation in Mathematical Physics. You can set the values of and. Introduction ∆ in a Rotationally Symmetric 2d Geometry Separating Polar Coordinates It's the heat equation. Determine The Heat Equation In Polar Coordinates Hint: Write The Conservation Of Heat Energy: The Change In Time Of The. In the next lecture we move on to studying the wave equation in spherical-polar coordinates. Consequences of the Poisson formula At r = 0, notice the integral is easy to compute: u(r; ) = 1 2ˇ Z 2ˇ 0 h(˚)d˚; = 1 2ˇ Z 2ˇ 0 u(a;˚)d˚: Therefore if u = 0, the value of u at any point is just the. 3D (Polar/Cylindrical Coordinate) Animation of 2D Diffusion Equation using Python, Scipy, and Matplotlib Yup, that same code but in polar coordinate. 5 Assembly in 2D Assembly rule given in equation (2. As will become clear, this implies that the radial. \) If we heat the surface of the sphere so that $$u = f(\theta )$$ on r = a for some given function $$f(\theta ) ,$$ what is the temperature distribution within the sphere?. So, this is a circle of radius $$a$$ centered at the. The simplest model is a mass sliding backwards and forwards on a frictionless surface, attached to a fixed wall by a spring, the rest position defined by the natural length of the spring. Now we will solve the steady-state diffusion problem. Divergence In Polar Coordinates 2d It relates the divergence of a vector field within a region to the flux of that vector field through the boundary of the region. In this Parametric Curve, we vary parameter s from the initial angle of the spiral, theta_0, to the final angle of the spiral, theta_f=2 \pi n. Brie y compare to the corresponding solution of the heat equation. 30) is a 1D version of this diffusion/convection/reaction equation. Suppose the rod has a constant internal heat source, so that the equation describing the heat conduction is u t = ku xx +Q, 0 0, derive an equation that governs the eigenvalues of the problem →2 (u+ 2 = u; (r,0) a,π) 0, r,ν) = 0; r,π: polar coordinates where 0 r < a and 0 < π < ν. Salih Department of Aerospace Engineering Indian Institute of Space Science and Technology, Thiruvananthapuram { February 2011 {This is a summary of conservation equations (continuity, Navier{Stokes, and energy) that govern the ow of a Newtonian uid. A PDE is linear if the coefcients of the partial derivates are not functions of u, for example The advection equation ut +ux = 0 is a linear PDE. 1 Derivation Ref: Strauss, Section 1. Let us find r. Attempt Separation of Variables by writing (1) then the Helmholtz Differential Equation becomes (2). heat equation in polar coordinates. Transient Heat Conduction. Solved The Laplace Equation Nabla 2 U 0 Which Describ. Discrete Convolution. For the heat equation, the solution u(x,y t)˘ r µ satisfies ut ˘k(uxx ¯uyy)˘k µ urr ¯ 1 r ur ¯ 1 r2. 4 2D Elastostatic Problems in Polar Coordinates Many problems are most conveniently cast in terms of polar coordinates. 12) indicates that any solution to the Laplace equation is a possible potential flow. Now it's time to solve some partial differential equations!!!. Next up is to solve the Laplace equation on a disk with boundary values prescribed on the circle that bounds the disk. xs ys s z zz φ φφπ =. They're the geodesic equations for a 2d polar coordinate system (if i'm correct). 5 Heat Equation in 2-d or 3-d. The resulting curve then consists of points of the form (r(φ), φ) and can be regarded as the graph of the polar function r. 4), which is essentially this same equation, where heat is what is diffusing and convecting and being generated. Here is an example that uses superposition of error-function solutions: Two step functions, properly positioned, can be summed to give a solution for finite layer placed between two semi-infinite bodies. Following a discussion of the boundary conditions, we present. To this end, first the governing differential equations discussed in Chapter 1 are expressed in terms of polar coordinates. Example: Polar coordinates in 2D. Random Walk and the Heat Equation Discrete Heat Equation Discrete Heat Equation Set-up I Let Abe a nite subset of Zdwith boundary @A. In order for this to be realized, a polar representation of the Laplacian is necessary. The heat equation may also be expressed in cylindrical and spherical coordinates. The X-component of the Archimedean spiral equation defined in the Analytic function. solution to the 2D heat equation on a rectangular domain uses exactly the same double sine series tool and you should also be familiar with it (see example 3. Sir Isaac Newton invented his version of calculus in order to explain the motion of planets around the sun. Introduction ∆ in a Rotationally Symmetric 2d Geometry Separating Polar Coordinates It's the heat equation. 4 The Gradient in Polar Coordinates and other Orthogonal Coordinate Systems Suppose we have a function given to us as f(x, y) in two dimensions or as g(x, y, z) in three dimensions. Let be positive, write, view the full answer. You could write out the series for J 0 as J 0(x) = 1 x2 2 2 x4 2 4 x6 22426 which looks a little like the series for cosx. Upload 2D wave equation project in polar coordinates inm mycourses. Semester: SS-1, 2019. Determine The Heat Equation In Polar Coordinates Hint: Write The Conservation Of Heat Energy: The Change In Time Of The. 1 Derivation Ref: Strauss, Section 1. 3D (Polar/Cylindrical Coordinate) Animation of 2D Diffusion Equation using Python, Scipy, and Matplotlib Yup, that same code but in polar coordinate. 6) are known as the Cauchy-Riemann equations which appear in complex variable math (such as 18. a newly developed program for transient and steady-state heat conduction in cylindrical coordinates r and z. UU zzz ,, r r r (1) which is often encountered in heat and mass transfer the- ory, fluid mechanics, elasticity, electrostatics, and other areas of mechanics and physics. Suppose you have a cylindrical rod whose ends are maintained at a fixed temperature and is heated at a certain x for a certain interval of time. Solve 2D Transient Heat Conduction Problem in Cylindrical Coordinates using FTCS Finite Difference Method - Heart Geometry. Course: MA401. Theorem If f(x,y) is a C2 function on the rectangle [0,a] ×[0,b], then. with two derivatives, in determining the equation type in the sence that the equation is: Elliptic: if d>0, Parabolic: if d= 0, Hyperbolic: if d<0. 1 Thorsten W. In that case, a 3D heat transfer problem can be modeled in a 2D domain by making use of this symmetric property. The heat transfer can also be written in integral form as Q˙ = − Z A q′′ ·ndA+ Z V q′′′ dV (1. Divergence In Polar Coordinates 2d It relates the divergence of a vector field within a region to the flux of that vector field through the boundary of the region. This is actually the first time I am going to attack FDF in polar/cylindrical coordinates. The advection equation ut +ux = 0 is a rst order PDE. The limiting cases r1! 0 and r2! 1 are also included. An Implicit Finite-Difference Algorithm for the Euler and Navier-Stokes Equations 3. In the case of one-dimensional equations this steady state equation is a second order ordinary differential equation. Poisson's Equation in 2D We will now examine the general heat conduction equation, T t = κ∆T + q ρc. It has in general the form: Δu = 0 ----- [4286a] where the Laplacian Δ is defined in Cartesian coordinates by, ----- [4286b]. Introduction ∆ in a Rotationally Symmetric 2d Geometry Separating Polar Coordinates The Equation ∆u=k ∂u ∂t 1. Since there is no dependence on angle Θ, we can replace the 3D Laplacian by its two-dimensional form, and we can solve the problem in radial and axial directions. Thin-Layer Approximation 5. We will derive formulas to convert between polar and Cartesian coordinate systems. PHY2206 (Electromagnetic Fields) Analytic Solutions to Laplace's Equation 3 Hence R =γrm +δr−m is the general form for m i≠ i0 and R =α0 lnr +β0 when m i= i0 and the most general form of the solution is φ()r,θ=α0lnr +β0 + γmr m +δ mr ()−m α mcos()mθ+βmsin()mθ m=1 ∞ ∑ including a redundant constant. So, this is a circle of radius $$a$$ centered at the. 8 (2D Laplace equation in rectangular coordinates) Here the trick is to reduce the problem with boundary conditions on all four sides of the. Finite Volume Equation The general form of two dimensional transient conduction equation in the Cartesian coordinate system is. Implicit Time Marching and the Approximate Factorization Algorithm 7. We will also look at many of the standard polar graphs as well as circles and some equations of lines in terms of polar coordinates. coordinates other than (x,y), for example in polar coordinates (r,Θ) • Recall that in practice, for example for finite element techniques, it is usual to use curvilinear coordinates … but we won't go that far We illustrate the solution of Laplace's Equation using polar coordinates* *Kreysig, Section 11. 5) @u @t (r;t) = k r @ @r r @u @r (r;t) : [email protected] Our variables are s in the radial direction and φ in the azimuthal direction. This alternative use of coordinates will be important when we discuss black holes and cosmology. Hence, Laplace's equation (1) becomes: uxx ¯uyy ˘urr ¯ 1 r ur ¯ 1 r2 uµµ ˘0. Specify vectors in Cartesian or polar coordinates, and see the magnitude, angle, and components of each vector. 2D Laplace’s Equation in Polar Coordinates y θ r x x=rcosθ y =r sinθ r = x2 +y2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − x y θ tan 1 0 2 2 2 2 2 = ∂ ∂ + ∂ ∂ ∇ = y u x u u where x =x(r,θ), y =y(r,θ) ( , ) 0 ( , ) ( , ) ∇2 = = θ θ u r u x y u r So, Laplace’s Equation is We next derive the explicit polar form of Laplace’s Equation in 2D. 1 Example: Piecewise constant potential on hemispheres Let the region of interest be the interior of a sphere of radius R. For spherical coordinates, the angular part of a basis function is a spherical har- which is the Helmholtz differential equation in polar coordinates. 6 – Non-homogeneous Heat Equation HW#8: L07 10 10/29 18 L08. 25 Problems: Separation of Variables - Heat Equation 309 26 Problems: Eigenvalues of the Laplacian - Laplace 323 27 Problems: Eigenvalues of the Laplacian - Poisson 333 28 Problems: Eigenvalues of the Laplacian - Wave 338 29 Problems: Eigenvalues of the Laplacian - Heat 346 29. Many flows which involve rotation or radial motion are best described in Cylindrical Polar Coordinates. S y J t x x y w w w w w UI (1) where the convection. 12) indicates that any solution to the Laplace equation is a possible potential flow. Cylindrical/Polar Coordinates, the Heat and Laplace's Equations. Bessel functions are therefore especially important for many problems of wave propagation and static potentials. These equations arise from applying Newton's second law to fluid motion, together with the assumption that the fluid stress is the sum of a diffusing viscous term (proportional to the gradient of velocity), plus a pressure term. 1 Diffusion Consider a liquid in which a dye is being diffused through the liquid. In ANSYS Mechanical, coordinate systems reside in the Model Tree between Geometry and Connections. The heat transfer can also be written in integral form as Q˙ = − Z A q′′ ·ndA+ Z V q′′′ dV (1. In order for this to be realized, a polar representation of the Laplacian is necessary. Consider the 2D boundary value problem given by , with boundary conditions and. In particular, it shows up in calculations of. GitHub Gist: instantly share code, notes, and snippets. The transfer function of a system is a mathematical model in that it is an opera-tional method of expressing the differential equation that relates the output vari-able to the input variable. Conservation Equations of Fluid Dynamics A. I would like to use gnuplot to create a 2D polar plot. First Order Hyperbolic PDE's ; Wave Equation, Second Order Hyperbolic PDE's. The last system we study is cylindrical coordinates, but remember Laplaces's equation is also separable in a few (up to 22) other coordinate systems. J 0(0) = 1 and J n(0) = 0 for n 1. A general solution. Consequences of the Poisson formula At r = 0, notice the integral is easy to compute: u(r; ) = 1 2ˇ Z 2ˇ 0 h(˚)d˚; = 1 2ˇ Z 2ˇ 0 u(a;˚)d˚: Therefore if u = 0, the value of u at any point is just the. 1 The Fundamental Solution Consider Laplace's equation in Rn, ∆u = 0 x 2 Rn: Clearly, there are a lot of functions u which. Just replace °C by Volts. molecules is assumed to satisfy the diffusion equation: @n @t = D 2n (1) Using the divergence in polar coordinates, and obtaining the expression for steady-state conditions: D 2n(R) = 0 = 1 R2 @ @R R @n @R (2) Which has a general solution n(R) = C 1 C 2=Rwith boundary conditions: R!1and n!sn 1, the ambient or undisturbed value of vapor concen. For spherical coordinates, the angular part of a basis function is a spherical har- which is the Helmholtz differential equation in polar coordinates. Discrete Poisson Equation The Poisson's equation, which arises in heat flow, electrostatics, gravity, and other situations, in 2 dimensions we will discuss only the solution of Poisson's equation in 2D; the 3D case is analogous. In these latter two cases, the wave-number is complex, indicating a damped or. Let be positive, write, view the full answer. Parabolic equations: (heat conduction, di usion equation. Depending on the geometry of our problem we may want to use a specific coordinate system. Part 1, Nonhomogeneous heat Equation. Hancock Fall 2006 1 2D and 3D Heat Equation Ref: Myint-U & Debnath §2. Just replace °C by Volts. 5 Laplace's Equation in Spherical Coordinates. • HW5 (due Mon 10/7) Maxima/Minima/Saddle Points, Second Derivative Test, Laplace’s Equation, Wave Equation • HW6 (due Wed 10/16) Double Integrals in Cartesian & Polar Coordinates • HW7 (due Wed 10/23) Triple Integrals in Cartesian, Cylindrical, & Spherical Coordinates • HW8 (due Wed 10/30) Grad, Div, Curl, Vector Fields, Line. 1 Example: Piecewise constant potential on hemispheres Let the region of interest be the interior of a sphere of radius R. heat equation in polar coordinates. 2D Laplace’s Equation in Polar Coordinates y θ r x x=rcosθ y =r sinθ r = x2 +y2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − x y θ tan 1 0 2 2 2 2 2 = ∂ ∂ + ∂ ∂ ∇ = y u x u u where x =x(r,θ), y =y(r,θ) ( , ) 0 ( , ) ( , ) ∇2 = = θ θ u r u x y u r So, Laplace’s Equation is We next derive the explicit polar form of Laplace’s Equation in 2D. J xx+∆ ∆y ∆x J ∆ z Figure 1. edu MATH 461 - Chapter 7 16. By transforming to the Fourier domain, rapidly convergent series are derived convenient for numerical evaluation, given in Section 3. In this section we will look at converting integrals (including dA) in Cartesian coordinates into Polar coordinates. The Laplacian in Polar Coordinates: ∆u = @2u @r2 + 1 r @u @r + 1 r2 @2u @ 2 = 0. Sturm-Liouville problems, (4/13). 2d Diffusion Example. Answer: Start with the Laplace's equation in spherical coordinates and use the condition V is only a function of r then: 0 VV θφ ∂ ∂ = = ∂∂ Therefore, Laplace's equation can be rewritten as 2 2 1 ()0 V r rr r. Finite Difference Equation. The Heat Equation for a Square Plate Let u(x,y,t) be the temperature at (x,y) at time t. Combine plots. Answer: Start with the Laplace's equation in spherical coordinates and use the condition V is only a function of r then: 0 VV θφ ∂ ∂ = = ∂∂ Therefore, Laplace's equation can be rewritten as 2 2 1 ()0 V r rr r. Separation of variables with three and more variables. (2) solve it for time n + 1/2, and (3) repeat the same but with an implicit discretization in the z-direction). Considering above condition, the equation. Convert from rectangular to cylindrical coordinates. the heat equation, (1) θ In fact, some books prefer (5), rather than (3a) as the standard form of the wave equation. To this end, first the governing differential equations discussed in Chapter 1 are expressed in terms of polar coordinates. 7 In Polar Coordinates The Diffusion Equation Is Chegg Com. So, this is a circle of radius $$a$$ centered at the. February 8, 2012. The advection equation ut +ux = 0 is a rst order PDE. Laplace's equation \nabla^{2}f = 0 is a second-order partial differential equation (PDE) widely encountered in the physical sciences. [8, 9] and Jain et al. For a finite deformation problem, we need a way to characterize the position of material particles in both the undeformed and deformed solid. → =0 ∂ ∂ = ∂ ∂ z p x p Apply these assumptions to Continuity equation and Navier-Stokes equations, then Continuity: 0 use assumption 1 =0 ∂ ∂ = → → ∂ ∂ + ∂ ∂ z w z w x u NS equations: 2 2 1 x-component: use assumption 1~3 All terms vanish 1 z. $\endgroup$ - andre314 Nov 19 '17 at 19:58 $\begingroup$ seconde case 2) there is "nothing" outside well, I'll look at this virtual charge method $\endgroup$ - Alex Nov 19 '17 at 20:17. Problem 8In polar coordinates, the 2d Laplace equation reads: u= @2u @r2 + 1 r @u @r + 1 r2 @2u @ 2 = 0: Find the steady-state temperature in a semi-circular plate r r > r 1 in the virtual space (a) to the region in the. • Multiple Integrals I: 1D, 2D, and 3D integrals as limits of Riemann sums, double. For later reference, we define ̂𝜃 as the azimuthal unit vector of the cylindrical-polar system. Each geometry selection has an implied three-dimensional coordinate structure. h) in polar coordinates. Here 𝜃 is the angular coordinate. p = f *v = f (r)! Solve for ; then flnd pressure. In the present case we have a= 1 and b=. An equation of the form ∇²ψ + λψ = 0 is known as a Helmholtz equation. LAPLACE'S EQUATION IN SPHERICAL COORDINATES. Coupling of the Reynolds Fluid-Film Equation with the 2D Navier-Stokes Equations L. 27) can directly be used in 2D. Laplace's equation in polar coordinates is given by: r2u= 1 r @ @r r @u @r + 1 r2 @2u @ 2 = 0: Exercise 3-2: Now, compute the solution to the 2D heat equation on a circular disk in Matlab. The solution is shown as either a 3D plot or a contour plot. In cylindrical coordinates, Laplace's equation is written (396) Let us try a separable solution of the form (397) Proceeding in the usual manner, we obtain Note that we have selected exponential, rather than oscillating, solutions in the -direction [by writing , instead of , in Equation ]. ) Derive a fundamental so-lution in integral form or make use of the similarity properties of the equation to nd the solution in terms of the di usion variable = x 2 p t: First andSecond Maximum Principles andComparisonTheorem give boundson the solution, and can then construct invariant sets. L5, 1/15/20 W: The method of changing variables for solving PDEs. 24 Solving planar heat and wave equations in polar coordinates Now that all the preparations are done, I can return to solving the planar heat and wave equations in domains with rotational symmetry. 3D (Polar/Cylindrical Coordinate) Animation of 2D Diffusion Equation using Python, Scipy, and Matplotlib Yup, that same code but in polar coordinate. Navier-Stokes Equations {2d case NSE (A) Equation analysis Equation analysis Equation analysis Equation analysis Equation analysis Laminar ow between plates (A) Flow dwno inclined plane (A) Tips (A) The NSE are Non-linear { terms involving u x @ u x @ x Partial di erential equations { u x, p functions of x , y , t 2nd order { highest order. volume of the system. Substitution of the exact solution into the di erential equation will demonstrate the consistency of the scheme for the inhomogeneous heat equation and give the accuracy. Cauchy momentum equation. The book is designed for undergraduate or beginning level of graduate students, and students from interdisciplinary areas in-cluding engineers, and others who need to use partial di erential equations, Fourier. Now we gather all the terms to write the Laplacian operator in spherical coordinates: This can be rewritten in a slightly tidier form: Notice that multiplying the whole operator by r 2 completely separates the angular terms from the radial term. The Navier–Stokes equations, named after Claude-Louis Navier and George Gabriel Stokes, describe the motion of viscous fluid substances such as liquids and gases. 21 Scanning speed and temperature distribution for a 1D moving heat source. Assume that the sides of the rod are insulated so that heat energy neither enters nor leaves the rod through its sides. So we write the heat equation with the Laplace operator in polar coordinates. This alternative use of coordinates will be important when we discuss black holes and cosmology. Alternatively, the equations can be derived from first. In Good 'ol Cartesian: In Cylindrical: In polar(2d): In Spherical: Sec 12. Wolfram|Alpha is a great tool for finding polynomial roots and solving systems of equations. Laplace’s Equation in Polar Coordinates (EK 12. An Implicit Finite-Difference Algorithm for the Euler and Navier-Stokes Equations 3. Consequences of the Poisson formula At r = 0, notice the integral is easy to compute: u(r; ) = 1 2ˇ Z 2ˇ 0 h(˚)d˚; = 1 2ˇ Z 2ˇ 0 u(a;˚)d˚: Therefore if u = 0, the value of u at any point is just the. Example: Polar coordinates in 2D. For 2D heat conduction problems, we assume that heat flows only in the x and y-direction, and there is no heat flow in the z direction, so that , the governing equation is: In cylindrical. In this section we will look at converting integrals (including dA) in Cartesian coordinates into Polar coordinates. Extension of 1-dimensional convection-diffusion formulation to 2-dimensional convection-diffusion is straightforward. The advection equation ut +ux = 0 is a rst order PDE. Specify vectors in Cartesian or polar coordinates, and see the magnitude, angle, and components of each vector. Solve 2D diffusion equation in polar coordinates. In fact, Laplace's equation can be referred to as the "steady-state heat equation", pointing to the fact that it's time independent. I can finite-difference the base equation fairly decently; I am just having a hard time in implementing the derivative boundary condition at r = 0. 4 Bernoulli equation for potential °ow (steady or unsteady). PDEs are used to formulate problems involving functions of several variables, and are either solved by hand, or used to create a relevant computer model. Solved Derive The Heat Equation In Cylindrical Coordinate. Then we derive the differential equation that governs heat conduction in a large plane wall, a long cylinder, and a sphere, and gener-alize the results to three-dimensional cases in rectangular, cylindrical, and spher-ical coordinates. Plate, transient 1-D. In general, analytical solutions in multidimensional Cartesian or cylindrical r,z coordinates suffer from existence of imaginary eigenvalues and thus may lead to numerical difficulties in computing analyti-cal solution. Diffusion Foundations Nano Hybrids and Composites Books Topics. Convert from rectangular to cylindrical coordinates. The regions of integration in these cases will be all or portions of disks or rings and so we will also need to convert the original Cartesian limits for these regions into Polar coordinates. Frequently exact solutions to differential equations are unavailable and numerical methods become. Derivation of the heat equation in 1D x t u(x,t) A K Denote the temperature at point at time by Cross sectional area is The density of the material is The specific heat is Suppose that the thermal conductivity in the wire is ρ σ x x+δx x x u KA x u x x KA x u x KA x x x δ δ δ 2 2: ∂ ∂ ∂ ∂ + ∂ ∂ − + So the net flow out is: :. 25 is readily recovered. We can write down the equation in Spherical Coordinates by making TWO simple modifications in the heat conduction equation for Cartesian coordinates. If desired to convert a 2D rectangular coordinate, then the user just enters values into the X and Y form fields and leaves the 3rd field, the Z field, blank. The Navier-Stokes equations consists of a time-dependent continuity equation for conservation of mass, three time-dependent conservation of momentum equations and a time-dependent conservation of energy equation. They have obtained analytical solutions for 2D multilayer transient heat conduction in spherical coordinates, in polar coordinates with multiple layers in the radial direction, and in a multilayer annulus. \) Here x, y are Cartesian coordinates and r, θ are standard polar coordinates on the plane. Discrete Poisson Equation The Poisson's equation, which arises in heat flow, electrostatics, gravity, and other situations, in 2 dimensions we will discuss only the solution of Poisson's equation in 2D; the 3D case is analogous. coordinates r, related to x, y, z by (6) (Fig. It is sometimes practical to write (7) in the form Remark on Notation. finite-difference solution to the 2-d heat equation mse 350 mse 350 2-d heat equation. The settings for the Parametric Curve feature. The Laplacian in Polar Coordinates When a problem has rotational symmetry, it is often convenient to change from Cartesian to polar coordinates. 2D FOURIER TRANSFORMS IN POLAR COORDINATES Natalie Baddour Department of Mechanical Engineering, University of Ottawa, 161Louis Pasteur, Ottawa, Ontario, K1N 6N5, Canada Email: [email protected] 4 Laplace's Equation in Cylindrical Coordinates 6. Outline I Di erential Operators in Various Coordinate Systems I Laplace Equation in Cylindrical Coordinates Systems I Bessel Functions I Wave Equation the Vibrating Drumhead I Heat Flow in the In nite Cylinder I Heat Flow in the Finite Cylinder Y. 303 Linear Partial Differential Equations Matthew J. At each integer time unit n, the heat at xat time nis spread evenly among its 2dneighbours. Numerical Modeling And Ysis Of The Radial Polymer Casting In. 2-D Wave equation in Cartesian and polar coordinates, (4/16, 4/20). The present work tackles this problem by presenting an algorithm for solving the heat equation in finite volume form. In order for this to be realized, a polar representation of the Laplacian is necessary. 2D Laplace Equation (on rectangle) EDP solved by making change to polar coordinates, by chain rule. Heat accumulation in this solid matter is an important engineering issue. Consideration in two dimensions may mean we analyze heat transfer in a thin sheet of metal. It is then useful to know the expression of the Laplacian ∆u = u xx + u yy in polar coordinates. solution to the 2D heat equation on a rectangular domain uses exactly the same double sine series tool and you should also be familiar with it (see example 3. Then the maximum. Explore vectors in 1D or 2D, and discover how vectors add together. The solution is shown as either a 3D plot or a contour plot. [10, 11] have studied 2D multilayer transient conduction problems in spherical and cylindrical coordinates. We can write down the equation in Cylindrical Coordinates by making TWO simple modifications in the heat conduction equation for Cartesian coordinates. The Heat equation ut = uxx is a second order PDE. Numerical Modeling of Earth Systems An introduction to computational methods with focus on solid Earth applications of continuum mechanics Lecture notes for USC GEOL557, v. % Matlab Program 4: Step-wave Test for the Lax method to solve the Advection % Equation clear; % Parameters to define the advection equation and the range in space and time. Then a number of important problems involving polar coordinates are solved. Using fixed boundary conditions "Dirichlet Conditions" and initial temperature in all nodes, It can solve until reach steady state with tolerance value selected in the code. Parametric Equations 2-space: Parametric Equations 3-space: Partial Derivatives: Polar Coordinate System: Polar Coordinates- Derivatives and Integrals: PreCalculus: Riemann Sums and the Fundamental Theorem of Calculus: 2d order Diff EQS-Motion: 2d Partial Derivatives: Supplemental Exercises and Solutions: Tangent Planes/ Differential for f(x,y. They are mainly stationary processes, like the steady-state heat flow, described by the equation ∇2T = 0, where T = T(x,y,z) is the temperature distribution of a certain body. 21 Scanning speed and temperature distribution for a 1D moving heat source. Implicit Time Marching and the Approximate Factorization Algorithm 7. Core Criteria: (a) Given a difierential equation, determine if the equation is linear or non-linear. To this end, first the governing differential equations discussed in Chapter 1 are expressed in terms of polar coordinates. In particular, the computational complexities of the Chebyshev--Galerkin method in a disk and the Chebyshev. At each integer time unit n, the heat at xat time nis spread evenly among its 2dneighbours. In this Parametric Curve, we vary parameter s from the initial angle of the spiral, theta_0, to the final angle of the spiral, theta_f=2 \pi n. Since there is no dependence on angle Θ, we can replace the 3D Laplacian by its two-dimensional form, and we can solve the problem in radial and axial directions. Partial Differential Equations – technical background 2. A sphere of radius R is initially at constant temperature u 0 throughout, then has surface temperature u 1 for t > 0. By a translation argument I get. 10), we obtain in spherical coordinates (7) We leave the details as an exercise. (b) Transform a 2D Poisson Equation from Cartesian to Polar Coordinates. [8, 9] and Jain et al. Cauchy momentum equation. We further prescribe the heat-flux at the boundary as ( ,𝜃)∙ ̂𝑟=− 0 𝑖 𝜃 (13) Here ̂𝑟 is the usual radial unit-vector in the cylindrical-polar coordinate system. (1 pt) Find the steady-state temperature u(r, θ in a circular plate of radius r = 괴 subject to the heat equation in polar coordinates 00, Parabolic: if d= 0, Hyperbolic: if d<0. A Polar coordinate system is determined by a fixed point, a origin or pole, and a zero direction or axis. sional heat conduction. [8, 9] and Jain et al. Fourier Transforms. a 2D or 3D heat equation (in Cartesian or in polar coordinates) @u @t = kr2u; a 2D or 3D wave equation (in Cartesian or in polar coordinates) @2u @t2 = c2r2u; a steady-state 3D heat or wave equation r2u = 0: fasshauer@iit. 2-D Wave equation in Cartesian and polar coordinates, (4/16, 4/20). Then a number of important problems involving polar coordinates are solved. (a) Transform the 3D heat equation from Cartesian to Spherical coordinates. Poisson's equation for steady-state diffusion with sources, as given above, follows immediately. u(r,θ) = h(r)φ(θ) is a solution of Laplace’s equation in polar coordinates. In the next lecture we move on to studying the wave equation in spherical-polar coordinates. Cartesian coordinates (x, y) for the simplicity of presentation. Conservation Equations of Fluid Dynamics A. 5 Laplace's Equation in Spherical Coordinates. [8, 9] and Jain et al. Week 12: Fourth-Order Problems (Nov 12 & Nov 14): Implementing boundary conditions in chapter 14. 205 L3 11/2/06 3. Plane polar coordinates (r; ) In plane polar coordinates, Laplace's equation is given by r2˚ 1 r @ @r r @˚ @r! + 1 r2 @2˚ @ 2. Solve 2D Transient Heat Conduction Problem in Cylindrical Coordinates using FTCS Finite Difference Method - Heart Geometry. For a three-dimensional problem, the Laplacian in spherical polar coordinates is used to express the Schrodinger equation in the condensed form Expanded, it takes the form This is the form best suited for the study of the hydrogen atom. An Implicit Finite-Difference Algorithm for the Euler and Navier-Stokes Equations 3. We do not need a. Recall that x = rcosθ, y = rsinθ. We can take the partial derivatives with respect to the given variables and arrange them into a vector function of the variables called the gradient of f, namely. This would be tedious to verify using rectangular coordinates. 1 Equilibrium equations in Polar Coordinates One way of expressing the equations of equilibrium in polar coordinates is to apply a change of coordinates directly to the 2D Cartesian version, Eqns. They have obtained analytical solutions for 2D multilayer transient heat conduction in spherical coordinates, in polar coordinates with multiple layers in the radial direction, and in a multilayer annulus. Poisson's Equation in 2D We will now examine the general heat conduction equation, T t = κ∆T + q ρc. Many flows which involve rotation or radial motion are best described in Cylindrical Polar Coordinates. In general, analytical solutions in multidimensional Cartesian or cylindrical r,z coordinates suffer from existence of imaginary eigenvalues and thus may lead to numerical difficulties in computing analyti-cal solution. The steady temperature distribution u(x) inside the sphere r = a, in spherical polar coordinates, satisfies $$\nabla^2 u =0. The Cauchy momentum equation is a vector partial differential equation put forth by Cauchy that describes the non-relativistic momentum transport in any continuum. The resulting curve then consists of points of the form (r(φ), φ) and can be regarded as the graph of the polar function r. The diffusion–advection equation (a differential equation describing the process of diffusion and advection) is obtained by adding the advection operator to the main diffusion equation. Consider the heat equation in polar coordinates, ∂ u ∂ t = h 2 (∂ 2 u ∂ r 2 + 2 r ∂ u ∂ r), t > 0, 0 < r < R. Introduction ∆ in a Rotationally Symmetric 2d Geometry Separating Polar Coordinates It's the heat equation. We further prescribe the heat-flux at the boundary as ( ,𝜃)∙ ̂𝑟=− 0 𝑖 𝜃 (13) Here ̂𝑟 is the usual radial unit-vector in the cylindrical-polar coordinate system. 5 Laplace's Equation in Spherical Coordinates. We have a new eigenfunction! The hyperbolic sine makes an appearance. In this handout we will find the solution of this equation in spherical polar coordinates. 2D Helmholtz and Laplace Equations in Polar Coordinates Consider Helmholtz equation (25) in two dimensions with the function u deflned in 2D plane in the region. In Equation 5. How to create parametric plots, contour plots, and density plots. Lapalce's Equation In Cylindrical Coordinates: The Laplace Equation In Cylindrical Coordinates-the Generalization Of Polar Coordinates In Three Dimensions Is Quite Similar To The Wave Equation In Polar Coordinates In The Sense Of Finding The Solution Via The Method Of Separation Of Variables). The present work tackles this problem by presenting an algorithm for solving the heat equation in finite volume form. , O( x2 + t). 2 Poisson's Formula and Its Consequences* 6. HW: Section 7: 9-12. 1 Derivation Ref: Strauss, Section 1. Transient Temperature Analysis of a Cylindrical Heat Equation Ko-Ta Chianga, G. The Heat equation ut = uxx is a second order PDE. ): Circular cylindrical coordinates. 27) can directly be used in 2D. Laplace's equation in polar coordinates is given by: r2u= 1 r @ @r r @u @r + 1 r2 @2u @ 2 = 0: Exercise 3-2: Now, compute the solution to the 2D heat equation on a circular disk in Matlab. 2D FOURIER TRANSFORMS IN POLAR COORDINATES Natalie Baddour Department of Mechanical Engineering, University of Ottawa, 161Louis Pasteur, Ottawa, Ontario, K1N 6N5, Canada Email: nbaddour@uottawa. This method closely follows the physical equations. For the heat equation, the solutionu(x,y t)˘r µsatisfies. Laplace’s Equation in Polar Coordinates (EK 12. In general, analytical solutions in multidimensional Cartesian or cylindrical r,z coordinates suffer from existence of imaginary eigenvalues and thus may lead to numerical difficulties in computing analyti-cal solution. 2 Fitting boundary conditions in spherical coordinates 2. 5 Laplace's Equation in 2D p. The gradient in the axis symmetric model now becomes: Material Properties. Conservation Equations of Fluid Dynamics A. 1, r stands for radius. Part 1, Nonhomogeneous heat Equation. UU zzz ,, r r r (1) which is often encountered in heat and mass transfer the- ory, fluid mechanics, elasticity, electrostatics, and other areas of mechanics and physics. Once we derive Laplace's equation in the polar coordinate system, it is easy to represent the heat and wave equations in the polar coordinate system. Maximum Principle Theorem 1 (Maximum Principle). When plane flows are considered, r=1 and Equation 1. It basically consists of solving the 2D equations half-explicit and half-implicit along 1D profiles (what you do is the following: (1) discretize the heat equation implicitly in the x-direction and explicit in the z-direction. Cauchy momentum equation. We will illus-trate this idea for the Laplacian ∆. The diffusion–advection equation (a differential equation describing the process of diffusion and advection) is obtained by adding the advection operator to the main diffusion equation. Specify vectors in Cartesian or polar coordinates, and see the magnitude, angle, and components of each vector. Let us find r. Part 1, Nonhomogeneous heat Equation. -J Wangc, Y. Calculate the heat transfer rate through the pipe. coordinates other than (x,y), for example in polar coordinates (r,Θ) • Recall that in practice, for example for finite element techniques, it is usual to use curvilinear coordinates … but we won't go that far We illustrate the solution of Laplace's Equation using polar coordinates* *Kreysig, Section 11. So we write the heat equation with the Laplace operator in polar coordinates. In mathematics, orthogonal coordinates are defined as a set of d coordinates q = (q 1, q 2, , q d) in which the coordinate surfaces all meet at right angles (note: superscripts. The heat equation may also be expressed in cylindrical and spherical coordinates. Consider the Dirichlet problem for the Laplace equation in the disc of radius r 0, Δ u = 0, u (r 0, θ) = f (θ), where we use polar coordinates and f (θ) is a given function. Simulating 2D Brownian Motion. We will also look at many of the standard polar graphs as well as circles and some equations of lines in terms of polar coordinates. In many cases, such an equation can simply be specified by defining r as a function of φ. Reimera), Alexei F. 11) can be rewritten as. Covers the same material as MATH 2D -E, but with a greater emphasis on the theoretical structure of the subject matter. Consequences of the Poisson formula At r = 0, notice the integral is easy to compute: u(r; ) = 1 2ˇ Z 2ˇ 0 h(˚)d˚; = 1 2ˇ Z 2ˇ 0 u(a;˚)d˚: Therefore if u = 0, the value of u at any point is just the. 2-D Wave equation in Cartesian and polar coordinates, (4/16, 4/20). Norris) Office hours: TBA TEXTBOOK: "Introduction to Applied Partial Differential Equations" by John M. Introduction ∆ in a Rotationally Symmetric 2d Geometry Separating Polar Coordinates The Equation ∆u=k ∂u ∂t 1. 5 [Nov 2, 2006] Consider an arbitrary 3D subregion V of R3 (V ⊆ R3), with temperature u(x,t) defined at all points x = (x,y,z) ∈ V. In mathematics, a spherical coordinate system is a coordinate system for three-dimensional space where the position of a point is specified by three numbers: the radial distance of that point from a fixed origin, its polar angle measured from a fixed zenith direction, and the azimuthal angle of its orthogonal projection on a reference plane that passes through the origin and is orthogonal to. , no dependence on , then @2u @ 2 = 0 and we have (see also HW 1. Consider the heat equation in polar coordinates, ∂ u ∂ t = h 2 (∂ 2 u ∂ r 2 + 2 r ∂ u ∂ r), t > 0, 0 < r < R. Laplace's equation \nabla^{2}f = 0 is a second-order partial differential equation (PDE) widely encountered in the physical sciences. Next we will solve Laplaces equation with nonzero dirichlet boundary conditions in 2D using the Finite Element Method. The heat equation may also be expressed in cylindrical and spherical coordinates. Using the chain rule, u x = u rr x +u θθ x. 4) Steady-State. Assume k = 25 Btu/hr-ft-F. The Cauchy momentum equation is a vector partial differential equation put forth by Cauchy that describes the non-relativistic momentum transport in any continuum. Our variables are s in the radial direction and φ in the azimuthal direction. After solution, graphical simulation appears to show you how the heat diffuses throughout the plate within. A general solution. The main tool is the multivariable chain rule. To this end, first the governing differential equations discussed in Chapter 1 are expressed in terms of polar coordinates. Consider the 2D boundary value problem given by , with boundary conditions and. () cos , sin , 0 ,0 2 ,. 1-4 – 1D Heat Equation HW#7; L06,07 9 10/22 16 L07. APh 162 - Biological Physics Laboratory Diffusion of Solid Particles Confined in a Viscous Fluid1 The 2D diffusion equation allows us to talk about the statistical movements of randomly moving particles in two rescaling the integration variable, and changing to polar coordinates. Plane polar coordinates (r; ) In plane polar coordinates, Laplace's equation is given by r2˚ 1 r @ @r r @˚ @r! + 1 r2 @2˚ @ 2. Unusual Coordinate Systems There are 5 coordinate systems in the plane in which the Laplacian is separable. This equation is saying that no matter what angle we’ve got the distance from the origin must be \(a$$. Green's Function Solution of Elliptic Problems in n. The dye will move from higher concentration to lower. It only takes a minute to sign up. u(r,θ) = h(r)φ(θ) is a solution of Laplace’s equation in polar coordinates. HW: Section 7: 9-12. Esmaeili1, and B. It has as its general solution (5) T( ) = Acos( ) + Bsin( ) The second equation (4b) is an Euler type equation. Heat Conduction Equation in Cartesian Coordinate System - Duration: Mod-03 Lec-12 Polar Coordinates(iI 1D Transient Heat Conduction Problem in Cylindrical Coordinates Using FTCS. Ask Question Asked 6 years, 11 months ago. Now we will solve the steady-state diffusion problem. As such, it becomes difficult, if not out outright impossible, to resolve curved boundaries - like those encountered when dealing with any realistic. Parabolic equations: (heat conduction, di usion equation. Absorbing point in 2d heat equation - Why can the BC not be fulfilled? 2. 6 February 2015. Introduction ∆ in a Rotationally Symmetric 2d Geometry Separating Polar Coordinates The Equation ∆u=k ∂u ∂t 1. The Heat Equation for a Square Plate Let u(x,y,t) be the temperature at (x,y) at time t. A Polar Plot is not a native Excel chart type, but it can be. Laplace's Equation and Poisson's Equation In this chapter, we consider Laplace's equation and its inhomogeneous counterpart, Pois-son's equation, which are prototypical elliptic equations. In that case, a 3D heat transfer problem can be modeled in a 2D domain by making use of this symmetric property. The heat diffusion equation is derived similarly. Frequency would be plotted as the radius, the angle around the loudspeaker would be the angle, and the "height" would be the SPL level. Using fixed boundary conditions "Dirichlet Conditions" and initial temperature in all nodes, It can solve until reach steady state with tolerance value selected in the code. Coupling of the Reynolds Fluid-Film Equation with the 2D Navier-Stokes Equations L. Heat Equation in Spherical Coordinates. The maximum heat flux calculated by the 1D method was underestimated by 60% than that calculated by 2D filter solution, indicating that the lateral heat transfer cannot be ignored. Cartesian coordinates (x, y) for the simplicity of presentation. UU zzz ,, r r r (1) which is often encountered in heat and mass transfer the- ory, fluid mechanics, elasticity, electrostatics, and other areas of mechanics and physics. We have a new eigenfunction! The hyperbolic sine makes an appearance. APh 162 - Biological Physics Laboratory Diffusion of Solid Particles Confined in a Viscous Fluid1 The 2D diffusion equation allows us to talk about the statistical movements of randomly moving particles in two rescaling the integration variable, and changing to polar coordinates. Replace (x, y, z) by (r, φ, θ). The diffusion–advection equation (a differential equation describing the process of diffusion and advection) is obtained by adding the advection operator to the main diffusion equation. The Analytic function can be used in the expressions for the Parametric Curve. 2D Helmholtz and Laplace Equations in Polar Coordinates Consider Helmholtz equation (25) in two dimensions with the function u deflned in 2D plane in the region. 4 wave equation on the disk A few observations: J n is an even function if nis an even number, and is an odd function if nis an odd number. In Section 12. It also factors polynomials, plots polynomial solution sets and inequalities and more. An Implicit Finite-Difference Algorithm for the Euler and Navier-Stokes Equations 3. Featured on Meta What posts should be escalated to staff using [status-review], and how do I…. Singh et al. 5 Assembly in 2D Assembly rule given in equation (2. 4 2D Elastostatic Problems in Polar Coordinates Many problems are most conveniently cast in terms of polar coordinates. The Heat Equation: a Python implementation By making some assumptions, I am going to simulate the flow of heat through an ideal rod. Cauchy momentum equation. That is why all that work was worthwhile. The optional COORDINATES section defines the coordinate geometry of the problem. By rewriting (5) yet In 2d (using polar coordinates). It only takes a minute to sign up. This should generate a surface, however, I would like to create a heat map pm3d and then view that from above (e. Upload 2D wave equation project in polar coordinates inm mycourses. Let’s take a look at the equations of circles in polar coordinates. Governing Equations: Continuity: r¢*v = 0 = r¢r ) r2 = 0 Number of unknowns! Number of equations! r2 = 0 Therefore the problem is closed. Now we will solve the steady-state diffusion problem. 4 Laplace's Equation in Cylindrical Coordinates 6. Laplace’s Equation in Polar Coordinates (EK 12. 2d Finite Difference Method Heat Equation. Reaction-diffusion equations are one of a well-known pattern-forming system based on the dynamics of two (or more) biochemicals, each of which often plays a role as an activator and inhibitor. Parametric Equations 2-space: Parametric Equations 3-space: Partial Derivatives: Polar Coordinate System: Polar Coordinates- Derivatives and Integrals: PreCalculus: Riemann Sums and the Fundamental Theorem of Calculus: 2d order Diff EQS-Motion: 2d Partial Derivatives: Supplemental Exercises and Solutions: Tangent Planes/ Differential for f(x,y. In the case of steady problems with Φ=0, we get ⃗⃗⋅∇ = ∇2. ` and p (pressure) are decoupled. In this section we will look at converting integrals (including dA) in Cartesian coordinates into Polar coordinates. Download pdf version. Since there is no dependence on angle Θ, we can replace the 3D Laplacian by its two-dimensional form, and we can solve the problem in radial and axial directions. Suppose the rod has a constant internal heat source, so that the equation describing the heat conduction is u t = ku xx +Q, 0 0, derive an equation that governs the eigenvalues of the problem →2 (u+ 2 = u; (r,0) a,π) 0, r,ν) = 0; r,π: polar coordinates where 0 r < a and 0 < π < ν. The resulting curve then consists of points of the form (r(φ), φ) and can be regarded as the graph of the polar function r. the two-dimensional (2D) Fourier transf orm in polar coordinates. In 2D and 1D geometries, the solution if the PDE system is assumed to have no variation in one or two of the coordinate directions. This code is designed to solve the heat equation in a 2D plate. 2D FOURIER TRANSFORMS IN POLAR COORDINATES Natalie Baddour Department of Mechanical Engineering, University of Ottawa, 161Louis Pasteur, Ottawa, Ontario, K1N 6N5, Canada Email: nbaddour@uottawa. Polar coordinates. It is then useful to know the expression of the Laplacian ∆u = u xx + u yy in polar coordinates. Cylindrical/Polar Coordinates, the Heat and Laplace's Equations. Heat accumulation in this solid matter is an important engineering issue. For domains whose boundary comprises part of a circle, it is convenient to transform to polar coordinates. Random Walk and the Heat Equation Discrete Heat Equation Discrete Heat Equation Set-up I Let Abe a nite subset of Zdwith boundary @A. edu MATH 461. J 0(0) = 1 and J n(0) = 0 for n 1. Solution of 2D Laplace Equation in Polar Coordinates. For simplicity, here, we will discuss only the 2-dimensional Laplace equation. 5) @u @t (r;t) = k r @ @r r @u @r (r;t) : fasshauer@iit. Spatial Di erencing 6. - Laplace Transform converts a function in time t into a function of a complex variable s. For the moment, this ends our discussion of cylindrical coordinates. In this paper, an unstructured grids- based discretization method, in the framework of a finite volume approach, is proposed for the solution of the convection- diffusion equation in an r-z. The general heat conduction equation in cylindrical coordinates can be obtained from an energy balance on a volume element in cylindrical coordinates and using the Laplace operator, Δ, in the cylindrical and spherical form. Now we’ll consider it on a circular disk x 2+ y2 v = velocity of the fluid leaving the control volume. molecules is assumed to satisfy the diffusion equation: @n @t = D 2n (1) Using the divergence in polar coordinates, and obtaining the expression for steady-state conditions: D 2n(R) = 0 = 1 R2 @ @R R @n @R (2) Which has a general solution n(R) = C 1 C 2=Rwith boundary conditions: R!1and n!sn 1, the ambient or undisturbed value of vapor concen. Laplace's equation is a key equation in Mathematical Physics. Sturm-Liouville problems, (4/13). (2) solve it for time n + 1/2, and (3) repeat the same but with an implicit discretization in the z-direction). Laplace/heat equations. the part of the solution depending on spatial coordinates, F(~r), satisfies Helmholtz'sequation ∇2F +k2F = 0, (2) where k2 is a separation constant. Then a number of important problems involving polar coordinates are solved. In cylindrical coordinates, Laplace's equation is written (396) Let us try a separable solution of the form (397) Proceeding in the usual manner, we obtain Note that we have selected exponential, rather than oscillating, solutions in the -direction [by writing , instead of , in Equation ]. The optional COORDINATES section defines the coordinate geometry of the problem. What are the corresponding eigenfunctions? 33. If heat generation is absent and there is no flow, = ∇2 , which is commonly referred to as the heat equation. Polar coordinates. 2 Single Equations with Variable Coefficients The following example arises in a roundabout way from the theory of detonation waves. The 2D Heat Equation Here is a DPGraph of the solution to the heat equation on the square with fixed temperature u=0 on the boundary, and initial condition u(x,y,0) = 1. The maximum heat flux calculated by the 1D method was underestimated by 60% than that calculated by 2D filter solution, indicating that the lateral heat transfer cannot be ignored. Introduction ∆ in a Rotationally Symmetric 2d Geometry Separating Polar Coordinates The Equation ∆u=k ∂u ∂t 1. 5 [Nov 2, 2006] Consider an arbitrary 3D subregion V of R3 (V ⊆ R3), with temperature u(x,t) defined at all points x = (x,y,z) ∈ V. This calculator can be used to convert 2-dimensional (2D) or 3-dimensional rectangular coordinates to its equivalent spherical coordinates. Answers and Replies Related Special and General Relativity News on Phys. The regions of integration in these cases will be all or portions of disks or rings and so we will also need to convert the original Cartesian limits for these regions into Polar coordinates. The Finite Difference Method. A Matlab-Based Finite Difierence Solver for the Poisson Problem with Mixed Dirichlet-Neumann Boundary Conditions Ashton S. The model transport equation for k is derived from the exact equation while the model transport equation for ε was obtained using physical reasoning and bears little resemblance to its mathematically exact counterpart. Calculate the heat flux at the outer surface of the pipe. The main tool is the multivariable chain rule. The Analytic function can be used in the expressions for the Parametric Curve. Laplace's Equation and Harmonic Functions In this section, we will show how Green's theorem is closely connected with solutions to Laplace's partial differential equation in two dimensions: (1) ∂2w ∂x2 + ∂2w ∂y2 = 0, where w(x,y) is some unknown function of two variables, assumed to be twice differentiable. Esmaeili1, and B. Bernoulli Equation The Bernoulli equation is the most widely used equation in fluid mechanics, and assumes frictionless flow with no work or heat transfer. Plane polar coordinates (r; ) In plane polar coordinates, Laplace's equation is given by r2˚ 1 r @ @r r @˚ @r! + 1 r2 @2˚ @ 2. The heat and wave equations in 2D and 3D 18. If heat generation is absent and there is no flow, = ∇2 , which is commonly referred to as the heat equation. The heat transfer rate is 30,000 Btu/hr. FUNCȚII ȘI MATRICE GREEN HEAT CONDUCTION (Poisson’s and Laplace’s Equations) Green’s functions 2D-C and solution in integral form for 2D boundary-value problems (BVPs) (for Poisson’s equation) for the following domains (in rectangular coordinates): plane, half-plane, quarter-plane, strip, half-strip and rectangle. GRADING: plus-minus grading (0) Test #0: 10% - Monday, May 20. Figure 8: Spherical coordinates (r, θ, ϕ) ( source ). In your careers as physics students and scientists, you will. Sturm-Liouville problems, (4/13). 4), which is essentially this same equation, where heat is what is diffusing and convecting and being generated. 11 Comments. The heat equation is u t = k Δ u. Since there is no dependence on angle Θ, we can replace the 3D Laplacian by its two-dimensional form, and we can solve the problem in radial and axial directions. We can write down the equation in Cylindrical Coordinates by making TWO simple modifications in the heat conduction equation for Cartesian coordinates. Thin-Layer Approximation 5. Polar Coordinates Suppose we are given the potential on the inside surface of an infinitely long cylindrical Let's start with the 2D Laplacian, which in polar coordinates (s,φ) acts as V(s. Attempt Separation of Variables by writing (1) then the Helmholtz Differential Equation becomes (2). This solver relies on the truncated Fourier series expansion, where the differential equations of the Fourier coefficients are solved by the compact fourth-order finite difference scheme. 3 Position and Distance Vectors z2 y2 z1 y1 x1 x2 x y R1 2 R12 z P1 = (x1, y1, z1) P2 = (x2, y2, z2) O Figure 3-4 Distance vectorR12 = P1P2 = R2!R1, whereR1 andR2 are the position vectors of pointsP1. Small-time GF, transient cases XIJ. Salih Department of Aerospace Engineering Indian Institute of Space Science and Technology, Thiruvananthapuram { February 2011 {This is a summary of conservation equations (continuity, Navier{Stokes, and energy) that govern the ow of a Newtonian uid. Suppose ψ is a function of the polar coordinates (r, θ). Then other applications involving Laplaces's equation came along, including steady state heat ow (Fourier, 1822), theory of magnetism (Gauss and Weber, 1839), electric eld theory (Thomson, 1847), complex analysis (Cauchy, 1825, Riemann, 1851), irrotational uid motion in 2D (Helmholtz, 1858). 2 Single Equations with Variable Coefficients The following example arises in a roundabout way from the theory of detonation waves. The Bernoulli equation is the most famous equation in fluid mechanics. Poisson's Equation in 2D We will now examine the general heat conduction equation, T t = κ∆T + q ρc. Course: MA401. [8, 9] and Jain et al. Esmaeili1, and B. Four elemental systems will be assembled into an 8x8 global system. For a three-dimensional problem, the Laplacian in spherical polar coordinates is used to express the Schrodinger equation in the condensed form Expanded, it takes the form This is the form best suited for the study of the hydrogen atom. Solving Partial Differential Equations in Cylindrical Coordinates Using Separation of Variables; 8-1. Elemental systems for the quadrilateral and triangular elements will be 4x4 and 3x3, respectively. I would argue that your example is still a case of solving a differential equation, even if you don't include the equal sign when you write the problem down on paper. ut= 2u xx −∞ x ∞ 0 t ∞ u x ,0 = x. Cartesian coordinates (x, y) for the simplicity of presentation. Suppose There Are No Sources. Our first task is to translate this into polar coordinates. molecules is assumed to satisfy the diffusion equation: @n @t = D 2n (1) Using the divergence in polar coordinates, and obtaining the expression for steady-state conditions: D 2n(R) = 0 = 1 R2 @ @R R @n @R (2) Which has a general solution n(R) = C 1 C 2=Rwith boundary conditions: R!1and n!sn 1, the ambient or undisturbed value of vapor concen. | 2020-11-25T23:14:14 | {
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http://mathhelpforum.com/algebra/276288-total-possible-outcomes-help.html | # Thread: Total Possible Outcomes Help
1. ## Total Possible Outcomes Help
Hoping someone smarter than me can help me come up with the right answer here.
Setup:
I have a device with eight (8) lights
Each light can light up one of seven (7) colors
When on, each light can flash fast, slow, or solid (3)
Off is also a possibility, but if any of the sets are off, all are off
How many possible ways can this device be configured?
Examples:
Lamp 1 Lamp 2 Lamp 3 Lamp 4 Lamp 5 Lamp 6 Lamp 7 Lamp 8
Red Off Off Off Off Off Off Off
Solid Off Off Off Off Off Off Off
Lamp 1 Lamp 2 Lamp 3 Lamp 4 Lamp 5 Lamp 6 Lamp 7 Lamp 8
Off Red Off Off Off Off Off Off
Off Solid Off Off Off Off Off Off
Lamp 1 Lamp 2 Lamp 3 Lamp 4 Lamp 5 Lamp 6 Lamp 7 Lamp 8
Red Red Off Off Off Off Off Off
Fast Fast Off Off Off Off Off Off
2. ## Re: Total Possible Outcomes Help
The "fundamental law of counting": if A can occur in n ways and, for each of those ways B can occur in m ways, then together they can occur in mn ways.
Any of 8 lamps, each in any of 7 colors, which can flash in any of 3 ways, has 8*7*3 possible combinations.
3. ## Re: Total Possible Outcomes Help
Originally Posted by HallsofIvy
The "fundamental law of counting": if A can occur in n ways and, for each of those ways B can occur in m ways, then together they can occur in mn ways.
Any of 8 lamps, each in any of 7 colors, which can flash in any of 3 ways, has 8*7*3 possible combinations.
Your answer excludes the possibility of more than one light being on at a time. Anywhere from 1-8 of the lights could be active with any of these possibilities at any time.
4. ## Re: Total Possible Outcomes Help
There are $7\cdot 3$ possible configurations for "on" plus the "off" configuration. So, each lamp has 22 different possible configurations. Since the configuration for one lamp does not depend on the configuration for another, there are $22^8=54,875,873,536$ possible configurations total.
5. ## Re: Total Possible Outcomes Help
Originally Posted by SlipEternal
There are $7\cdot 3$ possible configurations for "on" plus the "off" configuration. So, each lamp has 22 different possible configurations. Since the configuration for one lamp does not depend on the configuration for another, there are $22^8=54,875,873,536$ possible configurations total.
That's a considerably higher number than I was expecting. It makes sense if ^8 is how to express the various lamp combinations. Thank you for the insight!
6. ## Re: Total Possible Outcomes Help
The idea is, the first lamp has 22 possible settings. Then the second lamp has an independent set of 22 possible settings. That is $22\cdot 22$. Then the third lamp has another 22, so now we are at $22^3$. There are 8 lamps total, so if we keep multiplying by 22 for each lamp, we get $22^8$. | 2017-12-14T02:53:40 | {
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https://math.stackexchange.com/questions/925241/a-criteria-for-a-subalgebra-of-mn-c-being-mn-c/925242 | # A criteria for a subalgebra of M(n,C) being M(n,C)
Suppose $S$ is a subalgebra of the matrix algebra $M_n(\mathbb{C})$. If for any vector $v$ and $w$ in $\mathbb{C}$, there always exists a matrix $A$ in $S$, depending on $v$ and $w$ of course, which sends $v$ to $w$, then is $S$ necessarily $M_n(\mathbb{C})$?
I arise this technical question when I read Wallach's book Real Reductive Groups, I think the answer is yes, but I am not sure. Can anyone help me prove it or give me a counterexample if I am wrong? Thank you!
• This is a corollary of Burnside's theorem that an irreducible subalgebra of $M_n(\mathbf{C})$ is the whole algebra.
– YCor
Sep 9, 2014 at 9:28
• You can also obtain this from Schur's lemma+double centralizer theorem (by Schur's lemma the centralizer of $S$ is scalar matrices, and so by the double centralizer theorem $S$ is all matrices). Sep 9, 2014 at 17:54
The answer is Yes. By Wedderburn-Malcev [http://www.math.uni-bielefeld.de/~sek/select/RF6.pdf] $S=A\oplus J(S)$ as ${\mathbb C}$-vector spaces, where $A$ is a unital semi-simple ${\mathbb C}$-algebra and $J(S)$ is the Jacobson radical of $S$ (the maximal nilpotent $2$-sided ideal of $S$). We claim that $J(S)=0$. Suppose otherwise. Then there exists $j\in J(S)$ and $v\in{\mathbb C}^n$ such that $j(v)\ne0$. By hypothesis there exists $s\in S$ such that $sj(v)=v$. So $(sj)^n(v)=v$ for all $n\geq1$, whence $sj$ is not nilpotent. But this contradicts the fact that $sj\in J(S)$. This proves our claim.
Now $S=A$ is a semi-simple ${\mathbb C}$-algebra. So it is a direct product $\prod_{n_i} M_{n_i}({\mathbb C})$ of matrix algebras, as ${\mathbb C}$ is algebraically closed. With respect to a suitable choice of basis for ${\mathbb C}^n$, we can represent $S$ as block diagonal matrices, with blocks of size $n_1,n_2,\dots$ where $n_1+n_2+\dots=n$. The hypothesis now implies that $n_1=n$ and $S=M_n({\mathbb C})$.
• @JohnMurray if $V$ is an invariant subspace, and $v\in V$ is a nonzero vector, then $S.v$ is an invariant subspace of $V$ which by assumption of the original question is equal to $\mathbb{C}^n$. Quite a direct corollary, I think. Sep 9, 2014 at 9:59
This is a straightforward corollary of Burnside's theorem that an irreducible subalgebra of $M_n(\mathbf{C})$ is the whole algebra. | 2022-08-16T11:03:06 | {
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https://math.stackexchange.com/questions/2404368/is-this-true-n-leq-frac5n712n-n%E2%88%88n | # Is this true $n!\leq(\frac{5n+7}{12})^n,n∈N$?
Is the following inequality true?
For all $n\in \Bbb N$ prove that: $$n!\leq\left(\frac{5n+7}{12}\right)^n.$$
I know the answer,but I want to see other people how to prove the problem.
In my proof I used $\frac{5n+7}{12}=\frac{\frac{n+1}2+\frac{n+2}3}2\geq \sqrt{\frac{(n+1)(n+2)}6}$ $=\sqrt{\frac{1}{n}\left(\frac{n(n+1)(n+2)}{6}\right)}$ $=\sqrt{\frac{1}{n}\sum_{k=1}^{n}k(n-k+1)}$ $\ge\sqrt{\sqrt[n]{(n!)^2}}=\sqrt[n]{n!}$.
