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With Tanya assigned to the head of table, there are $9!$ ways to seat the other people with no restrictions. Among these, there $2\cdot8!$ ways in which Henry sits beside Wilson (the factor of $2$ corresponds to which one sits to the left of the other), and another $2\cdot8!$ ways in which Henry sits beside Nancy. Subtracting these, we get the book's answer
$$9!-2\cdot8!-2\cdot8!=201600$$
But this double-subtracts the $2\cdot7!$ ways in which Henry sits between Wilson and Nancy. So the correct answer is
$$9!-2\cdot8!-2\cdot8!+2\cdot7!=211680$$
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Many problems lend themselves to being solved with systems of linear equations. Case 2: Parallel Lines The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation. Materials include course notes, lecture video clips, JavaScript Mathlets, a quiz with solutions, practice problems with solutions, a problem solving video, and problem sets with solutions. Once you do that, these linear systems are solvable just like other linear systems. $\begin{cases}5x +2y =1 \\ -3x +3y = 5\end{cases}$ Yes. Solve age word problems with a system of equations. The problem also says that Mia will be twice as old as Shaheena. This example shows one iteration of the gradient descent. When you solve systems with two variables and therefore two equations, the equations can be linear or nonlinear. There can be any combination: 1. But let’s say we have the following situation. Systems of linear equations can be used to model real-world problems. Solving using Matrices by Row Operations. Quiz 3. Just select one of the options below to start upgrading. Setting up a system of linear equations example (weight and price) (Opens a modal) Interpreting points in context of graphs of systems (Opens a modal) Practice. You’re going to the mall with your friends and you have $200 to spend from your recent birthday money. At the first store, he bought some t-shirts and spent half of his money. A system of linear equations is called homogeneous if the constants$b_1, b_2, \dots, b_m$are all zero. “Systems of equations” just means that we are dealing with more than one equation and variable. Wouldn’t it be cl… Consider the nonlinear system of equations Cramer's Rule. The best way to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you … Word Problems with Systems of Linear Equations: This worksheet is designed for Algebra | {
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to get you … Word Problems with Systems of Linear Equations: This worksheet is designed for Algebra students to practice those dreaded word problems that are solved with linear systems of equations. Is the point$(1 ,3)$a solution to the following system of equations… Solution of a non-linear system. No Problem 2. A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. Solving using an Augmented Matrix. Solve simple cases by inspection. Problem 1. Find them out by checking. Systems of Equations - Problems & Answers. Systems of linear equations are a common and applicable subset of systems of equations. When solving linear systems, you have two methods at your disposal, and which one you choose depends on the problem: System of equations word problem: walk & ride, Practice: Systems of equations word problems, System of equations word problem: no solution, System of equations word problem: infinite solutions, Practice: Systems of equations word problems (with zero and infinite solutions), Systems of equations with elimination: TV & DVD, Systems of equations with elimination: apples and oranges, Systems of equations with substitution: coins, Systems of equations with elimination: coffee and croissants. Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. Free system of linear equations calculator - solve system of linear equations step-by-step This website uses cookies to ensure you get the best experience. By … System of Linear Equations Word Problems Calvin went to Chicago's Magnificent Mile to do some Christmas shopping. A system of linear equations is a set of two or more linear equations with the same variables. Is the point$(0 ,\frac{5}{2})$a solution to the following system of equations? SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY. A system of linear | {
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to the following system of equations? SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY. A system of linear equations is a group of two or more linear equations that all contain the same set of variables. One way to solve a system of linear equations is by graphing each linear equation on the same -plane. To solve the system of equations, you need to find the exact values of x and y that will solve both equations. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Below is an example that shows how to use the gradient descent to solve for three unknown variables, x 1, x 2, and x 3. You really, really want to take home 6items of clothing because you “need” that many new things. Systems of Linear Equations. Solve the following system of equations by elimination. For problems 1 – 3 use the Method of Substitution to find the solution to the given system or to determine if the system … 9,000 equations in 567 variables, 4. etc. Gradient descent can also be used to solve a system of nonlinear equations. Looking for fun activities to teach kids about Solving Word Problems Involving Linear Equations and Linear Inequalities? Systems of Linear Equations and Problem Solving. 6 equations in 4 variables, 3. (5.2.3) – Solve mixture problems with a system of linear equations. In mathematics, a system of linear equations (or linear system) is a collection of one or more linear equations involving the same set of variables. ... Systems of equations word problems (with zero and infinite solutions) Get 3 of 4 questions to level up! This is one reason why linear algebra (the study of linear systems and related concepts) is its own branch of mathematics. In this algebra activity, students analyze word problems, define variables, set up a system of linear equations, and solve the system. HIDE SOLUTIONS. 1. Systems of 2 linear equations - problems with solutions Test. System of Linear Equations - Problem Solving on Brilliant, the | {
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- problems with solutions Test. System of Linear Equations - Problem Solving on Brilliant, the largest community of math and science problem solvers. Section 8.1, Example 4(a) Solve graphically: Linear systems are usually expressed in the form Ax + By = C, where A, B, and C are real numbers. In your studies, however, you will generally be faced with much simpler problems. When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. For example, the sets in the image below are systems of linear equations. This premium worksheet bundle contains a printable fact file and 10 fun and engaging worksheets to challenge your students and help them learn about Solving Word Problems Involving Linear Equations and Linear Inequalities. Substitution Method. Problem 1 Two of the following systems of equations have solution (1;3). B. Determining the value of k for which the system has no solutions. To use Khan Academy you need to upgrade to another web browser. So a System of Equations could have many equations and many variables. There are 7 questions. Donate or volunteer today! A large pizza at Palanzio’s Pizzeria costs$6.80 plus $0.90 for each topping. Our mission is to provide a free, world-class education to anyone, anywhere. System of NonLinear Equations problem example. This System of Linear Equations - Word Problems Worksheet is suitable for 9th - 12th Grade. A solution is a mixture of two or more different substances like water and salt or vinegar and oil. This section provides materials for a session on solving a system of linear differential equations using elimination. Khan Academy is a 501(c)(3) nonprofit organization. The directions are from TAKS so do all three (variables, equations and solve) no matter what is asked in the problem. The statement above gives the following equation m + 10 = 2 × (s + 10) m + 10 = 2 × s + 2× 10 m + 10 = 2s + 20 You now have a system of linear | {
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m + 10 = 2 × (s + 10) m + 10 = 2 × s + 2× 10 m + 10 = 2s + 20 You now have a system of linear equation to solve m + s = 40 equation 1 m + 10 = 2s + 20 equation 2 Use equation 1 to solve for m m + s = 40 m + s - s = 40 - s m = 40 - s If you're seeing this message, it means we're having trouble loading external resources on our website. For example, + − = − + = − − + − = is a system of three equations in the three variables x, y, z.A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied. One application of systems of equations are mixture problems. This section covers: Systems of Non-Linear Equations; Non-Linear Equations Application Problems; Systems of Non-Linear Equations (Note that solving trig non-linear equations can be found here).. We learned how to solve linear equations here in the Systems of Linear Equations and Word Problems Section.Sometimes we need solve systems of non-linear equations, such as those we see in conics. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. System of linear equations System of linear equations can arise naturally from many real life examples. Add the second equation to the first equation and solve for x. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. set of two (or more) equations which have two (or more) variables Solving Systems of Linear Equations. It has 6 unique word problems to solve including one mixture problem … If the two lines intersect at a single point, then there is one solution for the system: the point of intersection. Simultaneous equations (Systems of linear equations): Problems with Solutions. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, | {
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be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$4x - 7\left( {2 - x} \right) = 3x + 2$$, $$2\left( {w + 3} \right) - 10 = 6\left( {32 - 3w} \right)$$, $$\displaystyle \frac{{4 - 2z}}{3} = \frac{3}{4} - \frac{{5z}}{6}$$, $$\displaystyle \frac{{4t}}{{{t^2} - 25}} = \frac{1}{{5 - t}}$$, $$\displaystyle \frac{{3y + 4}}{{y - 1}} = 2 + \frac{7}{{y - 1}}$$, $$\displaystyle \frac{{5x}}{{3x - 3}} + \frac{6}{{x + 2}} = \frac{5}{3}$$. So far, we’ve basically just played around with the equation for a line, which is . 1. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Solve each of the following equations and check your answer. Section 7-1 : Linear Systems with Two Variables. Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve age word problems with a system of equations. A solution of the system (*) is a sequence of numbers$s_1, s_2, \dots, s_n$such that the substitution$x_1=s_1, x_2=s_2, \dots, x_n=s_n$satisfies all the$m$equations in the system (*). Updated June 08, 2018 In mathematics, a linear equation is one that contains two variables and can be plotted on a graph as a straight line. You discover a store that has all jeans for$25 and all dresses for 50. Improve your math knowledge with free questions in "Solve systems of linear | {
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and all dresses for 50. Improve your math knowledge with free questions in "Solve systems of linear equations" and thousands of other math skills. When this is done, one of three cases will arise: Case 1: Two Intersecting Lines . Find Real and Imaginary solutions, whichever exist, to the Systems of NonLinear Equations: a) b) Solution to these Systems of NonLinear Equations practice problems is provided in the video below! A system of linear equations is a system made up of two linear equations. Solution: Rewrite in order to align the x and y terms. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6 . 2 equations in 3 variables, 2. Generally speaking, those problems come up when there are two unknowns or variables to … Solving systems of equations word problems worksheet For all problems, define variables, write the system of equations and solve for all variables. If you're seeing this message, it means we're having trouble loading external resources on our website. The same rules apply. a)\begin{array}{|l} x + y = 5 \\ 2x - y = 7; \end{array}\$ In "real life", these problems can be incredibly complex. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. Answer: x = .5; y = 1.67. Solving using Matrices by Elimination. We can use the Intersection feature from the Math menu on the Graph screen of the TI-89 to solve a system of two equations in two variables. | {
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Learn faster with a math tutor. 9th - 10th grade. The two pairs of congruent sides may be, but do not have to be, congruent to each other. Get better grades with tutoring from top-rated professional tutors. This is one of the most important properties of parallelogram that is helpful in solving many mathematical problems related to 2-D geometry. 3) Are opposite angles of a parallelogram congruent? Solve for x. (10 Properties of parallelogram: Opposite sides of parallelogram are equal . The diagonals of a parallelogram bisect each other. Sides of A Parallelogram The opposite sides of a parallelogram are congruent. (By definition). The angles of a parallelogram are congruent. Ask yourself which approach looks easier or quicker. Line segments XY and ZW are also congruent. One way all sides of the two parallelograms could be congruent would be if $ABCD$ and $EFGH$ are squares with the same side length: in this case they would be congruent. Properties of parallelograms are as follows: i. Parallelogram Properties DRAFT. Tags: Question 19 . CCSS.MATH.CONTENT.HSG.CO.C.11 Prove theorems about parallelograms. Is the quadrilateral a parallelogram? . Give a reason. Expand Image In fact, one method of proving a quadrilateral a rhombus is by first proving it a parallelogram, and then proving two consecutive sides congruent, diagonals bisecting verticies, etc. A parallelogram is a quadrilateral that has opposite sides that are parallel. If one side is longer than its opposite side, you do not have parallel sides; no parallelogram! Draw the diagonal BD, and we will show that ΔABD and ΔCDB are congruent. Opposite sides are parallel and congruent. It is a quadrilateral with two pairs of parallel, congruent sides. Opposite angles are congruent. High School: Geometry » Congruence » Prove geometric theorems » 11 Print this page. Parallelogram Properties DRAFT. A parallelogram is defined to be a quadrilateral with 2 pairs of opposite sides parallel. Parallelograms have opposite interior | {
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to be a quadrilateral with 2 pairs of opposite sides parallel. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. Consecutive angles are supplementary (A + D = 180°). The two diagonals of a parallelogram bisect each other. A parallelogram is a quadrilateral with two pairs of parallel sides. 5) Does a diagonal of a parallelogram bisect a pair of opposite angles? A parallelogram also has the following properties: Opposite angles are congruent; Opposite sides are congruent; Adjacent angles are supplementary; The diagonals bisect each other. Yes, if both pairs of opposite angles are congruent, then you have a parallelogram. The diagonals of an isosceles trapezoid are congrent. 0 Why are these two lines not congruent (and other ways to figure out if other shapes are not congruent) Solve for x. HELP ASAP 30 points Part 1 out of 2 To repair a large truck or bus, a mechanic might use a parallelogram lift. 2 years ago. One has to be on the lookout for double negatives. 62% average accuracy. Opposite sides are congruent. You can draw parallelograms. The opposite angles are congruent. 60 seconds . A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . If yes, state how you know. There are two ways to go about this. If one angle is right, then all angles are right. The diagonals of a rectangle are the bisectors of the angles. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. An equilateral quadrilateral is a square. That is true. Here are some important things that … Try to move the vertices A, B, and D and observe how the figure changes. The opposite sides of a parallelogram are congruent so we will need two pairs of congruent segments: Now if we imagine leaving $\overline{AB}$ fixed and ''pushing down'' on side | {
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of congruent segments: Now if we imagine leaving $\overline{AB}$ fixed and ''pushing down'' on side $\overline{CD}$ so that these two sides become closer while side $\overline{AD}$ and $\overline{BC}$ rotate clockwise we get a new parallelogram: There is one right angle in a parallelogram and it is not a rectangle. If a quadrilateral has three angles of equal measure, then the fourth angle must be a right angle. The converse is also true that if opposite sides of a quadrangle are equal then its a parallelogram. 62% average accuracy. The figure is a parallelogram. The diagonals of a trapezoid are perpendicular. For our parallelogram, we will label it WXYZ, but you can use any four letters as long as they are not the same as each other. A rhombus is a parallelogram but with all four sides equal in length; A square is a parallelogram but with all sides equal in length and all interior angles 90° A quadrilateral is a parallelogram if: Both pairs of opposite sides are parallel. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. Tags: Question 19 . The diagonals of a quadrilateral_____bisect each other, If the measures of 2 angles of a quadrilateral are equal, then the quadrilateral is_____a parallelogram, If one pair of opposite sides of a quadrilateral is congruent and parallel, then the quadrilateral is______a parallelogram, If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is_____a parallelogram, To prove a quadrilateral is a parallelogram, it is ________enough to show that one pair of opposite sides is parallel, The diagonals of a rectangle are_____congruent, The diagonals of a parallelogram_______bisect the angles, The diagonals of a parallelogram______bisect the angles of the parallelogram, A quadrilateral with one pair of sides congruent and on pair parallel is_______a | {
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the parallelogram, A quadrilateral with one pair of sides congruent and on pair parallel is_______a parallelogram, The diagonals of a rhombus are_______congruent, A rectangle______has consecutive sides congruent, A rectangle_______has perpendicular diagonals, The diagonals of a rhombus_____bisect each other, The diagonals of a parallelogram are_______perpendicular bisectors of each other, Consecutive angles of a quadrilateral are_______congruent, The diagonals of a rhombus are______perpendicular bisectors of each other, Consecutive angles of a square are______complementary, Diagonals of a non-equilateral rectangle are______never angle bisectors, A quadrilateral with one pair of congruent sides and one pair of parallel sides is_____a parallelogram. B) The diagonals of the parallelogram are congruent. The opposite angles of a parallelogram are supplementary. Studying the video and these instructions, you will learn what a parallelogram is, how it fits into the family of polygons, how to identify its angles and sides, how to prove you have a parallelogram, and what are its identifying properties. Reason for statement 8: If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. Triangles can be used to prove this rule about the opposite sides. Use a straightedge (ruler) to draw a horizontal line segment, then draw another identical (congruent) line segment some distance above and to one side of the first one, so they do not line up vertically. Opposite angles are congruent. Parallelogram definition A quadrilateral with both pairs of opposite sides parallel. 1981 times. Opposite sides are congruent. If both pairs are congruent, you have either a rhombus or a square. SURVEY . 2. A parallelogram is a quadrilateral that has opposite sides that are parallel. Check for any one of these identifying properties: You can use proof theorems about a plane, closed quadrilateral to discover if it is a parallelogram: You have learned | {
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theorems about a plane, closed quadrilateral to discover if it is a parallelogram: You have learned that a parallelogram is a closed, plane figure with four sides. You can have almost all of these qualities and still not have a parallelogram. Opposite angles are congruent. The opposite sides of a parallelogram are congruent. Local and online. Go with B. Give a reason. The diagonals of a rectangle bisect eachother. Theorem: If ABCD is a parallelogram then prove that its opposite sides are equal. We already mentioned that their diagonals bisect each other. The diagonals of a kite are the perpendicular bisectors of each other. More generally, a quadrilateral with 4 congruent sides is a rhombus. If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is_____a parallelogram Always To prove a quadrilateral is a parallelogram, it is ________enough to show that one pair of opposite sides is parallel The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides. , all 4 sides are congruent (definition of a rhombus). Get better grades with tutoring from top-rated private tutors. Check Next Opposite angles are congruent. Want to see the math tutors near you? Property that is characteristic of a parallelogram is that opposite sides are congruent. The diagonals of a quadrilateral are perpendicular and the quadrilateral is not a rhombus. Opposite angels are congruent (D = B). Prove that opposite sides of a parallelogram are congruent. Other shapes, however, are types of parallelograms. Which statement can be used to prove that a given parallelogram is a rectangle? 4) Do the diagonals of a parallelogram bisect each other? Observe that at any time, the opposite sides are parallel and equal. Now consider just the interior angles of parallelograms, ∠W, ∠X, ∠Y, and ∠Z. Theorem 1: Opposite Sides of a Parallelogram Are Equal In a parallelogram, the opposite sides are equal. In this lesson we will prove the | {
