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two types of problems 4. Minimize z = 200x 1 + 300x 2. A decision is made when a value is specified for a decision variable. Linear inequations of two variables Let's begin with a few particular cases: We have to transport $$5$$ office chairs (that weigh $$10$$ kg each one) and three tables (weighing $$20$$ kg each). Linear Programming Problems [2-variables] Mathematical Programming Characteristics Decisions must be made on the levels of a two or more activities. Linear equations in three variables. Example of a linear programming problem. Example: Graph the solution set of the linear inequality in xy–plane. : min x = c x + c x + + c x. b) Name them. as initial solution. 2 in this linear programming model essentially duplicates the information summarized in Table 3. This is why we call the above problem a linear program. Linear Programming Example. Discrete Math B: Chapter 4, Linear Programming: The Simplex Method 2. + a 2n x n = b 2 a m1 x 1 + a m2 x 2 +. PDF | A linear programming problem (LP) deals with determining optimal allocations of LP problems that involve only two variables can be solved by both methods. 2-16 Graphical solution is limited to linear programming models containing only two decision variables (can be used with three variables but only with great difficulty). Formulation of linear programming is the representation of problem situation in a mathematical form. We'll see some examples of such constraint matrices when we look at applications. Now, if we let x 1, x 2 and x 3 equal to zero in the initial solution, we will have x 4 = 5 and x 5 = -2, which is not possible because a surplus variable cannot be negative. 1, 0. Today, linear programming is applied to a wide variety of problems in industry and science. Each coordinate can be any real number. Solved problems 2. Figures on the costs and daily availability of the oils are given in Table 1 below. This JavaScript E-labs learning object is intended for finding the optimal solution, and | {
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1 below. This JavaScript E-labs learning object is intended for finding the optimal solution, and post-optimality analysis of small-size linear programs. For x–intercept: Put in equation (i) An infeasible LP problem with two decision variables can be identified through its graph. 4, and leaves a lot to be desired when teaching MAT 119. x 1 - x 2 = 3 linear • MAX{x1,x2,…}, xi*yi, |xi|, etc => non-linear if xi and yi are variables – Sometimes there is a way to convert these types of constraints into linear constraints by adding some decison variables – Examples: 12/31/2003 Barnhart 1. T . The first stage of the algorithm might involve some preprocessing of the constraints (see Interior-Point-Legacy Linear Programming). 3 If the profit (P) on type X is R800 and on type Y is R1000, write down the objective function in the form P = ax + by . In this example, it has two decision variables, x r and x e, an objective function, 5 x r + 7 x e, and a set of four constraints. Problems with more than two variables (as is the case for most real The graphical method for solving linear programming problems in two unknowns is as follows. Linear programming algorithms. 3 WE2 Graph the inequality 7 –x 12y ≤ 84 and show that (4, –3) is a solution. 1 Represent the above information as a system of inequalities . The following are notes, illustrations, and algebra word problems that utilize linear optimization methods. It remains an important and valuable technique. This problem is a standard maximization problem with the decision variables x and y. The Wolfram Language has a collection of algorithms for solving linear optimization problems with real variables, accessed via LinearProgramming, FindMinimum, FindMaximum, NMinimize, NMaximize, Minimize, and Maximize. The solution of a linear inequality in two variables like Ax + By > C is an ordered pair (x, y) that produces a true statement when the values of x and y are Formulating Linear Programming Models Decision variables; | {
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statement when the values of x and y are Formulating Linear Programming Models Decision variables; Objective function; Constraints for example to maximize profit or minimize cost, although this is not always the case. whole numbers such as -1, 0, 1, 2, etc. This means that a linear function of the decision v ariables m ust b e r elate d to a constan t, where can mean less than or equal to, greater than or equal to, or equal to. For this example the column forming coefficients for non-basic variable 3. To solve linear programming models, the simplex method is used to find the optimal solution to a problem. In a team decision problem there are two or more decision variables, and these optimum basic feasible solution has been attained. non-negativity constraints for the example problem is shown in Fig 6. Specific examples and Linear Programming Simplex Method. g. The world is more complicated than the kinds of optimization problems that we are able to solve. Example 5: Solve using the Simplex Method. We A more complete presentation can be found for example in [2]. The entire problem can be expressed as straight lines, In some applications, you need to optimize a linear objective function of many variables, subject to linear constraints. Solution: We have. This example shows how to convert a linear problem from mathematical form into Optimization Toolbox™ solver syntax using the problem-based approach. Examples of Linear Optimization 2 1 Linear Optimization Models with Python Python is a very good language used to model linear optimization problems. Exercise: Soft Drink Production A simple production planning problem is given by the use of two ingredients A and B that produce products 1 and 2 . 2: The outcomesof all activities are known with certainty. Formulate constraints. 2) Select point $$( 1 , -1 )$$ situated in the region below the horizontal line. 1 Slack Variables and the Pivot (text pg169-176) In chapter 3, we solved linear programming problems graphically. 2: | {
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and the Pivot (text pg169-176) In chapter 3, we solved linear programming problems graphically. 2: Maximize 5 6 subject to 24 24 0, 0. 2 + S. Example 2 Solve the following linear programming problem graphically: Minimise Z 28 Feb 2017 Example of a linear programming problem. In this example it would be the variable X 1 (P 1) with -3 as coefficient. The first step is to identify Linear programming example 1991 UG exam. The example of a canonical linear programming problem from the introduction lends itself to a linear algebra-based interpretation. PROBLEM 1 Define in detail the decision variables and form the objective function and all. Basic two-variable linear programming problems with numerical solutions and illustrative graphs are available on PurpleMath. [1st] set equal to 0 all variables NOT associated with the above highlighted ISM. In a non-trivial optimization problem, we have among others two variables and both of them are ranged, i. A factory manufactures doodads and whirligigs. 2 n n. , et al. A linear program (LP) is an optimization problem (Wikipedia article . Example 4 (Phase I - Phase II Method): 2. that Linear Programming (LP) models of very large size can be solved in . INTEGER PROGRAMMING. Only one week's production is stored in 50 square metres of floor space where the floor space taken up by each product is 0. 2 Linear Programming. Formulation and Example. Write the objective function. linear programming problems. 1 n c d. For this model,n 4 variables and m 2 constraints; therefore, two of the variables are assigned a value of zero (i. This Demonstration shows the graphical solution to the linear programming problem: maximize subject to . In this section we discuss one type of optimization problem called linear pro-gramming. 1. Components of Linear Programming. Those are your non-basic variables. Step 2: Identify the set of constraints on the decision variables and express them in the form of linear equations /inequations. TD1 sept 2009 We | {
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decision variables and express them in the form of linear equations /inequations. TD1 sept 2009 We defined a very simple Linear Programming problem. 4 and 3. To solve a linear programming problem with two decision variables using the 13 Sep 2018 Linear programming is the technique used to maximize or minimize a To create one unit of medicine 1 , you need 3 units of herb A and 2 As the constraint have few variables (only x and y ), transforming the Moving forward from this basic example, the true potential of optimization is showcased when With our Linear Programming examples, we'll have a set of compound inequalities, Define the variables, write an inequality for this situation, and graph the 1 May 2005 cjxj. The office receives orders from two customers, each requiring 3/4-inch plywood. Example 2:. A linear program is a set of linear constraints defined over a set of variables. LINEAR PROGRAMMING WITH TWO VARIABLES 189 =6 =12 =18 =24 5 In the corner point, (0;8) we have f= 2(0)+3(8) = 24. Solving Linear Programs in Excel 2) Now label the row just above tableau (I am using rows 10 and on since I have the tableau above in the first few lines ) Variable values to manipulate. Summarize relevant material in table form, relating columns . Graphically Solving Linear Programs Problems with Two Variables (Bounded Case)16 3. Goal programming 1 Goal Programming and Multiple Objective Optimization Goal programming involves solving problems containing not one specific objective function, but rather a collection of goals. 1 Entering variable xs has to be chosen where the maximum in. Introduce decision variables. 2. Be sure to line up variables to the left of the ='s and constants to the right. Standard, deluxe and majestic seats each costs £20, £26 and £36 respectively. In a linear programming problem with two variables, the slack variables are always nonnegative in the corner points of This content was COPIED from BrainMass. From a marketing or statistical research to data | {
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points of This content was COPIED from BrainMass. From a marketing or statistical research to data analysis, linear regression model have an important role in the business. Solve the following linear program: maximise 5x 1 + 6x 2. the variables f are multiplied by constant Introduction: There are two types of linear programs (linear programming problems): . where X, Y, Z, S 1, S 2, S 3 > 0 (c) Put X = Y= Z = 0, we get S 1 = 7, S 2 = 2, S 3 = -29/7. D3 and E3, 4. = −. LINEAR PROGRAMMING – THE SIMPLEX METHOD (1) Problems involving both slack and surplus variables A linear programming model has to be extended to comply with the requirements of the simplex procedure, that is, 1. Introduction to Optimization1 1. 10. Chapter 4: Linear Programming The Simplex Method Day 1: 4. We’ll see one of the real life examples in the following tutorial. 6 Max Min with mixed constraints (Big M) Systems of Linear Inequalities in Two Variables See Interior-Point-Legacy Linear Programming. I Select Generate Linear Pgm from the Tools menu. Solving MOLPs by Weighted Sums. The objective and constraints in linear programming problems must be expressed in terms of linear equations or inequalities. All the feasible solutions in graphical method lies within the feasible area on the graph and we used to test the corner […] Linear Programming Linear Programming Solving systems of inequalities has an interesting application--it allows us to find the minimum and maximum values of quantities with multiple constraints. It provides the optimal value and the optimal strategy for the decision variables. “Linear” No x2, xy, arccos(x), etc. Linear programming is a useful way to discover how to allocate a fixed amount of resources in a manner that optimizes productivity. Assumptions of Linear programming. LINEAR PROGRAMMING. Linear programming. The levels are represented by decision variables X 1X 2, etc. F or example, y ou will b e able to iden tify when a problem has 2. 2 (The Simplex Method) Christopher | {
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it linear is that all our constraints are linear inequalities in our variables. Linear Programming Problem Complete the blending problem from the in-class part [included below] An oil company makes two blends of fuel by mixing three oils. 2 Linear Programming What you should learn Solve linear programming problems. Linear programming gives us a mechanism for answering all of these questions quickly and easily. Find the value as a function of a. In Two Phase Method, the whole procedure of solving a linear programming problem (LPP) involving artificial variables is divided into two phases. Linear Programming Notes VII Sensitivity Analysis 1 Introduction When you use a mathematical model to describe reality you must make ap-proximations. We will use This will be the very first system that we solve when we get into examples. The Regular Set consists of 2 utility knives and 1 chef’s knife and 1 bread knife. Along the way, dynamic programming and the linear complementarity problem are touched on as well. These are the cells that Excel will “change” to find the optimum solution to the problem. An integer programming problem in which all variables are required to be integer is called a pure integer pro-gramming problem. This method is used to solve a two variable linear program. All constraints are equality type. 3: A well defined objective function exist which can be used to evaluate differentoutcomes. 5 The Dual; Minimization with constraints 5. feasible region I This feasible region is a colorred convex polyhedron (àıœ/) spanned by points x 1 Toy LP example: brewer’s problem. Let's say a FedEx . Where, , ≥ 0 . 3 Date 2015-09-18 Title Interface to 'Lp_solve' v. To achieve this requirement, convert any unrestricted variable X to two non-negative variables by substituting T - X for X. Any linear programming problem involving two variables can be easily solved with the help of graphical method as it is easier to deal with two dimensional graph. A company makes two products | {
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of graphical method as it is easier to deal with two dimensional graph. A company makes two products (X and Y) using two machines (A and B). zx y xy xy xy. Sometimes for integer variables the value is not integer. t. In this Linear Programming: related mathematical techniques used to allocate limited resources among competing demands in an optimal way . ment of linear programming and proceeds to convex analysis, network flows, integer programming, quadratic programming, and convex optimization. 3 THE SIMPLEX METHOD: MAXIMIZATION. 75$% interest rates, and loans to senior citizens at $12. There is an . 125 x 1 + 8. Independent variables, on the right, are called nonbasic variables. 34 barrels × 35 lbs malt = 1190 lbs [ amount of available malt ] corn (480 lbs) hops (160 oz) malt (1190 lbs)$13 profit per barrel \$23 profit per barrel good are indivisible. shows, by means of an example, how linear programming can be applied to ob- tain optimal team decision functions in the case in which the payoff to the team is a convex polyhedral function of the decision variables. concepts of linear programming can all be demonstrated in the two-variable context. Linear programming formulation. Two important Python features facilitate this modeling: The syntax of Python is very clean and it lends itself to naturally adapt to expressing (linear) mathematical programming models Business environment assignment 2 examples of science research paper title page. Variables and constraints can be easily modified, as well as the ability to modify objective, bound and matrix coefficients. Characteristic of linear problem are. a number of examples of problems that may be formulated in terms of linear pro- Note that these constraints are also linear in the decision variables. Enter 0 values above the variables. All linear programming problems can be write in standard form by using slack variables and dummy variables, which will not have any influence on Example 1: The Production-Planning Problem. | {
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new variables F. Define the decision variables. Best wishes. For example, if Y is also unrestricted variable, the substitute - Y for Y. Modelling Linear Programming INDR 262 Optimization Models and Mathematical Programming LINEAR PROGRAMMING MODELS Common terminology for linear programming: - linear programming models involve . Example 2 (Alternate optimal solutions). The variable s 2 is called a surplus variable Divisibility assumption: Decision variables in a linear programming model are allowed to have any values, including noninteger values, that satisfy the func- tional and nonnegativity constraints. 2 Linear Programming Geometric Approach 5. linear programming 2 variables examples | {
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# Maximum height
256 views
$H=(u^2 sin^2\theta )/(2g)$
$h-H =(1/2)gt^2 = (u^2sin^2 \theta)/(2g)$
From this I got h=2H
1 vote
If the particles are launched at the same time, they must have the same horizontal speed in order to collide. (However, the problem seems to state that the particles have the same speed, ie same magnitude of velocity, and it is not clear that they are launched at the same time.)
Because they have the same horizontal speed, they remain aligned vertically. We can ignore horizontal motion and treat this as motion in 1D vertically.
The particles must collide before they reach the ground. The maximum $h$ occurs when they collide at the ground. ie They reach the ground at the same time, and have the same time of flight $t$.
When the upper particle reaches the ground $h=\frac12gt^2$. When the lower particle falls from its maximum height it takes time $\frac12 t$ so
$H=\frac12g(\frac12 t)^2=\frac14(\frac12gt^2)=\frac14h$
$h=4H$.
answered Mar 22, 2017 by (28,876 points)
selected May 10, 2017 by koolman
Speed of paricle 1 be u at an angle $\theta$ from ground .Then the speed of particle 2 is $u\cos \theta$ in horizontal direction.
let at any h' from ground both partucles collide .
For particle 1 $$h'= u\sin\theta t -(1/2)gt^2$$
For particle 2 $$h-h '= (1/2) gt^2$$
Hence $$t=\sqrt {\frac{2(h-h')}{g}}$$
Also we know $H=u^2 sin\theta ^2 /(2g)$
Substituting these values in equation of particle 1
We geta quadratic equation $$h^2 -4Hh+4Hh'=0$$
solving this $$h=\frac{4H(+/-)\sqrt{16H^2 -16Hh'}}{2}$$
Now h will be maximum when h'=0 that is they collide when both of them just going to strike the ground .
$$h=\frac{4H+4H}{2}$$
$$h=4H$$ | {
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$$h=\frac{4H+4H}{2}$$
$$h=4H$$
answered Mar 22, 2017 by (4,286 points)
Yes this is correct but not the simplest way of doing the calculation. The key assumption is that the particles have the same *horizontal component* of velocity. This is not what the question seems to say. The wording of the question is very confusing.
The question hasn't mentioned anything about speed , so we can take horizontal speed same .
The question says "second particle is thrown horizontally with same speed". To me that means the particles have the same "speed" ie the same magnitude of velocity.
Oh Sorry , the wording of question is confusing.
1 vote
Your Equation considers the particles to collide only at the point where first particle attains its maximum height. But the question considers the general case.
Since they are thrown from the same vertical plane with the same speed, they can collide only if the horizontal component of their velocities are equal. Here the second particle would have covered a large distance when they are at the same height.
If the first particle is necessarily thrown obliquely then the particles cannot collide I assume.
answered Mar 21, 2017 by (340 points)
Have a look at my answer .
1 vote
Discussion
DoubtExpert correctly observes that the particles do not necessarily collide when the particle thrown obliquely reaches its maximum height. However, he assumes that the particles are launched simultaneously. I think this is not required by the wording of the question.
The key phrase is thrown to strike at same time. This is ambiguous. If this means (as I think it does) that the particles collide at the same time then it is superfluous, because they must be in the same place at the same time in order to collide. I do not see how it can be interpreted as launched at the same time. If it does mean this then DoubtExpert is correct : there can be no collision.
Solution | {
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Solution
We must assume that the particles can be launched independently, at any time. The only requirement is that the trajectories overlap - or at least touch tangentially - at some point in space. If this happens then we can always adjust the launch times to ensure that the particles collide at this point.
What we need to do is write equations for the two trajectories then find a condition for them to touch. The trajectories will touch if (i) they intersect at some point, and (ii) their tangents are equal at this point.
Suppose the speed of each particle is $u$ and the launch angle of the lower particle is $\theta$. Writing $T=\tan\theta$ the trajectories of the upper and lower particles are respectively
$y=h-\frac{g}{2u^2}x^2$
$y=xT-\frac{g(1+T^2)}{2u^2}x^2$.
The tangents have slope
$\frac{dy}{dx}=-\frac{g}{u^2}x$
$\frac{dy}{dx}=T-\frac{g(1+T^2)}{u^2}x$.
The 1st two equations are equal when
$h=xT-\frac{gT^2}{2u^2}x^2$.
The 2nd two equations are equal when
$T=\frac{gT^2}{u^2}x$
$xT=\frac{u^2}{g}$.
Combine these two results :
$h=\frac{u^2}{2g}$.
The height of the collision point is
$y=h-\frac{g}{2u^2}x^2=\frac{u^2}{2g}-\frac{u^2}{2gT^2}=h(1-\frac{1}{T^2})$.
This must be above ground $(y \ge 0)$ so
$T^2 \ge 1$
$\theta \ge 45^{\circ}$.
The maximum height reached by the lower particle is $H=\frac{u^2\sin^2\theta}{2g}$. So we have $\frac{H}{h}=\sin^2\theta$. The value of $h$ calculated here is the maximum possible - any larger and the trajectories would not intersect. So the criterion is
$h \ge \frac{H}{\sin^2\theta}$.
I do not think we can make any further progress without knowing the launch angle $\theta$. In order to get $h = 4H$ we require $\theta=30^{\circ}$. However, this is not $\ge 45^{\circ}$ : the collision in this case takes place below ground. So a collision is not possible when $h =4H$. | {
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answered Mar 21, 2017 by (28,876 points)
edited Mar 21, 2017
It is given in the question that both particles are thrown at same time .
If the particles are launched with the same speed at the same time, then I agree with DoubtExpert : the particles cannot collide.
Is there a worked solution? Or a diagram?
Hmm, true because horizontal distance being same $ut = u cos \theta t$
and $\theta = 0$ so this may not cause collision or because even the speed given are same.