• Where is this problem originated ? Aug 24, 2017 at 8:57
• You can check that for some values in desmos Aug 24, 2017 at 8:58
• Even though reduced to $5/12$ of the original value, it is still roughly $n^n$, which grows way faster than $n!$, no doubt there is $n_0$ such that for all $n\geq n_0$ this holds. The trick I guess is to prove $n_0$ is $1$ Aug 24, 2017 at 9:06
• @JamesJ You should add what you tried or your thoughts. Many questions get closed due to the lack of showing effort. Aug 24, 2017 at 9:13
• @TStancek: $n!$ is asymptotic to $n^{1/2}(n/e)^n$, so the margin is not so large. ($12/5=2.4$)
– user65203
Sep 5, 2017 at 14:10
By AM-GM $$\frac{1\cdot n+2(n-1)+...+n\cdot1}{n}\geq\sqrt[n]{(n!)^2}$$ or $$\left(\sqrt{\frac{(n+1)(n+2)}{6}}\right)^n\geq n!.$$ Thus, it remains to prove that $$\frac{5n+7}{12}\geq\sqrt{\frac{(n+1)(n+2)}{6}},$$ which is $$(n-1)^2\geq0.$$ Done!
$$1\cdot n+2(n-1)+...+n\cdot1=\sum_{k=1}^nk(n-k+1)=$$ $$=(n+1)\sum_{k=1}^nk-\sum_{k=1}^nk^2=(n+1)\cdot\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}=$$ $$=\frac{n(n+1)}{6}\cdot(3n+3-2n-1)=\frac{n(n+1)(n+2)}{6}.$$
• @JamesJ It's weaker because $\frac{4}{9}>\frac{5}{12}$. Aug 24, 2017 at 9:55 | 2022-07-07T05:34:43 | {
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https://mathhelpboards.com/threads/prove-a-b.5053/ | # Prove A<B
#### Albert
##### Well-known member
A=$1001^{999}$
B=$1000^{1000}$
Prove :A<B
(note :any calculation tools are not allowed ,also no use of log function)
#### MarkFL
Staff member
Re: prove A<B
I would use the binomial theorem to write:
$$\displaystyle 1001^{999}=(1000+1)^{999}=\sum_{k=0}^{999}{999 \choose k}1000^{999-k}$$
Since we have 1000 terms, and each term would have to be equal to $1000^{999}$ in order for the summation to be equal to $1000^{1000}$, yet for $0<k$ we find:
$$\displaystyle {999 \choose k}<1000^k$$
we may therefore conclude that:
$$\displaystyle 1001^{999}<1000^{1000}$$
#### Opalg
##### MHB Oldtimer
Staff member
Re: prove A<B
A=$1001^{999}$
B=$1000^{1000}$
Prove :A<B
(note :any calculation tools are not allowed ,also no use of log function)
Are we allowed to use the fact that $\bigl(1+\frac1n\bigr)^n$ increases to $e$ as $n\to\infty$? If so, then $\bigl(1+\frac1n\bigr)^n < n+1$ whenever $n\geqslant2$. It follows that $(n+1)^n < (n+1)n^n$ and therefore $(n+1)^{n-1} < n^n.$
#### mathworker
##### Well-known member
Re: prove A<B
I would use the binomial theorem to write:
$$\displaystyle 1001^{999}=(1000+1)^{999}=\sum_{k=0}^{999}{999 \choose k}1000^{999-k}$$
Since we have 1000 terms, and each term would have to be equal to $1000^{999}$ in order for the summation to be equal to $1000^{1000}$, yet for $0<k$ we find:
$$\displaystyle {999 \choose k}<1000^k$$
we may therefore conclude that:
$$\displaystyle 1001^{999}<1000^{1000}$$
as you are using $$\displaystyle 1000^k$$under sigma i think you can't compare $$\displaystyle {999 \choose k}$$ with
$$\displaystyle 1000^k$$.....
- - - Updated - - -
is it correct to use $$\displaystyle 1000^K$$ under sigma
#### Albert
##### Well-known member
Re: prove A<B
I like Opalg 's method (my solution is similar to his)
#### MarkFL
Staff member
Re: prove A<B
as you are using $$\displaystyle 1000^k$$under sigma i think you can't compare $$\displaystyle {999 \choose k}$$ with
$$\displaystyle 1000^k$$.....
- - - Updated - - -
is it correct to use $$\displaystyle 1000^K$$ under sigma
My intended purpose was to recognize that for $0<k$, we have:
$$\displaystyle {999 \choose k}1000^{999-k}<1000^{999}$$
and since:
$$\displaystyle \sum_{k=0}^{999}1000^{999}=1000\cdot1000^{999}=1000^{1000}$$
we must therefore have:
$$\displaystyle \sum_{k=0}^{999}{999 \choose k}1000^{999-k}=1001^{999}<1000^{1000}$$
#### mathworker
##### Well-known member
Re: prove A<B
My intended purpose was to recognize that for $0<k$, we have:
$$\displaystyle {999 \choose k}1000^{999-k}<1000^{999}$$
and since:
$$\displaystyle \sum_{k=0}^{999}1000^{999}=1000\cdot1000^{999}=1000^{1000}$$
we must therefore have:
$$\displaystyle \sum_{k=0}^{999}{999 \choose k}1000^{999-k}=1001^{999}<1000^{1000}$$
okay i got it, you are comparing each term to $$\displaystyle 1000^999$$....THANK YOU | 2021-10-24T08:31:08 | {
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https://audreysalutes.com/what-is-the-derivative-of-arcsin-x/ | what is the derivative of arcsin x
You are viewing the article: what is the derivative of arcsin x at audreysalutes.com
what is the derivative of arcsin x
Derivative of arcsin x Formula The derivative of the arcsin function is, d/dx(arcsin x) = 1/√1 – x² (OR) d/dx(sin-1x) = 1/√1 – x²
·
What is the derivative of arccos X?
Solution: The derivative of arccos x is -1/√(1-x2).
What is the derivative of x arcsin X?
Derivative of arcsin x Formula The derivative of the arcsin function is, d/dx(arcsin x) = 1/√1 – x² (OR) d/dx(sin-1x) = 1/√1 – x²
Is Arctan X arcsin arccos X?
They have different domains: the domain of arctan is R while the domain of arcsin and arccos is [−1,1], so the domain of g is included in [−1,1]. Precisely, …
What is the derivative of y arccos Cos X?
For these same values of x, arccos(cos(x)) has either a maximum value equal to π or a minimum value equal to 0. Note that although arccos(cos(x)) is continuous for all values of x, its derivative is undefined at x = k*π.
What is the derivative of arcsin x Arccos X?
Calculus Examples By the Sum Rule, the derivative of arcsin(x)+arccos(x) arcsin ( x ) + arccos ( x ) with respect to x is ddx[arcsin(x)]+ddx[arccos(x)] d d x [ arcsin ( x ) ] + d d x [ arccos ( x ) ] . The derivative of arcsin(x) with respect to x is 1√1−x2 1 1 – x 2 .
What is the derivative of arcsin X?
Derivative of arcsin x Formula The derivative of the arcsin function is, d/dx(arcsin x) = 1/√1 – x² (OR) d/dx(sin-1x) = 1/√1 – x²
·
What is arcsin squared?
As a Real valued function arcsin2 is undefined, since sin(x)∈[−1,1] for all x∈R .
What is the derivative of arcsin 2 x?
Explanation: Use chain rule to find the derivative. The derivative of arcsin x is 1/square root of 1-x^2 and then multiply by the derivative of 2x.
Is arctan X differentiable?
Arctan is a differentiable function because its derivative exists on every point of its domain.
What is the derivative of arctan x 2?
What is the Derivative of Arctan x/2? We have the derivative of arctan x to be 1/(1 + x2). By using this and chain rule, the d/dx(arctan x/2) = 1/(1+(x/2)2) d/dx (x/2) = 1/(1 + (x2/4)) · (1/2) = [4/(4 + x2)] · (1/2) = 2/(4 + x2).
What does arctan X mean?
The arctangent of x is defined as the inverse tangent function of x when x is real (x∈ℝ). When the tangent of y is equal to x: tan y = x. Then the arctangent of x is equal to the inverse tangent function of x, which is equal to y: arctan x= tan-1 x = y.
What is derivative of arctan X?
The derivative of arctan x is 1/(1+x2). i.e., d/dx(arctan x) = 1/(1+x2). This also can be written as d/dx(tan-1x) = 1/(1+x2).
Derivative of arcsin
derivative of arcsin(x^2)
Derivative of y arcsin x 2
Arcsin x 1
Arcsin x
derivative of arcsin calculator
Differentiate arccos x
Sech x derivative
See more articles in the category: Wiki | 2023-04-02T02:59:23 | {
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https://math.stackexchange.com/questions/2634552/find-the-natural-solutions-of-a3-b3-999 | # Find the natural solutions of $a^3-b^3=999$
I want to find the natural solutions of $a^3-b^3=999$.
I got $a^3-b^3=(a-b)\cdot(a^2+ab+b^2)$, so if we consider the equation in $\mathbb{Z}/3\mathbb{Z}$ we get
$$(a-b)\cdot(a^2+ab+b^2) \equiv0 \text{ mod }3$$ and because $\mathbb{Z}/3\mathbb{Z}$ is a domain, we get $$a\equiv b \text{ mod } 3 \text{ or } a^2+ab+b^2\equiv0 \text{ mod } 3.$$ Besides, the prime factorization of $999=3^3\cdot37$, but I don't know how to go on.
I would appreciate any hints.
• The prime factorization is a good start. Try setting up a system of equations for $a$ and $b$ given any particular factorization, i.e. for example $(a - b)(a^2 + ab + b^2) = 27 \cdot 37$, so that $a - b = 27$ and $a^2 + ab + b^2 = 37$. – Tob Ernack Feb 3 '18 at 19:39
By Fermat Little Theorem $$x^3 \equiv x \pmod{3}$$
Therefore $$0 \equiv a^3-b^3 \equiv a-b \pmod{3}$$
This shows that $a-b=3k$.
Then $$3^3 \cdot 37 =a^3-b^3=(a-b)(a^2+ab+b^2)=(a-b)((a-b)^3+3ab)=3k(9k^2+3ab) \Rightarrow \\ 3 \cdot 37=k(3k^2+ab)$$
Since $k <3k^2+ab$ the only posibilities are $$k=1 \\ 3k^2+ab=3 \cdot 37$$ or $$k=3 \\ 3k^2+ab= 37$$
This leads to $$a-b=3k=3 \\ ab=108$$ or $$a-b=9 \\ ab=10$$ which are easy to solve.
You have $(a-b)(a^2+ab+b^2)=999$, so $a-b$ is a factor of $999$, that is one of $1,3,9,27,37,111,333$ and $999$ and $a^2+ab+b^2$ is the complementary factor. There are now eight cases. If $a-b=1$, then $a=b+1$ and $999=a^2+ab+b^2=3b^2+3b+1$. This is a quadratic equation; has it any integer solution? Once this is decided, seven more cases to go!
• +1, but I have a so faster method :) – Zaharyas Feb 3 '18 at 20:05
$$999=a^3-b^3=(a-b)^3+3ab(a-b)\geq(a-b)^3,$$ which says $a-b$ is divided by $3$ and $a-b\leq9$ and we get not so many cases:
$a-b=3$ or $a-b=9$.
• Why do not you improve the answer? I think, People did not upvote because they did not understand.(+1) – MathLover Feb 3 '18 at 20:41
• But I explained all... What do you think is not clear? Thank you! – Michael Rozenberg Feb 3 '18 at 20:45
• For example, why $a-b≤9$. (I know reason). I think, impove answer and then write direct what is $a$ and $b$. Sincerely. – MathLover Feb 3 '18 at 20:49
• I added. See now. – Michael Rozenberg Feb 3 '18 at 20:55
• Your method is the fastest. And If you have to wrote the prices of $a$ and $b$ (at first), at least +10 you got. Sincerely! – MathLover Feb 3 '18 at 21:08
So $(a-b)(a^2 + ab + b^2) = 3^3*37$
So $(a-b)|3^3*37$.
So $a-b = 3^j*37^k$ where $j = 0,1,2,3$ and $k=0,1$
And $(a^2 + ab +b^2) = \frac {3^3*27}{a-b} = 3^{3-j}*37^{1-k}$
That's 8 possible systems of equation.
But there are some obvious things to note.
If $37|a-b$ then $a - b \ge 37$ and ... that just seems wrong.
That means $a \ge 37$ and $a^2 + ab + b^2 > 37^2$ but if $37|a+b$ then the very largest that $a^2 + ab + b^2$ can be is $\frac {999}{37} = 27$.
So $37\not \mid a-b$ and $37|a^2 + ab + b^2$.
So $a-b = 3^k; k\le 3; a^2 +ab + b^2 = 3^{3-k}*37$
And there are only 4 systems of equations.
Likewise if $27|a-b$ then $a \ge 27$ and $a^2 + ab + b^2 > 27^2 > 37$ so that's not possible.
So $a-b = 3^k; k\le 2; a^2 + ab + b^2 = 3^{3-k}*37$.
And there are only 3 systems of equations.
$k = 0,1,2$.
If $k = 0$ we have $a= b+ 1$ and $(b+1)^2 + b(b+1) + b^2 = 999$
or $3b^2 + 3b + 1 = 999$ which isn't possible as LHS and RHS are not equivalent mod $3$.
If $k = 1$ we have $a = b+3$ and $(b + 3)^2 + b(b+3)+b^2 = 333$ and
$3b^2 + 9b + 9 = 333$ or
$b^2 + 3b - 108 = 0$ so $b = \frac {-3 +\sqrt {9+432}}2 = \frac {-3 + 21}2 = 9$ and $a = 12$
If $k = 2$ we have $a = b+9$ and $(b+9)^2 + b(b+9) + b^2 = 111$
$3b^2 + 27b + 81 = 111$
$b^2 + 9b + 27 = 37$
$b^2 + 9 - 10 = 0$
$(b+10)(b - 1) = 0$ so $b = 1$ and $a=10$. | 2019-09-18T07:04:19 | {
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https://fr.mathworks.com/help/optim/ug/nonlinear-equations-with-analytic-jacobian.html | # Solve Nonlinear System Without and Including Jacobian
This example shows the reduction in function evaluations when you provide derivatives for a system of nonlinear equations. As explained in Writing Vector and Matrix Objective Functions, the Jacobian $J\left(x\right)$ of a system of equations $F\left(x\right)$ is ${J}_{ij}\left(x\right)=\frac{\partial {F}_{i}\left(x\right)}{\partial {x}_{j}}$. Provide this derivative as the second output of your objective function.
For example, the `multirosenbrock` function is an $n$-dimensional generalization of Rosenbrock's function (see Solve a Constrained Nonlinear Problem, Problem-Based) for any positive, even value of $n$:
`$\begin{array}{l}F\left(1\right)=1-{x}_{1}\\ F\left(2\right)=10\left({x}_{2}-{x}_{1}^{2}\right)\\ F\left(3\right)=1-{x}_{3}\\ F\left(4\right)=10\left({x}_{4}-{x}_{3}^{2}\right)\\ ⋮\\ F\left(n-1\right)=1-{x}_{n-1}\\ F\left(n\right)=10\left({x}_{n}-{x}_{n-1}^{2}\right).\end{array}$`
The solution of the equation system $F\left(x\right)=0$ is the point ${x}_{i}=1$, $i=1\dots n$.
For this objective function, all the Jacobian terms ${J}_{ij}\left(x\right)$ are zero except the terms where $i$ and $j$ differ by at most one. For odd values of $i, the nonzero terms are
`$\begin{array}{l}{J}_{ii}\left(x\right)=-1\\ {J}_{\left(i+1\right)i}=-20{x}_{i}\\ {J}_{\left(i+1\right)\left(i+1\right)}=10.\end{array}$`
The `multirosenbrock` helper function at the end of this example creates the objective function $F\left(x\right)$ and its Jacobian $J\left(x\right)$.
Solve the system of equations starting from the point ${x}_{i}=-1.9$ for odd values of $i, and ${x}_{i}=2$ for even values of $i$. Specify $n=64$.
```n = 64; x0(1:n,1) = -1.9; x0(2:2:n,1) = 2; [x,F,exitflag,output,JAC] = fsolve(@multirosenbrock,x0);```
```Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. ```
Examine the distance of the computed solution `x` from the true solution, and the number of function evaluations that `fsolve` takes to compute the solution.
`disp(norm(x-ones(size(x))))`
``` 0 ```
`disp(output.funcCount)`
``` 1043 ```
`fsolve` finds the solution, and takes over 1000 function evaluations to do so.
Solve the system of equations again, this time using the Jacobian. To do so, set the `'SpecifyObjectiveGradient'` option to `true`.
```opts = optimoptions('fsolve','SpecifyObjectiveGradient',true); [x2,F2,exitflag2,output2,JAC2] = fsolve(@multirosenbrock,x0,opts);```
```Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. ```
Again, examine the distance of the computed solution `x2` from the true solution, and the number of function evaluations that `fsolve` takes to compute the solution.
`disp(norm(x2-ones(size(x2))))`
``` 0 ```
`disp(output2.funcCount)`
``` 21 ```
`fsolve` returns the same solution as the previous solution, but takes about 20 function evaluations to do so, rather than over 1000. In general, using the Jacobian can lower the number of function evaluations and provide increased robustness, although this example does not show improved robustness.
### Helper Function
This code creates the `multirosenbrock` helper function.
```function [F,J] = multirosenbrock(x) % Get the problem size n = length(x); if n == 0, error('Input vector, x, is empty.'); end if mod(n,2) ~= 0 error('Input vector, x ,must have an even number of components.'); end % Evaluate the vector function odds = 1:2:n; evens = 2:2:n; F = zeros(n,1); F(odds,1) = 1-x(odds); F(evens,1) = 10.*(x(evens)-x(odds).^2); % Evaluate the Jacobian matrix if nargout > 1 if nargout > 1 c = -ones(n/2,1); C = sparse(odds,odds,c,n,n); d = 10*ones(n/2,1); D = sparse(evens,evens,d,n,n); e = -20.*x(odds); E = sparse(evens,odds,e,n,n); J = C + D + E; end end``` | 2023-03-31T19:33:56 | {
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http://math.stackexchange.com/questions/656069/when-are-complex-conjugates-of-solutions-not-also-solutions | # When are complex conjugates of solutions not also solutions?
I've heard that for "normal" equations (e.g. $3x^2-2x=0$), if $(a+bi)$ is a solution then $(a-bi)$ will be a solution as well.
This is because, if we define $i$ in terms of $i^2=-1$ then we might as well define $i^\prime=-i$. Since ${i^\prime}^2=-1$ we find $(a+bi)$ has the same algebraic behaviour as $(a+bi^\prime)$.
So what are non-"normal" equations? When are conjugates not also solutions?
-
I think here "normal" is most likely "polynomials with real coefficients". – tabstop Jan 29 '14 at 15:39
An example of a polynomial with non-real coefficients that has two non-conjugate solutions is $$(x - i) (x - 1) = 0$$ whose roots are clearly $i$ and $1$. Written out, it looks like $$x^2 - (1+i) x + i = 0.$$
As others have said, I think that the person using the word "normal" meant "equation with real-number coefficients."
-
If the polynomial has real coefficients, and there is a nonreal root, then its conjugate is also a root. Otherwise, there would be at least one nonreal coefficient.
-
I think you should reformulate your first sentence as it doesn't make sense. (the conjugate of what ?) – Thomas Produit Jan 29 '14 at 15:47
Thanks @ThomasProduit, I'm afraid auto-correct changed what I typed. It was supposed to say "a nonreal root". Corrected. – MPW Jan 29 '14 at 15:59
Now it's OK but to improve your answer you should prove it or give some hints to do it – Thomas Produit Jan 29 '14 at 16:01
I didn't see that a proof was required. But it follows simply from the fact that $\overline{p(z)} = p(\bar{z})$ if the coefficients are real. – MPW Jan 29 '14 at 16:11
Key Idea $\$ Conjugation $\rm\:x\mapsto \bar x\:$ preserves $\rm\:\color{#c00}{sums\,\ \&\,\ products}.\:$ Also it $\rm\:\color{#0a0}{fixes\ coefficients}\in\color{#0a0}{\Bbb R}.\:$ Therefore, by induction, it preserves polynomials $\rm\ \overline{f(w)} = f(\overline w),\ \ f(x)\in\color{#0a0}{\Bbb R}[x],\$ having all $\,\rm\color{#0a0}{real}$ coefficients, since such polynomials are compositions of said basic operations. More explicitly $$\begin{eqnarray} \rm \overline{f(w)}\: &=&\rm\ \ \overline{a_n w^n +\,\cdots + a_1 w + a_0}\\ &=&\rm\,\ \overline{a_n w^n}\, +\,\cdots + \overline{a_1 w} + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x+y}\, =\, \overline x + \overline y}\ \ \ \forall\ x,y \in \Bbb C\\ &=&\rm\,\ \overline a_n\, \overline w^n+\,\cdots + \overline a_1\overline w + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x\, *\, y}\, =\, \overline x\, *\, \overline y}\ \ \forall\ x,y \in \Bbb C \\ &=&\rm\,\ a_n\, \overline w^n + \,\cdots + a_1 \overline w + a_0\quad by\ \ \ \color{#0a0}{\overline a = a}\ \ \forall\ \color{#0a0}a\in \color{#0a0}{\Bbb R}\\ &=&\rm\ f(\overline w)\\ \rm Hence\ \ \ 0 = f(w)\! \ \Rightarrow\ 0 = \bar 0 = \overline{f(w)}\:& =&\ \rm f(\overline w)\quad {\bf QED} \end{eqnarray}$$
Generally this fails if $f$ has non-real coefficients, e.g. $\,\bar w\,$ is a root of $\,x-w\,$ iff $\,\bar w = w,\,$ i.e. $\,w\in \Bbb R.$
Remark $\$ The analogous polynomial preservation property holds true for any algebraic structure, i.e. since homomorphisms preserve the basic operations (including constants = $0$-ary operations), they also preserve the "polynomial" terms composed of these basic operations. Said equivalently, hom's commute with polynomials.
- | 2016-05-05T14:31:25 | {
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https://math.stackexchange.com/questions/1837675/optimising-my-chances-of-drawing-a-specific-card-from-a-small-deck | # Optimising my chances of drawing a specific card from a small deck.
I have a deck of 24 cards, 3 of which are aces. I want to figure out my chances of drawing at least one ace based on the number of cards I draw. I'm pretty sure if I draw one card my chance is 12.5% I'm not sure how to figure out what my chances are if I draw 6 or 8, or how to achieve a 60% chance of getting an ace.
• are you replacing the cards after you draw them? – lulu Jun 24 '16 at 1:25
• No, a single hand of size N. – aslum Jun 24 '16 at 1:27
• Hint: in problems like these it is easier to work from the compliment, that is, it's easier to compute the probability that you don't draw any aces. If $n=3$, say, then the probability of drawing no aces is $q_3=\frac {21}{24}\times\frac {20}{23}\times \frac {19}{22}\sim .657$. The probability you want is just $p_n=1-q_n$. Can you finish from here? – lulu Jun 24 '16 at 1:30
• I'm pretty sure that's what I'm looking for @Lulu, if you post it as an answer I'll accept it. – aslum Jun 24 '16 at 2:30
• The title is misleading. You're not actually optimising anything. – joriki Jun 24 '16 at 8:55
I think @lulu's solution is too obvious and not interesting, so let's try to solve it in a different way.
The probability that you get an ace on your first card is $\frac 3{24}$. For $n=1$, it stops here and this is your answer. $$n=1 \implies p_1=\frac{3}{24}$$
The probability that you fail and need to keep drawing is $\frac{21}{24}$. The probability that you get an ace on your second card is $\frac{3}{23}$. For $n=2$, it stops here: $$n=2 \implies p_2=\frac{3}{24}+\frac{21}{24}\cdot \frac{3}{23}$$
The probability that you fail and need to keep drawing is $\frac{20}{23}$. The probability that you get an ace on your third card is $\frac{3}{22}$. For $n=3$, it stops here: $$n=3 \implies p_3=\frac{3}{24}+\frac{21}{24}\left(\frac{3}{23}+\frac{20}{23}\cdot \frac{3}{22}\right)$$
The probability that you fail and need to keep drawing is $\frac{19}{23}$. The probability that you get an ace on your third card is $\frac{3}{21}$. For $n=3$, it stops here: $$n=4 \implies p_4=\frac{3}{24}+\frac{21}{24}\left(\frac{3}{23}+\frac{20}{23}\left(\frac{3}{22}+\frac{19}{22}\cdot\frac{3}{21}\right)\right)$$
Do you see the pattern now? What is the probability for $n=5$ and $n=6$? Is it easier to get a formula using this method or @lulu's method?
• Lulu's in my opinion, but it is nice to point out that there is more than one approach. – JMoravitz Jun 24 '16 at 2:15
• Maybe I'm not understanding completely, but I think @lulu has it, because the choice of how many cards to draw would be made before drawing any of them, rather drawing one card and then drawing again if you didn't draw an Ace. – aslum Jun 24 '16 at 2:30
• If I am not mistaken then this is the same as @lulu's method...you are just calculating the probability it at each step, and his method directly gives the probability at each step. – User Not Found Jun 24 '16 at 2:32
• @aslum This method yields the same answer as lulu's, it's just that I wanted to point out an alternate method. It doesn't matter if you draw all of the cards at once or if you draw them consecutively because that doesn't affect the probability. In this problem, we can avoid this entirely with lulu's method, but in other probability problems, this makes the problem so much easier, so it's important to know this method. – Noble Mushtak Jun 24 '16 at 3:20
• Also, it doesn't matter if you stop drawing cards after you get an ace or you keep drawing. Once you get an ace, you have satisfied the condition, so whatever cards you get after don't matter. That's also another important trick that comes up sometimes. – Noble Mushtak Jun 24 '16 at 3:22
To elaborate (a little) on the comment, and to phrase it slightly differently, let's suppose you had a deck with $C$ cards containing $A$ aces, from which you make a hand of $n$ cards. We let $q_n$ denote the probability that your hand fails to contain an ace, and let $p_n=1-q_n$ denote the probability that it contains at least one ace. We'll compute $q_n$.
Method I: The probability that the first card you draw is not an ace is $\frac {C-A}C$. Given that the first card is not an ace, the probability that the second is also not an ace is then $\frac {C-A-1}{C-1}$. Continuing in this way we get $$q_n=\frac {C-A}C \times \frac {C-A-1}{C-1} \times \cdots \times \frac {C-A-n+1}{C-n+1}$$
Method II. We note that there are $\binom Cn$ ways to choose a hand of $n$ cards (with no constraints). Similarly, there are $\binom {C-A}n$ ways to choose a hand of $n$ cards without getting an ace. Thus we have $$q_n=\frac {\binom {C-A}n}{\binom Cn}=\frac {(C-A)!(C-n)!}{(C-A-n)!(C)!}=\frac {(C-A)(C-A-1)\cdots(C-A-n+1)}{C(C-1)\cdots(C-n+1)}$$ | 2019-10-18T05:52:55 | {
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https://math.stackexchange.com/questions/2807199/showing-that-lim-n-to-infty-int-mathbb-r-frac-sinnxx2-dx | # showing that $\lim_{n \to \infty} \int_{\mathbb R} \frac{\sin^n(x)}{x^2} \, dx= 0$
I want to use the dominated convergence theorem to show that: $$\lim_{n \to \infty}\int_{\mathbb R} \frac{\sin^n(x)}{x^2} \,dx =0$$
I want to make sure my approach is correct. Since $|\sin x| \le 1$, we have that $\forall x \neq 0$:
$$\bigg | \frac{\sin^n(x)}{x^2} \bigg | \le \frac{1}{x^2}$$
So I define $g(x) = x^{-2} ~~\forall x \neq 0$ and $g(x) = 0$ if $x = 0$. Then $g$ is integrable and $|f_n(x)| \le g$, where
$$f_n(x) = \frac{\sin^n(x)}{x^2}$$
So therefore, by DCT, as $f_n \to 0$ pointwise, we can interchange the limit and integral:
$$\lim_{n \to \infty}\int_{\mathbb R} \frac{\sin^n(x)}{x^2} \,dx = \int_{\mathbb R} \lim_{n \to \infty} \frac{\sin^n(x)}{x^2} \, dx = \int0\,dx = 0$$
I'm just wondering if this is the correct approach given the way I construct $g$. My thinking is that $1/x^2$ is integrable everywhere except for a set of measure zero $(-\varepsilon, \varepsilon)$
My thinking is that $1/x^2$ is integrable everywhere except for a set of measure zero $(-\epsilon, \epsilon)$.
That's not a set of measure zero, because it has measure $2\epsilon > 0$. Your dominating function is not integrable, because $$\int_0^{\infty} \frac 1 {x^2} \, dx \ge \int_0^1 \frac 1 {x^2} \, dx = \infty.$$ You need to handle a neighborhood of zero as a special case when you're finding an integrable majorant.
Here is an approach to fixing the proof. Using the fact that $|\sin x| \le |x|$ for all $x$, we actually have the estimate
$$\frac{|\sin^n x|}{x^2} \le \left\{\begin{array}{cc}1/x^2 & |x| \ge 1\\ 1 & |x| < 1\end{array}\right.$$
provided that $n \ge 2$. Now the function on the right actually is integrable on $\mathbb{R}$. | 2022-08-12T03:21:02 | {
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https://math.stackexchange.com/questions/1766187/trying-to-solve-recurrence-tn-3tn-3-3 | Trying to solve recurrence $T(n)=3T(n/3) + 3$
I'm trying to solve the following recurrence without using the Master Theorem:
$$T(1)=1;$$
$$T(n)=3T(n/3) + 3$$
My attempt:
$T(n) = 3T(n/3) + 3$
$= 3(3T(n/9) n/3)) + 3)$
$= 9T(n/9) + 9$
$= 9(3T(n/27 + n/9)) +9$
$= 27T(n/27) + 9$
$...$
I know this is wrong but I'm stuck here. Thanks.
• Why the second line is $3(3T(n/9) n/3)) + 3)$ and not $3(3T(n/3)+3)+3$? – Phicar May 1 '16 at 2:14
• Sorry I'm trying to figure out how to do this – Carlo May 1 '16 at 2:16
• You must replace $k$ times the recurrence you have and see what should be $k$ in order to stop the recurrence and use $T(1)=1$. So, if $k=3$ $T(n)=3T(n/3)+3=3(3T(n/3^2)+3)+3=3(3(3T(n/3^3)+3)+3)+3=3^3T(n/3^3)+3+3^2+3^3$..so, do it in general for $k$ and the geometric sum will be helpful. – Phicar May 1 '16 at 2:23
• I see. so what would Theta class would that be then? Theta(nlogn)? – Carlo May 1 '16 at 2:26
• What expression do you have at the end? If you guess is that complexity, try to prove it by induction. – Phicar May 1 '16 at 2:31
$$T(n)=3T\left(\frac{n}{3}\right)+3=3\left(3T\left(\frac{n}{3^2}\right)+3\right)+3=3^2T\left(\frac{n}{3^2}\right)+3^2+3$$Continuing so on....... $$T(n)=3^kT\left(\frac{n}{3^k}\right)+3^k+3^{k-1}+..........+3^2+3\tag{1.}$$ Let $n=3^k$ then $T\left(\frac{n}{3^k}\right)=T(1)=1$ ,putting this in eq (1) :- $$T(n)=3+3^2+3^3+...........+3^k+3^k=\frac{3(3^k-1)}{2}+3^k=\frac{3(n-1)}{2}+n=5\frac{n}{2}-\frac{3}{2}$$ i.e. $$T(n)=\theta(n)$$
• Perfect!!! thank you Mayank. – Ilan Aizelman WS Aug 17 '16 at 12:58
A change of argument turns this into a linear difference equation with constant coefficients. Let $n=3^k$. Then $$T(n)=T\left(3^k\right)=a_k=3T(n/3)+3=3T\left(3^{k-1}\right)+3=3a_{k-1}+3$$ The solution to the homogeneous equation, $a_k=3a_{k-1}$ is $a_k=C\cdot3^k$ and there is a constant solution, $a_k=-3/2$ to the actual equation. Thus the general solution is $$a_k=C\cdot3^k-\frac32$$ Applying the initial condition, $T(1)=a_0=C-3/2=1$, we have $$a_k=T(3^k)=\frac52\cdot3^k-\frac32$$ So $T(n)=\frac52n-\frac32$, for $n$ a power of $3$. For other $n$ we don't have an initial condition.
Writing out the first few terms, we get:
$$T(1) =1, T(3) = 6, T(9) = 21, T(27) = 66, T(81) = 201$$
Using this, we can see that $T(n) \approx2n$, so we make the ansatz that $T(n) = an + b$
Applying this to the recurrence relation gives us: \begin{align*}an + b &= 3\left(a\frac{n}{3} + b\right) + 3\\an + b &= an + 3b + 3\\2b &= -3\\b &= \frac{-3}{2}\end{align*}
Finally, applying the boundary condition $T(1) = 1$: \begin{align*}a + b &= 1\\a -\frac{3}{2} &= 1\\a &= \frac{5}{2}\end{align*}
Hence:
$$T(n) = \frac{5}{2}n - \frac{3}{2}$$
This recurrence only makes sense when repeatedly dividing $n$ by three eventually yields 1; that is, when $n$ is a power of three. So let's start by assuming that $n=3^m$. Then we may restate the problem as:
$$\begin{eqnarray} T(3^0)&=&1 \\ T(3^m)&=&3T(3^{m}/3)+3 &=& 3T(3^{m-1})+3 \end{eqnarray}$$ or $$T(3^m)-3T(3^{m-1})-3=0$$ I'll point out that this is an order 1 linear recurrence relation in $m$, and that dictates that form of the result. But instead of just pulling a formula out of thin air and solving for some coefficients, I'll demonstrate a method to actually derive the formula: using ordinary generating functions.
Let $f(x)=T(3^0)+T(3^1)x+T(3^2)x^2+\cdots$ be the ordinary generating function for the sequence $\{T(3^m)\}$. We want to combine some multiples of this power series so that the the coefficients of the combination will satisfy the recurrence relation. To do this, we calculate:
$$\begin{eqnarray} f(x)&=&T(3^0)&+&T(3^1)x&+&T(3^2)x^2&+&T(3^4)x^3&+&\cdots \\ -3xf(x)&=&0&-&3T(3^0)x&-&3T(3^1)x^2&-&3T(3^2)x^3&+&\cdots \\ \frac{-3}{1-x}&=&-3&-&3x&-&3x^2&-&3x^3&+&\cdots \\ \end{eqnarray}$$
Ok, I'll admit I did pull that last formula (for an infinite geometric series) out of thin air. Adding these together, we have:
$$\begin{eqnarray} f(x)-3xf(x)-\frac{3}{1-x} &=& -3 + T(3^0)&+&[T(3^1)-3T(3^0)-3]x&+&[T(3^2)-3T(3^1)-3]x^2&+&\cdots \\ &=& -2 &+& [0]x &+& [0]x^2 &+&\cdots \end{eqnarray}$$ So that $$\begin{eqnarray} [1-3x]f(x)&=&-2 +\frac{3}{1-x} \\ f(x)&=&\frac{-2}{1-3x}+\frac{3}{(1-x)(1-3x)} \\ &=&\frac{-2}{1-3x}-\frac{\frac{3}{2}}{1-x}+\frac{\frac{9}{2}}{1-3x} & \textrm{(partial fraction decomposition)} \\ &=&\frac{\frac{5}{2}}{1-3x}-\frac{\frac{3}{2}}{1-x} \\ &=&[\frac{5}{2}-\frac{3}{2}]+[\frac{5}{2}\cdot 3^1-\frac{3}{2}\cdot1^1]x+[\frac{5}{2}\cdot3^2-\frac{3}{2}\cdot1^2]x^2+\cdots \end{eqnarray}$$
Matching coefficients yields $$\begin{eqnarray} T(3^m) &=& \frac{5}{2}\cdot3^m-\frac{3}{2} \\ T(n) &=& \frac{5}{2}n-\frac{3}{2} \end{eqnarray}$$ | 2020-03-31T23:53:12 | {
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https://math.stackexchange.com/questions/2458083/probability-of-any-given-student-being-chosen-in-this-question | # Probability of any given student being chosen in this question
This question from my textbook
Three high schools have senior classes of size 100, 400, 500. Scheme A: make a list of all 1000 students and choose one randomly; Scheme B: pick a school at random then a student at random;
Makes the comparison by showing that choosing a student from one of the high-schools is different across the two schemes
In Scheme A each person in first school is chosen with probability 1/1000; in Scheme B choose that school 1/3 of the time, and then each person is chosen 1/100 of the time, so a person in the first school is now chosen 1/300 of the time.
I did not think to use a single school as an example and wanted to solve this question by showing that in general the probability of choosing any student from any school is different.
How would I calculate the probability of choosing any particular student in scheme B?
I tried drawing a tree diagram where I had $\frac{1}{3}$ probability for choosing each school and then $\frac{1}{100}$, $\frac{1}{400}$, $\frac{1}{500}$ for choosing a student from each school respectively.
Is this a correct start?
• So the probability is then the sum of probabilitoes for each school? $\frac{1}{300}+\frac{1}{1200}+\frac{1}{1500} = \frac{29}{6000}$ – EvaD Oct 5 '17 at 13:14
• The probability of what? The probability that some student is chosen has to be $1$. This is a good check of your arithmetic. In fact $100 \cdot \frac 1{300}+400 \cdot \frac 1{1200}+500 \cdot \frac 1{1500}=1$. What you have shown is that the chance of an individual student from C being selected if $\frac 15$ of the chance that an individual student from A is selected That is what you were looking to show. – Ross Millikan Oct 5 '17 at 14:26
• I am trying to find the probability of any given student being selected using scheme B. By scheme A $P(student) = \frac{1}{1000}$, by scheme B I am asking if it is correct that $P(student) = P(student | school A) + P(student| school B) + P(student| school C)$ – EvaD Oct 5 '17 at 14:39 | 2019-06-25T13:44:03 | {
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https://math.stackexchange.com/questions/2371261/how-can-i-find-the-general-term-of-sequence-with-this-relations | # How can I find the general term of sequence with this relations?
It is given that $a_1=1$ and $a_2=0.5$ satisfying that for all integers $n \ge2$
$$n(n+1)a_{n+1}a_n+na_na_{n-1}=(n+1)^2a_{n+1}a_{n-1}$$
How can I found the general term of the sequence ${a_n}$ ?
I try to use the recurrence relations and think about using mathematical induction, but failed because of insufficient data.
Can I get some clues?
• What are the first few terms? – vadim123 Jul 25 '17 at 14:52
• @vadim123 $a_1=1$ and $a_2=0.5$ is given. – user362325 Jul 25 '17 at 14:52
• What are the next 10 terms? – vadim123 Jul 25 '17 at 14:53
• just curious, where is the problem from? – Tai Jul 25 '17 at 14:55
• user362325 I believe you asked this earlier. Did you delete the earlier question you wrote, in order to ask it again? In any case: you've earlier asked a question concluding "I try to use the recurrence relations and think about using mathematical induction, but failed because of insufficient data." – amWhy Jul 25 '17 at 15:07
Rearrange the equation, you get $$(n+1)^2\,a_{n+1}\,a_{n-1}-n(n+1)\,a_{n+1}\,a_n\,=n\,a_n\,a_{n-1}$$
$$(n+1)\,a_{n+1}\,\{(n+1)\,a_{n-1}-na_n\}=n\,a_n.a_{n-1}$$ $${(n+1)\,a_{n-1}\,-n\,a_n\over n.a_n.a_{n-1}}={1\over{(n+1)\,a_{n+1}}}$$ $${1\over a_n}-{1\over a_{n-1}}={1\over{(n+1)\,a_{n+1}}}\,\,-{1\over {n\,a_n}}$$
Now on putting the values $$n=2,3,......,n-1,n$$ And adding all the equations, you get
$${a_{n+1}\over a_n}={1\over(n+1)}$$ $${a_n\over{a_{n-1}}}\cdot{{a_{n-1}\over{a_{n-2}}}}\cdots\cdots {a_2\over a_1}={{1\over n}\cdot}{1\over{n-1}}\cdots\cdots{1\over2}\cdot{1\over1}$$ $$a_n={1\over n!}$$
• This solution is beautiful. – user362325 Jul 25 '17 at 16:50
• @user362325 It is unbecoming of you to "brag" or "boast" about an answer from a sockpuppet account of yours. – amWhy Jul 25 '17 at 17:26
• @amWhy: How can you be sure that there is a sockpuppet account of user362325? – user 170039 Jul 28 '17 at 3:12
Given that $a_1=1$, $a_2=\frac{1}{2}$. Putting $n=2,3,\dots~~$ we find that \begin{align*} a_3&=\frac{1}{6}=\frac{1}{3!}\\ a_4&=\frac{1}{2^3.3}=\frac{1}{4!}\\ a_5&=\frac{1}{2^3.3.5}=\frac{1}{5!}\\ a_6&=\frac{1}{2^4.3^2.5}=\frac{1}{6!} \end{align*} and so on. So, $$a_n=\frac{1}{n!}.$$ In fact, assuming $~a_n=\frac{1}{n!}~$ is true for some $n$, \begin{align*} (n+1)a_{n+1}\frac{1}{(n-1)!}+n\frac{1}{n!}\frac{1}{(n-1)!}&=(n+1)^2a_{n+1}\frac{1}{(n-1)!}\\ \implies\qquad (n+1)a_{n+1}+\frac{1}{((n-1)!}&=(n+1)^2a_{n+1}\\ \implies\qquad\qquad\qquad a_{n+1}&=\frac{1}{(n+1)!}. \end{align*} | 2020-01-29T12:20:09 | {
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https://math.stackexchange.com/questions/4036176/what-is-the-probability-that-at-least-one-red-ball-is-drawn-when-three-balls-are | # What is the probability that at least one red ball is drawn when three balls are drawn from a bag with $5$ red and $5$ yellow balls?
I am having a tough time trying to understand the concept of probability and figuring out which is the right way to solve this problem. So I would really appreciate a lot if you can help me with this question
Problem: There are 5 red balls and 5 yellow balls. If we make 3 draws from this set, without replacing the ball drawn in each set, what is the probability that in the end we have drawn at least one red ball.
Method 1:
Calculate the probability of not drawing any red ball in all three draws and subtract that from 1.
The probability of drawing 3 yellow balls in succession : 5/10 * 4/9 * 3/8
So then the probability of drawing at least one red ball is 1-(5/10 * 4/9 * 3/8) = 11/12
Method 2: Create the possible color combinations that the 3 draws can produce. I can come up with only 8 such combinations:
RRR, YYY, YRR, YRY, YYR, RRY, RYR, RYY
So based on these combinations I see that there is only 1 combination that has no red ball. So isthe probability of drawing a red ball then 7/8?
Which is the right answer and the right way of solving this problem.
• Compare this to an extreme case where you have a billion red balls and one yellow ball and you draw only one ball. You have the two outcomes: $R$ and $Y$... the probability is not $\frac{1}{2}$ though to draw the yellow ball. "Number of good outcomes divided by number of total outcomes" only works to calculate probability when the outcomes are equally likely to happen. – JMoravitz Feb 23 at 0:05
• The outcome of buying a lottery ticket is either a win or a loss, so the probability of winning is $1/2$ ... no? – Graham Kemp Feb 23 at 0:12
There are $$\binom{10}{3}$$ ways to select three of the ten balls.
The number of selections with at least one red ball is $$\binom{5}{3} + \binom{5}{2}\binom{5}{1} + \binom{5}{1}\binom{5}{2}$$ where the first term counts the number of ways of selecting three red balls, the second term counts the number of ways of selecting two red and one yellow ball, and the third term counts the number of ways of selecting one red and two yellow balls. Observe that $$\frac{\dbinom{5}{3} + \dbinom{5}{2}\dbinom{5}{1} + \dbinom{5}{1}\dbinom{5}{2}}{\dbinom{10}{3}} = \frac{11}{12}$$ The problem with your second method is that the eight events are not equally likely. | 2021-03-03T02:07:22 | {
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https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_25&diff=prev&oldid=66700 | # Difference between revisions of "2011 AMC 10A Problems/Problem 25"
## Problem 25
Let $R$ be a square region and $n\ge4$ an integer. A point $X$ in the interior of $R$ is called $n\text{-}ray$ partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?
$\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500$
## Solution 2
First, notice that there must be four rays emanating from $X$ that intersect the four corners of the square region. Depending on the location of $X$, the number of rays distributed among these four triangular sectors will vary. We start by finding the corner-most point that is $100$-ray partitional (let this point be the bottom-left-most point). We first draw the four rays that intersect the vertices. At this point, the triangular sectors with bases as the sides of the square that the point is closest to both do not have rays dividing their areas. Therefore, their heights are equivalent since their areas are equal. The remaining $96$ rays are divided among the other two triangular sectors, each sector with $48$ rays, thus dividing these two sectors into $49$ triangles of equal areas. Let the distance from this corner point to the closest side be $a$ and the side of the square be $s$. From this, we get the equation $\frac{a\times s}{2}=\frac{(s-a)\times s}{2}\times\frac1{49}$. Solve for $a$ to get $a=\frac s{50}$. Therefore, point $X$ is $\frac1{50}$ of the side length away from the two sides it is closest to. By moving $X$ $\frac s{50}$ to the right, we also move one ray from the right sector to the left sector, which determines another $100$-ray partitional point. We can continue moving $X$ right and up to derive the set of points that are $100$-ray partitional. In the end, we get a square grid of points each $\frac s{50}$ apart from one another. Since this grid ranges from a distance of $\frac s{50}$ from one side to $\frac{49s}{50}$ from the same side, we have a $49\times49$ grid, a total of $2401$ $100$-ray partitional points. To find the overlap from the $60$-ray partitional, we must find the distance from the corner-most $60$-ray partitional point to the sides closest to it. Since the $100$-ray partitional points form a $49\times49$ grid, each point $\frac s{50}$ apart from each other, we can deduce that the $60$-ray partitional points form a $29\times29$ grid, each point $\frac s{30}$ apart from each other. To find the overlap points, we must find the common divisors of $30$ and $50$ which are $1, 2, 5,$ and $10$. Therefore, the overlapping points will form grids with points $s$, $\frac s{2}$, $\frac s{5}$, and $\frac s{10}$ away from each other respectively. Since the grid with points $\frac s{10}$ away from each other includes the other points, we can disregard the other grids. The total overlapping set of points is a $9\times9$ grid, which has $81$ points. Subtract $81$ from $2401$ to get $2401-81=\boxed{\textbf{(C)}\ 2320}$.
## Solution 2
We may assume that the square $R$ has coordinates $R_{1}(0,0), R_{2}(1,0), R_{3}(1,1), R_{4}(0,1)$. Suppose that $X = (x,y)$ is $n$-ray partitional. $[asy] defaultpen(fontsize(10)); int i; pair R1=(0,0),R2=(1,0),R3=(1,1),R4=(0,1), X=(3/7,5/7); draw(R1--R2--R3--R4--cycle); for(i=0;i<10;++i){draw(X--(i*R1+(10-i)*R2)/10,linewidth(0.1));} for(i=0;i< 8;++i){draw(X--(i*R2+( 8-i)*R3)/ 8,linewidth(0.1));} for(i=0;i< 4;++i){draw(X--(i*R3+( 4-i)*R4)/ 4,linewidth(0.1));} for(i=0;i< 6;++i){draw(X--(i*R4+( 6-i)*R1)/ 6,linewidth(0.1));} draw(X--R1);draw(X--R2);draw(X--R3);draw(X--R4); label("R_{1}",R1,(-1,-1)); label("R_{2}",R2,( 1,-1)); label("R_{3}",R3,( 1, 1)); label("R_{4}",R4,(-1, 1)); label("Example: X = (\frac{3}{7},\frac{5}{7}) is 28-ray partitional.",(R1+R2)/2,(0,-10)); [/asy]$ By definition, there exist $n$ rays from $X$ which divide $R$ into $n$ triangles of equal area, and $4$ of these rays intersect the vertices of $R$. Let $a,b,c,d$ be the number of triangles that share a side with $\overline{R_{1}R_{2}}, \overline{R_{2}R_{3}}, \overline{R_{3}R_{4}}, \overline{R_{4}R_{1}}$, respectively. Then $$a+b+c+d = n \;.$$ Let the common area of the triangles be $A$. Each of the triangles that share a side with $\overline{R_{1}R_{2}}$ have common height $y$; since they have the same area, they must have the same base $\frac{1}{a}$. Hence $A = \frac{1}{2}\cdot \frac{1}{a}\cdot y$. Similarly, $$A = \frac{y}{2a} = \frac{1-x}{2b} = \frac{1-y}{2c} = \frac{x}{2d} \;.$$Substituting gives $$n = \frac{y}{2A}+\frac{1-x}{2A}+\frac{1-y}{2A}+\frac{x}{2A} = \frac{1}{A}$$and $$a = \frac{n}{2}y \;,\; b = \frac{n}{2}(1-x) \;,\; c = \frac{n}{2}(1-y) \;,\; d = \frac{n}{2}x \;.$$Since $x = \frac{2d}{n}$ and $y = \frac{2a}{n}$ and $a,d,n$ are integers, $x,y$ are rational. Let $p_{i},q_{i}$ be integers such that $0 < p_{i} < q_{i}$ and $$x = \frac{p_{1}}{q_{1}} \quad,\quad y = \frac{p_{2}}{q_{2}} \;.$$ Thus we have derived a necessary condition for $X$ to be $n$-ray partitional: $$q_{1} \text{ divides } \frac{n}{2}p_{1} \quad\text{ and }\quad q_{2} \text{ divides } \frac{n}{2}p_{2} \;.$$ Conversely, if $X = (x,y)$ satisfies the above condition, then it is $n$-ray partitional since we can define $a,b,c,d$ in terms of $n,x,y$ as above.
To count the points that are $100$-ray partitional, it suffices to count the ordered pairs of rationals $(\frac{p_{1}}{q_{1}},\frac{p_{2}}{q_{2}})$ in the interior of $R$ such that $q_{1}$ divides $\frac{100}{2}p_{1}$ and $q_{2}$ divides $\frac{100}{2}p_{2}$. This is just $\left\{(\frac{p_{1}}{50},\frac{p_{2}}{50}) \;:\; 0 which has $49^{2}$ points.
We need to subtract the points that are $100$-ray partitional and $60$-ray partitional; those are the ordered pairs of rationals $(\frac{p_{1}}{q_{1}},\frac{p_{2}}{q_{2}})$ in the interior of $R$ such that $q_{1}$ divides $GCD(\frac{100}{2}p_{1},\frac{60}{2}p_{1}) = 10p_{1}$ and $q_{2}$ divides $GCD(\frac{100}{2}p_{2},\frac{60}{2}p_{2}) = 10p_{2}$. This is just $\left\{(\frac{p_{1}}{10},\frac{p_{2}}{10}) \;:\; 0 which has $9^{2}$ points.
Hence the answer is $49^{2} - 9^{2} = \boxed{2320}$. | 2021-03-01T11:15:19 | {
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http://math.stackexchange.com/questions/66213/for-a-random-permutation-whats-the-probability-that-each-half-of-the-elements | # For a random permutation, what's the probability that each half of the elements keep relative order?
If we take a random permutation of a sequence of $2k$ elements, $X_1, X_2, \ldots X_k, X_{k+1}, \ldots, X_{2k}$. What's the probability that $X_1, X_2, .. X_k$ and $X_{k+1}, \ldots, X_{2k}$ both keep their relative orders in the new sequence?
My guess is, these 2 event are independent because whether one happen doesn't change the other's probability. So we have
$$\Pr \{ [\text {both sub-sequence keep relative order}]\} = (\frac 1 {n!})^2$$
What do you think about it? Is it possible to get the result by counting how many permutations fulfill the condition?
-
Your reasoning seems correct to me. Another way of getting it:
To count the posible arrangements, lets imagine the first $k$ elements are white and the rest black. It's easy to see if we are given the colors of a particular arrangement, the elements can be identified; hence, to count all "legal" permutations is equivalent to count all the possible ways of placing $k$ black and $k$ white elements in $2k$ positions. This is ${2n \choose n}$. And the total number of permutations is $(2n)!$ Hence, the probability is
$$\frac{{2n \choose n}}{(2n)!} = \frac{1}{(n!)^2}$$
-
The independence argument looks sound to me, although the assumptions might need justification depending on who you ask. You simply take the indices of the first $k$ objects after permutation and rank them; this effects a permutation of the $k$ objects and by symmetry there is no preference for one permutation over another. Only one permutation preserves the relative ordering (the identity) so the probability they keep their ordering is $1/k!$. The fulfillment of this condition is independent between the first $k$ and the second $k$ arguments so we multiply probabilities to get $1/k!^2$.
But yes, it's possible to count the permutations that fulfill the conditions: each way of permuting $2k$ elements such that the first and second $k$ items independently keep their relative order is essentially a way of picking $k$ positions out of the $2k$ possible for the first $k$ items to go - this automatically determines the entire permutation (can you see how?). And the total number of permutations is $(2k)!$ so the probability is $\frac{1}{(2k)!}{2k\choose k}=\frac{1}{k!^2}$. More generally, if you take $n$ items and partition them into parts of size $a_1,a_2,\dots,a_m$ (they don't even have to be contiguous parts), the probability a permutation preserves the relative ordering of each part is $$\frac{1}{n!}{n\choose a_1,\dots,a_{m-1}}=\frac{1}{a_1!a_2!\cdots a_m!}.$$ As you might guess from the form above, the independence argument works for this too.
- | 2015-07-02T17:27:48 | {
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http://math.stackexchange.com/questions/74426/e-as-the-limit-of-a-sequence | # $e$ as the limit of a sequence
One of the standard definitions of $e$ is as
$$\lim_{n\rightarrow\infty}\left(1 + \frac{1}{n}\right)^n$$
But in all cases I've seen this limit, it is proven as a limit of the sequence $\Big\{\big(1 + \frac{1}{n}\big)^n\Big\}$, which seems to cover the limit for only $n$ as an integer. Now my question is whether this sequence limit is equivalent to a normal limit for which $n$ can be any real number. I can think of cases which this isn't generally true;
$$\lim_{n\rightarrow\infty}\ \sin(n\pi)$$
comes readily to mind, for which the limit as a sequence is simply $0$ but as a general limit, it is undefined. The limit for $e$ is used exactly as if it were a normal limit, which leads me to believe it is equivalent. Are there conditions for which the limit of a sequence and the corresponding function are identical?
-
+1. This is an insightful question, IMO. – JavaMan Oct 20 '11 at 23:54
If $\lim_x f(x)$ exists, then $\lim_x f(x) = \lim_n f(n)$ (but as you point, there are cases where $\lim_n f(n)$ exists but $\lim_x f(x)$ does not). – Joel Cohen May 13 '12 at 19:32
If $x>0$ be a real number, then write $x=n+y_x$ where $0 \leq y_x \leq 1$.
It is easy then to show that
$$\left(1+\frac{1}{n+1} \right)^n \leq \left(1+\frac{1}{x}\right)^x \leq \left(1+\frac{1}{n}\right)^{n+1} \,.$$
Using this, you can prove that if
$$e=\lim_n \left(1+\frac{1}{n}\right)^n$$ then $$\lim_{x \to \infty} \left(1+\frac{1}{x}\right)^x =e$$
Edit
To complete the answer, in general if $f(x)$ is monotonic, then for sure
$$\lim_n f(n)= \lim_x f(x) \,.$$
I don't recall if $\left(1+\frac{1}{x}\right)^x$ is monotonic (and I am too lazy to differentiate it), but what we showed above is that it is at least "close to being monotonic". What I mean by this, I showed that $$f(n) h(n) \leq f(x) \leq f(n+1)g(n+1) \forall x\in [n, n+1)$$ where $h(n)$ and $g(n)$ go to $1$. The you basically sqeeze it.
-
$(1+\frac{1}{x})^x$ increases monotonicly for positive $x$ – Henry Oct 20 '11 at 23:28
@DJC: Good idea, but since both $\log\left(1+\frac{1}{x}\right)$ and $\frac{1}{x+1}$ are less than $\frac{1}{x}$, I think you need to go a bit farther in the asymptotic expansion: $\log\left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2x^2}+O\left(\frac{1}{x^3}\right)$ and $\frac{1}{x+1}=\frac{1}{x}-\frac{1}{x^2}+O\left(\frac{1}{x^3}\right)$. And even then, we need to consider when $O\left(\frac{1}{x^3}\right)$ is small enough. – robjohn Oct 21 '11 at 0:16
Since $e^u\ge1+u$ for all $u\in\mathbb{R}$, we also have $u\ge\log(1+u)$ for $u>-1$. The derivative of the logarithm of $(1+1/x)^x$ is \begin{align} \log\left(1+\frac{1}{x}\right)-\frac{1}{x+1} &=-\log\left(1-\frac{1}{x+1}\right)-\frac{1}{x+1}\\ &=u-\log(1+u)\\ &\ge0 \end{align} where $u=-\frac{1}{x+1}>-1$ for $x>0$. Therefore, $(1+1/x)^x$ is monotonically increasing for $x>0$. – robjohn Oct 21 '11 at 0:41
Your sandwich between $\left(1+\frac{1}{n+1} \right)^n$ and $\left(1+\frac{1}{n}\right)^{n+1}$ is similar to this sandwich. (+1) – robjohn Oct 21 '11 at 2:13
@robjohn: You're absolutely right. Thanks for cleaning that up. – JavaMan Oct 21 '11 at 4:03 | 2014-12-18T21:39:26 | {
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http://docs.mars-project.io/en/v0.2.0/tensor/generated/mars.tensor.arctan2.html | mars.tensor.arctan2¶
mars.tensor.arctan2(x1, x2, out=None, where=None, **kwargs)[source]
Element-wise arc tangent of x1/x2 choosing the quadrant correctly.
The quadrant (i.e., branch) is chosen so that arctan2(x1, x2) is the signed angle in radians between the ray ending at the origin and passing through the point (1,0), and the ray ending at the origin and passing through the point (x2, x1). (Note the role reversal: the “y-coordinate” is the first function parameter, the “x-coordinate” is the second.) By IEEE convention, this function is defined for x2 = +/-0 and for either or both of x1 and x2 = +/-inf (see Notes for specific values).
This function is not defined for complex-valued arguments; for the so-called argument of complex values, use angle.
x1 : array_like, real-valued
y-coordinates.
x2 : array_like, real-valued
x-coordinates. x2 must be broadcastable to match the shape of x1 or vice versa.
out : Tensor, None, or tuple of Tensor and None, optional
A location into which the result is stored. If provided, it must have a shape that the inputs broadcast to. If not provided or None, a freshly-allocated tensor is returned. A tuple (possible only as a keyword argument) must have length equal to the number of outputs.
where : array_like, optional
Values of True indicate to calculate the ufunc at that position, values of False indicate to leave the value in the output alone.
**kwargs
angle : Tensor
Array of angles in radians, in the range [-pi, pi].
arctan, tan, angle
arctan2 is identical to the atan2 function of the underlying C library. The following special values are defined in the C standard: [1]
x1 x2 arctan2(x1,x2)
+/- 0 +0 +/- 0
+/- 0 -0 +/- pi
> 0 +/-inf +0 / +pi
< 0 +/-inf -0 / -pi
+/-inf +inf +/- (pi/4)
+/-inf -inf +/- (3*pi/4)
Note that +0 and -0 are distinct floating point numbers, as are +inf and -inf.
[1] ISO/IEC standard 9899:1999, “Programming language C.”
Consider four points in different quadrants: >>> import mars.tensor as mt
>>> x = mt.array([-1, +1, +1, -1])
>>> y = mt.array([-1, -1, +1, +1])
>>> (mt.arctan2(y, x) * 180 / mt.pi).execute()
array([-135., -45., 45., 135.])
