# Linear Functions Pdf

Prerequisite materials, detailed proofs, and deeper treatments of selected topics. Addition and Subtraction Properties of Equality: Let , , and represent algebraic expressions. The graph is a line through (9,. If C(x) is linear, then MC is the slope of the cost function. 3c+8= 10 c=6 9. The zero from solving the linear function above graphically must match solving the same function algebraically. Chapter 2 Linear time series Prerequisites • Familarity with linear models. An arbitrary wave function ψ(x) can be expanded as a linear combination of the functions un(x), that is, we can write ψ(x) = X n cnun(x), (4) where the expansion coeﬃcients cn are given by cn = Z dxun(x)∗ψ(x). ƒ(º2)= º(º2)2º 3(º2) + 5 Substitute º2 for x. pdf: Alg 2 03. The slope of a linear function is 3 and its y-intercept is -2. Write a new function P(n) for the team's profit. °  ® °¯ t 5 if 2-4 if 2 xx fx x 2. One Solution Infinite Solutions No Solution Only Reasoning: What the type. As previously described, a linear equation can be defined as an equation in which the highest exponent of the equation variable is one. The _____ is the most basic linear function. Their cost per lawn mowed is the same, however, Service A charges $20 per month even if it does not mow that month. Intro to Linear Equations Algebra 6. The set Xis called the domain of the function, and. Graph linear functions that represent real-world situations and give their domain and range. Improve your math knowledge with free questions in "Identify linear functions from graphs and equations" and thousands of other math skills. In each of these examples, students will be identifying the rate of change and y-intercept in context and using their mathematical model to answer a predictive question. 7) y = −2x − 2 8) y = 1 4 x + 4. Activity Cab Company Task #1) Create a name for your cab company. Linear functions may be graphed by plotting points or by using the y-intercept and slope. Solving formulas is much like solving general linear equations. All linear equations eventually can be written in the form ax + b = c, where a, b, and c are real numbers and a ≠ 0. The inverse function of f(x) is given by f^{-1}(x) , and it tells us how to go from an output of f(x) back to its input. Simple Linear Equations (A) Answers Solve for each variable. This is because y is dependent on what you plug-in for x. Chapter 2: Linear Equations and Inequalities Lecture notes Math 1010 Ex. y 2x 1 and y 2x 2 are parallel lines since both equations have a slope of 2. Suppose a sequence is given by a linear recurrence relation (*). Divide by the coefficient of the variable to determine its value. ³halving ´) use an exponential function. Identify functions using differences or ratios EXAMPLE 2 Use differences or ratios to tell whether the table of values represents a linear function, an exponential function, or a quadratic function. Linear Functions Test SHOW ALL WORK! CHECK OVER ANSWERS! 8. You can graph the equation easily, because all you need are two points. A) Find the domain of the function. † graphed functions. Linear Equations. Such a function can be used to describe variables that change at a constant rate. The ability to work comfortably with negative numbers is essential to success in. These notes were revised 10/22/17 with additional pr. You will be able to solve any of these problems by the same methods that you have just mastered. The following observations can be made about this simplest example. Graphs of polynomial functions We have met some of the basic polynomials already. ©K n2 v0r1 s45 SK Wupt 9a7 nS Xo uf htGwBaBrQeP nL1LOCR. What does the slope represent? How much would a party for 40 people cost? 8. Analyzing!linear! functions. Example: Find the zero of $y=\frac{1}{2}x+2$ algebraically. This linear equations workbook is divided into 20 sections. Graphpaper generator ; Measurement. the graphing and systems of linear equations reporting cluster The following four California content standards are included in the Graphing and Systems of Linear Equations reporting cluster and are represented in this booklet by 16 test questions. Exponential book pdf free download link or read online here in PDF. Math 10 C Name: _____ 1 ID: A Relations and Functions Unit Test Multiple Choice Identify the choice that best completes the statement or answers the question. In a linear equation, y is called the dependent variable and x is the independent variable. College Algebra Version p 3 = 1:7320508075688772::: by Carl Stitz, Ph. The support of is where we can safely ignore the fact that , because is a zero-probability event (see Continuous random variables and zero-probability events ). What is the multiplicity in the following: y = ? M = _____ What does the graph do if M is ODD? Compare this to y = M = _____ SKETCH THE FUNCTIONS. One Vertical Shift Up 4. Solving formulas is much like solving general linear equations. 