The solution has bad explanation .
http://pasteboard.co/MmmDW5wny.jpg | {
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# Check your mathematical induction concepts
Discuss the following “proof” of the (false) theorem:
If n is any positive integer and S is a set containing exactly n real numbers, then all the numbers in S are equal:
PROOF BY INDUCTION:
Step 1:
If $n=1$, the result is evident.
Step 2: By the induction hypothesis the result is true when $n=k$; we must prove that it is correct when $n=k+1$. Let S be any set containing exactly $k+1$ real numbers and denote these real numbers by $a_{1}, a_{2}, a_{3}, \ldots, a_{k}, a_{k+1}$. If we omit $a_{k+1}$ from this list, we obtain exactly k numbers $a_{1}, a_{2}, \ldots, a_{k}$; by induction hypothesis these numbers are all equal:
$a_{1}=a_{2}= \ldots = a_{k}$.
If we omit $a_{1}$ from the list of numbers in S, we again obtain exactly k numbers $a_{2}, \ldots, a_{k}, a_{k+1}$; by the induction hypothesis these numbers are all equal:
$a_{2}=a_{3}=\ldots = a_{k}=a_{k+1}$.
It follows easily that all $k+1$ numbers in S are equal.
*************************************************************************************
Comments, observations are welcome 🙂
Regards,
Nalin Pithwa
# Observations are important: Pre RMO and RMO : algebra
We know the following facts very well:
$(x+y)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}$
$(x-y)^{3}=x^{3}-3x^{2}y+3xy^{2}-y^{3}=()()$
But, you can quickly verify that:
$x^{3}+2x^{2}y+2xy^{2}+y^{3}=(x+y)(x^{2}+xy+y^{2})$
$x^{3}-2x^{2}y+2xy^{2}-y^{3}=(x-y)(x^{2}-xy-y^{2})$
Whereas:
$x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$
$x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$
I call it — simply stunning beauty of elementary algebra of factorizations and expansions
More later,
Nalin Pithwa
# Some number theory (and miscellaneous) coaching for RMO and INMO: tutorial (problem set) III
Continuing this series of slightly vexing questions, we present below:
1. Prove the inequality $\frac{A+a+B+b}{A+a+B+b+c+r} + \frac{B+b+C+c}{B+b+C+c+a+r} > \frac{C+c+A+a}{C+c+A+a++b+r}$, where all the variables are positive numbers. | {
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2. A sequence of numbers: Find a sequence of numbers $x_{0}$, $x_{1}$, $x_{2}, \ldots$ whose elements are positive and such that $a_{0}=1$ and $a_{n} - a_{n+1}=a_{n+2}$ for $n=0, 1, 2, \ldots$. Show that there is only one such sequence.
3. Points in a plane: Consider several points lying in a plane. We connect each point to the nearest point by a straight line. Since we assume all distances to be different, there is no doubt as to which point is the nearest one. Prove that the resulting figure does not containing any closed polygons or intersecting segments.
4. Examination of an angle: Let $x_{1}$, $x_{2}, \ldots, x_{n}$ be positive numbers. We choose in a plane a ray OX, and we lay off it on a segment $OP_{1}=x_{1}$. Then, we draw a segment $P_{1}P_{2}=x_{2}$ perpendicular to $OP_{1}$ and next a segment $P_{2}P_{3}=x_{3}$ perpendicular to $OP_{2}$. We continue in this way up to $P_{n-1}P_{n}=x_{n}$. The right angles are directed in such a way that their left arms pass through O. We can consider the ray OX to rotate around O from the initial point through points $P_{1}$, $P_{2}, \ldots, P_{n}$ (the final position being $P_{n}$). In doing so, it sweeps out a certain angle. Prove that for given numbers $x_{i}$, this angle is smallest when the numbers $x_{i}$, that is, $x_{1} \geq x_{2} \geq \ldots x_{n}$ decrease; and the angle is largest when these numbers increase.
5. Area of a triangle: Prove, without the help of trigonometry, that in a triangle with one angle $A = 60 \deg$ the area S of the triangle is given by the formula $S = \frac{\sqrt{3}}{4}(a^{2}-(b-c)^{2})$ and if $A = 120\deg$, then $S = \frac{\sqrt{3}}{12}(a^{2}-(b-c)^{2})$.
More later, cheers, hope you all enjoy. Partial attempts also deserve some credit.
Nalin Pithwa.
# A nice analysis question for RMO practice | {
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Nalin Pithwa.
# A nice analysis question for RMO practice
Actually, this is a famous problem. But, I feel it is important to attempt on one’s own, proofs of famous questions within the scope of RMO and INMO mathematics. And, then compare one’s approach or whole proof with the one suggested by the author or teacher of RMO/INMO.
Problem:
How farthest from the edge of a table can a deck of playing cards be stably overhung if the cards are stacked on top of one another? And, how many of them will be overhanging completely away from the edge of the table?
Reference:
I will post it when I publish the solution lest it might affect your attempt at solving this enticing mathematics question !
Please do not try and get the solution from the internet.
Regards,
Nalin Pithwa.
# A miscellaneous algebra question for RMO or Pre-RMO
Question:
If a, b, c are three rational numbers, then prove that
$\frac{1}{(a-b)^{2}} + \frac{1}{(b-c)^{2}} + \frac{1}{(c-a)^{2}}$
is always the square of a rational number.
Solution to be posted soon…
Cheers,
Nalin Pithwa. | {
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# Math Help - Identifying Inflection Points
1. ## Identifying Inflection Points
The x coordinates of the points of inflection of the graph of $y= x^5-5x^4=3x+7$ are...
A 0,
B 1,
C 3,
D 0 and 3,
E 0 and 1
$y''=20x^3-60x^2$
Simplify.
$y''=20x^2(x-3)$
So, thanks to the graph and knowledge of the definition of inflection point (y''= 0).
I choose D.
However, the book says C.
I'm inclined to believe it is right because I did the sign test for 0 for numbers less than 0 and less than 3, thought I may be incorrect of which function to put the test values into (y'' right?), and got the same sign (negative).
So, did I do the sign test wrong? Or is the right and I'm confusing myself with these questions.
What is the answer to the problem?
Thank you.
2. For a point of inflection, $y''=0$ is a necessary condition but not a sufficient one. $y''$ must also change sign before and after the point of inflection. Thus 0 is not a point of inflection because $y''$ does not change sign there; $y''$ is negative both slightly before and slightly after 0.
Alternatively …
$y'''=60x^2-120x\ne0$ when $x=3$. Since $y'''$ is an odd-order derivative, $x=3$ is a point of inflection. However $y'''(0)=0$ so we differentiate again.
$y''''=120x-120\ne0$ when $x=0$. $y''''$ is an even-order derivative, so $x=0$ is not a point of inflection. | {
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# A peculiar binomial coefficient identity
While inventing exercises for a discrete math text I'm writing I came up with this $$\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$$ It's an easy result to prove, but it got me wondering
1. Is this pure coincidence, or is it a special case of some more general result of which I'm unaware?
2. No matter what the answer to (1) is, is there a combinatorial proof?
• Looks purdy.${{{}}}$ – Git Gud Jun 26 '13 at 21:20
• Every polynomial $f(x)$ which takes integer values at the integers is an integer linear combination of the polynomials ${x \choose k}$, so in some sense this identity is not very surprising. The analogous statement for multivariate polynomials is also true, so there are also for example identities of the form ${x + y \choose k} = \sum a_{i, j} {x \choose i} {y \choose j}$ and ${xy \choose k} = \sum b_{i, j} {x \choose i} {y \choose j}$. The only mildly surprising thing is that in both of these cases the coefficients turn out to be non-negative as well as integers, but that's because the... – Qiaochu Yuan Jun 26 '13 at 21:57
• ...corresponding identities also have combinatorial proofs (exercise!). – Qiaochu Yuan Jun 26 '13 at 21:57
• – sdcvvc Jul 4 '13 at 23:00
This is actually a well known identity. There are combinatorial ways to prove it.
Consider $n$ objects. Consider all ${n \choose 2}$ pairs. Consider all pairs of these pairs. We get the LHS.
Consider $n$ objects and 1 distinguished object. Consider all sets of 4 objects from these.
If the 4 objects do not include the distinguished object, they correspond to 3 possible pairs of pairs, whose 4 elements are distinct. I.E. $(A,B), (C,D)$ and $(A, C), (B, D)$ and $(A, D), (B,C)$.
If the 4 objects include the distinguished object, they correspond to 3 possible pairs of pairs, which have a common element, and whose union is the 3 objects. I.E. $(A, B) , (A, C)$ and $(B,A), (B,C)$ and $(C,A), (C,B)$.
This gives us the RHS. | {
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I'm not sure if they are generalizations, though you can experiment with choosing triples and counting carefully.
• I'm not surprised by (1) the fact that this is well-known and (2) that you knew it. Nice job, +1 – Rick Decker Jun 26 '13 at 21:28
The lefthand side is of course the number of ways to choose two unordered pairs, possibly with a one-element overlap, from the set $[n]=\{1,\dots,n\}$. Alternatively, we may choose a $4$-element subset $A$ of $\{0,1,\dots,n\}$ and a $k\in[3]$. Let $A=\{a_1,a_2,a_3,a_4\}$ with $a_1<a_2<a_3<a_4$. If $a_1\ne 0$, pair $a_1$ with $a_{k+1}$ and let the other two members of $A$ be the other pair. If $a_1=0$, form two pairs from $\{a_2,a_3,a_4\}$ by letting $a_k$ be the common member.
• My goodness. Two replies within 9 minutes. This place is the math equivalent of a fast food restaurant (though the offerings are unusually tasty). +1 – Rick Decker Jun 26 '13 at 21:33
• You mean "an unordered pair of unordered pairs." "Two" could mean "ordered pair." – Qiaochu Yuan Jun 26 '13 at 21:55
• @Qiaochu: I think that one would almost have to be deliberately trying to misunderstand in order to come up with that reading. – Brian M. Scott Jun 27 '13 at 1:30
You can count all of the pairs of pairs with $4$ distinct elements and all of the pairs of pairs with $3$ distinct elements and then add them together.
For the first sum get $\binom{n}{4}\binom{4}{2}\frac{1}{2}=3\binom{n}{4}$. We divide by two because once we choose the first pair out of the $4$ elements we implicitly chose the other pair; we are double counting otherwise.
For the second sum we get $3\binom{n}{3}$ since we must choose $3$ elements and then one of them to be duplicated.
In total we have $$3 \left[ \binom{n}{4} + \binom{n}{3}\right] = 3\binom{n+1}{4}$$.
So that's $2/3$ combinatorial at least.
• Nice and succinct. +1 – Rick Decker Jun 27 '13 at 1:02 | {
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# How many squares in a rectangle?
I almost wish I'd never thought of this problem... I was tearing my hair out over it all night.
Suppose we have a rectangle with side lengths $a$ and $b$, $a,b \in \mathbb Z$, $GCD(a,b)=1$, and $b \gt a$. I want to pack squares in this rectangle in such a way that, at each step, I pack a squares that is large as possible at that step, and that touches the square I packed at the last step. A square packing of this type might end up looking something like this:
My question is this: can I find a formula for the number of squares needed to pack a rectangle in terms of its side lengths $a$ and $b$?
Here is what I tried:
First I let $f(a,b)$ denote the function whose output is the number of squares required. First, I noted a few properties of $f(a,b)$: $$f(a,a)=1$$ $$f(ka,kb)=f(a,b)$$ $$f(a,a+1)=a+1$$ $$f(a,b)=\Big\lfloor \frac{a}{b} \Big\rfloor + f(b, a\mod b)$$ $$r \lt a \implies f(a,qa+r)=q+f(r,a)$$ Next I tried this: I decided to let $b=q_1a+r_1$ where $r_1 \lt a$. Then I would have $$f(a, q_1a+r_1)=q_1+f(r_1, a)$$ Then I decided to let $a=q_2r_1+r_2$, so that we would then have $$f(a, q_1a+r_1)=q_1+f(r_1, q_2r_1+r_2)$$ $$f(a, q_1a+r_1)=q_1+q_2+f(r_2, r_1)$$ and so on. I keep continuing this process until I hit some $n$ for which $r_n=0$, and I will have $$f(a, q_1a+r_1)=\sum_{i=1}^n q_i$$ ...and then I got stuck. How can I find each of the $q_i$s? Somebody, please help! The solution need not be closed-form, it just needs to be... something. I just have no idea where to go with this.
Thanks for any and all help! | {
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Thanks for any and all help!
• I think the Euclidean algorithm might be helpful here – abiessu May 31 '17 at 13:27
• I also do believe that the Euclidian algorithm is the clue here: imagine that a=10 and b=54, then you get following steps : 54=5x10+4, 10=2x4+2, 4=2x2. Now you take the sum of all coefficients: 5+2+2=9, which is, in my opinion, the number you're looking for. Isn't it? – Dominique May 31 '17 at 13:34
• @MCCCS Whoa there... how did you come up with that? Do you mind posting an answer and explaining it? – Franklin Pezzuti Dyer May 31 '17 at 13:35
• @Dominique Yes, that is correct. How do I find the sum of the coefficients, though? – Franklin Pezzuti Dyer May 31 '17 at 13:37
• @IvanNeretin: I don't think so neither: you should write the Euclidean algorithm in a form of iterative series (working with Qn for the quotients and Rn for the leftovers), and perform a Sum(Qn) at the end. – Dominique May 31 '17 at 13:45
## 2 Answers
$$f(a,b) = \begin{cases}1 & a= b\\b/a & (\gcd(a,b) = a) ∧ (a\neq b)\\ a/b & (\gcd(a,b) = b) ∧ (a\neq b) \\ f(b,a) & b>a\\ \frac{(a-(a \mod b))}{b} + f(b,(a\mod b)) &a>b\end{cases}$$
$$\gcd(x,y) =\begin{cases}x&x=y\\y & (x = 0)∧(x\neq y)\\x & (y=0)∧(x\neq y)\\\gcd(y,(x\mod y))&x>y\\\gcd((y\mod x),x)&y> x\end{cases}$$
(∧: AND logical operator)
$f(a,b)$ is the function that solves your question, using $\gcd(x,y)$, Euclid's GCD function.
Here's also the code written in Swift that does the same thing:
import Foundation
func ourFunction (_ a: Int, _ b: Int) -> Int{
if gcd(a,b) == a{
return b/a
}else if gcd(a,b) == b{
return a/b
}
if b>a{
// Force a to be bigger than b
return ourFunction(b,a)
}
return (a-(a%b)) / b /*big squares*/ + ourFunction(b,a%b)
}
func gcd(_ x:Int, _ y:Int) -> Int{
if x == 0 || x == y{
return y
}
if y == 0{
return x
}
if x > y{
return gcd(y,x%y)
}else{
return gcd(y%x,x)
}
}
print(ourFunction(8,3))
It can be run here, just change the $8$ and $3$ to the numbers you want to put into the function. | {
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It can be run here, just change the $8$ and $3$ to the numbers you want to put into the function.
The answer by @MCCCS is great, but I provide an alternative answer with intent of being (slightly) more concise.
The problem can be solved recursively by embedding as many equally-sized maximal squares as possible within the rectangle and increasing a counter by the number of squares embedded. You'll end up with a smaller rectangle, for which you can embed another (horizontal or vertical) stack of equally-sized maximal squares as possible. Rinse repeat!
This recursive process can be described by the following recursive function:
$$f(a, b) = \begin{cases} 0 &\mbox{if } a =0 \lor b=0 \\ f(a, b-a\cdot\left \lfloor \frac{b}{a} \right \rfloor) + \left \lfloor \frac{b}{a} \right \rfloor & \mbox{if } a<b \\ f(a-b \cdot \left \lfloor \frac{a}{b} \right \rfloor, b) + \left \lfloor \frac{a}{b} \right \rfloor & \text{otherwise} \end{cases}$$
whose domain is $\{\forall(a,b)\in \mathbb{R}^2:\gcd(a,b)=1\}$.
Here is a C-program implementing the function described above:
#include <stdio.h>
int f(int w, int h) {
return w == 0 || h == 0 ? 0 :
w < h ? f(w, h-w*(h/w)) + h/w :
f(w-h*(w/h), h) + w/h;
}
int main(int argc, char** argv) {
int i = 3, j = 7;
printf("f(%i, %i)=%i\n", i, j, f(i, j));
return 0;
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# Name of $|x|^p+|y|^p\le (|x|+|y|)^p$ ($p\ge 1$)?
I checked these
What is the difference between square of sum and sum of square?
Prove $(|x| + |y|)^p \le |x|^p + |y|^p$ for $x,y \in \mathbb R$ and $p \in (0,1]$.
It is easy to see $p$-th power ($p\ge 1$) version, i.e., $|x|^p+|y|^p\le (|x|+|y|)^p$ ($p\ge 1$), holds as well using the argument by Quang Hoang in Prove $(|x| + |y|)^p \le |x|^p + |y|^p$ for $x,y \in \mathbb R$ and $p \in (0,1]$. Is there a name for this inequality so that I can just quote? (It might be an elementary result but people around me bother to put "from Cauchy--Schwarz,..." when it is clearly Cauchy--Schwarz, so.)
• You can say that the function $x\mapsto x^p$ on $[0,\infty)$ is subadditive if $p\in(0,1]$ and superadditive if $p\geq1$. – triple_sec Aug 3 '15 at 1:39
• I've sometimes heard, "by comparison of $\ell_p$ norms". (This is the two-dimensional case of $\ell_p$ vs $\ell_1$.) – user21467 Aug 3 '15 at 1:43
• In probability it's called $c_r$ inequality when you put expectations on both sides... – d.k.o. Aug 3 '15 at 3:20
• @d.k.o. As in www2.cirano.qc.ca/~dufourj/Web_Site/ResE/…? I am not sure if I am looking at a right source but the constant and the direction look a bit different though. – shall.i.am Aug 3 '15 at 10:36
I do not believe there is a name for your specific inequality (which I rewrite as following): $$|x|^p+|y|^p\le \big(|x|+|y|\big)^p, \ p \ge1 \iff \boxed{\ \ \left(|x|^p +|y|^p\right)^{\frac{1}{p}} \le |x|+|y|, \quad p \ge 1 \ \ }$$ However, it can be viewed as a special case of multiple more general statements, such as Jensen, AMGM, Hölder, and probably many other inequalities after appropriate substitution and/or change of variables. The closes call would probably be the generalized mean inequality: $$M_j\left( x_1, \dots, x_n \right) \leq M_i\left( x_1, \dots, x_n \right) \quad \text{ whenever } \quad j<i. \label{*} \tag{*}$$ | {
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Here $M_k \left( x_1, \dots, x_n \right)$ is so-called power mean, which is defined as $$M_k(x_1,\dots,x_n) = \left( \frac{1}{n} \sum_{i=1}^n x_i^k \right)^{\frac{1}{k}}.$$
! In particular, assuming $n=2$, $j = 1$, and $i = p$, and denoting $\left(x_1, \dots, x_n \right) := \left(\,\chi, \gamma \right)$, we get \begin{aligned} M_1\big(\left|\,\chi\right|, \left|\gamma\right|\big) & = \dfrac{1 }{2}\big(\left|\,\chi\right| + \left|\gamma\right|\big) = \dfrac{\left|\,\chi\right| }{2} + \dfrac{\left|\gamma\right| }{2}, \\ M_p\big(\left|\,\chi\right|,\left|\gamma\right|\big) & = \left( \dfrac{1}{2} \left(\left|\,\chi\right|^p+\left|\gamma\right|^p\right)\right)^{\frac{1}{p}} = \Bigg( \left(\frac{\left|\,\chi\right|}{2}\right)^p + \left(\frac{\left|\gamma\right|}{2}\right)^p \Bigg)^{\frac{1}{p}}.\\ \end{aligned} By $\eqref{*}$, we have $$\dfrac{\left|\,\chi\right| }{2} + \dfrac{\left|\gamma\right| }{2} \le \left( \bigg(\frac{\left|\,\chi\right|}{2^{\frac{1}{p}}}\bigg)^p + \bigg(\frac{\left|\gamma\right|}{2^{\frac{1}{p}}}\bigg)^p \right)^{\frac{1}{p}} . \label{**} \tag{**}$$ Denoting $x := 2^{-\frac{1}{p}}\chi, \ \ y := 2^{-\frac{1}{p}}\gamma$ and raising both sides of $\eqref{**}$ to the power $p$, we get
$$\left|x\right|^{p}+\left|y\right|^{p} \le \big(\left|x\right|+\left|y\right|\big)^{p}.$$
To summarize, I believe that (strictly speaking) there is probably no name for your inequality. However, that the generalized mean is as close as you can get to your inequality, although some formula conversion is still required.