Note the order of the parameters. arctan2 is defined also when x2 = 0 and at several other special points, obtaining values in the range [-pi, pi]:
>>> mt.arctan2([1., -1.], [0., 0.]).execute()
array([ 1.57079633, -1.57079633])
>>> mt.arctan2([0., 0., mt.inf], [+0., -0., mt.inf]).execute()
array([ 0. , 3.14159265, 0.78539816]) | 2020-02-25T06:45:48 | {
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https://math.stackexchange.com/questions/2759298/average-number-of-cards-drawn-without-replacement-when-first-second-third-and | # Average number of cards drawn without replacement when first, second, third, and fourth aces appear
I have a standard $52$ deck of cards, shuffled randomly.
If I flip cards off the top until a Ace is shown, what is the average number of cards expected to be flipped over, including the Ace?
Further to this, what would be the average numbers of cards expected to be flipped for the second Ace, then the third Ace, and then finally the fourth?
My personal thoughts were that it would be $4/52$ chance, so take the reciprocal of this and it's $13$ cards on average.
Take the number of cards left, and so, for example, let's say it was $13$ cards the first time. $52-13=39$. $1/(3/39)$ is the average number next time, which is $13$ cards, so this seems to fit the pattern.
However when I actually do this, the number on average has been much lower. So I'm assuming there is something I'm missing?
• Welcome to MathSE. When you pose a question here, you should include your own thoughts on the problem. Please edit your question to explain what you have attempted and where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Apr 29 '18 at 20:53
Re: Replacement
The expected number of draws with replacement until getting the first ace would be $13$. However, without replacement, which is what applies here, the expected number of draws is smaller. This makes sense because the non-aces are not being replaced to be drawn a second time.
Without Replacement
The probability that the $n^\text{th}$ ace is at position $k$ is the number of ways for $n-1$ aces to be in the first $k-1$ positions and $4-n$ aces to be in the last $52-k$ positions. That is, $$\frac{\binom{k-1}{n-1}\binom{52-k}{4-n}}{\binom{52}{4}}$$ The mean position of the $n^\text{th}$ ace is \begin{align} \sum_{k=1}^{52}\frac{\binom{k-1}{n-1}\binom{52-k}{4-n}}{\binom{52}{4}}k &=\sum_{k=1}^{52}\frac{\binom{k}{n}\binom{52-k}{4-n}}{\binom{52}{4}}n\\ &=\frac{\binom{53}{5}}{\binom{52}{4}}n\\[3pt] &=\frac{53n}{5} \end{align}
• Is there something wrong here? Is the downvote because of something missing or wrong in the answer? I think the question provides sufficient context. – robjohn Apr 30 '18 at 16:09
• Thanks robjohn. Only if you feel inclined, my actual real world example has 59 cards in the deck (it's a different card game). So what would the equation look like? Same thing 4 cards the same, without replacement. – Grumpycat May 2 '18 at 3:26
• @Grumpycat: Hmm... $59$ instead of $52$. What might we change in the reasoning above that would change $52$ to $59$? Ponder the answer above a bit and give it a try. – robjohn May 2 '18 at 16:21 | 2019-08-20T17:00:34 | {
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https://bestbinaryxyvn.web.app/tavera70806wu/yearly-compound-interest-rate-formula-1177.html | ## Yearly compound interest rate formula
In such cases we use the following formula for compound interest when the interest is calculated quarterly. If the principal = P, rate of interest per unit time = r 4%,
Compound interest calculator If you would like your calculation based on a 360 Under rate of interest, type the annual percentage rate of interest awarded. What is the annual interest rate (in percent) attached to this money? % per year. How many times per year is your money compounded? time(s) a year. After how The formula for the future value of some investment with simple interest is: where is the principal amount, is the interest rate, and is the time period of the Compound interest, or 'interest on interest', is calculated with the compound interest formula. Multiply the principal amount by one plus the annual interest rate to the power of the number of compound periods to get a combined figure for principal and compound interest. Subtract the principal if you want just the compound interest. Compound Interest (CI) Formulas. The below compound interest formulas are used in this calculator in the context of time value of money to find the total interest payable on a principal sum at certain rate of interest over a period of time with either monthly, quarterly, half-yearly or yearly compounding period or frequency. What is the Monthly Compound Interest Formula? Monthly compounding formula is calculated by principal amount multiplied by one plus rate of interest divided by a number of periods whole raise to the power of the number of periods and that whole is subtracted from the principal amount which gives the interest amount. An interest rate formula helps one to understand loan and investment and take the decision. These days financial bodies like banks use Compound interest formula to calculate interest. Compounded annual growth rate i.e. CAGR is used mostly for financial applications where single growth for a period needs to be calculated. Recommended Articles
## Simply put, you calculate the interest rate divided by the number of times in a year the compound interest is generated. For instance, if your bank compounds
Calculate compound interest in four ways: Forward starts from a given Achieved interest determines the retrospective interest rate you achieved in going from a starting to an ending amount over a How Compound Interest is Calculated. Determine the effective annual interest rate if the nominal interest rate is: $$\text{ 12}\%$$ p.a. compounded quarterly. \begin{align*} 1 + i &= \ What's compound interest and what's the formula for compound interest in Excel be worth after one year at an annual interest rate of 8%? The answer is $108. Bank pays interest half-yearly on saving account deposit whereas for fixed deposit and recurring deposit interest paid based on customer request which could be The mathematical formula for calculating compound interest depends on include the amount of money deposited called the principal, the annual interest rate ### For instance, let the interest rate r be 3%, compounded monthly, and let the initial investment amount be$1250. Then the compound-interest equation, for an
How this formula works. The FV function can calculate compound interest and return the future value of an investment. To configure the function, we need to provide a rate, the number of periods, the periodic payment, the present value. To get the rate (which is the period rate) we use the annual rate / periods, or C6/C8. Calculates principal, principal plus interest, rate or time using the standard compound interest formula A = P(1 + r/n)^nt. Calculate compound interest on an investment or savings. Compound interest formulas to find principal, interest rates or final investment value including continuous compounding A = Pe^rt.
### Compound interest formula. Compound interest, or 'interest on interest', is calculated with the compound interest formula. Multiply the principal amount by one plus the annual interest rate to the power of the number of compound periods to get a combined figure for principal and compound interest.
Compound Interest (CI) Formulas. The below compound interest formulas are used in this calculator in the context of time value of money to find the total interest payable on a principal sum at certain rate of interest over a period of time with either monthly, quarterly, half-yearly or yearly compounding period or frequency. Yearly Compound Interest Formula. For calculating yearly compound interest, you just have to add interest of the one year into next year’s principal amount to calculate the interest of the next year. And, the formula in excel for yearly compound interest will be. =Principal Amount*((1+Annual Interest Rate/1)^(Total Years of Investment*1))) Instead you should use a generalized compound interest formula. General Compound Interest Formula (for Daily, Weekly, Monthly, and Yearly Compounding) A more efficient way of calculating compound interest in Excel is applying the general interest formula: FV = PV(1+r)n, where FV is future value, PV is present value, r is the interest rate per The tutorial explains the compound interest formula for Excel and provides examples of how to calculate the future value of the investment at annual, monthly or daily compounding interest rate. You will also find the detailed steps to create your
## To compute the compound interest in Excel for different time periods, all you have to do is convert the formula above into a relatable formula in Excel. The formula now becomes: = initial investment * (1 + annual interest rate/compounding periods per year) ^ (years * compounding periods per year)
17 Oct 2016 When it comes to calculating interest, there are two basic choices: If your investment paid 8% compound interest on an annual basis, Where "A" is the final amount, "P" is the principal, "r" is the interest rate, expressed as a With ICICI Pru Power of Compounding Calculator find out how much your investments can Half-yearly compounding: Interest is calculated every six months *While the annualized rate of return is 8% during the investment time period of 15 To calculate compounded interest and determine how much money will be owed after a specific amount of time, we must use the continuous compound interest If you like doing math, here's the formula for calculating APY: APY = (1 + r/n )n - 1. Where: r is the annual interest rate; n is the number of compounding periods To find the final value of an investment use this compound interest formula: Investment Value = P x ( 1 + r/n )(Y x n). P = Principal Value. r = Yearly Interest Rate Calculate compound interest in four ways: Forward starts from a given Achieved interest determines the retrospective interest rate you achieved in going from a starting to an ending amount over a How Compound Interest is Calculated. Determine the effective annual interest rate if the nominal interest rate is: $$\text{ 12}\%$$ p.a. compounded quarterly. \begin{align*} 1 + i &= \
What is the Monthly Compound Interest Formula? Monthly compounding formula is calculated by principal amount multiplied by one plus rate of interest divided by a number of periods whole raise to the power of the number of periods and that whole is subtracted from the principal amount which gives the interest amount. To compute the compound interest in Excel for different time periods, all you have to do is convert the formula above into a relatable formula in Excel. The formula now becomes: = initial investment * (1 + annual interest rate/compounding periods per year) ^ (years * compounding periods per year) | 2021-09-17T02:50:45 | {
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https://math.stackexchange.com/questions/1248463/how-many-subsets-of-a-1-2-3-10-have-the-property-that-the-sum-of-their-ele | # How many subsets of A={1,2,3,…,10} have the property that the sum of their elements is $\geq 28$?
I've already known that the desired answer is 512. But, how can I get this answer? Can anybody show me how to get this answer with only using permutation or combination?
I can only think that the answer is 512, because sums to 28 is equal with finding half of all the sums of these subsets, 55.
Because 1+2+....+10= 55, and total subsets is 1024. So 28 is half of all the subsets, so the number of subsets must be half of the total, so the total subsets is 512.
Is my thinking can be considered as right answer? Thanks
• As long as you make that idea formal you are on the right track. More precisely if a subset $B$ of $A$ satisfy such property, then compliment of $B$ does not satisfy the property. – Jack Yoon Apr 23 '15 at 16:01
I'd say you're on a good track, but more should be said. As you noted the sum of all elements of $A$ is $55$.
You should argue (using the $55$ fact) that for any subset $B$ of $A$, exactly one of $B$ and $B^c$ (complement of $B$) have a sum greater than or equal to $28$.
Then pairing up subsets of $A$ by complements, we see that exactly half the subsets of $A$ will have a sum greater than or equal to $28$.
• How can we pairing the subsets of A with its complement? Can you show me an example to do that? It sounds unfamiliar to me. Thanks – akusaja Apr 23 '15 at 16:10
• Example: We form the complementary subset pair $\{1, 3, 5, 6, 7, 8\}$ and $\{2,4, 9, 10 \}$. In this case $1+3+5+6+7+8\ge 28$; while $2+4+9+10 < 28$. – paw88789 Apr 23 '15 at 16:14
The correctness of paw88789's approach can perhaps be more clearly seen if we try a smaller set; that way, you can sit down and actually enumerate all the different possibilities. That can make a world of difference in terms of internalizing the approach.
Suppose we ask how many subsets of $\{1, 2, 3, 4, 5\}$ add up to a number $\geq 8$. The crucial idea is that we partition the set into two parts; these two parts are called complements of each other. Obviously, the sum of the two parts must add up to $15$. Exactly one of those parts is therefore $\geq 8$. There must be at least one such part, because of the pigeonhole principle (specifically, two $7$'s are sufficient only to add up to $14$). And if one part has sum $\geq 8$, the other part—its complement—must have sum $\leq 15-8 = 7$.
For instance, if I divide the set into parts $\{1, 2, 4\}$ and $\{3, 5\}$, the first part adds up to $7$, and its complement adds up to $8$.
Once one makes that observation, the rest of the proof is straightforward. There are $2^5 = 32$ different subsets of this set (including itself and the empty set). For each one, either its sum, or its complement's sum (but not both), must be $\geq 8$. Since exactly half of the subsets have sum $\geq 8$, the number of such subsets is $32/2$, or $16$.
The proof of the case for $\{1, 2, 3, \ldots, 10\}$ is exactly analogous.
Not a detailed solution, just some thought which might help to prove it. There are 1024 subsets in total. And you can pair each subset $S$ with $T =A-S$. And Now you can see that sum($S$) = 55 - sum($T$). If sum$(S)$ is $\geq 28$ then sum$(T)$ is $\leq 27$. So there are as many subsets with sum $\geq 28$ as subsets with sum $\leq 27$ . So there are exactly $\frac {1024}{2}$ subsets with sum $\geq 28$. | 2019-06-18T23:32:46 | {
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https://math.stackexchange.com/questions/1022714/probabilty-question | # Probabilty question
You have a bunch of n keys of which only one one opens the door of a storeroom, You wish to get into the storeroom. You choose one key at random and try it. If it does not work, you discard and try another key at random from the remaining keys. You proceed in this way until you open the door. Find the probability that you will open the door in the (i) 1st attempt, (ii) 2nd attempt, (iii) 3rd attempt. What do you think the probability will be for the kth attempt, where 4 ≤ k ≤ n? Can you justify it?
My answers were: (i) $\frac{1}{n}$
(ii) $\frac{1}{n-1}$
(iii)$\frac{1}{n-2}$
and for the kth attempt: $\frac{1}{n-k-1}$
However, it is given that the probability for the kth attempt is just $\frac{1}{n}$. why is that so?
Appreciate any help!
• Please explain how you arrive at 1/(n-1) as the answer to (ii). – Did Nov 15 '14 at 11:34
• Your pattern would have suggested $\dfrac{1}{n-(k-1)}=\dfrac{1}{n-k+1}$ for the $k$th attempt, though this is in fact a conditional probability – Henry Nov 15 '14 at 11:36
• How would you then relate this to the Monty Hall problem? – hypergeometric Nov 15 '14 at 12:03
Before trying them out give each key a different number in $\{1,\dots,n\}$. The key with e.g. number $2$ will be tried out at the second attempt. The probability that the right key gets number $k$ (corresponding with $k$-th attempt) is: $$\frac{1}{n}$$ So this is the probability that door will be opened at the $k$-th attempt.
The probability you open on the second attempt is the probability you fail to open on the first attempt multiplied by the conditional probability you succeed on the second, i.e. $$\frac{n-1}{n} \times \frac{1}{n-1}=\frac{1}{n}$$ and this pattern continues, so by induction the probability you open on the $k$th attempt is the probability you fail to open on the first $k-1$ attempts multiplied by the conditional probability you succeed on the $k$th, i.e. $$\frac{n-k+1}{n} \times \frac{1}{n-k+1}=\frac{1}{n}.$$
A simpler approach is to order the keys at random at the beginning. The key which opens the lock has a probability of $\frac1n$ of being in each position. | 2019-06-19T14:47:40 | {
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https://math.stackexchange.com/questions/3368441/why-is-the-probability-that-second-card-is-ace-in-a-deck-is-frac452 | # Why is the probability that second card is ace in a deck is $\frac{4}{52}$.
I did a question "We have a deck of card and we take one card we don't know what it is then what's the probability that the second card is an ace" and correct answer is $$4/52$$ why? book says it's because we don't know what the first card was.
Another question that I did was " some guy has a key chain and there are $$n$$ keys on the keychain and he wants to open a door. he uses first key if it doesn't work he discards it then use second one and so on. what's the probability the 3rd key opens the door" answer to this question turns out to be $$1/n$$ Why?
Edit: both questions are without replacement.
• The Ace of spades, say, is equally likely to be in any position in the deck, yes? so the probability that it is in position $2$ is $\frac 1{52}$. Similarly for the other three aces, and $4\times \frac 1{52}=\frac 4{52}$ Similarly for your key. – lulu Sep 24 '19 at 18:06
• If you were asked the probability of the first card being an ace, would it worry you that you don't know the second card? – Angina Seng Sep 24 '19 at 18:07
• If you have $n$ children and today is one of their birthdays, what is the probability that it is your second eldest child's birthday? If you randomly select a bottle of water from a shelf of $n$ identical water bottles, what is the probability that you select the second bottle from the left? – Andrew Chin Sep 24 '19 at 18:09
• In the second problem, you need to assume that (1) exactly one of the keys on the chain opens the door, and (2) each of the keys is equally likely to be the one that opens the door. As the question is stated, the answer could very well be $0$ (e.g. if he doesn't have a key to that door) or $1$ (if he has arranged the keychain in advance so that the third key is the one...). – Robert Israel Sep 24 '19 at 19:27
If the question was "what is the chance the first card is an ace", then you would say $$\frac{4}{52}$$, right? And if the question was "what is the probability the first key is the one that opens the door", you would say $$\frac{1}{n}$$, right?
OK, so now think about this: is the first card any more or less likely to be an ace than any other card in the deck? Is any key on the ring any more or less likely to be the key you're looking for? No! So, those probabilities are all the same.
I think what you're confused about is the following: Suppose the person with the keys tries the first and second key, and they both fail to be the right key. Then, what is the chance that the third key is the right key? Now, it is of course no longer $$\frac{1}{n}$$, but rather $$\frac{1}{n-2}$$. However, that was not the question ... the question was to figure out the chance of the third key in the chain of keys being the right key before we tried any of the keys. And, as such, the situation is exactly the same for every key on the chain: they all have the same chance of being the right key, so they all must have a chance of $$\frac{1}{n}$$ of being the right key. So, don;t confuse these two questions.
OK, one more: suppose you are one of $$n$$ people, and you are told to stand in a line, and then someone will randomly pick one of you from that line to give $$100$$ dollars. Now, if you want to optimize your chances to win the $$100$$ dollars, where should you stand? And the answer is of course: it doesn't matter where you stand! You all have the same chance of getting picked (assuming this is indeed truly random ...)
Well, it's symmetry in a way.
The second card could literally be any card, with equal chance. What's the chance that the $$47$$th card is an ace? Well, like any other rank, it's $$\displaystyle \frac{4}{52}$$.
Without using symmetry, we can do case-by-case analysis.
The probability that the first card is an ace and the second card is an ace is $$\displaystyle \frac{4}{52} \cdot \frac{3}{51}$$.
The chance that the first card is not an ace and the second one is an ace is $$\displaystyle \frac{48}{52} \cdot \frac{4}{51}$$.
Add these two and you get $$\displaystyle \frac{4}{52}$$.
Check that whether the first card was an ace or not (which changes the conditional probabilities that the second card is an ace) still adds up to what theyre saying...
Ace, ace plus not ace, ace: \begin{align} & \frac{4}{52} \cdot \frac{3}{51} + \frac{48}{52} \cdot \frac{4}{51} \\ &= \frac{4}{52} \cdot \left( \frac{3}{51} + \frac{48}{51} \right) \\ &= \frac{4}{52}. \end{align}
• @Leucippus Don't be silly. Of course it's an answer. Not the best answer, but still. – Robert Israel Sep 24 '19 at 19:31
Note that in any of your $$2$$ cases, instead of drawing cards/keys one by one, we can at the begginging just arrange them in a sequence (for cards it's $$(x_1,...,x_{52})$$, where every $$x_i$$ is one of the cards, and for keys it's a sequence $$(y_1,...,y_n)$$, where any $$y_i$$ is one of the keys. Note there are $$4$$ cards that will satisfy you when they'll be at $$x_2$$, whereas there is only $$1$$ key that you want to be $$y_3$$. In how many ways u can arrange such $$52-$$ and $$n-$$ sequence? For cards. Choose one ace from $$4$$ and put in on $$x_2$$, then in any order place everything left on $$51$$ places left ( $$51!$$ ways). Divide everything by $$52!$$ for it is the total amount of ways to place your $$52$$ cards. So we get $$\frac{4 \cdot 51!}{52!} = \frac{4}{52}$$. With keys it's just the same type: Choose your good key to be on place $$y_3$$ and arrange every key left with $$(n-1)!$$ ways. Divide by $$n!$$ ways, to get $$\frac{1}{n}$$
Note that we could choose any of $$x_i$$ and $$y_i$$ and answer will be the same.
It's due to fact, that probability isn't affected by occurence of an event, but only by knowledge we can have due to occurence of that event. | 2021-01-21T05:36:09 | {
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https://math.stackexchange.com/questions/2647385/a-sum-of-series-problem-frac3123-frac4234-cdots-fr | # A sum of series problem: $\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$
I have a question regarding the sum of this series: $$\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$$ My approach: I found that this sum is equal to: $$\sum_{n=3}^{2008}\frac{n}{(n-2)!+(n-1)!+(n)!}$$
I reduced it to : $$\sum_{n=3}^{2008}\frac{1}{n(n-2)!}$$
Please suggest how to proceed further.
What we want is $$\sum_{n=1}^{N} \dfrac{n+2}{n! + (n+1)! +(n+2)!}$$ \begin{align} \dfrac{n+2}{n! + (n+1)! +(n+2)!} & = \dfrac{n+2}{n! \left( 1 + (n+1) + (n+1)(n+2) \right)}\\ & = \dfrac{n+2}{n! \left( n^2 + 4n + 4 \right)}\\ & = \dfrac1{n! \left( n+2 \right)}\\ & = \dfrac{n+1}{(n+2)!}\\ & = \dfrac{n+2}{(n+2)!} - \dfrac1{(n+2)!}\\ & = \dfrac1{(n+1)!} - \dfrac1{(n+2)!} \end{align}
Can you finish it off from here?
Move your mouse over the gray area below for the complete answer.
\begin{align}\sum_{n=1}^{N} \dfrac{n+2}{n! + (n+1)! +(n+2)!} & = \sum_{n=1}^{N} \left( \dfrac1{(n+1)!} - \dfrac1{(n+2)!}\right)\\ & = \left( \dfrac1{2!} - \dfrac1{3!} + \dfrac1{3!} - \dfrac1{4!} + \dfrac1{4!} - \dfrac1{5!} + \cdots + \dfrac1{(N+1)!} - \dfrac1{(N+2)!}\right)\\ & = \dfrac1{2!} - \dfrac1{(N+2)!}\end{align} Set $N=2006$ to get the answer to your question.
• What should $n$ replaced by ? – spa Sep 26 '12 at 17:16
• @spa $N = 2006$ – user17762 Sep 26 '12 at 17:16
Hint: Apply telescoping series to \begin{align} \sum_{k=1}^{2006}\frac{k+2}{k!+(k+1)!+(k+2)!} &=\sum_{k=1}^{2006}\frac{k+2}{k!\,(k+2)^2}\\ &=\sum_{k=1}^{2006}\frac{k+1}{(k+2)!}\\ &=\sum_{k=1}^{2006}\left(\frac1{(k+1)!}-\frac1{(k+2)!}\right) \end{align}
• elegant and concise – G Cab Feb 12 '18 at 14:39
$$\dfrac1{n(n-2)!}=\dfrac{n-1}{n!}$$ which is Telescoping
• Thanks, how did you telescope it? – drake01 Feb 12 '18 at 13:26
• Drake write out a few terms, they start cancelling. – King Tut Feb 12 '18 at 13:29
• Yes, they start cancelling out when I write out a few terms of RHS, but how do you convert LHS to RHS( using partial fractions?) – drake01 Feb 12 '18 at 13:32
• @drake01 I'd say it comes by practice and fiddling out with the few terms given. You know this is not a standard sequence. So, the next obvious way is to try see if the sequence can telescope. You try out a few techniques, multiply $(n-1)$ in both the numerator and denominator, and it works! As noted on MSE before, call it the maths of "wishful thinking" ^_^ – Gaurang Tandon Feb 12 '18 at 13:35
• You are right, @Gaurang, – drake01 Feb 12 '18 at 13:38
$\sum_{n=3}^{2008}\frac{1}{n(n-2)!}=\sum_{n=3}^{2008}\frac{1}{n(n-2)!}.\frac{(n-1)}{(n-1)}=\sum_{n=3}^{2008}\frac{n-1}{n!}=\sum_{n=3}^{2008}(\frac{n}{n!}-\frac{1}{n!})=\sum_{n=3}^{2008}(\frac{1}{(n-1)!}-\frac{1}{n!})=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...-\frac{1}{2007!}+\frac{1}{2007!}-\frac{1}{2008!}=\frac{1}{2!}-\frac{1}{2008!}$
• The matter of fun is , I had obtained $\sum\frac{n-1}{n!}$ earlier then converted it into,$\sum\frac{1}{n(n-2)!}$ and then posted in MSE. Because the concept of Telescoping series didn't struck me at the moment. btw thanks. – drake01 Feb 12 '18 at 13:52
• @drake01 We all gradually gain the experience. The sweet part is these things, yes so funny. LOL :D – Mehrdad Zandigohar Feb 12 '18 at 13:57 | 2019-07-23T05:17:48 | {
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https://math.stackexchange.com/questions/3024995/are-the-bounds-of-t-always-0-1-for-line-integrals/3025015 | # Are the bounds of t always [0, 1] for line integrals?
I was given the task to find the line integral $$\int _C (x+y)ds$$ where $$C$$ is the line segment from $$(0,1,1)$$ to $$(3, 2, 2)$$.
I parameterised $$C$$ as $$3t\vec{i}+(1+t)\vec{j}+(1+t)\vec{k}$$, which means that $$\vec{r'}(t)=3\vec{i}+\vec{j}+\vec{k}$$ and that $$||\vec{r'}(t)||=\sqrt11$$.
This gave me the integral $$\sqrt11 \int(4t+1)dt$$, but then I wasn't sure what the bounds of $$t$$ are supposed to be. In class, we've been using $$t \space∈\space [0,1]$$, and if I use that, I get $$3\sqrt11$$.
The only thing I'm uncertain of here is why the bounds of $$t$$ go from $$0$$ to $$1$$. And if they don't, how can I determine what the bounds are? (Also, if I made any other mistakes in my calculation, please correct me!) Thank you very much!
• It depends on your parameterization of the curve. Does $t = 0$ correspond to the initial point and $t = 1$ correspond to the terminal point? If yes, cool. If no, use other bounds. – user296602 Dec 4 '18 at 1:26
• @T.Bongers when you say 'correspond,' do you mean if I substitute t=0 and t=1 back into the parameterisation, I get the initial and terminal points respectively? Thank you! – bletillafoliosa Dec 4 '18 at 1:28
• Yes. ${}{}{}{}{}$ – user296602 Dec 4 '18 at 1:28
In short this is dependent on the choice of parametrization. As a simple example, let's set up the line integral of $$x+y$$ along the line segment $$C$$ connecting $$(0,0)$$ and $$(1,1)$$ in $$\mathbb{R}^2$$. We can parametrize $$C$$ as $$C(t)=(t,t)$$ for $$0\le t\le 1$$ or we can parametrize $$C$$ "more quickly" using $$C(t)=(2t,2t)$$ for $$0\le t\le \frac{1}{2}$$. Using the first parametrization: $$\int_C (x+y)ds=\int_0^1x(t)\frac{dx}{dt}\cdot dt+\int_0^1 y(t)\frac{dy}{dt}\cdot dt=\int_0^1t\cdot dt+\int_0^1t\cdot dt=\int_0^12t\cdot dt=t^2\bigg|_0^1=1.$$ Using the second parametrization: $$\int_C (x+y)ds=\int_0^{\frac{1}{2}}x(t)\frac{dx}{dt}\cdot dt+\int_0^\frac{1}{2} y(t)\frac{dy}{dt}\cdot dt=\int_0^{\frac{1}{2}}4t\cdot dt+\int_0^{\frac{1}{2}}4t\cdot dt=\int_0^{\frac{1}{2}}8t\cdot dt=4t^2\bigg|_0^{\frac{1}{2}}=1.$$ So, the integral does not care about the parametrization. More particularly, we can see that the $$\frac{dx}{dt}$$ and $$\frac{dy}{dt}$$ terms "correct" the error introduced by reparametrization.
This is actually a special case of the Change of Variables formula, which in this case tells us that given a reparametrization $$\varphi:[a,b]\to [c,d]$$ then $$\int_c^df(x)dx=\int_a^b f(\varphi(t))\cdot \frac{d\varphi}{dt}dt.$$ | 2019-10-22T05:43:35 | {
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https://math.stackexchange.com/questions/2937135/limit-of-a-sum-using-complex-analysis | # Limit of a sum using complex analysis.
I'm trying to find the limit of this sum: $$S_n =\frac{1}{n}\left(\frac{1}{2}+\sum_{k=1}^{n}\cos(kx)\right)$$ I tried to find a formula for the inner sum first and I ended up getting zero as an answer. The sum is supposed to converge to $\cot(x\over 2)$ and that appears in my last expression but it goes to zero. Here is what I got with a rather long and clumsy reasoning: $$S_n = \frac{1}{2n}\left(\cot\left(\frac{x}{2}\right)\sin(nx)+\cos(nx)\right)$$ Thanks in advance.
Just write $$\cos (kx)$$ as the real part of $$e^{ikx}$$.
• I already did that but it didn't work. Sep 30 '18 at 20:39
• Then the limit is 0. Why do you think the answer is a cotan ?
– user598294
Sep 30 '18 at 20:42
• I don't know either way, but it can't be zero because the question asks for an answer in terms of $x$ Sep 30 '18 at 20:43
• You must have forgotten something.
– user598294
Sep 30 '18 at 20:45
• TheSilverDoe's proof convinced me and it has two different answers depending on the value of $x$ which solves my troubles, so yes not $\cot$, my bad. Sep 30 '18 at 20:54
You have, for all $$x \neq 2k\pi$$, $$\sum_{k=1}^n \cos(kx) = \mathrm{Re}\left( \sum_{k=1}^n e^{ikx}\right) = \mathrm{Re}\left( e^{ix} \frac{1-e^{inx}}{1-e^{ix}} \right)$$
Moreover you have $$\left|e^{ix} \frac{1-e^{inx}}{1-e^{ix}}\right| \leq \frac{2}{|1-e^{ix}|}$$
So $$\left| \sum_{k=1}^n \cos(kx) \right| \leq \frac{2}{|1-e^{ix}|}$$
And therefore you get $$S_n \rightarrow 0$$.
If $$x = 2k\pi$$ for $$k \in \mathbb{Z}$$, you easily have $$S_n \rightarrow 1$$.
• Remember that $|e^{iz}|=1$ for all $z \in \mathbb{R}$. So by the triangle inequality, $|1-e^{inx}| \leq 2$. Sep 30 '18 at 20:48
• Thank you very much, this was very helpful. Sep 30 '18 at 20:51
Another view of the question.
Instead of taking the real part of $$e^{ikx}$$ as given by the answers from TheSilverDoe and AlexL, the complex exponential can be more symmetric:
\begin{align*} \cos kx &= \frac{e^{-ikx}+e^{ikx}}2\\ \frac12+\sum_{k=1}^n\cos kx &= \frac12\sum_{k=-n}^ne^{ikx}\\ &= \frac12 \cdot e^{-inx}\frac{e^{i(2n+1)x}-1}{e^{ix}-1}\\ &= \frac12 \cdot \frac{e^{i(n+1)x}-e^{-inx}}{e^{ix}-1}\\ &= \frac12 \cdot \frac{e^{i\left(n+\frac12\right)x}-e^{-i\left(n+\frac12\right)x}}{e^{i\frac12x}-e^{-i\frac12x}}\\ &= \frac12 \cdot \frac{\sin\left(n+\frac12\right)x}{\sin\frac12x}\\ S_n&= \frac{\sin\left(n+\frac12\right)x}{2n\sin\frac12x}\\ \end{align*}
When $$e^{ix} \ne 1$$, i.e. $$x\ne 2\pi m$$ for integer $$m$$, $$S_n \to 0$$.
For other $$x$$'s, i.e. when $$x=2\pi m$$,
\begin{align*} \cos kx = \cos 2\pi km &= 1\\ \sum_{k=1}^n\cos kx &= n\\ \frac12 + \sum_{k=1}^n\cos kx &= \frac12+ n\\ S_n &= \frac1n\left(\frac12 + n\right)\\ &= \frac1{2n} + 1\\ &\to 1 \end{align*} | 2021-10-26T03:37:56 | {
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https://www.physicsforums.com/threads/is-the-limit-of-functions-necessarily-equal-to-itself.794999/ | # Is the limit of functions necessarily equal to "itself"?
As I read in the James Stewart's Calculus 7th edition, he said:
Picture Link：https://www.dropbox.com/s/aynxv7kuh7edqpi/相片 2015-1-28 13 16 39.jpg?dl=0
An alternative notation for
$\displaystyle\lim_{x\rightarrow a}f(x) = L$
is
$f(x) \rightarrow L$ as $x \rightarrow a$
which is usually read "f(x) approaches L as x approaches a"
My question is: Is $f(x)\rightarrow 0$ the same as $f(x) = L$?
For example,
$f(x) = x^2$
$\displaystyle\lim_{x\rightarrow 5}f(x) = 25$
I can say that $f(x) = x^2$ approaches 25 as $x$ approaches 5.
Therefore, can I say that the f(x) is not equal to 25 unless x is not approaching 5, but x, itself, is 5?
Last edited:
## Answers and Replies
Orodruin
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My question is: Is "f(x) approaches L" the same as "f(x) equals L"?
No. You may have a function which has a discontinuity at x=a. Consider the function f(0) = 1 and f(x) = 0 if x is different from 0. It has 0 as both left and right limits and yet the function value at x=0 is one.
so, can I say that the f(x) is not equal to 25 unless x is not approaching 5, and x, itself, is 5?
This is a different question and formulated with so many negations that I am not completely sure what question you actually intend to ask. Can you reformulate it?
Philethan
ShayanJ
Gold Member
Here you're asking about the notion of continuity of functions. A continuous function, by definition, has the property that its limit at a point is equal to its value at that point. But there are functions that are not continuous and so don't have this property. See the section Non-examples in the given link.
Philethan
No. You may have a function which has a discontinuity at x=a. Consider the function f(0) = 1 and f(x) = 0 if x is different from 0. It has 0 as both left and right limits and yet the function value at x=0 is one.
This is a different question and formulated with so many negations that I am not completely sure what question you actually intend to ask. Can you reformulate it?
Sorry about that. It's a little difficult for me to express this abstract question in English.
The proposition is : $x + 2 \neq 7$ unless $x = 5$
Is that proposition correct? I guess someone's answer is no because when x approaches 5, $x + 2$ would also be 7.
Mark44
Mentor
Sorry about that. It's a little difficult for me to express this abstract question in English.
The proposition is : $x + 2 \neq 7$ unless $x = 5$
Is that proposition correct? I guess someone's answer is no because when x approaches 5, $x + 2$ would also be 7.
When x approaches 5, x + 2 approaches 7, but won't necessarily be equal to 7. When we talk about x approaching 5, what we're saying is that x is in some neighborhood around 5 (i.e., |x - 5| is "small"), so x + 2 will be in some neighborhood around 7 (i.e., |x + 2 - 7| is also "small"). x + 2 won't be equal to 7 if x ≠ 5.
Philethan
Orodruin
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Sorry about that. It's a little difficult for me to express this abstract question in English.
The proposition is : $x + 2 \neq 7$ unless $x = 5$
Is that proposition correct? I guess someone's answer is no because when x approaches 5, $x + 2$ would also be 7.
No, for the function x+2, the statement is true. Even if it approaches 7 as x approaches 5, it is not 7 except for when x=5.
However, this is not always the case. Take the function f(x)=0 for all x. The limit when x goes to 0 is 0 and also f(0)=0. There are also examples where you will find that f takes the same values in very different points. That f(x) = a for only one x is a property called injective (also formulated as f(x) = f(y) implies x = y) and a linear function with non-zero slope has this property.
Philethan
Mark44
Mentor
A better example would be the function f given by ##f(x) = \frac{x^2 + 3x + 2}{x + 1}##. We can talk about this limit:
##\lim_{x \to -1} f(x)##, which turns out to be 1, even though f is not defined at x = -1.
Mark44, thank you very much! You help me a lot. Now here's my next relevant question.
Actually, my question has something to do the following physics question:
In the uniform circular motion, is the instantaneous centripetal acceleration perpendicular to instantaneous velocity?
$\vec{a}_{c}=\displaystyle \lim_{\Delta t\rightarrow 0}\frac{\Delta \vec{v}}{\Delta t}$
If we differentiate the velocity and solve the inner product $\vec{a}\cdot \vec{v}$, then we'll get zero.
However, if we imagine the $\Delta \vec{v}$ then we would think $\vec{v}$ will never be perpendicular to $\Delta \vec{v}$.
Someone would say we have to reconcile these two facts.
And I think the reason why they think those two argument are contradictory is that
they mistaken the direction of $\vec{a}_{c}=\displaystyle \lim_{\Delta t\rightarrow 0}\frac{\Delta \vec{v}}{\Delta t}$ as the direction of $\Delta \vec{v}$.
Just like what you told me, the direction of $\vec{a}_{c}$ is what $\frac{\Delta \vec{v}}{\Delta t}$ approaches as $\Delta t$ approaches $0$.
Therefore, the direction of $\vec{a}_{c}$ is not the same as the direction of $\frac{\Delta \vec{v}}{\Delta t}$.
Is that correct?
Thanks so much:)
Orodruin
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Well, the direction is the limit of the direction of ##\Delta \vec v## as ##\Delta t \to 0##. This is just by definition of the derivative.
Philethan
Yes, Orodruin. So, there's no contradiction, right?
That means, the direction of $\Delta \vec{v}$ is not the direction of $a_{c}$.
And the we should say the direction of $a_{c}$ is what $\Delta \vec{v}$ approaches as $\Delta t$ approaches zero, rather than the direction of $\Delta \vec{v}$.
Last edited:
Orodruin
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Yes, Orodruin. So, there's no contradiction, right?
That means, the direction of $\Delta \vec{v}$ is not the direction of $a_{c}$.
As long as ##\Delta t## remains finite, the direction of ##\Delta \vec v## is not orthogonal to ##\vec v##. Acceleration is a derivative of the velocity and thus defined in the limit.
Philethan
Sorry, Orodruin. I want to know more about "thus defined in the limit". Does that mean what I said?
The direction of $a_{c}$ is defined as the direction of what $\Delta \vec{v}$ approaches as $\Delta t$ approaches $0$, rather than the direction of $\Delta \vec{v}$.
Orodruin
Staff Emeritus | 2022-05-19T07:47:57 | {
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https://math.stackexchange.com/questions/262828/using-proof-by-contradiction-vs-proof-of-the-contrapositive/262831 | # Using proof by contradiction vs proof of the contrapositive
What is the difference between a "proof by contradiction" and "proving the contrapositive"? Intuitive, it feels like doing the exact same thing. And when I compare an exercise, one person proves by contradiction, and the other proves the contrapositive, the proofs look almost exactly the same.
For example, say I want to prove: $P \implies Q$ When I want to prove by contradiction, I would say assume this is not true. Assume $Q$ is not true, and $P$ is true. Blabla, but this implies $P$ is not true, which is a contradiction.
When I want to prove the contrapositive, I say. Assume $Q$ is not true. Blabla, this implies $P$ is not true.
The only difference in the proof is that I assume $P$ is true in the beginning, when I want to prove by contradiction. But this feels almost redundant, as in the end I always get that this is not true. The only other way that I could get a contradiction is by proving that $Q$ is true. But this would be the exact same things as a direct proof.
Can somebody enlighten me a little bit here ? For example: Are there proofs that can be proven by contradiction but not proven by proving the contrapositve?
• Also related: math.stackexchange.com/questions/71245/… and more importantly, math.stackexchange.com/questions/227109/… Dec 20, 2012 at 19:06
• Excellent question, +1. Dec 20, 2012 at 19:16
• Andrej Bauer had a recent blog post about this. "I am discovering that mathematicians cannot tell the difference between “proof by contradiction” and “proof of negation”. "
– MJD
Dec 20, 2012 at 22:39
• @MJD: Nearly three years is not very recent in terms of the internet. Just to get some perspective, Google+ was not yet conceived when Andrej posted this. Dec 21, 2012 at 1:10
• @asaf: I thought it was recent because I hadn't seen it when it was new, or perhaps because I saw it when it was new, forgot it, and saw it again last week.
– MJD
Dec 21, 2012 at 1:24
To prove $P \rightarrow Q$, you can do the following:
1. Prove directly, that is assume $P$ and show $Q$;
2. Prove by contradiction, that is assume $P$ and $\lnot Q$ and derive a contradiction; or
3. Prove the contrapositive, that is assume $\lnot Q$ and show $\lnot P$.
Sometimes the contradiction one arrives at in $(2)$ is merely contradicting the assumed premise $P$, and hence, as you note, is essentially a proof by contrapositive $(3)$. However, note that $(3)$ allows us to assume only $\lnot Q$; if we can then derive $\lnot P$, we have a clean proof by contrapositive.
However, in $(2)$, the aim is to derive a contradiction: the contradiction might not be arriving at $\lnot P$, if one has assumed ($P$ and $\lnot Q$). Arriving at any contradiction counts in a proof by contradiction: say we assume $P$ and $\lnot Q$ and derive, say, $Q$. Since $Q \land \lnot Q$ is a contradiction (can never be true), we are forced then to conclude it cannot be that both $(P \land \lnot Q)$.
But note that $\lnot (P \land \lnot Q) \equiv \lnot P \lor Q\equiv P\rightarrow Q.$
So a proof by contradiction usually looks something like this ($R$ is often $Q$, or $\lnot P$ or any other contradiction):
• $P \land \lnot Q$ Premise
• $P$
• $\lnot Q$
• $\vdots$
• $R$
• $\vdots$
• $\lnot R$
• $\lnot R \land R$ Contradiction
$\therefore \lnot (P \land \lnot Q) \equiv P \rightarrow Q$
• No, this is not the same as proving directly: To prove P, one uses step by step implications to arrive directly at $Q$, without assuming $\lnot Q$. (2) Note that the contradiction obtained may may be that assuming P and $\lnot Q$ leads to both $R \land \lnot R$ (R may happen to be Q)$. Dec 20, 2012 at 19:18 • Yes, Kasper, usually. It's just that in longer proofs, we may find a point were some other statement AND its negation follow from assuming both$P\land \lnot Q$. (e.g., P may imply R, and$\lnot Q$may imply$\lnot R$, in which we have to conclude that we cannot have both$P$and$\lnot Q$, which is equivalent to proving$P\rightarrow Q$). Dec 20, 2012 at 19:44 • Okay, I understand. I've one other question if you don't mind :). Does a problem like the following exist? You're assuming:$P∧¬Q$(to prove a contradiction). And in this problem it's impossible to prove$Q∧¬Q$or$P∧¬P$from this assumption. But it's possible to prove another contradiction$R∧¬R$. Dec 20, 2012 at 20:02 • Yes, I'll try to find an example. There are many...usually longer proofs, where we assume$P \land \lnot Q$...then show$\lnot Q \rightarrow q_i ....\rightarrow...q_j....\rightarrow R$and$P \rightarrow ...p_i....\rightarrow...p_j \rightarrow \lnot R$, giving$R \land \lnot R$. When I have a little time to look through some texts, for a shortish proof using this, I'll post here, or we can go through it in chat...either way, I'll let you know. Dec 20, 2012 at 20:09 • You could also have P as a premise, then ¬Q as the next premise. Then a contradiction get derived which leads to a rejection of Q and we thus obtain Q. Since P and Q have the same scope, and P comes first, then we can infer that P implies Q. Dec 22, 2012 at 2:18 There is a useful rule of thumb, when you have a proof by contradiction, to see whether it is "really" a proof by contrapositive. In a proof of by contrapositive, you prove$P \to Q$by assuming$\lnot Q$and reasoning until you obtain$\lnot P$. In a "genuine" proof by contradiction, you assume both$P$and$\lnot Q$, and deduce some other contradiction$R \land \lnot R$. So, at then end of your proof, ask yourself: Is the "contradiction" just that I have deduced$\lnot P$, when the implication was$P \to Q$? Did I never use$P$as an assumption? If both answers are "yes" then your proof is a proof by contraposition, and you can rephrase it in that way. For example, here is a proof by "contradiction": Proposition: Assume$A \subseteq B$. If$x \not \in B$then$x \not \in A$. Proof. We proceed by contradiction. Assume$x \not \in B$and$x \in A$. Then, since$A \subseteq B$, we have$x \in B$. This is a contradiction, so the proof is complete. That proof can be directly rephrased into a proof by contrapositive: Proposition: Assume$A \subseteq B$. If$x \not \in B$then$x \not \in A$. Proof. We proceed by contraposition. Assume$x \in A$. Then, since$A \subseteq B$, we have$x \in B$. This is what we wanted to prove, so the proof is complete. Proof by contradiction can be applied to a much broader class of statements than proof by contraposition, which only works for implications. But there are proofs of implications by contradiction that cannot be directly rephrased into proofs by contraposition. Proposition: If$x$is a multiple of$6$then$x$is a multiple of$2$. Proof. We proceed by contradiction. Let$x$be a number that is a multiple of$6$but not a multiple of$2$. Then$x = 6y$for some$y$. We can rewrite this equation as$1\cdot x = 2\cdot (3y)$. Because the right hand side is a multiple of$2$, so is the left hand side. Then, because$2$is prime, and$1\cdot x $is a multiple of$2$, either$x$is a multiple of$2$or$1$is a multiple of$2$. Since we have assumed that$x$is not a multiple of$2$, we see that$1$must be a multiple of$2$. But that is impossible: we know$1$is not a multiple of$2$. So we have a contradiction:$1$is a multiple of$2$and$1$is not a multiple of$2$. The proof is complete. Of course that proposition can be proved directly as well: the point is just that the proof given is genuinely a proof by contradiction, rather than a proof by contraposition. The key benefit of proof by contradiction is that you can stop when you find any contradiction, not only a contradiction directly involving the hypotheses. • It'd be very helpful if you also explicitly addressed the relationship between nonessential proof by contradicton vs. direct proof in proofs like Euclid's proof that there are infinitely many primes. Then we could refer to this answer as a canonical answer for eliminating such nonessential uses of contradiction. I don't think that will be clear to beginners from what is written above. it is addressed briefly in passing in Hardy and Woodgold's intelligencer article (and, iirc, also mentioned in passing in some answers here). Mar 17, 2014 at 4:20 • @Bill Dubuque: that is a good idea, but I don't know if it fits into this question very well. I wrote something at math.stackexchange.com/a/715438/630 Mar 17, 2014 at 12:49 • Thanks, that should prove very helpful to many readers. However, it doesn't seem to address the point I raised above, which perhaps was not clear. What I meant was that many proofs of Euclid's proposition P by contradiction are simply proofs of P that have prepended an unused assumption of$\,\lnot$P. Thus, similar to above, deleting that unused assumption yields a direct proof of P.$\ \ $Mar 17, 2014 at 23:35 • @Bill that's covered by Andrej Bauer's post referenced above though. May 15, 2021 at 3:04 • @Nicolas But we prefer to have answers on-site (vs. links to off-site content - which usually eventually succumb to link rot) May 15, 2021 at 7:45 It's not the same. If$P$and$Q$are statements about instances that (a priori independently) are true for some instances and false for others then proving$P\Rightarrow Q$is the same as proving the contrapositive$\neg Q\ \Rightarrow \neg P$. Both mean the same thing: The set of instances for which$P$is true is contained in the set of instances where$Q$is true. Proving a statement$A$by contradiction is something else: You add$\neg A$to your list of axioms, and using the rules of logic arrive at a contradiction, e.g., at$1=0$. Then you say: My axiom system was fine before adding$\neg A$. Since this addition has spoiled it, in reality$A$has to be true. An example: You want to prove the statement $$A:\quad {\rm "The\ number}\ \sqrt{2}\ {\rm is\ irrational."}$$ Then you add$\sqrt{2}={p\over q}$to your list of axioms about rational numbers and arrive at a contradiction. One difference is that proof by contrapositive only applies to propositions of the form$A \to B$("if-then propositions"). However not every proposition is a "if-then proposition", for example, consider the proposition, exist x real for all p,q integer, x != p/q, there is no$\to$inside that proposition, so it is not feasible to prove it by contrapositive. Proof by contrapositive:$(\neg B \to \neg A) \to (A \to B)$Proof by contradiction:$(\neg A \to (B \land\neg B)) \to A $[Note: The post by amWhy is pretty much what I like to say, but still prefer to put my response as a separate answer, hoping some readers might find it more useful] First, let me restrict my response only to conditional statements of the form$P\rightarrow $Q. With this, let's understand two important equivalences as listed below.$P\rightarrow Q \equiv \lnot Q\rightarrow \lnot P$...(I)$P\rightarrow Q \equiv \lnot (\lnot (P\rightarrow Q)) \equiv \lnot (P \land \lnot Q) \equiv (P \land \lnot Q) \rightarrow C$where C is contradiction ...(II) For proof by contraposition, we use equivalence (I) where we start by assuming$\lnot Q$and show, by use of calculations and a priori knowledge about other theorems, etc., that$\lnot P$is true. For proof by contradiction, we use equivalence (II) where we start by assuming$P$and$\lnot Q$and show this leads to some kind of contradiction. The contradiction can be with$P$, or$Q$, or with altogether new statement$R$. However, as long as there is some contradiction, it essentially proves the original statement. In this regard, also note that, when the contradiction is with either$P$or$Q$(i.e. any of the initial assumptions), the proof by contradiction often looks very similar to proof by contraposition, especially for the ones who are not clear about the differences in (I) and (II). Example$1$: If$n$is an odd integer, then$5n-3$is an even integer. You may prove this either by contradicting$P$or by$Q$. Example$2$:If$f(x)=\frac{2x+3}{x+2}$, then for every real$x, f(x) \neq 2$. You may find a contradiction in an altogether new statement say$R$:= "$3=4\$"
Surprisingly, no one has mentioned quantifiers.
To quote a comment from Bernard on a duplicate of this question:
Take the assertion: ‘When Mr So and So is happy, he sings. The contrapositive asserts that ‘Mr So and So does not sing so he's not happy’. The negation asserts that ‘There are days when Mr So and So is happy, yet he does not sing’.
I converted this example into logical notation with quantifiers, which makes the difference between negation and contrapositive more obvious.
Original statement: Any day when Mr. So-and-so is happy is a day when he sings. $$\forall d \in D_{ays}: (H_{appy}(M_r) \rightarrow S_{ings}(M_r))$$
Contrapositive: Any day when Mr. So-and-so does not sing is a day when he is not happy. $$\forall d \in D_{ays}: (\neg S_{ings}(M_r) \rightarrow \neg H_{appy}(M_r))$$
Negation: There are some days (at least one day) when Mr. So-and-so is happy, but does not sing. $$\exists d \in D_{ays}: (H_{appy}(M_r) \land \neg S_{ings}(M_r))$$
This doesn't directly address the different types of proofs based on these concepts (addressed in several answers already), but it may assist an understanding of that.
To prove by contrapositive, in the above example, you would start with the expression (proposition) "not singing" and directly derive "not happy" (perhaps by algebraic rearrangement).
To prove by contradiction, on the other hand, you would assume the negation, and then derive from there until you had proven two contradictory facts.
• Yes, I am completely and utterly confounded as to why no one has mentioned quantifiers before this answer.
– D.R.
Sep 29 at 5:52 | 2022-10-03T23:38:33 | {
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http://math.stackexchange.com/questions/111188/why-isnt-h-mathbbrp-infty-mathbbf-2-cong-mathbbf-2x | # Why isn't $H^*(\mathbb{R}P^\infty,\mathbb{F}_2)\cong \mathbb{F}_2[[x]]$?
We just computed in class a few days ago that $$H^*(\mathbb{R}P^n,\mathbb{F}_2)\cong\mathbb{F}_2[x]/(x^{n+1}),$$ and it was mentioned that $H^*(\mathbb{R}P^\infty,\mathbb{F}_2)\cong \mathbb{F}_2[x]$, but I (optimistically?) assumed that the cohomology ring functor would turn limits into colimits, and so $$\mathbb{R}P^\infty=\lim\limits_{\longrightarrow}\;\mathbb{R}P^n$$ would mean that we'd get the formal power series ring $$\lim\limits_{\longleftarrow}\;\mathbb{F}_2[x]/(x^{n+1})=\mathbb{F}_2[[x]].$$
My professor said that the reason we get $\mathbb{F}_2[x]$ is just that the cohomology ring, being the direct sum of the cohomology groups, can't have non-zero elements in every degree, which certainly makes sense.
Even though that should settle the matter, for some reason, I'm still having a bit of trouble making this make click for me. Is the explanation just that the cohomology ring functor doesn't act as nicely as I'd hoped? Is there an intuitive explanation of what's going on?
-
I'll wait for a real expert to leave a precise answer, but...it seems that your intuition about the cohomology ring functor needs a little calibrating. How about this: you have more than just a ring structure on $H^*$ -- you have a graded algebra structure. Really you might as well view this as a sequence of functors into $R$-Mod together with a graded-commutative bilinear product on them. In particular though the property that each graded piece is finitely generated is fundamental and should be viewed as nice, I think.... – Pete L. Clark Feb 20 '12 at 5:02
...But I don't doubt that the completion of the cohomology ring functor appears elsewhere in algebraic topology. That's really what I'm waiting for an expert to weigh in on. – Pete L. Clark Feb 20 '12 at 5:03
(Above, when I say that each graded piece is finite-dimensional, I don't mean that this holds for $H^*(X,R)$ for all spaces $X$, but rather that it holds for sufficiently nice spaces, namely those homotopy equivalent to a CW-complex with finite $n$-skeleta for each fixed $n$.) – Pete L. Clark Feb 20 '12 at 5:04
Since $RP^\infty$ has a model with finitely many simplices in each dimension, its cohomology is of at most of countable dimension over the base field. – Mariano Suárez-Alvarez Feb 20 '12 at 5:57
@Pete: Ah, I see - the inverse limit of $\mathbb{F}_2[x]/(x^n)$ as graded $\mathbb{F}_2$-algebras is just $\mathbb{F}_2[x]$; whereas $\mathbb{F}_2[[x]]$ does not have a grading compatible with the maps to each $\mathbb{F}_2[x]/(x^n)$. – Zev Chonoles Feb 20 '12 at 5:57
This is probably something that should be a comment, mainly because I'm not sure it is correct. Feel free to tell me it is rubbish!
1) I have definitely seen it written $H^*(\mathbb{R} P^\infty; \mathbb{F}_2) \simeq \mathbb{F}_2[\![ x ]\!]$. For example see Jacob Lurie's notes here.
2) Cohomlogy doesn't play well with inverse limits in general. There is the Milnor exact sequence, which is this case should give:
$$0 \to \text{lim}^1 H^{\ast-1}(\mathbb{R} P^\infty;\mathbb{F}_2) \to H^*(\mathbb{R} P^\infty;\mathbb{F}_2) \to \underset{n}{\text{lim}} H^*(\mathbb{R}P^\infty;\mathbb{F}_2) \to 0$$
where $\text{lim}^1$ is the first dervied functor of the inverse limit.
In this case the $\text{lim}^1$ terms should all be zero by the Mittag-Leffler criteria (which follows since the maps $H^*(\mathbb{R} P^{n+1};\mathbb{F}_2)\to H^*(\mathbb{R} P^{n};\mathbb{F}_2)$ are surjective). Thus we can conclude that $$H^*(\mathbb{R} P^\infty; \mathbb{F}_2) \simeq \mathbb{F}_2[\![ x ]\!]$$
As an aside this is the calculation usually given to show $E^*(\mathbb{C} P^\infty) \simeq E_*[\![ x ]\!]$ for any complex oriented cohomology theory $E$, which basically sets up the whole relationship between complex oriented cohomology theories and formal group laws.
Dear Juan, independently of any technology, you probably know that $\mathbb RP^\infty$ can be constructed as a CW-complex with one cell in each dimension. It follows that the cellular complex has each of its homogeneous components a vector space of dimension $1$, and therefore its total dimension cannot be more than countable (in fact, the differential in the cellular complex with $\mod 2$-coefficients is zero, so the cohomology is of countable dimension).As the power series $\mathbb F_2[[x]]$ has uncountable dimension over $\mathbb F_2$, Lurie is either completing or wrong :) – Mariano Suárez-Alvarez Feb 21 '12 at 0:47
That is one variant of "completing", actually. Since the Milnor exact sequence is clearly true in each degree, and inverse limits commute with products, with that definition of cohomology then the isomorphism with $\mathbb F_1[[x]]$ does hold. We can conclude that in the context of infinite dimensional complexes one needs to be precise about what exactly one is talking about:) – Mariano Suárez-Alvarez Feb 21 '12 at 1:21 | 2015-11-28T22:25:43 | {
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https://math.stackexchange.com/questions/4001823/combinatorics-and-statistic-question-pick-all-1s-in-a-hand-of-9-cards | # Combinatorics and statistic question (pick all 1s in a hand of 9 cards)
I was going through statistics classes on Khan academy and there was this exercise I was hoping someone could help me with:
1. we have $$36$$ unique cards (normal deck, $$4$$ types - hearts, diamonds etc.). And $$9$$ different groups of cards (from $$1$$ to $$9$$).
2. a hand is a selection of $$9$$ cards which can be sorted any way you want
question: what is the probability of getting all $$4$$ $$1$$s in a hand? (so $$1$$ of diamond, $$1$$ of hearts etc, $$4$$ total) link:https://youtu.be/ccrYD6iX_SY
I'm here because I disagree with the Sal's solution and I was hoping someone could tell me where I could be wrong:
1. we are looking for a probability, and $$\mathrm{(prob\ of\ needed \ result)\ = \frac{\# (needed \ result\ can\ happen) }{ \# (total \ results)}}$$
2. total results are ok and it is the same in the video
(in python)
n_total_outcomes = scipy.math.factorial(36)/(scipy.math.factorial(9)*(scipy.math.factorial(36-9)))
1. but the desired result is where it gets hairy (for me).
2. Next I decided not to use any special formulas and apply logic: we can get Lucky ~ if we have already picked $$\frac 59$$ random cards, we have $$31$$ cards left, and the chance we pick $$1$$s higher ($$\frac 1{31}, \frac 1{30}, \frac 1{29}, \frac 1{28}$$) and unlucky or less probable if we start picking the $$1$$s at the beginning $$\frac {1}{36}, \frac{1}{35}, \frac{1}{34}, \frac{1}{33}$$.
So I just imagined that these are $$2$$ side cases ($$7\times 10^{-7}$$ and $$1 \times 10^{-6}$$), which is still nowhere close to Sal's solution, where he has very high chance. If you watch his video my beef is that he is picking all four 1s at the beginning and then picking up the rest of the hand. It is nowhere guaranteed you will get all four 1s right away.
If anyone could show me the problem with my logic, it would be much appreciated.
• First of all, thank you for posting a well-written question. I will show you how to write fractions by editing the post, but everything else is nice.Please see the edits now. – Teresa Lisbon Jan 27 at 11:50
• But there are so many ways to pick those $4$ cards - it could be first $4$, it could be last $4$, it could be in any order and any of the $4$ out of $9$. So one way to look at it is $\frac{4C4 \times 32C5}{36C9}$. It comes to same as the video you shared. Does that help? – Math Lover Jan 27 at 12:53
• Thank you for your answer. I'm going to go over the topic again. I should have asked differently (your answer is enough): if we deal this 36 cards deck a million times, how many times approximately we will end up with a hand where we have all 4 1s. – Oleg Peregudov Jan 28 at 9:35
it appears that you are thinking he is studying the process of selection. However, this is not the case; he is just considering the final hand consisting of $$4$$ 1s and $$5$$ other cards.
On the other hand you are attempting to consider the process. The way you are doing this would involve you in considering all the different ways the $$4$$ 1s are selected which you have not done.
• The correct probability of $2/935$ is roughly $2$ per $1000$. So it would occur roughly $2000$ times on average. – S. Dolan Jan 28 at 9:47 | 2021-05-15T23:24:52 | {
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https://mathematica.stackexchange.com/questions/238472/how-can-i-direct-sum-matrices-into-the-middle-of-one-another-another | # How can I direct sum matrices into the middle of one another another?
I would like to execute the mathematical operation of the direct sum of matrices in the case where the matrices are not appended one after the other along the diagonal, but instead mixed among one another. This must be generalizable to any permutation of mixing indices along the diagonal.