2x+8=22 x =7. The simplest linear relation is a proportional relation, which is a line with a vertical intercept of 0. 2 continued Domain and Range of Continuous Functions SShake, Rattle, and Roll h ak e, Rt lnd o Write your answers on notebook paper. Luckily, this is not because function problems are inherently more difficult to solve than any other math problem, but because most students have simply not dealt with functions as much as they have other SAT math topics. 5 Extreme points and optimality Notice that in problem P the optimum of c⊤x occurs at a ‘corner’ of the feasible set, regardless of what is the linear objective function. Pre-AP Algebra 2 Lesson 1-5 - Translating Graphs of Lines Discovery Worksheet (with Answers) Name Date Translating Graphs of Lines The following graphs are transformations of the parent function f(x) = x in the form f(x) = a(x h) k. Identify the parameter that determines the change and determine the function rule. Linear equations are of the form ax + b = c. However, for the real part we get that (s)= 1 s−1 +C(s); where 0 0 and de ne the. Inflection points and golden ratio. Unit 4- Graphing Linear Equations. Teacher's notes: Connecting the pattern, the table, the graph and the equation. •Example: y = 3x + 2 •Solution: any ordered pair (x, y) that makes the equation true. 2 Notes – Linear vs. The second item is that none of the variables can have an. Linear Equations in Three Variables JR2 is the space of 2 dimensions. Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. x, y, and z coordinate. In order to use stochastic gradient descent with backpropagation of errors to train deep neural networks, an activation function is needed that looks and acts like a linear function, but is, in fact, a nonlinear function allowing complex relationships in the data to be learned. 1-3 Transforming Linear Functions Example 4A: Fund-raising Application The golf team is selling T-shirts as a fund-raiser. Duration: 0 hrs 35 mins Checkup: Practice Problems Complete a set of practice problems on linear functions. The company decides to add a one-time$10 fee for cleaning. Our first unit in Algebra 2 is an introduction to functions, function notation, domain and range, intercepts, maximums and minimums, intervals of increasing and decreasing, finding solutions, and transformations. So far, we have discussed how we can find the distribution of a function of a continuous random variable starting from finding the CDF. Elementary Algebra Skill Solving Linear Equations: Variable on Both Sides Solve each equation. This is also known as the “slope. Relations, Functions, and Graphs-A Review - YouTube #248397. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS Theorem 2. SOLVING LINEAR EQUATIONS Goal: The goal of solving a linear equation is to find the value of the variable that will make the statement (equation) true. It measured…. One Horizontal Shift Right 3. Write an equation of the line that passes through the point (1,2) and has a slope of 3. y = x2 Solution: a. The graph of any linear. However these techniques are not. Download CAT Linear equation formulas PDF. 58 Chapter 2 Linear Relations and Functions The table shows average and maximum lifetimes for some animals. Complete the table for and graph the resulting line. 1 Pre-Algebra - Integers Objective: Add, Subtract, Multiply and Divide Positive and Negative Numbers. We shall study these in turn and along the way ﬁnd some results which are useful for statistics. The seminal work ofKantorovich(1939) on such problems usually marks the birth of convex optimization as a distinct subject of mathematical inquiry. Download latest questions with multiple choice answers for Class 10 Linear Equations in pdf free or read online in online reader free. EQUATIONS OF LINES IN DIFFERENT FORMS. not a function 11. ƒ(º2)= º(º2)2º 3(º2) + 5 Substitute º2 for x. 3: Quadratic Functions and Their Properties Def: A quadratic function is a function of the form f(x) = ax2+bx+c, where a;b;c are real numbers and a 6= 0. Example 2 Linear Functions Which of the following functions are linear? a. Example: Find the zero of $y=\frac{1}{2}x+2$ algebraically. Lesson 3 Direct Variation. Comparing Linear Functions- The Greates Rate of Change Subject: SMART Board Interactive Whiteboard Notes Keywords: Notes,Whiteboard,Whiteboard Page,Notebook software,Notebook,PDF,SMART,SMART Technologies ULC,SMART Board Interactive Whiteboard Created Date: 1/30/2014 3:52:58 PM. This representational technique has succeeded at ﬁnding good policies for problems with high dimensional state-spaces such as simulated soccer [Stone et al. Practice: Relations & Functions Final corrections due: Use the given form of each relation to complete the other forms. For each function: (a) Find the vertex ( , )hk of the parabola by using the formulas 2 b a h and 2 b a kf. The linear attenuation coefficient can be considered as the fraction of photons that interact with the shielding medium per. The simplest linear relation is a proportional relation, which is a line with a vertical intercept of 0. indd 1 66/26/08 7:36:04 PM/26/08 7:36:04 PM. Systems of Linear Equations 0. You can choose from up to four types of equations depending on the sophistication of your students. 