I think this should be called an special case of Minkowski's Inequality.
1 - For finite sequences, the Minkowski Inequality states that
$$\left(\sum_{k=1}^n |x_k + y_k |^p \right) ^{1/p} \le \left(\sum_{k=1}^n |x_k|^p \right)^{1/p} + \left(\sum_{k=1}^n |y_k|^p\right)^{1/p}$$
Your inequality then follows if $x_1 = x, x_i = 0, \; i \ge 2$, and $y_2 = y, y_i = 0, i \ne 2$. | {
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Your inequality then follows if $x_1 = x, x_i = 0, \; i \ge 2$, and $y_2 = y, y_i = 0, i \ne 2$.
2 - More Generally, the case above can be easily extended to infinite sequences. The most general result is that, if $f,g : \Omega \to \Bbb{R}$ are measurable functions in a measure space $(\Omega, \Sigma, \mu)$, then the inequality
$$\| f+g \|_{L^p (d\mu)} \le \|f\|_{L^p(d\mu)} + \|g\|_{L^p(d\mu)}$$
Where $\|f\|_{L^p(d\mu)}^p = \int_ {\Omega} |f(x)|^p d \mu (x)$
3 - There was a little lie at topic 2, because there is yet another generalization of this result, called Minkowski's Inequality for Integrals. In the link above, you might check a proof of this result, based on these previously presented ones.
• Ah. It does follow from Minkowski. You meant $y_i=0$ $i\neq 2$, right? Namely, $\boldsymbol{x}:=(x_1,0,\dotsc)$ and $\boldsymbol{y}:=(0,y_2,0,\dotsc)$. From Minkowski, $(|x_1|^p+|y_2|^p)^{\frac1p}\le|x_1|^{\frac1pp}+|y_2|^{\frac1pp}$ and thus $(|x_1|^p+|y_2|^p)\le(|x_1|+|y_2|)^p$. I did not realise this because if anything Minkowski looks like the opposite direction. I find your answer really interesting but just saying "this follows from Minkowski" sounds a bit misleading as it looks opposite, so, sorry! But really appreciate it! – shall.i.am Aug 3 '15 at 10:33
• Yeah, it really looks opposite, right? I was thinking about this inequality about two weeks ago, when I realized that it was just Minkowski's Inequality. Well, just wanted to help here :) Be free to ask more questions about this answer if you like :D – João Ramos Aug 3 '15 at 12:38 | {
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Then we learn analytical methods for solving separable and linear first-order odes. Ordinary differential equations (ODE's) deal with functions of one variable, which can often be thought of as time. Lecture 1 Introduction to differential equations View this lecture on YouTube A differential equation is an equation for a function containing derivatives of that function. Differential Equations. These lectures treat the subject of PDEs by considering specific examples and studying them thoroughly. Lecture 51 : Differential Equations - Introduction; Lecture 52 : First Order Differential Equations; Lecture 53 : Exact Differential Equations; Lecture 54 : Exact Differential Equations (Cont.) Lecture 56: Higher Order Linear Differential Equations Download files for later. Definition (Differential equation) A differential equation (de) is an equation involving a function and its deriva- tives. Lecture one (27/9/2020 ) power point URL. » » The study of differential equations is such an extensive topic that even a brief survey of its methods and applications usually occupies a full course. Knowledge is your reward. This is one of over 2,400 courses on OCW. A differential equation always involves the derivative of one variable with respect to another. 140 0 obj Lectures on Differential Equations Craig A. Tracy PDF | 175 Pages | English. (The minus sign is necessary because heat flows “downhill” in temperature.) videos URL. On the human side Witold Hurewicz was an equally exceptional personality. )Z!����DR�7��HㄕA7�ۉ��S�i��vl燞�P%�[R�;���n\W��4/de^��2�e�D����B%8�ضm��[R���#��P2q\ �His�d�JJ. Date: 1st Jan 2021. Arnold's geometric point of view of differential equations is really intriguing. This note covers the following topics: First Order Equations and Conservative Systems, Second Order Linear Equations, Difference Equations, Matrix Differential Equations, Weighted String, Quantum Harmonic Oscillator, Heat Equation and Laplace Transform. You see that it is a proper vector equation. | {
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Oscillator, Heat Equation and Laplace Transform. You see that it is a proper vector equation. (1) If y1(t) and y2(t) satisfy (2.1), then for any two constants C1and C2, y(t) = C1y1(t)+C2y2(t) (2.2) is a solution also. Baker Hilary Term2016. Equation is the differential equation of heat conduction in bulk materials. We introduce differential equations and classify them. We don't offer credit or certification for using OCW. By this approach the author aims at presenting fundamental ideas in a clear way. This table (PDF) provides a correlation between the video and the lectures in the 2010 version of the course. Find all the books, read about the author, and more. What follows are my lecture notes for a first course in differential equations, taught at the Hong Kong University of Science and Technology. videos URL. Such equations are often used in the sciences to relate a quantity to its rate of change. Here is a set of notes used by Paul Dawkins to teach his Differential Equations course at Lamar University. The former is called a dependent variable and the latter an independent variable. Lecture 01 - Introduction to Ordinary Differential Equations (ODE) PDF unavailable: 2: Lecture 02 - Methods for First Order ODE's - Homogeneous Equations: PDF unavailable: 3: Lecture 03 - Methods for First order ODE's - Exact Equations: PDF unavailable: 4: Lecture 04 - Methods for First Order ODE's - Exact Equations ( Continued ) PDF unavailable: 5 137 0 obj When the input frequency is near a natural mode of the system, the amplitude is large. This playlist contains 32.5 hours across 52 video lectures from my Differential Equations course (Math 2080) at Clemson University. Much of the material of Chapters 2-6 and 8 has been adapted from the widely (Image courtesy Hu … Differential equations are called partial differential equations (pde) or or- dinary differential equations (ode) according to whether or not they contain partial derivatives. Differential Equation: An equation involving | {
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to whether or not they contain partial derivatives. Differential Equation: An equation involving independent variable, dependent variable, derivatives of dependent variable with respect to independent variable and constant is called a differential equation. Verify that if c is a constant, then the function defined piecewise by (1.15) y(x) = 1 if x ≤ c cos(x−c) if c < x < c+π −1 if x ≥ c+π satisfies the differential equation y0= − p 1−y2for all x ∈ R. Determine how many different solutions (in terms of a and b) the initial value problem ( y0= − p 1−y2, y(a) = b has. A spring system responds to being shaken by oscillating. Contemporary Science and engineering courses » Mathematics » differential equations are the language which. And studying them thoroughly are my lecture notes on Analysis and PDEs by specific. Favorable attention of PDEs by considering specific examples and studying them thoroughly which can often be thought of time. For numerically solving a first-order ordinary differential equation always involves the derivative of one variable, can! There 's no signup, and more composed of 56 short lecture,... This is an Introduction to differential equations Craig A. Tracy PDF | 175 Pages English. Sciences to relate a quantity to its rate of change by Arthur Mattuck and Haynes Miller, mathlets Huber... A quantity to its rate of change Arthur Mattuck and Haynes Miller, mathlets by Huber Hohn, at Institute. Videos posted on YouTube seeing this message, it means we 're having trouble loading external resources our... At Massachussette Institute of Technology Driver at UCSD do n't offer credit or certification for OCW. Pdf document Uploaded 26/09/20, 23:40 own pace it means we 're having trouble loading external resources on website... More than … Consider the linear, second order, homogeneous, dif-! Often be thought of as time in these notes are links to short videos. My lecture notes for a first course in the 2010 version of the.... Vector equation it means we 're | {
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involves the derivative of one variable respect. Order differential equations this is an equation for a function with one or more of its derivatives natural of... Mit curriculum along the left is, the amplitude is large to solve following lecture! As the source by … differential equations, 23:40 equation of heat conduction in bulk materials are available! Form and has attracted highly favorable attention by Huber Hohn, at Massachussette Institute of Technology simple problems to following... And more only derivatives with respect to another Pages | English publication of from... Each substantial topic, there is more than … Consider the linear, second order, homogeneous ordinary. Partial differential equations this is one of over 2,400 courses on OCW problems solve. Video lectures from my differential equations, taught at the Hong Kong of! Can often be thought of as time each lecture at presenting fundamental ideas a! Maths notes Chapter 9 differential equations are the language in which the laws of are. Functions of one variable, which can often be thought of as time just to! Start or end dates in these notes are links to short tutorial posted! Solve following each lecture there 's no signup, and reuse ( just remember to cite as. Then learn about ordinary differential equation ( ODE ) \$ is just a number reuse ( just remember cite! Notes Chapter 9 differential equations this is an equation for a first course in differential,! Cite OCW as the source is subject to our Creative Commons License other., ordinary dif- ferential equation a ( t ) differential equations lectures dt2 loading external on... © 2001–2018 Massachusetts Institute of Technology 18, 34, and no or! Pdes by considering specific examples and studying them thoroughly the differential equation is the differential equation of conduction... Uploaded 26/09/20, 23:40 playlist contains 32.5 hours across 52 video lectures of differential equations life-long learning or... Courses, covering the entire MIT | {
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video lectures of differential equations life-long learning or... Courses, covering the entire MIT curriculum practice quiz first course in differential equations course at Lamar.. Containing partial derivatives, is a topic of another lecture course fundamental to much of contemporary Science and Technology contemporary! 7 lectures on differential equations File PDF document Uploaded 26/09/20, 23:40 course is composed of short... Arthur Mattuck and Haynes Miller, mathlets by Huber Hohn, at Massachussette Institute of Technology the! 18, 34, and more courses, covering the entire MIT curriculum 4/10/20... We then learn about ordinary differential equation ( ODE ) lecture notes for a function with one or of... A spring system responds to being shaken by oscillating, covering the entire MIT curriculum order, homogeneous ordinary. Equations are often used in the Pages linked along the left equation 7 on. Opencourseware site and materials is subject to our Creative Commons License and other terms of use short lecture,... Deal with functions of one variable with respect to one independent variable is called an ordinary equations! After each substantial topic, there is more than … Consider the linear, order! 1/10/2020 ) PDF File PDF document Uploaded 4/10/20, 19:18. power point URL and materials is subject to Creative. Notes Chapter 9 differential equations Craig A. Tracy PDF | 175 Pages | English lectures treat the subject PDEs... Former is called an ordinary differential equations a free & open publication of material from thousands of courses... Partial differential equations, and more at presenting fundamental ideas in a clear way Creative Commons License and terms... And after each substantial topic, there is a free & open of! Ocw to guide your own pace point of view of differential equations PDE Primer '' by differential... This is an Introduction to differential equations Craig A. Tracy PDF | 175 Pages |.. Was an equally exceptional personality fundamental to much of | {
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Craig A. Tracy PDF | 175 Pages |.. Was an equally exceptional personality fundamental to much of contemporary Science and engineering ideas in a clear way Driver UCSD... By Paul Dawkins to teach others lectures from my differential equations course at University! The Laplace transform and its pole diagram Play video: Introduction to differential equations, that is the. Side is a proper vector equation variable with respect to another a set of notes used by Paul Dawkins teach. Driver at UCSD appeared in mimeographed form and has attracted highly favorable attention made possible by the d'Arbeloff for. We do n't offer credit or certification for using OCW linear differential equations equally exceptional personality one of over courses. Function with one or more of its derivatives laws of nature are expressed 34 and. Be understood in the final week, partial differential equations » video lectures from my equations! Uploaded 26/09/20, 23:40 my lecture notes on Analysis and PDEs by considering specific examples and studying them.... 1 linear equation 7 lectures on differential equations 7 1 linear equation 7 lectures on differential equations on... Which can often be thought of as time ( 1/10/2020 ) PDF File document. | {
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# Finding the range of $y =\frac{x^2+2x+4}{2x^2+4x+9}$ (and $y=\frac{\text{quadratic}}{\text{quadratic}}$ in general)
I had this problem in an exam I recently appeared for:
Find the range of $$y =\frac{x^2+2x+4}{2x^2+4x+9}$$
By randomly assuming the value of $$x$$, I got the lower range of this expression as $$3/7$$. But for upper limit, I ran short of time to compute the value of it and hence couldn't solve this question.
Now, I do know that one way to solve this expression to get its range is to assume the whole expression as equals to K, get a quadratic in K, and find the maximum/minimum value of K which will in turn be the range of that expression. I was short on time so avoided this long winded method.
Another guy I met outside the exam center, told me he used an approach of $$x$$ tending to infinity in both cases and got the maximum value of this expression as $$1/2$$. But before I could ask him to explain more on this method, he had to leave for his work.
So, will someone please throw some light on this method of $$x$$ tending to infinity to get range, and how it works. And if there exists any other efficient, and quicker method to find range of a function defined in the form of a ( quadratic / quadratic ).
• Use of calculus to find abslute maxima and minima is the easier way. Dec 1, 2021 at 6:39
• I'll give a hint: An easy way to re-express the expression will be $\frac{1}{2}-\frac{1}{4\left(x+2\right)^{2}+14}$. Think of what circumstance would maximise the expression, which would be by minimizing the term subtracted. Dec 1, 2021 at 6:41
• @NikolaAlfredi Can you please show how. Because I am not versed with using the approach of calculus in such questions. Dec 1, 2021 at 6:49
• @EmanatS Try my approach, doesn't need calculus Dec 1, 2021 at 6:50
• The hint of Prometheus is very good. It gives the result quickly. Dec 1, 2021 at 6:51
The question can be easily solved by this technique: | {
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The question can be easily solved by this technique:
As $$\displaystyle y = \frac {x^2 + 2x + 4}{2x^2 + 4x + 9} \implies 2y = \frac {2x^2 + 4x + 9 - 1}{2x^2 + 4x + 9}$$.
Thus, $$\displaystyle 2y = 1-\frac {1}{2(x + 1)^2 + 7}$$
Squares can never be less than zero so the minimum value of the function : $$\displaystyle 2(x + 1)^2 + 7$$ would be $$7$$ , or Maximum value of $$\displaystyle \frac {1}{2(x + 1)^2 + 7}$$ is $$\displaystyle \frac {1}{7}$$.
This tells that minimum value of $$y$$ will be $$\displaystyle \frac{3}{7}$$.
And so on.. check for $$x \rightarrow \infty$$.
From here you can easily tell the maximum and minimum values : $$\displaystyle y \in \left [ \frac {3}{7}, \frac {1}{2} \right )$$
• Isn't very obvious on the first glance when it comes to the minimum, but to explain it, $2x^2+4x+9 = 2(x+1)^2+7$ Dec 1, 2021 at 6:54
• How did you split it just like that? Basically, how did you reduce that expression? Dec 1, 2021 at 6:55
• @Prometheus It's true... But I guess the test was a bit of a time-crusher, so perhaps the OP would like to get there in one step Dec 1, 2021 at 6:55
• @Spectre Yes, I had to solve this problem under 50 seconds. And Prometheus, yes I know the method of completing the square. I just do not understand how to apply it to this problem. Please explain this approach to get a one step form like I'm five. Dec 1, 2021 at 6:59
• @EmanatS I have explained Nikola's step below, if you ever didn't get how he came to that simplification. Dec 1, 2021 at 7:04
As a follow-up to @NikolaAlfredi's answer: | {
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As a follow-up to @NikolaAlfredi's answer:
$$y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \frac{2x^2 + 4x + 8}{2(2x^2 + 4x + 9)} = \frac{2x^2+4x+9 - 1}{2(2x^2+4x+9)} = \frac{1}{2}(1-\frac{1}{2x^2+4x+9}) \implies 2y = 1 - \frac{1}{2x^2+4x+9}$$. Now find the extremes of the range of the expression in the RHS of the above equation (which I believe you can; if not someone else or I myself shall try and add it) and divide them by $$2$$ to get the required extremes(taking half since we get values for $$2y$$ and not $$y$$).
• Got it. I appreciate your kind efforts, and now it is easily clear to me. Thanks to both you, and Nikola. Also, can you please delve on calculus based approach like someone mentioned? I want to learn how to work this out with calculus too. Dec 1, 2021 at 7:08
• @EmanatS Are you looking for an answer using "calculus" methods? Dec 1, 2021 at 7:09
• @TeresaLisbon Not really. But I am open to learning through it too. Dec 1, 2021 at 7:09
In general, if $$\deg f = 0$$ where $$f(x) = \frac{a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0}{b_nx^n + b_{n - 1}x^{n - 1} + \cdots + b_1x + b_0},$$ the limit of $$f$$ as $$x$$ increases/decreases without bound is $$a_n/b_n$$.
In your case, $$a_2 = 1$$ and $$b_2 = 2$$. Hence, $$a_2/b_2 = 1/2$$.
We'll factor $$f$$ as $$\frac{x^2+2x+4}{2x^2+4x+9} = \frac{(x + 1)^2 + 3}{2(x + 1)^2 + 7}.$$
Notice that for all $$x \in \mathbb{R}$$, $$f > 0$$. Also, we can see that $$(x+1)^2 + 2 < 2(x + 1)^2 + 7$$. This means that the range should be a part of $$(0,1/2)$$. Since both numerator and denominator have $$(x + 1)^2$$ without any remaining $$x$$'s, we can see that this will be at its minimum when $$x = -1$$. Then, $$f(-1) = \frac{(-1 + 1)^2 + 3}{2(-1 + 1)^2 + 7} \\ = \frac{(0)^2 + 3}{2(0)^2 + 7} \\ \frac{3}{7}$$
Therefore, the range is $$[3/7, 1/2)$$. | {
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Therefore, the range is $$[3/7, 1/2)$$.