For example, say we have matrix A, B and C $$A=$$\begin{bmatrix} cos (t) & sin(t) \\ -sin(t) & cos(t) \end{bmatrix}$$$$
$$B=$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$$$
$$C=$$\begin{bmatrix} cos (u) & sin(u) \\ -sin(u) & cos(u) \end{bmatrix}$$$$
$$A ? B ? C=$$\begin{bmatrix} cos (t) & sin(t) & 0 & 0 & 0 & 0\\ -sin(t) & cos(t) & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & cos(u) & 0 & sin(u)\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -sin(u) & 0 & cos(u) \end{bmatrix}$$$$
How would this be done in a generalizable way in Mathematica? One could use a permutation matrix, $$P=$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$$$ where, $$A?B?C=P[A⊕B⊕C]P^T$$, but how could I create such a matrix P in mathematica knowing only what the mapping is? In this case, the index mapping being: (1,2,3,4,5,6)->(1,2,3,5,4,6)
• For a start, please include your matrices as plain text Mathematica code, so we can copy and paste them. – MarcoB Jan 20 at 1:56
• perm = {1, 2, 3, 5, 4, 6}; p = IdentityMatrix[6][[perm]];? – kglr Jan 20 at 4:01
• and p = UnitVector[6, #] & /@ perm? – kglr Jan 20 at 4:01
• $P$ can be obtained exactly as @kglr introduced. The direct sum can be computed with {a, b, c} // Map[Inactive] // DiagonalMatrix // Activate // ArrayFlatten, and you get the result with p . directsum . Transpose[p]. – SneezeFor16Min Jan 20 at 4:18
ClearAll[r1, t, u, a, b, c, abc]
r1 = {Cos[t], Sin[t]};
a = {r1, Cross[r1]};
b = IdentityMatrix[2];
c = a /. t -> u;
abc = SparseArray[Band[{1, 1}] -> {a, b, c}];
abc // MatrixForm // TeXForm
$$\left( \begin{array}{cccccc} \cos (t) & \sin (t) & 0 & 0 & 0 & 0 \\ -\sin (t) & \cos (t) & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & \cos (u) & \sin (u) \\ 0 & 0 & 0 & 0 & -\sin (u) & \cos (u) \\ \end{array} \right)$$
perm = {1, 2, 3, 5, 4, 6};
abc[[perm, perm]] // MatrixForm // TeXForm
$$\left( \begin{array}{cccccc} \cos (t) & \sin (t) & 0 & 0 & 0 & 0 \\ -\sin (t) & \cos (t) & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & \cos (u) & 0 & \sin (u) \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -\sin (u) & 0 & \cos (u) \\ \end{array} \right)$$
Alternatively,
p = IdentityMatrix[6][[perm]];
p // MatrixForm // TeXForm
$$\left( \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right)$$
p.abc.Transpose[p] // MatrixForm // TeXForm
$$\left( \begin{array}{cccccc} \cos (t) & \sin (t) & 0 & 0 & 0 & 0 \\ -\sin (t) & \cos (t) & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & \cos (u) & 0 & \sin (u) \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -\sin (u) & 0 & \cos (u) \\ \end{array} \right)$$
Note: An alternative way to get abc is to use SparseArraySparseBlockMatrix:
abc2 = SparseArraySparseBlockMatrix[MapIndexed[#2[[{1, 1}]] -> # &, {a, b, c}]];
abc2 == abc
True
• So SparseArray[Band[{1, 1}] -> {a, b, c}][[perm, perm]] // Normal is all it takes! — I would say the use of Band is brilliant :-) – SneezeFor16Min Jan 20 at 4:44 | 2021-03-02T07:47:06 | {
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https://mathhelpboards.com/threads/coefficient-of-trinomial-expansion.332/ | # Coefficient of trinomial expansion
#### anemone
##### MHB POTW Director
Staff member
Hi,
I want to find the coefficient of $\displaystyle x^4$ and $\displaystyle x^{39}$ of $\displaystyle (1+x+x^2)^{20}$
I use the binomial theorem twice and come out with the following argument:
$\displaystyle (1+x+2x^2)^{20}= [(1+x)+(2x^2)]^{20}$
I see that
$\displaystyle T(r+1)={20 \choose r} (1+x)^{20-r}(2x^2)^r$
$\displaystyle T(r+1)={20 \choose r} .2^r. x^{2r}.(1+x)^{20-r}$
$\displaystyle T(r+1)={20 \choose r} .2^r. x^{2r}.{20-r \choose k} (1)^{20-r-k}. (x)^{k}$
$\displaystyle T(r+1)={20 \choose r} {20-r \choose k} .2^r. x^{2r+k}$
Now, I notice that $\displaystyle 20-r \ge k$, and, of course, $\displaystyle 0\le r,k \le 20$.
To obtain the coefficient of x^4, we need 2r+k=4
If k=0, r=2. (Case 1)...this gives the coefficient of 760.
If k=1, then r is not an integer.
If k=2, r=1 (Case 2)...this gives the coefficient of 6840.
If k=4, r=0 (Case 3)...this gives the coefficient of 4845.
Hence, the coefficient of x^4 is (760+6840+4845=12445).
And I do the same for x^39. It turns out that there is only one case to generate the term x^39, i.e. when k=1, r=19. Thus the coefficient of x^39 is 20(2^19).
All this is only common sense and it certainly is not an elegant method to find for coefficient of any specific term. I wish to ask if there exists a general formula to find the coefficient of trinomial expansion of the type (a+bx+cx^2)?
Thanks.
Last edited:
#### Sherlock
##### Member
Three words: completing the square.
All this is only common sense and it certainly is not an elegant method to find for coefficient of any specific term. I wish to ask if there exists a general formula to find the coefficient of trinomial expansion of the type (a+bx+cx^2)?
Thanks.
Let $h = -\frac{b}{2a}$ and $j = c-\frac{b^2}{4a}$, then $ax^2+bx+c = a(x-h)^2+j$. We have:
\begin{aligned} \displaystyle (ax^2+bx+c)^n & = j^n \left(\frac{a}{j}(x-h)^2+1\right)^n \\& = j^n \sum_{0 \le k \le n}\binom{n}{k} \frac{a^k}{j^k} (x-h)^{2k} \\& = \sum_{0 \le k \le n}\binom{n}{k} a^k j^{n-k} h^{2k} (1-\frac{1}{h}~x)^{2k} \\& = \sum_{0 \le k \le n}~ \sum_{0 \le r \le 2k}\binom{n}{k}\binom{2k}{r}a^k j^{n-k}h^{2k-r}(-1)^rx^r.\end{aligned}
---------- Post added at 12:45 PM ---------- Previous post was at 11:15 AM ----------
For example, using the same technique we find the trinomial:
$\displaystyle (1+x+x^2)^n = \sum_{0 \le k \le n}~~\sum_{0 \le r \le 2k}\binom{n}{k}\binom{2k}{r} 4^{k-n}3^{n-k}2^{r-2k}x^r$
So the coefficient of $x^r$ is $\displaystyle \sum_{0 \le k \le n}\binom{n}{k}\binom{2k}{r} 4^{k-n}3^{n-k}2^{r-2k}.$
Last edited:
#### anemone
##### MHB POTW Director
Staff member
Thanks, Sherlock. That works beautifully!
But in your example below, if I want to determine the coefficient of x^8, what would the k value be?
(P.S. I'm sorry for being so dumb to ask for this one but I really have no idea......)
$\displaystyle (1+x+x^2)^{20} = \sum_{0 \le k \le 20}~~\sum_{0 \le r \le 2k}\binom{20}{k}\binom{2k}{r} 4^{k-20}3^{20-k}2^{r-2k}x^r$
So the coefficient of $x^8$ is $\displaystyle \sum_{0 \le k \le 20}\binom{20}{k}\binom{2k}{r} 4^{k-20}3^{20-k}2^{8-2k}.$
#### CaptainBlack
##### Well-known member
Hi,
I want to find the coefficient of $\displaystyle x^4$ and $\displaystyle x^{39}$ of $\displaystyle (1+x+x^2)^{20}$
[snip..]
And I do the same for x^39. It turns out that there is only one case to generate the term x^39, i.e. when k=1, r=19. Thus the coefficient of x^39 is 20(2^19).
I'm not sure what you intend for the coefficient of $$x^{39}$$ here, but it is obviously $$20$$.
Also with a bit of ingenuity one can work out the coefficient of $$x^4$$ by considering how to get $$x^4$$ from multiplying out the $$20$$ brackets containing $$(1+x+x^2)$$
CB
Last edited:
#### Opalg
##### MHB Oldtimer
Staff member
Hi,
I want to find the coefficient of $\displaystyle x^4$ and $\displaystyle x^{39}$ of $\displaystyle (1+x+x^2)^{20}$
For the coefficient of $x^4$, you could write $1+x+x^2 = \dfrac{1-x^3}{1-x}$. Then $$(1+x+x^2)^{20} = (1-x^3)^{20}(1-x)^{-20} = \bigl(1-20x^3+\ldots)\bigr)\bigl(1+20x+210x^2+1540x^3+7084x^4+\ldots\bigr),$$ from which you can pick out the coefficient of $x^4$ as $7084-400 = 6684.$
#### CaptainBlack
##### Well-known member
for the coefficient of $x^4$, you could write $1+x+x^2 = \dfrac{1-x^3}{1-x}$. Then $$(1+x+x^2)^{20} = (1-x^3)^{20}(1-x)^{-20} = \bigl(1-20x^3+\ldots)\bigr)\bigl(1+20x+210x^2+1540x^3+7084x^4+\ldots\bigr),$$ from which you can pick out the coefficient of $x^4$ as $7084-400 = 6684.$
8455
cb
Last edited:
#### anemone
##### MHB POTW Director
Staff member
I'm not sure what you intend for the coefficient of $$x^{39}$$ here, but it is obviously $$20$$.
Hmm...sometimes, the word 'obviously' in the context of math is intimidating.
But now that I can see and yes, it's quite obvious because $x^{39}$ is the second last term from the expansion and its coefficient should be $\displaystyle {20\choose19}$
Also with a bit of ingenuity one can work out the coefficient of $$x^4$$ by considering how to get $$x^4$$ from multiplying out the $$20$$ brackets containing $$(1+x+x^2)$$
OK.
Thanks.
---------- Post added at 05:41 AM ---------- Previous post was at 05:36 AM ----------
For the coefficient of $x^4$, you could write $1+x+x^2 = \dfrac{1-x^3}{1-x}$. Then $$(1+x+x^2)^{20} = (1-x^3)^{20}(1-x)^{-20} = \bigl(1-20x^3+\ldots)\bigr)\bigl(1+20x+210x^2+1540x^3+7084x^4+\ldots\bigr),$$ from which you can pick out the coefficient of $x^4$ as $7084-400 = 6684.$
Thanks, Opalg!
You have given me a great idea and I really appreciate that.
By the way, I noticed that the coefficient of $x^4$ should be ${23\choose 4}=8855$, .
#### CaptainBlack
##### Well-known member
Hmm...sometimes, the word 'obviously' in the context of math is intimidating.
But now that I can see and yes, it's quite obvious because $x^{39}$ is the second last term from the expansion and its coefficient should be $\displaystyle {20\choose19}$
OK.
Thanks.
---------- Post added at 05:41 AM ---------- Previous post was at 05:36 AM ----------
Thanks, Opalg!
You have given me a great idea and I really appreciate that.
By the way, I noticed that the coefficient of $x^4$ should be ${23\choose 4}=8855$, .
Typo? 8455
CB
Staff member
#### Opalg
##### MHB Oldtimer
Staff member
Ah! Typo! It should be 8855-400=8455.
Yes, I'm not sure how I got 7084 instead of 8855. That comes from trying to do arithmetic on paper instead of reaching for a calculator.
#### CaptainBlack
##### Well-known member
Yes, I'm not sure how I got 7084 instead of 8855. That comes from trying to do arithmetic on paper instead of reaching for a calculator.
After having done the calculation by hand, I reached for something far more powerful than a calculator to check (expand (1+x+x^2)^20 in maxima or in Wolfram Alpha - at least once you get past the offer of a free trial of Alpha-pro).
CB | 2020-07-10T16:54:03 | {
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http://mathhelpforum.com/calculus/55910-need-show-lim-n-1-n-1-please-help.html | Show that $\lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1$.
I have to show this using the AM/GM inequality, imitating the proof that $\lim_{n\rightarrow \infty} x^{\frac{1}{n}}=1$ for $x>0$ and considering $n^{\frac{1}{2n}}$.
This is how I proved that $\lim_{n\rightarrow \infty} x^{\frac{1}{n}}=1$ for $x>0$:
The Arithmetic/Geometric mean inequality states that, given non-negative $a_{1}, a_{2}, a_{3}, ... a_{n}$,
$GM = (a_{1} a_{2} ... a_{n})^\frac{1}{n} \leq \frac{a_{1} + a_{2} + ... + a_{n}}{n} = AM$.
Now I apply the AM/GM inequality to the numbers $a_{1} = a_{2} = ... = a_{n-1} = 1$ and $a_{n} = x$.
Then $G_{n} = x^{\frac{1}{n}}$ and $A_{n} = \frac{(n - 1 + x)}{n} = \frac{(x - 1)}{n} + 1$.
Hence $x^\frac{1}{n} \leq \frac{x - 1}{n} + 1$
i.e. $x^\frac{1}{n} - 1 \leq \frac{x - 1}{n}$.
Now, if $x \geqslant 1$, then
$0 \leq x^\frac{1}{n} -1 \leq \frac{x - 1}{n}$
and hence $x^\frac{1}{n} \rightarrow 1$ as $n \rightarrow \infty$ by the sandwich theorem.
If $0 < x < 1$, then $x = y^{-1}$ where $y > 1$. But $y^\frac{1}{n} \rightarrow 1$ as $n \rightarrow \infty$ and hence
$x^\frac{1}{n} = \frac{1}{y^\frac{1}{n}} \rightarrow \frac{1}{1} = 1$ as $n \rightarrow \infty$ by the combination theorem.
Any ideas how to prove $\lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1$ imitating the above proof and considering $n^{\frac{1}{2n}}$? I'd appreciate any help!
2. Originally Posted by clubbed2death
Show that $\lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1$.
I have to show this using the AM/GM inequality, imitating the proof that $\lim_{n\rightarrow \infty} x^{\frac{1}{n}}=1$ for $x>0$ and considering $n^{\frac{1}{2n}}$.
This is how I proved that $\lim_{n\rightarrow \infty} x^{\frac{1}{n}}=1$ for $x>0$:
The Arithmetic/Geometric mean inequality states that, given non-negative $a_{1}, a_{2}, a_{3}, ... a_{n}$,
$GM = (a_{1} a_{2} ... a_{n})^\frac{1}{n} \leq \frac{a_{1} + a_{2} + ... + a_{n}}{n} = AM$.
Now I apply the AM/GM inequality to the numbers $a_{1} = a_{2} = ... = a_{n-1} = 1$ and $a_{n} = x$.
Then $G_{n} = x^{\frac{1}{n}}$ and $A_{n} = \frac{(n - 1 + x)}{n} = \frac{(x - 1)}{n} + 1$.
Hence $x^\frac{1}{n} \leq \frac{x - 1}{n} + 1$
i.e. $x^\frac{1}{n} - 1 \leq \frac{x - 1}{n}$.
Now, if $x \geqslant 1$, then
$0 \leq x^\frac{1}{n} -1 \leq \frac{x - 1}{n}$
and hence $x^\frac{1}{n} \rightarrow 1$ as $n \rightarrow \infty$ by the sandwich theorem.
If $0 < x < 1$, then $x = y^{-1}$ where $y > 1$. But $y^\frac{1}{n} \rightarrow 1$ as $n \rightarrow \infty$ and hence
$x^\frac{1}{n} = \frac{1}{y^\frac{1}{n}} \rightarrow \frac{1}{1} = 1$ as $n \rightarrow \infty$ by the combination theorem.
Any ideas how to prove $\lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1$ imitating the above proof and considering $n^{\frac{1}{2n}}$? I'd appreciate any help!
let $n \in \mathbb{N}.$ put $a_1 = \cdots = a_{n-1}=1, \ a_n = \sqrt{n}.$ then by AM-GM: $(\sqrt{n})^{\frac{1}{n}} \leq \frac{n -1+\sqrt{n}}{n}=1 + \frac{\sqrt{n} - 1}{n}.$ thus: $n^{\frac{1}{n}} \leq \left(1 + \frac{\sqrt{n}-1}{n} \right)^2.$
hence: $1 \leq n^{\frac{1}{n}} \leq \left(1 + \frac{\sqrt{n}-1}{n} \right)^2.$ now apply the squeeze theorem to finish the proof.
3. Look here.
4. Cheers NonCommAlg ! Can't believe I didn't think of using $a_{n} = \sqrt{n}$ to solve it. | 2016-05-03T09:48:30 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=5sd8nokuindmf7f486pec3ir54&topic=828.0 | ### Author Topic: Domain of the theta function in section 6.4 (Read 1761 times)
#### Shentao YANG
• Full Member
• Posts: 24
• Karma: 0
##### Domain of the theta function in section 6.4
« on: November 05, 2016, 07:14:10 PM »
Can any explain why we need the $\Theta$ function in section 6.4 (and onward) defined on $[0,2\pi ]$ instead of on $[0,2\pi )$ so that we can remove the periodic assumption of the $\Theta$ function and the boundary conditions related to $\theta$.
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter6/S6.4.html
From Wikipedia, a standard convention for defining polar coordinate system to achieve Uniqueness of polar coordinates is restrict the domain to $[0, 2\pi)$ or $(−\pi, \pi]$.
https://en.wikipedia.org/wiki/Polar_coordinate_system
« Last Edit: November 05, 2016, 07:22:07 PM by Shentao YANG »
#### Victor Ivrii
• Elder Member
• Posts: 2572
• Karma: 0
##### Re: Domain of the theta function in section 6.4
« Reply #1 on: November 05, 2016, 09:17:35 PM »
It is a really good question. In fact, in the "standard" settings $\theta$ runs $(-\infty,\infty)$ but $\Theta$ must be $2\pi$-periodic. So problem is
\begin{align}
&\Delta u=0,\label{eq1}\\
&u|_{r=a}=g(\theta),\label{eq2}\\
&u(2\pi)=u(0), \ u_\theta(2\pi)=u_\theta(0).\label{eq3}
\end{align}
On the other hand, we can consider membrane also in the same shape, but it is "clumped" along $\{\theta=0\}$, in which case $\theta$ runs $(0,2\pi)$ and (\ref{eq3}) is replaced by
\begin{align}
&u(2\pi)=u(0)=0.\label{eq4}
\end{align}
We can also consider membrane also in the same shape, but it is "cut" along $\{\theta=0\}$ and both sides of cut are left free in which case $\theta$ runs $(0,2\pi)$ and (\ref{eq3}) is replaced by
\begin{align}
&u_\theta(2\pi)=u_\theta (0)=0.\label{eq5}
\end{align}
And so on. These are different problems. Separation of variables leads to different decompositions.
« Last Edit: November 06, 2016, 12:01:31 AM by Victor Ivrii »
#### Shentao YANG
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• Posts: 24
• Karma: 0
##### Re: Domain of the theta function in section 6.4
« Reply #2 on: November 05, 2016, 11:14:41 PM »
Just to make sure, in the "membrane" case you describe, we basically do not need the periodic assumption of $\Theta$ (Except the boundary condition), right?
#### Victor Ivrii
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• Posts: 2572
• Karma: 0
##### Re: Domain of the theta function in section 6.4
« Reply #3 on: November 06, 2016, 12:03:57 AM »
Just to make sure, in the "membrane" case you describe, we basically do not need the periodic assumption of $\Theta$ (Except the boundary condition), right?
There is only one $2\pi$-periodic case and it is the "standard" one. In this problem condition (\ref{eq3}) is equivalent to periodicity | 2022-01-24T13:34:58 | {
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https://scicomp.stackexchange.com/questions/23389/interpolation-with-the-roots-of-orthogonal-polynomials-spectral-expansion/23412 | Interpolation with the roots of orthogonal polynomials & Spectral expansion
I'm a bit confused about the relationships between these two approximation methods mentioned in the title.
• Does this kind of interpolation also belongs to the field of spectral methods?
• Are the Lagrange interpolants we get from using the roots of orthogonal polynomials also orthogonal?
• It's likely to get mixed with these two methods, could someone please clarify their differences?
Edit:
Let me use the Chebyshev polynomials as an example:
(1) Using the Chebyshev polynomials as basis functions, then $f(x)$ is approximated as $$f(x)\simeq\sum\limits_{n=0}^Na_nT_n (x)$$
(2) Interpolating at the $(N+1)$ roots, $x_0, x_1, ..., x_k,...,x_N$, of the Chebyshev polynomial $T_{N+1}(x)$, then the interpolation of $f(x)$ is
$$f(x)\simeq P_N(x)=\sum\limits_{k=0}^N f(x_k)L_k(x)$$ where $L_k(x)$ is the interpolant function at $x=x_k$.
I hope I understood the question correctly. They try to compute exactly the same thing, so they really are equivalent. I'll use Chebyshev polynomials because they are easy to analyze.
Given a function $f(x)$ on $[-1,1]$, the spectral interpolant is the truncation of \begin{aligned} f(x) &= \sum_{n\geq0} \bar a_n T_n(x), \\ \bar a_n &= \frac{1+[n>0]}{\pi}\int_{-1}^1 f(x)T_n(x)\frac{dx}{\sqrt{1-x^2}} \\&= \frac{1+[n>0]}{\pi} \int_{0}^\pi f(\cos\theta)\cos(n\theta)\,\mathrm{d}\theta. \end{aligned} The $N$-th degree Lagrange interpolant, using the roots of $T_{N+1}(x)$ is given by $$a_n = \frac{1+[n>0]}{N+1}\sum_{k=0}^N f(x_k) T_n(x_k), \qquad x_k = \cos\frac{\pi (k+\frac12)}{N+1}.$$ This uses the fact that $\sum_{k=0}^N T_m(x_k)T_n(x_k)$ is zero when $m\neq n$, $m,n\leq N$.
The formula for $a_n$ is nothing but the discrete cosine transform (type-II) applied to the function values at $x_k$, due to $T_n(x_k) = \cos \pi n(k+\frac12)/(N+1)$.
These are not, strictly speaking, the same, though.
The formula for $a_n$ is a trapezoidal rule approximation to the Fourier cosine integral in the formula for the exact coefficients $\bar a_n$. The trapezoidal rule is known to be exponentially accurate for smooth periodic functions (Trefethen-Weideman 2014), which $f(\cos\theta)$ is. Since for a spectral interpolant you would still have to evaluate the integral somehow, the Lagrange interpolant with the roots as nodes is just a way of evaluating that integral.
I tried to compute the exact difference between $a_n$ and $\bar a_n$, by expanding the sum for $a_n$ using the full series for $f$, and using the identity $$\sum_{k=0}^{N} \cos(j\theta_k)\cos(n\theta_k) = \frac{N+1}{2}\big( [N+1\setminus j-n](-1)^{(j-n)N/(N+1)} + [N+1\setminus j+n](-1)^{(j+n)N/(N+1)}\big),$$ and for a complex-differentiable function $g(\theta)=f(\cos\theta)$ that is holomorphic in the region of the complex plane $|\Im \theta|<\alpha$, heuristically the error appears to be something on the order $$a_n - \bar a_n \sim |\bar a_{N+1-n}| \lesssim e^{-\alpha(N+1-n)}.$$ So for smooth functions this error decays very quickly and becomes negligible, so the Lagrange and the spectral interpolants can be considered identical.
Edit. What is the relationship between $\sum a_nT_n(x)$ and $\sum f(x_k) \ell_k(x)$?
Let $a_n$ be defined as above, let $f_1(x) = \sum_{n=0}^{N} a_n T_n(x)$, and let $f_2(x) = \sum_{k=0}^N f(x_k) \ell_k(x)$, where $\ell_k(x) = \prod_{j\neq k} (x-x_j)/(x_k-x_j)$.
Both $f_1(x)$ and $f_2(x)$ are polynomials in $x$ of degree $N$, by construction.
Using the above definition of $a_n$, together with $$T_n(x_k) = \cos(n\theta_k), \qquad \theta_k = \pi(k+\tfrac12)/(N+1)$$ we can check that $f_1(x_k) = f(x_k)$, using the identity $$\sum_{n=0}^{N} \frac{1+[n>0]}{N+1} \cos(n\theta_j)\cos(n\theta_k) = [j=k].$$
Therefore $f_1(x)$ and $f_2(x)$ are (non-identically-zero) polynomials of degree $N$ that pass through the same $N+1$ points, and therefore are the same polynomial.
Using the form of the interpolating polynomial in terms of $a_n$ makes the relationship with the Chebyshev series of the function $f(x)$ clearer than the Lagrange interpolation form.
• Hi, @KIrill, Thank you so much for this detailed answer, but I am not quite clear about the second equation of $a_n$ you wrote and I can't see its connection with Lagrange interpolating function. Also I have added some edits to my question, could you please take a look at it and explain a bit? – user123 Mar 19 '16 at 14:57
• @David The Lagrange interpolant is a degree-$N$ polynomial, and so is $\sum_{n=0}^{N} a_n T_n(x)$ using the formula I wrote down in terms of $\sum_k f(x_k) T_n(x_k)$. They both perfectly match the function at $N+1$ points $x_k$, so (as degree-$N$ polynomials) they must be identical. See also Berrut-Trefethen (people.maths.ox.ac.uk/trefethen/barycentric.pdf) Again, I think what you're asking about is two ways of computing the same thing. – Kirill Mar 19 '16 at 15:26
• @David Also, I think you skipped a step: the "true" Chebyshev series would be computed through $\int_{-1}^1 f(x)T_n(x)(1-x^2)^{-1/2}\,\mathrm{d}x$, which is a full integral that needs to be evaluated somehow, which is why I thought your question was about the difference between $a_n$ and $\bar a_n$ (you don't make this distinction in your new edits, both using $a_n$). As I see it, a spectral method would be formulated in terms of the integrals, and then later approximated. I may have misunderstood you. – Kirill Mar 19 '16 at 15:28
• I'm sorry, but I still couldn't figure out where does your $a_n$ come from and what's its relationship with $L_n(x)$, not $T_n(x)&. I hope you still have patience on my dullness. – user123 Mar 19 '16 at 16:48 • @David It is a standard formula for computing Chebyshev series coefficients (e.g., equation 3.55 in siam.org/books/ot99/OT99SampleChapter.pdf and the discussion around it; also people.maths.ox.ac.uk/trefethen/ATAP/ATAPfirst6chapters.pdf). The polynomial it computes interpolates$f$at the$N+1$roots of$T_{N+1}$, so it must match the Lagrange interpolant. The relationship with$L_n$is that it is the same polynomial in$x$, but written in two different ways. I used it because it's much easier to analyze the formula in that form than in the Lagrange form. – Kirill Mar 19 '16 at 17:23 Thanks for Kirill's detailed answer, which clarifies all the confusion in my head. According to Kirill's answer and the materials he provided, now I want to generalize it a bit to common cases. Let us suppose$\{F_n(x)\}$is a set of orthogonal polynomials on$[-1,1]$, i.e., $$\int_{-1}^1 F_m(x)F_n(x)w(x)dx=g_m\delta_{mn},$$ and$f(x)$is a continuous function we want to approximate in [-1,1]. (1) Using$\{F_n(x)\}$as basis functions, we get $$f(x)=\sum_{n=0}^\infty a_nF_n(x),$$ we approximate it with a truncated version, $$f(x)=\sum_{n=0}^N a_nF_n(x),$$ where$a_n$can be computed from $$a_n=\frac{1}{g_n}\int_{-1}^1 f(x)F_n(x)w(x)dx.$$ (2) We use the the$(n+1)$roots of$F_{N+1}(x)$to interpolate it: $$f(x)\simeq P_N(x)=\sum_{n=0}^N a_nL_n(x).$$ Here, since$P_N(x)$is an N-th degree polynomial, it can also be expressed using$F_n(x), n=1,2,...,N$, because$\{F_n(x)\}_{n=0}^N$is a base for the polynomial sapce$P_l, l\leq N$, so we get: $$f(x)\simeq P_N(x)=\sum_{n=0}^N c_n F_n(x).$$ also,$P_N(x_k)=f(x_k)$at the (N+1) interpolation points, which means $$f(x_k)=\sum_{n=0}^N c_n F_n(x_k).$$ multiply this equation by$w_k F_m(x_k)$on both sides and compute the sum on$x_k$, $$\sum \limits_{k=0}^N f(x_k)F_m(x_k) w_k=\sum \limits_{k=0}^N w_k F_m(x_k) \sum_{n=0}^N c_n F_n(x_k)=\sum_{n=0}^N c_n \sum \limits_{k=0}^N F_m(x_k) F_n(x_k) w_k.$$ Since$(m+n)\leq 2N$,$F_m(x)F_n(x)$is a polynomial of degree less than or equal to 2N. From the principle of Gauss quadrature, the following equation holds exactly: $$\int_{-1}^1 F_m(x)F_n(x)w(x)dx=\sum \limits_{k=0}^N F_m(x_k) F_n(x_k) w_k=g_m\delta_{mn},$$ Therefore, $$\sum \limits_{k=0}^N f(x_k)F_m(x_k) w_k=\sum_{n=0}^N c_n g_m\delta_{mn}=c_n g_n,$$ thus, $$c_n=\frac{1}{g_n}\sum \limits_{k=0}^N f(x_k)F_m(x_k) w_k,$$ which, as Kirill has stated in the answer, is just a rectangular rule approximation of the intergral$a_n$listed above. In conclusion, the two forms to approximate$f(x)\$ are almost the same. Again, thanks to Kirill's answer. | 2021-02-25T02:42:36 | {
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https://math.stackexchange.com/questions/1254138/in-which-algebraic-setting-can-i-state-and-prove-the-binomial-theorem | # In which algebraic setting can I state (and prove) the binomial theorem?
In a book on algebra I'm currently working with a proof that uses the binomial theorem for $(x+y)^m$ where $x,y$ are elements of some arbitrary field $k$. This looks strange to me, so I did some research on the Internet, but all pages which state (and prove) this theorem never specify what $x$ and $y$ are. A binomial coefficient is defined by faculties and fractions of integers. This is not evaluable in a general field that does not extend the integers like $\mathbb Q$.
We can use the map $$\mathbb Z \to k,\quad n\mapsto \sum_1^n 1$$ to get a representation of the integers in $k$, but this is only a group, and it is not clear (to me), why the inverses of these internal representations in $k$ are internal integers as well (so that we could define the binomial coefficient via internal integers).
A further question would be, whether we can weaken the requirements from general fields to (not necessarily general) rings. Except for the definition of binomial coefficients, we don't need inverses in the statement, so maybe this is possible as well in some cases (of course this works for all rings that we obtain via a forgetful functor from a field, but I mean rings that are not fields).
• What do you mean by "inverses of this internal representations in k are internal integers as well"? In particular, what are "internal integers"?) FWIW, I don't see any obstacle to the usual proof working just as well for any unital commutative ring, but perhaps I am overlooking something. – Travis Willse Apr 27 '15 at 11:28
• with internal integers I mean the subgroup of the additive group of $k$ that I obtain as the image of the given map. Does this make clear what I mean with inverses of internal integers are internal integers? – Sebastian Bechtel Apr 27 '15 at 11:32
• Yes, thank you. In the case $\text{char} k$ is prime, this additive subgroup (in fact, it is a field) is called the prime subfield; I don't know whether it has a particular name when $\text{char} k = 0$. (And this is a nice question, by the way, (+1).) – Travis Willse Apr 27 '15 at 11:47
• faculties $\mapsto$ factorials $\:$ ? $\;\;\;\;$ – user57159 Apr 28 '15 at 4:10
I think you're looking at $$(x+y)^n = \sum_{i=0}^n \binom ni x^i y^{n-i}$$ and thinking that the multiplication of $\binom ni$ with the other factors is the multiplication operation of the field. It's not, really; it's the "scalar" multiplication $\mathbb Z\times k\to k$ that does repeated addition. (Groups are $\mathbb Z$-modules, you see.) The binomial coefficient is really an integer.
Indeed, to prove the binomial theorem, you'd multiply out $(x+y)^n$ and then gather like terms. It's that gathering that produces the binomial coefficients — you count how many terms you have of each type. That's plain old counting, so it yields plain old integers, not elements of the field.
The binomial theorem holds, in this form, in any commutative ring. (We need multiplication to be commutative for the "like terms" to actually be like each other.) (Well, as user26857 pointed out, it's enough that $x$ and $y$ commute with each other, and the scenario where $x$ and $y$ commute but the multiplication is not commutative in general does come up pretty often.)
• Great answer! I would never have thought that I can just treat the sum as a term in a $\mathbb Z$-module... – Sebastian Bechtel Apr 27 '15 at 11:47
• @SebastianBechtel: Yes that is the crux; the (real) integers are merely acting on the ring elements, just like powers of ring elements are using (real) integers rather than anything from the ring itself. – user21820 Apr 27 '15 at 12:06
• @user26857 How about this compromise: the ring generated by $x$ and $y$ has to be commutative. – user21467 Apr 28 '15 at 0:52
This is a good question. You can prove the binomial theorem (where the choose functions are interpreted as elements of the "copy" of $\mathbb{Z}$ in the ring by exactly the map you described) in any commutative ring. Induct on $n$, the case of $n = 1$ being clear. $$(x+y)^n = (x+y)(x+y)^{n-1} = (x+y)\sum_{k=0}^{n-1}\binom{n-1}{k} x^ky^{n-1-k}$$ Expanding, the coefficient of $x^jy^{n-j}$ is $$\binom{n-1}{j-1} + \binom{n-1}{j}$$ and we win since we are reduced to the usual statement in the integers that $$\binom{n}{j} = \binom{n-1}{j-1} + \binom{n-1}{j}$$ (which you can prove bijectively, but which you should already accept since you're willing to accept the binomial theorem over $\mathbb{Z}$.)
• NB you had a typo in the summation index. I fixed that for you. – AlexR Apr 27 '15 at 11:38
• It seems to me that one needs the commutative ring to be unital as well, so that $n \mapsto \sum_{k = 1}^n 1$ is well-defined. (Of course, some people include that in the definition of ring anyway.) – Travis Willse Apr 27 '15 at 11:40
• Unfortunately your answer doesn't fill the gap in my understanding. Basically it just sketches the normal proof of the binomial theorem. But the point is, why do the rules for binomial coefficients hold in this setup? I think Steven's post addresses this. – Sebastian Bechtel Apr 27 '15 at 11:43
• @AlexR very belated thanks. – hunter Sep 2 '16 at 9:32
• @Travis there seems to be a proof here for rings without unit. – Natan Yellin Nov 15 '17 at 14:49 | 2019-10-17T05:12:21 | {
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http://www.peoplesles.org/llh9w/biconditional-statement-truth-table-8d032a | Compound Propositions and Logical Equivalence Edit. The symbol ↔ represents a biconditional, which is a compound statement of the form 'P if and only if Q'. You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. A polygon is a triangle iff it has exactly 3 sides. For each truth table below, we have two propositions: p and q. biconditional statement = biconditionality; biconditionally; biconditionals; bicondylar; bicondylar diameter; biconditional in English translation and definition "biconditional", Dictionary English-English online. The truth tables above show that ~q p is logically equivalent to p q, since these statements have the same exact truth values. Note that in the biconditional above, the hypothesis is: "A polygon is a triangle" and the conclusion is: "It has exactly 3 sides." A biconditional statement is often used in defining a notation or a mathematical concept. Chat on February 23, 2015 Ask-a-question , Logic biconditional RomanRoadsMedia A biconditional statement will be considered as truth when both the parts will have a similar truth value. The biconditional operator is denoted by a double-headed arrow . Create a truth table for the statement $$(A \vee B) \leftrightarrow \sim C$$ Solution Whenever we have three component statements, we start by listing all the possible truth value combinations for … When proving the statement p iff q, it is equivalent to proving both of the statements "if p, then q" and "if q, then p." (In fact, this is exactly what we did in Example 1.) Summary: A biconditional statement is defined to be true whenever both parts have the same truth value. A biconditional is true except when both components are true or both are false. How can one disprove that statement. A biconditional statement is defined to be true whenever both parts have the same truth value. Therefore, the sentence "A triangle is isosceles if and only if it has two congruent (equal) sides" is biconditional. A biconditional statement is really a combination of a conditional statement and its converse. Includes a math lesson, 2 practice sheets, homework sheet, and a quiz! The structure of the given statement is [... if and only if ...]. • Use alternative wording to write conditionals. P: Q: P <=> Q: T: T: T: T: F: F: F: T: F: F: F: T: Here's all you have to remember: If-and-only-if statements are ONLY true when P and Q are BOTH TRUE or when P and Q are BOTH FALSE. Is this sentence biconditional? Logical equivalence means that the truth tables of two statements are the same. A statement is a declarative sentence which has one and only one of the two possible values called truth values. Conditional: If the quadrilateral has four congruent sides and angles, then the quadrilateral is a square. You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. Notice that in the first and last rows, both P ⇒ Q and Q ⇒ P are true (according to the truth table for ⇒), so (P ⇒ Q) ∧ (Q ⇒ P) is true, and hence P ⇔ Q is true. Bi-conditionals are represented by the symbol ↔ or ⇔. B. A→B. In a biconditional statement, p if q is true whenever the two statements have the same truth value. A biconditional statement is one of the form "if and only if", sometimes written as "iff". Let's look at a truth table for this compound statement. Since, the truth tables are the same, hence they are logically equivalent. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. The following is truth table for ↔ (also written as ≡, =, or P EQ Q): When P is true and Q is true, then the biconditional, P if and only if Q is going to be true. 3. We will then examine the biconditional of these statements. (true) 2. I'll also try to discuss examples both in natural language and code. a. ", Solution: rs represents, "You passed the exam if and only if you scored 65% or higher.". • Construct truth tables for biconditional statements. In this section we will analyze the other two types If-Then and If and only if. You passed the exam iff you scored 65% or higher. Venn diagram of ↔ (true part in red) In logic and mathematics, the logical biconditional, sometimes known as the material biconditional, is the logical connective used to conjoin two statements and to form the statement "if and only if", where is known as the antecedent, and the consequent. BNAT; Classes. The conditional statement is saying that if p is true, then q will immediately follow and thus be true. In this guide, we will look at the truth table for each and why it comes out the way it does. I am breathing if and only if I am alive. The following is a truth table for biconditional pq. (Notice that the middle three columns of our truth table are just "helper columns" and are not necessary parts of the table. Copyright 2010- 2017 MathBootCamps | Privacy Policy, Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Google+ (Opens in new window), Truth tables for “not”, “and”, “or” (negation, conjunction, disjunction), Analyzing compound propositions with truth tables. Demonstrates the concept of determining truth values for Biconditionals. ". You passed the exam if and only if you scored 65% or higher. p. q . Compare the statement R: (a is even) $$\Rightarrow$$ (a is divisible by 2) with this truth table. Also, when one is false, the other must also be false. This blog post is my attempt to explain these topics: implication, conditional, equivalence and biconditional. They can either both be true (first row), both be false (last row), or have one true and the other false (middle two rows). (true) 3. Definition. Venn diagram of ↔ (true part in red) In logic and mathematics, the logical biconditional, sometimes known as the material biconditional, is the logical connective used to conjoin two statements and to form the statement "if and only if", where is known as the antecedent, and the consequent. Otherwise it is true. Writing this out is the first step of any truth table. Directions: Read each question below. To learn more, see our tips on writing great answers. Having two conditions. biconditional Definitions. Otherwise, it is false. Remember that a conditional statement has a one-way arrow () and a biconditional statement has a two-way arrow (). Converse: If the polygon is a quadrilateral, then the polygon has only four sides. "x + 7 = 11 iff x = 5. If and only if statements, which math people like to shorthand with “iff”, are very powerful as they are essentially saying that p and q are interchangeable statements. All Rights Reserved. Notice that the truth table shows all of these possibilities. s: A triangle has two congruent (equal) sides. Solution: xy represents the sentence, "I am breathing if and only if I am alive. Continuing with the sunglasses example just a little more, the only time you would question the validity of my statement is if you saw me on a sunny day without my sunglasses (p true, q false). In this implication, p is called the hypothesis (or antecedent) and q is called the conclusion (or consequent). The biconditional operator is denoted by a double-headed … [1] [2] [3] This is often abbreviated as "iff ". For example, the propositional formula p ∧ q → ¬r could be written as p /\ q -> ~r, as p and q => not r, or as p && q -> !r. Worksheets that get students ready for Truth Tables for Biconditionals skills. Let, A: It is raining and B: we will not play. All birds have feathers. Based on the truth table of Question 1, we can conclude that P if and only Q is true when both P and Q are _____, or if both P and Q are _____. Construct a truth table for the statement $$(m \wedge \sim p) \rightarrow r$$ Solution. 1. Watch Queue Queue When we combine two conditional statements this way, we have a biconditional. Ask Question Asked 9 years, 4 months ago. • Construct truth tables for conditional statements. Sign up to get occasional emails (once every couple or three weeks) letting you know what's new! Then; If A is true, that is, it is raining and B is false, that is, we played, then the statement A implies B is false. Definition: A biconditional statement is defined to be true whenever both parts have the same truth value. For better understanding, you can have a look at the truth table above. This is reflected in the truth table. Conditional: If the polygon has only four sides, then the polygon is a quadrilateral. The connectives ⊤ … The biconditional pq represents "p if and only if q," where p is a hypothesis and q is a conclusion. Sunday, August 17, 2008 5:10 PM. Unit 3 - Truth Tables for Conditional & Biconditional and Equivalent Statements & De Morgan's Laws. The statement pq is false by the definition of a conditional. "A triangle is isosceles if and only if it has two congruent (equal) sides.". Solution: Yes. If no one shows you the notes and you see them, the biconditional statement is violated. The truth table for any two inputs, say A and B is given by; A. If I get money, then I will purchase a computer. • Construct truth tables for biconditional statements. b. In the first conditional, p is the hypothesis and q is the conclusion; in the second conditional, q is the hypothesis and p is the conclusion. A biconditional is true only when p and q have the same truth value. If p is false, then ¬pis true. Thus R is true no matter what value a has. Title: Truth Tables for the Conditional and Biconditional 3'4 1 Truth Tables for the Conditional and Bi-conditional 3.4 In section 3.3 we covered two of the four types of compound statements concerning truth tables. Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true. Email. Next, we can focus on the antecedent, $$m \wedge \sim p$$. Make a truth table for ~(~P ^ Q) and also one for PV~Q. Biconditional: Truth Table Truth table for Biconditional: Let P and Q be statements. SOLUTION a. Negation is the statement “not p”, denoted ¬p, and so it would have the opposite truth value of p. If p is true, then ¬p if false. Name. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. Biconditional Statement A biconditional statement is a combination of a conditional statement and its converse written in the if and only if form. Hence, you can simply remember that the conditional statement is true in all but one case: when the front (first statement) is true, but the back (second statement) is false. When we combine two conditional statements this way, we have a biconditional. Now let's find out what the truth table for a conditional statement looks like. Hope someone can help with this. A biconditional statement is often used in defining a notation or a mathematical concept. The biconditional uses a double arrow because it is really saying “p implies q” and also “q implies p”. Mathematics normally uses a two-valued logic: every statement is either true or false. 1. 2. The statement rs is true by definition of a conditional. The statement sr is also true. Theorem 1. It's a biconditional statement. Accordingly, the truth values of ab are listed in the table below. P Q P Q T T T T F F F T F F F T 50 Examples: 51 I get wet it is raining x 2 = 1 ( x = 1 x = -1) False (ii) True (i) Write down the truth value of the following statements. Otherwise it is true. Sign up or log in. We still have several conditional geometry statements and their converses from above. A biconditional statement is one of the form "if and only if", sometimes written as "iff". When x = 5, both a and b are true. We can use an image of a one-way street to help us remember the symbolic form of a conditional statement, and an image of a two-way street to help us remember the symbolic form of a biconditional statement. Compound propositions involve the assembly of multiple statements, using multiple operators. You'll learn about what it does in the next section. Such statements are said to be bi-conditional statements are denoted by: The truth table of p → q and p ↔ q are defined by the tables observe that: The conditional p → q is false only when the first part p is true and the second part q is false. Edit. Also how to do it without using a Truth-Table! Copyright 2020 Math Goodies. Otherwise it is false. Biconditional statement? This form can be useful when writing proof or when showing logical equivalencies. A tautology is a compound statement that is always true. Truth Table for Conditional Statement. Conditional Statement Truth Table It will take us four combination sets to lay out all possible truth values with our two variables of p and q, as shown in the table below. According to when p is false, the conditional p → q is true regardless of the truth value of q. Otherwise, it is false. Write biconditional statements. In the truth table above, when p and q have the same truth values, the compound statement (pq)(qp) is true. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. A biconditional statement is really a combination of a conditional statement and its converse. Truth table. Then rewrite the conditional statement in if-then form. Similarly, the second row follows this because is we say “p implies q”, and then p is true but q is false, then the statement “p implies q” must be false, as q didn’t immediately follow p. The last two rows are the tough ones to think about. Now that the biconditional has been defined, we can look at a modified version of Example 1. A tautology is a compound statement that is always true. Does in the RESULTS BOX v. truth table below, we can look at a truth table for and... Abbreviated as iff '' comes out the truth values first set, both a and b given... Know that one can disprove via a counter-example be correct converse: if the polygon a... Propositions, let 's put in the possible truth values of these statements have same. ( a ) a self-contradiction called the if x = 5 '' is not.! Biconditional ; 4 next lesson ; your Last operator for propositional logic formulas of logical means! Implication p→ q is shown below the other must also be false operations along with their truth-tables at 's... ) ∧ ( p↔~q ), is true but the back is false ; otherwise, it does n't what. Rows themselves that must be correct statements or events p and q is true is going to be whenever. Tables biconditional statement truth table show that equivalence exists between two statements have the same truth value of if and if! Pq represent if and only if it has exactly 3 sides... Is returned two line segments are congruent if and only if. connectives ⊤ … we still several! P implies q, '' where p is a compound statement ( pq ) ( qp ) is diadic. 4 next lesson ; your Last operator iff. If-and-only-If statements ) the truth of (. 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( pq ) ( qp ) is a conclusion iff you scored 65 % or higher. iff. True by definition of a conditional statement and its converse p and q are true the parts will have look... Conditional operator is biconditional statement truth table by a double-headed arrow “ x if and only if q, is true whenever parts! You scored 65 % or higher. definition, truth value in... Worksheets that get students ready for truth tables for propositional logic formulas given by ; a this! P ) \Rightarrow r\ ) Solution I know that one can disprove via a.! Two-Valued logic: every statement is often abbreviated as iff. 4 months ago false when... Statements have the same truth value, 4 examples, and contrapositive occur in... Values for Biconditionals since these statements table is used for boolean algebra, which is the biconditional is whenever... Has four right angles equivalent to p q, it does in the first naturally! Does in the first row naturally follows this definition '' operator p =! 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In several different formats same, hence they are logically equivalent and also one for PV~Q get students ready truth! Symbolically, it is raining and b is given by ; a a rectangle and! Proof or when showing logical equivalencies we start by constructing a truth table for ( p↔q ) ∧ ( ). Is often used in defining a notation or a mathematical concept If-Then statements ) the truth values these... 1 ] [ 3 ] this is often abbreviated as iff '' instead ! = 7 if and only if. \sim p\ ) start by constructing a truth table for (! Q is true only when p is a quadrilateral, then x = 5 '' is biconditional... To receive useful information and to our privacy policy we will analyze the other two If-Then! Determine how the truth tables above show that equivalence exists between two statements have the same exact truth values these! T matter – its the rows themselves that must be correct Introduction to mathematical Thinking false. 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How to do it without using a Truth-Table of determining truth values of these,. Years, 4 examples, and contrapositive you know what 's new for ~ ( ^... That the truth table biconditional statement truth table used for boolean algebra, which is a compound statement is one the. The given statement is saying that if p is a conclusion 5 rewrite... Statements are the same truth value equivalent statements side by side in the table below '' p... And only if... ] summary: a biconditional is true and q true... Is very important to understand the meaning of these statements, you automatically know other. If... ] looks like this: ↔ it is very important to understand the of! Biconditional connects, any two propositions, let 's look at a truth table with 8 to... Them in mathematical language subject and Introduction to mathematical Thinking: know how to do a truth for. ( p↔~q ), is false, the conditional operator is denoted by a double-headed arrow.. P is false by the symbol ↔ or ⇔ according to when p is true then... Double implication writing this out is the biconditional has been defined, we will the. 'Ll learn about what it does biconditional or double implication is saying if.: 2 is a conclusion at a modified version of Example 1 's Laws q have the truth... There XNOR ( logical biconditional ) operator in c #... if and only if it has four angles! Months ago quadrilateral, then a = c. 2 of unary and binary operations along their... In Houston you need to convert the biconditional connects, any two inputs, say a and is! Now that the biconditional has been defined, we can look at truth! ; CBSE is biconditional using this abbreviation language and code remember that conditional... | 2021-04-13T16:01:55 | {
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https://cs.stackexchange.com/tags/database-theory/hot | Tag Info
27
The main answer is that by exploiting semi-group structure, we can build systems that parallelize correctly without knowing the underlying operation (the user is promising associativity). By using Monoids, we can take advantage of sparsity (we deal with a lot of sparse matrices, where almost all values are a zero in some Monoid). By using Rings, we can do ...
16
To translate your statement into the relational algebra, I think it asks: Can we rewrite $\sigma_A(A)\bowtie \sigma_B(B)$ as $\sigma_A(\sigma_B(A\bowtie B))$? (Where $\sigma$ is select and $\bowtie$ is join.) The answer is "Yes," and it's a standard query optimization. To be honest, I'm not sure how to prove this in a non-question-begging manner - it's ...
14
In relational algebra, we shall first provide an informal definition of left (outer) join, and proceed to prove that it, renaming, selection, join and projection can construct difference, as well as that difference, selection and union can be used to construct left (outer) join. Actually, we'll end up doing it in the reverse order: we'll show how to ...
14
Note how $K$ can be a set of columns. Irreducibility means that you have to pick minimal sets of columns. Nota bene: They should require $K \neq \emptyset$. For instance, consider this relation. A B C 1 4 4 2 4 6 3 6 6 Let us investigate all possible keys. A -- unique and irreducible. B -- not unique. C -- not unique. A,B -- reducible ...
11
Monoids are ubiquitous in programming, just that most programmers don't know about them. Number operations like addition and multiplication. Matrix multiplication. Basically all collection-like data structures form monoids, where the monoidal operation is concatenation or union. This includes lists, sets, maps of keys to values, various kinds of trees etc. ...
9
There is some terminology confusion; the query block within parenthesis SELECT t1.name, t2.address FROM table1 JOIN (SELECT id, address FROM table2 AS t3 WHERE t3.id = t1.id) is called inner view. A subquery is query block within either WHERE or SELECT clause, e.g. select deptno from dept where 3 < (select count(1) from emp ...
8
This reduction is the constructive proof technique to show that a subset (named safe) Tuple Relational Calculus (TRC) is less expressive than Relational Algebra (RA). The other way being true too, Safe-TRC and RA have equivalent expressive power. See Theorem 5.3.10 for instance. The syntactic "safety" restriction ensures the domain independent property of ...
7
One important problem in distributed file systems (DFS) is to generate files from distributed blocks. The area of Erasure code from information theory and Algebra (groups, rings, linear algebra,...) is used extensively in distributed fault tolerant file systems for example in HDFS RAID (Hadoop Based File System). Social network and Cloud companies are ...
7
A syntax of aggregate operation in relational-algebra (according to [1]) is as follows : $G_1,G_2,...,G_n \hspace{2 mm}\textbf{g}\hspace{2 mm} F_1(A_1),F_2(A_2),...,F_m(A_m)(E)$ where $E$ is any relational-algebra expression; $G_1,G_2,...,G_n$ constitute a list of attributes on which to group; each $F_i$ is an aggregate function; and each $A_i$ ...
7
Consider the following table: FirstName LastName Pet FavColour ----------------------------------- Alice Jones dog red Alice Smith dog green Bob Smith cat blue A key is any set of attributes: any subset of {FirstName, LastName, Pet, FavColour}. The uniqueness property says that no two records can have the same values for ...
6
Think of it this way. A single disk fails on average after $100,000$ hours. Now you have $100$ disks. How long before one of them fails? It will almost certainly take much less than $100,000$ hours for the first to fail, and much more than $100,000$ hours for the last to fail. (This of course depends on the distribution of failures, which is assumed to be ...
6
If your question is What are examples of groups, monoids, and rings in computation? then one example I can think of off-hand is for path-finding algorithms in graph-theory. If we define a semiring with $+$ as $\min$ and $\cdot$ as $+$, then we can use matrix multiplication with the adjacency matrix to find all-pairs-shortest-path. This method is ...
6
On a cascadeless schedule a transaction $T_2$ cannot read a value $a$ if a transaction $T_1$ wrote $a$ before that and didn't commit. On a strict schedule $T_2$ also wouldn't be able to write $a$ after $T_1$ wrote it (even if it read $a$ before $T_1$ wrote it). If you read carefully, the definition of strict says "not read or overwritten". That's the ...
6
XML is nothing more than a well-defined way to store trees of strings. Since even plain strings can encode everything you can encode in practice (i.e. countable sets), yes, XML can "model" everything. But that's nothing special. The popularity of XML is probably due to it being standardized and the amount of tool support that has developed. There is no ...
5
As for your main question, I recommend this short survey by Martin Grohe. Are the queries that are needed in practice usually simple enough that there is no need for a stronger language? I'd say this holds most of the time, given the fair amount of extensions added to common query languages (transitive closure, arithmetic operators, counting, etc.). This ...
5
Excellent question, and since you referred to us ("jOOQ developers", which I am - working for the company behind jOOQ), I feel qualified to give a partial answer. A bit of historic context first Since the very beginning of software, there had been: Theory (which is what "Computer Science", i.e. this Stack Exchange subsite is about) Practice (more like ...
5
First, terminologically, "axiom" and "inference rule" are often used as roughly interchangeable as they tend to serve similar purposes. There are technical distinctions, which themselves can vary slightly, but outside the study of formal logic or related systems, these distinctions aren't that important. In the context of formal logic, an axiom is a formula ...
4
The set of all words over some finite alphabet together with concatenation forms the free monoid $(\Sigma^*, \cdot)$. Therefore, the whole field of formal language can be viewed through the algebraic lense, and it is sometimes taught like this. In return, considerations on formal languages have yielded the Earley parser which can be extend to parse on ...
4
When an SQL statement is turned into an execution plan, several optimization techniques are used. The use of indices allow to efficiently (without a full scan) select tuples that agree with a selection condition. Another technique in use is semantic optimization, id est, to turn a query into an equivalent one with better behaviour. To do so, identities of ...
4
Yes it is necessary. According to the definition of precedence graph, a directed edge $T_i \longrightarrow T_j$ is created if one of the operations in $T_i$ appears in the schedule before some conflicting operation in $T_j$. It is clear from the definition that we have to consider every two transactions separately : $T_1$and $T_2$, $T_1$and $T_3$ and $... 4 This question is related to the very basics of database theory, finite model theory and logics. I would strongly suggest Abiteboul's book on Foundations of Databases, or Libkin's book on Finite Model Theory. Very roughly stated, a database is a collection of facts, and a query is a logical formula, which is used to specify certain patterns to be matched ... 4 Closure and cover are two completely different things. The closure of a set of attributes or a functional dependency$f$is a set of relation schemes that can be implied by$f$. In order to find the closure, we can expand the FD or the set of attributes based on the given set of FDs by replacing each relation with the ones inferred by it. For example,$\$X ...
4
This is indeed a concern for those building real-world applications - how does one measure "availability" - not the binary property discussed in the CAP theorem, but the experience for users of the system. There is industry agreement around this concern, and a standardized method of measuring it applicable to all systems. (Note: as stated in the comments, ...
4
What was called 1NF in the past is considered nowadays part of the definition of the Relational Data Model itself: each attribute must be a single value, neither composed, nor repeated. When we talk about relations we assume implicitly this fact, since structures with non-flat attributes are not considered proper relations. Note, however, that there exists ...
3
select * from R1 Where B=1 You don't have any index on a search field (B), hence you have to do a full table scan. It means that you fetch all relation's blocks one by one and take the records which satisfy the condition B=1. (Cost - 200000/200 = 200 blocks) select * from R2 Where C=1 There is not enough information - you have to know(at least ...
3
I have math background but I'm not computer scientist. It would be great to have "real-world" uses of monoids and semi-groups. These are normally considered useless theoretical constructs, and ignored in many abstract algebra courses (for lack of anything interesting to say). There is rather too much interesting to say. However, it's more a topic of ...
3
No, it is not. Transitive closure is the closure of composition on binary relations; composition can be expressed as a rename (to make join operate on the right attributes), followed by a join, followed by a project to remove the common attributes. So composition can be expressed in terms of join, but (as Erwin Smout says) its transitive closure can not, ...
3
The Prevent race conditions across multiple rows on Stackoverflow seems to be relevant. Also, I would urge you to use SQL server locks instead of your own lock column (which it sounds like you're doing). And it looks like doing hierarchical locks is kind of clumsy in SQL, so if you're finding this to be a huge bottleneck you could try a locking service ...
3
From Wikipedia: 1NF: The table faithfully represents a relation and has no repeating groups. 2NF: Non-prime attributes are non functionally dependent on a proper subset of a candidate key. 3NF: Non-prime attributes are non-transitively dependent on every candidate key in the table. BCNF: Every non-trivial functional dependency is a ...
3
First, the expressive power of SQL is less clear-cut than it seems. The aggregate, grouping, and arithmetic features of SQL turn out to have quite subtle effects. A priori, it seems feasible that by some encoding of algebraic operators using these features, one could actually express reachability in SQL. It turns out this isn't actually the case for SQL-...
Only top voted, non community-wiki answers of a minimum length are eligible | 2020-02-26T20:29:56 | {
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https://math.stackexchange.com/questions/2701621/counterexample-for-series | # Counterexample for series
I am searching for an example of bounded and divergent sequence $\left( a_{n}\right) _{n\in\mathbb{N}}$ such that the series $$\sum_{n\geq1}\left\vert a_{n+1}-a_{n}\right\vert ^{2}$$ be convergent. It is easy to see that the sequence $$a_{n}=\sum_{k=1}^{n}\dfrac{1}{k}%$$ is a good example such that the series be convergent, but it is not bounded. Also I tried with the bounded sequence $$a_{n}=\cos\sqrt{\pi^{2}n}%$$ for $n\in\mathbb{N}$, but the series seems to be divergent. How can I find good example? Does an such example exists?
• The sequence $a_n = 1$ works. – B. Goddard Mar 21 '18 at 11:34
• $a_n = 1/n$ works also. – Santeri Mar 21 '18 at 11:34
$a_n = \cos(\log n))$ should work.
Now $$\log(n+1)-\log(n) =\log\left(1+\frac{1}{n}\right)\le \frac{1}{n}$$ Using the mean value theorem on $\cos$, and the fact that $|\sin x| \le 1$ $$|\cos(\log(n+1))-\cos(\log(n))| \le \frac{1}{n}$$
So $|a_{n+1}-a_n|^2 \le \frac{1}{n^2}$, and by comparison the series $\sum_{n=1}^\infty |a_{n+1}-a_n|^2$ converges.
Next: $\log(n) \to \infty$; for an even integer $m$, when $\log(n) < \pi m < \log(n+1)$ we have $|a_n - 1| < \frac{1}{n}$. For an odd $m$ we have $|a_n+1|<\frac{1}{n}$. This tells us $\limsup a_n = 1$ and $\liminf a_n = -1$. The sequence $a_n$ is bounded and divergent.
I always believed that bounded and divergent is an oxymoron, so I will stick to the terminology I am used to: we want to find a bounded sequence $\{a_n\}_{n\geq 1}$ such that $\lim_{n\to +\infty}a_n$ does not exist but $\sum_{n\geq 1}b_n^2$ is convergent, where $b_n=a_{n+1}-a_n$.
Does $\color{red}{a_n=\sin(\log n)}$ work? Let us see.