2b ( 7)=11 b= 2 6. Match the formulas, graphs, and equations that go together. Example: Determine whether f(x) 3x 9 and 3 3 1 g(x) x are inverse functions. “Linear Equations In Two Variables” is chapter 4 of the NCERT Textbook, and it falls under the unit 2 Algebra. These pdf worksheets provide ample practice in plotting the graph of linear functions. •Example: y = 3x + 2 •Solution: any ordered pair (x, y) that makes the equation true. The reason is that then the equality (5) does not provide enough information due to the fact that the multiplicative commutators all have determinant 1. It is the interaction between linear transformations and linear combinations that lies at the heart of many of the important theorems of linear algebra. Linear Functions Worksheets This collection of linear functions worksheets is a complete package and leaves no stone unturned. Equations of nonconstant coefficients with missing y-term If the y-term (that is, the dependent variable term) is missing in a second order linear equation, then the equation can be readily converted into a first order linear equation and solved using the integrating factor method. (a) f(x) = (x 4)2 (b) f(x) = p x+ 4 (c) f(x) = 1 x+ 1 (d) f(x) = x x+ 1 (e) f(x) = p x3 + 9 (f) f(x) = 1 p x2 + 1 3. The possible values of x approach a chosen value (e. This product is suitable for Preschool, kindergarten and Grade 1. The pdf worksheet under this section has sets of table values. Linear Attenuation Shielding Formula: x B A I I e = * −μ. so, matrix multiplication is a linear function converse: every linear function y =f(x), with y an m-vector and x and n-vector, can be expressed as y =Ax for some m×n matrix A you can get the coeﬃcients of A from A ij =y i when x =e j Linear Equations and Matrices 3–3. Then you can draw a line through those two points. f(x)=−(x+2) 2 −7 Describe the transformation performed on each function g(x) to result in m(x). If the items are complex numbers, en is used to denote the vector space. Answers found at https://www. Take the Schoology Quiz (Concept 7 – Level 2) Score of 4 or higher move to level 3 Score of 3 or less, complete the Level 2 Review Level 3 1. Carefully graph each equation on the same coordinate plane. Graphing linear function: Type 1 - Level 2. We are going to use this same skill when working with functions. 58 Chapter 2 Linear Relations and Functions The table shows average and maximum lifetimes for some animals. The impulse response of a linear system, usually. 1) Write Down the Basic Linear Function. Linear Cost Function 2. The slopes are represented as fractions in the level 2 worksheets. Write an equation of the line that passes through the point (2, 2) and has a slope of 4. It is 76º F at the 6000-foot level of a mountain, and 49º F at the 12,000-foot level of the mountain. 8 of 8) 12. 5 Partial Correlation 100. Chapter & Page: 42–4 Nonhomogeneous Linear Systems which, in matrix/vector form, is x′ = Ax + g with A = 1 2 2 1 and g(t) = 3 0 t + 0 2. Question #14: This question will assess if students understand the domain of a function as the inputs and the range as the outputs in an abstract scenario. It also publishes articles that give significant applications of matrix theory or linear algebra to other. In your Differential Equations course, you will see that every solution to the differential equation above is a linear combination of cos and(k t) sin. A function fis a map f: X!Y (1. In its most basic form, a linear supply function looks as follows: y = mx + b. Several questions on functions are presented and their detailed solutions discussed. Show your work. One way is to find the x-intercept and the y-intercept and. Before graphing linear equations, we need to be familiar with slope intercept form. To Graphing Linear Equations The Coordinate Plane A. 1-3 Transforming Linear Functions Example 4A: Fund-raising Application The golf team is selling T-shirts as a fund-raiser. The Next-Generation Advanced Algebra and Functions placement test is a computer adaptive assessment of test-takers’ ability for selected mathematics content. ƒ(x) is not a linear function because it has an x2-term. This section is a step-by-step presentation of how to use algebra formulae on all the topics covered in this site which include formulae on -linear equations, inequalities, decimals, fractions, exponents, graphing linear equations, binomial theorem, pythagoras theorem, quadratic equations, algebraic expressions, factorisation, ratios, geometry. Solve to find the x- and y-intercepts. iterative methods for linear systems have made good progress in scientiﬁc an d engi- neering disciplines. 