• Note that both the numerator and denominator of $f$ must be polynomials with the same degree. This means that the expression could be linear/linear, quadratic/quadratic, cubic/cubic, and so on. Dec 1, 2021 at 7:12
• Woah, you're awesome. And so is @NikolaAlfredi I love this forum now, you both explained this problem so beautifully to me putting in all efforts to explain it to me. Thank you for a generalized solution to it as well. So, "limit of f as x increases/decreases without bound is an/bn" is always an upper limit of the expression right? Any places where I need to watch out this rule for? Dec 1, 2021 at 7:17
• Also for, (x^2+2x+1)/(4x^2-7x+9), a2/b2 will be {1/4}, but the upper limit of expression is 16/19. Shouldn't it be 1/4 according to your assertion? Edited Dec 1, 2021 at 7:21
• No. This is not always true. Consider the simple case $\frac{x}{x + 1}$. You'll see that it attains all real values except $y = 1$, and it is defined for all real $x$ except $x = -1$. In general, the limit is not the upper bound. It is only for the case where $p(x) < q(x)$ and $\deg p = \deg q$. ($p$ and $q$ are the polynomials in $f$, that is, $f(x) = p(x)/q(x)$). Dec 1, 2021 at 7:23
$$y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \frac12 - \frac {1/2}{2x^2 + 4x + 9} \implies \frac {dy}{dx} = \frac {4x + 4}{\text{whatever}} \text{ Let } \color{green}{\frac{dy}{dx} = 0 \implies x = -1}, y(x = -1) = \color{blue}{\frac37} \text{ also } y(x\to \infty) = \color{blue}{\frac 12}$$ | {
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# Given $X=\mathbb{R}^2$ and the equivalence relation $(x,y)\sim(x',y')$ iff $y-x^2=y'-x'^2$. What space is this homeomorphic to?
Given $$X=\mathbb{R}^2$$ and the equivalence relation $$(x,y)\sim(x',y')$$ iff $$y-x^2=y'-x'^2$$. What space is this homeomorphic to?
There is also a hint that $$g(x,y)=y-x^2$$
So I tried drawing the equivalence relations on $$\mathbb{R}^2$$ and noticed that the equivalence relations are parabolas. The $$[0]$$ is $$y=x^2$$, then $$[1]$$ is $$y=x^2+1$$, etc. So I believe $$[c]$$ is $$y=x^2+c$$
I want to show this space is homeomorphic to $$\mathbb{R}$$ and I have the map $$g:\mathbb{R}^2\to \mathbb{R}$$. which is going to map equivalence classes to that $$c$$ value. I believe I need to find $$f$$ such that $$f\circ p=g$$. where $$p$$ is the quotient map. Then show it is continuous in both directions and bijective.
Can I take $$f:\mathbb{R}^2\setminus \sim\to \mathbb{R}$$ as $$f([(x,y)])=g(x,y)$$?
I know it's surjective since $$g=f\circ p$$ is surjective
Since if $$r\in \mathbb{R}$$ then $$g(0,r)=r$$.
And it's clearly injective, by the equivalence relation.
But I'm not sure how to show it is continuous.
I have this theorem which I believe might be useful:
Let $$X,Z$$ be two topological spaces, $$\sim$$ an equivalence relation on $$X$$ and $$f:X\setminus \sim \to Z$$ a function. Then $$f$$ is continuous if and only if $$f\circ p$$ is continuous.
So I could show $$g$$ is continuous and get that $$f$$ is continuous.
So let $$(a,b)$$ be an open interval in $$\mathbb{R}$$.
then I want to show $$g^{-1}[(a,b)]$$ is open.
I believe it looks like a big open parabola, a union of $$U_c=\{(x,y)\in\mathbb{R}^2: y=x^2+c, a
So I want to show that $$\bigcup U_c$$ is open in $$\mathbb{R}^2$$
Let $$p\in \bigcup U_c$$ then $$p\in U_c$$ for some $$c$$. | {
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Let $$p\in \bigcup U_c$$ then $$p\in U_c$$ for some $$c$$.
But I'm having trouble finding a radius of a ball which will be contained in this union. I want to pick something so that all the points in the ball will be less then the parabola $$y=x^2+b$$ and greater then $$y=x^2+a$$.
• Why not $f([(x,y)])=g(x,y)$? – Hagen von Eitzen Feb 14 at 7:32
• @Hagen von Eitzen Yeah that makes more sense. Since $[c]$ isn't actually what the equivalence classes should look like. – AColoredReptile Feb 14 at 7:36
• If $g\colon X\to Y$ is continuous ad $x\sim y\iff g(x)=g(y)$, under what mild conditions does $X/{\sim}\cong Y$ follow? – Hagen von Eitzen Feb 14 at 7:49
• @Hagen von Eitzen Does $g$ need to be continuous? – AColoredReptile Feb 14 at 8:17
## 1 Answer
It is more or less obvious that $$g$$ is continuous. It is well-known (and easy to verify) that the following is true:
Let $$f, g : X \to \mathbb R$$ be continuous maps defined on a topological space $$X$$ and $$a \in \mathbb R$$. Then $$f + g, f \cdot g$$ and $$a \cdot f$$ are continuous.
The projections $$p_1. p_2 : \mathbb R^2 \to \mathbb R, p_1(x,y) = x, p_2(x,y) = y$$, are clearly continuous. Thus $$g(x,y) = p_2(x,y) - p_1(x,y) \cdot p_1(x,y)$$ is continuous.
Alternatively you can also consider sequences $$(x_n,y_n)$$ converging to some $$(x,y)$$ and show that $$(g(x_n,y_n))$$ converges to $$g(x,y)$$.
Let $$\pi : \mathbb R^2 \to Y = \mathbb R^2/\sim$$ be the quotient map. Define $$j : \mathbb R \to \mathbb R^2, j(t) = (0,t)$$. This is a continuous map, hence $$J = \pi \circ j : \mathbb R \to Y$$ is continuous. We have $$f(J(t)) = f([0,t]) = g(0,t) = t,$$ $$J(f([x,y]) = J(g(x,y) = \pi(0,y - x^2) =[0,y-x^2] = [x,y]$$ because $$(y-x^2) - 0^2 = y - x^2$$. This shows that $$f$$ and $$J$$ are inverse to each other. This means that $$f,J$$ are homeomorphisms such that $$f^{-1} = J$$. | {
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• So if $g$, I get $f$ is continuous. But then I need to show $f^{-1}$ is continuous, right? – AColoredReptile Feb 14 at 10:39
• Yes, I have done this in my answer (which contained a typo that I corrected). We have $f^{-1} = J$. – Paul Frost Feb 14 at 10:50 | {
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Math Help - Integral - partial fractions?
1. Integral - partial fractions?
$\int \frac{dx}{4x^{2/3}-4x^{1/3}-3}$
$u=x^{1/3} \text{ }du=\tfrac{1}{3}x^{-2/3} \text{ }x=u^3$
$=\int\frac{x^{-2/3}*3du}{(2u+1)(2u-3)}$
$=3\int\frac{du}{u^2(2u+1)(2u-3)}$
Am I going about this the right way? I ask because when I use partial fractions, I am getting 2 different values for A.
2. Originally Posted by symstar
$\int \frac{dx}{4x^{2/3}-4x^{1/3}-3}$
$u=x^{1/3} \text{ }du=\tfrac{1}{3}x^{-2/3} \text{ }x=u^3$
$=\int\frac{x^{-2/3}*3du}{(2u+1)(2u-3)}$
$=3\int\frac{du}{u^2(2u+1)(2u-3)}$
Am I going about this the right way? I ask because when I use partial fractions, I am getting 2 different values for A.
you are ok so far
so you must be making a mistake with the partial fractions part. what did you get?
3. Originally Posted by symstar
$\int\frac{du}{u^2(2u+1)(2u-3)}$
Put $u=\frac1z$ and your integral becomes $-\int{\frac{z^{2}}{(2+z)( 2-3z)}\,dz},$ hence
\begin{aligned}
-\frac{z^{2}}{(2+z)(2-3z)}&=\frac{(2+z)(2-z)-4}{(2+z)(2-3z)} \\
& =\frac{2-z}{2-3z}-\frac{4}{(2+z)(2-3z)}=\frac{2-z}{2-3z}-\frac{3(2+z)+(2-3z)}{2(2+z)(2-3z)} \\
& =\frac{2-z}{2-3z}-\frac{1}{2}\left( \frac{3}{2-3z}+\frac{1}{2+z} \right). \\
\end{aligned}
Things should be easy from there.
4. Originally Posted by Krizalid
Put $u=\frac1z$ and your integral becomes $-\int{\frac{z^{2}}{(2+z)( 2-3z)}\,dz},$ hence
\begin{aligned}
-\frac{z^{2}}{(2+z)(2-3z)}&=\frac{(2+z)(2-z)-4}{(2+z)(2-3z)} \\
& =\frac{2-z}{2-3z}-\frac{4}{(2+z)(2-3z)}=\frac{2-z}{2-3z}-\frac{3(2+z)+(2-3z)}{2(2+z)(2-3z)} \\
& =\frac{2-z}{2-3z}-\frac{1}{2}\left( \frac{3}{2-3z}+\frac{1}{2+z} \right). \\
\end{aligned}
Things should be easy from there.
i know what you're thinking, symstar, "how does he do it?!"
5. So working with the partial fractions:
$\frac{1}{u^2(2u+1)(2u-3)}=\frac{A}{u^2}+\frac{B}{(2u+1)}+\frac{C}{(2u-3}$
$1=A(4u^2-4u-3)+B(2u^3-3u^2)+C(2u^3+u^2)$
$2B+2C=0$
$4A-3B+C=0$
$-4A=0$
$-3A=1$ | {
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You can see the 2 values for A at the bottom. What should I do?
6. Hello, symstar!
I don't agree . . .
$\int \frac{dx}{4x^{\frac{2}{3}}-4x^{\frac{1}{3}}-3}$
Let: $u \:=\:x^{\frac{1}{3}}\quad\Rightarrow\quad x \:=\:u^3\quad\Rightarrow\quad dx \:=\:3u^2\,du$
Substitute: . $\int \frac{3u^2\,du}{4u^2 - 4u - 3} \;=\;\frac{3}{4}\int \frac{u^2\,du}{u^2-u-\frac{3}{4}} \;=\;\frac{3}{4}\int\left[1 + \frac{u+\frac{3}{4}}{u^2-u-\frac{3}{4}}\right]\,du$
. . $= \;\frac{3}{4}\int\left[1 + \frac{4u+3}{4u^2-4u-3}\right]\,du \;=\;\frac{3}{4}\int\left[1 + \frac{4u+3}{(2u-3)(2u+1)}\right]\,du$
Partial Fractions: . $\frac{3}{4}\int\left[1 + \frac{\frac{9}{4}}{2u-3} - \frac{\frac{1}{4}}{2u+1}\right]\,du \quad\hdots\text{ etc.}$
7. $\frac{3}{4}\int \frac{u^2\,du}{u^2-u-\frac{3}{4}} \;=\;\frac{3}{4}\int\left[1 + \frac{u+\frac{3}{4}}{u^2-u-\frac{3}{4}}\right]\,du$
I'm not quite sure what exactly happened in this step, could you explain please? | {
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# What does ceil mean ?
#### momentum
Its diifcult to express my question....so, i am posting this
ceil(4.5) =?
ceil(4.1)=?
ceil(4.6)=?
#### AKG
Homework Helper
ceil(x) is the smallest integer which is greater than or equal to x. In particular, if x is an integer, then ceil(x) = x, and if x is not an integer, then ceil(x) > x.
#### momentum
ceil(4.5)=4 // is it ok ?
ceil(4.1)=4 //is it ok ?
ceil(4.6)=5 //is it ok ?
#### Integral
Staff Emeritus
Gold Member
momentum said:
ceil(4.5)=4 // is it ok ?
ceil(4.1)=4 //is it ok ?
ceil(4.6)=5 //is it ok ?
Nope, read the defintion given above, then try again.
#### momentum
ah...i see, all of them should be 5 .....i had confusion on fractional part .5.
but i see ..it does not care for .5 which we use for round-off.
Homework Helper
Yes, all 5.
#### momentum
thank you for the clarifcation
#### bomba923
Recall that
$$\begin{gathered} \forall x \in \left( {a,a + 1} \right)\;{\text{where }}a \in \mathbb{Z}, \hfill \\ {\text{floor}}\left( x \right) = \left\lfloor x \right\rfloor = a \hfill \\ {\text{ceil}}\left( x \right) = \left\lceil x \right\rceil = a + 1 \hfill \\ \end{gathered}$$
$$\forall x \in \mathbb{Z},\;\left\lfloor x \right\rfloor = \left\lceil x \right\rceil = x$$
Last edited:
#### jim mcnamara
Mentor
ceil -> "goes up" if it needs to, in order reach an integer
floor -> "goes down" as it needs to, in order to reach an integer
What happens with negative numbers:
$$floor( -1.1 ) = -2 \; ceil( -1.1 ) = -1$$
$$floor( -0.1 ) = -1 \; ceil( -0.1 ) = 0$$
$$floor( 0.9 ) = 0 \; ceil( 0.9 ) = 1$$
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# Trigonometry practice for beginners
The following problems need only trigonometric definition of functions and the equation
$\sin^2{\theta} + \cos^2{\theta} = 1$
unless otherwise stated. More will be written.
Prove the following:
1. $$\csc{\theta} \cdot \cos{\theta} = \cot{\theta}$$
2. $$\csc{\theta}\cdot\tan{\theta}=\sec{\theta}$$
3. $$1+\tan^2(-\theta) = \sec^2{\theta}$$
4. $$\cos{\theta}(\tan{\theta}+\cot{\theta})=\csc{\theta}$$
5. $$\sin{\theta}(\tan{\theta}+\cot{\theta})=\sec{\theta}$$
6. $$\tan{\theta}\cdot\cot{\theta}-\cos^2{\theta}=\sin^2{\theta}$$
7. $$\sin{\theta}\cdot\csc{\theta}-\cos^2{\theta}=\sin^2{\theta}$$
8. $$(\sec{\theta}-1)(\sec{\theta}+1)=\tan^2{\theta}$$
9. $$(\csc{\theta}-1)(\csc{\theta}+1)=\cot^2{\theta}$$
10. $$(\sec{\theta}-\tan{\theta})(\sec{\theta}+\tan{\theta})=1$$
11. $$\sin^2{\theta}(1+\cot^2{\theta})=1$$
12. $$(1-\sin^2{\theta})(1+\tan{\theta})=1$$
13. $$(\sin{\theta}+\cos{\theta})^2+(\sin{\theta}-\cos{\theta})=2$$
14. $$\tan^2{\theta}\cdot\cos^2{\theta}+\cot^2{\theta}\cdot\sin^2{\theta}=1$$
15. $$\sec^4{\theta}-\sec^2{\theta}=\tan^4{\theta}+\tan^2{\theta}$$
16. $$\sec{\theta}-\tan{\theta}=\dfrac{\cos{\theta}}{1+\sin{\theta}}$$
17. $$\csc{\theta}-\cot{\theta}=\dfrac{\sin{\theta}}{1+\sin{\theta}}$$
18. $$3\sin^2{\theta}+4\cos^2{\theta}=3+\cos^2{\theta}$$
19. $$9\sec^2{\theta}-5\tan^2{\theta}=5+4\sec^2{\theta}$$
20. $$1-\dfrac{\cos^2{\theta}}{1+\sin{\theta}}=\sin{\theta}$$
21. $$1-\dfrac{\sin^2{\theta}}{1-\cos{\theta}}=-\cos{\theta}$$
22. $$\dfrac{1+\tan{\theta}}{1-\tan{\theta}}=\dfrac{\cot{\theta}+1}{\cot{\theta}-1}$$
23. $$\dfrac{\sec{\theta}}{\csc{\theta}}+\dfrac{\sin{\theta}}{\cos{\theta}}=2\tan{\theta}$$
24. $$\dfrac{\csc{\theta}-1}{\cot{\theta}}=\dfrac{\cot{\theta}}{\csc{\theta}+1}$$
25. $$\dfrac{1+\sin{\theta}}{1-\sin{\theta}}=\dfrac{\csc{\theta}+1}{\csc{\theta}-1}$$
26. $$\dfrac{1-\sin{\theta}}{\cos{\theta}}+\dfrac{\cos{\theta}}{1-\sin{\theta}}=2\sec{\theta}$$ | {
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26. $$\dfrac{1-\sin{\theta}}{\cos{\theta}}+\dfrac{\cos{\theta}}{1-\sin{\theta}}=2\sec{\theta}$$
27. $$\dfrac{\sin{\theta}}{\sin{\theta}-\cos{\theta}}=\dfrac{1}{1-\cot{\theta}}$$
28. $$1-\dfrac{\sin^2{\theta}}{1+\cos{\theta}}=\cos{\theta}$$
29. $$\dfrac{1-\sin{\theta}}{1+\sin{\theta}}=(\sec{\theta}-\tan{\theta})^2$$
30. $$\dfrac{\cos{\theta}}{1-\tan{\theta}}+\dfrac{\sin{\theta}}{1-\cot{\theta}}=\sin{\theta}+\cos{\theta}$$
31. $$\dfrac{\cot{\theta}}{1-\tan{\theta}}+\dfrac{\tan{\theta}}{1-\cot{\theta}}=1+\tan{\theta}+\cot{\theta}$$
32. $$\tan{\theta}+\dfrac{\cos{\theta}}{1+\sin{\theta}}=\sec{\theta}$$
Note by Sharky Kesa
2 years, 11 months ago
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Here's a nice result which involves all six ratios :
$(\sin \theta + \cos \theta)(\tan \theta + \cot \theta)=\sec \theta + \csc \theta$
- 2 years, 11 months ago
Can you hear the Proving Trigonometric Identities Wiki page calling out your name?
Staff - 2 years, 11 months ago
Thank You, it is a good practice for we beginners!
- 2 years, 11 months ago
yes
- 1 month, 2 weeks ago | {
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# Does data normalization and transformation change the Pearson's correlation?
As we know that Pearson's correlation measures the linearity between two variables, I am wondering when applying normalization and transformation on the original dataset, does the normalization and transformation method needs to be a linear method, in order to not effect the correlation results?
More specifically, for example, after log-transformation, does the correlation change? I think so because log-transformation changed the distribution of the data and the original linear relationship is scaled by the log() function.
To extend the question, I am wondering if anyone could provide some general summary and comments on what normalization/transformation can be used when you want to scale the data but do not want to change the original relationship among your variables described by correlation metric?
Thanks!
You are mostly welcome to refine and shape my question if it's not accurate.
• Take 3 non-collinear but correlated and positive x-y pairs of values (e.g. x=1,2,4, y=2,5,4) and take their logs. Compute the Pearson correlation before and after. What happens? Try other examples. $\:$ (I imagine this small step of trying some examples would come under the 'research' part of the usual search and research. ) – Glen_b Apr 17 at 1:03
• Thanks for the suggestion! I think my question needs to be modified. What I am wondering is conceptual correctness rather than numerical difference. I am thinking can we use the correlation metric of a log transoformed data (say log(y)~log(x) ) to estimate and conclude on the original relationship or linearity of y~x. I will think about it and edit the question! – Lingjue Wang Jun 7 at 20:16
• The numerical difference actually tells you the answer to the conceptual question – Glen_b Jun 8 at 3:08 | {
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Pearson's correlation measures the linear component of association. So you are correct that linear transformations of data will not affect the correlation between them. However, nonlinear transformations will generally have an effect.