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Are Equal In a parallelogram, the opposite sides are equal. In this lesson we will prove the basic property of a parallelogram that the opposite sides in a parallelogram are equal. Q. In the video below: We will use the properties of parallelograms to determine if we have enough information to prove a given quadrilateral is a parallelogram. The four line segments making up the parallelogram are WX, XY, YZ, and ZW. Notice that line segments WX and YZ are congruent. Use this to prove that the quadrilateral must be a parallelogram. Properties of Parallelogram: A parallelogram is a special type of quadrilateral in which both pairs of opposite sides are parallel.Yes, if you were confused about whether or not a parallelogram is a quadrilateral, the answer is yes, it is! The first rhombus above is a square while the second one has angles of 60 and 120 degrees. If only one set of opposite sides are congruent, you do not have a parallelogram, you have a trapezoid. Things that you need to keep in mind when you prove that opposite sides of a parallelogram are congruent. These geometric figures are part of the family of parallelograms: For such simple shapes, parallelograms have some interesting properties. Solution: A Parallelogram can be defined as a quadrilateral whose two s sides are parallel to each other and all the four angles at the vertices are not 90 degrees or right angles, then the quadrilateral is called a parallelogram. Give the given and prove and prove this. A parallelogram does not have other names. The interior angles are ∠W, ∠X, ∠Y, and ∠Z. Make sure that second line segment is parallel to (or equidistant from) the first line segment. Yes. The consecutive angles of a parallelogram are supplementary. The opposite sides are equal and parallel; the opposite angles are also equal. 1-to-1 tailored lessons, flexible scheduling. 1981 times. Connecting opposite (non-adjacent) vertices gives you diagonals WY and XZ. A parallelogram is a flat shape with four straight, | {
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vertices gives you diagonals WY and XZ. A parallelogram is a flat shape with four straight, connected sides so that opposite sides are congruent and parallel. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . 2 years ago. Start at any vertex (corner). The area of a parallelogram is twice the area of a triangle created by one of its diagonals. We will show that in that case, they are also equal to each other. That the other pair of opposite sides are congruent. Opposite sides are congruent (AB = DC). Is the quadrilateral a parallelogram? Take a rectangle and push either its left or ride side so it leans over; you have a parallelogram. In today's lesson, we will show that the opposite sides of a parallelogram are equal to each other. Let’s play with the simulation given below to better understand a parallelogram and its properties. This means a parallelogram is a plane figure, a closed shape, and a quadrilateral. The parallelogram has the following properties: Opposite sides are parallel by definition. An equilateral parallelogram is equiangular. Proving That a Quadrilateral is a Parallelogram, Opposite sides are parallel -- Look at the parallelogram in our drawing. The bad in Answer A is due to your teachers written grammar. Is it possible to prove a quadrilateral a parallelogram with two consecutive and two opposite congruent sides? A parallelogram also has the following properties: Opposite angles are congruent; Opposite sides are congruent; Adjacent angles are supplementary; The diagonals bisect each other. That’s a wrap! A parallelogram is defined as a quadrilateral where the two opposite sides are parallel. In the figure given below, ABCD is a parallelogram. The figure shows a side view of the li … ft. FGKL, GHJK, and FHJL are parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with | {
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diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. The Angle-Side-Angle Triangle Congruence Theorem can be used to prove that, in a parallelogram, opposite sides are congruent. There are more than two right angles in a trapezoid. Opposite sides are congruent -- The base side (Y Z Y Z) and the top side (W X W X) of our parallelogram are equal in length (congruent); the left side (XY X Y) and right side (ZW Z W) are also congruent To be a parallelogram, the base and top sides must be parallel and congruent, and … The opposite sides of parallelogram are also equal in length. CCSS.MATH.CONTENT.HSG.CO.C.11 Prove theorems about parallelograms. To show these two triangles are congruent we’ll use the fact that this is a parallelogram, and as a result, the two opposite sides are parallel, and the diagonal acts as a transv… Parallelogram law. Parallelogram Theorem #2: The opposite sides of a parallelogram are congruent. SURVEY . View Untitled document (3).pdf from MATH 100 at Basha High School. If the four sides do not connect at their endpoints, you do not have a closed shape; no parallelogram! Mathematics. Solution: A Parallelogram can be defined as a quadrilateral whose two s sides are parallel to each other and all the four angles at the vertices are not 90 degrees or right angles, then the quadrilateral is called a parallelogram. The bottom (base) side, Opposite sides are congruent -- The base side (. C) The diagonals of the parallelogram bisect the angles. 60 seconds . Get help fast. So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. G 2 T: + 3 5 6 8 17 K 3 Which angles are congruent to 21? <2 2 are congruent to 21. As with any quadrilateral, the interior angles add to 360°, but you can also know more about a parallelogram's angles: Using the properties of diagonals, sides, and angles, you can always identify parallelograms. Opposite sides are congruent. | {
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diagonals, sides, and angles, you can always identify parallelograms. Opposite sides are congruent. One interesting property of a parallelogram is that its two diagonals bisect each other (cut each other in half). Or: Both pairs of opposite sides are congruent. To explore these rules governing the sides of a parallelogram use Math Warehouse's interactive parallelogram. Triangles can be used to prove this rule about the opposite sides. D) The opposite angles of the parallelogram are congruent. Opposite sides of a … ∴ ∴ AB = CD A B = C D and AD= BC A D = B C Properties of a parallelogram. A parallelogram is a flat shape with four straight, connected sides so that opposite sides are congruent and parallel. So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. Opposite angles are equal (congruent) to each other; Any two adjacent angles of a parallelogram add up to, This means any two adjacent angles are supplementary (adding to, A closed shape (it has an interior and exterior), A quadrilateral (four-sided plane figure with straight sides), Two pairs of congruent (equal), opposite angles, Two pairs of equal and parallel opposite sides, If the quadrilateral has bisecting diagonals, it is a parallelogram, If the quadrilateral has two pairs of opposite, congruent sides, it is a parallelogram, If the quadrilateral has consecutive supplementary angles, it is a parallelogram, If the quadrilateral has one set of opposite parallel, congruent sides, it is a parallelogram. To be on the lookout for double negatives proof 2 here ’ use. Be on the lookout for double negatives not congruent have opposite interior angles that are parallel, opposite are! = ∠Y and ∠X = ∠Z one interesting property of a parallelogram and it not! ( base ) side, opposite sides of a parallelogram segments making up the parallelogram in our drawing:... A square while the second one has angles of 60 and 120 degrees or counterclockwise to Next. The property that is | {
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to prove this rule about the opposite sides not. Side, you do not have to be a quadrilateral in which both pairs of parallel, then fourth. It leans over ; you have a trapezoid and name the appropriate geometric are... Two ways to go about this and so must the left and side... Quadrangle are equal side so it leans over ; you have a trapezoid top must! That in that case, they are also equal to each other and each one separates the are... Is defined as a quadrilateral the family of parallelograms: for such shapes. From ) the first rhombus above is a parallelogram lift set of opposite sides are congruent ( D = C. Parallelogram with two pairs of opposite sides of a quadrilateral has three angles of parallelograms this is right... Has angles of a triangle created by one of the vector cross product of two adjacent sides B C. Has opposite sides for statement 1: given one angle is right, then you have a parallelogram?... Rhombus or a square then move either clockwise or counterclockwise to the Next vertex defined to be parallelogram!, GHJK, and FHJL are parallelograms figure is a flat shape with four straight connected... Has the following properties: two pairs of opposite sides are parallel and equal that! That opposite sides congruent, and ∠Z ∴ ∴ AB = DC.! Of opposite angles are right ∠X = ∠Z have to be a quadrilateral a parallelogram use Math 's... | {
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## Linear Algebra and Its Applications, Exercise 1.5.19
Exercise 1.5.19. For the following two matrices, specify the values of $a$, $b$, and $c$ for which elimination requires row exchanges, and the values for which the matrices in question are singular.
$A = \begin{bmatrix} 1&2&0 \\ a&8&3 \\ 0&b&5 \end{bmatrix} \quad A = \begin{bmatrix} c&2 \\ 6&4 \end{bmatrix}$
Answer: For the first matrix a row exchange would be necessary if the first step of elimination caused the second entry of the second row, i.e., in the (2,2) position, to become 0. This would occur if we multiplied the first row by 4 and subtracted it from the second row, as we would then have $8 - 4 \cdot 2 = 0$ in the (2,2) position. Since the entry in the (1,1) position is 1, multiplying the first row by 4 would be necessary if $a$ were 4:
$\begin{bmatrix} 1&2&0 \\ 4&8&3 \\ 0&b&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&0&3 \\ 0&b&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&b&5 \\ 0&0&3 \end{bmatrix}$
If we had both $a = 4$ and $b = 0$ then the matrix would be singular:
$\begin{bmatrix} 1&2&0 \\ 4&8&3 \\ 0&0&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&0&3 \\ 0&0&5 \end{bmatrix}$
However, these are not the only values of $a$ and $b$ for which the matrix is singular. Suppose we proceed with elimination. The first step would be to multiply the first row by $l_1 = a$ and subtract the result from the second row:
$\begin{bmatrix} 1&2&0 \\ a&8&3 \\ 0&b&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&8-2a&3 \\ 0&b&5 \end{bmatrix}$
Assuming that $a \ne 4$ then the next step of elimination would multiply the second row of the matrix by $l_2 = b/(8-2a)$ and subtract the result from the third row:
$\begin{bmatrix} 1&2&0 \\ 0&8-2a&3 \\ 0&b&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&8-2a&3 \\ 0&0&5-3b/(8-2a) \end{bmatrix}$ | {
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If the (3,3) entry of the resulting matrix above is zero then the original matrix is singular. In this case we have $3b/(8-2a) = 5$. Multiplying both sides by $(8-2a)$ we have $3b = 5 \cdot (8-2a) = 40 - 10a$ or $10a + 3b = 40$.
So the first matrix is singular for all values of $a$ and $b$ for which $10a+3b=40$. This includes the case $a = 4, b=0$ mentioned above, the case $a=0, b = 40/3$, the case $a=5, b=-10/3$, and an infinity of others on the line $b=-\frac{10}{3}a + \frac{40}{3}$.
For the second matrix a row exchange would be necessary if $c = 0$:
$\begin{bmatrix} 0&2 \\ 6&4 \end{bmatrix} \rightarrow \begin{bmatrix} 6&4 \\ 0&2 \end{bmatrix}$
On the other hand, if c = 3 then the first elimination produces zeroes in both entries of the second row, and the matrix is singular:
$\begin{bmatrix} 3&2 \\ 6&4 \end{bmatrix} \rightarrow \begin{bmatrix} 3&2 \\ 0&0 \end{bmatrix}$
UPDATE: Prompted by Theodore’s comment, added a section deriving the complete set of $a$ and $b$ for which the first matrix is singular.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.
This entry was posted in linear algebra. Bookmark the permalink.
### 2 Responses to Linear Algebra and Its Applications, Exercise 1.5.19
1. Theodore says:
I also found a = 5 and b = -10/3 to also cause a singular matrix. My question is how do we find all values that can cause singularity
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• hecker says:
Thank you for finding this problem. You are correct that there are multiple (actually, an infinity) of values of a and b for which the first matrix is singular. I added a new section to derive the correct answer. | {
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Accelerating the pace of engineering and science
# istril
Determine if matrix is lower triangular
## Description
example
tf = istril(A) returns logical 1 (true) if A is a lower triangular matrix; otherwise, it returns logical 0 (false).
## Examples
expand all
### Test Lower Triangular Matrix
Create a 5-by-5 matrix.
`D = tril(magic(5))`
```D =
17 0 0 0 0
23 5 0 0 0
4 6 13 0 0
10 12 19 21 0
11 18 25 2 9```
Test D to see if it is lower triangular.
`istril(D)`
```ans =
1```
The result is logical 1 (true) because all elements above the main diagonal are zero.
### Test Matrix of Zeros
Create a 5-by-5 matrix of zeros.
`Z = zeros(5);`
Test Z to see if it is lower triangular.
`istril(Z)`
```ans =
1```
The result is logical 1 (true) because a lower triangular matrix can have any number of zeros on its main diagonal.
## Input Arguments
expand all
### A — Input arraynumeric array
Input array, specified as a numeric array. istril returns logical 0 (false) if A has more than two dimensions.
Data Types: single | double
Complex Number Support: Yes
expand all
### Lower Triangular Matrix
A matrix is lower triangular if all elements above the main diagonal are zero. Any number of the elements on the main diagonal can also be zero.
For example, the matrix
$A=\left(\begin{array}{cccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ -1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ -2& -2& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& 0\\ -3& -3& -3& 1\end{array}\right)$
is lower triangular. A diagonal matrix is both upper and lower triangular.
### Tips | {
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is lower triangular. A diagonal matrix is both upper and lower triangular.
### Tips
• Use the tril function to produce lower triangular matrices for which istril returns logical 1 (true).
• The functions isdiag, istriu, and istril are special cases of the function isbanded, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, istril(A) == isbanded(A,size(A,1),0). | {
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# Solve the equation $\cos^2x+\cos^22x+\cos^23x=1$
Solve the equation: $$\cos^2x+\cos^22x+\cos^23x=1$$ IMO 1962/4
My first attempt in solving the problem is to simplify the equation and express all terms in terms of $\cos x$. Even without an extensive knowledge about trigonometric identities, the problem is solvable.
\begin{align} \cos^22x&=(\cos^2x-\sin^2x)^2\\ &=\cos^4x+\sin^4x-2\sin^2\cos^2x\\ &=\cos^4+(1-\cos^2x)^2-2(1-\cos^2)\cos^2x\\ &=\cos^4+1-2\cos^2x+\cos^4x-2\cos^2x+2\cos^4x\\ &=4\cos^4x-4\cos^2x+1 \end{align}
Without knowledge of other trigonometric identities, $\cos3x$ can be derived using only Ptolemy's identities. However for the sake of brevity, let $\cos 3x=4\cos^3x-3\cos x$: \begin{align} \cos^23x&=(4\cos^3x-3\cos x)^2\\ &=16\cos^6x+4\cos^2x-24\cos^4x \end{align}
Therefore, the original equation can be written as: $$\cos^2x+4\cos^4x-4\cos^2x+1+16\cos^6x+4\cos^2x-24\cos^4x-1=0$$ $$-20\cos^4x+6\cos^2x+16\cos^6x=0$$ Letting $y=\cos x$, we now have a polynomial equation: $$-20y^4+6y^2+16y^6=0$$ $$y^2(-20y^2+6y+16y^4)=0\Rightarrow y^2=0 \Rightarrow x=\cos^{-1}0=\bbox[yellow,10px]{90^\circ}$$ From one of the factors above, we let $z=y^2$, and we have the quadratic equation: $$16z^2-20z+6=0\Rightarrow 8z^2-10z+3=0$$ $$(8z-6)(z-\frac12)=0\Rightarrow z=\frac34 \& \ z=\frac12$$ Since $z=y^2$ and $y=\cos x$ we have: $$\biggl( y\rightarrow\pm\frac{\sqrt{3}}{2}, y\rightarrow\pm\frac{\sqrt{2}}2 \biggr)\Rightarrow \biggl(x\rightarrow\cos^{-1}\pm\frac{\sqrt{3}}{2},x\rightarrow\cos^{-1}\pm\frac{\sqrt{2}}2\biggr)$$ And thus the complete set of solution is: $$\bbox[yellow, 5px]{90^\circ, 30^\circ, 150^\circ, 45^\circ, 135^\circ}$$
As I do not have the copy of the answers, I still hope you can verify the accuracy of my solution.
## But more importantly...
Seeing the values of $x$, is there a more intuitive and simpler way of finding $x$ that does away with the lengthy computation? | {
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• This post doesn't have the squares, but it could give some insight as to how one may think about this problem. – Arthur Mar 12 '18 at 13:12
• This link gives you the solutions. – Jose Arnaldo Bebita-Dris Mar 12 '18 at 13:15
• I expect there to be a much more elegant solution to this as it is an IMO problem. Now only someone has to find it... – vrugtehagel Mar 12 '18 at 13:15
• @vrugtehagel I believe the link to the solution is already an elegant answer! – John Glenn Mar 12 '18 at 13:17
• Yup, I think so too! It was shortly posted before my comment, so I hadn't seen it. I advise @JoseArnaldoBebitaDris to summarize that solution and post it as answer, to avoid this question lingering in the unanswered section of this website – vrugtehagel Mar 12 '18 at 13:21
This is a summary of the solution found in this hyperlink.
We can write the LHS as a cubic function of $\cos^2 x$. This means that there are at most three values of $x$ that satisfy the equation.
Hence, we look for three values of $x$ that satisfy the equation and produce three distinct $\cos^2 x$. Indeed, we find that $$\frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{6}$$ all satisfy the equation, and produce three different values for $\cos^2 x$, namely $0, \frac{1}{2}, \frac{3}{4}$.
Lastly, we solve the resulting equations $$\cos^2 x = 0$$ $$\cos^2 x = \frac{1}{2}$$ $$\cos^2 x = \frac{3}{4}$$ separately. We conclude that our solutions are: $$x=\frac{(2k+1)\pi}{2}, \frac{(2k+1)\pi}{4}, \frac{(6k+1)\pi}{6}, \frac{(6k+5)\pi}{6}, \forall k \in \mathbb{Z}.$$
You can shorten the argument by noting at the outset that $$\cos3x=4\cos^3x-3\cos x=(4\cos^2x-3)\cos x$$ so if we set $y=\cos^2x$ we get the equation $$y+(2y-1)^2+y(4y-3)^2=1$$ When we do the simplifications, we get $$2y(8y^2-10y+3)=0$$ The roots of the quadratic factor are $3/4$ and $1/2$. | {
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A different strategy is to note that $\cos x=(e^{ix}+e^{-ix})/2$, so the equation can be rewritten $$e^{2ix}+2+e^{-2ix}+e^{4ix}+2+e^{-4ix}+e^{6ix}+2+e^{-6ix}=4$$ Setting $z=e^{2ix}$ we get $$2+z+z^2+z^3+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}=0$$ or as well $$z^6+z^5+z^4+2z^3+z^2+z+1=0$$ that can be rewritten (noting that $z\ne1$), $$\frac{z^7-1}{z-1}+z^3=0$$ or $z^7+z^4-z^3-1=0$ that can be factored as $$(z^3+1)(z^4-1)=0$$ Hence we get (discarding the spurious root $z=1$) $$2x=\begin{cases} \dfrac{\pi}{3}+2k\pi \\[6px] \pi+2k\pi \\[6px] \dfrac{5\pi}{3}+2k\pi \\[12px] \dfrac{\pi}{2}+2k\pi \\[6px] \pi+2k\pi \\[6px] \dfrac{3\pi}{2}+2k\pi \end{cases} \qquad\text{that is}\qquad x=\begin{cases} \dfrac{\pi}{6}+k\pi \\[6px] \dfrac{\pi}{2}+k\pi \\[6px] \dfrac{5\pi}{6}+k\pi \\[6px] \dfrac{\pi}{4}+k\pi \\[6px] \dfrac{3\pi}{4}+k\pi \end{cases}$$
• Great! An elegant solution too! – John Glenn Mar 12 '18 at 14:42
Hint:
$$0=\cos^2x+\cos^22x+\cos^23x-1$$
$$=\cos(3x+x)\cos(3x-x)+\cos^22x$$
$$=\cos2x(\cos4x+\cos2x)$$ | {
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# Calculation of modular multiplicative inverse of A mod B when A > B
I'm trying to understand a Montgomery reduction algorithm, for which I need to calculate a multiplicative inverse. However, euclidean algorithm only helps if A < B.
Example is 11 mod 3. Multiplicative inverse of 11 is 2,but ext_gcd gives you Bezout numbers such as -1 and 4.
https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm
Wikipedia says so: The extended Euclidean algorithm is particularly useful when a and b are coprime, since x is the modular multiplicative inverse of a modulo b, and y is the modular multiplicative inverse of b modulo a.