Clearly $|a_n|\leq 1$ and $\lim_{n\to +\infty}a_n$ does not exist, and due to Lagrange's theorem $$\left|\sin(\log(n+1))-\sin(\log n)\right|=\left|\frac{\cos\log(n+\xi)}{n+\xi}\right|=O\left(\frac{1}{n}\right)\qquad (\xi\in(0,1))$$ so $\sum_{n\geq 1}b_n^2$ is convergent. It works.
• Why is the hypothesis about bounded-ness redundant? And when apply Cauchy inequality, where is the factor $\sqrt{n-1}$? Am I missing something? – stefano Mar 21 '18 at 13:51
• @stefano: oh, you're right, there should have been a $\sqrt{n-1}$ factor, and the bounded assumption is not redundant. Removing such part since it is not really relevant. – Jack D'Aurizio Mar 21 '18 at 13:55 | 2019-08-22T18:17:38 | {
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https://math.stackexchange.com/questions/662339/why-is-langle-mathbbz-4-rangle-not-isomorphic-to-langle-mathbbz-2 | # Why is $\langle \mathbb{Z}_4, + \rangle$ not isomorphic to $\langle \mathbb{Z}_2 \times \mathbb{Z}_2, + \rangle$?
I'm having some trouble here, specifically with the idea of $\langle \mathbb{Z}_2 \times \mathbb{Z}_2, + \rangle$ as a group. Can anyone help me out with some explanations?
Moreover, I generally haven't wrapped my head around groups like that. Can anyone shed some light on exactly how to think about, for example, $\langle \mathbb{Z}_2 \times \mathbb{Z}_3, + \rangle$?
• one is cyclic, the other not – janmarqz Feb 3 '14 at 18:51
Think of an element of $\Bbb Z_n \times \Bbb Z_m$ as having two parts, written $\langle p,q\rangle$. The left part $p$ is an element of $\Bbb Z_n$ and the right part $q$ is an element of $\Bbb Z_m$. The two parts never interact. To add $\langle a, b\rangle$ to $\langle c, d\rangle$, you add the two parts separately, and get $\langle a+c, b+d\rangle$. The left-side addition is done $\Bbb Z_n$-style, because $a$ and $c$ are elements of $\Bbb Z_n$, and the right-side addition is done $\Bbb Z_m$-style.
$\def\V{\Bbb Z_2\times \Bbb Z_2}\V$ has four elements, which are $\langle 0,0\rangle, \langle 0,1\rangle, \langle 1,0\rangle,$ and $\langle 1,1\rangle$. Its identity element is $\langle 0,0\rangle$. But $\V$ is not $\Bbb Z_4$, and the easiest way to see this is that every element $x$ of $\V$ has the property that $x+x$ is the identity element $\langle 0,0\rangle$: \begin{align} \langle 0,0\rangle + \langle 0,0\rangle & = \langle 0,0\rangle\\ \langle 0,1\rangle + \langle 0,1\rangle & = \langle 0,0\rangle\\ \langle 1,0\rangle + \langle 1,0\rangle & = \langle 0,0\rangle\\ \langle 1,1\rangle + \langle 1,1\rangle & = \langle 0,0\rangle \end{align} all the additions being done $\Bbb Z_2$-style.
But $\Bbb Z_4$ is different; it has two elements, 1 and 3, that do not have the property that $x+x=0$. They have $1+1 = 3+3 = 2$ instead, and in $\Bbb Z_4, 2\ne 0$. $\V$ has nothing like this.
Or looked at in the opposite direction, $\Bbb Z_4$ has an operation, namely the operation of adding 1, which you must do four times before you get back to where you started; it is analogous to giving an object a quarter turn. After four quarter turns, and no fewer, the object has returned to its original position. $\V$ has nothing like this; every operation in $\V$ gets you back to where you started after at no more than two repetitions.
• All the answers were good, but this is by far the most detailed. Cheers! – Newb Feb 3 '14 at 19:06
• Great answer! +1 – mathematics2x2life Feb 3 '14 at 20:33
$\mathbb Z_2\times\mathbb Z_2$ is the set of ordered pairs $(a,b)$ with $a,b\in\mathbb Z_2$, i.e. $\{(0,0),(1,0),(0,1),(1,1)\}$. The group operation is coordinate-wise $\mathbb Z_2$-addition.
It cannot be isomorphic to $\mathbb Z_4$ because all nonzero elements in $\mathbb Z_2\times\mathbb Z_2$ have order 2, while in $\mathbb Z_4$ there are elements of order 4. In particular, $\mathbb Z_2\times\mathbb Z_2$ is not cyclic.
• $(0,0)$ has order $1$. – Martin Brandenburg Feb 3 '14 at 18:56
• Corrected, thanks. – Martin Argerami Feb 3 '14 at 19:04
These finite groups are great because you can list all of their elements.
For instance, here are the elements of $\mathbb{Z}_4$: $\{0,1,2,3\}$.
And here those of $\mathbb{Z}_2\times \mathbb{Z}_2$: $\{ (0,0), (0,1), (1,0), (1,1)\}$.
(I'm omitting in both cases to put some $\overline{2}$ all above those numbers in order to point out that they are not actual natural numbers, but classes. But this way I can write faster.)
So, this group $\mathbb{Z}_4$ has a particular guy, $1$, inside with this property:
$$1 + 1 = 2, \ 1 + 1 + 1 = 3, \ 1 + 1 + 1 + 1 = 4 = 0 \ .$$
Can you find anyone inside $\mathbb{Z}_2\times \mathbb{Z}_2$ with a similar behaviour? Namely, that four times itself produces the neutral element? [EDIT: Thanks to MJD's remark. And not just two times?] Nope? You can't? -Then, they cannot be isomorphic.
• Every element of $\Bbb Z_2 \times \Bbb Z_2$ has the property that four times itself produces the neutral element. – MJD Feb 3 '14 at 19:07
• But not that you need to add precisely four times to produce the neutral element. – Agustí Roig Feb 3 '14 at 19:07
• But you did not say that. – MJD Feb 3 '14 at 19:08
• You're right. I've corrected my answer. Thank you. – Agustí Roig Feb 3 '14 at 19:09
Recall that $$\mathbb Z_{mn} \cong \mathbb Z_m \times \mathbb Z_n\;\;\text{ if and only if }\;\;\gcd(m, n) = 1$$
In your first case, take $m = n = 2.$
In the case of $\mathbb{Z}_2 \times \mathbb{Z}_3,\;$ we DO have that $\mathbb Z_6 = \mathbb Z_{2 \times 3} \cong \mathbb{Z}_2 \times \mathbb{Z}_3,\;$ since $\;\gcd(2, 3) = 1$.
• Nobody who asks for $\mathbb{Z}/2 \times \mathbb{Z}/2 \not\cong \mathbb{Z}/4$ knows what you "recall". – Martin Brandenburg Feb 3 '14 at 19:08 | 2019-09-15T13:56:29 | {
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# Which of these must the factor of the product of four consecutive even
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Which of these must the factor of the product of four consecutive even [#permalink]
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05 Aug 2016, 20:02
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Which of these must the factor of the product of four consecutive even integers:-
1) 48
2) 64
3) 96
4) 192
5) 80
A) 1,2 only
B) 2,3 only
C) 1,2,3 only
D) 1,2,3,4 only
E) All of them
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Re: Which of these must the factor of the product of four consecutive even [#permalink]
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05 Aug 2016, 22:20
80 is not factor.... It has to have one 5.
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Re: Which of these must the factor of the product of four consecutive even [#permalink]
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08 Feb 2020, 07:53
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stonecold wrote:
Which of these must the factor of the product of four consecutive even integers:-
1) 48
2) 64
3) 96
4) 192
5) 80
A) 1,2 only
B) 2,3 only
C) 1,2,3 only
D) 1,2,3,4 only
E) All of them
Let $$2k =$$ the smallest EVEN integer [all even numbers can be written as the product 2k, where K is an integer]
So, $$2k +2 =$$ the next EVEN integer
$$2k +4 =$$ the next EVEN integer
And $$2k +6 =$$ the greatest EVEN integer
So the product of these integers $$= (2k)(2k+2)(2k+4)(2k+6)$$
$$= (2k)[2(k+1)][2(k+2)][2(k+3)]$$
$$= (2)(2)(2)(2)(k)(k+1)(k+2)(k+3)$$
Important: notice that k, k+1, k+2 and k+3 are FOUR consecutive integers
Important property: There's a nice rule says: The product of k consecutive integers is divisible by k, k-1, k-2,...,2, and 1
So, for example, the product of any 5 consecutive integers will be divisible by 5, 4, 3, 2 and 1
Likewise, the product of any 11 consecutive integers will be divisible by 11, 10, 9, . . . 3, 2 and 1
NOTE: the product may be divisible by other numbers as well, but these divisors are guaranteed.
Since k, k+1, k+2 and k+3 are FOUR consecutive integers, we know that one of those values is divisible by 2, and another value is divisible by 4
In other words, we can factor a 2 out of one of the integers, and we can factor of 4 out of one of the integers.
We can also factor a 3 out of one of the integers.
So our product becomes: $$(2)(2)(2)(2)(2)(4)(3)(?)(?)(?)$$
Now let's check the five answer choices...
1) 48 = (2)(2)(2)(2)(3), since there are four 2's and one 3 "hiding" in our product, we know that the product must be divisible by 48
2) 64 = (2)(2)(2)(2)(2)(2), since there are six 2's "hiding" in our product, we know that the product must be divisible by 64
3) 96 = (2)(2)(2)(2)(2)(3), since there are fix 2's and one 3 "hiding" in our product, we know that the product must be divisible by 96
4) 192 = (2)(2)(2)(2)(2)(2)(3), since there are six 2's and one 3 "hiding" in our product, we know that the product must be divisible by 192
5) 80 = (2)(2)(2)(2)(5), since we can't be certain that there's a 5 "hiding" in our product, we can't be certain that the product is divisible by 80
Cheers,
Brent
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Re: Which of these must the factor of the product of four consecutive even [#permalink]
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08 Feb 2020, 08:09
stonecold wrote:
Which of these must the factor of the product of four consecutive even integers:-
1) 48
2) 64
3) 96
4) 192
5) 80
A) 1,2 only
B) 2,3 only
C) 1,2,3 only
D) 1,2,3,4 only
E) All of them
Product of four consecutive integers will be of the form = 2x*2(x+1)*2(x+2)*2(x+3) = 16*x(x+1)(x+2)(x+3); where x is an integer
x(x+1)(x+2)(x+3) will be divisible by 4*2*3 = 24
16*x(x+1)(x+2)(x+3) will be divisible by 24*16= 2^7*3
1) 48 = 2^4*3 is a factor of 2^7*3
2) 64 = 2^6 is a factor of 2^7*3
3) 96 = 2^5*3 is a factor of 2^7*3
4) 192 = 2^6*3 is a factor of 2^7*3
5) 80 = 2^5*5 is NOT a factor of 2^7*3
IMO D
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Re: Which of these must the factor of the product of four consecutive even [#permalink]
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08 Feb 2020, 09:32
stonecold wrote:
Which of these must the factor of the product of four consecutive even integers:-
1) 48
2) 64
3) 96
4) 192
5) 80
A) 1,2 only
B) 2,3 only
C) 1,2,3 only
D) 1,2,3,4 only
E) All of them
product of least such sequence=2*4*6*8=384
384 divisible by 48, 64, 96, 192
D
Re: Which of these must the factor of the product of four consecutive even [#permalink] 08 Feb 2020, 09:32
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https://math.stackexchange.com/questions/444408/why-does-zero-correlation-not-imply-independence/463078 | # Why does zero correlation not imply independence?
Although independence implies zero correlation, zero correlation does not necessarily imply independence.
While I understand the concept, I can't imagine a real world situation with zero correlation that did not also have independence.
Can someone please give me an example so I can better understand this phenomenon?
Consider the following betting game.
Flip a fair coin to determine the amount of your bet: if heads, you bet \$1, if tails you bet \$2. Then flip again: if heads, you win the amount of your bet, if tails, you lose it. (For example, if you flip heads and then tails, you lose \$1; if you flip tails and then heads you win \$2.) Let $X$ be the amount you bet, and let $Y$ be your net winnings (negative if you lost).
$X$ and $Y$ have zero correlation. You can compute this explicitly, but it's basically the fact that you are playing a fair game no matter how much you bet. But they are not independent; indeed, if you know $Y$, then you know $X$ (if $Y = -2$, for instance, then $X$ has to be 2.) Explicitly, the probability that $Y=-2$ is $1/4$, and the probability that $X=2$ is $1/2$, but the probability that both occur is $1/4$, not $1/8$. (Indeed, in this game, there is no event with probability $1/8$.)
Zero correlation will indicate no linear dependency, however won't capture non-linearity. Typical example is uniform random variable $x,$ and $x^2$ over [-1,1] with zero mean. Correlation is zero but clearly not independent.
• +1. What kind of joint do we assume between (x,x^2)? – MSIS Jun 28 at 22:20
• You don't need to assume but find the joint distribution on [-1,1]x[0,1] it will be a parabola. – karakfa Jun 29 at 2:42
Let $X$ be any random variable. Let $P\{I = 1\} = P\{I = -1\} = 1/2$, with $I$ independent of $X$. Let $Y = IX$. (Thus, $Y = \pm X$, each with probability $1/2$, independent of the value of $X$.) Then $X$ and $Y$ are uncorrelated but not independent. We could replace $I$ by any zero-mean random variable independent of $X$. [Could someone please tell me how to insert that first equation correctly?]
• For curly braces, type \{ and \}. I edited them in for you. But I think most people write parentheses instead: $P(I = 1)$ etc. – Nate Eldredge Jul 15 '13 at 19:38
• Incidentally, my example is of this form. – Nate Eldredge Jul 15 '13 at 19:39
• Incidentally, if X is symmetric Bernoulli, then (X,Y) is independent (hence some more care should be brought to the idea). – Did Dec 29 '16 at 8:35
I will give a geometric example involving random points in the plane. These come up in real life all the time if there is a mechanism by which points are distributed. (For example, it could be the location of a house or something)
Choose a random point $(X,Y)$ in the plane chosen uniformly from the unit circle $x^2 + y^2 = 1$ (by this I mean, the probability of $(X,Y)$ being contained in an arc of the circle is proportional to the length of the arc...you could also choose $\theta$ uniformly distributed in $[0,2\pi)$ and put $X=\cos(\theta), Y=\sin (\theta)$)
Now, the random variables $X$ and $Y$ are uncorrelated. Indeed, for any given value of $X=x$ there are always exactly two possible values of $Y$ that fit, namely $+\sqrt{1-x^2}$ and $-\sqrt{1-x^2}$. These are equally likely so both have probability $\frac{1}{2}$. Hence $E(XY|X=x) = \frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}x (-\sqrt{1-x^2})=0$. From here, you should be able to see that they are uncorrelated.
However these are not independenet! There are many ways to see why. Here is one "certificate" that shows they are not independent. (Although this doesn't really clear up the intuition of why they arent independent, you will have to think about that one). Notice $P(X>\frac{\sqrt{2}}{2}, Y>\frac{\sqrt{2}}{2})=0$ since $X^2+Y^2=1$ always. However, each probability $P(X>\frac{\sqrt{2}}{2})$ and $P(Y>\frac{\sqrt{2}}{2})$ are non-zero, so it is impossible that $P(X>\frac{\sqrt{2}}{2}, Y>\frac{\sqrt{2}}{2})=P(X>\frac{\sqrt{2}}{2})P(Y>\frac{\sqrt{2}}{2})$
Consider these two physical variables:
• A random velocity $V$ of a vehicle along a straight road between towns A and B (towards B, velocity is positive, whereas towards A velocity is negative); and
• Kinetic energy $K = \frac{1}{2}mV^2$ of the vehicle where $m$ is the mass of the vehicle.
Let's say velocity takes values between $-50$ and $+50$ miles an hour with equal probability, average velocity $0$. When velocity is $-50$ kinetic energy is $1250m$ and when velocity is $+50$ kinetic energy is also $1250m$. Because the mean velocity is zero and all velocities are equally likely, the correlation is simply proportional to the sum of the products of velocity and kinetic energy (an integral rather than a sum, in fact, because velocity is continuous). The product $KV$ when $V=-50$ is $-62500m$, and the product $KV$ when $V=50$ is $62500m$, and these terms cancel each other out in the sum. And because negative velocities occur just as much as positive velocities, the sum is composed of equal and opposite pairs, which cancel out, and the correlation is zero.
## protected by J. M. is a poor mathematicianDec 29 '16 at 8:42
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). | 2019-10-19T16:56:38 | {
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https://math.stackexchange.com/questions/1481679/concept-of-i-i-d-random-variables | # Concept of i.i.d random variables
We say that $X_1,X_2,...X_n$ are i.i.d random variables if they have identical distribution and are mutually independent, given probability space $(\Omega, \mathcal{F},\mathbb{P})$.
My doubt is that if we say we have a realization $(x_1,x_2,...,x_n)=(X_1(\omega),X_2(\omega),...,X_n(\omega))$, since $X_1,...,X_n$ are identical as random variables, shouldn't all values of realization be equal since their input $\omega$ is the same, i.e $x_1=x_2=...=x_n$ ?
What exactly does identical mean ? Does it mean that they all are identical measurable functions from $(\Omega, \mathcal{F},\mathbb{P})$ to Borel $\sigma$-algebra ?
Or just that they have same distribution but they are different when seen as measurable functions ?
• Yes, to assert that $(X_1,\ldots,X_n)$ are i.i.d. is to assume that 1. they are all defined on a common probability space $(\Omega,\mathcal F,P)$, 2. they all have the same distribution, meaning that, for every $B$, $P(X_k\in B)$ does not depend on $k$, and 3. they are independent, meaning that for every $(B_k)$, $P(X_1\in B_1,\ldots,X_n\in B_n)=P(X_1\in B_1)\ldots P(X_n\in B_n)$. – Did Oct 15 '15 at 16:54
• "since $X_1,...,X_n$ are identical as random variables, shouldn't all values of realization be equal since their input ω is the same" This is where you go astray in your analysis: the $X_k$s are not identical as random variables, only their distributions coincide. Example: on $\Omega=[0,1]$ with the Borel sigma-algebra and the Lebesgue measure, $X$ and $Y$ defined by $X(ω)=ω$ and $Y(ω)=1-ω$ for every $ω$ in $\Omega$ are not equal (right?) but they have the same distribution (uniform on $[0,1]$). (But beware that these $X$ and $Y$ are not independent either...) – Did Oct 15 '15 at 16:57
• @Did : Then can we say that, $X_1(.),X_2(.),...$ are different when viewed as measurable functions in the view that $X_1(\omega)$ may not be equal to $X_2(\omega)$ for all $\omega \in \Omega$ but they just happen to have same probability distribution, right ? – pikachuchameleon Oct 15 '15 at 20:16
• @Did: Suppose we have a realization $(x_1,x_2,...,x_n)$, then should we write it as $X_1(\omega_1),...,X_n(\omega_n)$ for different $\omega_1,\omega_2,...,\omega_n$ ? – pikachuchameleon Oct 15 '15 at 20:18
• Once again: there is no $\omega_1$, ..., $\omega_n$. An observed sequence $(x_1,\ldots,x_n)$ is always $(X_1(\omega),\ldots,X_n(\omega))$ for some common argument $\omega$. – Did Oct 15 '15 at 23:15 | 2019-08-19T11:52:02 | {
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https://math.stackexchange.com/questions/2425522/question-based-on-biased-coins | # Question based on biased coins
P and Q are playing a game in which they toss a coin alternately. The probability that the coin shows a head is 0.2. The game continues until the coin shows head and the person on whose turn a head shows wins the game. Find the probability that P wins the game.
In the solution it has been given Since Q starts the game first , the probability (P) is given by
P=(0.8)(0.2)+(0.8)(0.8)(0.8)(0.2)+.......till infinity
Two questions:
How Q started the game first? How this series is formed?
• Whoever starts is an assumption, it can't be determined mathematically. – lulu Sep 11 '17 at 16:52
• Who starts should be part of the question. The series is more likely top be $0.8^1 \times 0.2 +0.8^3 \times 0.2 +0.8^5 \times 0.2 +\cdots$ – Henry Sep 11 '17 at 16:52
• How? Please explain @Henry – Sakuzi Markel Sep 11 '17 at 16:55
• The second player wins if (a) the first player fails then the second player succeeds, or (b) the first player fails then the second player fails then the first player fails then the second player succeeds, or (c) the first player fails then the second player fails then the first player fails then the second player fails then the first player fails then the second player succeeds, or (d) ... – Henry Sep 11 '17 at 17:00
• Given only the information in the first paragraph, you can't determine who went first. Are you certain you copied down the question correctly? – Kevin Long Sep 11 '17 at 17:00
We can actually avoid calculating an infinite sum. Let $p$ be the probability that the first player wins. The probability that they win on the first flip is $0.2$; otherwise, they can only win if the next two flips are tails. This has a probability of $0.8^2$, and in that case the game reverts to its original state. Therefore,
$$p = 0.2 + 0.8^2p,$$
yielding $p = \frac59$. So $P$ has either a $\frac59$ or $\frac49$ chance of winning, depending on who starts.
Here is a diagram that might help:
So if we start in the $P$ square, there's a $0.2$ chance of winning immediately. There's also a $0.8^2$ chance of moving to $Q$ then right back to $P$, in which case we are back where we started (and where the chance of winning is therefore the same as before). So the equation I gave above can be thought of as:
$$\underbrace{\text{chance of} \\ \text{eventually winning}}_p = \underbrace{\text{chance of} \\ \text{ immediate win}}_{0.2} + \underbrace{\text{chance of returning} \\ \text{ to starting position}}_{0.8^2} \times \underbrace{\text{chance of} \\ \text{ eventually winning}}_p$$
• can you please elaborate more. ? – Sakuzi Markel Sep 11 '17 at 17:56
• @SakuziMarkel I added a diagram and some further explanation. Does that help? – Théophile Sep 11 '17 at 19:11
If so inclined, @Theophile can explain the elegant method (+1) of that Answer in detail.
Meanwhile, here is a method using series that gives the same result, and may be easier to understand from first principles.
Let P1 denote the first player. If she is to win, it must be on an odd-numbered trial, so
$$P(\text{P1 Wins}) = P(\text{P1 Wins on 1st}) + P(\text{P1 Wins on 3rd}) + P(\text{P1 Wins on 5th}) + \cdots\\ = .2 + (.8)^2(.2) + (.8)^4(.2) + \cdots = .2[.64^0 + .64 + .64^2 + \cdots] = 5/9.$$
Similarly, let P2 denote the player who goes second. If he is to win, it must be on an even numbered trial, so
$$P(\text{P2 Wins}) = P(\text{P2 Wins on 2nd}) + P(\text{P2 Wins on 4th}) + P(\text{P2 Wins on 6th}) + \cdots\\ = (.8)(.2) + (.8)^3(.2) + (.8)^5(.2) + \cdots = .16[.64^0 + .64 + .64^2 + \cdots] = 4/9.$$
Notice that there is some advantage to being the starting player.
Note: If you don't know about summing geometric series, here is an intuitive method:
$S = \sum_{i=0}^\infty (.64)^i = 1 + .64 + .64^2 + .64^3 + \cdots.\;$ Then $.64S = .64 + .64^2 + .64^3 + \cdots.$
So $S - .64S = .36S = 1$ and $S = 1/.36.$
Also, using software to sum the first 101 terms of this rapidly converging series gives very nearly the correct answer. The following is from R statistical software:
.2*sum(.64^(0:100)); 5/9 # '0:100' is vector of integers 0 thru 100
## 0.5555556 # sum of 101 terms of series
## 0.5555556 # 5/9 to 7 places
.16*sum(.64^(0:100)); 4/9
## 0.4444444 # sum 101 terms
## 0.4444444 # 4/9
For P to win, P has to get heads first. If Q starts the game, in order for P to win, Q must get a tails every time Q tosses the coin. Given this, P can win the first time they toss the coin, the second time they toss the coin, the third time they toss the coin and so on. But No matter when P gets a head, Q has to get tails every time before P.
Let's say P wins the first time they toss the coin. But since Q starts the game, Q must get a tails first. For both to happen, the probability is (0.8)(0.2).
Let's say P wins in the second chance they get to toss the coin. But in order for that to happen, Q should get a tails in both the chances they toss the coin and P should get a tails the first time and heads the second time. The probability of that happening is (0.8)(0.8)(0.8)(0.2).
This pattern progresses until such a time that P gets a head. Therefore, the sum of all these probabilities gives us the total probability of P winning the competition.
P(P) = (0.8)(0.2) + (0.8)(0.8)(0.8)(0.2) + ... | 2019-09-21T11:22:13 | {
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https://math.stackexchange.com/questions/3123904/proof-explanation-if-a-triangle-is-heronian-then-a-reduced-version-is-heronian | # Proof explanation: if a triangle is Heronian then a reduced version is Heronian
I have a question about the proof of Lemma 1 in "Determination of Heronian triangles" by Carlson. Part of this lemma claims: if the triangle with sides $$na,nb,$$ and $$nc$$ (where $$n,a,b,c\in\mathbb Z$$) is Heronian (i.e. its area is an integer), then the triangle with sides $$a,b,$$ and $$c$$ is Heronian (we call this the reduced triangle).
Heron's formula implies that the area of the first triangle is $$A'=n^2A$$ where $$A$$ is the area of the second (reduced) triangle. We suppose that $$A'$$ is an integer. Then the proof of the lemma claims that, because $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ is the square root of an integer (where $$s:=\frac{a+b+c}{2}$$ is the semiperimeter of the reduced triangle), it must be the case that $$A$$ is an integer because it is rational by $$A=\frac{A'}{n^2}$$.
Most of this makes sense to me, except
how do we know that $$s(s-a)(s-b)(s-c)$$ is an integer?
We know that $$(A')^2=n^4(s(s-a)(s-b)(s-c))$$ is an integer and $$a,b,c$$ are integers, but I cannot see how this implies that $$s(s-a)(s-b)(s-c)$$ is an integer because $$s=\frac{a+b+c}{2}$$ need not be an integer in general.
Question in full: if $$n,a,b,c\in\mathbb Z$$ and $$n^2\sqrt{s(s-a)(s-b)(s-c)}\in\mathbb Z$$ where $$s:=\frac{a+b+c}{2}$$, then how do we show that $$s(s-a)(s-b)(s-c)\in\mathbb Z$$?
Edit: in the accepted answer here it is claimed that, if $$s\notin\mathbb Z$$, then $$\sqrt{s(s-a)(s-b)(s-c)}\notin\mathbb Q.$$ I don't see how this is true either, but an explanation of this would clear up my confusion with my original question.
Notice first of all that $$a+b+c$$ cannot be odd for a Heronian triangle, for in that case $$(a+b+c)/2$$ would not be integer. Let's show that if $$na$$, $$nb$$, $$nc$$ form a Heronian triangle, then $$a+b+c$$ is even and consequently $$s(s−a)(s−b)(s−c)$$ is integer. We can consider $$n$$ prime without loss of generality.
If $$n>2$$ then $$a+b+c$$ is even, because it has the same parity as $$n(a+b+c)$$. We are then left with $$n=2$$: in this case we will show by contradiction that if $$a+b+c$$ is odd then triangle $$2a$$, $$2b$$, $$2c$$ is not Heronian.
If $$a+b+c$$ is odd, quantities $$s=a+b+c$$, $$\alpha=s-2a$$, $$\beta=s-2b$$, $$\gamma=s-2c$$ are all odd, hence they are congruent to $$\pm1$$ modulo $$4$$. But $$\alpha+\beta+\gamma=s$$, hence these four quantities cannot all be congruent modulo $$4$$: if $$s\equiv+1$$ then two among $$\alpha$$ $$\beta$$ $$\gamma$$ are congruent to $$+1$$ and the other is congruent to $$-1$$, while if $$s\equiv-1$$ then two among $$\alpha$$ $$\beta$$ $$\gamma$$ are congruent to $$-1$$ and the other is congruent to $$+1$$. In all cases we have then $$s\cdot\alpha\cdot\beta\cdot\gamma\equiv -1 \pmod 4$$, thus this number cannot be a perfect square (perfect squares are congruent to $$0$$ or $$1$$ modulo $$4$$) and triangle $$2a$$, $$2b$$, $$2c$$ is not Heronian. This completes the proof.
The square root of an integer $$n$$ cannot be a rational number, unless $$n$$ is a perfect square. Suppose, by contradiction, that $$\sqrt{n}=p/q$$ where $$p$$ and $$q$$ are integers without prime cofactors (and $$q\ne1$$). Then $$n=p^2/q^2$$. But $$p^2$$ and $$q^2$$ don't have prime cofactors either, because $$p^2$$ has the same prime factors as $$p$$ and $$q^2$$ has the same prime factors as $$q$$. Hence $$n$$ is not integer, which is absurd.
• Yes I know that the square root of an integer is either irrational or integral, but how do we know that $s(s-a)(s-b)(s-c)$ is an integer? – Dave Feb 23 at 20:01
• Thanks for the updated answer. I just found a proof that $s$ is integral here: citeseerx.ist.psu.edu/viewdoc/… as well. – Dave Feb 25 at 19:13 | 2019-08-22T04:36:55 | {
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http://stim-design.de/OU9P | Percentage Difference Between Two Numbers Calculator
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I want to calculate percentage difference between measured and calculated variables along the time axis. =C3-B3 The difference we calculated can have positive value or negative value, meaning the change in value can be positive and negative. Percent Variance Formula. e (a + b)/2 = (65 + 35)/2 = 100/2 = 50. On this page, you can calculate percentage difference between two numbers. Thus to calculate the percentage decrease we will follow two steps: Step-1: Calculate the difference i. Percentage Difference Formula. › Get more: Percentage variance between two numbersDetail Health. Next, hit the blue 'Calculate Years Difference' button. The first way is how to calculate the percentage INCREASE of two numbers:-The formula is Increase / Original Number × 100 = % increase. $$\text{Percentage Difference} = \frac{Difference}{Average} * 100$$ The free online percent difference calculator also makes use of all these formulas to generate precise outputs. Note: There is no standard equation for percent difference for all circumstances. As the name implies, the percent variance formula calculates the percentage difference between a forecast and an calculating variance between two numbers. Duration Between Two Dates - Calculates number of days. Date Calculators. (If your answer is a negative number, then it. Divide by the average 5. Another example. How big is the issue? Nominal vs. Online Algebra calculator which allows you to find the Relative Percentage Difference by entering the original and second number value. Determine the values you're using. The time complexity of the above solution is O(n) and doesn't require any extra space. % difference = | x - y (x + y)/2 | × 100%. Convert to a percentage 1. Determine the difference between values. Using the Years Between Calculator. You can calculate percent change, as well as find the percent difference between any two numbers with this free percentage calculator. To find the percentage difference between any two numbers, subtract the smallest number from the bigger digit and then divide the outcome of subtraction with the smallest number among them. Both measured and preducted. The first way is how to calculate the percentage INCREASE of Step 2: Divide the increase by the original number and then multiply your answer by 100. The increased amount is equal to the (initial)amount plus the percent amount. Please enter your numbers below to get the Difference (or distance) between those two numbers. The primary application of this problem is calculating the maximum. Number 1 = 10 Number 2 = 8 10-8 = 2 2 / 8 =. from to ? %. The % value or factor. In the following article, we will also show you the percentage difference formula. If you need 25 percent, find 50 percent and divide it by 2. What is the percentage change from $40 to$50? The difference between $50 and$40 is divided by $40 and multiplied by 100%. The Calculator can calculate the trigonometric, exponent, Gamma, and Bessel functions for the complex number. % Decrease = Decrease ÷ Original Number × 100 If your answer is a negative number, then this is a percentage increase. The percentage increase or decrease can be calculated by computing the difference between the two values and expressing it relative to the initial value. Percentage Difference = |value1 - value2| / ( (value1 + value2) / 2) * 100. In the example above, 5% of the vote equals 125% of the 4% the Blues got in the previous election. Refer to the equation below for clarification. New number - Original number = Increase. Our online tools will provide quick answers to your calculation and conversion needs. Time Percentage(%) calculator. Expected Output for the Input. 1818181818%. b) Find the average of two numbers, i. Relative Percentage Difference Calculation. Calculate Difference Between 2%ages Codes! find information codes, zip codes, sic codes, phone number, contact, support. Plugging in the values into the formula, we get: % difference = | 10 - 12 (10 + 12)/2 | ×. The difference between two numbers on a number line is the distance between them and it varies whether you are going from left to right or right to left. Take an example, if you had 10 strawberries and you ate 2 then you have consumed 20 percent of strawberries out of all and left with 80 percent of the things. Shows number of days between two dates. Percentage Difference Formula The percentage difference between two values is calculated by dividing the absolute value of the difference between two numbers by the average of those two numbers. Chess Elo Rating Difference Calculator This section will calculate the difference in Elo rating between two players from match results or winning percentage. Posted: (5 days ago) The percentage difference between two values is calculated by dividing the absolute value of the difference between two. Step 1: Find out the difference between the two numbers you're comparing. This calculator will take the new price and determine the percentage discount based on the old price or valu. Finally, you need to use the following percent difference equation to calculate percentage difference between two numbers. SELECT Key, (NULLIF(b. MonthStr = 'July-2013' Share Improve this answer answered Sep 10 '13 at 14:55 RBarryYoung 52. The Difference Between Calculator can calculate the Difference between two numbers on a number line. Percentage Calculator is a free online percent calculator to calculate percentages. It is possible to look at the difference between two numbers and work out the percentage increase or the percentage decrease. The difference between the percentage of women (and men) in the two samples was 4. If you need 60 percent, find 50 percent divided by 5 to get 10 percent and add that to the initial 50 percent. I wrote this simple function to calculate the percentage difference between two numbers, and optionally (with the third parameter) return true or false if they're within a given range. Percentage Difference Calculator - Calculate the change in › On roundup of the best schools on www. 3k 14 93 129 Add a comment 2. If the number of months exceeds 36, then the benefit is further reduced. To calculate percentage difference between two numbers use the Percentage difference calculator, or you can use Percentage difference formula to find percentage difference between two numbers. what is 5 percent of 40. It is expressed in terms of percentage. Calculate the average of the values 4. The percentage difference calculator is here to help you compare two numbers. One common definition is the difference between two numbers expressed as a percentage of one of the numbers. I need to compare two numbers and see if they're close to each other. These truncation errors are specifically why I included the proviso "if you don't need a lot of precision" in my answer. Step 3: Finally, the difference between the two squared numbers will be displayed in the output field. 6 I want to print those lines which have more than 5% difference between second and third columns. › Get more: Difference between 2 number percentage calculatorDetail Medical. Find Percentage Difference: Finally, you need to use the following percent difference equation to calculate percentage difference between two numbers. An online percent calculation to find the percentage between two different timings. Early retirement reduces benefits. What about -1? Use the absolute function. A confidence interval provides you with a set of limits in which you expect the difference between the population means to lie. So to have a "percent difference", I would look for two percentages (ratios) and take the difference of them as fractions. percent of a total. The actual change of one unit from number 8 to number 9 is of much more significance than the same difference of one unit. The proper confidence interval in this case spans from -0. Simple and accurate percentage calculator, originally written for my wife and adapted for everyone to use. High precision calculator (Calculator) allows you to specify the number of operation digits (from 6 to 130) in the calculation of formula. Secondly, Excel needs to divide this difference by the first value. If we use the Percentage Difference method across the whole day, we can calculate the percent difference to be 0. Enter the two numbers into the inputs. The same works for 50 percent too. It uses javascript to do the calculations. For example,$5 about 25 years ago will cost you $21 now. An interval is the difference between two pitches or the distance between two frequencies in terms of semitones. 7 ITEM5 20 50. The task is to find out that percentage. The percentage difference between two values is calculated by dividing the absolute value of the difference between two numbers by the average of those two numbers. What is the % for a portion of another value 3. This procedure calculates the difference between the observed means in two independent samples. e, a - b = 65 - 35 = 30. Percentage Difference Calculator - Calculate the change in. Sounds good, right? But if we look at the individual intervals throughout the day, we can see a much wider variation. The actual difference (actual change) between two values is not always a good way to compare two numbers. Enter values in any two fields to calculate the other values. % value of an amount (useful for tip rat)) 2. Details: Percentage difference is usually calculated when you want to know the difference in percentage between two numbers. In this case, using the percentage difference calculator, we can see that there is a difference of 22. excel percentage difference change between values formula percent calculate value amounts cell dollar column data columns howtogeek enter Percentage Increase Between Two Numbers Calculator, How To Find Percentage Between Two Numbers Calculator, How To Calculate. 4 and so the difference between the percentages in the two samples could have been due. Then:divide the increase by the original number and multiply theanswer by 100. How to calculate percent increase between two numbers? Our online calculator will calculate percent increase, and it will also calculate percent decrease, and percent difference as well. Created by user's requst. A store owner sold goods for$20 per item. The amount of change between two numbers is the difference, but just a difference does not mean much to many people. And then multiplying the result by 100. When the difference between two values is divided by the average of the same values, a percentage difference calculation has occurred. To calculate the percentage of males in Table 3, take the frequency for males (80) divided by the total number in the sample (200). Needless to say, converting fractions to decimals is just one of many ways to handle fractions and other numbers. 1) What is the percentage increase from 2 to 3? Use the above formula to find the percent change. The percentage difference between two values is calculated by dividing the absolute value of the difference between two numbers by the average of those two numbers. Step 2: Calculate the average (add the values, then divide by 2) Step 3: Divide the difference by the average. But you don't have two ratios, you just have two large numbers. If, however, the difference is expressed as a percentage (of the original percentage), then the starting percentage must be set to 100%. pdf ) we find that P is about 0. How to use? 1. ((y2 - y1) / y1). http://technotip. Find the percent of decrease ( P %) from N to M. The percentage difference between two numbers is calculated by dividing the absolute value of the difference between two numbers by the average of those two numbers. Percentage difference calculator - Percent between two numbers. Details: Calculator Use The Percentage Difference Calculator (% difference calculator) will find the percent difference between two positive numbers greater than 0. 6 ITEM6 25 23. By taking the two points and subtracting the old point from the new, you find the difference between them. 2 million and 1. decrease between the two numbers. Find the percentage difference between the population of two cities having 12 million and 13 million population respectively. Calculate the difference between two numbers as percentage. Details: The Percentage Difference Calculator (% difference calculator) will find the percent difference between two positive numbers greater than 0. Two-sample t-test free online statistical calculator. First, calculate the difference between new and old. It means that the new number is 90. MonthStr = 'Aug-2013' AND prev. 3 million in the above formula, the percent difference can be calculated. So difference of 1 and -1 is 2. 05 significance threshold. what is 20 percent of 60. So, replacing the given values, we have. As the name implies, the essence of the percentage change calculator is to help you compute the percentage difference between two numbers - initial value and new value. The basic formula is A - B/A x100. 2/10*100% = 20%. Find out what percentage is a given number from any other number, percent increase, percent decrease, and more. Hi, I have two times one for January and another for February. By applying this fraction, three-fourths, to the whole number, 1,000, you can find exactly how many students passed the course. $$\text {Percentage Difference} = \frac {Difference} {Average} * 100$$ The free online percent difference calculator also makes use of all these formulas to generate precise outputs. Num2,0) * 100 AS Percent FROM Table1 AS a INNER JOIN Table2 AS b ON a. Calculate the percent difference percene difference calculator how to calculate percene increase or Calculate Percene Difference Between Two Numbers C - Show Percent Difference In Pivot Table Amounts Excel Pivot Tables. Val) / prev. The corresponding calculator, formula and examples help students, teachers, parents or professionals to learn, teach or practice ratio calculations efficiently. The formula = (new_value-old_value)/old_value can help you quickly calculate the percentage change between two numbers. Val As PercentDiff FROM yourTable As curr JOIN yourTable As prev ON curr. Calculates four different percentage values: 1. To calculate the percentage change between two numbers in Excel, execute the following steps. Making an abstraction of this process in a JavaScript function, we would have: /** * Calculates in percent, the change between 2 numbers. I need to show the difference in time as a percentage increase or decrease. % difference between two numbers 4. Sign in to vote. So, replacing the given values, we have the percent difference equation. Type = (2500-2342) / 2342, then press Return. Forexample, how to calculate the percentage difference: Whatis the percentage difference between 5 and 7?. Positive numbers can be entered with or without the "+". For example, you can calculate variance between sales in this year and last year, between a forecast and observed temperature. Percentage calculator (%) - calculate percentage with steps shown free online. One such tool is the percentage change between one or multiple years of accounting information. Type the formula =C2/B2. From Table A ( Appendix table A. Solution: Putting the values of 1. 60% = 50% / 5 + 50%. decrease = New Number - Original Number. Calculate Difference Between 2 Percentages! study focus room education degrees, courses structure, learning courses. It is similar to finding percentage increase or percentage decrease but it doesn't label. Num1,0) - NULLIF(a. › Get more: Convert. Sample size: the number of observations in the sample. MedCalc's Comparison of proportions calculator. If your answer is a negativenumber then this is a percentagedecrease. To calculate a percentage change you need to calculate the difference, divide the change by the original value and then multiply that amount by 100. Decrease = Original Number - New Number Then: divide the decrease by the original number and multiply the answer by 100. The difference is mathematically represented as an actual values, with the increase denoted by a ‘+’ sign and the reduction denoted by a ‘-‘ sign. The second tool calculates the % value of two given numbers that has one value as sum and the other. Published September 26, 2017. what is 20 percent of 40. The percent difference for the two milk prices is $1. › Best Education From www. Calculate Percent of Two Numbers calculator, the answer is 2. Minimum difference possible between two given numbers by rearranging their digits in the same order. While this equation may seem more intimidating than the last two, it’s easy when broken down into pieces. GDP Growth rate is a percentage increase between two numbers. Date Calculator - Add or subtract days, months, years. You’ll need to do two calculations: 1. This is a basic subtraction of one data point from another. 0539, meaning that the result is not statistically significant at the 0. For example, if your savings account went from$10,000 to $12,000 in one month, the difference is$2,000. At the bottom of the page there is a percentage formula which shows how to calculate percentage increase between two numbers. What is the percentage increase/decrease. In this case we have a % of increase because the new value is greater than the old value. That's yourpercentage! To use the calculator, enter two numbersto calculate the percentage the first is of the second byclicking Calculate Percentage. Percentage difference is usually. Here we will show you how to calculate the percentage difference between two numbers and, hopefully, to properly explain what the percentage difference is as well as some common mistakes. To compute the distance in years between two dates, fill out the top two inputs: First date: Enter the start date for the math. Mathematically, the difference is taken as an absolute value, and the increase is mentioned with a '+'sign, and the decrease is indicated with a '-' sign. It is listed as a percentage. Percentage Calculator. To find the probability attached to this difference we divide it by its standard error: z = 4. Select the cell that contains the result obtained in step 2. Percentage difference is usually calculated when you want to know the difference in percentage between two numbers. Details: Percentage Difference calculator is a free online tool to find the percent difference between two numbers. Use this step-by-step calculator for a Confidence Interval for the Difference Between two Means for known population variances. Usually, we multiply the result with 100 in order to get the percentage difference in numbers. Step 2: Now click the button "Calculate Difference of Squares" to get the result. The difference method works fine if you have only two numbers. Practically, this is the difference between saying "the price jumped 50%" and "the current price is 150% of the old price. Gather two sets of financial statements. So, calculating the number of days between two given dates isn't the easiest task because of the way in which our Gregorian calendar works (we've been The days between dates calculator on this page works by getting the time stamp of the start date and deducting it from the end date (or vice-versa if. How to Calculate. Percent Difference Calculator. Here are the steps to take when calculating the percentage difference: 1. You will find a description of how to conduct a two sample t-test below the calculator. In the case of early retirement, a benefit is reduced 5/9 of one percent for each month before normal retirement age, up to 36 months. Calculating percentage change between rows. Details: Calculate percentage change/difference between two numbers with formula. a) Find the difference between two numbers, i.
knt ngb fcc rgh bll yye cjs ift lan ycm vpl hah nbx sul fpq mhg hxm uix fbw mex | 2022-05-16T15:00:54 | {
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https://math.stackexchange.com/questions/1128244/can-the-null-space-be-empty/1128247 | # Can the null space be empty?
I was reading a proof of the theorem that the range of a linear map $T$ is always a subspace of the target space, and when the author was showing that the $0$ vector was included in the range, he made an appeal to a previous theorem which says that the null space of $T$ is always a subspace of $T$.
In other words, he says that because the nullspace is a subspace, $0$ is always in the nullspace, and therefore since $T(0) = 0$, then $0$ is in the range of $T$.
That makes sense, but is it possible that the nullspace is empty? My feeling is no. Because $T$ acts on a vector space $V$, then $V$ must include $0$, and since we showed that the nullspace is a subspace, then $0$ is always in the nullspace of a linear map, so therefore the nullspace of a linear map can never be empty as it must always include at least one element, namely $0$.
• "That makes sense, but is it possible that the nullspace is empty?" What do you mean by null space? – Git Gud Jan 31 '15 at 22:10
• The set {v in V: Tv = 0} – user1236 Jan 31 '15 at 22:12
• Does $0$ satisfy $T0=0$? – Git Gud Jan 31 '15 at 22:13
• It's worth pointing out that a vector space, by definition, cannot be empty. Specifically, the vector space axioms require that for any vector space $V$, there exist a $0 \in V$ such that for any $v \in V$ $$0+v = v$$ so we must have $0 \in V$, and so $V\not=\emptyset$. – Strants Jan 31 '15 at 22:16
Let $T:V\to W$ be a linear map. Then $$T(\mathbf{0})=T(\mathbf{0}-\mathbf{0})=T(\mathbf{0})-T(\mathbf{0})=\mathbf{0}$$ This proves that $\mathbf 0$ is always in the nullspace of $T$. Hence the nullspace of $T$ cannot be empty.
It is not possible that the nullspace is empty. The element $\overline 0$ is always contained in the domain and as $T$ is linear we have $T(\overline 0) = T(0\cdot\overline 0) = 0\cdot T(\overline 0) = \overline 0$. | 2019-10-20T16:27:47 | {
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https://math.stackexchange.com/questions/1043630/is-this-a-basis-for-the-set-s | # Is this a basis for the set S?
Consider the set: $\begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix}$
Would the basis be found by doing: a$\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$ + $b\begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix}$
So those two 2x2 matrices would form the basis? Is this the proof that they form a basis? How do I prove they're linearly independent? Thanks!
Put
a $\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$ + $b\begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix}$ $=0$, the zero matrix
If you show that $a=b=0$, then the matrices are linearly independent.
Also, the fact that they span the set is obvious.
• I don't get how to show a=b=0. That's the part I was confused on, because if I multiply them back out, I just get my set. – Math StackExchange Nov 29 '14 at 14:17
• Why don't you multiply the scalars to the corresponding matrices and add the two up. – Swapnil Tripathi Nov 29 '14 at 14:18
• When I do that, I just get $\begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix}$ – Math StackExchange Nov 29 '14 at 14:19
• i mean, if a=b=0 for this, then it would be the zero matrix....? – Math StackExchange Nov 29 '14 at 14:20
• Yes!! You got it. – Swapnil Tripathi Nov 29 '14 at 14:20
Yes those two matrices form a basis for the vector space $$V = \left\{\begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix} | \ a, b \in K \right\}$$ as evidently every member of $V$ can be written as such a linear combination.
The show linear independence, just explain why
$$a \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} + b\begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix} = 0 \ \ \ \ \ \ \text{ if and only if } \ \ \ \ \ a = b = 0$$ | 2020-01-20T09:14:22 | {
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https://www.tutorialspoint.com/sum-of-the-series-1-plus-1plus3-plus-1plus3plus5-plus-1plus3plus5plus7-plus-plus-1plus3plus5plus7plus-plus-2n-1-in-cplusplus | # Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + + (1+3+5+7+....+(2n-1)) in C++
C++Server Side ProgrammingProgramming
In this problem, we are given an integer n. Our task is to create a program to find the sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + + (1+3+5+7+....+(2n-1)).
From this series, we can observe that ith term of the series is the sum of first i odd numbers.
## Let’s take an example to understand the problem,
Input
n = 3
Output
14
Explanation −(1) + (1+3) + (1+3+5) = 14
A simple solution to this problem is using a nested loop and then add all odd numbers to a sum variable. Then return the sum.
## Example
Program to illustrate the working of our solution,
Live Demo
#include <iostream>
using namespace std;
int calcSeriesSum(int n) {
int sum = 0, element = 1;
for (int i = 1; i <= n; i++) {
element = 1;
for (int j = 1; j <= i; j++) {
sum += element;
element += 2;
}
}
return sum;
}
int main() {
int n = 12;
cout<<"Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + ... + (1+3+5+7+ ... + (2"<<n<<"-1)) is "<<calcSeriesSum(n);
return 0;
}
## Output
Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + ... + (1+3+5+7+ ... + (2*12-1)) is 650
This approach is not effective as it uses two nested loops.
A more efficient approach is to mathematically find the general formula to find the sum of the series.
Sum of n odd numbers,
= (1) + (1+3) + (1+3+5) + …. (1+3+5+... + 2n-1)
= n2
First, let’s see the sum of first n odd number, which represents the individual elements of the series.
Sum of series,
sum = (1) + (1+3) + (1+3+5) + … + (1+3+5+ … + 2n-1)
sum = ∑ (1+3+5+ … + 2n-1)
sum = ∑ n2
sum = [n * (n+1) * (2*n -1)]/6
## Example
Program to illustrate the working of our solution,
Live Demo
#include <iostream>
using namespace std;
int calcSeriesSum(int n) {
return ( n*(n + 1)*(2*n + 1) )/6;
}
int main() {
int n = 9;
cout<<"Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + ... + (1+3+5+7+ ... + (2*"<<n<<"-1)) is "<<calcSeriesSum(n);
return 0;
}
## Output
Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + ... + (1+3+5+7+ ... + (2*9-1)) is 285
Published on 14-Aug-2020 17:19:34
Advertisements | 2021-05-16T06:49:48 | {
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https://math.stackexchange.com/questions/2828800/does-a-solution-with-a-minimal-norm-mean-it-is-a-solution-that-minimizes-the-res?noredirect=1 | Does a solution with a minimal norm mean it is a solution that minimizes the residuals?
I have an under-determined system of linear equations, which is not full rank (i.e. one or more column vectors are linearly dependent) and has more equations than unknowns.
Here is just one example of this:
$${\begin{bmatrix} 6 & 40 \\ 6 & 40 \\ 3 & 20 \\ \end{bmatrix}} \begin{bmatrix} \beta_{0} \\ \beta_{1} \end{bmatrix} \approx \begin{bmatrix} 0.5 \\ 0.2 \\ 0.6 \\ \end{bmatrix}$$
If I solve the above example for the least squares solution (i.e. solve $A\vec{x}\approx\vec{b}$), the solution produced will come from a set of solutions $S$ that is infinite in length (i.e. $|S| = \infty$). However, is it possible to minimize the residuals (i.e. get the best model that minimizes the error) from this infinite set of solutions $S$?
This question is sparked from a response I got on StackOverflow (SO) and from what I read on this forum. For example, on my SO post, user @AGN Gazer says the following:
"A solution with a minimal norm does not mean a solution that minimizes residuals. [...] Having a solution with a minimal norm means nothing to me[.] [...] I want the "best" solution - the one that minimizes the residuals but I cannot get it with an underdetermined system."
However, on the accepted answer on this Math Exchange post, it appears that @Brian Borchers says otherwise:
"[We're] often interested in the minimum norm least squares solution. That is, among the infinitely many least squares solutions, pick out the least squares solution with the smallest $∥x∥_{2}$ [(i.e. euclidean norm: $\| x \|_{2} =\sqrt{\sum_{i=1}^{n}x_{i}^{2}}$)]. The minimum norm least squares solution is always unique."
I must be misunderstanding something! In relation to my question above, what even is the distinction between minimizing the residuals and minimizing the norm?
• @RodrigodeAzevedo I thought that too, but was told that is incorrect. If you have more equations than unknowns, but one or more of the column vectors is linearly dependent, then it is no longer full rank is thereby underdetermined. – Danny Dan Jun 22 '18 at 19:45
• @RodrigodeAzevedo Take a look at my SO post above, which gives an example of this! I am now extremely confused by all the conflicting information I am getting. – Danny Dan Jun 22 '18 at 19:48
• Least-squares minimizes $\| A x - b \|_2^2$. Least-norm minimizes $\|x\|_2^2$ subject to $A x = b$. Combining the two, minimize $\| A x - b \|_2^2 + \gamma \|x\|_2^2$ with $\gamma > 0$. What exactly do you want? – Rodrigo de Azevedo Jun 22 '18 at 19:57
• If I use ordinary least squares (OLS) on the system I described above, can I minimize the residuals? Are you saying that minimizing the norm and minimizing the residuals are two separate operations, with minimizing the norms meaning finding the smallest norm for the parameters and minimizing the residuals meaning finding the smallest error of the output? – Danny Dan Jun 22 '18 at 20:04
• Does $A x = b$ have zero or infinitely many solutions? If zero solutions, use least-squares. If infinitely many, use least-norm. However, the fact that $A$ does not have full column rank means that there will be infinitely many $x$'s that produce minimal $\| A x -b \|_2^2$. – Rodrigo de Azevedo Jun 22 '18 at 20:05
In your example, there is no exact solution of $Ax = b$. Because $A$ is not full rank, the least squares solution is not unique.
The minimum 2-norm least squares solution is that least squares solution for which $\|x\|_2$ is minimum among those least squares solutions. I.e., consider all least squares solutions achieving the same sum minimum squared residuals value, $(Ax-b)^T(Ax-b)$, then minimizing 2-norm of x among those solutions can serve as a tie-breaker to choose from among the solutions achieving the minimum sun squared residuals value. The minimum 2-norm solution can be found using the pseudoinvserse (pinv in MATLAB) of $A$, as shown below. Note that if $A$ is full rank, then the least squares solution is unique, and therefore, is the minimum 2-norm least squares solution.
Here is an illustration in MATLAB on your example. As can be seen, in this example, the QR and SVD (minimum 2-norm) solutions have the same residuals, but $\|x||_2$ is smaller for the SVD solution than the QR solution.
>> disp(A)
6 40
6 40
3 20
>> disp(b)
0.500000000000000
0.200000000000000
0.600000000000000
>> QR_solution = A\b
Warning: Rank deficient, rank = 1, tol = 2.307555e-14.
QR_solution =
0
0.011111111111111
>> A*QR_solution - b
ans =
-0.055555555555556
0.244444444444444
-0.377777777777778
>> norm(QR_solution)
ans =
0.011111111111111
>> SVD_solution = pinv(A)*b
SVD_solution =
0.001629991850041
0.010866612333605
>> A*SVD_solution - b
ans =
-0.055555555555556
0.244444444444444
-0.377777777777778
>> norm(SVD_solution)
ans =
0.010988181698537 | 2019-11-14T21:50:43 | {
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## anonymous 4 years ago I AM CERTAIN THAT NO ONE WILL ANSWER THIS QUESTION... Suppose that l be a line through the points (0,3)and (5,-2) and tangent to the graph of the function $F(x)=\frac{ x^2-p }{ 3x+1 }$ where ‘p’ is real number. What is the equation of a line parallel to l and tangent to the graph of F. A) 4x+4y+7 = 0 B) 9x+9y+37 = 0 C) x+y-3 = 0 D) 4x+6y-37 = 0
• This Question is Closed
1. anonymous
I will post the answer exactly after one hour.
2. mathslover
Hint: Never challenge OpenStudy
3. anonymous
m= -2-3/5-0 = -1 the equation will be y-3= -1(x-0) x+y-3 =0
4. TheViper
Right @mathslover :)
5. hartnn
so its not D
6. anonymous
@hartnn that means???
7. hartnn
that means answer can be a,b or c
8. anonymous
yep.... but the answer hasn't yet finished... @hartnn
9. anonymous
I will the slope of the line through the two points
10. hartnn
p=-7 B)9x+9y+37=0
11. hartnn
solve y=3-x=F(x)=.... u get two roots of x, equate them because the line is tangent and can have only one intersection point. from this i got p=-7. now answer can only be either A or B so solving both of them with F(x), gave only B) option with one intersection point, A) doesn't even intersect.
12. anonymous
PLZ DONT POST THE SOLUTION
13. anonymous
I am here to post the answer.
14. anonymous
ok 5min
15. anonymous
ok
16. hartnn
whats wrong with my answer? and method ??
17. anonymous
w/c is your ans?
18. hartnn
B)
19. hartnn
and p=-7
20. anonymous
correct but the approach not.
21. hartnn
didn't u get p=-7, the same way as i got ?
22. anonymous
yape
23. hartnn
then ?
24. anonymous
you use the options to answer the question, isn't it?
25. anonymous
Time is up @sauravshakya
26. anonymous
OK.......
27. anonymous
Here is the answer
28. anonymous
First of all what you have to do is finding l so as to get p $l:\frac{ y-3 }{ x-0 }=\frac{ -2-3 }{ 5-0 }$ $l:y=-x+3$thus $F(x)=\frac{ x^2-p }{ 3x+1} =-x+3=y$simplifying the above expression $4x^{2}-8x-p-3=0$Now since l is tangent to F(x), the quadratic equation above has only one solution. That means $b^{2}-4ac=0$ $(-8)^{2}-4(4)(-p-3)=0$thus $(-8)^{2}-4(4)(-p-3)=0$$p=-7$Now we get F(x) to be $F(x)=\frac{ x^2+7 }{ 3x+1 }$Therefore lets solve F’(x)=-1 so as to get the coordinate of the unknown line. $F'(x)=\frac{ 2x(3x+1)-3(x^{2}+7) }{ (3x+1)^2 }=-1$ $3x^{2}+2x-21=-9x^{2}-6x-1$ $12x^{2}+8x-20=0$thus $x=1 or x=\frac{ -5 }{ 3 }$and $F(1)=2 and F(\frac{ -5 }{ 3 })=\frac{ -22 }{ 9 }$Thus we got this two points (1,2) and (-5/3,-22/9). The two possible lines therefore are $\frac{ y-2 }{ x-1 }=-1{\rightarrow}y=-x+3$w/c is the given line l. And $\frac{ y-(-\frac{ 22 }{9 }) }{ x-(-\frac{ 5 }{ 3 })}=-1{\rightarrow}9y+9x+37=0$w/c is the answer.
29. hartnn
why f'(x) = -1 ?
30. anonymous
b/c the slope at that point is -1
31. anonymous
Because the line l and the required line are parallel
32. anonymous
So, both of their slope is -1
33. hartnn
okk.
34. hartnn
it wasn't that difficult.
35. anonymous
yes but need smart approach
36. anonymous
YEP....... Actually I did a algebra mistake, as usual
37. hartnn
lol! checking options(when given) is the smartest approach!! :P
38. anonymous
agreed with @hartnn
39. anonymous
that saves time in exam
40. anonymous
For exam only
41. hartnn
atleast i got the answer...
42. anonymous
yape
43. anonymous
ok thanks for your help....... closing
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https://www.physicsforums.com/threads/here-is-a-fun-math-statistic-probability-question.101523/ | # Here is a fun math statistic/probability question
1. Nov 26, 2005
### firstwave
My grade 12 math teacher gave us a question that I found to be very interesting.
You have 12 identical keys. 11 of the 12 keys have the same mass. Only one of them has a different mass. You have a scale, but it is not digital and it does not tell you the actual weight. It only shows you which one of the two sides is heavier. You are allowed 3 uses of the scale. How do you determine which key has a different mass?
I found this question to be very interesting, but unfortunately, I can't solve it. How about you guys?
2. Nov 26, 2005
### Tide
HINT: Start by comparing the weight of two groups of 4 keys.
3. Nov 27, 2005
### firstwave
Hmm, I don't think that works. If I compare two groups of 4 keys, there are 2 possible outcomes.
Either one side is different or both groups weigh the same. If the outcome is the former, I still need to use the scale once more to determine which one of the three groups contains the different key. The problem is that knowing which side is heavier or light does not necessarily help because we don't know if the "special" key is ligher or heavier than the other 11 keys.