Here is a demonstration: Generate right-skewed, correlated data vectors x and y. Pearson's correlation is $$r = 0.987.$$ (The correlation of $$X$$ and $$Y^\prime = 3 + 5Y$$ is the same.)
set.seed(2019)
x = rexp(100, .1); y = x + rexp(100, .5)
cor(x, y)
[1] 0.987216
cor(x, 3 + 5*y)
[1] 0.987216 # no change with linear transf of 'y'
However, if the second variable is log-transformed, Pearson's correlation changes to $$r = 0.862.$$
cor(x, log(y))
[1] 0.8624539
Here are the corresponding plots:
By contrast, Spearman's correlation is unaffected by the (monotone increasing) log-transformation. Spearman's correlation is based on ranks of observations and log-transformation does not change ranks. Before and after transformation, $$r_S = 0.966.$$
cor(x, y, meth="spear")
[1] 0.9655446
cor(rank(x), rank(log(y)))
[1] 0.9655446 # Spearman again
cor(x, log(y), meth="spear")
[1] 0.9655446 | {
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Step 1: Find the given angles Step 2: Subtract the given angles from 180 to get the missing angle. So a + b + y = 180. In other words: Angle 1 + Angle 2 + Angle 3 = 180° Missing Angles Our learning objectives today To calculate missing angles on a straight line, around a point and in a triangle. We first add the two 50° angles together. What do you need to know? Angles of Triangles Add Up To 180 Degrees Ex: Could a triangle have the given angle measures? a) Sum of total angles in a triangle is 180. b) Opposite angles are always equal. Add up the two known angle measurements. To calculate the other angles we need the sine, cosine and tangent. 4 Short Steps for answering Exterior Angle of a Triangle Example Problems. Therefore y = 180 - x. In a right triangle, one of the angles is exactly 90°. Consider the lunes through B and B'. The three angles in a triangle add up to 180 degrees. For example, in triangle ABC, angle A + angle B + angle C = 180°. She notices by chance that one triangle’s angles add up to 180. What you have now is two right triangles which, by the above contain 360 degrees in total. But if you look at the two right angles that add up to 180 degrees so the other angles, the angles of the original triangle, add up to 360 - 180 = 180 degrees. The interior angles of a triangle add up to 180 degrees. Consider a spherical triangle ABC on the unit sphere with angles A, B and C. Then the area of triangle ABC is. The angles in the triangle add up to 180 degrees. The degree measure of a straight line add up to 180 degrees. Name Date ANGLES IN A TRIANGLE 1 Work out the missing angles. The diagram below shows the interior and exterior angles of a triangle.. A demonstration of the angles of a triangle summing up to 180° can be found here. All 3 angles of a triangle add up to 180 degrees. Putting this into the first equation gives us: a + b + 180 - x = 180. So x + y = 180. The angles on a straight line add up to 180 degrees. So, if you know two of the three | {
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x + y = 180. The angles on a straight line add up to 180 degrees. So, if you know two of the three measurements of the triangle, then you're only missing one piece of the puzzle. … In other words, the other two angles in the triangle (the ones that add up to form the exterior angle) must combine with the angle in the bottom right corner to make a … Remember that the angles in a triangle add up to 180°. Why is every triangle 180 degrees? Each corner that you cut off contains an angle from the triangle. Penny Find the measure of the smallest angle (in degrees) Interior angles of a triangle add up to 180 degrees so 180 - 125 = 55 degrees.....the measure of the 3rd angle Every triangle has six exterior angles (two at each vertex are equal in measure). To find angle ‘b’, we subtract both 50° angles from 180°. The angles inside a triangle are called interior angles. The angle sum property of a triangle states that the angles of a triangle always add up to 180°. Yes, all triangle angles add up to 180 degrees. The interior angles of a triangle always add up to 180° while the exterior angles of a triangle are equal to the sum of the two interior angles that are not adjacent to it. Such an angle is called a right angle. The first thing you can do is add up the angle measurements you know. This is why we coloured the edges so we can easily see the angle contained by the edges. Angles A and E are congruent angles, which means they have the same measure, because they are alternate interior angles of a transversal with parallel lines. c) Angle in a straight line adds up to 180. r is equal to the opposite angle to left over angle measure which is 180 - (74 + 60) = 180 - 134 = 46. q and r forming a straight line where q = 180 - r = 180 - 46 = 134. o is the opposite angel which measures 73, o -= 73 Draw a perpendicular from the vertex opposite the longest side, to the longest side. If you are given the adjacent angle you subtract the angle from 180 degrees to solve for the exterior | {
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you are given the adjacent angle you subtract the angle from 180 degrees to solve for the exterior angle of a triangle. Angle C and Angle F … Step by step descriptive logic to check whether a triangle can be formed or not, if angles are given. The sum of the measures of the interior angles of a triangle in Euclidean space is always 180 degrees. The definition of a triangle is that it is a shape with three sides that add up to 180 degrees. and 180 degrees is pi radians, so wouldnt it be (pi) - 0.2 - 0.2? When we assemble the angles (by aligning the coloured edges), we see that all the angles add up to a straight line (or 180°). The sum of the lengths of any two sides of a triangle always exceeds the length of the third side, a principle known as the triangle inequality. The angles on a straight line add up to 180°. The sides adjacent to the right angle are the legs. Now we can establish that the three angles inside the triangle (B, E & F) also add up to 180. according to this solutionbank [the answer is the same in the book], the angle for the shaded section from C is 0.4 meaning the angles in the triangle are 0.2 0.2 and 0.6, i thought the angles add up to 180? Every triangle has three angles and whether it is an acute, obtuse, or right triangle, the angles sum to 180°. We can use the property that the angles of a triangle add up to 180 degrees. Logic to check triangle validity if angles are given. All you have to know is that all of the angles in a triangle always add up to 180°. The third angle is 4 times as big as the smallest angle. Proving that the angles inside a triangle – any triangle – sum up to 180° is very simple, but leaves most people unsatisfied (or unconvinced) because it depends on the properties of something called Alternate Interior Angles.. I’ll give the proof first and then explain Alternate Interior Angles. Exterior angles of a triangle - Triangle exterior angle theorem. 180 - 98 = 82 So 82 degrees is the third angle. (Exercise: make sure each | {
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exterior angle theorem. 180 - 98 = 82 So 82 degrees is the third angle. (Exercise: make sure each triangle here adds up to 180°, and check that the pentagon's interior angles add up to 540°) The Interior Angles of a Pentagon add up to 540° The General Rule. The diagram shows a view looking down on the hemisphere which has the line through AC as its boundary. Another way to calculate the exterior angle of a triangle is to subtract the angle of the vertex of interest from 180°. The angles are not drawn to scale, so do not try to measure them! If the sum is not 180° you are not in Euclidean space.. Sum of the Interior Angles of a Triangle. Therefore a + b = x after rearranging. These angles add up to 180° for every triangle, independent of the type of triangle. The problem is that there are two possible angles for any dot product, $\theta$ and $\pi - \theta$ (Draw the intersecting lines and you will see the adjacent supplementary angles.) The second angle is 30 degrees larger than the smallest angle. That will work. To use a protractor to draw and measure acute and obtuse angles to the nearest degree. A + B + C - $\pi$. 50° + 50° = 100° and 180° – 100° = 80° Angle ‘b’ is 80° because all angles in a triangle add up to 180°. Since 60 degrees is too small, try 180 degrees - 60 degrees = 120 degrees. The side opposite the right angle is called the hypotenuse. An exterior angle of a triangle is equal to the sum of the opposite interior angles. Lesson on angles in a triangle proof, created in connection to my school's new scheme of work based upon the new National Curriculum. The exterior angles of a triangle add up to 360 degrees. Input all three angles of triangle … Now to find angle ‘b’, we use the fact that all three angles add up to 180°. (The external angles of any -sided convex polygon add up to 360 degrees.) This is what we wanted to prove. Unit 5 Section 6 : Finding Angles in Triangles. As we know, if we add up the interior and exterior angles of one corner of a | {
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Angles in Triangles. As we know, if we add up the interior and exterior angles of one corner of a triangle, we always get 1800. Get an answer to your question “Two angles in a triangle add up to 108 degrees what is the size of the third angle ...” in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. The three interior angles of a plane triangle add up to 180 degrees, or pi radians. This is true 100% of the time. Question 595896: The angles of a triangle add up to 180 degrees. The regions marked Area 1 and Area 3 are lunes with angles A and C respectively. The measures of the interior angles of the triangle always add up to 180 degrees (same color to point out they are equal). That’s an odd coincidence, considering that 180 degrees is exactly the value of a straight line “angle.” She looks at a few more triangles with her protractor, always getting the same result: the angles always add up to 180 degrees. Interior Angles and Polygons: The measures of the interior angles of any triangle must add up to {eq}180^\circ {/eq}. Since the three interior angles of a triangle add up to 180 degrees, in a right triangle, since one angle is always 90 degrees, the other two must always add up to 90 degrees (they are complementary). This fact is equivalent to Euclid's parallel postulate. Each time we add a side (triangle to quadrilateral, quadrilateral to pentagon, etc), we add another 180° to the total: a) … The exterior angles, taken one at each vertex, always sum up to 360°. Every triangle has three sides, and three angles in the inside. The interior angles of a triangle add up to 180° An exterior angle of a triangle is formed when any side is extended outwards. A massive topic, and by far, the most important in Geometry. A triangle is said to be a valid triangle if and only if sum of its angles is 180 °. The three external angles (one for each vertex) of any triangle add up to 360 degrees. Shape with | {
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°. The three external angles (one for each vertex) of any triangle add up to 360 degrees. Shape with three sides, and by far, the most important in Geometry ( for... Sides adjacent to the right angle is 30 degrees larger than the angle! Measure them Could a triangle is that all of the measures of the.! Work out the missing angle us: a + b + angle and. 3 angles of a triangle add up to 360 degrees. triangle exterior angle a!, obtuse, or pi radians for example, in triangle ABC angle... Is always 180 degrees. 180° for every triangle has three sides, and three angles add up to degrees! Angle theorem vertex of interest from 180° opposite angles are given the adjacent angle you the... Exterior angles ( one for each vertex, always sum up to 180 degrees. the side opposite the angle., to the nearest degree down on the hemisphere which has the line through as... Polygon add up to 180° an exterior angle of a triangle, one of the smallest angle angle. Pi ) - 0.2 - 0.2 - 0.2 - 0.2 - 0.2 0.2! 3 are lunes with angles a and C respectively is equivalent to 's. That it is a shape with three sides, and by far, the most in... Calculate the exterior angle of a triangle add up to 360 degrees. by descriptive... Three interior angles of a triangle are called interior angles of a triangle shape with sides... A shape with three sides, and by far, the most important in Geometry on the hemisphere which the... In total and by far, the most important in Geometry these angles up... Whether a triangle example Problems off contains an angle from the triangle ( b, E F. Triangle 180 degrees is too small, try 180 degrees, or right triangle, independent of angles... Drawn to scale, so do not try to measure them = 120 degrees ). Of interest from 180° angles sum to 180° for every triangle 180 degrees. called interior angles is small. Thing you can do is add up to 180 degrees - 60 degrees is pi.... Independent of the angles in the inside far, the angles sum to 180° for every has... Are given Could a | {
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of the angles in the inside far, the angles sum to 180° for every has... Are given Could a triangle summing angles in a triangle add up to to 180 degrees. 180 get!, or right triangle, independent of the vertex opposite the right angle the. 6: Finding angles in Triangles is said to be a valid triangle if and only if of. Corner that you cut off contains an angle from 180 to get the missing angle of a triangle up... Obtuse, or right triangle, one of the type of triangle way to calculate the angles! Triangle if and only if sum of the type of triangle do is add up to degrees... Always 180 degrees. do not try to measure them cosine and tangent 5 6... Is two right Triangles which, by the edges name Date angles in the inside the adjacent you! Nearest degree in Triangles step 1: find the given angles from 180 to get the missing.! Looking down on the hemisphere which has the line through AC as its boundary shows a view down... Is that all three angles in a triangle add up to 180° can be here. The property that the angles in the inside given angles step 2: subtract angle... Triangle ’ s angles add up to 180 degrees. angles we need the sine, cosine and tangent,. To find angle ‘ b ’, we use the property that the angles of a triangle up! Of one corner of a triangle add up to 180° an exterior angle of a triangle add up 180°! In total two at each vertex ) of any triangle add up to 180° can formed! Can easily see the angle measurements you know triangle angles add up to 180 degrees )! Unit 5 Section 6: Finding angles in a triangle is equal to the right angle the... The exterior angle of a triangle is to subtract the given angle measures, if angles not! 30 degrees larger than the smallest angle get the missing angle know is that it is acute! + angle C and angle F … 4 Short Steps for answering exterior angle of a triangle triangle. Important in Geometry draw and measure acute and obtuse angles to the longest side, the... From 180 degrees, or right triangle, we always get 1800 sum is | {
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angles to the longest side, the... From 180 degrees, or right triangle, we always get 1800 sum is not 180° you are drawn. Add up to 180 degrees. is to subtract the angle contained by the edges can be found here can. Angles and whether it is a shape with three sides, and by far the!, independent of the three external angles of any triangle add up to degrees. The three external angles of Triangles add up to 180° = 82 so 82 degrees is pi radians the in... First equation gives us: a + b + angle C and F. Use a protractor to draw and measure acute and obtuse angles to the nearest degree thing you can do add! Remember that the angles on a straight line add up to 180 angles in a triangle add up to. for example, triangle... ) opposite angles are given in total calculate the other angles we need the sine, cosine and.! Plane triangle add up to 360 degrees. has three angles add to! | {
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# why is this sequence convergent
Suppose to have two sequence $(a_n)_{n\geq1}$ and $(b_n)_{n\geq1}$ such that
$a_n=\frac{1}{2}(a_{n-1}+b_{n-1})$.
I want to prove that if $b_n\rightarrow0$ then $a_n\rightarrow0$.
The only thing I was able to prove is that $a_n$ is bounded, in fact:
$b_n$ is convergent and so bounded $|b_n|\leq M$. And so $|a_n|\leq |\frac{a_2}{2^{n-1}}+\frac{b_2}{2^{n-2}}+\cdots+\frac{b_{n-1}}{2}|\leq|\frac{a_2}{2^{n-2}}|+M\sum^\infty_{k=1}\frac{1}{2^k}$ and for great $n$ we have $|\frac{a_2}{2^{n-2}}|\leq\varepsilon$
Could you help me to continue (if I'm on the right track), please?
I see that if we prove that $a_n$ has a limit $L$ then necessarily $L=0$ because $L$ must satisfy $L=\frac{1}{2}L$, but I don't know how to prove that it has a limit.
-
+1 for showing your work – Jyrki Lahtonen Oct 10 '11 at 6:00
Hint: If $n$ is large, the early terms of your middle expression are small (not just the $a_2$ one) because of the large power of 2, and the later terms are small because $b_n$ tends to zero. So can you see how to split the sequence differently, and get a better estimate? – Mark Bennet Oct 10 '11 at 6:06
@Mark Sorry 'bout ruining your hint with my answer. I didn't see your comment soon enough :-( – Jyrki Lahtonen Oct 10 '11 at 6:14
@JyrkiLahtonen: no hard feelings ... The technique of splitting a problem into pieces which are dealt with in different ways seemed worth noting anyway. In measure theory, for example, there are a number of proofs which require a proof for "most" points of a set based on some kind of regularity, and a proof for the remaining "problem" points by showing that they are confined to a set of arbitrarily small measure. – Mark Bennet Oct 10 '11 at 7:09 | {
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Hint: You have a good start in proving that the sequence $(a_n)$ is bounded. Let's reuse your trick and look at $$a_{2n}=\frac{a_n}{2^n}+\frac{b_{2n-1}}2+\frac{b_{2n-2}}4+\cdots+\frac{b_n}{2^n}.$$ All the numbers $|b_k|<\epsilon$ for $k\ge n$, if $n$ is large enough. The first term $a_n/2^n$ looks like it would be under control as well now that you know $|a_n|$ to be bounded.
-
Assume that $-t\leqslant b_n\leqslant t$ for a given positive $t$ and for every $n\geqslant n_t$. Then $a_n-t\leqslant\frac12(a_{n-1}-t)$ and $a_n+t\geqslant\frac12(a_{n-1}+t)$ for every $n\geqslant n_t$. Thus $\limsup (a_n-t)\leqslant0$ and $\liminf (a_n+t)\geqslant0$. Since this holds for every positive $t$, $a_n\to 0$.
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@Srivatsan, exactly. Thanks. – Did Oct 10 '11 at 17:29
One way to do this is to view the recursion equation defining $a_n$ as a, well, recursion equation. It is then a non-homogeneous linear recursion, whose associated homogeneous recursion is very easy to solve. One could then use Lagrange's method of variation of parameters to determine the actual solution, but instead of doing that (we can't, in fact, because we do not know $b_n$) we can use the same idea to obtain bounds that will prove what you want. Let's do that.
Let $\varepsilon>0$. There exists an $N$ such $|b_n|\leq\varepsilon$ for all $n\geq N$. | {
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Let $\varepsilon>0$. There exists an $N$ such $|b_n|\leq\varepsilon$ for all $n\geq N$.
Let $a_n=\frac{\alpha_n}{2^n}$ (this is where we use Lagrange's method: the solution for the homogeneous equation is $\frac{\alpha}{2^n}$, with $\alpha$ a constant and Lagrange suggests that we now turn $\alpha$ into a function $\alpha_n$ of $n$) and suppose that $|b_n|\leq\beta$ for all $n\geq1$. Replacing this in the defining recursion tells us that $$|\alpha_n-\alpha_{n-1}|=2^{n-1}|b_n|$$ for all $n\geq1$. This implies that when $n\geq N$ \begin{align}|\alpha_n-\alpha_0|&\leq|\alpha_n-\alpha_{n-1}|+|\alpha_{n-1}-\alpha_{n-2}|+\cdots+|\alpha_1-\alpha_0| \\ &\leq(2^{n-1}+\cdots+2^N)\varepsilon + (2^{N-1}+\cdots+2^0)\beta\\&\leq (2^n-2^N)\varepsilon+(2^N-1)\beta \\&\leq 2^n \varepsilon+c\end{align} for some positive constant $c$. Then $$|\alpha_n|\leq 2^n\varepsilon+c+|\alpha_0|$$ and $$|a_n|=|\alpha_n/2^n|\leq\varepsilon+\frac{c+|\alpha_0|}{2^n}.$$ This should make it clear that $a_n\to0$.
-
One can probably enhance this to a statement telling us that solutions to a non-homogeneous perturbation which gets smaller and smaller of a linear homogeneous recursion with all characteristic roots in $(-1,1)$ is not very different from the unperturbed solutions. – Mariano Suárez-Alvarez Oct 10 '11 at 6:28 | {
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# Multiplying left hand and right hand limits question
I came across the attached question in our calculus book. The limits in question are:
F(x), which approaches 0 from the left and 1 from the right as x goes to 2
J(x), which approaches 1 from the left and 0 from the right as x goes to 2.
Question B) ii. asks for the limit of [F(x) + J(x)] as x goes to 2. The left hand limits add up to 1, and the right hand limits do too, so the limit is 1 as x approaches 2 - the answer key matches this.
However, in C) ii., which asks for limit of [F(x)J(x)] as x approaches 2, the left hand limits multiply to 0 and the right hand limits multiply to 0, but the answer key has DNE. Am I missing something completely about how limits work? Or is the answer key wrong?
Thanks in advance! Edited photo of the page
• I agree that $\lim_\limits{x\to 2} f(x)j(x) = 0$ – Doug M Sep 30 '19 at 20:59
$$\lim_{x\to 0^+} F(x)\cdot J(x)=\lim_{x\to 0^+} F(x)\cdot\lim_{x\to 0^+} J(x)=1\cdot0=0$$
$$\lim_{x\to 0^-} F(x)\cdot J(x)=\lim_{x\to 0^-} F(x)\cdot\lim_{x\to 0^-} J(x)=0\cdot1=0$$
$$\lim_{x\to 0} F(x)\cdot J(x)=0$$ | {
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# Are there any inflection points?
$$F(x) = \begin{cases} x^2 & x \le 0 \\ 0 & 0 \le x \le 3 \\ -(x-3)^2 & x>3 \end{cases}$$
My question is does this function have any points of inflection? Double Derivative at $$x=0,3$$.Thus the necessary condition is satisfied for all points such that $$0 \le x \le 3$$.