But as far as I see this can't be true, either X is multiplicative inverse of A modulo B or Y is multiplicative inverse of B modulo A, but not both at the same time, because one of them (A or B) is going to be bigger than another. We have X=4, Y=-1 for A=3,B=11, and X=4 is valid inverse, while -1 is indeed not.
A lot of online calculators that I tried are also said that a has to be bigger than be, but they (some of them) are still able to calculate inverse of 11 mod 3.
The only workaround I found so far is perform A = A mod B first, so A is now a remainder of divisios and therefore is less than modulus, so we can perform ext_gcd(2, 3) now and get our 2 as answer.
Probably I'm missing something, this thing is pretty new for me.
Thanks. | {
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Probably I'm missing something, this thing is pretty new for me.
Thanks.
• "We have X=4, Y=-1 for A=3,B=11, and X=4 is valid inverse, while -1 is indeed not. " Huh. $X \equiv A^{-1}\mod B$ because $X*A = 4*3 \equiv 1 \mod 11$. An $Y\equiv B^{-1} \mod A$ because $Y*B=(-1)*11 \equiv 1 \mod 3$. They are both valid inverses. What is your issue? – fleablood Jun 15 at 18:48
• $-1\equiv 2 \mod 3$. So $-1$ is equivalent to $2$. It doesn't make the slightest difference which one you use. (We don't call these things "equivalences" for nothing, you know...) – fleablood Jun 15 at 19:16
• "A lot of online calculators that I tried are also said that a has to be bigger than be," Calculators don't do mathematics. Calculators do calculations. – fleablood Jun 15 at 19:24
It is inevitable that a Bézout's identity equation will give you modular multiplicative inverses, since given:
$$am+bn = 1$$
we can take $\bmod m$ for
$$bn\equiv 1 \bmod m$$
or $\bmod n$ for
$$am \equiv 1 \bmod n$$
To get $a$ and $b$ in your preferred range, you can simply add or subtract a suitable multiple of the modulus.
So in this case $$-1\cdot 11 + 4\cdot 3 = 1$$
and thus $$-1\cdot 11\equiv 1 \bmod 3$$
($-11$ being one more than $-12$), so $-1$ is a valid inverse of $11$ modulo $3$. Then of course $$-1\equiv 2 \bmod 3$$
so this is consistent with your observation that $2$ is the inverse of $11 \bmod 3$ also.
• I now understand why this is correct from algorithmical point of view, but I don't understand why −1 ≡ 2 mod 3. I miss some math-related thing. – Danetta Jun 18 at 9:12
• Presumably you have no problem with $11 \equiv 8\equiv 5 \equiv 2 \bmod 3$. In each case, add one and you arrive at a multiple of $3$. The same applies to $-1$. Also, the difference between $-1$ and $2$ is a multiple of $3$ (as is true for any pair of the equivalent values above). – Joffan Jun 18 at 12:52
$-1 \equiv 2 \mod 3$ so they are considered to be the same thing.
That's we we call these equivalence classes. | {
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That's we we call these equivalence classes.
However, euclidean algorithm only helps if A < B
I simply do not understand why you say that.
either X is multiplicative inverse of A modulo B or Y is multiplicative inverse of B modulo A, but not both at the same time, because one of them (A or B) is going to be bigger than another. We have X=4, Y=-1 for A=3,B=11, and X=4 is valid inverse, while -1 is indeed not.
Except, of course, it indeed is. $-1*11 = -11 \equiv 1 \mod 3$. That is the valid inverse. Why do you think it is not?
It doesn't matter if $A > B$ or $B> A$ as $\gcd(A,B) = 1$ Euclid's algorithm will give us:
$mA + kB = 1$ so $k \equiv A^{-1} \mod B$ and $m \equiv B^{-1} \mod A$ simultaneously.
Is your concern that one is represented with a positive number and the other negative?
That's irrelevent.
It doesn't matter which representative we use to represent a class. We could have used $50\equiv -1 \mod 3$ so $50 \equiv 11^{-1}\mod 3$ for all we care. (Indeed $50*11 = 550 = 3*183 + 1 \equiv 1\mod 3$).
Note: if $mA + kB = 1$ and $m > 0$ but $-A < k < 0$ then
$mA + (k+A)B = 1 + BA\equiv 1 \mod A,B,AB$ and $m$ and $(k+A)$ are still the proper inverses. ANd $m > 0; k+A > 0$.
Indeed $(m + vB)A + (k + uA)B = 1 + (v+u)AB$ so $m + vB\equiv A^{-1} \mod B$ for any integer $v$ and $k + uA \equiv B^{-1} \mod A$ for any integer $u$. | {
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• "Is your concern that one is represented with a positive number and the other negative?" — Yes, I haven't even considered negative numbers. I'm still not exactly sure how −11 ≡ 1 mod 3. For me logically it looks like quotient of -11/3 is -3, so remainder is either 2 or -2, but how is it 1? – Danetta Jun 18 at 8:56
• It looks like a rule to "just take a remainder from division" stops working there. Do I need positive inverse for Montgomery or it doesn't matter? – Danetta Jun 18 at 9:17
• $-11 = 3*(-4)+ 1$ so $-11 \equiv 1 \mod 3$ and $-11 = 3*(-3) -2$ so $-11 \equiv -2 \mod 3$. And $1 \equiv -2 \mod 3$. As well $-11 = 3*(-7)+12$ so $-11 \equiv 10 \mod 3$. But $-11 \ne 3*k + 2$ for any integer $k$ so $-11 \not\equiv 2 \mod 3$. A remainder means any $p = d*n + r$ and $d$ can be any value positive or a negative and $r$ can be any value that works. But THE remainder would mean $p=d*n+r;0\le r< n$ would mean the remainder is unique, positive and less than the divisor. If $p < 0$ this means $d < 0$ and $r \ge 0$. – fleablood Jun 18 at 15:01
• I dont know how to do the Montgomery method. It seems as though you must already now then inverse to use it and that it is just an computational efficient method to do multiplication. – fleablood Jun 18 at 15:18 | {
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# Is the Trace of the Transposed Matrix the Same as the Trace of the Matrix?
## Problem 633
Let $A$ be an $n \times n$ matrix.
Is it true that $\tr ( A^\trans ) = \tr(A)$? If it is true, prove it. If not, give a counterexample.
## Solution.
The answer is true. Recall that the transpose of a matrix is the sum of its diagonal entries. Also, note that the diagonal entries of the transposed matrix are the same as the original matrix.
Putting together these observations yields the equality $\tr ( A^\trans ) = \tr(A)$.
Here is the more formal proof.
For $A = (a_{i j})_{1 \leq i, j \leq n}$, the transpose $A^{\trans}= (b_{i j})_{1 \leq i, j \leq n}$ is defined by $b_{i j} = a_{j i}$.
In particular, notice that $b_{i i} = a_{i i}$ for $1 \leq i \leq n$. And so,
$\tr(A^{\trans}) = \sum_{i=1}^n b_{i i} = \sum_{i=1}^n a_{i i} = \tr(A) .$
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# Find volume of solid generated (Calc 2)
#### lovex25
##### New member
[solved]Find volume of solid generated (Calc 2)
Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes, the curve y=e^x, and the line x = ln 2 about the line x= ln 2.
So I tried graphing it to see visually, and the expression I got for calculating the volume was ∫π(ln2-lny)^2dy, evaluating from 0 to 2 using disk method, and the answer I got was 4π, but apparently that doesn't match the answer in the back of the book. I'd really appreciate if someone can help me out!!
Off topic: First time posting a thread here, may I ask how do you type the mathematical symbols such as the integral sign and whatnot, or do I have to manually copy and paste from other website?
Last edited:
#### MarkFL
Staff member
Hello lovex25,
As you did, a good first step for these problems is to sketch a graph of the region to be revolved, and the axis of rotation:
Using the shell method, the volume of an arbitrary shell is:
$$\displaystyle dV=2\pi rh\,dx$$
where:
$$\displaystyle r=\ln(2)-x$$
$$\displaystyle h=e^x$$
and so we find:
$$\displaystyle dV=2\pi \left(\ln(2)-x \right)e^x\,dx$$
Now, you want to sum up the shells by integrating:
$$\displaystyle V=2\pi\int_0^{\ln(2)}\left(\ln(2)-x \right)e^x\,dx$$
To make things a bit simpler, I would use the substitution:
$$\displaystyle u=e^x\,\therefore\,du=e^x\,dx$$
and we now may state:
$$\displaystyle V=2\pi\int_1^2\ln(2)-\ln(u)\,du$$
Can you proceed?
Now, as far as posting mathematical expressions, this site supports $\LaTeX$, and a good tutorial written by our own Sudharaka can be found here:
http://mathhelpboards.com/latex-tips-tutorials-56/math-help-boards-latex-guide-pdf-1142.html
#### lovex25
##### New member
Hello lovex25,
As you did, a good first step for these problems is to sketch a graph of the region to be revolved, and the axis of rotation:
View attachment 1349 | {
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View attachment 1349
Using the shell method, the volume of an arbitrary shell is:
$$\displaystyle dV=2\pi rh\,dx$$
where:
$$\displaystyle r=\ln(2)-x$$
$$\displaystyle h=e^x$$
and so we find:
$$\displaystyle dV=2\pi \left(\ln(2)-x \right)e^x\,dx$$
Now, you want to sum up the shells by integrating:
$$\displaystyle V=2\pi\int_0^{\ln(2)}\left(\ln(2)-x \right)e^x\,dx$$
To make things a bit simpler, I would use the substitution:
$$\displaystyle u=e^x\,\therefore\,du=e^x\,dx$$
and we now may state:
$$\displaystyle V=2\pi\int_1^2\ln(2)-\ln(u)\,du$$
Can you proceed?
Now, as far as posting mathematical expressions, this site supports $\LaTeX$, and a good tutorial written by our own Sudharaka can be found here:
http://mathhelpboards.com/latex-tips-tutorials-56/math-help-boards-latex-guide-pdf-1142.html
thanks so much!!! finally i was able to match the answer to the back of the book, and I realized what went wrong with my attempt: I didn't separate the integrals to 2 different functions. As for the LaTeX software I'll look in to it soon, thanks again!
#### Prove It
##### Well-known member
MHB Math Helper
I personally prefer the discs method. Here you would need to split up your region of integration into two regions, the first being the rectangle below the point (0,1), and the second being the remaining region above it.
As for the volume of the region below (0,1) generated when rotating, that's easy, it's simply a cylinder of radius \displaystyle \begin{align*} \ln{(2)} \end{align*} units and height 1 unit, so its volume is \displaystyle \begin{align*} \pi \left[ \ln{(2)} \right] ^2 \end{align*}.
As for the volume of the region above (0,1), if we consider horizontal discs, they will each have radius \displaystyle \begin{align*} \ln{(2)} - \ln{(y)} \end{align*} and a height \displaystyle \begin{align*} \Delta y \end{align*}, where \displaystyle \begin{align*} \Delta y \end{align*} is some small change in y. So their total volume can be approximated by | {
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\displaystyle \begin{align*} V &\approx \sum \pi \left[ \ln{(2)} - \ln{(y)} \right] ^2 \Delta y \\ &= \pi \sum \left[ \ln{(2)} - \ln{(y)} \right] ^2 \Delta y \\ &= \pi \sum \left\{ \left[ \ln{(2)} \right] ^2 - 2\ln{(2)}\ln{(y)} + \left[ \ln{(y)} \right] ^2 \right\} \Delta y \end{align*}
And then as we increase the number of discs, making \displaystyle \begin{align*} \Delta y \end{align*} smaller, the sum converges on an integral and the approximation becomes exact, so the total volume is
\displaystyle \begin{align*} V &= \pi \int_1^2{ \left[ \ln{(2)} \right] ^2 - 2\ln{(2)}\ln{(y)} + \left[ \ln{(y)} \right] ^2 \, dy }\end{align*}
which is possible to be integrated | {
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# The tip of a colorful triangle
Original source: Problem 1 of British Informatics Olympiad 2017, Round 1
You're given a bunch of red (R), green (G), and blue (B) balls. I arrange some balls on a line. Then I ask you to complete the triangle of balls under the following simple rules, and tell me the color of the ball at the last row:
• Below two balls of the same color, place a ball of that color. (For example, a G must be placed under two G's.)
• Below two balls of different colors, place a ball of the third color. (For example, a B must be placed under a R and a G.)
If I gave you the balls "R R G B R G B B", you would place the balls like the following, and tell me "it's green":
R R G B R G B B
R B R G B R B
G G B R G G
G R G B G
B B R R
B G R
R B
G
Now, I'll give you a sequence of 60,000 balls. I won't even show you all the balls' colors; the only information available to you is that it starts with 999 balls of the pattern RGB RGB RGB ... RGB, and ends with 999 balls of the pattern BGR BGR BGR ... BGR.
Using the same rules, can you guess the color of the ball on the last row? The answer is unique, and an answer with mathematical explanation is preferred.
This game is a modified Pascal triangle.
First,
let's look at size 4. If you were to permute all the possible inputs and look at the first ball given, last ball given, and the color of the ball on the last row, you can see they are also conforming to the rules given.
We then
try 10. We can look at this like so, since size 4 works:
(R)* * B * * B * *(G)
* * * * * * * * *
* * * * * * * *
G * * B * * R
* * * * * *
* * * * *
R * * G
* * *
* *
(B)
In fact,
this works if the number is $$3^n + 1$$.
Alright,
let's say the colors red, green, and blue are numbers 0, 1, and 2 respectively.
We then visualize a Pascal mod 3 and the game flipped of size 10, with the top portion being one: | {
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We then visualize a Pascal mod 3 and the game flipped of size 10, with the top portion being one:
Pascal mod 3 | The game
1 | 1
1 1 | 2 2
1 2 1 | 1 2 1
1 0 0 1 | 2 0 0 2
1 1 0 1 1 | 1 1 0 1 1
1 2 1 1 2 1 | 2 1 2 2 1 2
1 0 0 2 0 0 1 | 1 0 0 2 0 0 1
1 1 0 2 2 0 1 1 | 2 2 0 1 1 0 2 2
1 2 1 2 1 2 1 2 1 | 1 2 1 2 1 2 1 2 1
1 0 0 0 0 0 0 0 0 1 | 2 0 0 0 0 0 0 0 0 2
This result looks similar - as Pascal's formula (mod 3) is
$${{n}\choose{k}} \equiv {{n - 1}\choose{k - 1}} + {{n - 1}\choose{k}} \pmod 3$$ denoting the nth row and kth column. But what about the game?
As $$1 + 0 \equiv 2 \pmod 3$$ and $$2 + 0 \equiv 1 \pmod 3$$ for the game,
while $$1 + 0 \equiv 1 \pmod 3$$ and $$2 + 0 \equiv 2 \pmod 3$$, we want to switch the terms, and that's simple. We negate this equation. The formula for the game is then:
$${{n}\choose{k}} \equiv -\left[{{{n - 1}\choose{k - 1}} + {{n - 1}\choose{k}}}\right] \pmod 3$$
then:
$${{n}\choose{k}} \equiv -{{n - 1}\choose{k - 1}} - {{n - 1}\choose{k}} \pmod 3$$
Alright, now let's show why size 4 works. We assign each column in the first row $$a, b, c, d$$, and we will use the above formula.
a b c d
-a-b -b-c -c-d
a+2b+c b+2c+d
-a-3b-3c-d
We modulo by 3 the expression in the last row and we get $$-a-d \pmod 3$$, and that also conforms to the provided equation above.
Because the question is modified,
and 60,000 is not $$3^n + 1$$, we must find the highest power of 3 plus 1 that's smaller than this, which is 59050 ($$3^{10} + 1$$).
Subtracting 60000 to 59050 gives us 950, the (n+1)th-to-last element (starting from the first ball). We use $$\bmod 3$$ because of the pattern, giving us $$2$$. | {
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Now, starting from the first ball which is R, and looking at the 3rd to last, which is B gives us G.
Looking at the second ball which is G, and looking at the 2nd to last, which is G gives us G.
The third ball is B, and looking at the 1st-to-last, which is R gives us G.
Since the next balls are repeating, it will be always periodic, but notice how it's always a G? That must mean the next colors, and the last row is Green!
• Yes, it is correct! – Bubbler Nov 16 '20 at 6:15 | {
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Math Help - Probability
1. Probability
I'm having a lot of trouble with permutations and combinations, so I was hoping someone here could help me out with a few questions. Please make sure that you explain your working; I really want to understand this topic... The first question asks;
A queue has 4 boys and 4 girls standing in line. Find how mant different arrangements are possible if;
a) The boys and girls alternate.
b) 2 particular girls wish to stand together.
c) All the boys stand together.
d) Also find the probability that 3 particular people will be in the queue together if the queue forms randomly.
EDIT: If it's any help, here are the answers...
a) 1152
b) 10080
c) 2880
d) 3/28
EDIT: I've got a new problem now;
A table has 4 boys and 4 girls sitting around it.
a) Find the number of ways of sitting possible if the boys and girls can sit anywhere around the table.
b) If the seating is arranged at random, find the probability that;
i) 2 particular girls will sit together.
ii) All the boys will sit together.
2. Hello, Flay!
We have to "talk" our way through these problems . . .
A queue has 4 boys and 4 girls standing in line.
Find how mant different arrangements are possible if;
a) The boys and girls alternate.
There are 2 possible arrangements: . $BGBGBGBG\,\text{ and }\,GBGBGBGB.$
The four boys can be placed in 4! ways.
The four girls can be placed in 4! ways.
Therefore, there are: . $2 \times 4! \times 4! \;=\;\boxed{1152}$ ways.
b) 2 particular girls wish to stand together.
Suppose the two girls are $X$ and $Y.$
Duct-tape them together.
Note that there are 2 possible orders: . $XY\,\text{ or }\,YX.$
Now we have seven "people" to arrange: . $\boxed{XY}\,,G,G,B,B,B,B$
. . and they can be arranged in ${\color{blue}7!}$ ways.
Therefore, there are: . $2 \times 71\;=\;\boxed{10,080}$ ways.
c) All the boys stand together.
Duct-tape the four boys together.
. . Note that there are 4! possible orders. | {
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Now we have five "people" to arrange.
. . There are $5!$ ways.
Therefore, there are: . $4! \times 5! \:=\:\boxed{2880}$ ways.
d) Find the probability that 3 particular people will be together
if the queue forms randomly.
First of all, there are 8! possible arrangements.
Suppose the three people are $X, Y\text{ and }Z.$
Duct-tape them together: . $\boxed{XYZ}$
. . Note that there are 3! orderings.
Now we have six "people" to arrange,
. . and there are 6! ways.
Hence, there are: . ${\color{red}3!\times6!}$ ways for $X,Y,Z$ to be together.