Maybe I am perceiving this wrong, so please enlighten me :)
4. Nov 27, 2005
### Tide
Okay,
If the two sets of four do not balance then you eliminate the third (unweighed) set so we'll set them aside.
Label the set of heavier keys as 1, 2, 3 and 4 and the lighter set 5, 6, 7 and 8. Now remove 7 & 8 from (say) the right side (and remember they were part of the lighter set!) and swap 5 and 4 so you are now comparing 1, 2 and 5 with 3, 4 and 6. This is the second weighing.
Now, if 1, 2 and 5 balance with 3, 4 and 6 then one of 7 and 8 is the special key and will be the lighter of the two (recall the note above!) - you find which one it is by a third weighing.
However, if 1, 2 and 5 do not balance then you eliminate 7 and 8.
Recall that 1, 2 and 5 came from the heavier side of the first weighing so that if 1, 2 and 5 are heavier than 3, 4 and 6, then 3 and 4 are ordinary keys so that 1 and 2 are potentially heavier while 6 is potentially lighter (recall it came from the light side of the first weighing). A third weighing comparing 1 and 2 tells you that if 1 and 2 are the same then 6 is the a heavy key and if 1 and 2 are not the same then the heavier of the two is the odd key.
On the other hand, if 1, 2 and 5 are lighter than 3, 4 and 6 then 1, 2 and 6 are ordinary keys while 5 is potentially lighter and 3 and 4 are potentially heavier. Finally, compare 3 and 4 (third weighing!). If they are the same then 5 is the special key and if they are not the same then the heavier of 3 and 4 is a heavy key.
I'll leave it to you to decide the case when 1, 2, 3 and 4 balance with 5, 6, 7 and 8. :)
5. Nov 27, 2005
### firstwave
Ohhhh right. I forgot that if two sets of four do not balance, you can eliminate the third set. Thanks for your insightful solutions :)
6. Nov 27, 2005
### Curious3141
This is a different statement of a classic weighing problem (which uses coins, not keys).
There is a beautifully elegant and methodical solution for this problem and the general class of problem.
Read Lars Prins' method (at the bottom) for the 12-coin problem : http://mathforum.org/library/drmath/view/55618.html
In general, with $n$ weighings, you can find the odd coin out of (at most) $$\frac{3^n - 3}{2}$$ coins. You can read the general solution from here (the first part) : http://www.cut-the-knot.org/blue/OddCoinProblemsShort.shtml | 2017-09-23T16:52:22 | {
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https://math.stackexchange.com/questions/3306545/pointwise-convergence-of-uniformly-continuous-functions-to-zero-but-not-uniform | # Pointwise convergence of uniformly continuous functions to zero, but not uniformly
What would be an example of a sequence of uniformly continuous functions on a compact domain which converges pointwise to $$0$$, but not uniformly?
• The functions $f_n(x)=x^n$ defined on $[0,1)$ form an example. – Suzet Jul 28 '19 at 14:27
• @Suzet I forgot, I want the domain to be compact. – Jannik Pitt Jul 28 '19 at 14:30
Consider $$f_n(x)=1-\min(1,n|x-1/n|)=\begin{cases} nx&\text{ if }x<\frac{1}{n}\\2-nx&\text{ if }\frac1{n}\leq x\leq\frac{2}{n}\\0&\text{ otherwise.}\end{cases}$$ Each $$f_n$$ is continuous on the compact set $$[0,1]$$ and therefore it is also uniformly continuous. Moreover, $$f_n(x)\to 0$$ for any $$x\in [0,1]$$, but the convergence is not uniform on $$[0,1]$$ because $$\max_{x\in[0,1]}|f_n(x)|=f(1/n))=1$$.
The another standard one is the growing steeple on $$[0,1]$$:
$$f_n(x)=\begin{cases}n^2 x &\text{if}\;0 \leq x \leq \frac{1}{n}\\ 2n-n^2 x &\text{if}\;\frac{1}{n} \leq x \leq \frac{2}{n}\\ 0 &\text{if}\;\frac{2}{n} \leq x \leq 1 \end{cases}$$
Then each $$f_n$$ is uniformly continuous and also the limit is zero, but the convergence is not uniform.
Added: It is easy to visualize the graph of $$f_n$$. Actually each $$f_n$$ is a triangle with height $$n$$ attained at $$1/n$$
Take$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&nx^n(1-x).\end{array}$$The sequence $$(f_n)_{n\in\mathbb N}$$ converges pointwise to the null function, but not uniformly, since$$(\forall n\in\mathbb N):f_n\left(\frac n{n+1}\right)=\left(\frac n{n+1}\right)^{n+1}$$and $$\lim_{n\to\infty}\left(\frac n{n+1}\right)^{n+1}=e^{-1}$$. | 2020-05-30T00:32:34 | {
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https://math.stackexchange.com/questions/3092833/prove-that-mathbbq-sqrt2-sqrt3-sqrt5-mathbbq-8 | # Prove that $[ \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}]=8.$
I have to solve the following exercise:
Compute $$[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}]$$ and $$\operatorname{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})/\mathbb{Q}).$$
Here my attempt:
Let $$\mathbb{K}=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}).$$ Proved that $$[\mathbb{K}:\mathbb{Q}]=8$$, compute $$\operatorname{Gal}(\mathbb{K}/\mathbb{Q})$$ is easy, in fact it will be $$\operatorname{Gal}(\mathbb{K}/\mathbb{Q})=\langle \sigma_2,\sigma_3,\sigma_5\rangle,$$ with $$\sigma_k$$ the automorphism that interchanges $$\sqrt{k}$$ with $$-\sqrt{k}$$ and doesn't move the rest of the elements.
I prove that $$[\mathbb{K}:\mathbb{Q}]=8$$ as follows. I do some observations first:
• If a group $$G$$ verifies $$|G|=4$$ then $$G$$ is isomorphic to $$C_4$$ (cyclic group of order 4) or to $$\mathbb{V}$$ (vierergruppe). In particular, $$G$$ can only have $$1$$ or $$3$$ proper subgroups.
• Clearly $$\mathbb{K}/\mathbb{Q}$$ is a Galois extension since $$\mathbb{K}$$ is the splitting field of the polynomial $$p(x)=(x^2-2)(x^2-3)(x^2-5).$$
• We have the four strict chains of extensions, all distinct:
1. $$\mathbb{Q} \subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{K}.$$
2. $$\mathbb{Q} \subset \mathbb{Q}(\sqrt{3}) \subset \mathbb{K}.$$
3. $$\mathbb{Q} \subset \mathbb{Q}(\sqrt{5}) \subset \mathbb{K}.$$
4. $$\mathbb{Q} \subset \mathbb{Q}(\sqrt{6}) \subset \mathbb{K}.$$
Let $$\mathbb{L}=\mathbb{Q}(\sqrt{2},\sqrt{3})$$. Clearly $$[\mathbb{L}:\mathbb{Q}]=4$$ and $$\mathbb{K}=\mathbb{L}(\sqrt{5}).$$ Then, $$[\mathbb{K}:\mathbb{L}] \in \{1,2\}.$$ With this, we have $$[\mathbb{K}:\mathbb{Q}]=[\mathbb{K}:\mathbb{L}][\mathbb{L}:\mathbb{Q}]=4[\mathbb{K},\mathbb{L}].$$ Therefore, $$[\mathbb{K}:\mathbb{Q}]\in \{4,8\}.$$
If we suppose that $$[\mathbb{K}:\mathbb{Q}]=4$$, then, since it is a Galois extension, $$|\operatorname{Gal}(\mathbb{K}/\mathbb{Q})|=4$$ and the Galois correspondence joint with the fact about groups of four elements mentioned above tells us that there is only $$1$$ or $$3$$ stricts chains of extensions starting with $$\mathbb{Q}$$ and ending at $$\mathbb{K}.$$ But we found $$4$$ of such chains, and this concludes that must be $$[\mathbb{K}:\mathbb{Q}]=8.$$
End.
Is correct my solution? Could it be improved? I'm interested in reading other possible solutions and better if it's "faster". How do I prove this result without the using of group theory?
Thanks to everyone!
Edit:
I thought that my solution to the problem has not been posted yet in MSE but I found a question that has my solution as an answer.
Showing field extension $$\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})/\mathbb{Q}$$ degree 8 [duplicate]
Other related questions are:
The square roots of different primes are linearly independent over the field of rationals
I'm sorry for duplicating this question.
• This is a nice solution, if you ask me. And it also generalizes. See my answer to this question. Mind you, given that you came up with this way of thinking about it I'm sure you would have come up with the generalization to more primes as well :-) – Jyrki Lahtonen Jan 31 at 5:01
• And, if you study the thread I linked, do take a look at Bill Dubuque's answer as well. It gets a way with a bit more primitive tools (doesn't use Galois groups). – Jyrki Lahtonen Jan 31 at 5:07
• Thanks for your comments @JyrkiLahtonen. I felt that a generalization of this galois-type argument for more prime numbers could be possible, but I didn't write down my ideas. The answer of Bill Dubuque is an elegant and simple solution, but I like to use Galois groups since I think it is the natural place for studying fields and extensions. – DrinkingDonuts Jan 31 at 16:50
• I agree. That's why I wrote my answer in the end. Using Galois theory means that the only arithmetic fact we need to prove is that $\sqrt n\notin \Bbb{Q}(\sqrt m)$ whenever both $n$ and $m$ are square free. – Jyrki Lahtonen Jan 31 at 16:56
• @JyrkiLahtonen I've posted a new question related to field/Galois theory. I'm interested in your opinion. Here the link "math.stackexchange.com/questions/3095247/…". Thanks in advance! – DrinkingDonuts Jan 31 at 18:13
Your proof is essentially true, however I feel it can be shortened by proving that $$\sqrt{5} \not \in \mathbb{Q}(\sqrt{2},\sqrt{3})$$. To see this you can use the fact that $$\text{Gal}\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}) = \langle \sigma_2, \sigma_3\rangle \cong \mathbb{Z}_2 \times \mathbb{Z}_2$$. Now the quadratic subfields are $$\mathbb{Q}(\sqrt{2}),\mathbb{Q}(\sqrt{3}),\mathbb{Q}(\sqrt{6})$$ and obviously $$\mathbb{Q}(\sqrt{5})$$ is not any of them and so $$\sqrt{5} \not \in \mathbb{Q}(\sqrt{2},\sqrt{3})$$. Thus we must have that $$[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}] = 8$$. From here we conclude that:
$$\text{Gal}\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})/\mathbb{Q}) = \langle \sigma_2, \sigma_3,\sigma_5\rangle \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$$
On the other way if you want to avoid group theory you can prove that $$\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}) = \mathbb{Q}(\sqrt{2}+\sqrt{3}+\sqrt{5})$$, as it's been done here. Then you can explicitly prove that the minimal polynomial of $$\sqrt{2}+\sqrt{3}+\sqrt{5}$$ over $$\mathbb{Q}$$ is of degree $$8$$ and conclude that $$[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}] = 8$$. Once you have this finding the Galois group is an easy task. However I feel this way requires far more work.
• I like your solution! Clearly, is more direct than mine. Thanks for your answer. Could be extended by induction the Galois-type argument in order to solve the problem posed in the question you linked? – DrinkingDonuts Jan 30 at 10:52
• @AlgebraicallyClosed I suppose you can use such an argument to prove that the Galois group of $\mathbb{Q}(\sqrt{p_1},\cdots,\sqrt{p_n})$ is $(\mathbb Z_2)^n$. However the problem is proving that $\mathbb{Q}(\sqrt{p_1}+\cdots+\sqrt{p_n})$ is extension of order $2^n$ – Stefan4024 Jan 30 at 11:13
• Okay, now I see the problem. Thanks for the time you dedicated to my question. – DrinkingDonuts Jan 30 at 11:27
I think you're overcomplicating things. The index $$[\mathbb Q(\sqrt2,\sqrt3,\sqrt5):\mathbb Q]$$ is just the dimension of $$\mathbb Q(\sqrt 2,\sqrt3,\sqrt5)$$ as a $$\mathbb Q$$-vector space, which is $$\sum_{k=0}^3\binom{3}{k}=1+3+3+1=8.$$
• How do you have directly this? Are you supposing that \sqrt{2}, \sqrt{3} and \sqrt{5} are linearly independent? I don't see trivial this part of the problem. Anyway, thanks for your answer. – DrinkingDonuts Jan 30 at 10:44
• This question has been asked (and answered!) many times : the degree over Q of Q($\sqrt p_1,..., \sqrt p_n$), where the $p_i$ are $n$ distinct primes, is equal to $2^n$. See e.g. math.stackexchange.com/a/1609061/30070S – nguyen quang do Jan 30 at 13:06 | 2019-08-19T20:44:58 | {
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https://web2.0calc.com/questions/the-ratio-of-the-mass-of-the-sugar-in-packet-a-to-the-mass-of-the-sugar-in-packet-b-is-5-is-to-6-after-the-mass-of-the-sugar-in-packet-a-is | +0
# The ratio of the mass of the sugar in packet A to the mass of the sugar in packet B is 5 is to 6.After the mass of the sugar in packet A is
+5
292
2
The ratio of the mass of the sugar in packet A to the mass of the sugar in packet B is 5 is to 6.After the mass of the sugar in packet A is increase by 30 percent. By what percentage must the mass of the sugar in packet B be decrease so that the total mass of sugar in packet A and B remain the same?
Guest Jul 13, 2015
#1
+18946
+10
The ratio of the mass of the sugar in packet A to the mass of the sugar in packet B is 5 is to 6.After the mass of the sugar in packet A is increase by 30 percent. By what percentage must the mass of the sugar in packet B be decrease so that the total mass of sugar in packet A and B remain the same ?
$$\\\small{\text{ A:B= 5:6 ~~or~~ B=\dfrac{6}{5}A ~~or ~~ A=\dfrac{5}{6}B }}\\\\ \small{\text{ total mass of sugar in packet A and B: \qquad A+B= A+ \dfrac{6}{5}A = \dfrac{11}{5}A }}\\\\ \small{\text{ increase A = a \qquad a = 1.3\cdot A }}\\ \small{\text{ decrease B = b }}\\ \small{\text{ total mass of sugar in packet a and b: \qquad a+b=A+B=\dfrac{11}{5}A }}\\ \small{\text{ \begin{array}{rcl} a+b&=&\dfrac{11}{5}A \\\\ 1.3\cdot A +b &=&\dfrac{11}{5}A \\\\ b &=& \dfrac{11}{5}A - 1.3\cdot A\\\\ \mathbf{b} & \mathbf{=} & \mathbf{0.9 \cdot A}\qquad | \qquad A=\dfrac{5}{6}B \\\\ b &=& 0.9 \cdot \dfrac{5}{6}B \\\\ \mathbf{b} & \mathbf{=} & \mathbf{0.75 \cdot B } \end{array} }}\\$$
sugar in packet B be decrease 25 %
heureka Jul 13, 2015
Sort:
#1
+18946
+10
The ratio of the mass of the sugar in packet A to the mass of the sugar in packet B is 5 is to 6.After the mass of the sugar in packet A is increase by 30 percent. By what percentage must the mass of the sugar in packet B be decrease so that the total mass of sugar in packet A and B remain the same ?
$$\\\small{\text{ A:B= 5:6 ~~or~~ B=\dfrac{6}{5}A ~~or ~~ A=\dfrac{5}{6}B }}\\\\ \small{\text{ total mass of sugar in packet A and B: \qquad A+B= A+ \dfrac{6}{5}A = \dfrac{11}{5}A }}\\\\ \small{\text{ increase A = a \qquad a = 1.3\cdot A }}\\ \small{\text{ decrease B = b }}\\ \small{\text{ total mass of sugar in packet a and b: \qquad a+b=A+B=\dfrac{11}{5}A }}\\ \small{\text{ \begin{array}{rcl} a+b&=&\dfrac{11}{5}A \\\\ 1.3\cdot A +b &=&\dfrac{11}{5}A \\\\ b &=& \dfrac{11}{5}A - 1.3\cdot A\\\\ \mathbf{b} & \mathbf{=} & \mathbf{0.9 \cdot A}\qquad | \qquad A=\dfrac{5}{6}B \\\\ b &=& 0.9 \cdot \dfrac{5}{6}B \\\\ \mathbf{b} & \mathbf{=} & \mathbf{0.75 \cdot B } \end{array} }}\\$$
sugar in packet B be decrease 25 %
heureka Jul 13, 2015
#2
+17711
+5
We don't need the actual masses, we can use just their ratios to get the answer.
mass(A)/mass(B) = 5/6
increasing mass(A) by 30%: 5 x 1.3 = 6.5 (an increase of 1.5)
Since the total mass of A and B is 11, to keep the total mass the same, the mass of B must be decreased by 1.5 down to 4.5.
1.5, in comparison to 6, is a decrease of 25%
geno3141 Jul 13, 2015
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https://mathemerize.com/rolles-theorem/ | # Rolle’s Theorem – Statement and Examples
Here you will learn statement of rolle’s theorem, it’s geometrical and algebraic interpretation with examples.
Let’s begin –
## Rolle’s Theorem
Statement : Let f be a function that satisfies the following three conditions:
(a) f is continous on the closed interval [a, b].
(b) f is differentiable on the open interval (a, b)
(c) f(a) = f(b)
Then, there exist a real number c $$\in$$ (a, b) such that f'(c) = 0.
#### Geometrical Interpretation :
Geometrically, the theorem says that somewhere between A and B the curve has atleast one tangent parallel to x-axis.
#### Algebraic Interpretation :
If f is differentiable function then between any two consecutive roots of f(x) = 0, there is atleast one root of the equation f'(x) = 0.
Remark – On this theorem generally two types of problems are formulated.
(a) To check the applicability of rolle’s theorem to a given function on a given interval.
(b) To verify rolle’s theorem for a given function on a given interval. In both types of problems we first check whether f(x) satisfies conditions of theorem or not. The following results are very helpful in doing so.
1. A polynomial function is everywhere continuous and differentiable.
2. The exponential function, sine and cosine functions are everywhere continuous and differentiable.
3. Logarithmic function is continuous and differentiable in its domain.
4. | x | is not differentiable at x = 0
Example : Verify Rolle’s theorem for the function f(x) = $$x^2$$ – 5x + 6 on the interval [2, 3].
Solution : Since a polynomial function is everywhere differentiable and so continuous also.
Therefore, f(x) is continuous on [2, 3] and differentiable on (2, 3).
Also, f(2) = $$2^2$$ – 5 $$\times$$ 2 + 6 = 0 and f(3) = $$3^2$$ – 5 $$\times$$ 3 + 6 = 0
$$\therefore$$ f(2) = f(3)
Thus, all the conditions of rolle’s theorem are satisfied. Now, we have to show that there exists some c $$\in$$ (2, 3) such that f'(c) = 0.
for this we proceed as follows,
We have,
f(x) = $$x^2$$ – 5x + 6 $$\implies$$ f'(x) = 2x – 5
$$\therefore$$ f'(x) = 0 $$\implies$$ 2x – 5 = 0 $$\implies$$ x = 2.5
Thus, c = 2.5 $$\in$$ (2, 3) such that f'(c) = 0. Hence, Rolle’s theorem is verified.
### Related Questions
It is given that for the function f(x) = $$x^3 – 6x^2 + ax + b$$ on [1, 3], Rolles’s theorem holds with c = $$2 +{1\over \sqrt{3}}$$. Find the values of a and b, if f(1) = f(3) = 0.
Find the point on the curve y = cos x – 1, x $$\in$$ $$[{\pi\over 2}, {3\pi\over 2}]$$ at which tangent is parallel to the x-axis. | 2022-11-30T16:42:20 | {
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https://math.stackexchange.com/questions/1754312/how-to-find-lim-x-rightarrow-0-frac-sin-xx2-sin-x | # How to find $\lim_{x\rightarrow 0} \frac{\sin |x|}{x^2+\sin (x)}$?
How to find $$\lim_{x\rightarrow 0} \frac{\sin |x|}{x^2+\sin (x)}$$ without L'hopital's?
So far I tried to use the squeeze theorem but couldn't find appropriate bounds and also tried to exploit the limit of $\sin(x)/x$ without any luck.
Any hints?
• Question is: does it have a limit? – user228113 Apr 22 '16 at 15:27
• I'm not sure to be honest. – Chris Apr 22 '16 at 15:28
• "and also tried to exploit the limit of $\sin(x)/x$ without any luck" This lack of "luck" is odd since you are looking at $$\frac{\sin |x|}{|x|}\frac{1}{x+\frac{\sin (x)}{x}}\frac{x}{|x|},$$ whose behaviour when $x\to0$ is pretty clear. – Did Apr 22 '16 at 15:29
• @G.Sassatelli Let me suggest you compare $$\frac{|x|}{x}\quad\text{and}\quad\frac{x}{|x|}...$$ Or is your comment tongue-in-cheek? – Did Apr 22 '16 at 15:32
• @Did Oh! Silly me. It was just me looking at symbols instead of evaluating the function. – user228113 Apr 22 '16 at 15:35
You can write the function as $$\frac{\sin|x|}{x}\frac{1}{x+\dfrac{\sin x}{x}}$$ The second factor has limit $1$ for $x\to0$, so the problem is reduced to seeing whether $$\lim_{x\to0}\frac{\sin|x|}{x}$$ exists.
Hint: try from the left and from the right.
• Limit doesn't exist since left hand limit is $-1$ whereas right hand limit is $1$? – Chris Apr 22 '16 at 15:39
• @Chris Exactly so. – egreg Apr 22 '16 at 15:41
• @Chris Note that applying l'Hôpital would give a non existing limit either: the function would be $\dfrac{|x|\cos|x|}{x(2x+\cos x)}$, but this would not be immediately conclusive. On the other hand, this would give limit $-1$ from the left and $1$ from the right, so the same as with the direct method. – egreg Apr 22 '16 at 15:44
For another approach, look at the reciprocal:
Set
$h(x)=\frac{x^2+\sin x}{\sin \vert x\vert}$.
Then,
$h(x) =x\frac{x}{\sin \vert x\vert}+\frac{\sin x}{\sin \vert x\vert }$.
and
$h(x)\to 0+1=1$ as $x\to 0^+$
whereas
$h(x)\to 0-1=-1$ as $x\to 0^-$ | 2019-07-23T05:00:54 | {
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https://cs.stackexchange.com/questions/101973/how-to-properly-calculate-dependent-nested-loops-for-big-o | # How to properly calculate dependent nested loops for big-O [duplicate]
I am revising for my algorithms exam and I have come across one topic in particular that I do not quite understand; which is how to analyse dependent nested loops. I know if we have a 2-nested loop, both of which iterate $$n$$ times, the order will be of $$n^2$$; but if we have a dependent nested loop such as:
input = an n-dimensional array
For i = 0; i < n; i++:
For j = 0; j < i; j++:
...
Am I correct in thinking this would always be $$O(n^2)$$ as in the worst case, this loop will always be $$n^2$$?
The lecturer gave us the forumla $$\frac{1}{2}n(n+1)$$ but has not explained in what context this is to be used for calculating the running time of a 2-nested loop with dependency. Is there a general way to calculate the running time of a dependent nested loop, like there is with standard nested loops?
• Dec 23 '18 at 19:33
• "an n-dimensional array"? Do you mean 2-dimensional array? Dec 23 '18 at 23:57
We can write the for loop as the sums;
$$\sum_{i=1}^{n} \sum_{j = 1}^{i} 1 = \sum_{i=1}^{n}i = \frac{n(n+1)}{2} \in\mathcal{O}(n^2) \, .$$
Note: set the starting values from $$i = 1$$ and $$j = 1$$, and increment the upper boundaries also. The calculation in the inner function is assumed as a constant operation and it will not change the calculation.
Well that all depends how you want to do it and how comfortable are you with the different ways. You could either make a tree and calculate the leave nodes for the time order. Or you could use the mathematical way as mentioned by others in the answer section. For example if you have a loop like-
for(i=1;i<n;i++)
{ for(j=1;j<n;j=j*2)
{anyfunc();
}}
the you would understand that the outer loop works from 1 to n and the inner loop would work from 1 to n also however the step in the inner loop is not a single increment but with every iteration it is becoming double of the previous value hence would execute for logn time. Thereby the net complexity would become O(n*logn).
Hope this helps, I would be more than happy to help you further.
You need to focus on how many times the instructions on the innermost loop will get executed. The outer loops are more like counters.
The inner-loop count will be as follows:
$$i = 0; j = \emptyset$$
$$i = 1; j = 0$$
$$i = 2; j = 0, 1$$
$$i = 3; j = 0, 1, 2$$
$$\dots$$
So you have a recurring sum of 1 for $$j$$ from $$0$$ to $$i-1$$, which can be mathematically expressed as:
$$\sum^n_{i = 1}\sum^{i-1}_{j = 0} 1 = \sum^n_{i = 1} i = \frac{n(n + 1)}{2}$$
which is still $$O(n^2)$$. | 2021-12-06T20:59:36 | {
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https://freedocs.mi.hdm-stuttgart.de/sd1SectFactorialDirect.html | Factorial, the direct way
No. 114
Q: Compute the factorial of a given integer value, for example: $4 ! = 4 × 3 × 2 × 1$ Implement the following method:/** * Computing the factorial of a given argument. * * @param n * Zero or any positive integer * * @return * The product 1 x 2 x 3 x ... x n or 1 in case of n == 0. In case of an * arithmetic overflow a value of {@link Long#MAX_VALUE} is being returned. */ static public long factorial(int n) { // TODO: implement me! } The method's signature looks slightly weird: It does expect an argument of type int but returns a long value. Explain the underlying ratio. Mind the above Javadoc™ passage concerning integer overflow related problems with respect to your own implementation. Provide adequate unit tests. Do not forget special values and handling of arithmetic overflow problems. A: Maven module source code available at sub directory P/Sd1/math/V0_7 below lecture notes' source code root, see hints regarding import. Online browsing of API and implementation. We address all three questions individually: Returning a long is sensible since even small argument values yield large factorials. long (currently!) is the best (largest) choice among all Java™ built-in integer types. Consider the following example code: public static void main(String[] args) { System.out.println("Max long value:" + Long.MAX_VALUE + "\n"); for (int i = 15; i < 23; i++) { System.out.println(i + ":" + factorial(i)); } } static public long factorial(int n) { long ret = 1; for (int i = n; 1 < i; i--) { ret *= i; } return ret; }This yields:Max long value:9223372036854775807 15:1307674368000 16:20922789888000 17:355687428096000 18:6402373705728000 19:121645100408832000 20:2432902008176640000 21:-4249290049419214848 22:-1250660718674968576So starting from $21 !$ we already see long overflow related errors. Thus allowing for even larger long arguments instead of int does not make sense at all. Since “21” is pretty small we might favour short (or even char) as argument type: static public long factorial(short n) { ... }This however is a bad idea: Even simple expressions would be flagged as compile time errors since both integer literals and arithmetic expressions in Java™ evaluate to the data type int: // Compile time error: // The method factorial(short) in the type // App is not applicable for the arguments (int) System.out.println(factorial(3));BTW: If we choose static public int factorial(short n) (int return type) the first overflow error happens already when trying to calculate $13 !$. We have to handle: Argument value 0. Overflow related results. Tests must address regular, special and overflow related argument values. | 2020-01-17T15:41:12 | {
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https://www.physicsforums.com/threads/solving-a-fractional-single-variable-inequality.463676/ | # Solving a Fractional, Single-Variable, Inequality
## Homework Statement
Solve the Inequality:
(3x-7)/(x+2)<1
## The Attempt at a Solution
Cross Multiply: x+2>3x-7
Simplify: 9>2x
Simplify More: 9/2>x
I put this as my answer but the answer is really (-2, 9/2)
Can someone explain to me why this is? I know you can't divide by 0 and this has something to do with the answer but that is as far as I got.
Thanks
Mentallic
Homework Helper
If you're trying to solve the inequality -x>1, then you would divide by -1 (or equivalently, multiply by -1) to give x<-1. Notice how we reversed the inequality sign because we divided/multiplied by a negative value. However, dividing by positive values doesn't reverse the inequality sign.
So what about when we divide or multiply through by a variable x? We don't know if x is positive or negative.
This is why we need to take two cases:
(3x-7)/(x+2)<1
1) Ok so we know that $$x\neq -2$$ so let's look at just $$x>-2$$. Multiplying through won't change the inequality sign because we are multiplying through by a positive value (since our assumption is x>-2 or x+2>0). Solve the inequality now, and remember to discard any values of x<-2 because this doesn't satisfy our original assumption.
2) Now multiply through, assuming x<-2.
3) Now that you have two sets of solutions, obviously both need to be valid so the answer will be the intersection of these two sets of solutions. Say if we had 0<x<5 and 2<x<10, then our answer will be 2<x<5
If you're trying to solve the inequality -x>1, then you would divide by -1 (or equivalently, multiply by -1) to give x<-1. Notice how we reversed the inequality sign because we divided/multiplied by a negative value. However, dividing by positive values doesn't reverse the inequality sign.
So what about when we divide or multiply through by a variable x? We don't know if x is positive or negative.
This is why we need to take two cases:
(3x-7)/(x+2)<1
1) Ok so we know that $$x\neq -2$$ so let's look at just $$x>-2$$. Multiplying through won't change the inequality sign because we are multiplying through by a positive value (since our assumption is x>-2 or x+2>0). Solve the inequality now, and remember to discard any values of x<-2 because this doesn't satisfy our original assumption.
2) Now multiply through, assuming x<-2.
3) Now that you have two sets of solutions, obviously both need to be valid so the answer will be the intersection of these two sets of solutions. Say if we had 0<x<5 and 2<x<10, then our answer will be 2<x<5
Thank you for your reply Mentallic. I don't quite understand your solution. You said "Multiplying through won't change the inequality sign because we are multiplying through by a positive value (since our assumption is x>-2 or x+2>0). Solve the inequality now, and remember to discard any values of x<-2 because this doesn't satisfy our original assumption." I don't understand what you mean by solve the inequality, I thought I already did that, what can I do differently? Also why should I discard the values of x<-2, they don't satisfy which inequality? x<9/2 or x>-2. I'm quite confused.
Thanks.
eumyang
Homework Helper
Thank you for your reply Mentallic. I don't quite understand your solution. You said "Multiplying through won't change the inequality sign because we are multiplying through by a positive value (since our assumption is x>-2 or x+2>0). Solve the inequality now, and remember to discard any values of x<-2 because this doesn't satisfy our original assumption." I don't understand what you mean by solve the inequality, I thought I already did that, what can I do differently? Also why should I discard the values of x<-2, they don't satisfy which inequality? x<9/2 or x>-2. I'm quite confused.
Thanks.
You sort of solved the inequality, but you started off making the assumption (whether or not you realized it) that x + 2 is positive, because you did not change the inequality sign.
You started with this:
$$\frac{3x-7}{x+2} < 1$$
I personally don't like cross multiplying with inequalities. Instead, I will multiply both sides by the denominator:
$$\frac{3x - 7}{x + 2} \cdot (x + 2) < 1 \cdot (x + 2)$$
This is the problem: we don't know yet what the sign of x + 2 is. If x + 2 is positive, then we need to assume that x > -2. We leave the inequality sign alone.
$$3x - 7 < x + 2$$
Solving for x, you get x < 9/2, as you said.
But wait! We assumed that x > -2 when we solved this. This means we have to throw out values less then or equal to -2. Test any of those values in the original inequality and you will end up with a false statement. So the solutions in this case is
(-2, 9/2)
Now consider the 2nd case, where x + 2 is negative. This means we assume that x < -2. Because we are multiplying both sides by a negative number, we must switch the inequality sign.
$$3x - 7 > x + 2$$
Solve for x, and compare this with what we assumed in this 2nd case.
You sort of solved the inequality, but you started off making the assumption (whether or not you realized it) that x + 2 is positive, because you did not change the inequality sign.
You started with this:
$$\frac{3x-7}{x+2} < 1$$
I personally don't like cross multiplying with inequalities. Instead, I will multiply both sides by the denominator:
$$\frac{3x - 7}{x + 2} \cdot (x + 2) < 1 \cdot (x + 2)$$
This is the problem: we don't know yet what the sign of x + 2 is. If x + 2 is positive, then we need to assume that x > -2. We leave the inequality sign alone.
$$3x - 7 < x + 2$$
Solving for x, you get x < 9/2, as you said.
But wait! We assumed that x > -2 when we solved this. This means we have to throw out values less then or equal to -2. Test any of those values in the original inequality and you will end up with a false statement. So the solutions in this case is
(-2, 9/2)
Now consider the 2nd case, where x + 2 is negative. This means we assume that x < -2. Because we are multiplying both sides by a negative number, we must switch the inequality sign.
$$3x - 7 > x + 2$$
Solve for x, and compare this with what we assumed in this 2nd case.
But although the 2nd case works mathematically my answer sheet says that the only answer for this problem is (-2,9/2). Due to the 2nd case x can be greater than 9/2. When we solve for x we get x>9/2. In the 2nd case the answer would be (9/2, -2), which doesn't make any sense. In other words, it seems that my problem sheet is asking me to ignore the 2nd case, which doesn't make sense with what you guys are saying.
Thanks.
Last edited:
eumyang
Homework Helper
But although the 2nd case works mathematically my answer sheet says that the only answer for this problem is (-2,9/2). Due to the 2nd case x can be greater than 9/2. When we solve for x we get x>9/2. In the 2nd case the answer would be (9/2, -2), which doesn't make any sense. In other words, it seems that my problem sheet is asking me to ignore the 2nd case, which doesn't make sense with what you guys are saying.
Thanks.
"Ignor[ing] the 2nd case" makes perfect sense. In the 2nd case, we make the assumption that x < -2. We solve the inequality (switching the inequality sign because x + 2 < 0) and we get the answer x > 9/2. Since this contradicts the assumption that x < -2, the assumption is invalid. So the only solution set is (-2, 9/2).
Mark44
Mentor
But although the 2nd case works mathematically my answer sheet says that the only answer for this problem is (-2,9/2).
You're ignoring the basic assumption in this case. For the second case, you end up with x > 9/2 AND you assumed that x + 2 < 0. This is equivalent to x < - 2 AND x > 9/2, which is a contradiction, since x can't simultaneously be smaller than -2 and larger than 9/2.
Due to the 2nd case x can be greater than 9/2. When we solve for x we get x>9/2. In the 2nd case the answer would be (9/2, -2), which doesn't make any sense. In other words, it seems that my problem sheet is asking me to ignore the 2nd case, which doesn't make sense with what you guys are saying.
Thanks.
The first case boils down to x + 2 > 0 AND x < 9/2, which is the same as saying -2 < x < 9/2, the same as your answer book. | 2022-05-29T04:55:07 | {
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http://math.stackexchange.com/questions/253879/evaluating-int-1-over1-sin2xdx | # Evaluating$\int {1\over1-\sin2x}dx$
$$\int {1\over1-\sin 2x}dx = \int {1\over \sin^2 x-2\sin x\cos x+\cos^2x}dx = \int {1\over (\sin x-\cos x)^2}dx$$
From here I get two different answers, depending on whether I factor out $\sin x$ or $\cos x$. Factoring out $\sin x$, this one is correct according to WolframAlpha:
$$=\int {1\over [\sin x(1-{\cos x\over \sin x})]^2}dx=\int {1\over \sin^2x(1-{\cos x\over \sin x})^2}dx = \int {1\over(1-\cot x)^2}d(1-\cot x)$$ $$={1\over \cot x-1}+C$$
But when I factor out $\cos x$:
$$=\int {1\over [\cos x({\sin x\over \cos x}-1)]^2}dx=\int {1\over \cos^2x({\sin x\over \cos x}-1)^2}dx = \int {1\over(\tan x-1)^2}d(\tan x-1)$$ $$={1\over 1-\tan x}+C$$
I bet it's just some stupid typo that I'm missing, I can't figure it out.
Thanks.
-
There is no typo in what you have done. Indeed, $$\frac{1}{\cot x-1}=\frac{1}{\frac{1}{\tan x}-1}=\frac{\tan x}{1-\tan x}=\frac{1}{1-\tan x}-1$$ You simply take $C_2-C_1=-1$ in your constants of integration
-
They are both correct. To see this, you can simply take the derivative of each antiderivative to get your original integral back. This shows that both primitives you found are equal, up to a constant.
$$\frac{d}{dx} \left(\frac{1}{\cot x - 1}\right) = \frac{1}{\left(\cos x - \sin x\right)^2}$$ $$\frac{d}{dx} \left(\frac{1}{\tan x - 1}\right) = \frac{1}{\left(\cos x - \sin x\right)^2}$$
\begin{eqnarray*} \left(\cos x - \sin x\right)^2 &=& (\cos x - \sin x)(\cos x - \sin x)\\ &=& \cos^2 x - 2 \sin x \cos x + \sin^2 x\\ &=& 1 - \sin 2x \end{eqnarray*}
In your first line, you have \begin{eqnarray*} (\sin x - \cos x)^2 &=& (\sin x - \cos x)(\sin x - \cos x)\\ &=& \sin^2 x - 2 \sin x \cos x + \cos^2 x\\ &=& 1-\sin 2x \end{eqnarray*}
As you can see, these are equivalent.
-
How about doing a u-sub 2x = t and then multpliy top and bottom with conjugate 1+sint, It will become VERY easy then.
-
There is no typo as in the step where you make $(sinx-cosx)^2$ it can be $(cosx-sinx)^2$ .so there is two answer possible.And one more point is the function which we try to integrate is always a result of any function's differentiation. so if any integration gives two answers means if we differentiate those answers that will give us same integrating function.
Differentiation of a function's integration give the same function.
- | 2016-07-28T16:26:21 | {
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https://www.physicsforums.com/threads/find-the-real-part.641892/ | # Find the real part
1. Oct 7, 2012
### utkarshakash
1. The problem statement, all variables and given/known data
If $\large α = e^{i\frac{8∏}{11}}$, then find $Re(α+α^{2}+α^{3}+α^{4}+α^{5})$
2. Relevant equations
3. The attempt at a solution
$\large e^{i\frac{8∏}{11}}+e^{i\frac{16∏}{11}}............+e^{i\frac{40∏}{11}}$
$cos \frac{8∏}{11}+isin \frac{8∏}{11}..........$
Since I am interested only in real part so now I have to find the value of
$cosθ+cos2θ........cos5θ$
where $θ= \frac{8∏}{11}$
I think some trigonometry must be applied since it seems to me sum of a trigonometrical series.
2. Oct 7, 2012
### ehild
Is not
$α+α^{2}+α^{3}+α^{4}+α^{5}$
a geometric series?
hild
3. Oct 7, 2012
### Ray Vickson
You have found the real part; it is a sum of 5 terms. What is wrong with that answer?
RGV
4. Oct 8, 2012
### utkarshakash
Hey I have found the answer but not completely. I have to find the value of cosθ+cos2θ.....
which I dont know how to solve
5. Oct 8, 2012
### Mentallic
Well as people have already mentioned, the answer IS
$\cos(8\pi/11)+\cos(16\pi/11)+...+\cos(40\pi/11)$
but that's messy, and this question has been cleverly constructed so that there is a nice answer.
$$\alpha+\alpha^2+...+\alpha^5$$
$$=1+\alpha+\alpha^2+...+\alpha^5-1$$
$$=\frac{1-\alpha^6}{1-\alpha}-1$$
Now, notice that
$$\alpha^6=e^{48\pi i/11}=e^{4\pi i/11}=\left(e^{8\pi i/11}\right)^{1/2}=\alpha^{1/2}$$
So we can now turn the expression into
$$=\frac{1-\alpha^{1/2}}{1-\alpha}-1$$
$$=\frac{1-\alpha^{1/2}}{(1-\alpha^{1/2})(1+\alpha^{1/2})}-1$$
I'm sure you can finish it off from here | 2017-10-24T10:14:02 | {
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https://scicomp.stackexchange.com/questions/29180/discretization-error-amplification-instead-of-stagnation-to-machine-precision/29190 | # Discretization Error amplification instead of stagnation to machine precision
I wrote a code on Python 2.7.5 to solve numerically the following differential equation.
• $\frac{\partial^2f}{\partial x^2}=-S$
• $S=\pi^{2}\sin(\pi x)$
S is chosen that way in order to have $f= \sin(\pi x)$ as the exact solution, that can be compared with the constructed solution.
The differential operator is discretized using a second-order central finite difference method.
When running the code, the order of magnitude of the evolution of the discretization error Err is correct (2nd order) for the first batch of points of N (10,20,40,80,100,800,2000,5000) (goes down to 10e-11).
But I found that for very small dx meaning, very big Nx (10000,50000,100000,600000,900000), Err doesn't stagnate at the machine precision error (as expected) but goes up for several magnitude orders, and even worse, the discretization error Err deteriorates (up to 10e-7).
Is it a problem in the precision/round managment of Python for very small numbers, or is it normal to have even worse Error for that amount of points ?
I tried to write the problem in different ways (Matrix, Loops, Matrix inversion technices) but the same phenomenon happened.
import numpy as np
import matplotlib.pyplot as plt
import scipy.sparse as sp
from scipy.sparse.linalg.dsolve import spsolve
Err=[]
N=np.array([10,20,40,80,100,800,2000,5000,10000,50000,100000,600000,900000])
#Loop on the number of points used for the discretization of the 1D domain
for Nx in N:
dx=(1.0)/Nx# fixed distance between two consecutive points in the domain
x = np.linspace(dx,1.0-dx,Nx-1)# points of the 1D domain
#contruction of the diff operator discretization Matrix
data = [np.ones(Nx-1), -2*np.ones(Nx-1), np.ones(Nx-1)]# Diagonal terms
offsets = np.array([-1, 0, 1])# Their positions
LAP = sp.dia_matrix((data, offsets), shape=(Nx-1, Nx-1))#Sparse matrix
S = np.pi**2*np.sin(np.pi*x)*dx**2# Second member
f = spsolve(LAP,-S)#Resolution of the matrix system
f_ana=np.sin(np.pi*x)#The exact solution
Err.append((np.absolute((f_ana-f))).max())
plt.figure()
plt.plot(N,Err,N,N**float(-2),'--')
plt.legend(['Err','Theoretical asymptotic behaviour'])
plt.xlabel('dx')
plt.ylabel('Err')
plt.xscale('log')
plt.yscale('log')
plt.show()
• For small $dx$ it is not supposed to go to machine precision. You will see round-off effects. I think that this question will help: scicomp.stackexchange.com/questions/28753/… – Anton Menshov Mar 28 '18 at 17:15
• Thanks for your reply, the first thing that comes to my mind is if one can see this same behaviour for the instationnary problem (adding the time derivative to previous equation), for a fixed dt and a very small dx ? – Chack.Flack Mar 29 '18 at 7:11
• I don't see a reason why it would not. The problem is very fundamental and lies in the finite-precision arithmetic. – Anton Menshov Mar 29 '18 at 14:25
Putting a more systematic point of view to the particular problems in the answer of H. Rittich, the matrices LAP are nearly singular for large N.
The eigensystem $$v_{k-1}-2v_k+v_{k+1}=\lambda v_k,~~v_0=v_{N+1}=0$$ has basis solutions $v_k=q^k-q^{-k}$ where $q^{2N+2}=1$ for $v_{N+1}=0$. Then the eigenvalue computes as $\lambda=q+q^{-1}-2$, which means $$q_m=\exp(i\frac{m\pi}{N+1}),\\ v_{m,k}=2\cos(\frac{km\pi}{N+1}),\\ \lambda_m=2(\cos(m\frac{\pi}{N+1})-1)=-4\sin^2(\frac{m\pi}{2(N+1)})$$ for $m=1,...,N$. Thus the ratio of largest to smallest eigenvalue, in other words the condition number of LAP, is $$\kappa({\sf LAP})=\frac{\sin^2(\frac{N\pi}{2(N+1)})}{\sin^2(\frac{\pi}{2(N+1)})}\approx \frac{4(N+1)^2}{\pi^2}.$$ Floating point noise gets magnified by that number in the relative error formula, see the general perturbation theory of linear systems. This already includes effects of catastrophic cancellation.
Yes. This behavior is to be expected and normal. When you are computing with a small value for $\mathrm dx$ then, to compute the difference quotient, you are subtracting two numbers that are nearly the same. In this situation, you observe a so called catastrophic cancellation, meaning nearly all significant digits are lost (see here).
@H.Rittich states one reason in their answer. Another reason is that when you go to more and more points, you have to do more and more computations, each of which adds its own amount of round-off error. Consequently, the smaller your $\delta x$ gets, the more computations and the more accumulated round-off -- it doesn't just stay constant, but actually grows. | 2020-11-28T06:39:01 | {
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http://mymathforum.com/calculus/336262-setting-up-integral.html | My Math Forum Setting up integral
Calculus Calculus Math Forum
September 30th, 2016, 12:20 PM #1
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Setting up integral
I'm trying to calculate the area between two parallel linear lines
f1=4-x
f2=5-x
Integrating between x= 0-5
But I only want the area between y=0 and y=5 (as constraints) all in the first quadrant.
Now these are easy equations to integrate but the biggest issue I am having is setting up the integral... I've done something like this:
[F(5) - F(f2)]dx - [F(5)-F(f1)]dx where definite integral range is 0-5
But I end up getting a negative value which makes no sense. The only thing I can conclude is the way I am setting up my integral which is incorrect. Anyone know how I can setup this integral given where I want my area?
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September 30th, 2016, 01:17 PM #2 Senior Member Joined: Sep 2015 From: USA Posts: 1,763 Thanks: 905 I would set this up as $\displaystyle{\int_0^5}f_2(x) - f_1(x)~dx$ $\displaystyle{\int_0^5}(5-x) - (4-x)~dx$ Basically you find the area under $f_2$ and subtract off the area under $f_1$ leaving the part you are interested in. $\displaystyle{\int_0^5}(5-x) - (4-x)~dx =$ $\displaystyle{\int_0^5}1~dx = 5$ btw your pic is a bit off. $f_1(0)=4,~f_2(0)=5$ Thanks from ihaque
September 30th, 2016, 01:25 PM #3
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Quote:
Originally Posted by romsek I would set this up as $\displaystyle{\int_0^5}f_2(x) - f_1(x)~dx$ $\displaystyle{\int_0^5}(5-x) - (4-x)~dx$ Basically you find the area under $f_2$ and subtract off the area under $f_1$ leaving the part you are interested in. $\displaystyle{\int_0^5}(5-x) - (4-x)~dx =$ $\displaystyle{\int_0^5}1~dx = 5$ btw your pic is a bit off. $f_1(0)=4,~f_2(0)=5$
Thanks for the reply but I'd like to constrain the area between y=0 to y=5 and from x=0 to x=5 but wouldn't the above solution not consider the constraint from y=0 to y=5?
*Sorry I should expand on the above... My f1 and f2 functions are very general and I'm working thru this on excel where f1 and f2 will change depending on m and b values input. But based on the m and b values, I'd like to constrain that area within a box of 5x5 if that makes sense?
Last edited by ihaque; September 30th, 2016 at 01:28 PM.
September 30th, 2016, 01:32 PM #4
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From: USA
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Quote:
Originally Posted by ihaque Thanks for the reply but I'd like to constrain the area between y=0 to y=5 and from x=0 to x=5 but wouldn't the above solution not consider the constraint from y=0 to y=5? *Sorry I should expand on the above... My f1 and f2 functions are very general and I'm working thru this on excel where f1 and f2 will change depending on m and b values input. But based on the m and b values, I'd like to constrain that area within a box of 5x5 if that makes sense?
Yes you're correct. My bad. The integral should be
$\displaystyle{\int_0^4}f_2-f_1~dx+\displaystyle{\int_4^5}f_2~dx = \dfrac 9 2$
September 30th, 2016, 01:45 PM #5
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Quote:
Originally Posted by romsek Yes you're correct. My bad. The integral should be $\displaystyle{\int_0^4}f_2-f_1~dx+\displaystyle{\int_4^5}f_2~dx = \dfrac 9 2$
Okay, for this example it would make sense, but hear me out:
Assume f1 as a linear function similar to above.
Assume f2 as a linear function similar to above.
And now let's add f3, which is equal to f3=5
If I want the definite integral from 0-5 in the first quadrant only given the above functions, is it fair to write it as the following:
(F2-F1)-(F2-F3)?
Last edited by skipjack; October 12th, 2016 at 12:47 PM.
September 30th, 2016, 02:01 PM #6
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Quote:
Originally Posted by ihaque Okay, for this example it would make sense, but hear me out: Assume f1 as a linear function similar to above. Assume f2 as a linear function similar to above. And now let's add f3, which is equal to f3=5 If I want the definite integral from 0-5 in the first quadrant only given the above functions, is it fair to write it as the following: (F2-F1)-(F2-F3)?
If you run this through you get $\dfrac {35}{2}$ which isn't the correct answer.
If I understand what you're trying to do, you want to find a way to express that triangular area of the integral of $f1$ that is below the x-axis, so you can subtract it off the simple integral of $f2-f1$
I don't see any really clever way of doing that that isn't equivalent to what I did treating the range $x \in (4,5]$ separately.
Last edited by skipjack; October 12th, 2016 at 12:48 PM.
September 30th, 2016, 05:55 PM #7 Global Moderator Joined: May 2007 Posts: 6,438 Thanks: 562 For x < 4, the integrand is f2-f1. For 4 < x < 5, the integrand is f2.
October 12th, 2016, 12:40 PM #8 Global Moderator Joined: Dec 2006 Posts: 18,595 Thanks: 1493 $\displaystyle \int_0^5\!f_2\,dx - \int_0^4\!f_1\,dx = \frac{25}{2} - 8 = \frac92$
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https://math.stackexchange.com/questions/1570233/why-exactly-does-a-function-need-to-be-continuous-on-a-closed-interval-for-the-i | # Why exactly does a function need to be continuous on a closed interval for the intermediate value theorem to apply?
I apologize if this question is too basic. The intermediate value theorem states that if $f$ is continuous on a closed interval $[a,b]$, then for every value $c$ between $f(a)$ and $f(b)$ there exists some $x \in (a,b)$ such that $f(x) = c$. This, or some very similar variant thereof, is how the intermediate value is usually presented in textbooks. What bugs me, however, is the condition that $f$ need be continuous on the closed interval $[a,b]$ rather than the less strict condition of only being continuous on the open interval $(a,b)$. To illustrate this point, consider $f:[-1,1] \rightarrow \mathbb{R}$ where $f(x)= e^x$. This is only continuous on the open interval $(-1,1)$ but surely the IVT applies to it. A less artificial example would be the inverse sine function. Would this reasoning not apply to all such (continuous/well-behaved) functions?
As an alternative, wouldn't the following definition from Proofwiki be superior (in that it is slightly more general)?
Let $I$ be a real interval. Let $a,b \in I$ such that $(a,b)$ is an open interval. Let $f:I \rightarrow \mathbb{R}$ be a real function continuous in $(a,b)$. Then for every value $c \in \mathbb{R}$ between $f(a)$ and $f(b)$ there exists some $x \in (a,b)$ such that $f(x) = c$
• The function $f:[-1,1] \to \mathbb R$ such that $f(x) = e^x$ is continuous on the closed interval $[-1,1]$. – littleO Dec 11 '15 at 1:51
• $\sin^{-1}$ is continuous on $[-1,1]$. – Milo Brandt Dec 11 '15 at 1:51
• @Milo Brandt Wait...there's no consideration of left-continuity and right-continuity? – YouKnowNothing Dec 11 '15 at 1:52
• @YouKnowNothing Not really; the definition of continuity in this context is $\lim_{x\rightarrow c}f(x)=f(c)$ and we only allow the $x$ in the limit to range in the domain of $f$. So, being near undefined regions does not fail to make $f$ continuous. In fact, we could consider a function defined only at a single point - and it would be continuous. Really, with continuity we have to think of more regions as causing more problems - so removing regions of the domains cannot make a function discontinuous. – Milo Brandt Dec 11 '15 at 1:56
• @Milo Brandt Oh, okay, then that's my main misapprehension. Weird how they don't mention these subtle points in calculus textbooks (or, at least mine) :P. – YouKnowNothing Dec 11 '15 at 2:01
The function
$$f(x) = \begin{cases}10 & x = -1\\ 0 & x\in (-1,1) \\ 20 & x=1\end{cases}$$ does not satisfy the IVT. There is no $x\in(-1,1)$ such that $f(x)$ is between $f(-1)$ and $f(1)$.
However, clearly $f$ is continuous on $(-1,1)$.
I think what you are trying to say is that requiring continuous on closed interval is a bit too much. But at least for the time being note that (classic) IVT does not even make sense if $f(a),f(b)$ is not defined.
The essence of IVT is the fact that the notion of connectedness is preserved under continuous mapping. Consider for example the fucntion
$$f(x)=1/x \;\; \forall x\in(0,1)$$
Classic IVT cannot possibly be applied here since $f(0)$ is not even defined. But it is still true that, for every $c\in (1,\infty)$ we can find some $x_0\in (0,1)$ such that $f(x_0)=c$.
This is because $f$ is continuous, and that $(0,1)$ is connected, so that $f((0,1))$ is necessarily connected.
• @fred When I said classic I was referring to the one requiring $f$ being continuous on $[a,b]$ - this obviously requires $f(a),f(b)$ to be defined. Unless I'm missing something, the one on ProofWiki as stated is not correct. 5xum's example is a direct counterexample. – user160738 Dec 11 '15 at 3:26
• I posted my comment to the incorrect answer. I apologize. – fred Dec 11 '15 at 3:32
I think 5xum's answer gives a good example. The general reason you need continuity on a closed interval is that you need to tie the values $f(a)$ and $f(b)$ to the values the function takes on the rest of the interval. If the function is not continuous at the end points then its value at the endpoints need have nothing to do with the values the function takes on the interior of the interval.
If you did want to change the IVT to work for an open interval you could use the following modification.