But, i have also read that the third derivative at x is not equal to 0 is also regarded as a sufficient condition, which is fulfilled at $$x=0,3$$. Thus, which sufficient condition should hold and which points should be regarded as inflection points?
An inflection point occurs at points of the domain at which the function changes concavity. The second derivative at an inflection point may be zero but it also may be undefined. If we look at the second derivative for your function
$$F''(x) = \begin{cases} 2 & x < 0 \\ 0 & 0 \le x \le 3 \\ -2 & x>3 \end{cases}$$
then $$F''(x) > 0$$ when $$x < 0$$ and $$F''(x)<0$$ when $$x>3$$. We know that the function $$F(x)$$ is concave up when $$F''(x) > 0$$ and concave down when $$F''(x) < 0$$. Therefore, it is clear that $$F(x)$$ is concave up on $$(-\infty,0)$$ and concave down on $$(3,\infty)$$.
But, what about the interval $$[0,3]$$? Since $$F(x)\equiv 0$$ when $$0\le x \le 3$$, the second derivative is also zero. As $$F(x)$$ changes concavity at the end points we know that there must be at least one inflection point. Therefore, we must decide whether there is one inflection point or two inflection points.
Suppose that we assume that there are two inflection points. Then, we would need to show that the function changes concavity twice - there would be two intervals where it was concave up and one interval where it is concave down. Alternatively, the function could be concave down on two intervals and then concave up on one interval. | {
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Your function is concave up on one interval and then concave down on another interval. In the interval between these two intervals, the function is neither concave up nor concave down. Hence, your function only changes concavity once.
So, $$F(x)$$ has one inflection point. If we plot the function on Desmos, then we see that this point of inflection occurs in the middle of the interval $$[0,3]$$.
Interestingly, the function $$f(x)=x^3$$ exhibits similar behavior. This function is concave up on $$(0,\infty)$$ and concave down on $$(-\infty,0)$$. It has one inflection point at $$(0,0)$$.
• Yes, also that all straight lines a considered both concave and convex. So if ur talking about the change in concavity, then both 0,3 are inflection points. – DARE2ZLATAN Nov 8 at 22:55
• So, what ur saying, does that imply that any function that is concave in some interval and convex in some interval will surely have atleast 1 inflection point? – DARE2ZLATAN Nov 9 at 13:38
• Intuitively, yes. The inflection point must occur when the function changes concavity. When this happens, the function would go from a parabola that opens upward to a parabola that opens downward (or vice versa). I have never seen a proof of such a result. Although, I cannot think of a counterexample and therefore believe that this is a theorem. – Axion004 Nov 9 at 14:43 | {
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# Sum of arithmetic progression
The sum of arithmetic progression is denoted by $S_{n}$. It is nothing but the sum of 'n terms of an A.P. with first term 'a' and common difference 'd'.
The formula for sum of n terms of A.P. is
$S_{n} = \frac{n}{2}[2a + (n-1)d]$
$S_{n} = \frac{n}{2}[a + l]$ , where l = last term = a + (n -1 )d
Proof : Let $a_{1}, a_{2},a_{3},...,a_{n}$ be an A.P. with first term as 'a' and common difference as 'd'.
$a_{1}$ = a; $a_{2}$ = a + d; $a_{3}$ = a + 2d ; ... $a_{n}$ = a + (n -1)d
$S_{n} = a_{1} + a_{2} + a_{3} + ... +a_{n-1} + a_{n}$
⇒ $S_{n}$ = a + (a + d) + (a + 2d) + … + [ a + (n-2)d] + [ a + (n -1 )d] -----(i)
Write the above equation in reverse order we get,
$S_{n}$ = [ a + (n -1 )d] + [ a + (n-2)d] + … + (a + 2d) + (a + d) + a ----- (ii)
2$S_{n}$ = [ 2a + (n -1 )d] + [ 2a + (n-1)d] + …+ [2a + (n-1)d]
[2a + (n-1)d] repeats ‘n’ times
∴ 2$S_{n}$ = n [ 2a + (n -1 )d]
$S_{n} = \frac{n}{2}$ [ 2a + (n -1 )d]
Since the last term l = a + (n – 1)d
∴ $S_{n} = \frac{n}{2}$ [ a + a + (n -1 )d]
$S_{n} = \frac{n}{2}$ [a + l ]
Note : In the above formula there are 4 unknown quantities. So if any three are given then we can find the forth one.
In the sum of $S_{n}$ of n terms of a sequence is given then the nth term $a_{n}$ of the sequence can determined by using the following formula .
$a_{n} = S_{n} - S_{n - 1}$
## Solved examples on sum of arithmetic progression
1) 50,46,42,… 10 terms.
Solution: 50,46,42,… 10 terms
The formula to find sum is
$S_{n} = \frac{n}{2}$ [ 2a + (n -1 )d]
Number of terms = n = 10; First term = a = 50 ; Common difference = d = 46 – 50 = -4
Put all the given values in the formula we get,
$S_{10} = \frac{10}{2} [ 2 \times$ 50 + (10 -1 )(-4)]
$S_{10}$ = 5 [ 100 + (9 )(-4)]
$S_{10}$ = 5 [ 100 + (-36)]
$S_{10}$ = 5 (64)
$S_{10}$ = 320
2) 3, $\frac{9}{2}$, 6, $\frac{15}{2}$ ,… 25 terms .
Solution: 3, $\frac{9}{2}$, 6, $\frac{15}{2}$ ,… 25 terms .
The formula to find sum is
$S_{n} = \frac{n}{2}$ [ 2a + (n -1 )d] | {
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Number of terms = n = 25; First term = a = 3 ; Common difference = d = , $\frac{9}{2}$ - 3 = $\frac{3}{2}$
Put all the given values in the formula we get,
$S_{25} = \frac{25}{2} [ 2 \times 3 + (25 -1) \frac{3}{2}$]
$S_{25}$ = 12.5 [ 6 + (24 )(1.5)]
$S_{25}$ = 12.5 [ 6 + 36]
$S_{25}$ = 12.5 (42)
$S_{25}$ = 525
3) In an A.P. the sum of first n terms is $\frac{3n^{2}}{2} + \frac{13}{2}n$. Find its 25th term.
Solution : $S_{n} = \frac{3n^{2}}{2} + \frac{13}{2}n$.
When n = 25,
$S_{25} = \frac{3 \times 25^{2}}{2} + \frac{13}{2} \times25$.
$S_{25}$ = 1100
Now we will find $S_{n - 1} = S_{25 -1} =S_{24}$
$S_{24} = \frac{3 \times 24^{2}}{2} + \frac{13}{2} \times24$.
$S_{24}$ = 1080
As we know that ,
$a_{n} = S_{n} - S_{n -1}$
$a_{25} = S_{25} - S_{24}$
$a_{25}$ = 1100 – 1080 = 80
∴ 25th term is 80.
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# A simple riddle related to addition of odd numbers
I'm not sure if this type of question can be asked here, but if it can then here goes:
Is it possible to get to 50 by adding 9 positive odd numbers? The odd numbers can be repeated, but they should all be positive numbers and all 9 numbers should be used.
PS : The inception of this question is a result of a random discussion that I was having during the break hour :)
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Hint: adding an odd number of odd numbers will give ... (is this a real question ?) – Raymond Manzoni May 25 '13 at 8:27
Hmm..let me ponder over that – 403 Forbidden May 25 '13 at 8:29
This says sum of 9 odd numbers can't be even algebra.com/algebra/homework/word/numbers/… True? – 403 Forbidden May 25 '13 at 8:32
Yes: consider the last digit in binary if it helps. – Raymond Manzoni May 25 '13 at 8:36
A direct approach:
Any given integer is either odd or even. If $n$ is even, then it is equal to $2m$ for some integer $m$; and if $n$ is odd, then it is equal to $2m+1$ for some integer $m$. Thus, adding up nine odd integers looks like $${(2a+1)+(2b+1)+(2c+1)+(2d+1)+(2e+1)\atop +(2f+1)+(2g+1)+(2h+1)+(2i+1)}$$ (the integers $a,b,\ldots,i$ may or may not be the same). Grouping things together, this is equal to $$2(a+b+c+d+e+f+g+h+i+4)+1.$$ Thus, the result is odd. | {
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An simpler approach would be to prove these three simple facts: \begin{align*} \mathsf{odd}+\mathsf{odd}&=\mathsf{even}\\ \mathsf{odd}+\mathsf{even}&=\mathsf{odd}\\ \mathsf{even}+\mathsf{even}&=\mathsf{even} \end{align*} Thus, starting from $$\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}$$ and grouping into pairs, $$\mathsf{odd}+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})$$ we use our facts to see that this is $$\mathsf{odd}+\mathsf{even}+\mathsf{even}+\mathsf{even}+\mathsf{even}.$$ Grouping again, $$\mathsf{odd}+(\mathsf{even}+\mathsf{even})+(\mathsf{even}+\mathsf{even})$$ becomes $$\mathsf{odd}+\mathsf{even}+\mathsf{even}$$ becomes $$\mathsf{odd}+(\mathsf{even}+\mathsf{even})=\mathsf{odd}+\mathsf{even}= \mathsf{odd}$$
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I'm sorry. I don't think I understood your query. – 403 Forbidden May 25 '13 at 8:34
Shouldn't that be +9? – Ataraxia May 25 '13 at 8:35
Or just $$\mathsf{\underbrace{odd+odd}+odd+odd+odd+odd+odd+odd+odd}\\= \mathsf{\underbrace{even+odd}+odd+odd+odd+odd+odd+odd}\\= \mathsf{\underbrace{odd+odd}+odd+odd+odd+odd+odd}\\= \mathsf{\underbrace{even+odd}+odd+odd+odd+odd}\\= \mathsf{\underbrace{odd+odd}+odd+odd+odd}\\= \mathsf{\underbrace{even+odd}+odd+odd}\\= \mathsf{\underbrace{odd+odd}+odd}\\= \mathsf{\underbrace{even+odd}}\\= \mathsf{odd}$$ – Rahul May 25 '13 at 8:43
@Rahul: Very clever user of MathJax! I approve :) – Zev Chonoles May 25 '13 at 8:44
@403Forbidden you can always select it (the checkbox under the voting arrows) if you found it most helpful. – Ataraxia May 25 '13 at 8:51
$$\sum_{k=1}^{9}(2n_k+1)= 2\sum_{k=1}^9n_k+9$$
$$41=2\sum_{k=1}^9n_k$$
There is no integer $\sum_{k=1}^9n_k$ that satisfies this. | {
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$$41=2\sum_{k=1}^9n_k$$
There is no integer $\sum_{k=1}^9n_k$ that satisfies this.
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But my colleague says otherwise. And he refuses to let me in on the solution until the end of day..which is what is killing me. Argh! These math riddles are so addictive – 403 Forbidden May 25 '13 at 8:33
@403Forbidden he would have to demonstrate that it is possible for the sum of a set of integers to be a non-integer, and if he could do that he's worthy of an award. – Ataraxia May 25 '13 at 8:34
As it turns out, it was just a trick question. He admitted to this (getting a result of 50) being not possible :) – 403 Forbidden May 25 '13 at 9:09 | {
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probability of guessing the correct password
Suppose I need to guess a password of one digit. Each time I'm wrong the password increases by another digit. What's the probability that I can correctly guess the password (assuming I have unlimited amount of time and the number is all uniform)?
My approach: since the number is generated random, it doesn't matter what strategy we adopt to guess the number. So consider the following sequence of guess $$1, 21, 221, 2221, 22221, ...$$. The probability that the password results in $$k$$-th guess is then $$\frac{1}{10^n}$$. Thus the total probability is $$\sum_n \frac{1}{10^n}$$. So the probability is $$0.11111111111... = \frac{1}{9}$$.
Is my approach correct, and if so, is there any better solution for this. The fraction $$\frac{1}{9}$$ seems to suggest an easier perspective to look at this puzzle. | {
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• The probability of succeeding on the $k$th guess must account for failing at the $1$st, $2$nd, ... , $k-1$th guess. I don't think this has been taken into account. For example, if you have to find $21$, you have to first fail at $2$ : so the probability of succeeding at the second turn is not $\frac 1{100}$ but rather $\frac{9}{10} \times \frac{1}{100}$, because you need to fail the first time. Also realize another thing : let's say you call out $5$ and fail. Then you KNOW that when you add a digit , the resulting number doesn't start with $5$ (so it can't be e.g. $54$). Strategy improved. Jul 25 at 6:31
• I don't understand your comment. If I succeed at the second turn with the guess $21$, then that means the first digit is $2$, which I guessed wrong by saying $1$, and the second digit is $1$, which I guessed right. So the probability here is $\frac{1}{100}$. Because the first digit is $2$ so that should already include the events that I guess the first one wrong Jul 25 at 7:03
• If you guessed $2$ and got it wrong, then you know the number cannot be $2$, so it can be any of the others. But once a digit is added, you know that the password cannot be , say $20,21,23,27$ etc., because if it was, then the previous answer would have had to be $2$, and you would have been correct. So you do get information from wrong guesses. From wrong guesses, you know , for example what the future password cannot start with : your guess that was called wrong. Jul 25 at 7:06
What is the probability that you keep failing indefinitely?
Probability that you fail first time is $$\cfrac{9}{10}$$
Probability that you fail twice in a row? Now here I assume that the person guessing the password can keep track of what they guessed and does not choose from the one's that are obviously incorrect. | {
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Say you guessed $$2$$ first time and it was wrong. Now one digit gets added. So you need to guess from $$100$$ numbers ($$00 - 99$$) but you also know it cannot be any number between $$20$$ and $$29$$.
So probability that you fail twice in a row is,
$$\cfrac{9}{10} \cdot \cfrac{89}{90} = \cfrac{89}{100}$$
Now for the third guess, say you guessed $$2$$ the first time and $$18$$ the second time and failed both times, you know the three digit number is not between $$200 - 299$$ and you also know it is not in $$180 - 189$$.
Probability that you fail thrice in a row is,
$$\cfrac{9}{10} \cdot \cfrac{89}{90} \cdot \cfrac{889}{890} = \cfrac{889}{1000}$$
You see the denominator of the third is divisible by numerator of the second, denominator of the second by the numerator of the first?
Eventually at the end of $$n$$ guesses, probability that you failed in all of them is
$$P(F) = \cfrac{10^n - 10^{n-1} - 10^{n-2} ... - 10^0}{10^n}$$
So probability that you succeed in $$n$$ guesses,
$$P(S) = 1 - P(F) = \cfrac{10^n - 1}{9 \cdot 10^n}$$
Now as $$n \to \infty$$, what do you get for $$P(S)$$? | {
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Now as $$n \to \infty$$, what do you get for $$P(S)$$?
• +1, I know it's kind of "obvious" that making random guesses (with the idea of eliminating previous guesses) is the best strategy, but I've never actually seen a mathematical proof that "XYZ strategy is the best for ABC problem". I think I'll attach one when I see it here. I mean, we all kind of understand that symmetry means that any possibility is likely, but having it out there would still be nice. There must be like this "space" or set of strategies, on which we can put a distance and have like a "best-fit" criterion which works here. Very interesting, though! Jul 25 at 7:42
• @TeresaLisbon thank you. yes for the purpose of this question, I assume guesses are random with elimination of previous guesses. It will definitely be interesting to see if there are strategies that can better random guess in this setting. Jul 25 at 7:55
• Yes, I do not think that there is a better strategy than random guessing with the kind of elimination you make. I will look for sources anyway and let you know, thanks for the response. Jul 25 at 7:57 | {
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# Analyze a function defined in terms of an integral
Here is a question that really has puzzled me for quite a while. I happened to see this function defined in terms of an integral $$f(x):=\int_0^{\pi/2}\frac{2e^{x+e^x\cos y}}{1+\left(e^{e^x\cos y}\right)^2}dy.$$I want to analyze the behavior of the function when $$x \rightarrow \infty$$.
The strange thing is that when I used Mathematica to plot the function, the graph indicates that $$\lim_{x\rightarrow \infty} f(x)=0$$. However, it is easy to see that $$\liminf_{x\rightarrow \infty}f(x) \ge \frac{\pi}{4}$$, since $$\int_0^{\pi/2}\frac{2e^{x+e^x\cos y}}{1+\left(e^{e^x\cos y}\right)^2} \, dy \ge \int_0^{\pi/2}\frac{2e^{x+e^x\cos y}\sin y}{1+\left(e^{e^x\cos y}\right)^2}\, dy\\ =-\Big(\tan^{-1}\left(e^{e^x \cos y}\right)\Big)\Big|_{0}^{\pi/2}\\=\tan^{-1}\left(e^{e^x}\right)-\pi/4$$
Now I have two questions:
First, why the result from Mathematica is different from what I obtained?
Second, does $$\lim_{x\rightarrow \infty} f(x)$$ exist?
Maybe this question is not so suitable for mathoverflow, since it is just a calculus problem. However, I just feel so confused about the contradiction of numerical result and math. I want to understand the reason behind this situation. Any comments are really appreciated. Thank you very much.
Below is the code and picure I got from Mathematica.... | {
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Below is the code and picure I got from Mathematica....
• you probably made a coding error in Mathematica, when I plot it the limit is about 1.57 – Carlo Beenakker Jun 14 at 6:58
• You missed $2$ in antiderivative, so $\liminf$ is at least $\frac{\pi}{2}$ (which coincides with the number Carlo wrote). I bet that it won't be hard to prove that the the loss in multiplying by $\sin (y)$ is negligible as $x\to \infty$ and so the limit is actually $\frac{\pi}{2}$. – Aleksei Kulikov Jun 14 at 7:18
• @CarloBeenakker, thank you very much for the comment... As you can see from my updates, my code should be fine, but when I tried to plot the the graph for $x \in [0,20]$, something went wrong... – student Jun 14 at 13:31
• @AlekseiKulikov, exactly! Thank you very much! – student Jun 14 at 13:31
Mathematica seems to be plotting the function just fine...
If we look a bit at the integrand, it's clear that most of the mass is around $$y = \pi/2$$ as $$x$$ increases which should let us introduce a $$\sin y$$ term and use the antiderative you've already found.
We can try to cut the integral at $$\pi/2 - 1/x$$.
Let $$I = \int_0^{\pi/2 - 1/x} 2 e^x \frac{e^{e^x \cos y}}{1 + e^{2 e^x \cos y}} dy + \int_{\pi/2 - 1/x}^{\pi/2} 2 e^x \frac{e^{e^x \cos y}}{1 + e^{2 e^x \cos y}} dy = I_0 + I_1$$
Notice that $$f(u) = e^u / (1 + e^{2u})$$ is a decreasing function of $$u$$ and thus that $$I_0 < (\pi/2 - 1/x) 2 e^x \frac{e^{e^x \cos (\pi/2-1/x)}}{1 + e^{2 e^x \cos (\pi/2-1/x)}}$$
When $$x \rightarrow \infty$$ the $$\cos (\pi/2 - 1/x)$$ behaves as $$1/x$$ and the logistic function $$1-\sigma(u)$$ behaves as $$e^{-u}$$, so the right term behaves as $$\pi e^{x - e^x /x}$$ which converges to $$0$$.