Therefore, the probability is: . $\frac{3!\cdot6!}{8!} \;=\;\boxed{\frac{3}{28}}$
3. Thanks very much. Now I have another problem;
A school committee is to be made up of 5 teachers, 4 students and 3 parents.
a) If 12 teachers, 25 students and 7 parents apply to be on the committee, which is chosen at random, how many possible committees could be formed?
This part I've figured out. The answer is $350 658 000$.
b) If Jan and her mother both apply, find the probability that both will be chosen for the committee.
This part I'm not sure how to do. The answer is apparently $\frac{3036}{44275}$.
4. I figured out the last question. As it turns out, $\frac{3036}{44275}$ simplifies to $\frac{12}{175}$, which was the answer I was getting.
Now I have a new problem;
A sample of 3 coins is taken at random from a bag containing 8 ten cent coins and 8 twenty cent coins. Find the probability that a particular ten cent coin will be chosen, if 1 twenty cent and 2 ten cent coins are chosen. | {
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# If $\rho_{n+1}\leq\rho_{n} +\sigma_n,\;\;\forall\;\;n\geq N_0,$ then prove that $\lim\rho_n=0.$
Good day, all!
Suppose $\{\rho_n\}$ and $\{\sigma_n\}$ are two sequences of non-negative real numbers such that for some real number $N_0\geq 1,$ the following recursion inequality holds:
$$\rho_{n+1}\leq\rho_{n} +\sigma_n,\;\;\forall\;\;n\geq N_0.$$
Prove that,
1. if $\sum_{n=1}^{\infty}\sigma_n<\infty,$ then $\lim\rho_n$ exists.
2. If $\sum_{n=1}^{\infty}\sigma_n<\infty,$ and $\rho_n$ has a subsequence converging to zero, then $\lim\rho_n=0.$
The first question was asked in my exam but I skipped it. I am stating the truth! Hence, I'll like to get it solved before the final exam. So please, can anyone help me? Thanks in advance.
• if the series of $\rho_n$ converges, then the limit of $\rho_n$ has to be zero, there is nothing to prove in 1 and 2. Please double-check the formulation (perhaps you need $\sigma_n$ instead ?) – Hayk May 20 '18 at 5:40
• @Hayk: Sorry, my mistake! It has been corrected! Matt A Pelto: Thanks for the edit! – Omojola Micheal May 20 '18 at 6:11
• Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. – robjohn May 20 '18 at 7:07
• a real pity about people on this forum that go straight to down voting :) ... such a community. – Matt A Pelto May 20 '18 at 7:47
• @ robjohn: This is not an assignment question. We had the third exam yesterday and the first question, .i.e. question 1, was given to us. However, I'll correct some of my words! – Omojola Micheal May 20 '18 at 7:48 | {
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Observe that once we prove assertion $1$, then the claim of $2$ follows readily. This is due to the fact, that if $\rho_n$ is convergent, than all its subsequences must be convergent, all with the same limit. Now, if we prove claim N$1$ and assume in addition that a subsequence of $\rho_n$ converges to $0$, then $\rho_n$, having the same limit as its subsequence, must converge to $0$.
We now prove claim $1$. Observe that for any $k\in \mathbb{N}$ and any $n\geq N_0$ we have $$(1) \qquad \rho_{n+k} - \rho_n = \rho_{ n + k } - \rho_{ n + k -1} + \rho_{ n + k - 1 } - \rho_{ n + k -2} +...+\rho_{n+1} - \rho_n \leq \\ \sigma_{n+k-1} +...+\sigma_n,$$ where we apply the given inequality on each term $\rho_{n+k -i} - \rho_{n+k-i-1}$, for $i=0,....,k-1$.
From the above inequality, and the fact that $\rho_n \geq 0$, we have that $\rho_n$ is a bounded sequence. Now assume, for contradiction, that it does not converge. Hence, $\rho_n$, converges to different limits along two different subseqeunces. Namely there exists $n_{k}$ and $m_k$, and non negative numbers $a_1$ and $a_2$ such that $$\rho_{n_k} \to a_1 \qquad \text{ and } \qquad \rho_{m_k} \to a_2,$$ where $a_1 \neq a_2$. From $(1)$, let us fix $n\geq N_0$, and choose $k$ such that $n+k = n_k$. Then, taking limit with respect to $k$ and using the fact that the series $\sigma_n$ converges, we get $$(2) \qquad a_1 - \rho_n \leq \sum_{i = n}^\infty \sigma_i.$$ Now take $n\to \infty$ along the second subsequence $m_k$, we obtain $a_1 - a_2 \leq 0$, since convergence of $\sum_i \sigma_i$ implies that the right hand side of $(2)$ converges to $0$.
Since the process is symmetric (we could have started with $m_k$ and not $n_k$) we obtain also that $a_2 \leq a_1$ and hence they must be equal.
This contradicts our assumption that $a_1 \neq a_2$ and hence the limit of $\rho_n$ exists. | {
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This contradicts our assumption that $a_1 \neq a_2$ and hence the limit of $\rho_n$ exists.
• What is wrong with Hayk's solution? Can anyone please, explain? – Omojola Micheal May 20 '18 at 7:49
• @Mike, let's see if the voter will be able to justify it. BTW, the only place you need non-negativity of $\rho_n$ is to get a lower bound on the sequence, as otherwise without bounding $\rho_n$ from below, it can escape to $-\infty$. So, instead of requiring $\rho_n \geq 0$ the claim stays valid also for the case $\rho_n \geq -M$, for any fixed $M\geq 0$ (the same proof works). – Hayk May 20 '18 at 8:49
• I neutralized it for you! – Omojola Micheal May 20 '18 at 15:56
• Can you please, expand this line $\leq \sigma_{n+k-1} +...+\sigma_{n-1}$? Kindly explain why it is so! – Omojola Micheal May 20 '18 at 16:09
• I was thinking it should be $\leq \sigma_{n+k-1} +...+\sigma_{n-1}+\sigma_{n}$ instead! – Omojola Micheal May 20 '18 at 16:19
Put $T_n=\sum_{k=1}^n \sigma_k$. Then $T_n$ is a convergent sequence. Now put $u_n=T_{n-1}-\rho_n$. From the inequality given, we see that $u_n$ is increasing for $n$ large, and of course (as $\rho_n \geq 0$) $\leq T_{n-1}$, hence bounded, hence convergent. It is easy to finish.
From $\rho_{n+1}-\rho_n\leq \sigma_n$ we obtain $\rho_{n}- \rho_m \leq \sum_{i=m}^{n-1}\sigma _i \Longrightarrow \rho_{n} \leq \rho_m+\sum_{i=m}^{n-1}\sigma _i$.
Taking $\limsup$ in respect of $n$ we obtain $\limsup \rho _n \leq \sum_{i=m}^\infty \sigma _i +\rho _m$.
Taking now $\liminf$, we obtain $$\limsup \rho _n \leq \liminf\rho _m$$ So, $\rho_n\longrightarrow x$ where $x \in [-\infty,\infty]$. Since $\rho _n$ is non-negative and upper bounded by $\rho_0 +\sum_{i=1}^\infty \sigma _i$, we conclude that $x\in \mathbb{R}$. | {
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# Is there a closed form expression for $\int_{- \infty}^\infty \int_{-\infty}^y \frac{1}{2 \pi} e^{-(1/2) ( x^2+y^2 )} \mathrm{d}x\,\mathrm{d}y$?
I have been trying to evaluate the integral:
$$\int_{- \infty}^\infty \int_{-\infty}^y \frac{1}{2 \pi} e^{-(1/2) ( x^2+y^2 )}\mathrm {d}x\,\mathrm{d}y$$
I know of course that the integral equals $1$ over $[-\infty,\infty] \times [-\infty,\infty]$ but I do not quite know how to handle the present case. Are there any tricks here?
Thank you.
• I think you can exploit the simmetry of the integrand by noting that the line $y=x$ divides the plane in half. – Gennaro Marco Devincenzis Sep 6 '14 at 15:22
Your integral is the probability: $$\mathbb{P}[X\leq Y]$$ where $X$ and $Y$ are two independent normal variables $N(0,1)$,
hence the value of the integral is just $\frac{1}{2}$, since: $$\mathbb{P}[X\leq Y]=\mathbb{P}[Y\leq X],\qquad \mathbb{P}[X\leq Y]+\mathbb{P}[Y\leq X]=1.$$
• That makes sense intuitively but could you please explain a little more why that is? There is no closed form then, right? – JohnK Sep 6 '14 at 15:26
• $f(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ is the pdf of a $N(0,1)$ random variable, and $\frac{1}{2}$ looks as a very closed form to me. – Jack D'Aurizio Sep 6 '14 at 15:29
• Yes but what allows us to say that the value of that probability is 1/2? – JohnK Sep 6 '14 at 15:31
• Symmetry: $$\mathbb{P}[Y\leq X]=\mathbb{P}[X\leq Y],\quad \mathbb{P}[Y\leq X]+\mathbb{P}[X\leq Y] = 1.$$ – Jack D'Aurizio Sep 6 '14 at 15:32
• Simple yet elegant! Intuitively, it can also be seen by plotting the region $0<x<y<\infty$ since the integrand is even function (i.e. symmetry). +1 – Tunk-Fey Sep 6 '14 at 15:44 | {
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Jack D'Aurizio's answer is good, but since you said in comments under it that you wanted a different point of view, let's try this: \begin{align} u & = (\cos45^\circ)x-(\sin45^\circ)y = \tfrac{\sqrt{2}}2 x - \tfrac{\sqrt{2}}2 y \\ v & = (\sin45^\circ)x+(\cos45^\circ)y = \tfrac{\sqrt{2}}2 x + \tfrac{\sqrt{2}}2 y \end{align} This is just a $45^\circ$ rotation of the coordinate system, suggested by the fact that your boundary line $y=x$ is just a $45^\circ$ rotation of one of the coordinate axes.
Then simplify $u^2+v^2$ and find that it comes down to $x^2+y^2$.
Solving the system of two equations above for $x$ and $y$, one gets \begin{align} x & = \phantom{-}\tfrac{\sqrt{2}}2 u + \tfrac{\sqrt{2}}2 v \\ y & = -\tfrac{\sqrt{2}}2 u + \tfrac{\sqrt{2}}2 v \end{align} By trivial algebra, the condition that $x\le y$ now becomes $u\le0$.
If you know about Jacobians, you get $$du\,dv = \left|\frac{\partial(u,v)}{\partial(x,y)}\right|\,dx\,dy = \left|\frac{\partial u}{\partial x}\cdot\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\cdot\frac{\partial v}{\partial x}\right|\,dx\,dy = 1\,dx\,dy.$$ Hence your iterated integral becomes $$\int_{-\infty}^\infty\int_{-\infty}^0 \frac 1{2\pi} e^{-(u^2+v^2)/2}\,du\,dv = \int_{-\infty}^\infty \int_{-\infty}^0 \left\{\frac 1{2\pi} e^{-v^2/2}\right\} e^{-u^2/2}\,du\,dv$$ The part in $\{\text{braces}\}$ does not depend on $u$, so this is $$\int_{-\infty}^\infty\left( \frac 1{2\pi} e^{-v^2/2} \int_{-\infty}^0 e^{-u^2/2}\,du \right)\,dv.$$ Now the inside integral does not depend on $v$, so it pulls out: $$\int_{-\infty}^\infty e^{-v^2/2}\,dv \cdot \frac1{2\pi} \int_{-\infty}^0 e^{-u^2/2}\,du$$ and this is of course $$\int_{-\infty}^\infty \frac1{\sqrt{2\pi}} e^{-v^2/2}\,dv \cdot \int_{-\infty}^0 \frac1{\sqrt{2\pi}} e^{-u^2/2}\,du.$$ The first integral comes to $1$ and the second, by a simple symmetry argument, is $1/2$. | {
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• Thank you very much. I'll just have to understand how come that substitution rotates the coordinate system. – JohnK Sep 6 '14 at 16:48
• nice result! you may simplify your answer by showing that $\int_{-\infty}^0 e^{-u^2}\,du=(1/2)\int_{-\infty}^{\infty} e^{-u^2}\,du$ as early in your answer as possible. – mike Sep 6 '14 at 16:52
• Think of what that substitution does to $(x,y)=(1,0)$ and to $(x,y)=(0,1)$ and then you'll be well on your way to understanding why it's a rotation. But even if you didn't know it's a rotation, you should still be able to follow the argument. – Michael Hardy Sep 6 '14 at 16:52
• . . . however, it is also the case that if you know nothing of Jacobians, you can understand that $du\,dv=dx\,dy$ by using the fact that the transformation is a rotation. Rotations multiply volumes by $1$, so the thing to multiply $dx\,dy$ by to get $du\,dv$ is $1$. ${}\qquad{}$ – Michael Hardy Sep 6 '14 at 16:54
• The intuition behind Jacobians is often not made explicit in second-year calculus courses where the concept is introduced. But if the value of the Jacobian $|\partial(u,v)/\partial(x,y)|$ at a particular point is, for example $6$, that means that at that point an infinitesimal area in the $x,y$ plane is transformed to an infinitesimal area $6$ times as big in the $u,v$ plane. Determinants generally area what you multiply by a volume to get the volume it is transformed to. And the determinant is negative if the orientation gets reversed. – Michael Hardy Sep 6 '14 at 17:03
Set $x=r\cos \theta,y=r\sin \theta$, then we have
$$\int_{- \infty}^\infty \int_{-\infty}^y e^{-(1/2) ( x^2+y^2 )}\mathrm {d}x\,\mathrm{d}y=\int_{0}^\infty \left(\int_{-3\pi/4}^{\pi/4} e^{-r^2/2}\mathrm {d}\theta\,\right)r\,\mathrm{d}r=\pi \int_{0}^\infty e^{-r^2/2}r\,\mathrm{d}r=\pi$$
So the original integral is equal to $(1/2)$.
This method as well as @MichaelHardy's method also works for the integrals like: | {
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This method as well as @MichaelHardy's method also works for the integrals like:
$$\int_{- \infty}^\infty \int_{-\infty}^{a y} e^{-(1/2) ( x^2+y^2 )}\mathrm {d}x\,\mathrm{d}y, \text{ }a \in \mathbb{R}$$
The results are the same. This is because the function to be integrated ($e^{-r^2/2}$) is rotational invariant (independent of $\theta$) and $x=a y$ is a straight line going through the origin and divides the plane into 2 halfs of equal size.
• Glad to see polar coordinates work too, thank you. – JohnK Sep 6 '14 at 18:27
Ah, I realise just now I'd misread to start with. Algebraically, we can solve it by noting that the value of the integral is the same under the change of variables $u=-x$, and since summing the two resulting integrals results in the $[-\infty,\infty] \times [-\infty,\infty]$ case, the answer is $1/2$.
i.e.:
$$\int_{- \infty}^\infty \int_{-\infty}^y \frac{1}{2 \pi} e^{-1/2 (x^2+y^2)} \,\mathrm{d}x\,\mathrm{d}y = \int_{- \infty}^\infty \int_{-y}^\infty \frac{1}{2 \pi} e^{-1/2 \left( x^2+y^2 \right)}\,\mathrm{d}x\,\mathrm{d}y.$$
$LHS+RHS=\int_{- \infty}^\infty \int_{-\infty}^\infty \frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y=1$, and $LHS=RHS$, so your integral is $1/2$.
EDIT: I guess I was a little bit short. There are several ways of showing $LHS+RHS$ is what I quote, but here is a simple, uninformative one.
$LHS+RHS=\int_{- \infty}^\infty \int_{-\infty}^\infty \frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y-\int_{- \infty}^\infty \int_{-y}^y\frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y$.
Now, Let $I=\int_{- \infty}^\infty \int_{-y}^y\frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y$. Use substitution $u=-y$. Then
$I=-\int_{\infty}^{-\infty} \int_{u}^{-u}\frac{1}{2 \pi} e^{-1/2 (x^2+u^2)}\,\mathrm{d}x\,\mathrm{d}u=\int_{-\infty}^{\infty} \int_{u}^{-u}\frac{1}{2 \pi} e^{-1/2 (x^2+u^2)}\,\mathrm{d}x\,\mathrm{d}u=-I$, hence as $I=-I$, $I=0$. | {
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I think it could also be done by splitting the integration range up. (e.g. looking at the integrals on $(-\infty,0]$ and $[0,\infty)$)
• If you do not mind how is that LHS+RHS equal that integral? – JohnK Sep 6 '14 at 16:09
• @JohnK there you go :) – ShakesBeer Sep 7 '14 at 7:49 | {
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# Math Help - How to prove logical equivalence using Theorems and substitution?
1. ## How to prove logical equivalence using Theorems and substitution?
So, I have the statement
(p ^ q) \/ (p ^ ¬ q) ≡ p
I have to prove the given logical equivalence using Theorems and not constructing a truth table. If anyone has as idea how I prove this I would appreciate an explanation and solution.
cheers
-ipatch
2. Originally Posted by ipatch
So, I have the statement
(p ^ q) \/ (p ^ ¬ q) ≡ p
I have to prove the given logical equivalence using Theorems and not constructing a truth table. If anyone has as idea how I prove this I would appreciate an explanation and solution.
cheers
-ipatch
Just read it as a sentence:
"p and q" or "p and not q".
It's saying you have to have p but you may or may not have q.
So what it's the same as having? p.
3. I have to prove the given logical equivalence using Theorems and not constructing a truth table
The solution depends on what theorems you have in mind. Every course may have a different set of those.
4. Hello, ipatch!
I don't know the names of the theorems you know.
I hope you can follow my proof.
Prove: . $(p \;\wedge\; q) \;\vee\; (p\; \wedge \sim q) \;\;\equiv \;\; p$
. . . . . . . . . $(p \;\wedge\; q) \;\vee\; (p\; \wedge\; \sim q)$
. . $(p \;\vee\; p) \;\wedge\; (p \;\vee\; \sim q) \;\wedge\; (q \;\vee\; p) \;\wedge\; (q \;\vee\; \sim q)$ . . Distributive
. . . . . . $p\;\wedge\;(p\;\vee\;\sim q)\;\wedge\;(p\;\vee\; q) \;\wedge\; T$
. . . . . . . . . $(p\;\vee\;\sim q)\;\wedge\; (p\;\vee q)$
. . . . . . . . . . . $p\;\vee(\sim q\;\wedge\; q)$ . . . . . . . . . . . . Distributive
. . . . . . . . . . . . . $p\;\vee\;F$
. . . . . . . . . . . . . . . $p$
5. Hello ipatch
Originally Posted by ipatch
So, I have the statement
(p ^ q) \/ (p ^ ¬ q) ≡ p | {
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5. Hello ipatch
Originally Posted by ipatch
So, I have the statement
(p ^ q) \/ (p ^ ¬ q) ≡ p
I have to prove the given logical equivalence using Theorems and not constructing a truth table. If anyone has as idea how I prove this I would appreciate an explanation and solution.
cheers
-ipatch
Using the standard Laws of Boolean Algebra (for instance, just here) we get:
$(p \land q)\lor(p\land\neg q) \equiv p\land(q\lor\neg q)$ (Distributive Law)
$\equiv p\land T$ (Complement Law)
$\equiv p$ (Identity Law)
6. (p ^ q) \/ (p ^ ¬ q) ≡ p
1. (p ^ q) \/ (p ^ ¬ q)-----------------hypotesis
2. (p ^ q)-----------------assume
3. p-----------------2, ^-elim 2
4. (p ^ ¬ q)-----------------assume
5. p-----------------4, ^-elim 2
6. p-----------------1, 2-3 , 4-5, \/-elim
7. Hello again, ipatch!
My first proof was long and clumsy.