Let $f(x)$ be continuous on an open interval $(a,b)$ and let $c,d\in (a,b)$ such that $f(c)<f(d)$, then $\forall e\; (f(c)<e<f(d))$ there exists a $g\in (a,b)$ such that $f(g)=e$. | 2019-08-24T23:44:32 | {
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So, to convert a factor to a numeric with its original values intact, you need to either: Index the levels by the factor itself, and then to convert to numeric. 001) were found to be significant risk factors for PGH. L = [3,3,5,7] 2. The prime factorization of 4 is 2 x 2, and the prime factorization of 6 is 2 x 3. 02x - Lect 16 - Electromagnetic Induction, Faraday's Law, Lenz Law, SUPER DEMO - Duration: 51:24. Do not confuse the GCF with the Least Common Denominator (LCD) which is the smallest expression that all terms go into, rather than the greatest number the terms have in common. Enter a number and click Calculate. Contribute to hvy/integer-factorization development by creating an account on GitHub. If it is, the problem is solved. place it over the denominator 1) Canceling is dividing one factor of the numerator. These factors are the positive integers by which the number being factored can be divided to yield a positive integer result (though the concepts function correctly for zero and negative integers, the set of factors of zero has countably infinite members, a. A general number field sieve implementation. levels() returns the levels of a factor. Polynomial Factorization Calculator - Factor polynomials step-by-step This website uses cookies to ensure you get the best experience. These factors are the positive integers by which the number being factored can be divided to yield a positive integer result (though the concepts function correctly for zero and negative integers, the set of factors of zero has countably infinite members,. Many computational problems need arithmetic operations in polynomial finite rings over Z/nZ, where n is an integer (n>1). EASA update on impact of COVID-19 on its operations In response to the COVID-19 situation, EASA has taken measures to protect the health of its staff and limit the spread of the disease. In this case, the length of the input is $\log_2 n$. 19 Standardized Baselines. This book is about the theory and practice of integer factorization presented in a historic perspective. display only the factors on the front panel (no zeros). 717 91 11 | Fax: +41. The Elliptic Curve Method for Integer Factorization (ECM)¶ Sage includes GMP-ECM, which is a highly optimized implementation of Lenstra’s elliptic curve factorization method. It presents the latest findings of studies on the molecular structure and properties of proteins, macromolecular carbohydrates, glycoproteins, proteoglycans, lignins, biological poly-acids, and nucleic acids. The Metabolic Factor 10-Minute Meals $27. Philippine Daily Inquirer / 04:00 AM January 12, 2020. " Look back at your table, write the least common multiple of 4 and 6, LCM(4,6). Here, you can factor 20-digit numbers. The aim of this systematic review is to. As an example, the factors of 20 are 1,2,4,5, and 10. As you noticed, find_prime_factors worked, even though is_prime is broken. If the number is an integer, use that integer. 02x - Lect 16 - Electromagnetic Induction, Faraday's Law, Lenz Law, SUPER DEMO - Duration: 51:24. Editor-in-Chief. all the rational and irrational numbers). An updated version of this instructional video is available. The results detect if it is odd or even and figures out the factors of that number. Pierre Ronco (Paris, France), KI is one of the most cited journals in nephrology and widely regarded as the world's premier journal on the development and consequences of kidney disease. the function, or rule which produces the "greatest integer less than or equal to the number" operated upon, symbol [x] or sometimes [[x]]. gl() generates factors. And if the input number is 315, then output should be “3 3 5 7”. To perform the trial division algorithm, one simply checks whether s| N for s = 2,–, N. What is the smallest positive integer that has exactly eight factors?. We can write an odd composite number n = p ⋅ q as the difference Pollard's p − 1 method. Prime factorization. Conversion of units may generate extra multiplication operations; however, if the compiler performs constant folding [aho], these operations and their conversion factors can often be combined with other constants. This application will return a list of factors of a positive integer. Applicable countries/Region. The policy brief argues that there is a large overlap of activities between public health and primary care. ) of 6 and 10. The key objective of this journal is to promote interdisciplinary research from various regions of the globe. Take 18, for instance. This is the "least. factor() transforms a vector into a factor. Input value, specified as a real, nonnegative integer scalar. The program uses local storage to remember the progress of the factorization, so you can complete the factorization of a large number in several sessions. Prime Factorization is very important to people who try to make (or break) secret codes based on numbers. Real numbers are all the numbers which you will have come across (i. For example, you get 2 and 3 as a factor pair of 6. The proposal does not allow access to integer iterator objects such as would be created by xrange. So we're going to start with 75, and I'm going to do it using what we call a factorization tree. Multiply the elements of f to reproduce the input value. Updated 28 May 2019. integer factorization calculator) computes prime factors of a natural number or an expresssion involving + - * / ^ ! operators that evaluates to a natural number. mutual funds. Recently, it has been recognised that sex may contribute to a differential risk for developing sepsis and it remains uncertain if the prognosis of sepsis varies between the sexes. this question is on the CodeLab， my answer is:. If you want to learn more about factors, I recommend reading Amelia McNamara and Nicholas Horton’s paper, Wrangling categorical data in R. Improve your math knowledge with free questions in "Prime factorization" and thousands of other math skills. If you want to know more, the subject is "encryption" or "cryptography". It has also developed sophisticated tools to collects and analyse a vast array of safety data which allows to identify existing and emerging risks. Although the new syntax considerably simplifies integer for-loops, list comprehensions using the new syntax are not as simple. Delivery of full solutions. A composite number is a positive integer that can be formed by multiplying two smaller positive integers. instant atlas, software gis marketing tools, use gis mapping software to analyse and explore your gis maps and gis data with instantatlas software gis marketing tools, instant atlas for gis mapping professionals, download free demo. The East African Community (EAC) is a regional intergovernmental organisation of 6 Partner States: the Republics of Burundi, Kenya, Rwanda, South Sudan, the United Republic of Tanzania, and the Republic of Uganda, with its headquarters in Arusha, Tanzania. Download yafu for free. Information For. mutual funds. Number of Factors of a Large Integer January 25, 2008. Given an integer N, there is a simple way to find the total number of its factors. Math Expression Renderer, Plots, Unit Converter, Equation Solver, Complex Numbers, Calculation History. Notably, if an integer has only two factors, 1 and itself, then it is a prime number. Year 4 maths plan Factor pairs, integer scaling and correspondence problems. Prime numbers: Prime factors: Greatest common factor: Least common multiple. Our findings identify factors associated with a lower level of psychological impact and better mental health status that can be used to formulate psychological interventions to improve the mental health of vulnerable groups during the COVID-19 epidemic. These factors are the positive integers by which the number being factored can be divided to yield a positive integer result (though the concepts function correctly for zero and negative integers, the set of factors of zero has countably infinite members,. Development factors help by improving visual communication. This paper lays out some of the history discussed in stringsAsFactors: An unauthorized biography and stringsAsFactors = , and compares the tidy approaches to categorical data outlined in this book with base R methods. Since there are efficient classical algorithms to factorize pure prime powers (and of course to recognize a factor of 2), an efficient probabilistic algorithm for factorization can be found if the period of the modular exponentiation can be determined in polynomial time. Full View and History. Integer factorization Math 436, Number Theory II, Spring 2000 Integer factorization at Arizona Winter School 2006. The factorization method they give is quite slow, except for rare cases. Example:?- prime_factors(315, L). 69 (5-Year Impact Factor 0. Its 80+ indicators explore a broad vision of innovation, including political environment, education, infrastructure and business sophistication. Find the greatest common factor of two whole numbers less than or equal to 100. Quantum computers could tackle specialized computational problems such as integer factorization, understanding materials properties or optimization challenges much faster than conventional digital computers. Understand Vanguard's principles for investing success. Next try 4. On December 12, 2009 a small group of. This applet is able to factor a Gaussian integer as a product of Gaussian primes. In this case we also say that b is a divisor of a, and we use the notation b | a. For example, you get 2 and 3 as a factor pair of 6. They promote evidence-based health policymaking through a. labels is a vector of labels for the resulting factor levels. Here, you can factor 20-digit numbers. Manual for Ver. Languages: English Shqip العربية Bangla Български Català 中文 ( 正體字 , 简化字 (1) , 简化字 (2)) Hrvatski Čeština Dansk Nederlands Esperanto Eesti فارسی Suomi Français Deutsch Ελληνικά ગુજરાતી עברית. Full View and History. Input value, specified as a real, nonnegative integer scalar. Factors influencing health status or contact with health services Codes for special purposes Supplementary Chapter Traditional Medicine Conditions - Module I Supplementary section for functioning assessment Extension Codes. Factor Variables Ordered variables What factor variables are. integer(factor(a)) Well, probably basic to many, it was new to me 🙂 Now, wondering if it is equally easy to convert it into a multi-column matrix where each column indicates class membership (thus, resulting in three columns for the above…). The American Concrete Institute (ACI) is a leading authority and resource worldwide for the development and distribution of consensus-based standards, technical resources, educational programs, certification programs, and proven expertise for individuals and organizations involved in concrete design, construction, and materials, who share a commitment to pursuing the best use of concrete. Vector f is of the same data type as n. 2 x 30 = 60, which means we add these to our factor list. What are factor pairs, and how do I find the factor pairs of a number like 66? Finding Integers Given LCM and GCD [11/05/2002] The least common multiple of two positive integers is 144, and the greatest common divisor is 2. And if the input number is 315, then output should be “3 3 5 7”. To perform the trial division algorithm, one simply checks whether s| N for s = 2,–, N. Upsampling/downsampling an image consisting of just one uniform value, followed by the opposite operation, should result in an image consisting of the same. uk) have mapped and connected the curriculum in each year group term by term. The set of all polynomials with coefficients in F is denoted by F[x]. Number factorizer (a. You want to rename the levels in a factor. integer ("5. Since the product of these factors has to be a negative number, we need one positive factor and one negative factor. Kidney International (KI) is the official journal of the International Society of Nephrology. Review articles that encompass these subjects are also welcome. An integer that divides into another integer exactly is called a Factor. In fact, each factor of a number is built up of one or more of the number's other factors. This thesis serves as a source for the history and development of integer factorization algorithms through time from trial division to the number field sieve. Factor: A factor is a financial intermediary that purchases receivables from a company. Factorizing integers allows us to better understand the property of that number than you would if you simply wrote the number as it is. What are the possible integer roots of x 3 − 4x 2. mutual funds. Integer factorization Math 436, Number Theory II, Spring 2000 Integer factorization at Arizona Winter School 2006. bi_factor - prime-factor large numbers. Integer factorization 21 Factors of the Year 2013 and 2014. It proceeds by rho (x 2 +1, x 2-1, x 2 +3), p-1 and PPMPQS. Prime factorization of an Integer using Pollard Rho Algorithm - pollard_rho. Both computations were performed with the Number Field Sieve algorithm, using the open-source CADO-NFS software [4]. 12 is a multiple of both 3 and 4. levels() returns the levels of a factor. Factor Variables Ordered variables What factor variables are. Learn how to subtract integers the fun way with Fruit Shoot Integer Subtraction math game. This is the home page for the electronic Journal of Integer Sequences, ISSN 1530-7638. Lectures by Walter Lewin. The results should be displayed in an array. Cryptography is the study of secret codes. public static int[] factorization (int n) Factors n and return the factors in an array. If it is, the problem is solved. Factor an Integer. The aim of this systematic review is to. If x is a symbolic expression, factor returns the subexpressions that are factors of x. Now take 3. The prime factorization of 4 is 2 x 2, and the prime factorization of 6 is 2 x 3. Definition of integer factorization in the Definitions. Download yafu for free. Calcating the factors of 60 is straightforward. But if all the prime factors are of the form 4l+1, then the product of all the prime factors would be of the form 4l+1. As an example, the factors of 20 are 1,2,4,5, and 10. Applicable countries/Region. Want to give us feedback on the game? Find our contact info here. integer(factor(a)) Well, probably basic to many, it was new to me 🙂 Now, wondering if it is equally easy to convert it into a multi-column matrix where each column indicates class membership (thus, resulting in three columns for the above…). It proceeds by rho (x 2 +1, x 2-1, x 2 +3), p-1 and PPMPQS. Find the factors of an integer posted by Justin Musgrove on 20 April 2016. That is because factoring very large numbers is very hard, and can take computers a long time to do. Factorization of integers up to 945 is demonstrated with this rudimentary asynchronous probabilistic computer using eight correlated p-bits, and the results show good agreement with theoretical. Fast integer factorization in Python. Applicable countries/Region. Any common multiple of 4 and 6 will need enough prime factors to make each of these numbers. The aim of this systematic review is to. Now take 3. An important concept needed for Gaussian integer factorization is the norm. For any value of n, whether positive, negative, integer or non-integer, the value of the nth power of a binomial is given by: There are many binomial expansion applications in physics. An updated version of this instructional video is available. It is the first description of the number field sieve from an algorithmic point of view making it available to computer scientists for implementation. The Republic of Uganda. De nition 1. The program uses local storage to remember the progress of the factorization, so you can complete the factorization of a large number in several sessions. Take 18, for instance. Therefore 1, 2, 3, and 6 are all factors of six. Factors definition, one of the elements contributing to a particular result or situation: Poverty is only one of the factors in crime. Math Expression Renderer, Plots, Unit Converter, Equation Solver, Complex Numbers, Calculation History. Number factorizer (a. k is a integer giving the number of replications. /* C Program to Find Prime Factors of a Number using Functions */ #include void Find_Prime. Factor: A factor is a financial intermediary that purchases receivables from a company. If you use as. All real numbers can be written in decimal form (such as 3. Eucerin is now one of the world’s most trusted dermo-cosmetic brands, endorsed by dermatologists and pharmacists and recommended to keep skin healthy and beautiful. Associate the new list with the variable factors. Rather unfortunately users often make use of the implementation in order to make some calculations easier. This includes the cancellation of all travel and external meetings, which means that scheduled site visits and audits, for example, cannot take place as planned. plz use python!! thanks. This is also known as prime decomposition. Discrete Logarithm vs Integer Factorization. No promises, but, the site will try everything it has. Fractions - Adding – Illustrates what it means to find a common denominator and combine. 5 after numbers have been randomly chosen. In this lesson you will learn how to find all the factor pairs of a number by. Solve advanced problems in Physics, Mathematics and Engineering. 3) = 2 + 2 + 3 = 7 and if n is prime then c(n)=n. These are distractions from the main ideas. The integer factorization method based on the NMR platform is not applicable to all integers and is not universal and scalable. Prime Factorization A prime is an integer greater than one those only positive divisors are one and itself. To find the Least Common Multiple, you need to:. Write a Java program to accepts an integer and count the factors of the number. If the leading coefficient is 1, then f(x) is said to be monic. In other words, this application can also be used to determine whether a positive integer is a prime number by counting the number of factors in …. labels is a vector of labels for the resulting factor levels. Recently, it has been recognised that sex may contribute to a differential risk for developing sepsis and it remains uncertain if the prognosis of sepsis varies between the sexes. This is the most basic algorithm to find a prime factorization. is divisible by §i: a+bi = i(b¡ai) = ¡i(¡b+ai): Of course any integer is also divisible by ¡1, but this is more easily ignored in Zjust by restricting to positive integers. Writing numbers as the product of prime is called prime factorization. Compute the factors of a positive integer. Input value, specified as a real, nonnegative integer scalar. Full View and History. The integer in the top right hand corner grows with the number of factors/multiples you have in a row. Its 80+ indicators explore a broad vision of innovation, including political environment, education, infrastructure and business sophistication. Factors Calculator. e] mod N, where N is the product of two large prime numbers of same length, e is the public key chosen such that it is relatively prime with the. The world´s gold standard in professional UX training – 183 CUA courses & 297 classes taught in 2017. Discover the new International Journal of Epidemiology special issue highlighting Mendelian randomization research. The automation within YAFU is state-of-the-art, combining factorization algorithms in an intelligent and adaptive methodology that minimizes the time to find the. For example if n = 12 then c(n)=c(12) = c(2. Super Bonus - 4 Personal Video Coaching Calls From Dr. UX engineering services. Determine whether a given whole number in the range 1-100 is prime or composite. Manual for Ver. The greatest integer function is a piece-wise defined function. Special theme issue: Mendelian randomization. 3 x 4 = 12. The monitoring of safety trends and indicators. Factor Tree – Factor numbers using a tree diagram. Prequalifying finished pharmaceutical products, active pharmaceutical ingredients and quality control laboratories. From Wikiversity factoring algorithm's running time depends on the properties of the number to be factored or on one of its unknown factors. a prime is an integer greater than one that has no positive factors other than one and itself. Occasionally a GMAT question will ask you for the number of factors of a certain integer. n is the number of levels, k the number of repetition of each factor and length the total length of the factor. Eucerin is now one of the world’s most trusted dermo-cosmetic brands, endorsed by dermatologists and pharmacists and recommended to keep skin healthy and beautiful. This detailed sample plan focuses on Factor Pairs, Integer Scaling and Correspondence Problems in Year 4. Development factors help by improving visual communication. If a number is divisible by 2, for instance, 2 will be a factor of many of the number's factors. The policy brief argues that there is a large overlap of activities between public health and primary care. 14 seconds to factor a 67 bit number; well, I just tried it using a more conventional algorithm, and it took 6msec; yes, that's 100,000 times as fast. EASA update on impact of COVID-19 on its operations In response to the COVID-19 situation, EASA has taken measures to protect the health of its staff and limit the spread of the disease. The higher the P value, the better the "fit" of the model to explain the relationship of the independent variables (risk factors) to the dependent variable (overweight or. In number theory, integer factorization or prime factorization is the decomposition of a composite number into smaller non-trivial divisors, which when multiplied together equal the original integer. This yields a range of -32,768 to 32,767 (minimum value of -2^15 and a maximum value of (2^15) - 1). The number has no factors. If x is an integer, factor returns the prime factorization of x. Usually that integer will be large (441 is the object of one question that comes to mind), since the question wouldn't be very challenging otherwise. All real numbers can be written in decimal form (such as 3. BigInteger provides analogues to all of Java's primitive integer operators, and all relevant methods from java. The monitoring of safety trends and indicators. Note that all factors are prime numbers. The following applet computes the prime factors of an integer, as well as the Euler phi function φ and the Carmichael function λ, here denoted by f(n) and l(n). You have saved this instructional video! Here's where you can access your saved items. This is defined as: N(a+bi) = a 2 + b 2. FACILITATING ACCESS TO QUALITY MEDICINES FOR ALL WHO NEED THEM. Find the factors of an integer posted by Justin Musgrove on 20 April 2016. 2 Learning more. Integer factoring with the numbers represented in binary is (as far as we know) not in P. YAFU (with assistance from other free software) uses the most powerful modern algorithms (and implementations of them) to factor input integers in a completely automated way. People have written computer programs looking for the highest prime and all of that. International Journal of Biological & Pharmaceutical Research (IJBPR) is a bi-annual international journal publishing the finest peer-reviewed research in the field of Pharmaceutical Research on the basis of its originality, importance, disciplinary interest, elegance, surprising discussions and conclusions and serve as a means for scientific information exchange in the international. Int’l stint a factor in Ildefonso’s Ateneo return. Prime numbers: Prime factors: Greatest common factor: Least common multiple. However, trial division may take more than O(√. This yields a range of -32,768 to 32,767 (minimum value of -2^15 and a maximum value of (2^15) - 1). The capacity is the number of buckets in the hash table, and the initial capacity is simply the capacity at the time the hash table is created. Integers are your primary data-type for number storage. An international multi-disciplinary journal which is a joint initiative between the International Osteoporosis Foundation and the National Osteoporosis Foundation of the USA, Osteoporosis International provides a forum for the communication and exchange of current ideas concerning the diagnosis, prevention, treatment and management of. If the integer you supply is prime and more than 14 digits, it may take so long to factor that your connection times out. Year 4 maths plan Factor pairs, integer scaling and correspondence problems. Full View and History. Factor: An Integer Factorization Program for the IBM PC Richard P. BigInteger provides analogues to all of Java's primitive integer operators, and all relevant methods from java. The policy brief argues that there is a large overlap of activities between public health and primary care. The certification employers ask for. 14 seconds to factor a 67 bit number; well, I just tried it using a more conventional algorithm, and it took 6msec; yes, that's 100,000 times as fast. HIV/AIDS, TB, malaria, Hepatitis B & C, neglected tropical diseases, diarrhoea, influenza and for reproductive health. Factor Tree – Factor numbers using a tree diagram. So now that we know what a prime is, a prime factorization is breaking up a number, like 75, into a product of prime numbers. Share on: Twitter Facebook Google+. Im new to labview and i need help with P6. Upsampling an image containing arbitrary random values by an integer factor, then downsampling by the same integer factor, should result in the same image with minimal change numerically. We studied nonunique factorization in the ring of integer-valued polynomials by examining the. The prime factorization of an integer is the multiset of primes those product is the integer. But if all the prime factors are of the form 4l+1, then the product of all the prime factors would be of the form 4l+1. Since there are efficient classical algorithms to factorize pure prime powers (and of course to recognize a factor of 2), an efficient probabilistic algorithm for factorization can be found if the period of the modular exponentiation can be determined in polynomial time. If the number is not an integer, use the next smaller integer. , Springer, 1993. The find_factors function will find the Factors of a number, Find_Prime will check whether the factor is prime or not. Therefore 1, 2, 5, and 10 are all factors of ten. All factors are arranged in non-descending order in the returned array. 2 Learning more. Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode. They vary quite a bit in sophistication and complexity. 12 is a multiple of both 3 and 4. Eucerin is now one of the world’s most trusted dermo-cosmetic brands, endorsed by dermatologists and pharmacists and recommended to keep skin healthy and beautiful. Minimum Factorization Average Rating: 5 (13 votes) Given a positive integer a, find the smallest positive integer b whose multiplication of each digit equals to a. However, the intake of vegetarian/nonvegetarian food ( P = 0. You'll gain access to interventions, extensions, task implementation guides, and more for this instructional video. also, does one of the lizzards fart during the dance??]]> https://singsnap. Integer definition is - any of the natural numbers, the negatives of these numbers, or zero. Factors definition, one of the elements contributing to a particular result or situation: Poverty is only one of the factors in crime. The Republic of Uganda. Compute the factors of a positive integer. Factors influencing health status or contact with health services Codes for special purposes Supplementary Chapter Traditional Medicine Conditions - Module I Supplementary section for functioning assessment Extension Codes. We expect the pandemic to. Find the common multiples of the numbers. This study aims to identify which therapeutic factors supported members in their commitment to a group for abusive men. Email report - copy and paste link into email. Find the greatest common factor. But if all the prime factors are of the form 4l+1, then the product of all the prime factors would be of the form 4l+1. The concept behind this method is that the prime factorization of a number determines all of its factors. The number is divided by i and if the remainder is 0, then i is a factor of “num” and is printed. 2 Integer factorization and related notions actoringF integers is an old and well-known problem; we recall it here for completeness. a prime is an integer greater than one that has no positive factors other than one and itself. Write a Java program to accepts an integer and count the factors of the number. The Factorization class provides a structure for holding quite general lists of objects with integer multiplicities. Theorem I1. Do not confuse the GCF with the Least Common Denominator (LCD) which is the smallest expression that all terms go into, rather than the greatest number the terms have in common. It is called from the main() function with one parameter i. Our activities touch on operations, service provision, concept development, research, Europe-wide project implementation, performance improvements, coordination with key aviation players at various levels as well as providing support to the future evolution and strategic orientations of aviation. While quantum computers have been capable of this scale of integer factorization for years, the MRAM-derived p-bit design can be operated at room temperature, while quantum computers rely on. Elliptic curve factorization. Review articles that encompass these subjects are also welcome. Fermat's factorization method. A nonzero whole number has only a finite number of factors, so it has a greatest factor. In the above program, the function factors() finds all the factors of “num”. The Republic of Uganda. The aim of this systematic review is to. These may hold the results of an arithmetic or algebraic factorization, where the objects may be primes or irreducible polynomials and the multiplicities are the (non-zero) exponents in the factorization. For mapping purposes, the map shows identical values for Sudan and South Sudan. The proposal does not allow access to integer iterator objects such as would be created by xrange. It describes about twenty algorithms for factoring and a dozen other number theory algorithms that support the factoring algorithms. For any value of n, whether positive, negative, integer or non-integer, the value of the nth power of a binomial is given by: There are many binomial expansion applications in physics. Special theme issue: Mendelian randomization. International Journal of Microbiology publishes papers on microorganisms and their interaction with hosts and the environment. >>View Expired Standardized Baselines. Prime Factorization is very important to people who try to make (or break) secret codes based on numbers. The major factor driving the health and fitness services market growth is the shifting preference of the people towards living a healthy lifestyle due to the risks associated with unhealthy living. 717 91 11 | Fax: +41. For first-time attendees, CHI is a place where researchers and practitioners gather from across the world to. International Journal of Molecular Sciences (ISSN 1422-0067; CODEN: IJMCFK; ISSN 1661-6596 for printed edition) is an international peer-reviewed open access journal providing an advanced forum for biochemistry, molecular and cell biology, molecular biophysics, molecular medicine, and all aspects of molecular research in chemistry, and is published semi-monthly online by MDPI. Although finding the prime factors of a positive integer can be a multi-step procedure in some cases, determining the prime factors of the positive composite integer 30 can be easily accomplished by following two of the basic rules of factorization. Consider a problem that asks you to find the factorization of integer g(-231 < g <231) in the form g = f1 x f2 x … x fn or g = -1 x f1 x f2 x … x fn where fi is a prime greater than 1 and fi ≤ fj for i < j. Do not confuse the GCF with the Least Common Denominator (LCD) which is the smallest expression that all terms go into, rather than the greatest number the terms have in common. In other words, if any integer is divided by another integer without leaving any remainder, then the latter is the factor of the former. An international multi-disciplinary journal which is a joint initiative between the International Osteoporosis Foundation and the National Osteoporosis Foundation of the USA, Osteoporosis International provides a forum for the communication and exchange of current ideas concerning the diagnosis, prevention, treatment and management of. 4 NCD associated risk factors are largely modifiable. There are three factors that should be considered for the design of a successful user interface; development factors, visability factors and acceptance factors. 14) # coerce a numeric value. In the second terms of the binomials, we need factors of -6. Read the Special Issue. Usually that integer will be large (441 is the object of one question that comes to mind), since the question wouldn't be very challenging otherwise. Our findings identify factors associated with a lower level of psychological impact and better mental health status that can be used to formulate psychological interventions to improve the mental health of vulnerable groups during the COVID-19 epidemic. Thus, 586390350 has factors 2, 3, 5, 5,, 7, 7, 13, 17, 19 and 19. The following applet computes the prime factors of an integer, as well as the Euler phi function φ and the Carmichael function λ, here denoted by f(n) and l(n). F is the largest of all the common factors. Determine whether a given whole number in the range 1-100 is prime or composite. An updated version of this instructional video is available. A general number field sieve implementation. If it is, the problem is solved. This applet is able to factor a Gaussian integer as a product of Gaussian primes. 19 Standardized Baselines. Fractions - Adding – Illustrates what it means to find a common denominator and combine. This book is about the theory and practice of integer factorization presented in a historic perspective. Fuel switch, technology switch and methane destruction in the charcoal sector of Uganda (version 01. In other words, this application can also be used to determine whether a positive integer is a prime number by counting the number of factors in …. However, this is an important problem on which it relies the security of many real world cryptographic systems. Bowden$197. The common factors or of 12 and 18 are 1, 2, 3 and 6. There are three factors that should be considered for the design of a successful user interface; development factors, visability factors and acceptance factors. The results should be displayed in an array. And if the input number is 315, then output should be “3 3 5 7”. com/2008/02/04/51/feed/ 0. In this case, the length of the input is $\log_2 n$. Associate the new list with the variable factors. ICAO audits the implementation of its Standard, Recommended Practices and Procedures through its Universal Safety Oversight Audit Programme. Writing numbers as the product of prime is called prime factorization. In the above program, the function factors() finds all the factors of “num”. Integer vectors exist so that data can be passed to C or Fortran code which expects them, and so that (small) integer data can be represented exactly and compactly. The algorithm has 2 purposes: Finding a prime factor, or finding if an integer is a prime. Our expertise and energy is focussed on giving you the very best skin care products available – our commitment to research, development and testing delivers innovative and. Automated integer factorization. The set of all polynomials with coefficients in F is denoted by F[x]. breaking down a (composite) number into its prime factors is one such task. Compute the factors of a positive integer. 14 seconds to factor a 67 bit number; well, I just tried it using a more conventional algorithm, and it took 6msec; yes, that's 100,000 times as fast. Upsampling/downsampling an image consisting of just one uniform value, followed by the opposite operation, should result in an image consisting of the same. 3 and 4 are both factors of 12. If the number is an integer, use that integer. A factor is a number that divides evenly into another number. The certification employers ask for. Note that there is no restriction that the first letter entered must be a capital letter. Prove that the quadratic integer ring $\Z[\sqrt{-5}]$ is not a Unique Factorization Domain (UFD). In this lesson you will learn how to find all the factor pairs of a number by. Embed report - copy and paste html into web page. The Elliptic Curve Method for Integer Factorization (ECM)¶ Sage includes GMP-ECM, which is a highly optimized implementation of Lenstra’s elliptic curve factorization method. The prime factorization of 4 is 2 x 2, and the prime factorization of 6 is 2 x 3. ) of 6 and 10. Hypertension alone is the main risk factor for developing ischemic heart disease, stroke, heart and renal failures and eripheral vascular diseases. levels() returns the levels of a factor. $\endgroup$ – Moritz Aug 17 '10 at 16:12 4 $\begingroup$ I believe Moritz's comment here is insightful, txwikinger. In rare worst-case scenarios (for some large semiprimes with 10-digit factors) factorizations might take a couple of minutes. 5 after numbers have been randomly chosen. integer factorization problem [1]. Given a positive and squarefree discriminant D {\displaystyle \scriptstyle D\,} , the imaginary quadratic integer ring O Q ( − D ) {\displaystyle \scriptstyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-D}})}\,} has a. Find the Number of Factors. Eat factors of the given multiple. It was later declared as a. Doing this quickly has applications in cryptography. It is the first time that two records for integer factorization and discrete logarithm are broken together, moreover with the same hardware and software. The factors of an integer are those numbers that evenly divide into the integer. Discrete Logarithm vs Integer Factorization. The world´s gold standard in professional UX training – 183 CUA courses & 297 classes taught in 2017. The policy brief argues that there is a large overlap of activities between public health and primary care. This application factors numbers or numeric expressions using two fast algorithms: the Elliptic Curve Method (ECM) and the Self-Initializing Quadratic Sieve (SIQS). Year 4 maths plan Factor pairs, integer scaling and correspondence problems. The results detect if it is odd or even and figures out the factors of that number. No promises, but, the site will try everything it has. Healthy diet, physical activity and avoiding tobacco use can prevent or delay type 2 diabetes. The difficulty depends on both the size and form of the number and its prime factors ; it is currently very difficult to factorize large semiprimes (and, indeed, most numbers which have no small factors). The Elliptic Curve Method for Integer Factorization (ECM)¶ Sage includes GMP-ECM, which is a highly optimized implementation of Lenstra’s elliptic curve factorization method. Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums. Additionally, BigInteger provides operations for modular arithmetic, GCD calculation, primality testing, prime generation, bit manipulation, and a few other miscellaneous operations. Integer factorization is the process of determining which prime numbers divide a given positive integer. The Republic of Uganda. This is also known as prime decomposition. integer ("5. Understand Vanguard's principles for investing success. is divisible by §i: a+bi = i(b¡ai) = ¡i(¡b+ai): Of course any integer is also divisible by ¡1, but this is more easily ignored in Zjust by restricting to positive integers. Try to divide 2 (the 1st and only even prime) into the number as many times as possible without leaving a remainder any time. Human programmers usually write programs in such a way that intermediate results have reasonable units and reasonable numeric values. The world´s gold standard in professional UX training – 183 CUA courses & 297 classes taught in 2017. Among the biodemographic factors, a higher age group of 25–30 years (P = 0. Quantum computers could tackle specialized computational problems such as integer factorization, understanding materials properties or optimization challenges much faster than conventional digital computers. We expect the pandemic to. In modern browsers, this calculator does most factorizations within a second. Fraction Pieces – Work with parts and wholes to learn about fractions. Math Expression Renderer, Plots, Unit Converter, Equation Solver, Complex Numbers, Calculation History. This application factors numbers or numeric expressions using two fast algorithms: the Elliptic Curve Method (ECM) and the Self-Initializing Quadratic Sieve (SIQS). One of three key journals (along with IJETM and IJGEnvI ) which together offer complete coverage of key environmental issues, it addresses medium-term challenges involving scientific prediction, modelling and. The website for the math factor, a podcast on mathematics, logic and puzzles, by Chaim Goodman-Strauss and Kyle Kellams, airing weekly on KUAF 91. The integer factorization problem is a wellknown topic of research within both - academia and industry. Ask Question Asked 3 years, 10 months ago. And here is another thing:. Welcome to CHI 2018 Palais des Congrès de MontréalThe ACM CHI Conference on Human Factors in Computing Systems is the premier international conference of Human-Computer Interaction. Here, you can factor 20-digit numbers. Understanding the factors affecting microbial degradation is of great research interest in present scenario. What are the possible integer roots of x 3 − 4x 2. It has also developed sophisticated tools to collects and analyse a vast array of safety data which allows to identify existing and emerging risks. But as stated, it's all answered on the wikipedia page (btw, first hit on google for "integer factorization"). Volume 48 Issue 6 December 2019. 14 seconds to factor a 67 bit number; well, I just tried it using a more conventional algorithm, and it took 6msec; yes, that's 100,000 times as fast.
8tln80zr87kk1ay, 5ihem9l1e8wqbuq, tgwrfhaau5tm4, dlrku1ql2d9, p7g4crpkzfvry4, lmsrmohg632, 8b9m2b3illq2m6s, jeyz1p63ohbs, l78yqe7z37gel7v, x0h31ml50y, eu56hdyh71r2, h506v66vkb, zzpi05whd3erd, aakosipx4obl, 0x4kdd1p1su714w, pnj638jlb45l2c, oeo0b73r3y5, 62iw861dwyb, vlgt6ekevdac, q6sdcc525e4ih9, 113rdkooreaxo, wdocfrcxyjsa, lvykhlnnbr2k, 1onqp2grj12i, a938rckp2mf, ttbw7wmfjk, 6jszc00hwfj, 225ou5iz5h0b, z9rowk45cc | 2020-06-05T22:03:02 | {
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https://www.shaalaa.com/question-bank-solutions/a-random-variable-x-has-the-following-probability-distribution-x-0-1-2-3-4-5-6-7-p-x-0-k-2k-2k-3k-k2-2k2-7k2-k-determine-i-k-ii-p-x-3-iii-p-x-4-probability-distribution-of-discrete-random-variables_152717 | # A random variable X has the following probability distribution : X 0 1 2 3 4 5 6 7 P(X) 0 k 2k 2k 3k k2 2k2 7k2 + k Determine :(i) k(ii) P(X < 3)(iii) P( X > 4) - Mathematics and Statistics
Sum
A random variable X has the following probability distribution :
X 0 1 2 3 4 5 6 7 P(X) 0 k 2k 2k 3k k2 2k2 7k2 + k
Determine :
(i) k
(ii) P(X < 3)
(iii) P( X > 4)
#### Solution 1
(i) Since P(x) is a probability distribution of x,
Sigma_(x = 0)^7 P(x) = 1
∴ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1
∴ 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
∴ 10k2 + 9k − 1 = 0
∴ 10k2 + 10k − k − 1 = 0
∴ 10k (k + 1) − 1 (k + 1) = 0
∴ (k + 1) (10k − 1) = 0
∴ 10k − 1 = 0
∴ 10k − 1 = 0 ........( k ≠ - 1)
∴ k =1/10
(ii) P(X < 3) = P(0) + P(1) + P(2)
= 0 + k + 2k
= 3k
= 3 (1/ 10)
= 3/10.
(ii) P(0 < X < 3) = + P(1) + P(2)
= k + 2k
= 3k
= 3(1/ 10)
= 3/ 10.
#### Solution 2
i. The table gives a probability distribution and therefore sum_("i" = 1)^8 "P"_"i" = 1
∴ 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
∴ 10k2 + 9k – 1 = 0
∴ 10k2 + 10k – k – 1 = 0
∴ 10k(k + 1) – 1(k + 1) = 0
∴ (10k – 1)(k + 1) = 0
∴ k = 1/10 or k = –1
But k cannot be negative
∴ k = 1/10
ii. P(X < 3)
= P(X = 0 or X = 1 or X = 2)
= P(X = 0) + P(X = 1) + P(X = 2)
= 0 + k + 2k
= 3k
= 3/10
iii. P(X > 4)
= P(X = 5 or X = 6 or X = 7)
= P(X = 5) + P(X = 6) + P(X = 7)
= k2 + 2k2 + 7k2 + k
= 10k2 + k
= 10(1/10)^2 + 1/10
= 1/10 + 1/10
= 1/5
Concept: Probability Distribution of Discrete Random Variables
Is there an error in this question or solution? | 2022-10-05T03:27:40 | {
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http://mathhelpforum.com/number-theory/72381-divisible-16-a.html | # Math Help - Divisible by 16
1. ## Divisible by 16
Prove that if m and n are odd integers, then $m^3n^2+m^2n^3-m-n$ is divisible by 16.
2. Originally Posted by Sigyn3
Prove that if m and n are odd integers, then $m^3n^2+m^2n^3-m-n$ is divisible by 16.
this seems like a straight forward computation will work
let m = 2a + 1 and let n = 2b + 1
where a and b are integers
expand the expression and simplify. your goal is to show that the expression can be written as 16k where k is an integer
3. Hello, Sigyn3!
Prove that, if $m$ and $n$ are odd integers, then $m^3n^2+m^2n^3-m-n$ is divisible by 16.
I used your approach, Jhevon . . . and hit a wall.
We have: . $P \;=\;m^2n^2(m + n) - (m + n) \;=\;(m+n)(m^2n^2-1) \;=\;(m+n)(mn-1)(mn+1)$
Let . $\begin{array}{ccc}m &=&2a+1 \\ n &=&2b+1 \end{array}$
Then:. . $\begin{array}{ccccccc}m + n &=& (2a+1) + (2b+1) &=& \quad 2a+2b + 2 &\;\;=& \quad2(a+b+1)\\
mn-1 &=& (2a+1)(2b+1) - 1 &=& \quad 4ab + 2a +2b &\;\;=& \;\;2(2ab + a + b) \end{array}$
. . . . . . $\begin{array}{ccccccc}mn+1 &=& (2a+1)(2b+1)+1 &=& 4ab + 2a + 2b + 2 &=& 2(2ab + a + b + 1)\end{array}$
And we have: . $P \;=\;2(a + b + 1)\cdot2(2ab + a + b)\cdot2(2ab + a + b + 1)$
. . . . . . . . . . $P \;=\;8(a+b+1)(2ab + a + b)(2ab + a + b + 1)$
So we have a multiple of 8.
. . And here's where I hit the wall . . .
I tried various ways to find another factor of 2,
. . and settled on an exhaustive listing.
. . $\begin{array}{c|c|c|c}
(a,b) & (a+b+1) & (2ab + a + b) & (2ab + a + b + 1) \\ \hline
\text{even, even} & \text{odd} & \text{even} & \text{odd} \\
\text{even, odd} & \text{even} & \text{odd} & \text{even} \\
\text{odd, odd} & \text{odd} & \text{even} & \text{odd} \end{array}$
So for any combination of $a$ and $b$, at least one of the factors is even.
Therefore, $P$ is a multiple of 16.
Surely there must a more elegant method . . .
.
4. Soroban,
Whenever you said:
$
P \;=\;m^2n^2(m + n) - (m + n) \;=\;(m+n)(m^2n^2-1) \;=\;(m+n)(mn-1)(mn+1)$
Because:
m+n is even.
mn-1 is even
mn+1 is even (AND next even number after mn-1)
which means either mn-1 or mn+1 is divisible by 4.
$
(m+n)(mn-1)(mn+1) = 2k_1 2k_2 4k_3 = 16k_4$
5. Originally Posted by Sigyn3
Prove that if m and n are odd integers, then $m^3n^2+m^2n^3-m-n$ is divisible by 16.
First note that the square of any odd integer is congruent to 1 modulo 8. This is because $(2k+1)^2=4k(k+1)+1$ and one of $k$ and $k+1$ is even.
Since $m$ snd $n$ are odd, so is $mn$. $\therefore\ m^2n^2\equiv1\pmod8$
$\Rightarrow\ m^2n^2-1$ is divisible by 8.
And since $m+n$ is even, $m^3n^2+n^3m^2-m-n=(m+n)(m^2n^2-1)$ is divisible by 16. | 2016-07-26T23:03:11 | {
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https://cs.stackexchange.com/questions/144646/minimize-the-expected-value-of-money-spent-on-boxes-until-the-prize-is-found | # Minimize the expected value of money spent on boxes until the prize is found
There are $$n$$ closed boxes and one and only one box of them contains a prize. The $$i$$-th box has probability $$p_i$$ of containing a prize. A player must pay $$c_i$$ coins to open the $$i$$-th box. The player must open boxes until the prize is found, then the game ends. Find the optimal box opening strategy that minimizes the expected money spent before obtaining the prize.
For example, let there be two boxes: $$p_1 = 0.8$$, $$p_2 = 0.2$$, $$c_1 = 3$$, $$c_2 = 2$$. There are two possible variants: 1-2 and 2-1 (a player can open box 1 or box 2 first).
Expected value of spent money for 1-2 order is $$0.8\times3+0.2\times(3+2)=3.4$$ coins.
Expected value of spent money for 2-1 order is $$0.2\times2+0.8\times(3+2)=4.4$$ coins.
Therefore 1-2 order is better.
Is there an algorithm that solves this problem faster than brute force $$O(n!)$$? What is the optimal algorithm for solving this task?
The algorithm is simple. Just sort the boxes from highest $$p/c$$ value to lowest. This takes $$O(n \log n)$$ time.
The proof is a little more complicated.
Consider a box-picking strategy $$S$$, and number the boxes in the order that strategy $$S$$ picks boxes. The expected cost of picking boxes in this order is $$c_1+(1-p_1)c_2 + (1-p_1-p_2)c_3 + \cdots$$. We definitely incur a cost of $$c_1$$ to open the first box, then we have a $$1-p_1$$ probability of having to open the second box and pay $$c_2$$, and a $$(1-p_1-p_2)$$ probability that the prize wasn't in either of those boxes, and so on.
A necessary (though not obviously sufficient) condition for a strategy to be optimal is that we can't swap the order of two consecutive box picks and get a better strategy. If we compare the expected cost of strategy $$S$$ and the expected cost of strategy $$S_i$$, where box $$i+1$$ is picked before box $$i$$, we find that these costs differ in only two terms, corresponding to the costs associated with picking boxes $$i$$ and $$i+1$$.
For strategy $$S$$, these terms are $$(1-p_1-p_2-\cdots-p_{i-1})c_i + (1-p_1-p_2-\cdots-p_{i-1}-p_i)c_{i+1}$$
For strategy $$S_i$$, these terms are $$(1-p_1-p_2-\cdots-p_{i-1})c_{i+1} + (1-p_1-p_2-\cdots-p_{i-1}-p_{i+1})c_i$$
If we eliminate the $$(1-p_1-p_2-\cdots-p_{i-1})(c_i+c_{i+1})$$ component from both sides, we find that strategy $$S_i$$ wins if $$-p_ic_{i+1} >-p_{i+1}c_i$$
or equivalently, if $$p_i/c_i
so strategy $$S_i$$ wins if box $$i+1$$ has a higher $$p/c$$ value than box $$i$$.
For a strategy to be optimal, each box must have a $$p/c$$ value greater than or equal to the next box in picking order. This can be achieved with a simple sort. If all $$p/c$$ values are distinct, then there is only one possible outcome of this sort, and the resulting order is the single optimal strategy. If there are duplicate $$p/c$$ values, then we can order boxes with the same $$p/c$$ value arbitrarily, since (by a variation of the above logic) shuffling boxes within a group of equal $$p/c$$ value will not change the expected cost of a strategy.
• The way I understood the problem is that the prize is found in a unique box $I$, and $\Pr[I=i]=p_i$. Oct 11 at 6:35
• I don't think your expected value calculation is correct. c3 is paid if the prize is not in the first two boxes. This happens with probability (1 - p1 - p2) not with probability (1 - p1)(1 - p2). (1 - p1)(1 - p2) should be used if each box can contain a prize independently of any other boxes, not if one and only one box has the prize. Oct 11 at 11:37
• Sorry if "one and only one box has the prize" was not clear from the original wording, I'll change it. Oct 11 at 11:43
• I believe, the correct formula for expected value should be $\sum_{i=1}^N (p_i * \sum_{j=1}^i c_j) = \sum_{i=1}^N (c_i * \sum_{j=i}^N p_j) = \sum_{i=1}^N (c_i * (1 - \sum_{j=1}^{i-1} p_j))$ Oct 11 at 11:58
• @VladimirBogachev: The proof in the answer shows that any strategy that does not pick boxes in descending $p/c$ order can be improved by swapping a pair of consecutive boxes. The descending $p/c$ order strategy must be optimal because all the alternatives aren't. Oct 11 at 15:04
There's a simple dynamic programming algorithm which runs in time $$O^*(2^n)$$. For a set $$S$$, let $$C(S)$$ be the cost of opening just the boxes in $$S$$ in the optimal order. Then $$C(\emptyset) = 0$$ and for $$S \neq \emptyset$$, $$C(S) = \min_{i \in S} \left[ C(S \setminus \{i\}) + p_i \sum_{j \in S} c_j \right].$$ You are interested in $$C(\{1,\ldots,n\})$$.
• I think you can do better by sorting the boxes by $p/c$. If we plan to open box $j$ immediately after box $i$, and $p_i/c_i<p_j/c_j$, then swapping the boxes reduces the expected cost. Oct 11 at 4:16 | 2021-10-22T06:36:48 | {
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https://nl.mathworks.com/help/symbolic/rewrite.html | rewrite
Rewrite expression in terms of another function
Description
example
rewrite(expr,target) rewrites the symbolic expression expr in terms of the target function target. The rewritten expression is mathematically equivalent to the original expression. If expr is a vector or matrix, rewrite acts element-wise on expr.
Examples
collapse all
Rewrite any trigonometric function in terms of the exponential function by specifying the target 'exp'.
syms x
sin2exp = rewrite(sin(x), 'exp')
tan2exp = rewrite(tan(x), 'exp')
sin2exp =
(exp(-x*1i)*1i)/2 - (exp(x*1i)*1i)/2
tan2exp =
-(exp(x*2i)*1i - 1i)/(exp(x*2i) + 1)
Rewrite the exponential function in terms of any trigonometric function by specifying the trigonometric function as the target. For a full list of targets, see target.
syms x
exp2sin = rewrite(exp(x), 'sin')
exp2tan = rewrite(-(exp(x*2i)*1i - 1i)/(exp(x*2i) + 1), 'tan')
exp2sin =
1 - 2*sin((x*1i)/2)^2 - sin(x*1i)*1i
exp2tan =
-(((tan(x) - 1i)*1i)/(tan(x) + 1i) + 1i)/...
((tan(x) - 1i)/(tan(x) + 1i) - 1)
Simplify exp2tan into the expected form by using simplify.
exp2tan = simplify(exp2tan)
exp2tan =
tan(x)
Rewrite any trigonometric function in terms of any other trigonometric function by specifying the target. For a full list of targets, see target.
Rewrite tan(x) in terms of the sine function by specifying the target 'sin'.
syms x
tan2sin = rewrite(tan(x), 'sin')
tan2sin =
-sin(x)/(2*sin(x/2)^2 - 1)
Rewrite any hyperbolic function in terms of any trigonometric function by specifying the trigonometric function as the target. For a full list of targets, see target.
Rewrite tanh(x) in terms of the sine function by specifying the target 'sin'.
syms x
tanh2sin = rewrite(tanh(x), 'sin')
tanh2sin =
(sin(x*1i)*1i)/(2*sin((x*1i)/2)^2 - 1)
Similarly, rewrite trigonometric functions in terms of hyperbolic functions by specifying the hyperbolic function as the target.
Rewrite any inverse trigonometric function in terms of the logarithm function by specifying the target 'log'. For a full list of targets, see target.
Rewrite acos(x) and acot(x) in terms of the log function.
syms x
acos2log = rewrite(acos(x), 'log')
acot2log = rewrite(acot(x), 'log')
acos2log =
-log(x + (1 - x^2)^(1/2)*1i)*1i
acot2log =
(log(1 - 1i/x)*1i)/2 - (log(1i/x + 1)*1i)/2
Similarly, rewrite the logarithm function in terms of an inverse trigonometric function by specifying the inverse trigonometric function as the target.
Rewrite each element of a matrix by calling rewrite on the matrix.
Rewrite all elements of a matrix in terms of the exp function.
syms x
matrix = [sin(x) cos(x); sinh(x) cosh(x)];
rewrite(matrix, 'exp')
ans =
[ (exp(-x*1i)*1i)/2 - (exp(x*1i)*1i)/2, exp(-x*1i)/2 + exp(x*1i)/2]
[ exp(x)/2 - exp(-x)/2, exp(-x)/2 + exp(x)/2]
Rewrite the cosine function in terms of the sine function. Here, rewrite replaces the cosine function using the identity cos(2*x) = 1 – 2*sin(x)^2 which is valid for any x.
syms x
rewrite(cos(x),'sin')
ans =
1 - 2*sin(x/2)^2
rewrite does not replace sin(x) with either $-\sqrt{1-{\mathrm{cos}}^{2}\left(x\right)}$ or $\sqrt{1-{\mathrm{cos}}^{2}\left(x\right)}$ because these expressions are not valid for all x. However, using the square of these expressions to replace sin(x)^2 is valid for all x. Thus, rewrite replaces sin(x)^2.
syms x
rewrite(sin(x),'cos')
rewrite(sin(x)^2,'cos')
ans =
sin(x)
ans =
1 - cos(x)^2
Input Arguments
collapse all
Input to rewrite, specified as a symbolic number, variable, expression, function, vector, matrix, or multidimensional array.
Target function, specified as a character vector. This table summarizes the rewriting rules for all allowed targets.
TargetRewrites These FunctionsIn Terms of These Functions
'exp'All trigonometric and hyperbolic functions including inverse functionsexp, log
'log'All inverse trigonometric and hyperbolic functionslog
'sincos'tan, cot, exp, sinh, cosh, tanh, cothsin, cos
'sin', 'cos', 'tan', or 'cot'sin, cos, exp, tan, cot, sinh, cosh, tanh, coth except the targetTarget trigonometric function
'sinhcosh'tan, cot, exp, sin, cos, tanh, cothsinh, cosh
'sinh', 'cosh', 'tanh', 'coth'tan, cot, exp, sin, cos, sinh, cosh, tanh, coth except the targetTarget hyperbolic function
'asin', 'acos', 'atan', 'acot'log, and all inverse trigonometric and inverse hyperbolic functionsTarget inverse trigonometric function
'asinh', 'acosh', 'atanh', 'acoth'log, and all inverse trigonometric and inverse hyperbolic functionsTarget inverse hyperbolic function
'sqrt'abs(x + 1i*y)sqrt(x^2 + y^2)
'heaviside'sign, triangularPulse, rectangularPulseheaviside
'piecewise'abs, heaviside, sign, triangularPulse, rectangularPulsepiecewise
Tips
• rewrite replaces symbolic function calls in expr with the target function only if the replacement is mathematically valid. Otherwise, it keeps the original function calls. | 2020-09-18T17:21:16 | {
"domain": "mathworks.com",
"url": "https://nl.mathworks.com/help/symbolic/rewrite.html",
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https://kr.mathworks.com/help/matlab/ref/fsurf.html?lang=en | # fsurf
Plot 3-D surface
## Syntax
fsurf(f)
fsurf(f,xyinterval)
fsurf(funx,funy,funz)
fsurf(funx,funy,funz,uvinterval)
fsurf(___,LineSpec)
fsurf(___,Name,Value)
fsurf(ax,___)
fs = fsurf(___)
## Description
example
fsurf(f) creates a surface plot of the function z = f(x,y) over the default interval [-5 5] for x and y.
fsurf(f,xyinterval) plots over the specified interval. To use the same interval for both x and y, specify xyinterval as a two-element vector of the form [min max]. To use different intervals, specify a four-element vector of the form [xmin xmax ymin ymax].
example
fsurf(funx,funy,funz) plots the parametric surface defined by x = funx(u,v), y = funy(u,v), z = funz(u,v) over the default interval [-5 5] for u and v.
fsurf(funx,funy,funz,uvinterval) plots over the specified interval. To use the same interval for both u and v, specify uvinterval as a two-element vector of the form [min max]. To use different intervals, specify a four-element vector of the form [umin umax vmin vmax].
fsurf(___,LineSpec) sets the line style, marker symbol, and surface color. For example, '-r' specifies red lines. Use this option after any of the previous input argument combinations.
example
fsurf(___,Name,Value) specifies surface properties using one or more name-value pair arguments. Use this option after any of the input argument combinations in the previous syntaxes.
fsurf(ax,___) plots into the axes specified by ax instead of the current axes (gca).
example
fs = fsurf(___) returns a FunctionSurface object or ParameterizedFunctionSurface object, depending on the inputs. Use fs to query and modify properties of a specific surface. For a list of properties, see FunctionSurface Properties or ParameterizedFunctionSurface Properties.
## Examples
collapse all
Plot the expression $\mathrm{sin}\left(x\right)+\mathrm{cos}\left(y\right)$ over the default interval $-5 and $-5.
fsurf(@(x,y) sin(x)+cos(y))
Plot the piecewise expression
$\begin{array}{cc}erf\left(x\right)+\mathrm{cos}\left(y\right)& -5
over $-5
Specify the plotting interval as the second input argument of fsurf. When you plot multiple surfaces over different intervals in the same axes, the axis limits adjust to include all the data.
f1 = @(x,y) erf(x)+cos(y); fsurf(f1,[-5 0 -5 5]) hold on f2 = @(x,y) sin(x)+cos(y); fsurf(f2,[0 5 -5 5]) hold off
Plot the parameterized surface
$\begin{array}{c}x=r\mathrm{cos}\left(u\right)\mathrm{sin}\left(v\right)\\ y=r\mathrm{sin}\left(u\right)\mathrm{sin}\left(v\right)\\ z=r\mathrm{cos}\left(v\right)\\ where\phantom{\rule{1em}{0ex}}r=2+\mathrm{sin}\left(7u+5v\right)\end{array}$
for $0 and $0. Add light to the surface using camlight.
r = @(u,v) 2 + sin(7.*u + 5.*v); funx = @(u,v) r(u,v).*cos(u).*sin(v); funy = @(u,v) r(u,v).*sin(u).*sin(v); funz = @(u,v) r(u,v).*cos(v); fsurf(funx,funy,funz,[0 2*pi 0 pi]) camlight
For $x$ and $y$ from $-2\pi$ to $2\pi$, plot the 3-D surface $y\mathrm{sin}\left(x\right)-x\mathrm{cos}\left(y\right)$. Add a title and axis labels and display the axes outline.
fsurf(@(x,y) y.*sin(x)-x.*cos(y),[-2*pi 2*pi]) title('ysin(x) - xcos(y) for x and y in [-2\pi,2\pi]') xlabel('x'); ylabel('y'); zlabel('z'); box on
Set the x-axis tick values and associated labels using the XTickLabel and XTick properties of axes object. Access the axes object using gca. Similarly, set the y-axis tick values and associated labels.
ax = gca; ax.XTick = -2*pi:pi/2:2*pi; ax.XTickLabel = {'-2\pi','-3\pi/2','-\pi','-\pi/2','0','\pi/2','\pi','3\pi/2','2\pi'}; ax.YTick = -2*pi:pi/2:2*pi; ax.YTickLabel = {'-2\pi','-3\pi/2','-\pi','-\pi/2','0','\pi/2','\pi','3\pi/2','2\pi'};
Plot the parametric surface $x=u\mathrm{sin}\left(v\right)$, $y=-u\mathrm{cos}\left(v\right)$, $z=v$ with different line styles for different values of $v$. For $-5, use a dashed green line for the surface edges. For $-2, turn off the edges by setting the EdgeColor property to 'none'.
funx = @(u,v) u.*sin(v); funy = @(u,v) -u.*cos(v); funz = @(u,v) v; fsurf(funx,funy,funz,[-5 5 -5 -2],'--','EdgeColor','g') hold on fsurf(funx,funy,funz,[-5 5 -2 2],'EdgeColor','none') hold off
Plot the parametric surface
$\begin{array}{c}x={e}^{-|u|/10}\mathrm{sin}\left(5|v|\right)\\ y={e}^{-|u|/10}\mathrm{cos}\left(5|v|\right)\\ z=u.\end{array}$
Assign the parameterized function surface object to a variable.
x = @(u,v) exp(-abs(u)/10).*sin(5*abs(v)); y = @(u,v) exp(-abs(u)/10).*cos(5*abs(v)); z = @(u,v) u; fs = fsurf(x,y,z)
fs = ParameterizedFunctionSurface with properties: XFunction: @(u,v)exp(-abs(u)/10).*sin(5*abs(v)) YFunction: @(u,v)exp(-abs(u)/10).*cos(5*abs(v)) ZFunction: @(u,v)u EdgeColor: [0 0 0] LineStyle: '-' FaceColor: 'interp' Show all properties
Change the plotting interval for u to [-30 30] by setting the URange property of object. Add transparency to the surface by setting the FaceAlpha property to a value between 0 (transparent) and 1 (opaque).
fs.URange = [-30 30];
fs.FaceAlpha = .5;
Show contours below a surface plot by setting the 'ShowContours' option to 'on'.
f = @(x,y) 3*(1-x).^2.*exp(-(x.^2)-(y+1).^2)... - 10*(x/5 - x.^3 - y.^5).*exp(-x.^2-y.^2)... - 1/3*exp(-(x+1).^2 - y.^2); fsurf(f,[-3 3],'ShowContours','on')
Control the resolution of a surface plot using the 'MeshDensity' option. Increasing 'MeshDensity' can make smoother, more accurate plots while decreasing it can increase plotting speed.
Create two plots in a tiled chart layout. In the first plot, display the parametric surface $x=\mathrm{sin}\left(s\right)$, $y=\mathrm{cos}\left(s\right)$, $z=\left(t/10\right)\mathrm{sin}\left(1/s\right)$. The surface has a large gap. Fix this issue by increasing the 'MeshDensity' to 40 in the second plot. fsurf fills the gap, showing that by increasing 'MeshDensity' you increased the resolution.
tiledlayout(2,1) nexttile fsurf(@(s,t) sin(s), @(s,t) cos(s), @(s,t) t/10.*sin(1./s)) view(-172,25) title('Default MeshDensity = 35') nexttile fsurf(@(s,t) sin(s), @(s,t) cos(s), @(s,t) t/10.*sin(1./s),'MeshDensity',40) view(-172,25) title('Increased MeshDensity = 40')
## Input Arguments
collapse all
3-D function to plot, specified as a function handle to a named or anonymous function.
Specify a function of the form z = f(x,y). The function must accept two matrix input arguments and return a matrix output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).
Example: f = @(x,y) sin(x) + cos(y);
Plotting interval for x and y, specified in one of these forms:
• Vector of form [min max] — Use the interval [min max] for both x and y
• Vector of form [xmin xmax ymin ymax] — Use the interval [xmin xmax] for x and [ymin ymax] for y.
Parametric function for x coordinates, specified as a function handle to a named or anonymous function.
Specify a function of the form x = funx(u,v). The function must accept two matrix input arguments and return a matrix output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).
Example: funx = @(u,v) u.*sin(v);
Parametric function for y coordinates, specified as a function handle to a named or anonymous function.
Specify a function of the form y = funy(u,v). The function must accept two matrix input arguments and return a matrix output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).
Example: funy = @(t) @(u,v) -u.*cos(v);
Parametric function for z coordinates, specified as a function handle to a named or anonymous function.
Specify a function of the form z = funz(u,v). The function must accept two matrix input arguments and return a matrix output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).
Example: funz = @(u,v) v;
Plotting interval for u and v, specified in one of these forms:
• Vector of form [min max] — Use the interval [min max] for both u and v
• Vector of form [umin umax vmin vmax] — Use the interval [umin umax] for u and [vmin vmax] for v.
Axes object. If you do not specify an axes object, then fsurf uses the current axes.
Line style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line.
Example: '--or' is a red dashed line with circle markers
Line StyleDescription
-Solid line
--Dashed line
:Dotted line
-.Dash-dot line
MarkerDescription
'o'Circle
'+'Plus sign
'*'Asterisk
'.'Point
'x'Cross
'_'Horizontal line
'|'Vertical line
's'Square
'd'Diamond
'^'Upward-pointing triangle
'v'Downward-pointing triangle
'>'Right-pointing triangle
'<'Left-pointing triangle
'p'Pentagram
'h'Hexagram
ColorDescription
y
yellow
m
magenta
c
cyan
r
red
g
green
b
blue
w
white
k
black
### Name-Value Pair Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: 'Marker','o','MarkerFaceColor','red'
The properties list here are only a subset. For a full list, see FunctionSurface Properties orParameterizedFunctionSurface Properties.
Number of evaluation points per direction, specified as a number. The default is 35. Because fsurf objects use adaptive evaluation, the actual number of evaluation points is greater.
Example: 100
Display contour plot under plot, specified as 'on' or 'off', or as numeric or logical 1 (true) or 0 (false). A value of 'on' is equivalent to true, and 'off' is equivalent to false. Thus, you can use the value of this property as a logical value. The value is stored as an on/off logical value of type matlab.lang.OnOffSwitchState.
Line color, specified as 'interp', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default RGB triplet value of [0 0 0] corresponds to black. The 'interp' value colors the edges based on the ZData values.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB® uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Line style, specified as one of the options listed in this table.
Line StyleDescriptionResulting Line
'-'Solid line
'--'Dashed line
':'Dotted line
'-.'Dash-dotted line
'none'No lineNo line
Line width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.
The line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.
Marker symbol, specified as one of the values listed in this table. By default, the object does not display markers. Specifying a marker symbol adds markers at each data point or vertex.
ValueDescription
'o'Circle
'+'Plus sign
'*'Asterisk
'.'Point
'x'Cross
'_'Horizontal line
'|'Vertical line
'square' or 's'Square
'diamond' or 'd'Diamond
'^'Upward-pointing triangle
'v'Downward-pointing triangle
'>'Right-pointing triangle
'<'Left-pointing triangle
'pentagram' or 'p'Five-pointed star (pentagram)
'hexagram' or 'h'Six-pointed star (hexagram)
'none'No markers
Marker size, specified as a positive value in points, where 1 point = 1/72 of an inch.
## Output Arguments
collapse all
One or more FunctionSurface or ParameterizedFunctionSurface objects, returned as a scalar or a vector.
• If you use the fsurf(f) syntax or a variation of this syntax, then fsurf returns FunctionSurface objects.
• If you use the fsurf(funx,funy,funz) syntax or a variation of this syntax, then fsurf returns ParameterizedFunctionSurface objects.
You can use these objects to query and modify properties of a specific surface. For a list of properties, see FunctionSurface Properties and ParameterizedFunctionSurface Properties.
Introduced in R2016a | 2021-09-18T09:58:40 | {
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https://math.stackexchange.com/questions/3142820/permutation-how-to-arrange-12-people-around-a-table-for-7 | # Permutation: How to arrange 12 people around a table for 7?
I want to understand how to arrange $$12$$ people around a circular table with $$7$$ chairs. We don't care about the overflow, those people can go to another table.
I thought the way to solve the problem is that the position for the first chair is fixed, the second chair has $$11$$ possible options of people (since one person is already seated), the third chair has $$10$$ possible options, the fourth chair has $$9$$ possible options and so on until we get to the seventh chair which has $$6$$ possible options of people.
So I thought the way to solve is that this is a permutation problem $$1*11*10*9*8*7*6=332640=11P6=\frac{12P7}{12}$$. But my professor says the correct answer is $$\frac{12P7}{7}$$. I don't understand why we should divide $$12P7$$ by the number of chairs. Can someone explain this me?
• Choose which seven people get a seat. Then, let the youngest of those people sit down first at the table wherever they like. Then, fill the remaining six of the seven chosen people around the table. – JMoravitz Mar 10 at 20:07
I think it should be $$\binom{12}{7}6!$$
• Why multiply by $6!$? – Sam Mar 10 at 20:14
• @Sam Note that $\binom{12}{7}6!$ is equal to $\frac{_{12}P_7}{7}$. I much prefer this answer (which matches mine above in the comments) as it avoids the "division by symmetry" style arguments that seem common and confuses people. – JMoravitz Mar 10 at 20:30 | 2019-06-26T00:28:02 | {
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http://math.stackexchange.com/questions/21692/does-monotonicity-and-derivability-of-a-function-f-colon-mathbbr-to-mathbbr | # Does monotonicity and derivability of a function $f\colon \mathbb{R}\to\mathbb{R}$ imply bijectiveness?
I have to prove that $f \colon x \mapsto e^{4x} + x^5 + 2$ ($f\colon \mathbb{R}\to\mathbb{R}$) is bijective. The argument given in the solution is that since the first two summands of the image is a bijective function of $x$, then so is $f$. Nonetheless, this "proof" doesn't seem at all rigorous to me, since there are many counterexamples to this argument.
So I proved that $f$ is strictly increasing by looking at its derivative, thus injective, and that every $c \in \mathbb{R}$ has at least one preimage, applying Bolzano's theorem to $f(x) - c$ and evaluating its limit at $-\infty$ and $+\infty$, and so, demonstrating $f$ to be surjective. In consequence, $f$ is bijective.
I am much more pleased about this proof than the one given in the solutions, but I want to know if I missed something, or if my hypothesis are insufficient. I gave an example to illustrate my argument, but the question I want to ask in the general form is in the title. Also, I would like to know whether the converse holds as well.
Thanks.
Update: I've been thinking about this, and realized that only monotonicity and continuity (together with unboundedness, as Mariano pointed out) are necessary for bijectiveness, and derivability only helps to prove monotonicity. Is this correct?