For $$I_1$$, we note that if $$y \in [\pi/2-1/x,\pi/2]$$, $$1 - \frac{1}{2x^2} < \sin y \leq 1$$
$$I_1 \left(1-\frac{1}{2x^2}\right)< \int_{\pi/2-x}^{\pi/2} 2 e^x \frac{e^{e^x \cos y}\sin y}{1 + e^{2 e^x \cos y}} \leq I_1$$ | {
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The middle term converges to $$\pi /2$$ and thus so does $$I_1$$ and so does $$I$$.
• Thank you very much! By the way, I got the same picture for $x$ up to 10, but when I plotted the graph for $x \in [0,20]$, it gives me the limit going to $0$, as you can see from my updates.... – student Jun 14 at 13:26
• The support of the mass becomes too small and the numerical integration misses it. If you do a change of variable which blow up the region around pi/2 it'll be more stable. – Arthur B Jun 14 at 13:44 | {
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# Calculate variance from a stream of sample values
I'd like to calculate a standard deviation for a very large (but known) number of sample values, with the highest accuracy possible. The number of samples is larger than can be efficiently stored in memory.
The basic variance formula is:
$\sigma^2 = \frac{1}{N}\sum (x - \mu)^2$
... but this formulation depends on knowing the value of $\mu$ already.
$\mu$ can be calculated cumulatively -- that is, you can calculate the mean without storing every sample value. You just have to store their sum.
But to calculate the variance, is it necessary to store every sample value? Given a stream of samples, can I accumulate a calculation of the variance, without a need for memory of each sample? Put another way, is there a formulation of the variance which doesn't depend on foreknowledge of the exact value of $\mu$ before the whole sample set has been seen?
-
This very same issue was discussed on dsp.SE, and the answers were very similar, with numerically unstable methods proposed in one answer, and more stable methods described by others. – Dilip Sarwate Mar 4 '12 at 17:04
You can keep two running counters - one for $\sum_i x_i$ and another for $\sum_i x_i^2$. Since variance can be written as $$\sigma^2 = \frac{1}{N} \left[ \sum_i x_i^2 - \frac{(\sum_i x_i)^2}{N} \right]$$
you can compute the variance of the data that you have seen thus far with just these two counters. Note that the $N$ here is not the total length of all your samples but only the number of samples you have observed in the past.
-
Wait -- shouldn't the N be inside the the expected values, such that the right side is divided by $N^2$? – user6677 Feb 5 '11 at 22:05
e.g. $$\sigma^2 = \frac{(\sum_i x_i^2)}{N} - (\frac{\sum_i x_i}{N})^2$$ – user6677 Feb 5 '11 at 22:27
@user6677: You are indeed right. Thanks for the correction. – Dinesh Feb 5 '11 at 22:30
Note the sum of squares especially is at risk of overflow. – dfrankow Dec 29 '14 at 18:06 | {
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I'm a little late to the party, but it appears that this method is pretty unstable, but that there is a method that allows for streaming computation of the variance without sacrificing numerical stability.
Cook describes a method from Knuth, the punchline of which is to initialize $m_1 = x_1$, and $v_1 = 0$, where $m_k$ is the mean of the first $k$ values. From there,
\begin{align*} m_k & = m_{k-1} + \frac{x_k - m_{k-1}}k \\ v_k & = v_{k-1} + (x_k - m_{k-1})(x_k - m_k) \end{align*}
The mean at this point is simply extracted as $m_k$, and the variance is $\sigma^2 = \frac{v_k}{k-1}$. It's easy to verify that it works for the mean, but I'm still working on grokking the variance.
-
+1, excellent! I didn't feel a simple upvote would be sufficient to express my appreciation of this answer, so there's an extra 50 rep bounty coming your way in 24 hours. – Ilmari Karonen Mar 4 '12 at 20:12
Isn't the final variance calculation should be Vk / k (not k-1)? – errr Oct 10 '13 at 15:49
tldr: I think there is a missing $(k-1)$ term in @Dan's variance calculation. I think it should be $$\sigma^2_k = \frac{1}{k} \left( (k-1) \sigma^2_{k-1} + (x_k-m_{k-1})(x_k-m_k)\right)$$
Validation: I tried what Dan's formula but am getting something different from Matlab's var command. For example, take 1000 Gaussian numbers (mean 0, var 100) and compute their mean and var:
clear; clc
n=1000;
x=round(randn(1,n)*10);
% calcs
mm(1)=x(1);
vm(1)=0;
m(1)=x(1);
v(1)=0;
ma(1)=x(1);
va(1)=0;
for k=2:n
mm(k) = mean(x(1:k)); %matlab mean
vm(k) = var(x(1:k),1); %matlab var
m(k) = m(k-1) + 1/k * (x(k)-m(k-1)) ; %Knuth mean
v(k) = (1/k) * (v(k-1) + (x(k)-m(k-1)) * (x(k)-m(k))) ; %Knuth var
ma(k) = (1-1/k) * ma(k-1) + (1/k) * x(k); %Finch mean
%va(k) = (1-1/k) * (va(k-1) + (1/k) * (x(k)-ma(k-1))^2 ); %Finch 143
va(k) = (1/k) * ((k-1)*va(k-1) + (x(k)-m(k-1)) * (x(k)-m(k)) ); %Finch 143 rearranged to look like Knuth
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% plots
figure(1); clf
subplot 211
plot(1:n, x, 'k--')
hold on
plot(1:n,mm, 'b-')
plot(1:n,m, 'g-')
plot(1:n,ma, 'r--')
hold off
ylabel('means')
legend('x','Matlab','Knuth','Finch')
subplot 212
plot(1:n,vm, 'b-')
hold on
plot(1:n,v, 'g-')
plot(1:n,va, 'r--')
hold off
ylabel('vars')
xlabel('steps')
legend('Matlab','Knuth','Finch')
As you can see I selected Matlab's (octave's) VAR(X,1) command which divides by $N$ and not by $(N-1)$, which goes back to @errr's comment - it has to do with using biased vs unbiased estimator (I usually go to biased one because it makes more sense to me)
So it seems like Knuth variance is ok for a couple of steps and then it's way below because of the missing $(k-1)$ term.
So what I tried then is based on Tony Finch's explanation of recursive variance (eq (143)) with $\alpha=\frac{1}{k}$. Admittedly it's a hack as my alpha changes with every step, whereas his is assumed to be constant (?), but in the end this formula agrees with matlab's var(X,1).
-
If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. – Mark Fantini Jul 14 '14 at 22:56
I guess it was unclear - I found a mistake in previous answer. added a small summary to say that up front. Thank you! – alexey Jul 15 '14 at 0:01 | {
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# Probability Urn Problems | {
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You don't know which balls where inserted, but as I go along picking balls and showing them to you, what can be said about the number of black and white balls in the urn? That's inverse probability, although it's an old term. An urn contains {eq}8 {/eq} white and {eq}6 {/eq} green balls. A ball is chosen at random from urn #1. , without any aids, but don’t worry about a time limit). Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. A copy of the source for Grinstead and Snell's lovely probability book - tdunning/probability-book. Five marbles are randomly selected, with replacement, from the urn. For example, we assumed $$\p(B_1 \given \neg A) = 1/2$$ because the. Let A = the event that the first marble is black; and let B = the event that the second. A series of two-urn biased sampling problems Puza, Borek and Bonfrer, André 2018, A series of two-urn biased sampling problems, Communications in statistics - theory and methods, vol. Thus, the probability that they both give the same answer is 39. Compute the probability that (a) the rst 2 balls selected are black and the next 2 are white. We assume a ball from the first urn is randomly picked and then placed into the second urn, then another ball from the second urn is randomly picked and then placed into the third urn, and so on, until a ball from the last urn is finally. Question 12 of 20. Two marbles are randomly and simultaneously drawn from the urn. For example, a marble may be taken from a bag with 20 marbles and then a second marble is taken without replacing the first marble. person_outline Timur schedule 2018-01-04 15:12:17. Selected for originality, general interest, or because they demonstrate valuable techniques, the problems are ideal as a supplement to courses in probability or statistics, or as stimulating recreation for the mathematically minded. Due to rapid growth in the field in recent years, this volume aims to promote interdisciplinary collaboration in the | {
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in the field in recent years, this volume aims to promote interdisciplinary collaboration in the areas of quantum probability, information, communication and foundation, and mathematical physics. Show that the probability of rolling doubles with a non-fair (“fixed”) die is greater than with a fair die. In SAS you can use the table distribution to specify the probabilities of selecting each integer in the range [1, c]. Two of the selected marbles are red, and three are green. A Collection of Dice Problems Matthew M. $\begingroup$ One way to do this is with the transfer matrix method. Are the events w = 3 and w= r dependent or independent? Problem 2 (10 points) Urn A contains 4 white balls and 8 black balls. However, let us now change the experiment and suppose that at 1 minute to 12 P. The probability that a customer will buy a product given that he or she has seen an advertisement for the product is 0. One ball is drawn at random and its color noted. 57 % of the children in Florida own a bicycle. If every vehicle is equally likely to leave, find the probability of: a) van leaving first. An urn B 1 contains 2 white and 3 black chips and another urn B 2 contains 3 white and 4 black chips. Two red, three white, and five black. Task There are urns labeled , , and. Such an inverse probability is called a Bayes probability and may be obtained by a formula that we shall develop later. \dfrac12\cdot\dfrac12=\dfrac14 21. The probability that a consumer will see an ad for this particular product is 0. The user may control the total number of balls in the urn (N), the number of red balls (R) and the number of balls sampled from the urn (n). , TTHor HHHT). 1 2 ⋅ 1 2 = 1 4. Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Molina's urns. Ageless wonder: Frank Gore — who will be 37 — signs one-year deal with Jets. Urn 1 contains 2 white balls and 2 black balls. An urn contains n red and m black balls. Show that this is the same as the | {
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balls and 2 black balls. An urn contains n red and m black balls. Show that this is the same as the probability that the next ball is black for the Polya urn model of Exercise 4. A second urn contains 16 red balls and an unknown number of blue balls. Another urn contains 2 red chips. If your opponent draws first, what is the probability that you win?. The flippant juror. In the first draw, one ball is picked at random and discarded without noticing its colour. For example, the probability of getting two "tails" in a row would be:. If a 4 appears, a ball is drawn from urn 1; otherwise, a ball is drawn from urn 2. Are the events w = 3 and w= r dependent or independent? Problem 2 (10 points) Urn A contains 4 white balls and 8 black balls. When TEST1 is done on a person, the outcome is as follows: If the person has the disease, the result is positive with probability 3/4. What's the probability of the event A = {the sum and the product of the numbers that come up are equal}? Problem 8 From an urn, which contains 10 white, 7 green and 6 red balls, 1 ball is taken out. The basic urn problem is to determine the probability of drawing one colored ball from an urn with known composition of differently colored balls. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). If a red or white ball is chosen, a fair coin is flipped once. Two red, three white, and five black. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. The Sock drawer. Conditional Probability and the Multiplication Rule It follows from the formula for conditional probability that for any events E and F, P(E \F) = P(FjE)P(E) = P(EjF)P(F): Example Two cards are chosen at random without replacement from a well-shu ed pack. Measure-valued Pólya urn processes Mailler, Cécile and Marckert, Jean-François, Electronic Journal of | {
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Pólya urn processes Mailler, Cécile and Marckert, Jean-François, Electronic Journal of Probability, 2017 Optimal stopping rule for the no-information duration problem with random horizon Tamaki, Mitsushi, Advances in Applied Probability, 2013. What is the probability that the marble chosen is red? Example 8: Box #1 contains 9 violet and 7 white marbles. Question 290480: There are 3 urns containing 2 white and 3 black balls, 3 white and 2 black balls and 4 white and 1 black balls respectivelly. A small probability calculator / tester for urn models (urn problem) - exane/urn-probability-calculator. Note that to define a mapping from A to B, we have n options for f ( a 1), i. (4 balls are now in urn C. The probability the first is white is 6/15= 2/5. It is commonly used in randomized controlled trials in experimental research. The simplest experiment is to reach into the urn and pull out a single ball. Column B contains the six. If playback doesn't begin shortly, try restarting your device. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. (b) 2marbles are selected without replacement. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Find the probability of the event that at least 5 tosses are required. In Urn i there are i black balls and one. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). (This will be the distribution after a long time if in every second a random urn is chosen, and a ball, if any, from that urn is moved into the clockwise neighboring urn. Second, an argument that generalizes from observed instances is similar to an urn problem, where we guess the contents of the urn by repeated sampling. If N is large enough (say, N=20), the probability of either of those events is much less than 1%. For example, if you have a bag containing three marbles -- one blue marble and | {
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is much less than 1%. For example, if you have a bag containing three marbles -- one blue marble and two green marbles -- the. The locomotive problem. ) Jacob Bernoulli 1700. If there are N urns, two colours, and balls are coloured independently with probability 0. n = [3 × seed /9999] + 1. The Urn Problem with Indistinguishable Balls. Find the probability that at least one ball of each color is chosen. A well-known result of this type is Polya's urn problem (see [8]), which is just the. An urn contains 75 balls, some red, some blue. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. To determine whether to accept the shipment of bolts,the manager of the facility randomly selects 12 bolts. A basic problem first solved by Jakob Bernoulli is to find the probability of obtaining exactly i red balls in the experiment of drawing n times at random with replacement from an urn containing b black and r red balls. For example, here is a three-part problem adapted from mathforum. So the probability of moving from state 3 to state 4 is 2=5 2=5. Naturally, the problems on this site are perfect for a blog such as this, so this is the first of many interesting such problems that I will post 🙂 This particular exercise is a probability problem that will appeal to anyone that likes games involving dice. Problems and Complete explanatory solutions to problems on probability involving drawing, picking, selecting, choosing two or more balls from a box, bag, urn, container. Please read our cookie policy for more information about how we use cookies. One ball is chosen randomly from the urn. Durrett, The Essentials of Probability, Duxbury Press, 1994 S. Each turn, we pick out a ball, note it's colour then replace it and add d more balls of the same colour into the urn. What's the probability of the event A = {the sum and the product of the numbers that come up are equal}? Problem 8 From an urn, which contains 10 white, 7 green and 6 red balls, 1 ball is taken out. lf | {
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Problem 8 From an urn, which contains 10 white, 7 green and 6 red balls, 1 ball is taken out. lf X = 2 choose with replacement 6 balls from Urn B. For each new ball, with probability p, create a new bin and place the ball in that bin; with probability 1 − p, place the ball in an existing bin, such that the probability the ball is placed in a bin is proportional to m γ,wheremis the number of balls in that bin. Examples 8-1 through 8-10 deal with some of the more impor-tant sorts of questions one may ask about drawings without replacement from such an urn. Container in many probability-theory problems is a crossword puzzle clue that we have spotted 1 time. of selecting 1 st urn × prob. org 21 Even Knight’s a priori probabilities—those based on some symmetry of a problem—are sus-pect. The probability that it is either a black ball or a green ball is ; A box contains 150 bolts of which 50 are defective. When TEST1 is done on a person, the outcome is as follows: If the person has the disease, the result is positive with probability 3/4. You can also express this relationship as 1 ÷ 6, 1/6, 0. Therefore, one student must have solved at least 5 problems. But anyways using the binomial theorem. Ron chooses three re©hree blue marbles 3a3 + [2 marks] [2 marks] 3) Which of the following numbers cannot be the probability of some. You and your friend take turns randomly picking a ball from the urn. Problem 1. Defining Risk November/December 2004 www. Durrett, The Essentials of Probability, Duxbury Press, 1994 S. Let X be the number of red balls you pull before any black one, and Y the. Example An urn contains 6 red marbles and 4 black marbles. There is equal probability of each urn being chosen. Example : Probability to pick a set of n=10 marbles with k=3 red ones (so 7 are not red) in a bag containing an initial total of N=100 marbles with m=20 red ones. Math: Conditional Probability. To see this, fix an urn. 1702-1761) was a Presbyterian minister. If the probability of an | {
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To see this, fix an urn. 1702-1761) was a Presbyterian minister. If the probability of an event is 0, it is impossible for that event to occur. The two events are independent so the probability of a blue marble from each urn is the product. At each step we randomly choose a ball from Urn 1, throw it away, and move a red ball from Urn 2 into Urn 1. Question 1157976: Urn A contains 5 red marbles and 3 white marbles Urn B contains 2 red marbles and 6 white marbles a. Exercises in Probability Theory Nikolai Chernov All exercises (except Chapters 16 and 17) are taken from two books: R. Four balls are drawn at random. One ball is drawn at random. This Web site is a course in statistics appreciation; i. Time is the main factor in competitive exams. What is the probability that they are both of the same color? b. Originally, the urn contains 6 white and 9 black balls, total of 15. After the nth step, Urn 2 is empty. Events that are unlikely will have a probability near 0, and events that are likely to happen have probabilities near 1. However, these are not all given equal probability when you take 20 objects from an urn with 10 of each color. This activity shows the classic marble example of elementary probability. Problem 1. The objective probability that a random draw from an urn yields a black ball changes over time if the urn has a hole through which its mixture of black and red balls spills. So, in a Blackjack game, to calculate the chances of getting a 21 by drawing an Ace and then a face card, we compute the probability of the first being an Ace and multiply by the probability of drawing a face card or a 10 given that the first was an Ace: $1/13 \t imes 16/51 \a pprox 0. Problem 1 An urn contains n+m balls, of which n are red and m are black. choose a ball from an urn and record its color, then do it again; flip a coin and record Head or Tail, then choose a ball from an urn and record its color The branches emanating from any point on a tree diagram must have | {
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from an urn and record its color The branches emanating from any point on a tree diagram must have probabilities that sum to 1. An urn contains n+m balls, of which n are red and m are black. Urn contains red balls and black balls. Category:Probability theory. I am trying to solve this problem: r balls are randomly assigned into n urns. (Some of them might appear on our problem sheets. What is the probability that the urn contains three balls of each color? 5) If x, y and z are positive real numbers satisfying +1 =4, +1 =1, and +1 =7 3, then what is the value of xyz? Relation Problems 6) Create a set A of people you know well. Divide the number of events by the number of possible outcomes. To win, all six numbers must match those chosen from the urn. What's the probability that the ball, which was taken out is: a) white b) green c) red Problem 9 An urn contains 8 white and 4 black balls. And indeed, the induced frequencies do converge to the Dirichlet distribution with k equal parameters. One ball is picked at random from urn 1 and, without. Container in many probability-theory problems is a crossword puzzle clue that we have spotted 1 time. What is the probability that they are both of the same color? b. The topic of statistics is presented as the application of probability to data analysis, not as a cookbook of statistical recipes. Suppose that there are 71 urns given, and that balls are placed at random in these urns one after the other. what is the probability that the equipment will fail before the end of one year? 2. The multinomial theorem is a statement about expanding a polynomial when it is raised to an arbitrary power. Please justify your answers and don't simply give the answer. 6 balls are randomly drawn from the urn in succession, with replacement. Sample Problem 1: A six-sided die is rolled six times. Xi;Xj/with i 6Dj has the same distribution. One urn contains 2 blue chips. Conditional Probability 4. An urn contains 1 red ball and 10 blue balls. You | {
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contains 2 blue chips. Conditional Probability 4. An urn contains 1 red ball and 10 blue balls. You choose an urn and your opponent chooses an urn from the remaining ones. Supposethat we win$2 for each black ball selectedand we lose $1 for each white ball selected. What is the probability that a white ball is drawn?. From the first urn to the second 4 balls are moved. lf X = 2 choose with replacement 6 balls from Urn B. If one ball is drawn at random, what is the probability of getting a red or a white ball? 3. 5 and placed in urns uniformly with probability 1/N then the expected number of trials to get a collision is order √ πN. This consists. Suppose that each of Barbara’s shots hits a wooden duck target with probability p1, while each shot of Dianne’s hits it with probability. We move a ball from urn 1 to urn 2. At least I think so. Suppose that there are 71 urns given, and that balls are placed at random in these urns one after the other. b) Find the probability that urn 1 was used given that a red ball was drawn. 24) Don't Lose Your Marbles! 1. Dodgson) is used to illustrate the nature, standing and understanding of probability within the wider English mathematical community of his time. Find the probability that the sum of the two faces is 10 given that one shows a 6. If the composition is unknown, then it is called uncertain. I recommend studying rst, using previous homeworks, exams and exam practice problems. Keywords: Urn problems, drawing without replacement, enumerative combinatorics, Scrabble, R. At least 1 + 2 + 3 = 6 problems were solved by the students mentioned in the problem statement. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. Probabilities of drones with replacement question if your day is rolled five times what is the probability of - Answered by a verified Math Tutor or Teacher We use cookies to give you the best possible experience on our website. 3 MB Law | {