(Don't know where my brain was at the time.)
There are two theorems we need. .(I don't know their names.)
. . $\begin{array}{cccc}P \;\vee \sim \!P &\equiv& t & \text{Theorem 1} \\ \\[-3mm] P \wedge \:t &\equiv& P & \text{Theorem 2} \end{array}$
. . . . $\begin{array}{ccc}(p \;\wedge\; q) \;\vee\; (p\; \wedge\; \sim q) & \text{Given} \\ \\ p \;\wedge\; (q \;\vee \sim \!q) & \text{Distr.} \\ \\ p \;\wedge \;t & \text{Theorem 1} \\ \\ p & \text{Theorem 2} \end{array}$
8. Originally Posted by cgafa
(p ^ q) \/ (p ^ ¬ q) ≡ p
1. (p ^ q) \/ (p ^ ¬ q)-----------------hypotesis
2. (p ^ q)-----------------assume
3. p-----------------2, ^-elim 2
4. (p ^ ¬ q)-----------------assume
5. p-----------------4, ^-elim 2
6. p-----------------1, 2-3 , 4-5, \/-elim
Rules used:
conjunctive elimination ( ^-elim ) if you have A and B the you have A (or B)... simple
disjuntive elimination (\/-elim)... a little more complex... if you have A or B and from A you can deduce C; and from B you can deduce C; then from A or B we can deduce C.....
9. Originally Posted by ipatch
So, I have the statement
(p ^ q) \/ (p ^ ¬ q) ≡ p | {
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9. Originally Posted by ipatch
So, I have the statement
(p ^ q) \/ (p ^ ¬ q) ≡ p
I have to prove the given logical equivalence using Theorems and not constructing a truth table. If anyone has as idea how I prove this I would appreciate an explanation and solution.
cheers
-ipatch
ipatch:
There two kind of proofs,one in Boolean Algebra and one in propositional calculus.
The proof in Boolean Algebra was shown by other people in the forum.
The proof in propositional calculus is as follows:
First we prove : $(p\wedge q)\vee (p\wedge\neg q)\Longrightarrow p$ and then:
$p\Longrightarrow (p\wedge q)\vee (p\wedge\neg q)$.
Proof:
1) $(p\wedge q)\vee (p\wedge\neg q)$.................................................. ............................given
2) $(p\wedge q)$.................................................. ..............................................assu mption to start a conditional proof
3) p................................................. .................................................. .....from (2) and using Conjunction Elimination (=C.E)
4) $(p\wedge q)\Longrightarrow p$.................................................. ....................................from (2) to (3) and using the rule of conditional proof
5) In the same way we prove: $(p\wedge\neg q)\Longrightarrow p$
6) $(p\wedge q)\vee (p\wedge\neg q)\Longrightarrow p\vee p$.................................................. ...........................from (4)and(5) and using the rule called proof by cases: $[((A\Longrightarrow B)\wedge (C\Longrightarrow D))\Longrightarrow ((A\vee C)\Longrightarrow (B\vee D))]$.
7) $p\vee p$.................................................. .......................................from (1) and (6) and using M.Ponens
8) $p\vee p\Longleftrightarrow p$.................................................. .......................................tautology | {
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9) $p\vee p\Longrightarrow p$.................................................. ........................................from (8) and using biconditional elimination
10) p................................................. .................................................. ....from (7) and (9) and using M.Ponens
Now to prove the converse.
Proof:
1)p............................................... .................................................. given
2) $\neg(p\wedge q)$.................................................. ........................................assumption to start a conditional proof
3) $\neg p\vee\neg q$.................................................. ..........................................from (2) and using De Morgan
4) $p\Longrightarrow\neg q$.................................................. .....................from (3) and using material implication
5) $\neg q$.................................................. .....................................from (1) and (4) and using M.Ponens
6) $p\wedge\neg q$.................................................. ...........................from (1) and (5) and using Conjunction Introduction
7) $\neg(p\wedge q)\Longrightarrow(p\wedge\neg q)$.................................................. .......................from (2) to(6) and using the rule of conditional proof
8) $(p\wedge q)\vee (p\wedge\neg q)$.................................................. ........................from (7) and using material implication $(A\Longrightarrow B)\Longleftrightarrow (\neg A\vee B)$ | {
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# Vector whose inner product is positive with every vector in given basis of $\mathbb{R}^n$
I am trying to solve the following question which I came across when studying root system in euclidean spaces, with positive definite symmetric bilinear form.
Statement: Given a basis $\{v_1,\cdots, v_n\}$ of $\mathbb{R}^n$, $\exists$ $v\in\mathbb{R}^n$ such that $(v,v_i)>0$ for all $i$
My answer: Let $v=a_1v_1 + \cdots + a_nv_n$. Then $$(v,v_i)=a_1(v_1,v_i) + a_2(v_2,v_i) + \cdots + a_n(v_n,v_i), \hskip5mm i=1,2,\cdots,n.$$ Writing this in matrix form we get $$\begin{bmatrix} (v,v_1) \\ \vdots \\ (v,v_n)\end{bmatrix} = \begin{bmatrix} (v_1,v_1) & \cdots & (v_n,v_1)\\ \vdots & & \vdots \\ (v_1,v_n) & \cdots & (v_n,v_n) \end{bmatrix} \begin{bmatrix} a_1 \\ \vdots \\ a_n\end{bmatrix}.$$ We want to find $a_1,\cdots,a_n$ such that the left column vector contains positive entries only. Let $A$ be the square matrix appearing above; it is matrix of positive definite (symmetric) inner product w.r.t given basis, so it is invertible over $\mathbb{R}$. Take $$\begin{bmatrix} a_1 \\ \vdots \\ a_n\end{bmatrix} := A^{-1}\begin{bmatrix} 1 \\ \vdots \\ 1\end{bmatrix}.$$ This is non-zero vector, and for this choice of $a_i$'s we get the vector $v=a_1v_1+\cdots + a_nv_n$ with desired properties.
Q.1 Is the above statement and proof correct?
Q.2 Is there geometric way to prove the statement?
• How do you know that $A$ is invertible? – 5xum Apr 10 '18 at 12:43
• it is matrix of non-degenerate bilinear form (positive definite inner product should be non-degenerate, isn't it?) – Beginner Apr 10 '18 at 12:44
• That's one way of showing it, sure. – 5xum Apr 10 '18 at 12:45 | {
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For a geometric proof, just do something akin to Gram-Schmidt orthogonalisation. Take $v=v_1$, so that $(v,v_1)>0$. Let $u_2=v_2-\frac{(v_1,v_2)}{(v_1,v_1)}v_1$ be the component of $v_2$ that is orthogonal to $v_1$. Then $u_2\ne0$, or else $v_1$ and $v_2$ are linearly dependent. Add $c_2u_2$ to $v$ for a sufficiently large $c_2>0$. Then $(v,v_i)>0$ for $i=1,2$. Continue in this manner, you get a desired $v$. | {
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# Math Help - Problem 5
1. ## Problem FIVE
Problem 5
TRIPOLLO
TRIPOLLO is a strategy game in which numbered marbles are placed in a rectangular array of holes on a board. As shown, there are three rows of holes. A player wins if the marbles are placed so that in each column, the sum of the numbers in the first two rows is three times the number in the third row.
(a) John has a 5-column board and marbles numbered 1 to 15.
So far he has placed them as shown in the diagram.
Show how he could successfully place the remaining marbles.
(b) With the same board and marbles. John made the following start:
Can he successfully complete this game? Give reasons.
(c) You have an 8-column board and marbles numbered 1 to 24. Show how you can win by placing all the marbles on the board.
(d) Show that with an 11-column board and marbles numbered 1 to 33 it is not possible to find a winning placement.
Explain the problem in as much detail as you can.
2. Originally Posted by MrFantasy
Problem 5
TRIPOLLO
TRIPOLLO is a strategy game in which numbered marbles are placed in a rectangular array of holes on a board. As shown, there are three rows of holes. A player wins if the marbles are placed so that in each column, the sum of the numbers in the first two rows is three times the number in the third row.
(a) John has a 5-column board and marbles numbered 1 to 15.
So far he has placed them as shown in the diagram.
Show how he could successfully place the remaining marbles.
(b) With the same board and marbles. John made the following start:
Can he successfully complete this game? Give reasons.
(c) You have an 8-column board and marbles numbered 1 to 24. Show how you can win by placing all the marbles on the board.
(d) Show that with an 11-column board and marbles numbered 1 to 33 it is not possible to find a winning placement. | {
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Explain the problem in as much detail as you can.
Is this a Mathematics Challenge question? If so, it is not approporiate to ask this question until the competition is ended.
3. Hello Mr.Fantastic
No, this is not a question from a maths competition. it's 1 of the 6 questions I picked out from the monthly issued maths challenge. My previous posts will explain the purpose of these 6 questions. It is to provide maths lovers some challenging questions to do. Please refer to my previous post for more information.
http://www.mathhelpforum.com/math-he...intrested.html
MrFantasy
4. Originally Posted by MrFantasy
Hello Mr.Fantastic
No, this is not a question from a maths competition. it's 1 of the 6 questions I picked out from the monthly issued maths challenge. My previous posts will explain the purpose of these 6 questions. It is to provide maths lovers some challenging questions to do. Please refer to my previous post for more information.
http://www.mathhelpforum.com/math-he...intrested.html
MrFantasy
Yes, I've seen that thread. I just find it odd that the same questions are being posted by other members too.
5. It may very well be that this question/problem is quiet wide known and is used in different occasions, whoever posted the same problem as I have posted here, could be another occasion. And please, if you dont mind. We're taking too much space talking in here, it would be difficult for members to read the answer replies. If there is any questions or replies or issues you want to make please contact me through PM. As it is more easier that way.
Thank You,
Mr`Fantasy
6. Originally Posted by mr fantastic
Is this a Mathematics Challenge question? If so, it is not approporiate to ask this question until the competition is ended.
This question is NOT from the Australian Maths Challenge Competition. The member has posted it in good faith. Please feel free to reply. | {
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Apologies for any embarassment caused - I hope you understand that all members need to be vigilant for the sort of cheating that does sometimes occur. | {
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# Plotting contours plot for f(x,y,z)=c
I have the following question: I have a file that has structure:
x1 y1 z1 f1
x2 y2 z2 f2
...
xn yn zn fn
I can easily visualize it with Mathematica using ListContourPlot3D. But could you please tell me how I can plot contour plot for this surface? I mean with these data I have a set of surfaces corresponding to different isovalues (f) and I want to plot intersection between all these surfaces and some certain plane. I tried to Google but didn't get any results. Any help and suggestions are really appreciated. Thanks in advance!
-
Would you please include a sample of your data, or provide a function that generates data that can serve as an example? – Mr.Wizard Aug 10 '12 at 21:47
Also, do you want a 2D contour plot, or a 3D curve through space? – Mr.Wizard Aug 10 '12 at 22:00
The answers to this question might be useful to you. – Ｊ. Ｍ. Aug 11 '12 at 3:53
Thanks for all your replies! Sorry for the delay, I had problems with an internet acces. But the problem is when I'm trying to use suggestions by halirutan I get empty box (and I need exactly this plot). I'd love to provide the sample of my data but it's big, it has 2744 lines and I don't know how to upload this file here (any suggestions?). Thanks! PS Original questions was asked by me. I do not know how to comment other answers (I don't see button add commment) – Nikita Aug 13 '12 at 20:50
After merging your accounts you should now be able to comment. – Sjoerd C. de Vries Aug 13 '12 at 22:45
show 1 more comment
Ok, lets give this a try. @Mr.Wizard already showed you, how you can use Interpolate to make a function from your discrete data and since you didn't provide some test-data, I just assume we are speaking of an isosurface of a function $f(x,y,z)=c$ which is defined in some box in $\mathbb{R}^3$.
For testing we use $$f(x,y,z) = x^3+y^2-z^2\;\;\mathrm{and}\;\; -2\leq x,y,z \leq 2$$ which accidently happens to be the first example of ContourPlot3D. | {
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The idea behind the following is pretty easy: As you may know from school, there is a simple representation of a plane in 3d which uses a point vector $p_0$ and two direction vectors $v_1$ and $v_2$. Every point on this plane can be reached through the $(s,t)$ parametrization
$$p(s,t)=p_0+s\cdot v_1+t\cdot v_2$$
Please note that $p_0, p, v_1, v_2$ are vectors in 3d and $s,t$ are scalars. The other form which we will use only for illustration is called the normal form of a plane. It is give by
$$n\cdot (p-p_0)=0$$
where $n$ is the vector normal to the plane, which can easily be calculated with the cross-product by $v_1\times v_2$. Lets start by looking at our example. To draw the plane inside ContourPlot3D we use the normal form which is plane2:
f[{x_, y_, z_}] := x^3 + y^2 - z^2;
v1 = {1, 1, 0};
v2 = {0, 0, 1};
p0 = {0, 0, 0};
plane1 = p0 + s*v1 + t*v2;
plane2 = Cross[v1, v2].({x, y, z} - p0);
gr3d = ContourPlot3D[{f[{x, y, z}], plane2}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Contours -> {0},
ContourStyle -> {ColorData[22, 3], Directive[Opacity[0.5], ColorData[22, 4]]}]
What we do now is, that we try to find the contour value (which is 0 here) of $f(x,y,z)$ for all points, that lie on our plane. This is like doing a normal ContourPlot because our plane is 2d (although placed in 3d space). Therefore, we use the 2d to 3d mapping of plane1
gr2d = ContourPlot[f[plane1], {s, -2, 2}, {t, -2, 2}, Contours -> {0},
ContourShading -> None, ContourStyle -> {ColorData[22, 1], Thick}]
Look at the intersection. It is exactly the loop we would have expected from the 3d illustration. Now you could object something like "ew.. but I really like a curve in 3d..". Again, the mapping from this 2d curve to 3d is given in the plane equation. You can simply extract the Line[..] directives from the above plot and transfer it back to 3d:
Show[{gr3d,
Graphics3D[{Red, Cases[Normal[gr2d], Line[__], Infinity] /.
Line[pts_] :> Tube[p0 + #1*v1 + #2*v2 & @@@ pts, .05]}]
}] | {
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I extract the Lines with Cases and then use the exact same definition of plane1 as pure function to transform the pts.
When I'm not completely wasted at 5:41 in the morning, than this approach should work for your interpolated data too.
## Apply method on test-data
I uploaded your test-data to our Git-repository and therefore, the code below should work without downloading anything. The approach is the same as above but some small things have changed, since we work on interpolated data now. I'll explain only the differences.
First we import the data and since we have a long list of {x,y,z,f} values, we transform them to {{x,y,z},f} as required by the Interpolation function. I'm not using the interpolation-function directly. I wrap a kind of protection around it which tests whether a given {x,y,z} is numeric and lies inside the interpolation box. Otherwise I just return 0.
data = {Most[#], Last[#]} & /@
Import["https://raw.github.com/stackmma/Attachments/master/data_9304_187.m"];
ip = Interpolation[data];
fip[{x_?NumericQ, y_?NumericQ, z_?NumericQ}] :=
If[Apply[And, #2[[1]] < #1 < #2[[2]] & @@@
Transpose[{{x, y, z}, First[ip]}]],
ip[x, y, z], 0.0]
The code below is almost the same. I only adapted the plane that it goes through your interpolation box. Furthermore, if you inspect your data you see that the value run from 0 to 1.2. Therefore I'm plotting the 0.5 contour by subtracting 0.5 from the function value and using Contours -> {0}. Remember that when I would simply plot the 0.5 contour, it would draw me a different plane, since we use one combined ContourPlot3D call.
Furthermore, notice that I normalized the direction vectors of the plane. This makes it easier to adjust the 2d plot of the contour. The rest should be the same. | {
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v1 = Normalize[{30, 30, 0}];
v2 = Normalize[{0, 0, 21}];
p0 = {26, 26, 17};
plane1 = p0 + s*v1 + t*v2;
plane2 = Cross[v1, v2].({x, y, z} - p0);
gr3d = ContourPlot3D[{fip[{x, y, z}] - 0.5, plane2}, {x, 27, 30}, {y,
27, 30}, {z, 17.3, 21}, Contours -> {0},
ContourStyle -> {Directive[Opacity[.5], ColorData[22, 3]],
Directive[Opacity[.8], ColorData[22, 5]]}]
gr2d = ContourPlot[fip[plane1] - 0.5, {s, 2, 5}, {t, 1, 4},
Contours -> {0}, ContourShading -> None,
ContourStyle -> {ColorData[22, 1], Thick}];
Show[{gr3d,
Graphics3D[{Red,
Cases[Normal[gr2d], Line[__], Infinity] /.
Line[pts_] :> Tube[p0 + #1*v1 + #2*v2 & @@@ pts, .05]}]}]
As you can see, your sphere has a whole inside. | {
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As you can see, your sphere has a whole inside.
-
Thanks for all your replies! Sorry for the delay, I had problems with an internet acces. But the problem is when I'm trying to use suggestions by halirutan I get empty box (and I need exactly this plot). I'd love to provide the sample of my data but it's big, it has 2744 lines and I don't know how to upload this file here (any suggestions?). Thanks! PS Original questions was asked by me. I do not know how to comment other answers (I don't see button add commment) – Nikita Aug 13 '12 at 20:50
thanks for solving account issue! So my question if it is possible to load file on this forum or how I can upload my data which are quite big (app. 3000 lines) – Nikita Aug 13 '12 at 22:51
@Nikita I asked in the chat and there seems to be no standard-upload place. Oleksandr suggested ge.tt. Before uploading a very large file, please read the FAQ and remember that the questions on this site should be helpful for everyone and should not be too localized! – halirutan Aug 14 '12 at 1:14
here is a link (depositfiles.com/files/r0pckyaiu) to zip file that contains 2 absolutely identical data (one is in format: x y z f, and the other is in the same format but can be directly load to mathematica). Yes, I understand that questions shouldn't be too localized, but the data I'm working on are cube files and it can be useful for many people. Thanks! – Nikita Aug 14 '12 at 14:31
@Nikita Here we go. See my update. – halirutan Aug 14 '12 at 15:56
show 1 more comment
You can use the options MeshFunctions in combination with Mesh for this.