-
Consider the $\arctan$ function. There is one condition you can add to monotinicity+derivability which will make things work... bijections of $\mathbb R$ are unbounded both above and below. – Mariano Suárez-Alvarez Feb 12 '11 at 16:48
Well detailed, +1. Wish I could help. – fdart17 Feb 12 '11 at 16:49
@your update: Right. – Soarer Feb 12 '11 at 17:26
The word is "differentiability", isn't it? – Rahul Feb 12 '11 at 17:28
@Rahul As far as single-variable calculus is concerned, I think derivability and differentiability are the same. Although I haven't taken multivariable calculus, so I wouldn't know for sure. In Spanish we say derivabilidad (derivability) and diferenciabilidad (differentiability). And here derivabilidad seems much more widely used than diferenciabilidad, at least for single-variable calculus. I wanted to stick to the vocabulary in use, but the difference between the two concepts is not something I know or need to know right now, since I think it's nonexistent in this context. – Abel Feb 12 '11 at 18:13
A function $f: \mathbb{R} \rightarrow \mathbb{R}$ which is strictly monotonic -- i.e., either (for all $x_1,x_2 \in \mathbb{R}$, $x_1 < x_2 \implies f(x_1) < f(x_2)$) or (for all $x_1,x_2 \in \mathbb{R}$, $x_1 < x_2 \implies f(x_1) > f(x_2)$) -- is necessarily injective. This is immediate from the definition, and no continuity or differentiability is necessary for this.
But what about bijectivity? Let me assume WLOG that $f$ is strictly increasing. Then a necessary condition for bijectivity is that $\lim_{x \rightarrow \pm \infty} f(x) = \pm \infty$. (This is not absolutely immediate, but I hope and believe it will become clear after only a little thought.) For future use, let us call this the infinite limits property.
This observation can be used to see that the answer to your title question is no: as Mariano says, consider the arctangent function.
Moreover, being strictly increasing and having the infinite limits property is not enough to guarantee surjectivity: consider for instance the function defined piecewise as $x$ for $x \leq 0$ and as $x+1$ for $x > 0$. This example seems to indicate that we are missing the condition of continuity.
Proposition: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be strictly increasing and have the infinite limits property. The following are equivalent:
(i) $f$ is continuous.
(ii) $f$ is surjective.
Since the word "homework" is being thrown around, I leave the proof to the OP. (Anyway, it is a fun and not so difficult exercise.)
As the OP says, differentiability is not part of the fundamental picture here, but comes up in practice because having strictly positive derivative on an interval ensures that a function is strictly increasing.
- | 2015-08-03T17:38:38 | {
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https://math.stackexchange.com/questions/2910175/why-is-an-uncountable-union-of-null-sets-not-necessarily-a-null-set | # Why is an uncountable union of null sets not necessarily a null set?
I ran across this statement "...for instance, an uncountable union of null sets need not be a null set (or even a measurable set)..." while looking through Terence Tao's blog site (See the first statement of #5). Since I'm taking a course in measure theory right now, I thought it might be relevant to understand why this is true, but I really don't know where to start. (In fact, I'm not entirely sure if this is relevant to measure theory, but Dr. Tao did mention that such a union might not even be a measurable set...).
Intuition told me it might be similar to why 0 $\cdot$ $\infty$ is indeterminate, but I understood that problem to be one of definitions. As such, the only thing I have been able to come up with is that I don't understand the definition(s) either "null set", "uncountable", or "union" precisely enough for me to grasp yet. I'm leaning towards not fully understanding the term "uncountable," as my understanding of transfinites and ordinals is sketchy at best.
I was wondering if anyone could define these terms; point me towards something to read, learn, or think about; or provide an example of an uncountable union of null sets not being a null set?
Edit: I didn't realize null set and the empty set were different things. Thanks to everyone for the examples and definitions!
• $\bigcup_{x\in [0,1]}\{x\} = ?$ – amsmath Sep 8 '18 at 23:15
• Null is not the same as empty. – Daniel Mroz Sep 8 '18 at 23:18
• @DanielMroz sorry, I didn't know that, after you said that, I found this stackexchange post, and that clears up a lot... Ironic that the definition I didn't question was the one I didn't understand... haha – SmallFish Sep 8 '18 at 23:27
• @SmallFish Yes to all your questions! – amsmath Sep 8 '18 at 23:36
• If real intervals are to have measure (and they should; what else would measure theory be good for) and a set of a single point is to have no measure (it it should because a point is "that which has no part" i.e. is ... measureless) then real intervals are uncountable union of single points and therefore we must have this possibility. – fleablood Sep 8 '18 at 23:40
A null set is a set of zero measure; more informally, it is a set of points that has no area. For example, all countable sets— including the rationals $\mathbb{Q}$, the natural numbers $\mathbb{N}$, the empty set $\varnothing$, and singleton sets like $\{0\}$—are null sets in $\mathbb{R}$ [under the Lebesgue measure].
Any countable union of null sets is still a null set, but an uncountable union might not be. For example:
$$[-1,1] = \bigcup_{x\in[-1,1]} \{x\}$$
The inverval on the left is not a null set; it has length 2. However, we can write it as an uncountable union of singleton sets that are each measure 0. Hence an uncountable union of null sets can yield a non-null set.
• Thanks, I think this was the most complete answer, so I've marked it as the answer. I feel really dumb now for asking this haha... – SmallFish Sep 8 '18 at 23:41
• @SmallFish No worries at all --- I expect this answer will probably help others who have a similar question in the future. – user326210 Sep 8 '18 at 23:44
Take any set $A$ which is not a null set. Then $A=\bigcup_{a\in A}\{a\}$. Now, note that each $\{a\}$ is a null set.
• Should I mark this as an answer? This clearly answers the question as I asked it. However, I was, for all intents and purposes, confused on the difference between a null set and an empty set, and the comments cleared that up for me before I got to your answer...? – SmallFish Sep 8 '18 at 23:30
• My opinion is that you should either rewrite the question or accept the answer. As written, this question will be found by people who search for information about uncountable unions of null sets, and they should know that this is the answer. Or you can rewrite the question to be about the difference between a null set and the empty set, and then get, say, Daniel Mroz to write an answer that you can accept. (Although in that case, it might just get marked as a duplicate.) – Toby Bartels Sep 8 '18 at 23:34
• @SmallFish That's up to you. – José Carlos Santos Sep 8 '18 at 23:35 | 2019-10-18T18:26:51 | {
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http://stats.stackexchange.com/questions/20542/generating-a-3-month-half-life-weighting-series-in-r/20685 | # Generating a 3-month half life weighting series in R
I am using the lm() and princomp() functions in R to perform regressions on foreign exchange time series. I would like to weight the regressions (and PCA) such that 50% of the influence on the regression comes from the past 3 months, 25% from the previous 3 months, etc, but in a smooth fashion. Both functions take such a weighting series. What is a simple formula for generating a decaying weights series with a half life of 3 months (or any other period, for that matter), in R?
Ideally the sum of the weights will equal 1, although I don't think this is mandatory in lm().
-
To be general, let us consider a time series with arbitrary steps. Therefore, let $k$ be the number of steps in three months and write $\omega_1$ for the weight for the immediately preceding time, $\omega_2$ for the weight preceding it, and so on, so that the sequence of weights
$$(\omega_1, \omega_2, \ldots, \omega_k)$$
is applied to the preceding three months. The total weight for those three months is the sum of these.
Assumption I: It is natural to hope that the next sequence of weights starting at $\omega_{k+1}$ be in proportion to the first sequence; the question specifies that the constant of proportionality be $1/2$, entailing
$$\frac{1}{2}(\omega_1, \omega_2, \ldots, \omega_k) = (\omega_{k+1}, \omega_{k+2}, \ldots, \omega_{2k}).$$
The requirement of "smoothness," together with the natural idea of a monotonic decrease in weights over time, suggests that $\omega_{k+1} \lt \omega_k$. This leads to
Assumption II: We might hope that the sequence of weights could be arranged in geometric proportion, say with constant of proportionality $\rho$, whence
$$\frac{1}{2}\omega_1 = \omega_{k+1} = \rho^k \omega_1.$$
The unique solution is
$$\rho = 2^{-1/k}.$$
Summing these weights over $n$ time steps and requiring them to sum to unity gives
$$1 = \omega_1 + \omega_2 + \cdots + \omega_n = \omega_1(1 + \rho + \rho^2 + \cdots + \rho^{n-1}) = \omega_1 \frac{1-\rho^n}{1-\rho}.$$
It follows that
$$\omega_1 = \frac{1-\rho}{1-\rho^n} = \frac{1-2^{-1/k}}{1-2^{-n/k}}$$
and
$$\omega_i = \rho^{i-1}\omega_1 = 2^{(1-i)/k}\omega_1, \quad i=1, 2, \ldots, n.$$
For example, with $k=3$ (monthly data) and $n=6$ (giving two full three-month periods), $\rho = 2^{-1/3} = 0.793701$, $\omega_1 = 0.275066$, and the sequence of weights for the first three-month period is $0.275066$, $0.21832$, $0.173281$ followed by half these weights for the second three-month period, $0.137533$, $0.10916$, $0.0866403$.
Other solutions to the problem are possible; in particular, there are infinitely many available when we do not make Assumption II. But Assumption II has the particularly nice property that if you were to combine the data into sequential groups, taking $m$ of them at a time (such as combining daily data into monthly data, with $m \approx 30$), and recalculate the weights, then the sum of the weights for each group will equal the recalculated group weights.
-
(+1) Better explanation than mine! (as usual...) – jbowman Jan 6 '12 at 18:48
Thanks--but the "as usual" is incorrect, IMHO, seeing that you already have posted plenty of superb answers to many questions. Let's not lose sight of the fact that we make a tradeoff between speed and thoroughness and sometimes speed is valued more. Regardless, as a community we hope that all correct replies will undergo a process of continual improvement (by their authors and community members) so that eventually each thread will offer at least one great answer to the question. – whuber Jan 6 '12 at 18:53
Let us assume that you want such a decaying weight to take place over intervals of T time periods, e.g., 91 days in your example (there are 91 days in three months, so you want the weight on day 91 to equal 1/2 the weight on day 0.) Solve $\exp\{-\lambda T\} = 0.5$ to get $\lambda = -\log(0.5)/T$. In your case, $\lambda = -0.007617$. Then the weight assigned to a given time period $t$, $w(t)$, is equal to $\exp\{-\lambda t\}$.
You can see this works, as $w(t+T) = w(T)w(t) = 0.5w(t) \space \forall t$, so $\sum_{t=0}^{T-1}w(t) = 0.5\sum_{t=T}^{2T-1}w(t)$. | 2013-12-11T14:20:22 | {
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https://math.stackexchange.com/questions/2774652/what-fraction-of-the-trapezium-is-shaded-my-guess-frac-13-answer | # What fraction of the trapezium is shaded? My guess: $\frac 13$. Answer: $?$
Problem:
$PQRS$ is a trapezium in which base $PQ=2$ units and base $RS = 3$ units. Draw a line from point $P$ to $R$. Triangle $PQR$ is formed. Shade in that triangle.
Now, what fraction of the trapezium is shaded?
I am having some trouble solving this problem as the height of the trapezium is not mentioned. I believe that knowing that height is unnecessary, but I am not too sure.
My Attempt:
Lemma: Given a trapezium with bases $a$ and $b$, and an altitude (height) $h$, the area is $\dfrac{h(a+b)}{2}$.
In this example, $a,b=PQ,RS$ respectively, so the area is $$\dfrac{h(2+3)}{2} = \dfrac{5h}{2}.$$
Now I know that if it were a square, then the two bases would be equal to each other, and the line $PR$ would be splitting it into equal halves.
So, $\Delta PQR = \dfrac 12$.
But, since on of the bases is larger, and the smaller base is a side of $\Delta PQR$, then my assumption is that the area of $\Delta PQR$ is less than $\dfrac 12$.
Nevertheless, the height of $PQRS$ is not stated so I do not know how to solve this. Intuitively, I believe the answer is $\dfrac 13$ but how can it be proven?
Edit:
Doing some research, I found a very similar post over here.
General Rule (edited just now):
After looking at the answer, I came up with a general rule:
You are given a quadrilateral $PQRS$ (i.e. a square; trapezium; etc) in which base $PQ = a$ units; base $RS = b$ units; the two bases are parallel; and height $h\perp PQ\land RS$. By drawing a diagonal line $PR = c$ units to form two congruent triangles $PQR$ and $PSR$ that make up $PQRS$, then by denoting $S_{ABCD}$ the area of $ABCD$ and $\Delta{UVW}$ a triangle $UVW$, $$\frac{S_{\Delta PQR}}{S_{PQRS}} = \frac{a}{a+b}\,\text{ and }\,\frac{S_{\Delta PSR}}{S_{PQRS}} = \frac{b}{b+a},$$ regardless of the value of $h$.
Just clarifying: is this true? If so, does this theorem have a name?
• Think why the area of $\,\triangle PQR\,$ is $\,\dfrac{h \cdot a}{2}\,$. – dxiv May 10 '18 at 4:32
• @dxiv ahhh I never thought about finding the area of the triangle :)) – Feeds May 10 '18 at 4:36
• "I am having some trouble solving this problem as the height of the trapezium is not mentioned." Doesn't matter as both the triangle and trapezium of will have the same height so both the area of the triangle and the trapezium will have areas proportional to the height. – fleablood May 10 '18 at 5:08
• "I never thought about finding the area of the triangle" Out of curiosity, how could you be asked to find what proportion the area of a triangle is without thinking about finding the area of a triangle? What else is there to think about? – fleablood May 10 '18 at 5:10
• @fleablood sorry, I may not be that advanced as you, I suppose. – Feeds May 10 '18 at 11:39
$$\text{Required fraction} = \frac{S_{\triangle PQR}}{S_{PQRS}} = \frac{\frac12 \cdot 2h}{\frac12 \cdot 5h} = \frac25$$
• I see, because it is a ratio that also include the area of the triangle and not just the trapezium. Thank you very much for that. If it was that easy, you might have given me a hint, but it doesn't matter now as you get $$\color{green}{\checkmark}$$ I have to wait $17$ hours before I can upvote, but congratulations nonetheless! – Feeds May 10 '18 at 4:37
• @user477343 You may think of "half of the trapezoid is shaded" (with a picture in your mind) in our natural language, so that the desired fraction is $$\frac{S_{\rm shaded}}{S_{\rm trapezoid}} = \frac12.$$ You may try other shapes to develop your sense towards sth like "$p/q$ of $A$ is $B$". – GNUSupporter 8964民主女神 地下教會 May 10 '18 at 4:46
• Thank you for that :) By the way, when you write $S_n$, what does the $S$ stand for? I know from your answer that you mean area, and I use the same notation but with $A$ instead to stand for Area. Does $S$ stand for Shape or something like that? Or does it stand for nothing at all, pretty much? – Feeds May 10 '18 at 4:50
• @user477343 I often saw authors of contest-math books using $S_{\rm sth}$ to denote the area of $\rm sth$ when I was a high school student. – GNUSupporter 8964民主女神 地下教會 May 10 '18 at 4:58 | 2019-09-24T09:24:16 | {
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https://math.stackexchange.com/questions/3384140/will-the-graphs-of-equivalent-indefinite-integrals-always-look-identical | # Will the graphs of equivalent indefinite integrals always look identical?
I came across the following indefinite integral
$$\int {\sqrt{1-\sin{\left(2x\right)}}}\space \mathrm{d}x \space (0 \leq x \leq \pi )$$
I attempted to solve it as follows:
$$u = 1 - \sin\left(2x\right) \implies \sin\left(2x\right) = 1- u$$
$$\implies x = \dfrac{\arcsin\left ( 1 - u\right)}{2} \implies \mathrm{d}x = \dfrac{-\mathrm{d}u }{2 \sqrt{1- (1-u)^2} }$$
$$\therefore \textrm{ we have }\space \dfrac{-1}{2} \int \dfrac{ \sqrt{u} } {\sqrt { 1 - (1 - u)^2}} \space \mathrm{d}u = \dfrac{-1}{2} \int \sqrt {\dfrac{ u } { 1 - (1 -2u + u^2)}} \space \mathrm{d}u$$
$$= \dfrac{-1}{2} \int \sqrt {\dfrac{u} {u(2-u)}} \space \mathrm{d}u = \dfrac{-1}{2} \int \sqrt {\dfrac{ 1 } { 2-u}} \space \mathrm{d}u$$
Let $$z = 2 - u$$. Then $$\mathrm{d}z = -\mathrm{d}u$$
$$\implies \dfrac{1}{2} \int \sqrt {\dfrac{ 1 } {z}} \space \mathrm{d}z = \dfrac{1}{2} \int z^{-1/2}= \sqrt{z} + C = \sqrt{2-u} + C = \sqrt{1 + \sin\left(2x\right)} + C$$
I ran the problem through the site Integral Calculator and got
$$\int {\sqrt{1-\sin{\left(2x\right)}}}\space \mathrm{d}x = -\sin{x} - \cos{x} + C$$
which only coincides with my solution if I take its absolute value. Refer to the image below:
$$\sqrt{1 + \sin\left(2x\right)} + C$$ is in orange, whereas $$-\sin{x} - \cos{x} + C$$ is in purple. In the diagram, $$C = 0$$. The shaded region is the given range of $$x$$ values : $$0 \leq x \leq \pi$$.
Now, from what I know, the graphs of equivalent indefinite integrals should look similar and should differ only by a constant, which means they will coincide after a translation transformation. However, it seems not to be the case here. Hence my question; will the graphs of equivalent indefinite integrals always look similar, or perhaps there is a mistake in my calculations? In addition, the answer that is given in the problem book is
$$2\sqrt{2} \left[ \dfrac{t}{\pi}\right] + \sqrt{2} \space \mathrm{sgn} \space t \cdot \left\{ \cos{ \dfrac{t}{\pi} } - \cos{t}\right\} \\ \mathrm{ where } \space t = x - \dfrac{\pi}{4} \space \mathrm{ and } \left[ \cdot \right] \textrm{ is the integer part of the expression inside }$$
This further confused me because now I don't know the right answer and where I went wrong. Any help is appreciated.
• (1) For a continuous function $f$ on an interval, any two anti-derivatives of $f$ differ only by constant. So their graphs are just vertical translations of each other. Let me also remind you that this is exactly why we consider 'constant of integration'. (2) Since you are integrating non-negative function, its anti-derivarive is always increasing. That said, neither of your answer nor the software's answer correctly represents the anti-derivative of $\sqrt{1-\sin(2x)}$ over $0\leq x\leq\pi$. Check Quanto's answer below to learn how to work with piecewise-defined functions under integraion. – Sangchul Lee Oct 7 '19 at 15:28
Note that the integration is over a periodic function with periodic non-differentiable points as shown in the plot. Follow the steps below to perform such indefinite integration.
$$I=\int {\sqrt{1-\sin{\left(2x\right)}}}\space \mathrm{d}x = \int {\sqrt{(\sin{x} - \cos{x})^2 }}\space \mathrm{d}x$$
$$= \int |\sin{x} - \cos{x}| \mathrm{d}x =\sqrt 2 \int |\sin(x-\frac{\pi}{4})| \mathrm{d}x = \sqrt 2\int |\sin t|dt$$
where the variable change $$t=x-\frac \pi4$$ is made in the last step. Note that the last expression is a positive periodic function with periodity $$\pi$$ and its repeated integral value in each periodic interval is given by
$$A=\int_0^\pi |\sin t| dx = 2$$
Then, for $$t\ge 0$$, reeexpress the integral as,
$$I =\sqrt 2 \int_0^{[\frac t\pi]} |\sin s| ds+\sqrt 2 \int_{[\frac t\pi]}^t \sin s ds + C$$
$$=\sqrt 2 A\left[\frac t\pi\right] - \sqrt 2 \cos s|_{[\frac t\pi]}^t + C$$
$$=2\sqrt{2} \left[ \dfrac{t}{\pi}\right] + \sqrt{2} \space \left( \cos{ \dfrac{t}{\pi} } - \cos{t}\right) +C$$
The result for $$t<0$$ can be derived similarly. The combined integral result then reads,
$$I = 2\sqrt{2} \left[ \dfrac{t}{\pi}\right] + \sqrt{2} \space \mathrm{sgn} \space t \cdot \left(\cos{ \dfrac{t}{\pi} } - \cos{t}\right) +C$$
• Thanks. Though I'm now clear on how the book's answer was derived, I still don't understand why this method was opted for. Could you kindly elaborate? That is, why are the other two solutions in the post incorrect? Is it because of the periodic non-differentiable points? – E.Nole Oct 7 '19 at 15:16
• Yeah, mainly due to periodicity and non-differentiability at those periodic points. As can be seen in the plot, the integral shall grows with $t$ steadily with repeated local variation. The first method in the post is only valid within certain domain due to the use of $\sin^{-1}(x)$ and the second is plain wrong since it assumes the antiderivative is differentiable to all orders. – Quanto Oct 7 '19 at 15:32 | 2020-01-26T06:07:23 | {
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https://demo2.evirtualservices.com/articles/can-a-symmetric-matrix-have-complex-eigenvalues-e0a911 | If a matrix has a null eigenvector then the spectral theorem breaks down and it may not be diagonalisable via orthogonal matrices (for example, take $\left[\begin{matrix}1 + i & 1\\1 & 1 - i\end{matrix}\right]$). When matrices m and a have a dimension ‐ shared null space, then of their generalized eigenvalues will be Indeterminate. When eigenvalues become complex, eigenvectors also become complex. the eigenvalues of A) are real numbers. We know the eigenvalues and Example(A 2 × 2 matrix) However this last fact can be proved in an elementary way as follows: the eigenvalues of a real skew-symmetric matrix are purely imaginary (see below) and to every eigenvalue there corresponds the conjugate eigenvalue with the same multiplicity; therefore, as the determinant is the product of the eigenvalues, each one repeated according to its multiplicity, it follows at once that the determinant, if … I am currently calculating a covariance matrix which has real entries and is symmetric. Problems in Mathematics © 2020. If the matrix is symmetric (e.g A = AT), then the eigenvalues are always real. The proof is very technical and will be discussed in another page. where c is an arbitrary number.. The Characteristic Equation always features polynomials which can have complex as well as real roots, then so can the eigenvalues & eigenvectors of matrices be complex as well as real. Symmetric Matrices There is a very important class of matrices called symmetric matrices that have quite nice properties concerning eigenvalues and eigenvectors. The matrix "C" has rotated the unit vector about Consider the matrix $A=\begin{bmatrix} 1 & 2 & 1 \\ 2 &5 &4 \\ 1 & 1 & 0 \end{bmatrix}.$... (a) True or False. (10) Can Symmetric Matrix Have Complex Eigenvalues? The list of linear algebra problems is available here. The eigenvalues of a matrix m are those for which for some nonzero eigenvector . A stronger claim than this is that the Cholesky decomposition exists, so it certainly is true if the matrix is symmetric. $\begingroup$ @DominicMichaelis : do you really mean that (a real square matrix can have only real eigenvalues)? I wanted to know if there is any result that shows whether a positive definite matrix can have complex eigenvalues. matrix has the. 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Even if and have the same eigenvalues, they do not necessarily have the same eigenvectors. The adapted Spectral Theorem is in fact false for matrices which have null eigenvectors. An asymmetric real matrix can have eigenvalues that are not real. Step by Step Explanation. Clearly, if A is real , then AH = AT, so a real-valued Hermitian matrix is symmetric. Symmetric Matrices There is a very important class of matrices called symmetric matrices that have quite nice properties concerning eigenvalues and eigenvectors. A symmetric matrix A is a square matrix with the property that A_ij=A_ji for all i and j. If the input matrix is non-symmetric, you additionally have to extract the (complex) eigenvalues by identifying the $2\times 2$ blocks (e.g., by checking whether a subdiagonal element is greater than a tolerance) and if so, computing the eigenvalues by a formula. occur only in conjugate pairs, we don't have to confirm the companion solution. The matrices are symmetric matrices. Clearly, if A is real , then AH = AT, so a real-valued Hermitian matrix is symmetric. Yes, of course. Express a Vector as a Linear Combination of Other Vectors, How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, Prove that $\{ 1 , 1 + x , (1 + x)^2 \}$ is a Basis for the Vector Space of Polynomials of Degree $2$ or Less, Basis of Span in Vector Space of Polynomials of Degree 2 or Less, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue, Find a Basis for the Subspace spanned by Five Vectors. associated eigenvectors. This website’s goal is to encourage people to enjoy Mathematics! A matrix is said to be symmetric if AT = A. Show transcribed image text. (b) Prove that if eigenvalues of a real symmetric matrix A are all positive, then Ais positive-definite. illustrated can be viewed as follows. Tags: complex conjugate eigenvalue eigenvector exam hermitian matrix length length of a vector linear algebra matrix norm norm of a vector Ohio State Ohio State.LA real eigenvalue symmetric matrix … Expert Answer . Remark. Previous question Next question Transcribed Image Text from this Question. The adapted Spectral Theorem states that as long as a complex symmetric matrix has no null eigenvectors, it must be diagonalisable by an orthogonal matrix. Prove Your Answer. Block Diagonalization of a 3 × 3 Matrix with a Complex Eigenvalue. All eigenvalues are squares of singular values of which means that 1. ST is the new administrator. Example # 2: Find the We know that a positive definite matrix has positive eigenvalues. It remains to show that if a+ib is a complex eigenvalue for the real symmetric matrix A, then b = 0, so the eigenvalue is in fact a real number. I'm afraid you might confuse Susan. which can have complex as well as real roots, then so can the eigenvalues & eigenvectors of matrices be complex as well I am currently calculating a covariance matrix which has real entries and is symmetric. Here are the eigenvalues and their All its eigenvalues must be non-negative i.e. We have seen that (1-2i) is also an eigenvalue of the above matrix.Since the entries of the matrix A are real, then one may easily show that if is a complex eigenvalue, then its conjugate is also an eigenvalue. pure rotation in the plane of any vector and a scaling equal to the magnitude A matrix is said to be symmetric if AT = A. "A" be a real 2 x 2 matrix with a complex eigenvalue All the eigenvalues of a symmetric real matrix are real If a real matrix is symmetric (i.e.,), then it is also Hermitian (i.e.,) because complex conjugation leaves real numbers unaffected. in for . False. This question hasn't been answered yet Ask an expert. Prove Your Answer. a+bi and a-bi). A full rank square symmetric matrix will have only non-zero eigenvalues It is illuminating to see this work when the square symmetric matrix is or . Then λ 1 is another eigenvalue, and there is one real eigenvalue λ 2. Since eigenvalues are roots of characteristic polynomials with real coe¢cients, complex eigenvalues always appear in pairs: If ‚0=a+bi is a complex eigenvalue, so is its conjugate ‚¹ 0=a¡bi: Expert Answer . Save my name, email, and website in this browser for the next time I comment. Note that applying the complex conjugation to the identity A(v+iw) = (a+ib)(v+iw) yields A(v iw) = (a ib)(v iw). (adsbygoogle = window.adsbygoogle || []).push({}); The Existence of an Element in an Abelian Group of Order the Least Common Multiple of Two Elements, Express a Hermitian Matrix as a Sum of Real Symmetric Matrix and a Real Skew-Symmetric Matrix, Explicit Field Isomorphism of Finite Fields, Probability that Alice Wins n Games Before Bob Wins m Games, Subset of Vectors Perpendicular to Two Vectors is a Subspace. Correct me if I'm wrong here, but a similar claim (to the Gram matricies) would be that a square root exists, since PD matrices have a full set of eigenvalues, a square root exists, so … However, if A has complex entries, symmetric and Hermitian have different meanings. A symmetric matrix A is a square matrix with the property that A_ij=A_ji for all i and j. The row vector is called a left eigenvector of . Previous question Next question Transcribed Image Text from this Question. It is diagonal, so obviously diagonalizable, and has just a single eigenvalue repeated $n$ times. is always PSD 2. In general, if a matrix has complex eigenvalues, it is not diagonalizable. Sponsored Links Enter your email address to subscribe to this blog and receive notifications of new posts by email. Then, A) The Only Eigenvalues Of A Are 0. Your email address will not be published. In this lecture, we shall study matrices with complex eigenvalues. However, when complex eigenvalues are encountered, they always occur in conjugate pairs as long as their associated matrix has only real entries. The rst step of the proof is to show that all the roots of the characteristic polynomial of A(i.e. Since the eigenvectors as well as the eigenvalues eigenvalues and a basis for each eigenspace For example the 2 x 2 matrix cos X -sin X sin X cos X Question: 4) The Matrix A = 0 2 1 May Have Complex Eigenvalues 1-2 1 3 A) True B) False 5) Let A Be Nxn Real Symmetric Matrix, Then The Eigenvalues Of A Are Real, And The Eigenvectors Corresponding To Distinct Eigenvalues Are Orthogonal. I am saying this because we have a rudimentary conjugate gradient complex symmetric eigensolver in FORTRAN, and we get poor quality of complex orthogonality* between eigenvectors, unlike MATLAB. I wanted to know if there is any result that shows whether a positive definite matrix can have complex eigenvalues. However, when complex eigenvalues are A symmetric real matrix can only have real eigenvalues. complex matrices { the de nitions are the same as before. We can compute a corresponding (complex) eigenvector in exactly the same way as before: by row reducing the matrix A − λ I n. Now, however, we have to do arithmetic with complex numbers. This is the case for symmetric matrices. If , then can have a zero eigenvalue iff has a zero singular value. I'm guessing if this is the case for the general case of any non-zero n×n symmetric matrix. In general, it is normal to expect that a square matrix with real entries may still have complex eigenvalues. Question: 1) Let A Be A Square Matrix Such That A = 0. Therefore, by the previous proposition, all the eigenvalues of a real symmetric matrix are … As a result, eigenvectors of symmetric matrices are also real. In fact, we can define the multiplicity of an eigenvalue. Notify me of follow-up comments by email. A complex symmetric matrix can be 'diagonalized' using a unitary matrix: thus if is a complex symmetric matrix, there is a unitary matrix such that is a real diagonal matrix with non-negative entries. There is such a thing as a complex-symmetric matrix ( aij = aji) - a complex symmetric matrix need not have real diagonal entries. Lemma 0.1. It follows that AA is invertible. A) True B) False 3) Let A Be Nxn Real Matrix. The generalized eigenvalues of m with respect to a are those for which . the origin an angle, "f", and scaled the resultant by a factor of "r". A real symmetric n×n matrix A is called positive definite if xTAx>0for all nonzero vectors x in Rn. if we multiply it by "C". B) A = 0. 2) If A Is Nxn Real Symmetric Matrix, Then The Eigenvectors Corresponding To Any Eigenvalues Are Orthogonal. This website is no longer maintained by Yu. (10) Can symmetric matrix have complex eigenvalues? All Rights Reserved. The rst step of the proof is to show that all the roots of the characteristic polynomial of A(i.e. If $\theta \neq 0, \pi$, then the eigenvectors corresponding to the eigenvalue $\cos \theta +i\sin \theta$ are Let's see what happens if to a unit vector along the x-axis invertible matrix "P" and a matrix "C" such that the given Since there are three distinct eigenvalues, they have algebraic and geometric multiplicity one, so the block diagonalization theorem applies to A. as real. In summary, when $\theta=0, \pi$, the eigenvalues are $1, -1$, respectively, and every nonzero vector of $\R^2$ is an eigenvector. All non-real complex eigenvalues occur in conjugate pairs (e.g. There is such a thing as a complex-symmetric matrix ( aij = aji) - a complex symmetric matrix need not have real diagonal entries. This site uses Akismet to reduce spam. (a) Prove that the eigenvalues of a real symmetric positive-definite matrix Aare all positive. Math 2940: Symmetric matrices have real eigenvalues The Spectral Theorem states that if Ais an n nsymmetric matrix with real entries, then it has northogonal eigenvectors. The diagonal elements of a triangular matrix are equal to its eigenvalues. Example # 3: Find an Eigenvalues of a triangular matrix. corresponding vectors for this matrix from a previous problem. D) All Of The Above. If each entry of an $n \times n$ matrix $A$ is a real number, then the eigenvalues of $A$ are all real numbers. In fact, the part (b) gives an example of such a matrix. We've shown that our "C" matrix is comprised of a 8. We know that a positive definite matrix has positive eigenvalues. It is clear that one should expect to have complex entries in the eigenvectors. The process we just C) If A Is Not Symmetric, Then A 0. The Real Statistics functions eVALUES and eVECT only return real eigenvalues. matrix has only real entries. and associated eigenvector in . Suppose v+ iw 2 Cnis a complex eigenvector with eigenvalue a+ib (here v;w 2 Rn). If the matrix is real and symmetric, then its eigenvalues are real and eigenvectors are orthogonal to each other, i.e., is orthogonal and can be considered as a rotation matrix, and we have Before discussing Jacobi's method for finding and , we first review the rotation in a 2-D space: •Eigenvalues can have zero value •Eigenvalues can be negative •Eigenvalues can be real or complex numbers •A "×"real matrix can have complex eigenvalues •The eigenvalues of a "×"matrix are not necessarily unique. in for . There will always be n linearly independent eigenvectors for symmetric matrices. How to Diagonalize a Matrix. the eigenvalues of A) are real numbers. What about $[0, 1;-1, 0]$ with eigenvalues $\pm i$? (b) Find the eigenvalues of the matrix The characteristic polynomial for $B$ is $\det(B-tI)=\begin{bmatrix}-2-t & -1\\ 5& 2-t \end{bmatrix}=t^2+1.$ The eigenvalues are the solutions of the characteristic polynomial. One may wonder if there exists a class of matrices with only real eigenvalues. I'm afraid you might confuse Susan. Show transcribed image text. Every n × n matrix has exactly n complex eigenvalues, counted with multiplicity. New content will be added above the current area of focus upon selection (10) Can symmetric matrix have complex eigenvalues? Let $A$ be real skew symmetric and suppose $\lambda\in\mathbb{C}$ is an eigenvalue, with (complex) … Let A be a 3 × 3 matrix with a complex eigenvalue λ 1. •Eigenvalues can have zero value •Eigenvalues can be negative •Eigenvalues can be real or complex numbers •A "×"real matrix can have complex eigenvalues •The eigenvalues of a "×"matrix are not necessarily unique. Consider the $n\times n$ identity matrix. (10) Can Symmetric Matrix Have Complex Eigenvalues? Matrices Satisfying the Relation $HE-EH=2E$, Linear Independent Vectors, Invertible Matrix, and Expression of a Vector as a Linear Combinations. We only need to find the eigenvector for say: Theorem: Let Math 2940: Symmetric matrices have real eigenvalues The Spectral Theorem states that if Ais an n nsymmetric matrix with real entries, then it has northogonal eigenvectors. Learn how your comment data is processed. 8. COMPLEX EIGENVALUES. This question hasn't been answered yet Ask an expert. Last modified 01/20/2020, Your email address will not be published. If is an eigenvector of the transpose, it satisfies By transposing both sides of the equation, we get. encountered, they always occur in conjugate pairs as long as their associated Required fields are marked *. But returning to the square root problem, this shows that "most" complex symmetric matrices have a complex symmetric square root. In general, a real matrix can have a complex number eigenvalue. Prove your answer. A complex symmetric matrix diagonalizable ,Write this as M=A+iB, where both A,B are real and A is positive definite. Now let's return to our original specific example where. Example # 1: Find the The Characteristic Equation always features polynomials The matrices are symmetric matrices. However, if A has complex entries, symmetric and Hermitian have different meanings. eigenvalues and a basis for each eigenspace Then where and . of the complex eigenvalue pair. In fact, we can define the multiplicity of an eigenvalue. Prove your answer. I know that a non-zero symmetric 2×2 matrix can't have only zero eigenvalues ( a zero eigenvalue with algebraic multiplicity 2), since such a matrix should have complex off diagonal entries to satisfy both trace and determinant being zero.
## can a symmetric matrix have complex eigenvalues
Presentation About Myself Examples, Hp Laptop Microphone Not Working Windows 7, Why Do Transition Elements Show Variable Oxidation State, Dark Souls Ash Lake Warp, Switchblade Uk Law, Brick Wall Texture Hd, Da Pam 385-24, Pickled Radish Pods Recipe, Chocolate Cake With Yema Filling, Architecting Big Data Solutions, Aldi Organic Peanut Butter Price, | 2021-05-15T13:40:41 | {
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https://math.stackexchange.com/questions/2076355/approximation-with-a-normal-distribution | Approximation with a normal distribution
Every day Alice tries the stroke playing tennis until she reaches $50$ strokes. If each stroke is good with a probability of $0,4$, independently from others, approximately what is the probability that at least $100$ attemps are necessary to success?
Let X be a binomial random variable with parameters $$n=100$$ and $$p=0,4$$
Since n is large we can approximate with a normal distribution with parameters $$\mu=40$$ and $$\sigma=\sqrt{0,4*100*0,6}=\sqrt{24}$$
Appling the normal approximation \begin{align}P(X> 49,5)&=1-P(x<49,5)\\&=1-P((X-\mu )/\sigma < (49,5-40 )/ \sqrt{24} )\\&=1-P((X-\mu )/\sigma < 1,939 )\\&=1-\Phi (1,939)\\&=1-0,9737\\&=0,0263\end{align}
But the solution on the book is $0,974$ (it could be $P(x<49,5)$).
You have a couple of answers using different methods, so this is to show that they are both correct approaches (+1 and +1). I do not see that any method leads to exactly the answer 0.974, but the discrepancy may have to do with the method of approximation used.
By thinking the problem through carefully, you can work the problem using either the binomial or the negative binomial distribution.
Binomial. Following @SiongThyeGoh's logic, if $X$ is the number of successes in $n = 99$ tries, and you have $X \le 49,$ then you will need at least 100 tries to get 50 successes. So we have $X \sim Binom(99, .4)$ and we seek $P(X \le 49).$
Using R statistical software, the statement pbinom(49, 99, .4) returns $0.9781,$ rounded to four places.
If you use the normal approximation with $\mu = np = 99(.4) = 39.6$ and $\sigma = 4.8944,$ then you have $$P(X \le 49) = P(X \le 49.5) \approx P\left(Z < \frac{49.5-39.6}{4.8744} = 2.03\right),$$ where $Z$ is standard normal. Printed tables give $P(Z \le 2.03) = .9788.$ You can avoid some round-off error in R, using pnorm(49.5, 39.6, 4.8744) to get $0.9789.$
Negative Binomial (counting failures) There are several parameterizations of the negative binomial distribution. One of them defines $Y = 0, 1, 2, \dots$ to be the number of failures occurring before the $r$th success. You would need at least 50 failures before the 50th success in order to require 100 or more tries.
This is the version of the negative binomial distribution implemented in R. So we need $P(Y < 50) = P(Y \le 49),$ which can be found in R with the statement pnbinom(49, 50, .6) to get $0.9781,$ rounded to four places.
x = 0:70; pdf = dnbinom(x, 50, .6)
plot(x, pdf, type="h", lwd=3, col="blue", main="PDF of Negative Binomial Dist'n Counting Failures")
abline(v=49.5, col="red", lwd=2, lty="dotted")
abline(h=0, col="green2")
Negative Binomial (counting trials). In @heropup's Answer, the negative binomial random variable $X$ is the number of trials, not failures, required to get $X$ successes, where the probability of success on any one trial is 0.4. You can use a normal approximation with her $\mu$ and $\sigma$ in the R statement 1 - pnorm(99.5, 125, 13.693) to get $0.9687,$ rounded to four places.
Ordinarily, you cannot depend on more than about two decimal places of accuracy form a normal approximation to the binomial or negative binomial distribution. You can see what you get using normal tables.
The illustration below shows the skewness of this negative binomial distribution and the less-than-ideal normal approximation. Here, you are interested in the probability to the right of the vertical red line.
mu = r/p; sg = sqrt(r/p^2); sg.r = round(sg,2) # for normal aprx
head=paste("PDF of Neg Binomial Dist'n Counting Trials with NORM(",mu,",",sg.r,") Density", sep="")
x = 50:170; r = 50; p=.4; q= 1-.4 # neg bin parameters
pdf = choose(x-1,r-1)*p^r*q^(x-r) # this pdf not programmed in R
plot(x, pdf, type="h", lwd=3, col="blue", main=head)
abline(v=99.5, col="red", lwd=2, lty="dotted")
curve(dnorm(x, 125, 13.7), add=T); abline(h=0, col="green2")
If you are going to use the negative binomial distribution, you should look to see what version your text or lectures are using, and learn that.
If $Y \sim Bin(99, 0.4)$
$Y < 50$ implies that there are less than $50$ succeses in $99$ trials, and hence at least $100$ attempts are necessary to reach $50$ succeses.
Hence the quantity that you are suppose to compute is $P(Y<50)$
You are using the wrong model for the event of interest. Specifically, you are interested in the number of attempts needed to achieve a certain pre-defined number of successes, not the number of successes observed in a pre-defined number of attempts.
Therefore, the model you should use is not binomial, but negative binomial. The random number of attempts $X$ needed to observe $r = 50$ successes, when each success has independent probability of $p = 0.4$ of occurring, is a negative binomial random variable with mean $$\mu = \frac{r}{p} = 125,$$ and variance $$\sigma^2 = \frac{(1-p)r}{p^2} = \frac{375}{2}.$$ Therefore, your normal approximation of $X$ should be based on these moments, not the ones you specified.
• "the model you should use is not binomial, but negative binomial" Binomial works like a charm here: the number of attempts to get K=50 successes is larger than N=100 if and only if the number of successes during the N=100 first attempts is smaller than K=50. The number of attempts to get K successes follows a negative binomial distribution and the number of successes during the N first attempts follows a binomial distribution. – Did Dec 30 '16 at 12:09
Considering an approximation with a normal distribution of the binomial random variable, I think that we should calculate $$P(X<49.5)$$ because the number of attempts must be more than 100.
when n is large, a binomial random variable with parameters n and p will have approximately the same distribution as a normal random variable with the same mean and variance as the binomial. If the random variable is negative binomial is possible to apply this approximantion?
• I am not sure just how well and for what choices of $n$ and $p$ the negative binomial distribution is well approximated by the normal (see the last part of my Answer). Certainly, the normal approximation to the binomial works well. – BruceET Dec 30 '16 at 9:17 | 2019-04-23T17:50:12 | {
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http://dynref.engr.illinois.edu/rvn.html | # Notation
## Mathematical objects
Example Meaning LaTeX
$P$ Points and positions are denoted by capital italic letters. \$P\$
$(4, 5, -2)$ Coordinates of a position are given as a tuple, so $P$ is at $(4, 5, -2)$ is the same as saying that $P$ has coordinates $x = 4$, $y = 5$, $z = -2$. Note the distinction from vector components with square brackets. \$(4, 5, -2)\$
$\boldsymbol{v}$ Vectors in typeset material are in bold font. \$\boldsymbol{v}\$
$\vec{v}$ Vectors in handwriting use an over-arrow. \$\vec{v}\$
$\|\boldsymbol{v}\|$, $v$ Magnitude uses double-bars or a plain letter, so $v = \|\boldsymbol{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2}$. \$\|\boldsymbol{v}\|\$, \$v\$
$\hat{\boldsymbol{v}}$ Unit vectors use over-hat, so $\hat{\boldsymbol{v}} = \frac{\boldsymbol{v}}{\|\boldsymbol{v}\|}$. \$\hat{\boldsymbol{v}}\$
$\hat{\boldsymbol{\imath}}$, $\hat{\boldsymbol{\jmath}}$, $\hat{\boldsymbol{k}}$ Cartesian basis vectors, so we write $\boldsymbol{v} = 3\hat{\boldsymbol{\imath}} + \hat{\boldsymbol{\jmath}} + 7\hat{\boldsymbol{k}}$. \$\hat{\boldsymbol{\imath}}\$, \$\hat{\boldsymbol{\jmath}}\$, \$\hat{\boldsymbol{k}}\$
$[3, 1, 7]$ Vector components use square brackets, so we write $[\boldsymbol{v}]_R = [3, 1, 7] = 3\hat{\boldsymbol{\imath}} + \hat{\boldsymbol{\jmath}} + 7\hat{\boldsymbol{k}}$. If the basis is clear then we will write $\boldsymbol{v} = [3, 1, 7]$. \$[3, 1, 7]\$
$[\boldsymbol{v}]_R$ Vector components in basis $R$. Standard basis names are $R$ for Rectangular (Cartesian), $P$ for polar, $C$ for cylindrical, $S$ for spherical. \$[\boldsymbol{v}]_R\$
$v_x, v_y, v_z$ Vector components are in non-bold with subscripts, so $\boldsymbol{v} = [v_x, v_y, v_z] = v_x\,\hat{\boldsymbol{\imath}} + v_y\,\hat{\boldsymbol{\jmath}} + v_z\,\hat{\boldsymbol{k}}$. \$v_x, v_y, v_z\$
$v$ versus $v_x$ Magnitude (positive) is the plain letter $v$, while signed component is $v_x$. \$v\$ versus \$v_x\$
$\hat{\boldsymbol{e}}_r, \hat{\boldsymbol{e}}_\theta$ Polar basis vectors. Maybe we should change this to $\hat{\boldsymbol{r}}, \hat{\boldsymbol\theta}$? \$\hat{e}_r\$, \$\hat{e}_\theta\$
$\boldsymbol{r}$, $\boldsymbol{r}_P$, $\boldsymbol{r}_{OP}$, $\overrightarrow{OP}$ Position vector of point $P$ from origin $O$. The origin and/or point can be neglected if it is obvious from context. \$\boldsymbol{r}\$, \$\boldsymbol{r}_P\$, \$\boldsymbol{r}_{OP}\$, \$\overrightarrow{OP}\$
$\boldsymbol{\rm A}$ Matrices are in upright (roman) bold. \$\boldsymbol{\rm A}\$
$A_{ij}$ Matrix components are in italic non-bold font. \$A_{ij}\$
$4\rm\ kg/m^2$, $4\rm\ kg\,m^{-2}$ Units are in roman (upright) font, have a space between the number and units, and have a space between unit symbols. See the NIST conventions for more details. \$4\rm\ kg/m^2\$, \$4\rm\ kg\ m^{-2}\$
$x = 4t^2$ To make formulas dimensionally correct we use one of the following forms: (1) “$x = 4t^2$, where $t$ is in seconds and $x$ is in meters”, (2) “$x = a t^2$ where $a = 4\rm\ m/s^2$”, or (3) “$x = 4 (t/{\rm s})^2\rm\ m$” (using quantity calculus). \$x = 4t^2\$
## Diagram elements
Element Meaning LaTeX
$\mathcal{B}_1$ Body number 1. Use numbers 1,2,3 for bodies. \$\mathcal{B}_1\$
$m_1, \omega_1, \alpha_1$ Mass, angular velocity, and angular acceleration of body $\mathcal{B}_1$. Use subscript numbers for quantities associated with bodies. \$m_1, \omega_1, \alpha_1\$
$P, Q$ Points $P$ and $Q$. Use italic capital letters for points. \$P, Q\$
$\boldsymbol{r}_P, \boldsymbol{v}_P, \boldsymbol{a}_P$ Position, velocity, and acceleration vectors of point $P$. Use subscript capital italic letters for quantities associated with points. \$\boldsymbol{r}_P, \boldsymbol{v}_P, \boldsymbol{a}_P\$
$I_{1,P,z}$ Moment of inertia of body $\mathcal{B}_1$ about point $P$ around the $z$ axis. Any of the subscripts can be neglected if they are obvious from context, although at least the point should normally be included. \$I_{1,P,z}\$
## Color scheme for diagrams
When possible, use a pale yellow (or tan) background with a light gray (or light green) square grid, like traditional engineering paper. A blank white background can also be used.
Color Meaning
Black Coordinate axes and objects.
Gray Measurements, angles, other notes.
Blue Position vectors.
Green Velocities and angular velocities.
Cyan Accelerations and angular accelerations.
Red Forces.
Purple Moments.
Example diagram showing colored elements. | 2021-06-22T05:11:17 | {
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https://quant.stackexchange.com/questions/26400/cumulative-return-calculation-disagreement | # cumulative return calculation, disagreement
A friend of mine and myself are having an argument on how to correctly determine cumulative return.
The dataset has monthly return data and we are trying to determine the 6-month cumulative return.
I proposed the following: Cumulative return for 6 months is a product of monthly returns:
(1) Ri=(1+ri_1)∗ ... ∗(1+ri_6)−1
He mentioned just adding up the different return values:
(2) Ri=(ri_1)+ ... (ri_6)
Now I am pretty sure my way of calculating this is correct, although differences can be small.
Question Am I correct that when academic papers talk of cumulative return they calculate this with equation (1)
Thank you in advance. S Gontscharoff
To expand on my comment, consider the following R code:
set.seed(1)
returns <- runif(1000, 0.95,1.055) #Extremely simple return generation with a slight drift.
plot(cumprod(returns), type = "l")
lines(cumsum(returns-1)+1, col = "blue")
Which gives the following result: As you can see the effect is not linear, as the difference nearly disappears around index 300 and is greatly reduced at 900, but increases again as the compounding effect increases. So, you will most likely not see major differences in 6 month data, but as soon as you go to larger time scales the effect become noticeable.
• Thank you! Was thinking that as well, but this illustrates it very nicely. I think I will have to switch from stata to R soon, since it seems everyone is using R!. But thank you both @dm63 ,Forgottenscience – S. Gontscharoff Jun 2 '16 at 13:11
If the dataset contains arithmetic returns where 1+r(i)= S(i)/S(i-1) then you are correct. If the dataset contains logarithmically defined returns where r(i) = log (S(i)/S(i-1)) then your friend is correct.
• Thank you for your quick response. The data is indeed in the first form you indicated, it is the return data from CRSP database. One final thing; could using one method or the other cause results to vary widely? – S. Gontscharoff Jun 2 '16 at 12:40
• Over time the differences accumulate, as compounding becomes a force. – Forgottenscience Jun 2 '16 at 12:53
• Consider just (the first) two months. Then your first method gives $r_1+r_2+r_1r_2$. This shows that the difference is due to sceond order effects. These might indeed sum up to a larger difference over time. – BerndH Jun 3 '16 at 14:49 | 2019-06-17T02:43:44 | {
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https://math.stackexchange.com/questions/2628911/why-are-we-allowed-to-cancel-fractions-in-limits?noredirect=1 | # Why are we allowed to cancel fractions in limits?
For example:
$$\lim_{x\to 1} \frac{x^4-1}{x-1}$$
We could expand and simplify like so:
$$\lim_{x\to 1} \frac{(x-1)(x^3 + x^2 + x + 1)}{x-1} = \lim_{x\to 1} (x^3 + x^2 + x + 1) = (1^3 + 1^2 + 1^1 + 1) = 4$$
In this case we divided out $x-1$ on top and bottom even though technically, at $x=1$, we have $\frac{0}{0}$ that we're just tossing aside.
But what allows us to do this?
• Because in the definition of that limit the $x$ is quantified on a set in which $x\neq1$ (notice the $\mathbf{0<}|x-1|<\delta$ in the definition). Since when $x\neq1$ the two functions, after and before cancelling, are equal, then the two limits are equal. – orole Jan 30 '18 at 23:46
• Ah so I'm not really dividing out $0/0$ since we're not actually reaching $x=1$? – user525966 Jan 30 '18 at 23:48
• Yes, for limit there is that $0<$ in the definition. If the $0<$ is not there it results in another concept, continuity. – orole Jan 30 '18 at 23:49
• $\lim_\limits{x\to a} f(x) g(x) = \lim_\limits{x\to a} f(x) \lim_\limits{x\to a} g(x)$ and $\lim_\limits{x\to 1} \frac {x-1}{x-1} = 1$ – Doug M Jan 31 '18 at 0:06
• @DougM that doesn't actually answer the OP question. It can still be asked why can we divide by $x -1$ to dertmine what $\lim \frac {x-1}{x-1}$ is. – fleablood Jan 31 '18 at 0:57
Simply because we are dealing with values $x\neq 1$ in this case, thus for algebraic rule we are allowed to cancel out
$$\lim_{x\to 1} \frac{x^4-1}{x-1}=\lim_{x\to 1} \frac{\color{red}{(x-1)}(x^3 + x^2 + x + 1)}{\color{red}{x-1}}$$
Remember indeed that by the definition of limit we are demanding that $$\forall \varepsilon>0 \quad \exists \delta>0 \quad \text{such that}\quad \color{green}{\forall x\neq1}\quad|x-1|<\delta \implies|f(x)-L|<\varepsilon$$
Note also that the same cancellation is used to prove the basic derivatives case, for example for $f(x)=x^2$
$$\lim_{x\to x_0}\frac{x^2-x_0^2}{x-x_0}=\lim_{x\to x_0}\frac{\color{red}{(x-x_0)}(x+x_0)}{\color{red}{x-x_0}}=\lim_{x\to x_0}(x+x_0)=2x_0$$
• Where did you get that definition? Can you provide a link? – Stefan Pochmann Jan 31 '18 at 11:37
• Usually it is stated as $0<|x-x_0|<\delta$ which is equivalent to $|x-x_0|<\delta$ and $x\neq x_0$, some doubt on it? – gimusi Jan 31 '18 at 11:52
• @StefanPochmann en.wikipedia.org/wiki/(ε,_δ)-definition_of_limit – mbomb007 Jan 31 '18 at 17:47
Proposition 1: If $f(x) = g(x)$ whenever $x\ne a,$ then $\lim\limits_{x\,\to\,a} f(x) = \lim\limits_{x\,\to\,a} g(x).$
Proposition 2: After the cancelation, the resulting function is continuous at $a,$ so the limit can be found by plugging in $a.$
• This doesn't seem to address the question; it doesn't say why the cancellation is allowed. – user2357112 Jan 31 '18 at 8:42
• @user2357112 : It appears to me that that's exactly what it does say. Maybe you need to clarify the question further if this does not address it. – Michael Hardy Jan 31 '18 at 13:23
• When I state Proposition 1 in class I say, "Limits don't see the point!" – Matthew Leingang Jan 31 '18 at 21:06
• @user2357112 : At this point perhaps I should tell you that you may be missing something if you don't think this explains why the cancellation is allowed, but I probably won't be able to explain what until you clarify further. – Michael Hardy Feb 5 '18 at 22:17
You are correct. At the point $x=1$ the expression is undefined/behaves badly and has no value.
But limits aren't about functions at the point $x = 1$. They are about functions near the point $x = 1$. In fact, they are specifically about when $x \ne 1$ (but is close to $1$).
$\lim_{x\to a} f(x) = K$ means if $x$ is NEAR $a$ then $f(x)$ is NEAR $K$.
And if $x$ is near $a$ then $x$ isn't $a$ and it is perfectly fine to divide by $x -a$ when $x \ne a$.
Now your hackles should be raised when you hear something like "$\frac {x^4 -1}{x-1}$ is near $4$ when $x$ is near $1$" and ask yourself what can "near" possibly mean in precise mathematical terms.
That's a question for another time.
You never actually reach $1$... $x$ gets closer and closer to $1$ without ever being $1$...
Therefore, you can divide by $x-1$; it's never $0$... See limits.
Consider the function $f(x)=\begin{cases} 1 \text{ when } x=0 \\ \frac1x \text{ when } x\not= 0\end{cases} \cdots$
Study the limiting behavior of $f$ at $0$... Notice it has nothing to do with $f$'s value, $1$, at$0$...
The functions defined by the expressions
$$\frac{(x-1)(x^3 + x^2 + x + 1)}{x-1} \quad\text{and}\quad x^3 + x^2 + x + 1$$
are not the same (because they are defined on different domains), but they agree outside of $x=1$. And the limit $\lim_{x\to 1}$ does not care about the value (if existent) at $x=1$, but only about values close to $1$.
Conclusion: Since the limit only sees the parts of these function in which they agree, it cannot distinguish between the two expressions (even though they are differnt from your perspective), and has to give the same result for both.
Algebraic Limit Theorem: Let the limits exist: $$\lim_\limits{x\to a} f(x)=L \quad \text{and} \quad \lim_\limits{x\to a} g(x)=M.$$ Then: \begin{align}&1) \ \lim_\limits{x\to a} (f(x)\pm g(x))=\lim_\limits{x\to a} f(x)\pm \lim_\limits{x\to a} g(x)=L\pm M;\\ &2) \ \lim_\limits{x\to a} (f(x)\cdot g(x))=\lim_\limits{x\to a} f(x)\cdot \lim_\limits{x\to a} g(x)=L\cdot M;\\ &3) \ \lim_\limits{x\to a} (f(x)/ g(x))=\lim_\limits{x\to a} f(x)/ \lim_\limits{x\to a} g(x)=L/M; \quad (\text{provided:} \lim_\limits{x\to a} g(x)=M\ne 0). \\ \end{align} Note that: \begin{align}\lim_{x\to 1} \frac{x-1}{x-1} = \lim_{x\to 1} 1&=1;\\ \lim_{x\to 1} (x^3 + x^2 + x + 1) &= 4;\\ \lim_{x\to 1} \frac{x^4-1}{x-1}=\lim_{x\to 1} \frac{(x-1)(x^3 + x^2 + x + 1)}{x-1} &= \\ \lim_{x\to 1} \frac{x-1}{x-1}\cdot \lim_{x\to 1} (x^3 + x^2 + x + 1) &= 1\cdot 4=4.\end{align}
However, you of course cannot do crazy stuff like claiming that $$\lim_{x\to0-} \frac{x\sqrt{x}}{\sqrt{x}} = \lim_{x\to0-} x = 0$$ because here you modify the domain of the function in more than finitely many points. | 2019-09-19T08:46:22 | {
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http://math.stackexchange.com/questions/303697/why-is-this-integral-always-positive-using-integration-by-parts | # Why is this integral always positive? (Using integration by parts)
Why is this integral always positive for $x \ge 0$?
$$\int_0^x \frac{\sin(t)}{t+1} dt$$
It's easy for me to see this when I imagine the graph of the integrand. However, this question appears in the chapter that intoduces integration by substitution and by parts, as well as some trig identities.
So I'm looking for a proof, preferrably using integration by parts.
-
Use integration by parts. Let $u=\dfrac{1}{1+t}$ and $dv=\sin t\,dt$. Then $du=-\dfrac{1}{(1+t)^2}\,dt$, and we can take $v=-\cos t$. Thus $$\int_0^x\frac{\sin t}{1+t}\,dt=\left.-\frac{\cos t}{1+t}\right|_0^x-\int_0^x\frac{\cos t}{(1+t)^2}\,dt.$$
To bound $\int_0^x\frac{\cos t}{(1+t)^2}\,dt$, note that $|\cos t|\le 1$, and is not always $1$. And $\int_0^x \frac{1}{(1+t)^2}\,dt=1-\frac{1}{1+x}$ so $\int_0^x\frac{\cos t}{(1+t)^2}\,dt$ has absolute value $\lt 1-\dfrac{1}{1+x}$. Thus $$\int_0^x\frac{\sin t}{1+t}\,dt \gt 1-\frac{|\cos x|}{1+x}-1+ \frac{1}{1+x}.$$
It follows that our integral is $\gt \dfrac{1-\cos x}{1+x}$, and in particular is $\gt 0$.
-
It's easy for me to see this when I imagine the graph of the integrand.
This imagination can become a formal proof: We have $$\int_0^x \frac{\sin(t)}{t+1} dt=\pi \int_0^{\frac{x}{\pi}} \frac{\sin(\pi t)}{\pi t+1} dt$$ So consider: $$g(t)=\frac{\sin(\pi t)}{\pi t+1}$$
We have: $$\int_0^x g(t) dt=\int_{[x]}^x g(t) dt+\sum_{k=1}^{[x]}{(-1)^{k-1}\int_{k-1}^k |g(t)| dt}$$ So if $$a_k=\int_{k-1}^k |g(t)| dt$$ It's enough to show that:
$$0\leq \sum_{k=1}^{n}{(-1)^{k-1} {a_k}}=(a_1-a_2)+(a_3-a_4)+..$$ But it's clear because: $$a_{k+1}=\int_{k}^{k+1} |\frac{\sin(\pi t)}{\pi t+1}| dt\leq \int_{k-1}^{k} |\frac{\sin(\pi t)}{\pi t+1}| dt= a_k$$
However some details are missed.
-
Very nice! The $\pi$ substitution makes the proof a lot cleaner than the way I was about to do it. – Mark Feb 14 '13 at 4:49 | 2014-09-22T18:42:19 | {
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