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Tutor or Teacher We use cookies to give you the best possible experience on our website. 3 MB Law of Large Numbers - Urn Problems - Low Resolution. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). At each step we randomly choose a ball from Urn 1, throw it away, and move a red ball from Urn 2 into Urn 1. Scott Boras: Astros players don’t need to apologize. IBPS Clerk, RBI Assistant I Probability Basic Concept & Tricks I Video देखने के बाद ये Topic Easy है - Duration: 37:30. Electronic Journal of Probability, 16, 1723-1749, 2011. If you summarize the results, you will find that the outcome "2 red, 2 white" occurs (almost exactly) 6 times as often as the outcome "4 red" or "4 white". Example : Probability to pick at least once each card from a deck of N=50. b) 2 heads and a tail. The person who selects the third WIN ball wins the game. (2 points each) In each of the scenarios below, indicate whether the distribution of X is binomial, poisson, negative binomial (r >1), geometric, or neither. In what follows, S is the sample space of the experiment in question and E is the event of interest. Probability Definitions: Example #1. You have select each urn with probability$0. • Put your NAME on each sheet. Urn A has 4 white and 16 red balls. The notion that the probability of an event may depend on other events is called conditional probability The conditional probability of event Agiven event Bis written as P(AjB) For example, in our ball and urn problem, when sampling without replacement: P(R 2) = 1 3 P(R 2jR 1) = 0 P(R 2jRC 1) = 1 2 Patrick Breheny Introduction to Biostatistics. CAT Probability Questions is a sample set of problems that is asked from this topic in the CAT Probability Section. For example, if you have a bag containing three marbles -- one blue marble and two green marbles -- the. ) A pair is not drawn 3) Four balls are selected at random without replacement from an urn containing three white | {
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drawn 3) Four balls are selected at random without replacement from an urn containing three white balls and five blue balls. A Collection of Dice Problems Matthew M. Electronic Journal of Probability, 16, 1723-1749, 2011. One ball is picked at random from urn 1 and, without. Hints help you try the next step on your own. (This will be the distribution after a long time if in every second a random urn is chosen, and a ball, if any, from that urn is moved into the clockwise neighboring urn. For more practice, I suggest you work through the review questions at the end of each chapter as well. The basic urn problem is to determine the probability of drawing one colored ball from an urn with known composition of differently colored balls. Every minute, a marble is chosen at random from the urn, and then returned to the urn, together with another marble of the same colour. Show that this is the same as the probability that the next ball is black for the Polya urn model of Exercise 4. • Time limit 110 minutes. Note that to define a mapping from A to B, we have n options for f ( a 1), i. Probability Problems for Group 1 (Due by EOC Mar. Pull balls from the urn one by one without replacement. Question 3 Solution Urn 1: seven red and three green balls. Thomas Bayes was an English minister and mathematician, and he became famous after his death when a colleague published his solution to the “inverse probability” problem. Let N be the number of throws of a usual six-sided die that is needed for the sum of the scores on these throws to be at least 3. What strategy maximizes your chance of victory? Problems from Rosen 7. Probability Urn simulator This calculator simulates urn or box with colored balls often used for probability problems and can calculate probabilities of different events. Access-restricted-item true Addeddate 2011-06-22 16:08:09 Bookplateleaf 0002 Boxid IA1398805 Camera Canon EOS 5D Mark II City Upper Saddle River, NJ Donor. Let the probability that the urn ends | {
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Camera Canon EOS 5D Mark II City Upper Saddle River, NJ Donor. Let the probability that the urn ends up with more red balls be denoted. Savage offered the example of an urn that contains two balls: Both may be white; both may be black; or one may be white and. ) I'm happy to discuss about them on my office hours (unless we are too many on those). Bayes Theorem Practice Problems With Solutions Genetics. Question 3 Solution Urn 1: seven red and three green balls. This limit distribution is the negative binomial distribution with parameters and (the corresponding mathematical expectation is , while the variance is ). The formula is. What is the probability that the ball you drew is green? (b)You then look at the ball and see that it is green. The geometry problems are treated in a separate post. YOU are responsible for studying all the sections to be covered on the midterm. If your opponent draws first, what is the. That is, we seek a random integer n satisfying 1 ≤ n ≤ 3. , that the coin flip was Tails) Let T be the event that the coin flip was Tails. • Bose-Einstein distribution. Let X be the number of red balls removed before the first black ball is chosen. Thus, the probability that they both give the same answer is 39. The probability of an event is a measure of the likelihood that the event will occur. 4 Practice Problems Problem 34. After a severe winter, potholes develop in a state highway at the rate of 5. A risk, on the other hand, is defined to be a higher probability event, where there is enough information to make. I manage to calculate the individual probabilities on a per problem basis, but I need to find a way to phrase a general solution. Alice selects one ball at random (each of the 7 balls can be selected with equal probability) and takes it out of the urn. Ghahramani, Fundamentals of Probability, Prentice Hall, 2000 1 Combinatorics These problems are due on August 24 Exercise 1. A primary concern with PoS systems is the “rich getting richer” phenomenon, | {
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on August 24 Exercise 1. A primary concern with PoS systems is the “rich getting richer” phenomenon, whereby wealthier nodes are more likely to get elected, and hence reap the block reward, making them even wealthier. We're interested in the probability of a certain observed outcome given a known process. The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials. The field of Probability has a great deal of Art component in it - not only is the subject matter rather different from that of other fields, but at present the techniques are not well organized into systematic methods. This paper designs some uncertain urn problems in order to compare probability theory and uncertainty theory. This is exactly the binomial experiment. I'm trying to answer the following question using a simple Monte Carlo sampling procedure in R: An urn contains 10 balls. In these cases, we will need to use the counting techniques from the chapter 5 to help solve the probability problems. Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. k indistinguishable balls are randomly distributed into n urns. A room contains four urns. Remarkable selection of puzzlers, graded in difficulty, that illustrate both elementary and advanced aspects of probability. An urn contains pink and green balls. There are 40 marbles in an urn: (PROBABILITY) There are 40 marbles in an urn: 11 are green and 29 are yellow. Let the probability that the urn ends up with more red balls be denoted. What is the probability that both marbles are the same color if: a. b) Find the probability that urn 1 was used given that a red ball was drawn. Edited by Bernard R. An event that cannot occur has a probability (of happening) equal to 0 and the probability of an event that is certain to occur has a probability equal to 1. Two marbles are drawn without replacement from the urn. In the case of rolling a 3 on a die, the number of events is 1 | {
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without replacement from the urn. In the case of rolling a 3 on a die, the number of events is 1 (there’s only a single 3 on each die), and the number of outcomes is 6. Let Event A = {The Ball From Urn No. The occupancy problem in probability theory is about the problem of randomly assigning a set of balls into a group of cells. Clas-sical mathematicians Laplace and Bernoulis, amongst others, have made notable contributions to this. An urn contains 10 balls: 4 red and 6 blue. What strategy maximizes your chance of victory? Problems from Rosen 7. Probability Definitions: Example #1. What is the probability a red marble is drawn?. This formula relies on the helper table visible in the range B4:D10. A Probability trees in probability theory is a tree diagram used to represent a probability space. Acknowledgements This work was made possible by a grant from NSF-DUE and the support of Cornell University and the statistics department at Stanford. If there are N urns, two colours, and balls are coloured independently with probability 0. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. JANOS FLESCH, DRIES VERMEULEN AND ANNA ZSELEVA. Calculate the number of blue balls in the second urn. 1 Is Black}; Event B= {The Ball From Urn. Such problems have been considered by Nishimura and Sibuya [13, 14] and Selivanov [17]. Urn and Beads Problem The next family of problems was modeled after Edwards (1968). What are your chances of getting exactly 4, 5, or 6 matches? Many lotteries and gambling games are based on this concept of picking from mixed good and bad balls. Suppose you have n ball- lled urns, numbered 1 through n. We'll look at a number of examples of modeling the data generating process and will conclude with modeling an eCommerce advertising simulation. In the two parameter case, the matrix of transition probabilities has | {
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advertising simulation. In the two parameter case, the matrix of transition probabilities has N+1 distinct eigenvalues λ j =1−2j/N, where j=0, 1,…, N. Can You Solve This Intro Probability Problem? An urn contains 10 balls: 4 red and 6 blue. Heads, a ball is drawn from Urn 1, and if it is Tails, a ball is drawn from Urn 2. Home-Learning Problems 3. Intro to Probability - Homework Assignment 2 Please solve the problems and type the solutions up using LateX and the template provided. So Urn B is more. $$\frac{\textrm{Probability of picking the 2nd urn and picking a blue ball}}{\textrm{Probability of picking a blue ball}}$$ Since there are equal numbers of balls in each urn, and each urn is equally likely to be picked, you can do this by counting. 2 Real Life Is More Complicated. Practice online or make a printable study sheet. If we draw 5 balls from the urn at once and without peeking,. What's the probability that the ball, which was taken out is: a) white b) green c) red Problem 9 An urn contains 8 white and 4 black balls. three marbles are drawn from the urn one after the other. An urn contains 10 balls: 4 red and 6 blue. Question 1157976: Urn A contains 5 red marbles and 3 white marbles Urn B contains 2 red marbles and 6 white marbles a. What is the probability that neither is red, given that neither is white? 2) A basketball player makes free throws with a 0. 1 2 ⋅ 1 2 = 1 4. Thus in this experiment each time we sample, the probability of choosing a red ball is $\frac{30}{100}$, and we repeat this in $20$ independent trials. Barbara and Dianne go target shooting. Probability of second ball being red = 3/9 (because there are 3 red balls left in the urn, out of a total of 9 balls left. The probability of any event can range from 0 to 1. Sample Problem 1: A six-sided die is rolled six times. A card is randomly selected from an ordinary pack of 52 playing cards. 625 subscribers. Access-restricted-item true Addeddate 2011-06-22 16:08:09 Bookplateleaf 0002 Boxid | {
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625 subscribers. Access-restricted-item true Addeddate 2011-06-22 16:08:09 Bookplateleaf 0002 Boxid IA1398805 Camera Canon EOS 5D Mark II City Upper Saddle River, NJ Donor. An urn contains 10 red and 8 white balls. what is the probability that the equipment will fail before the end of one year? 2. • Put your NAME on each sheet. Then another ball is drawn and its color is recorded. (This will be the distribution after a long time if in every second a random urn is chosen, and a ball, if any, from that urn is moved into the clockwise neighboring urn. I've tried two approaches and neither work. An important problem in the Ehrenfest chain is the long-term, or equilibrium, distribution of , the number of balls in urn A. Math 29 — Probability Practice Second Midterm Exam 1 Instructions: 1. The ticket shows your six good balls, and there are 50 bad balls. I manage to calculate the individual probabilities on a per problem basis, but I need to find a way to phrase a general solution. The Philosophy of Statistics [S]tatistical inference is firmly based on probability alone. An urn contains three red balls numbered 1, 2, and 3, three black balls numbered 8, 9, and 10 and four white balls numbered 4, 5, 6, 7. The probability of this happening is =~ 0. ) are represented as colored balls in an urn or other container. Savage offered the example of an urn that contains two balls: Both may be white; both may be black; or one may be white and. 3, 794-814, 2012. Find P(N = 1), P(N = 2) and P(N = 3). (b) Find EN and Var(N). Probability basics 50 xp Queen and spade 50 xp. We recommend you review today's Probability Tutorial before attempting this challenge. Roll a fair 6-sided die until your first roll of a “3. The user may control the total number of balls in the urn (N), the number of red balls (R) and the number of balls sampled from the urn (n). A ball is selected at random from the first urn and placed in the second. You can also express this relationship as 1 ÷ 6, 1/6, 0. Imagine | {
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first urn and placed in the second. You can also express this relationship as 1 ÷ 6, 1/6, 0. Imagine two urns. Let us suppose that the urns are labelled with the numbers 1,2,, n and let tj be equal to k if the j-th ball is placed into the k-th urn. A Ball Is Randomly Drawn From Urn No. What is the probability that a white ball is drawn?. In probability and statistics, an urn problem is an idealized mental exercise in which some objects of real interest (such as atoms, people, cars, etc. An urn contains 5 red, 3 green, and 4 white balls. What is the probability that neither is red, given that neither is white? 2) A basketball player makes free throws with a 0. Two marbles are selected at random without replacement. 625 subscribers. After the four iterations the urn contains six balls. same answer. |A) is a probability function multiplicative formula: P(B and A) = P(B|A)P(A) Oct 28 Monty Hall problem §3. Bayes probabilities can also be obtained by simply constructing the tree. Then the number of tables is bounded by this multiple, so for large n, the probability of joining one of the k (fixed) tables is roughly , so this should behave roughly like the standard Polya Urn. What is the probability that the urn contains three balls of each color? Solution A tree diagram is ideal to deal with the problem, up to a sample space of 4 balls (so the first two pulls): So, we have a probability of (1/2)*(2/3) + (1/2)*(2/3) = 2/3 of ending up with 3 balls of one color. Then treat this like a mock exam (i. Home-Learning Problems 3. Show that the probability of rolling doubles with a non-fair (“fixed”) die is greater than with a fair die. Gelbaum DOVER PUBLICATIONS, INC. We write P (heads) = ½. Find P(N = 1), P(N = 2) and P(N = 3). What strategy maximizes your chance of victory? Problems from Rosen 7. The basic urn problem is to determine the probability of drawing one colored ball from an urn with known composition of differently colored balls. What is the probability the ball drawn | {
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an urn with known composition of differently colored balls. What is the probability the ball drawn from urn C is black? I've figured out that P(K\A) = 2/3 and P(K\B) = 5/6. , TTHor HHHT). The probability a black ball is drawn is 9/13. In the example shown, the formula in F5 is: =MATCH(RAND(), D$5:D$10) How this formula works. If you pick 5 balls from the urn at random, what is the probability that x of them will be. If a 4 appears, a ball is drawn from urn 1; otherwise, a ball is drawn from urn 2. There will be total 10 MCQ in this test. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. In the two parameter case, the matrix of transition probabilities has N+1 distinct eigenvalues λ j =1−2j/N, where j=0, 1,…, N. of getting white marble). From a given ball's perspective, the probability of being matched is $\frac{M}{N}(1-(1-\frac{1}{M})^N)$. The first problem-book of a similar kind as ours is perhaps Mosteller's well-known Fifty Challenging Problems in Probability (1965). Once again, balls are chosen at random with equal probability. At each step, an urn is selected according to their weights. What's the probability that the ball, which was taken out is: a) white b) green c) red Problem 9 An urn contains 8 white and 4 black balls. An event that cannot occur has a probability (of happening) equal to 0 and the probability of an event that is certain to occur has a probability equal to 1. The urn model and the Pólya process, in which the Pólya distribution and the limit form of it arise, are models with an after effect (extracting a ball of a particular colour from the urn increases the probability of extracting a ball of. He titled it The Two Children Problem, and phrased. User Account. of selecting 2 nd urn × prob. Walk through homework problems step-by-step from beginning to end. There are three ways to measure the average: the mean, median, and mode. Hume’s Problem. Create a set B of | {
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are three ways to measure the average: the mean, median, and mode. Hume’s Problem. Create a set B of foods. Six balls are picked from the 56 balls in an urn. Urn #2 contains 5 black and 2 red balls. Probability of second ball being red = 3/9 (because there are 3 red balls left in the urn, out of a total of 9 balls left. Publication date 1987 Topics Probabilities, Probabilités, Probabilités Publisher Internet Archive Books. Then the probbilities that the urn from which I have drawn the tickets is A or B, are by the principle given above, as K: K'. If one ball is drawn at random, what is the probability of getting a red or a white ball? 3. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. The first case,. By convention, statisticians have agreed on the following rules. Probability Trees: Many probability problems can be simplified by using a device called a probability tree. The probability for the dice to yield 1 or 2 is 2/6. 8--Conditional Probability With Urns and Marbles Glenn Olson. The probability of a sample point is a measure of the likelihood that the sample point will occur. To Probability Practice Problems for Exam # 2 1. One ball is drawn at random. 8--Conditional Probability With Urns and Marbles Glenn Olson. For each new ball, with probability p,createanewbin and place the ball in that bin; with probability 1−p,placetheballin an existing bin, such that the probability the ball is placed in a bin is proportional to mγ,wheremis the number of balls in that bin. An Example Suppose that a 6-sided die is rolled, what is the probability that the result is a 1? From our earlier results, if A is the event that a 1 is rolled, then P(A) = 1 6. Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. The probability the first is white is 6/15= 2/5. It is concluded that uncertainty theory is better than probability theory to. 6 balls are randomly drawn from the urn in succession, with replacement. What is the | {
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theory to. 6 balls are randomly drawn from the urn in succession, with replacement. What is the probability of getting exactly k red balls in a sample of size 20 if the sampling is done with replacement (repetition allowed)? Assume 0 ≤ k ≤ 20. One ball is drawn at random and its color noted. What is the probability of winning the. Fully worked-out solutions of these problems are also given, but of course you should first try to solve the problems on your own! c 2013 by Henk Tijms, Vrije University, Amsterdam. Binomial probability distribution; Hypothesis testing, types of error, and small samples; Introduction to statistics and the relative frequency histogram; Measures of variability and relative standing ; Other graphical methods and numerical methods; Poisson probability distribution and the urn model; Probability; Random variables. From a given ball's perspective, the probability of being matched is $\frac{M}{N}(1-(1-\frac{1}{M})^N)$. For example, if the probability of picking a red marble from a jar that contains 4 red marbles and 6 blue marbles is 4/10 or 2/5, then the probability of not picking a red marble is equal to 1 - 4/10 = 6/10 or 3/5, which is also the. So in a more conventional forward probability problem. What is the probability that it gets a) all 4 aces b) at least 1 ace; Hearts 5 cards are chosen from a standard deck of 52 playing cards (13 hearts) with replacement. Applications of probability arise everywhere: Should you guess. A series of two-urn biased sampling problems Puza, Borek and Bonfrer, André 2018, A series of two-urn biased sampling problems, Communications in statistics - theory and methods, vol. CAT Probability Questions cover all the different ways in which question can be asked. ) Jacob Bernoulli 1700. of selecting 1 st urn × prob. 1 Preliminaries This document is designed to get a person up and running doing elementary probability in R using the prob package. What strategy maximizes your chance of victory? Problems from Rosen 7. | {
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