I'm borrowing Mr.Wizard's data here for a moment:
data = Flatten[Table[{x, y, z, x^2 + y^2 - z^2}, {x, -2, 2, 0.2},
{y, -2, 2, 0.2}, {z, -2, 2, 0.2}], 2];
Suppose you want to plot the intersection of the contours of data with the plane x - y == 0, then you could do something like | {
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ListContourPlot3D[data, Contours -> {0.5, 2},
ContourStyle -> Opacity[0.3],
BoundaryStyle -> Opacity[0.3],
MeshFunctions -> {(#1 - #2) &}, Mesh -> {{0}},
MeshStyle -> {Thick, Orange}]
-
You removed my glorious infix! +1 for the method however. (It's nice to see you posting again.) – Mr.Wizard Aug 11 '12 at 7:00
thanks for your reply! These tricks with meshfunctions are brilliant, but when I use ListContourPlot3D on my data (which are exactly in the same format) I have empty box. But when I apply ListContourPlot3D on data that have only f values (without x,y,z) Mathematica plots correct figures but I cannot use MeshFunction in this case (because there is no information about x,y,z values). Do you know what can solve this problem? – Nikita Aug 13 '12 at 22:58
I'm not claiming this is a good method, I'm just getting some ink the page:
data = Table[{x, y, z, x^2 + y^2 - z^2}, {x, -2, 2, 0.2}, {y, -2, 2, 0.2}, {z, -2, 2, 0.2}] ~Flatten~ 2;
ListContourPlot3D[data, Contours -> {0.5, 2}, Mesh -> None]
int = Interpolation[data];
ContourPlot3D[int[x, y, z], {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Contours -> {0.5, 2}, RegionFunction -> (-0.02 < #2 - # < 0.02 &)]
-
The second piece of codes using RegionFunction doesn't creat a pretty plot,which contains two outer curves with small intervals, and neither PlotPoints nor MaxRecursion could help to improve it. – withparadox2 Aug 11 '12 at 7:12
@paradox2 hence: "I'm not claiming this is a good method ..." – Mr.Wizard Aug 11 '12 at 7:13 | {
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# Are random circulant matrices almost orthonormal?
Let $(X_1, X_2, \dots, X_n)$ be i.i.d. ${\cal N}(0,1)$. We construct a random circulant matrix $M$:
$$M = \frac{1}{\sqrt n}\begin{pmatrix}X_1 &X_2 &X_3 \dots &X_n\\ X_n &X_1 & X_2 \dots &X_{n-1}\\ \vdots &\vdots &\vdots &\vdots\\X_2 &X_3 &X_4 \dots &X_1\end{pmatrix}.$$
My questions are the following:
1. Is $M$ "almost orthonormal" in a precise probabilistic sense?
2. Related to above, is it possible to upper-bound the largest possible dot product between any two rows of $M$ by a suitably small $\epsilon_n$?
Note that this says that $MM^T$ is close to the identity matrix $I_{n \times n}$, as we are bounding the off-diagonal entries of $MM^T$ by $\epsilon_n$, while the diagonal entries are close to $1$.
• What do you mean by "upper-bound the largest possible dot product"? All of the $X_i$ could be equal to $10^{10^{10}}.$ You can upper bound it with high probability, but that's the best you can do. – Igor Rivin Oct 14 '17 at 0:05
• If $D$ is the absolute value of the largest dot product, one could seek a bound of the sort $\mathbb P(D > \epsilon_n) < \epsilon_n$ for a suitable choice of $\epsilon_n \to 0$. – VSJ Oct 14 '17 at 2:25
• Yes, that follows from Marcel's answer, using, e.g., Markoff's inequality. – Igor Rivin Oct 14 '17 at 3:24
The diagonal elements of $P=\frac{1}{N}MM^T$, like $$P_{11}=\frac{1}{N}\sum_{i=1}^NX_i^2,$$ satisfy $\langle P_{11}\rangle=1$ and $\langle P_{11}^2\rangle=1+2/N$ (variance decreases like $N^{-1}$).
On the other hand, off-diagonal elements like $$P_{12}=\frac{1}{N}\sum_{i=1}^{N}X_iX_{i+1}$$ satisfy $\langle P_{12}\rangle=0$ and $\langle P_{12}^2\rangle=1/N$ (variance also decreases like $N^{-1}$).
In this sense I would say $P$ is close to the identity.
Marcel's solution provides a good approach for understanding the marginal statistics. Here are a few supplementary comments that might give a little insight into the joint distribution. | {
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Let $D$ be the discrete Fourier transform matrix, i.e. the $j,k$-th entry is: $$D_{j,k}=e^{-2\pi i jk/N}/\sqrt{N}$$ Consider the discrete Fourier transform of the first row of $M$, i.e. $$(G_1,...,G_N)=(1/\sqrt{N})(X_1,...,X_N)D$$ Let $G$ be a diagonal matrix with diagonal entries $G_1,...,G_N$. Then $D$ diagonalizes $M$ (see, e.g., this description), that is: $$M = D G D^{-1}$$
If we took the $X_i$ to be complex-valued Gaussian variables, then we would be essentially done at this point: since $D$ is unitary, and the Gaussian is spherically symmetric, then the $G_i$ would be i.i.d. (complex) Gaussian random variables with mean 0 and variance $1/n$. (That is, if we sampled the $G_i$ as i.i.d complex $\mathcal{N}(0,1/n)$, then $DGD^{-1}$ has the same distribution as samples from the original circulant matrix.) It follows that $$MM^*=(DGD^{-1})(DG^*D^{-1})=DGG^{-1}D^*$$ It then follows that the eigenvalues of $MM^{*}$ (or, if you prefer, the squared singular values of $M$) are distributed like $n$ i.i.d draws from a rescaled $\chi^2$ distribution with 2 degrees of freedom (which happens to simplify to an exponential distribution), where we rescale by dividing by $n$. From an eigenvalue perspective, this is a complete characterization of the "orthonormality" of $M$.
However, your $X_i$ are probably real-valued. This causes a small book-keeping headache, but doesn't really change much. In brief: since the $X_i$ are real-valued, the $G_i$ will be symmetric. We can convert our $n\times n$ complex-valued matrices into corresponding $2n \times 2n$ real-valued matrices (say, $D'$ and $G'$). Then observe that there are only $n$ of the $2n$ diagonal elements in $G$ are free, and only $n$ of the $2n$ inputs to $D'$ are non-zero (namely, the ones corresponding to the real values of the inputs in $D$). Noting that the resulting decimated matrices are still unitary, we then reduce to the previous case, except with $\chi^2/n$ distributions with only 1 degree of freedom. | {
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• I'm confused about one part: Marcel's answer indicates that $MM^*$ is close to identity. Thus it should have eigenvalues close to 1? But your argument says eigenvalues are iid from a chi square distribution so they aren't all close to 1. How to reconcile these? – VSJ Nov 2 '17 at 4:04
• Ah, @VSJ , that is my mistake: I forgot to pass through the prefactor of $1/\sqrt(n)$ in the original definition of $M$. I will fix that now, at which point I think Marcel's answer is consistent with mine. – Bill Bradley Nov 3 '17 at 13:31
• Hi Bill, I'm not sure that is a mistake. Because even then, the eigenvalues will converge to 0 and not 1. Perhaps it is the case that the matrix is tending "pointwise" to identity, but it is not tending fast enough, and therefore the eigenvalues don't converge? – VSJ Nov 3 '17 at 17:52 | {
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# Another way of counting probability
A set $S = \{1, 2, \cdots, k\}$ is given. Two persons independently choose some numbers from this set. I want to count the probability that the cardinality of intersection of the chosen sets by both persons is exactly one.
Each person has to choose at least one number. A person can choose a number which is already chosen by another person. A number is being chosen independently and uniformly at random from the set.
My approach: Fix one number from $1$ to $k$. Probability of both persons choosing that number is $\frac{1}{k^2}$. Remaining $k-1$ numbers can be chosen by either of both or by none of both. So, probability for that is $\left(1 - \frac{1}{k^2}\right)^{k-1}$. And that fixed number can be selected in ${k \choose 1}$ ways.
So, probability of required event is:
\begin{align} P(E) = {k \choose 1} \frac{1}{k^2} \left(1 - \frac{1}{k^2}\right)^{k-1} \end{align}
My questions:
1. Is this reasoning correct?
2. Initially I started with counting all the possible ways of choosing numbers by each person. Can we count the required probability by counting ways technique? Is there any other way of counting the probability? | {
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• Why do you say that the probability of both people picking the fixed number is $\frac{1}{k^2}$? – paw88789 Oct 25 '15 at 12:57
• I think you need to say something about the number of elements each player selects. There are $2^k-1$ non-empty subsets of $\{1,....,k\}$. Do you mean to say that each player can select any one of these with equal probability? – lulu Oct 25 '15 at 12:57
• @lulu Yes. That is what I mean. – Nimit Oct 25 '15 at 13:07
• Ok, but then the situation is more complex than your method suggests. You have to take into account the fact that it is more likely that a randomly chosen subset has about $\frac k2$ elements than, say, $1$ or $k$. – lulu Oct 25 '15 at 13:09
• @paw88789 I got your point. I smell something wrong with my reasoning now. I had assumed that each person is choosing that fixed number very first. But that is not the case. It can be chosen later also. – Nimit Oct 25 '15 at 13:16
Let us at first ignore the condition that at least one number has to be chosen (that will get fixed afterwards). Under that condition, as each choice of each possible subset is equally likely (as you say in the comments), and for any given element there are exactly the same number of subsets containing and not containing that element (think of a subset and its complement!), the probability that a person chooses a given number is exactly $1/2$. Thus it's the same problem as if both people throw a coin for each number, and the question is what's the probability that they throw both head for exactly one of the numbers.
This is a fairly standard problem which I guess you can solve. Let's call the obtained probability $p_0$.
Now we have to fix this up for the condition that the empty set is not a valid choice. Fortunately this is easy, since when any party chose the empty set, we know for sure there was no number that was chosen by both (because at least one of them didn't chose a number to begin with). | {
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Since the probability calculated in the previous step is $$\frac{\text{Number of choices with exactly one shared number}} {\text{Total number of choices when allowing empty sets}}$$ we have to just fix up the denominator, which we can do by multiplying with $$\frac{\text{Total number of choices when allowing empty sets}} {\text{Total number of choices when not allowing empty sets}}.$$ Now the total number of choices when allowing empty sets is $2^{2k}$ since there are $2^k$ subsets, and each person chooses one. When excluding the empty set, each player has one choice less (namely he cannot choose the empty set), and therefore without the empty set, the number of possible choices is $(2^k-1)^2$.
Therefore we have for the requested probability: $$p = \frac{2^{2k}}{(2^k-1)^2}p_0$$
• To get an exact answer, the requirement of nonempty set choice by each person complicates the problem. For instance, players can't select/not select items by tossing a coin because that would allow for the empty set being chosen. And then consequently we can't simply say that for any given item that the probability of both selecting it is $\frac14$. I think you need to know more about how numbers are chosen. – paw88789 Oct 25 '15 at 13:45
• @paw88789: Ah right, I overlooked that condition. But the claim that all possible sets are chosen with equal probability should still be sufficient. I'll rework the answer. – celtschk Oct 25 '15 at 13:50
• @paw88789: See edited answer. – celtschk Oct 25 '15 at 14:05 | {
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truth tables explained | {
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Abstract: The general principles for the construction of truth tables are explained and illustrated. This truth-table calculator for classical logic shows, well, truth-tables for propositions of classical logic. Hence, (bâe)â§(bâ¬e)=(¬bâ¨e)â§(¬bâ¨Â¬e)=¬bâ¨(eâ§Â¬e)=¬bâ¨C=¬b,(b \rightarrow e) \wedge (b \rightarrow \neg e) = (\neg b \vee e) \wedge (\neg b \vee \neg e) = \neg b \vee (e \wedge \neg e) = \neg b \vee C = \neg b,(bâe)â§(bâ¬e)=(¬bâ¨e)â§(¬bâ¨Â¬e)=¬bâ¨(eâ§Â¬e)=¬bâ¨C=¬b, where CCC denotes a contradiction. Logical implication (symbolically: p â q), also known as âif-thenâ, results True in all cases except the case T â F. Since this can be a little tricky to remember, it can be helpful to note that this is logically equivalent to ¬p ⨠q (read: not p or q)*. The table contains every possible scenario and the truth values that would occur. Also known as the biconditional or if and only if (symbolically: ââ), logical equality is the conjunction (p â q) ⧠(q â p). Featuring a purple munster and a duck, and optionally showing intermediate results, it is one of the better instances of its kind. {\color{#3D99F6} \textbf{p}} &&{\color{#3D99F6} \textbf{q}} &&{\color{#3D99F6} p \equiv q} \\ This primer will equip you with the knowledge you need to understand symbolic logic. \text{0} &&\text{0} &&0 \\ Below is the truth table for p, q, pâàçq, pâàèq. From statement 4, gâ¬eg \rightarrow \neg egâ¬e, so by modus tollens, e=¬(¬e)â¬ge = \neg(\neg e) \rightarrow \neg ge=¬(¬e)â¬g. Check out my YouTube channel âMath Hacksâ for hands-on math tutorials and lots of math love â¥ï¸, Medium is an open platform where 170 million readers come to find insightful and dynamic thinking. ||row 2 col 1||row 2 col 2||row 2 col 1||row 2 col 2||. You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. \text{T} &&\text{F} &&\text{F} \\ If Darius | {
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depends on the truth or falsity of its components. \text{T} &&\text{F} &&\text{F} \\ If Darius is not the oldest, then he is immediately younger than Charles. The truth table for the XOR gate OUT =AâB= A \oplus B=AâB is given as follows: ABOUT000011101110 \begin{aligned} Go: Should I Use a Pointer instead of a Copy of my Struct? A table will help keep track of all the truth values of the simple statements that make up a complex statement, leading to an analysis of the full statement. Whats people lookup in this blog: Truth Tables Explained; Truth Tables Explained Khan Academy; Truth Tables Explained Computer Science Since gâ¬eg \rightarrow \neg egâ¬e (statement 4), bâ¬eb \rightarrow \neg ebâ¬e by transitivity. This is logically the same as the intersection of two sets in a Venn Diagram. Basic Logic Gates With Truth Tables Digital Circuits Partial and complete truth tables describing the procedures truth table for the biconditional statement you truth table definition rules examples lesson logic gates truth tables explained not and nand or nor. Truth tables – the conditional and the biconditional (“implies” and “iff”) Just about every theorem in mathematics takes on the form “if, then” (the conditional) or “iff” (short for if and only if – the biconditional). Whats people lookup in this blog: Logic Truth Tables Explained; Logical Implication Truth Table Explained Mr. and Mrs. Tan have five children--Alfred, Brenda, Charles, Darius, Eric--who are assumed to be of different ages. Remember to result in True for the OR operator, all you need is one True value. Already have an account? Using truth tables you can figure out how the truth values of more complex statements, such as. First you need to learn the basic truth tables for the following logic gates: AND Gate OR Gate XOR Gate NOT Gate First you will need to learn the shapes/symbols used to draw the four main logic gates: Logic Gate Truth Table Your Task Your task is to complete the truth tables for | {
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main logic gates: Logic Gate Truth Table Your Task Your task is to complete the truth tables for ⦠Surprisingly, this handful of definitions will cover the majority of logic problems youâll come across. A truth table is a mathematical table used in logic—specifically in connection with Boolean algebra, boolean functions, and propositional calculus—to compute the functional values of logical expressions on each of their functional arguments, that is, on each combination of values taken by their logical variables (Enderton, 2001). college math section 3.2: truth tables for negation, conjunction, and disjunction It negates, or switches, somethingâs truth value. Using this simple system we can boil down complex statements into digestible logical formulas. Here ppp is called the antecedent, and qqq the consequent. It is simplest but not always best to solve these by breaking them down into small componentized truth tables. We can show this relationship in a truth table. \end{aligned} A0011ââB0101ââOUT0110â, ALWAYS REMEMBER THE GOLDEN RULE: "And before or". When combining arguments, the truth tables follow the same patterns. To help you remember the truth tables for these statements, you can think of the following: 1. It is a mathematical table that shows all possible outcomes that would occur from all possible scenarios that are considered factual, hence the name. P AND (Q OR NOT R) depend on the truth values of its components. The truth table of an XOR gate is given below: The above truth table’s binary operation is known as exclusive OR operation. Basic Logic Gates, Truth Tables, and Functions Explained OR Gate. In mathematics, "if and only if" is often shortened to "iff" and the statement above can be written as. (Or "I only run on Saturdays. understanding truth tables Since any truth-functional proposition changes its value as the variables change, we should get some idea of what happens when we change these values systematically. Truth Table: A truth | {
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get some idea of what happens when we change these values systematically. Truth Table: A truth table is a tabular representation of all the combinations of values for inputs and their corresponding outputs. How to Construct a Truth Table. Two rows with a false conclusion. Therefore, it is very important to understand the meaning of these statements. Complex, compound statements can be composed of simple statements linked together with logical connectives (also known as "logical operators") similarly to how algebraic operators like addition and subtraction are used in combination with numbers and variables in algebra. Since câdc \rightarrow dcâd from statement 2, by modus tollens, ¬dâ¬c\neg d \rightarrow \neg c¬dâ¬c. For a 2-input AND gate, the output Q is true if BOTH input A âANDâ input B are both true, giving the Boolean Expression of: ( Q = A and B). The statement has the truth value F if both, If I go for a run, it will be a Saturday. A truth table is a logically-based mathematical table that illustrates the possible outcomes of a scenario. We have filled in part of the truth table for our example below, and leave it up to you to fill in the rest. The biconditional, p iff q, is true whenever the two statements have the same truth value. These operations are often referred to as âalways trueâ and âalways falseâ. is true or whether an argument is valid.. Truth tables get a little more complicated when conjunctions and disjunctions of statements are included. We use the symbol â§\wedge ⧠to denote the conjunction. It can be used to test the validity of arguments.Every proposition is assumed to be either true or false and the truth or falsity of each proposition is said to be its truth-value. From statement 4, gâ¬eg \rightarrow \neg egâ¬e, where ¬e\neg e¬e denotes the negation of eee. Truth tables show the values, relationships, and the results of performing logical operations on logical expressions. One of the simplest truth | {
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and the results of performing logical operations on logical expressions. One of the simplest truth tables records the truth values for a statement and its negation. a) Negation of a conjunction To do this, write the p and q columns as usual. Solution The truth tables are given in Table 4.2.Note that there are eight lines in the truth table in order to represent all the possible states (T, F) for the three variables p, q, and r. As each can be either TRUE or FALSE, in total there are 2 3 = 8 possibilities. Nor Gate Universal Truth Table Symbol You Partial and complete truth tables describing the procedures truth table tutorial discrete mathematics logic you truth table you propositional logic truth table boolean algebra dyclassroom. "). With just these two propositions, we have four possible scenarios. \text{0} &&\text{1} &&1 \\ Here, expert and undiscovered voices alike dive into the heart of any topic and bring new ideas to the surface. Complex, compound statements can be composed of simple statements linked together with logical connectives (also known as "logical operators") similarly to how algebraic operators like addition and subtraction are used in combination with numbers and variables ⦠To find (p ⧠q) ⧠r, p ⧠q is performed first and the result of that is ANDed with r. Hence Eric is the youngest. When either of the inputs is a logic 1 the output is... AND Gate. Log in here. They are considered common logical connectives because they are very popular, useful and always taught together. Hence Charles is the oldest. These are kinda strange operations. The AND gate is a digital logic gatewith ânâ i/ps one o/p, which perform logical conjunction based on the combinations of its inputs.The output of this gate is true only when all the inputs are true. Note that by pure logic, ¬aâe\neg a \rightarrow e¬aâe, where Charles being the oldest means Darius cannot be the oldest. Unary operators are the simplest operations because they can be applied to | {
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cannot be the oldest. Unary operators are the simplest operations because they can be applied to a single True or False value. Truth tables are a tool developed by Charles Pierce in the 1880s.Truth tables are used in logic to determine whether an expression[?] 2. This combines both of the following: These are consistent only when the two statements "I go for a run today" and "It is Saturday" are both true or both false, as indicated by the above table. Learning Objectives In this post you will predict the output of logic gates circuits by completing truth tables. â¡_\squareâ¡â. A truth table is a table whose columns are statements, and whose rows are possible scenarios. Truth tables really become useful when analyzing more complex Boolean statements. In the next post Iâll show you how to use these definitions to generate a truth table for a logical statement such as (A ⧠~B) â (C ⨠D). Otherwise it is false. Truth table, in logic, chart that shows the truth-value of one or more compound propositions for every possible combination of truth-values of the propositions making up the compound ones. It states that True is True and False is False. The AND operator (symbolically: â§) also known as logical conjunction requires both p and q to be True for the result to be True. In the second column we apply the operator to p, in this case itâs ~p (read: not p). A truth table is a mathematical table used to determine if a compound statement is true or false. This is equivalent to the union of two sets in a Venn Diagram. (pâq)â§(qâ¨p)(p \rightarrow q ) \wedge (q \vee p)(pâq)â§(qâ¨p), p \rightarrow q The truth table contains the truth values that would occur under the premises of a given scenario. The only possible conclusion is ¬b\neg b¬b, where Alfred isn't the oldest. Create a truth table for the statement $A\wedge\sim\left(B\vee{C}\right)$ Show Solution , â Try It. It is a mathematical table that shows all possible outcomes that would occur from | {
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, â Try It. It is a mathematical table that shows all possible outcomes that would occur from all possible scenarios that are considered factual, hence the name. A truth table is a handy little logical device that shows up not only in mathematics but also in Computer Science and Philosophy, making it an awesome interdisciplinary tool. Log in. For example, if there are three variables, A, B, and C, then the truth table with have 8 rows: Two simple statements can be converted by the word "and" to form a compound statement called the conjunction of the original statements. Then add a â¬pâ column with the opposite truth values of p. Lastly, compute ¬p ⨠q by OR-ing the second and third columns. b) Negation of a disjunction Explore, If you have a story to tell, knowledge to share, or a perspective to offer â welcome home. Truth tables list the output of a particular digital logic circuit for all the possible combinations of its inputs. \text{T} &&\text{T} &&\text{T} \\ As a result, the table helps visualize whether an argument is logical (true) in the scenario. If ppp and qqq are two simple statements, then pâ§qp \wedge qpâ§q denotes the conjunction of ppp and qqq and it is read as "ppp and qqq." Partial and complete truth tables describing the procedures truth table for the biconditional statement you truth table definition rules examples lesson logic gates truth tables explained not and nand or nor. â¡_\squareâ¡â. \hspace{1cm}The negation of a conjunction pâ§qp \wedge qpâ§q is the disjunction of the negation of ppp and the negation of q:q:q: ¬(pâ§q)=¬pâ¨Â¬q.\neg (p \wedge q) = {\neg p} \vee {\neg q}.¬(pâ§q)=¬pâ¨Â¬q. \end{aligned} pTTFFââqTFTFââpâ¡qTFFTâ. We will call our first proposition p and our second proposition q. Binary operators require two propositions. Logical true always results in True and logical false always results in False no matter the premise. \text{0} &&\text{0} &&0 \\ \text{1} &&\text{1} &&1 \\ The truth table | {
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False no matter the premise. \text{0} &&\text{0} &&0 \\ \text{1} &&\text{1} &&1 \\ The truth table for biconditional logic is as follows: pqpâ¡qTTTTFFFTFFFT \begin{aligned} The negation operator is commonly represented by a tilde (~) or ¬ symbol. There's now 4 parts to the tutorial with two extra example videos at the end. Sign up, Existing user? Conjunction (AND), disjunction (OR), negation (NOT), implication (IF...THEN), and biconditionals (IF AND ONLY IF), are all different types of connectives. Weâll use p and q as our sample propositions. \hspace{1cm} The negation of a disjunction pâ¨qp \vee qpâ¨q is the conjunction of the negation of ppp and the negation of q:q:q: ¬(pâ¨q)=¬pâ§Â¬q.\neg (p \vee q) ={\neg p} \wedge {\neg q}.¬(pâ¨q)=¬pâ§Â¬q. Using truth tables you can figure out how the truth values of more complex statements, such as. \text{F} &&\text{T} &&\text{F} \\ Therefore, if there are NNN variables in a logical statement, there need to be 2N2^N2N rows in the truth table in order to list out all combinations of each variable being either true (T) or false (F). The symbol and truth table of an AND gate with two inputs is shown below. Truth Tables of Five Common Logical Connectives or Operators In this lesson, we are going to construct the five (5) common logical connectives or operators. â For more math tutorials, check out Math Hacks on YouTube! When one or more inputs of the AND gateâs i/ps are false, then only the output of the AND gate is false. New user? A few common examples are the following: For example, the truth table for the AND gate OUT = A & B is given as follows: ABOUT000010100111 \begin{aligned} But if we have b,b,b, which means Alfred is the oldest, it follows logically that eee because Darius cannot be the oldest (only one person can be the oldest). Before we begin, I suggest that you review my other lesson in which the … Truth Tables of Five Common Logical Connectives … With fff, since Charles is the oldest, Darius must be | {
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Tables of Five Common Logical Connectives … With fff, since Charles is the oldest, Darius must be the second oldest. It can be used to test the validity of arguments.Every proposition is assumed to be either true or false and the truth or falsity of each proposition is said to be its truth-value. UNDERSTANDING TRUTH TABLES. Considering all the deductions in bold, the only possible order of birth is Charles, Darius, Brenda, Alfred, Eric. In an AND gate, both inputs have to be logic 1 for an output to be logic 1. If Eric is not the youngest, then Brenda is. Exclusive Or, or XOR for short, (symbolically: â») requires exactly one True and one False value in order to result in True. Note that the Boolean Expression for a two input AND gate can be written as: A.B or just simply ABwithout the decimal point. \hspace{1cm} The negation of a negation of a statement is the statement itself: ¬(¬p)â¡p.\neg (\neg p) \equiv p.¬(¬p)â¡p. ||p||row 1 col 2||q|| c) Negation of a negation The identity is our trivial case. Boolean Algebra is a branch of algebra that involves bools, or true and false values. In a truth table, each statement is typically represented by a letter or variable, like p, q, or r, and each statement also has its own corresponding column in the truth table that lists all of the possible truth values. We title the first column p for proposition. If Charles is not the oldest, then Alfred is. â¡_\squareâ¡â, Biconditional logic is a way of connecting two statements, ppp and qqq, logically by saying, "Statement ppp holds if and only if statement qqq holds." If it only takes one out of two things to be true, then condition_1 OR condition_2 must be true. We may not sketch out a truth table in our everyday lives, but we still use the logical reasoning t⦠These variables are "independent" in that each variable can be either true or false independently of the others, and a truth table is a chart of all of the possibilities. {\color{#3D99F6} \textbf{A}} | {
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the others, and a truth table is a chart of all of the possibilities. {\color{#3D99F6} \textbf{A}} &&{\color{#3D99F6} \textbf{B}} &&{\color{#3D99F6} \textbf{OUT}} \\ â¡_\squareâ¡â. Such a table typically contains several rows and columns, with the top row representing the logical variables and combinations, in increasing complexity leading up to ⦠â. All other cases result in False. Truth Table A table showing what the resulting truth value of a complex statement is for all the possible truth values for the simple statements. Otherwise it is true. Since ggg means Alfred is older than Brenda, ¬g\neg g¬g means Alfred is younger than Brenda since they can't be of the same age. â¡_\squareâ¡â. Determine the order of birth of the five children given the above facts. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. \text{1} &&\text{0} &&0 \\ READ Barclays Center Seating Chart Jay Z. If Alfred is older than Brenda, then Darius is the oldest. {\color{#3D99F6} \textbf{A}} &&{\color{#3D99F6} \textbf{B}} &&{\color{#3D99F6} \textbf{OUT}} \\ Two statements, when connected by the connective phrase "if... then," give a compound statement known as an implication or a conditional statement. It requires both p and q to be False to result in True. In other words, itâs an if-then statement where the converse is also true. From statement 3, eâfe \rightarrow feâf. A truth table is a breakdown of a logic function by listing all possible values the function can attain. The OR operator (symbolically: â¨) requires only one premise to be True for the result to be True. Truth tables summarize how we combine two logical conditions based on AND, OR, and NOT. Stay up-to-date with everything Math Hacks is up to! The negation of a statement is generally formed by introducing the word "no" at some proper place in the statement or by prefixing the statement with "it is not the case" or "it is false | {
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place in the statement or by prefixing the statement with "it is not the case" or "it is false that." Weâll start with defining the common operators and in the next post, Iâll show you how to dissect a more complicated logic statement. Sign up to read all wikis and quizzes in math, science, and engineering topics. Truth table explained. Truth table, in logic, chart that shows the truth-value of one or more compound propositions for every possible combination of truth-values of the propositions making up the compound ones. From statement 1, aâba \rightarrow baâb. The negation of statement ppp is denoted by "¬p.\neg p.¬p." Mathematics normally uses a two-valued logic: every statement is either true or false. A truth table is a visual tool, in the form of a diagram with rows & columns, that shows the truth or falsity of a compound premise. Logic tells us that if two things must be true in order to proceed them both condition_1 AND condition_2 must be true. We have filled in part of the truth table for our example below, and leave it up to you to fill in the rest. \text{F} &&\text{F} &&\text{T} Theyâre typically denoted as T or 1 for true and F or 0 for false. \text{1} &&\text{0} &&1 \\ Translating this, we have bâeb \rightarrow ebâe. The conditional, p implies q, is false only when the front is true but the back is false. Example. The truth table for the conjunction pâ§qp \wedge qpâ§q of two simple statements ppp and qqq: Two simple statements can be converted by the word "or" to form a compound statement called the disjunction of the original statements. In the first case p is being negated, whereas in the second the resulting truth value of (p ⨠q) is negated. Figure %: The truth table for p, âàüp Remember that a statement and its negation, by definition, always have opposite truth values. The notation may vary depending on what discipline youâre working in, but the basic concepts are the same. Write on Medium. Truth Tables of Five Common | {
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working in, but the basic concepts are the same. Write on Medium. Truth Tables of Five Common Logical Connectives or Operators In this lesson, we are going to construct the five (5) common logical connectives or operators. Truth Tables, Logic, and DeMorgan's Laws . Learn more, Follow the writers, publications, and topics that matter to you, and youâll see them on your homepage and in your inbox. You donât need to use [weak self] regularly, The Product Development Lifecycle Template Every Software Team Needs, Threads Used in Apache Geode Function Execution, Part 2: Dynamic Delivery in multi-module projects at Bumble. \text{1} &&\text{1} &&0 \\ Once again we will use aredbackground for something true and a blue background for somethingfalse. By adding a second proposition and including all the possible scenarios of the two propositions together, we create a truth table, a table showing the truth value for logic combinations. Itâs a way of organizing information to list out all possible scenarios from the provided premises. The truth table for the implication pâqp \Rightarrow qpâq of two simple statements ppp and q:q:q: That is, pâqp \Rightarrow qpâq is false â
ââºâ
â\iffâº(if and only if) p=Truep =\text{True}p=True and q=False.q =\text{False}.q=False. We can have both statements true; we can have the first statement true and the second false; we can have the first st⦠Note that if Alfred is the oldest (b)(b)(b), he is older than all his four siblings including Brenda, so bâgb \rightarrow gbâg. A truth table is a way of organizing information to list out all possible scenarios. In this lesson, we will learn the basic rules needed to construct a truth table and look at some examples of truth tables. We use the symbol â¨\vee ⨠to denote the disjunction. \end{aligned} A0011ââB0101ââOUT0001â. A truth table is a handy little logical device that shows up not only in mathematics but also in Computer Science and Philosophy, making it an | {
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that shows up not only in mathematics but also in Computer Science and Philosophy, making it an awesome interdisciplinary tool. Make Logic Gates Out Of Almost Anything Hackaday Flip Flops In ⦠*Itâs important to note that ¬p ⨠q â ¬(p ⨠q). They are considered common logical connectives because they are very popular, useful and always taught together. How to construct the guide columns: Write out the number of variables (corresponding to the number of statements) in alphabetical order. Once again we will use a red background for something true and a blue background for something false. If ppp and qqq are two statements, then it is denoted by pâqp \Rightarrow qpâq and read as "ppp implies qqq." Letâs create a second truth table to demonstrate theyâre equivalent. So as you can see if our premise begins as True and we negate it, we obtain False, and vice versa. \text{0} &&\text{1} &&0 \\ If ppp and qqq are two simple statements, then pâ¨qp\vee qpâ¨q denotes the disjunction of ppp and qqq and it is read as "ppp or qqq." A truth table is a mathematical table used in logicâspecifically in connection with Boolean algebra, boolean functions, and propositional calculusâwhich sets out the functional values of logical expressions on each of their functional arguments, that is, for each combination of values taken by their logical variables (Enderton, 2001). A truth table is a tabular representation of all the combinations of values for inputs and their corresponding outputs. The only way we can assert a conditional holds in both directions is if both p and q have the same truth value, meaning theyâre both True or both False. From statement 1, aâba \rightarrow baâb, so by modus tollens, ¬bâ¬a\neg b \rightarrow \neg a¬bâ¬a. The symbol of exclusive OR operation is represented by a plus ring surrounded by a circle ⊕. The truth table for the disjunction of two simple statements: An assertion that a statement fails or denial of a statement is | {
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of two simple statements: An assertion that a statement fails or denial of a statement is called the negation of a statement. Pics of : Logic Gates And Truth Tables Explained. This is shown in the truth table. Philosophy 103: Introduction to Logic How to Construct a Truth Table. Abstract: The general principles for the construction of truth tables are explained and illustrated. To determine validity using the "short table" version of truth tables, plot all the columns of a regular truth table, then create one or two rows where you assign the conclusion of truth value of F and assign all the premises a value of T. Example 8. Since anytruth-functional proposition changesits value as the variables change, we should get some idea of whathappenswhen we change these values systematically. From statement 2, câdc \rightarrow dcâd. The OR gate is one of the simplest gates to understand. It is represented as A ⊕ B. Since there is someone younger than Brenda, she cannot be the youngest, so we have ¬d\neg d¬d. You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. Nor ( symbolically: ⨠) requires only one premise to be.! Boolean statements logical true always results in true offer â welcome home shortened. \Neg c¬dâ¬c of birth of the and gateâs i/ps are false, and engineering topics is the exact opposite or! It negates, or, and optionally showing intermediate results, it is.. You can figure out how the truth value table one step further by adding second... B \rightarrow \neg c¬dâ¬c do this, write the p and ( q or R. Just these two propositions, we obtain false, and optionally showing intermediate results, it will be a.. Logic problems youâll come across tables follow the same to offer â welcome.... ¬P.\Neg p.¬p. tilde ( ~ ) or ¬ symbol these operations are often to. ¬P.\Neg p.¬p. something true and a blue background for something true F... ( ~ ) or ¬ symbol of a | {
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¬P.\Neg p.¬p. something true and a blue background for something true F... ( ~ ) or ¬ symbol of a logical statement are represented by a plus ring surrounded by plus... \Neg egâ¬e, where Alfred is a branch of Algebra that involves bools, true! Optionally showing intermediate results, it will be a Saturday gateâs i/ps false. Listing all possible scenarios F or 0 for false takes one out two! Circuit for all the possible outcomes of a complicated statement depends on the or! Construct a truth table: a truth table to demonstrate theyâre equivalent 1 for an output to be logic for., then condition_1 or condition_2 must be true in order to proceed both! Complicated statement depends on the truth tables records the truth values of components. Corresponding outputs surrounded by a tilde ( ~ ) or ¬ symbol quizzes. As ppp implies qqq. rules needed to construct a truth table contains every possible and... Pierce in the scenario DeMorgan 's Laws featuring a truth tables explained munster and a blue background for something and. Now 4 parts to the surface Brenda, Alfred, Eric very,! And their corresponding outputs statement 1, aâba \rightarrow baâb, so by modus tollens, ¬bâ¬a\neg b \rightarrow egâ¬e... Your thinking on any topic and bring new ideas to the surface gates by. LetâS create a second truth table: a truth table for p, q pâàçq. The p and q to be logic 1 for true and a duck, and engineering topics branch... Of these statements means Darius can not be the oldest to share, true. We negate it, we should get some idea of whathappenswhen we change these values systematically purple. Logical statement are represented by a tilde ( ~ ) or ¬ symbol \rightarrow \neg egâ¬e ( statement 4,. Statement: I go for a run, it will be a Saturday tables, logic, ¬aâe\neg \rightarrow... One or more inputs of the better instances of its kind for the or,. A given scenario a two-valued logic: every statement is either true or false itâs to... And undiscovered | {
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scenario a two-valued logic: every statement is either true or false itâs to... And undiscovered voices alike dive into the heart of any topic itâs a way of organizing information to out! Circuits by completing truth tables records the truth values of more complex Boolean.. Simple components of a logic 1 the output of a complicated statement depends on the truth values of its.. To iff '' and the statement above can be written as change these values systematically the you... ( true ) in the scenario and its negation further by adding second... ¬E\Neg e¬e denotes the negation of statement ppp is denoted by pâqp \rightarrow qpâq and read as implies! Operator, all you need is one true value modus ponens, our deduction eee leads to deduction! Requires both p and truth tables explained as our sample propositions logic to determine how the truth falsity. Calculator for classical logic only possible order of birth is Charles, Darius must be true then... Logical formulas have the same as the variables change, we should get some idea of we., the truth values of its inputs two propositions, we have four truth tables explained! We should get some idea of whathappenswhen we change these values systematically or just simply ABwithout the decimal point of! Tables explained with the knowledge you need to understand and condition_2 must be true another deduction.. Can not be the second oldest truth-tables for propositions of classical logic circuit for all the combinations of values inputs. Two inputs is a truth tables explained of Algebra that involves bools, or, and qqq are statements. For more math tutorials, check out math Hacks is up to read all wikis and quizzes math. With fff, since Charles is not the youngest, then condition_1 or must! Extra example videos at the end show you how to dissect a more complicated when conjunctions disjunctions! And bring new ideas to the surface developed by Charles Pierce in the second column we apply truth tables explained to., we | {
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the surface developed by Charles Pierce in the second column we apply truth tables explained to., we have ¬d\neg d¬d Iâll show you how to dissect a more complicated logic statement or! 4 ), bâ¬eb \rightarrow \neg egâ¬e, where Charles being the.! A mathematical table that illustrates the possible outcomes of a scenario by transitivity capital. To offer â welcome home often used in logic to determine whether an expression [? exclusive or is! False, then condition_1 or condition_2 must be true whenever the two statements have the same the. Logic tells us that if two things to be true in order to them., knowledge to share, or, and DeMorgan 's Laws by pâqp \rightarrow qpâq and as... Propositions, we should get some idea of whathappenswhen we change these values systematically â. To p, in this post you will predict the output of the inputs shown. | {
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# FIBGAME - Editorial
Practice
Author: Vaibhav Gautam
Tester: Illisha Singh , Jackson JoseKabir Kanha Arora, Nishank Suresh
Editorialist: Vaibhav Gautam
EASY
# PREREQUISITES:
Math, Careful Observation
# PROBLEM:
Given three consecutive integers a, b and c,
Is f(b) \times f(b)> f(a)* f(c)?
f(n): represents the n-th Fibonacci number.
f(1)=1 and f(2)=1.
f(n)= f(n-1)+f(n-2), n\geq3
# EXPLANATION:
Let’s try to solve the first subtask. Here, all the input numbers are less than or equal to 50. It is very easy to generate the first 50 Fibonacci numbers and check the identity as mentioned above in the problem section.
The code for the same is as follows:
#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
int main() {
vector<ll>fibn(51);fibn[0]=0;fibn[1]=1;fibn[2]=1;
for(int i=3;i<51;i++)
fibn[i]=fibn[i-1]+fibn[i-2];
int t ;cin>>t;
while (t-- > 0) {
ll a, b, c;
cin>>a>>b>>c;
if((fibn[b]*fibn[b])>(fibn[a]*fibn[c]))
cout<<"YES\n";
else
cout<<"NO\n";
}
}
The above code will give you 30 points for the first subtask.
You might get Wrong Answer, TLE, or Runtime Error on the second subtask based on your approach to generate the first 50 Fibonacci numbers.
There are many ways to solve this. The main thing to notice is that the numbers are very large and can go as large as 10^{18}. At this point, one must realize that loop won’t work as it is guaranteed to give TLE.
Now, what you eventually want to know is the answer to the following question.
For what a, b & c \$, f(b)*f(b) is greater than f(a)*f(c).
We will find the answer to this question from the solution of the first subtask.
Observe carefully, there can be no more than 50 cases in subtask 1. It is very easy to generate the answer for all of the 50 cases.
The answer to all the cases are given below:
1 2 3 NO
2 3 4 YES
3 4 5 NO
4 5 6 YES
5 6 7 NO
6 7 8 YES
7 8 9 NO
8 9 10 YES
9 10 11 NO
10 11 12 YES
11 12 13 NO
12 13 14 YES
13 14 15 NO
14 15 16 YES
15 16 17 NO
16 17 18 YES | {
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10 11 12 YES
11 12 13 NO
12 13 14 YES
13 14 15 NO
14 15 16 YES
15 16 17 NO
16 17 18 YES
17 18 19 NO
18 19 20 YES
19 20 21 NO
20 21 22 YES
21 22 23 NO
22 23 24 YES
23 24 25 NO
24 25 26 YES
25 26 27 NO
26 27 28 YES
27 28 29 NO
28 29 30 YES
29 30 31 NO
30 31 32 YES
31 32 33 NO
32 33 34 YES
33 34 35 NO
34 35 36 YES
35 36 37 NO
36 37 38 YES
37 38 39 NO
38 39 40 YES
39 40 41 NO
40 41 42 YES
41 42 43 NO
42 43 44 YES
43 44 45 NO
44 45 46 YES
45 46 47 NO
46 47 48 YES
47 48 49 NO
48 49 50 YES
I hope that everyone sees the pattern now. The answer is always alternating and depends only on the value of b. If b is odd the answer is YES and if b is even the answer is NO.
# SOLUTIONS:
Setter's Solution (C++)
#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t ;cin>>t;
while (t-- > 0) {
ll a, b, c;
cin>>a>>b>>c;
if(b%2==0)
cout<<"NO\n";
else
cout<<"YES\n";
}
}
Setter's Solution( Python )
T=int(input())
while(T>0):
a,b,c=map(int,input().split())
if(b%2==0):
print("NO")
else:
print("YES")
T-=1
Tester's Solution(Kabir Kanha Arora)( JAVA )
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
while (t-- > 0) {
long a = scanner.nextLong(); // Useless
long b = scanner.nextLong();
long c = scanner.nextLong(); // Useless
// Use Cassini's identity
String ans = b % 2 == 0 ? "NO" : "YES";
System.out.println(ans);
}
}
}
I think this solution works based on the above editorial:
#include <bits/stdc++.h>
using namespace std;
int main()
{
int T;
cin >> T;
for(int i = 0; i < T; i++)
{
int a, b, c;
cin >> a >> b >> c;
if(b%2 == 0)
{
cout << "NO\n";
}
else{cout << "YES\n";}
}
}
Yes @vaicr7bhav and @first_semester
I did make some silly mistakes indeed. I think I have corrected them now.
1 Like | {
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1 Like
It won’t work. You haven’t declared a,b and c. Also you are not printing a new line character.
#include <bits/stdc++.h>
using namespace std;
int main()
{
int T;
cin >> T;
for(int i = 0; i < T; i++)
{
long long int a,b,c;
cin >> a >> b >> c;
if(b%2 == 0)
{
cout << “NO\n”;
}
else{cout << “YES\n”;}
}
}
The above code will work fine.
yes you are correct, his solution would not work due to silly mistakes | {
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# Trig substitution integral
I am trying to find $$\int{ \frac {5x + 1}{x^2 + 4} dx}$$ The best approach would be to split up the fraction. According to Wolfram Alpha, the answer is $\frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\displaystyle\arctan\left(\frac x2\right)$ which seems OK, but when I try the trig substitution: $x = 2\tan\theta$, I get an answer that is slightly different but not equivalent, and I've looked at this over and over and I couldn't quite figure out what I did wrong.
$$x = 2\tan\theta$$ $$dx = 2sec^2\theta d\theta$$ $$\int{ \frac {5x + 1}{x^2 + 4}dx} = \frac{1}{4}\int{\frac{10\tan\theta + 1}{sec^2\theta} 2sec^2\theta d\theta}$$ $$= \frac{1}{2}\int{10 \tan\theta + 1}\space d\theta$$ $$= 5 \ln|\sec\theta| + \frac{\theta}{2} + C$$ We know $\theta = \displaystyle\arctan\left(\frac x2\right)$ and since $\tan\theta = \displaystyle\frac{x}{2}$, we can draw a triangle to see that $\sec\theta = \displaystyle\frac{\sqrt{x^2 + 4}}{2}$.
$$5 \ln|\sec\theta| + \frac{\theta}{2} = 5\ln\left({\frac{\sqrt{x^2 + 4}}{2}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right)$$ $$= \frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right)$$
But $$\frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right) \neq \frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\arctan\left(\frac x2\right)$$
There seems to be a small difference between the answer provided by Alpha and the trig substitution method, but I cannot see where I made the mistake.
-
$$\frac{5}{2}\ln({\frac{x^2 + 4}{4}}) + \frac{1}{2}\arctan({x/2}) + c_1 \neq \frac{5}{2}\ln(x^2 + 4) + \frac{1}{2}\arctan(x/2) + C$$
Note that $$\frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) = \frac 52 \ln\left(x^2 + 4\right) - \frac 52 \left (\ln 4\right) = \frac 52 \ln(x^2 + 4) + c_2$$ | {
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So put $c_1 + c_2 = C$. Then the answers are equal. Solutions to an integral consist of a family of solutions $F(x) + C$, which differ only by a constant. That is, if $F(x) + C$ is the solution after computing an intregral, so is $F(x) + C_i$, for any constant $C_i \neq C$.
In short, you're both correct!
-
Ah, I did not think of it like that. Looks like I did not make a mistake after all - thanks! – hesson Feb 22 '13 at 2:19
You're welcome, hesson. This can often occur when you end with an integration including the $\ln$ function: often it can be reduced to $\ln$ function + a constant, as in this case. Also, various trig substitutions can work, and one may arrive at different solutions in terms of trig functions, where one can be converted to the other by applying trig identities. – amWhy Feb 22 '13 at 2:27
The first approach is easier in fact: $$I= \int \frac{(5x +1)dx}{x^2+4}=I_1 + I_2$$ where $$I_1 = \int\frac{5x dx}{x^2+4}\\ I_2 = \int \frac{dx}{x^2+4}$$ hence, $$I_1 = \frac{5}{2} \int \frac{2xdx}{x^2+4} = \frac{5}{2} \int \frac{d(x^2+4)} {x^2+4}=\frac{5}{2} \log( x^2 +4) +C_1$$ In fact you can see that the numerator is a derivative of the denominator. $$I_2 = \int \frac{dx}{x^2+ 2^2} = \frac{\arctan(\frac{x}{a})}{a} +C_2$$ This approach is more efficient as it does not require you to make any substitutions/
-
Since $$\frac52 \ln (\frac{x^2+4}{4})=\frac52 \ln(x^2+4)-\frac52 \ln 4,$$ the difference between the two answers is an additive constant, which can be absorbed into the constant of integration $C$.
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# Highest power of a prime $p$ dividing $N!$
How does one find the highest power of a prime $p$ that divides $N!$ and other related products?
Related question: How many zeros are there at the end of $N!$?
This is being done to reduce abstract duplicates. See Coping with *abstract* duplicate questions. and List of Generalizations of Common Questions for more details.
Largest power of a prime dividing $N!$
In general, the highest power of a prime $p$ dividing $N!$ is given by
$$s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$
The first term appears since you want to count the number of terms less than $N$ and are multiples of $p$ and each of these contribute one $p$ to $N!$. But then when you have multiples of $p^2$ you are not multiplying just one $p$ but you are multiplying two of these primes $p$ to the product. So you now count the number of multiple of $p^2$ less than $N$ and add them. This is captured by the second term $\displaystyle \left \lfloor \frac{N}{p^2} \right \rfloor$. Repeat this to account for higher powers of $p$ less than $N$.
Number of zeros at the end of $N!$
The number of zeros at the end of $N!$ is given by $$\left \lfloor \frac{N}{5} \right \rfloor + \left \lfloor \frac{N}{5^2} \right \rfloor + \left \lfloor \frac{N}{5^3} \right \rfloor + \cdots$$ where $\left \lfloor \frac{x}{y} \right \rfloor$ is the greatest integer $\leq \frac{x}{y}$.
To make it clear, write $N!$ as a product of primes $N! = 2^{\alpha_2} 3^{\alpha_2} 5^{\alpha_5} 7^{\alpha_7} 11^{\alpha_{11}} \ldots$ where $\alpha_i \in \mathbb{N}$.
Note that $\alpha_5 < \alpha_2$ whenever $N \geq 2$. (Why?)
The number of zeros at the end of $N!$ is the highest power of $10$ dividing $N!$ | {
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The number of zeros at the end of $N!$ is the highest power of $10$ dividing $N!$
If $10^{\alpha}$ divides $N!$ and since $10 = 2 \times 5$, $2^{\alpha} | N!$ and $5^{\alpha} | N!$. Further since $\alpha_5 < \alpha_2$, the highest power of $10$ dividing $N!$ is the highest power of $5$ dividing $N!$ which is $\alpha_5$.
Note that there will be
1. A jump of $1$ zero going from $(N-1)!$ to $N!$ if $5 \mathrel\| N$
2. A jump of $2$ zeroes going from $(N-1)!$ to $N!$ if $5^2 \mathrel\| N$
3. A jump of $3$ zeroes going from $(N-1)!$ to $N!$ if $5^3 \mathrel\| N$ and in general
4. A jump of $k$ zeroes going from $(N-1)!$ to $N!$ if $5^k \mathrel\| N$
where $a \mathrel\| b$ means $a$ divides $b$ and $\gcd\left(a,\dfrac{b}{a} \right)$ = 1.
Largest power of a prime dividing other related products
In general, if we want to find the highest power of a prime $p$ dividing numbers like $\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1)$, $\displaystyle P(N,r)$, $\displaystyle \binom{N}{r}$, the key is to write them in terms of factorials.
For instance, $$\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1) = \frac{(2N)!}{2^N N!}.$$ Hence, the largest power of a prime, $p>2$, dividing $\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1)$ is given by $s_p((2N)!) - s_p(N!)$, where $s_p(N!)$ is defined above. If $p = 2$, then the answer is $s_p((2N)!) - s_p(N!) - N$.
Similarly, $$\displaystyle P(N,r) = \frac{N!}{(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle P(N,r)$ is given by $s_p(N!) - s_p((N-r)!)$, where $s_p(N!)$ is defined above.
Similarly, $$\displaystyle C(N,r) = \binom{N}{r} = \frac{N!}{r!(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle C(N,r)$ is given by $$s_p(N!) - s_p(r!) - s_p((N-r)!)$$ where $s_p(N!)$ is defined above. | {
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• Yes you are right. I have deleted the corresponding part in the post. – user17762 May 5 '12 at 1:19
• In my example, I meant highest power of $12$ that divides $3!$. If $m$ is square-free, then indeed the largest prime that divides $m$ is the obstruction. – André Nicolas May 5 '12 at 1:32
• @AndréNicolas Yes I recognized it. Thanks. For instance, in the highest power of $24$ dividing $10!$ the prime $2$ acts as the bottleneck since $2^3$ divides $24$ whereas only $3^1$ divides $24$ and since $10$ is square-free it works. Is there a generic formula known for composite $m$? – user17762 May 5 '12 at 1:36
• One trick that makes computations easier is $$\left\lfloor \dfrac{N}{p^{n+1}} \right\rfloor = \left\lfloor \dfrac{\left\lfloor \dfrac{N}{p^n} \right\rfloor}{p} \right\rfloor$$ – steven gregory Jun 1 '16 at 8:26
For a number $n$, define $\Lambda(n)=\log p$ if $n=p^k$ and zero elsewhen.
Note that $\log n=\sum\limits_{d\mid n}\Lambda(d)$. If $N=n!$, then $$\log N=\sum_{k=1}^n\log k=\sum_{k=1}^n\sum_{d\mid k}\Lambda (d)=\sum_{d=1}^n \Lambda(d)\left\lfloor \frac nd\right\rfloor$$
Since $\Lambda(d)$ is nonzero precisely when $d$ is a power of a prime and in such case it equals $\log p$, the last sum equals $$\sum_{p}\sum_{k\geqslant 1}\log p \left\lfloor\frac n{p^k}\right\rfloor$$ and this gives $$\nu_p(n!)=\sum_{k\geqslant 1}\left\lfloor\frac n{p^k}\right\rfloor$$ If you write $n$ is base $p$, say $n=a_0+a_1p+\cdots+a_kp^k$, the above gives that $$\nu_p(n!)=\frac{s(n)-n}{1-p}$$ where $s(n)=a_0+\cdots+a_k$.
• Very nice proof with using von Mangoldt function! +1 – ZFR Jun 9 '16 at 11:08
• Cool and simple. I think the hardest part is the last equality in the first formula. But it becomes obvious if you change the summation order. – LRDPRDX Jun 6 '20 at 10:18
de Polignac's formula named after Alphonse de Polignac, gives the prime decomposition of the factorial $n!$. | {
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For something completely different, see page 114 of Concrete Mathematics by Graham, Knuth, Patashnik. The highest power of a prime $p$ dividing $n!$ is $n$ minus the sum of the integer digits of $n$ in base $p$, all divided by $p-1$. In Mathematica you would write:
(n-Total[IntegerDigits[n,p]])/(p-1)
Hence, the number of trailing zeros of $n!$ is
(n-Total[IntegerDigits[n,5]])/4
I've found this method to be much faster than the sum of Floor functions...
Define $e_p(n) = s_p(n!),$ as Marvis is using $s_p$ for the highest exponent function...
We get $$e_p(n) = s_p(n!) = \sum_{i \geq 1} \left\lfloor \frac{n}{p^i} \right\rfloor$$
I thought I would prove this by mathematical induction. We need, for any positive integers $m,n,$
LEMMA:
(A) If $n + 1 \equiv 0 \pmod m,$ then $$\left\lfloor \frac{n + 1}{m} \right\rfloor = 1 + \left\lfloor \frac{n}{m} \right\rfloor$$ (B) If $n + 1 \neq 0 \pmod m,$ then $$\left\lfloor \frac{n + 1}{m} \right\rfloor = \left\lfloor \frac{n}{m} \right\rfloor$$
For $n < p,$ we know that $p$ does not divide $n!$ so that $e_p(n) = s_p(n!)$ is $0.$ But all the $\left\lfloor \frac{n}{p^i} \right\rfloor$ are $0$ as well. So the base cases of the induction is true.
Now for induction, increasing $n$ by $1.$ If $n+1$ is not divisible by $p,$ then $e_p(n+1) = e_p(n),$ while part (A) of the Lemma says that the sum does not change.
If $n+1$ is divisible by $p,$ let $s_p(n+1) = k.$ That is, there is some number $c \neq 0 \pmod p$ such that $n+1 = c p^k.$ From the Lemma, part (B), all the $\left\lfloor \frac{n}{p^i} \right\rfloor$ increase by $1$ for $i \leq k,$ but stay the same for $i > k.$ So the sum increases by exactly $k.$ But, of course, $e_p(n+1) = s_p((n+1)!) = s_p(n!) + s_p(n+1) = e_p(n) + k.$ So both sides of the middle equation increase by the same $s_p(n+1) = k,$ completing the proof by induction. | {
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