2 Linear Functions. The impulse response of a linear system, usually. Find the inverse function for the following function: 3 a. Input 6 7 7 14 Output 5 2 5 2 Find the slope and y-intercept of the line with the given equation. Linear Function Word Problems Exercise 1Three pounds of squid can be purchased at the market for $18. Learn More at mathantics. y 2x 1 and y 2x 2 are parallel lines since both equations have a slope of 2. The function R(n) = 7. 5, Solving Systems of Linear Equations by Graphing. Manipulate equations from one form to another[Lesson 7. We tried to explain the trick of solving word problems for equations with two variables with an example. Relations and functions review worksheet Collection. 4 Guided Notes; 4. A linear function is a function whose graph consists of segments of one straight line throughout its domain. Identify the slope and y-intercept of the line with the given equation. Linear transformations: Rank-nullity theorem, Algebra of linear transformations, Dual spaces. Solutions to Linear Equations in One Variable The _____ of an equation is the value(s) of the variable(s) that make the equation a true statement. † identified independent and dependent variables. (a) Graph the equation. Table 5-1 provides examples of common linear and nonlinear systems. The quantity mx is called the variable costand the intercept b is called the ﬁxed cost. Each customer can buy a maximum of four tickets. Prove: f(X2) f (x, ) = f(xy) Proof: I. x y -1 -4 0 -3 1 -2 -2 -5 x y -2 5 -1 -3 0 -1. Students recognize a proportion (y/x = k, or y = kx) as a special case of a linear equation of the form y = mx + b, understanding that the constant of proportionality (k) is the slope and. Then answer the questions at the bottom of the page. 04 Solve Systems of Linear Equations in Three Variables. An affine function is the composition of a linear function with a translation, so while the linear part fixes the origin, the translation can map it somewhere else. Then explore different ways to find the slope and y-intercept for a linear function. The y intercept is a point in itself and to complete our graph we only need one additional point. y 2x 1 and y 2x 2 are parallel lines since both equations have a slope of 2. Leave any comments, questions, or suggestions below. Worksheets for linear equations Find here an unlimited supply of printable worksheets for solving linear equations, available as both PDF and html files. functional relationship? A The number of cookies ordered. Linear functions are those that exhibit a constant rate of change, and their graphs form a straight line. Graphs of Linear Equations reviews the rectangular (Cartesian) coordinate system, and contains lessons on different methods of interpreting the lines and their applications, and has examples of solving different practice problems related to finding the slope and using different forms of writing the equation for a line. We use the rst equation to eliminate x1 from equations (2) through (m), leaving m 1 equations in n 1 unknowns. Review Of Linear Functions Lines Answer Key - Displaying top 8 worksheets found for this concept. Plan your 60-minute lesson in Math or linear functions with helpful tips from James Bialasik linear_functions1_investigation. 6 Linear Functions of Random Vectors 79 3. KEY Graphing Linear Functions WS 2. Linear Equations 1 1. Typically we consider B= 2Rm 1 ’Rm, a column vector. Slope of a linear function is the change in y for a unit change in x along the line and usually denoted by the letter "m" Slope is sometimes referred to as "rise over run". f (x) = a (x – h)2 + k (a ≠ 0). 4) y = 9(1/x) + 4 (x is in the denominator). The coordinate plane has 4 quadrants. Two or more products are usually produced using limited resources. What is the y-intercept? 4. Intro to Linear Equations Algebra 6. linear algebra, the so-called pivot operation. The cost to rent a piece of equipment is$27, plus $6. Outlines of concepts covered in this linear equations formulas pdf are: Definition and types of simple linear equations; General form of single, two and three variable linear equations; Tips for finding out number of solutions in two variable linear equations. Because of this prevalence of numerical linear algebra, we begin our treatment of basic. Matrices and Elementary Row Operations 6 1. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. 03 Linear Functions Zoey Ramsay. Profits, P, are equal to sales, S, minus expenses, E. GRAPHING LINEAR EQUATIONS VOCABULARY ( 3, -7 ) VOCABULARY coordinate plane coordinates diagonal line horizontal line ordered pair origin point quadrant vertical line x-axis y-axis x-coordinate y-coordinate x-intercept y-intercept. Homework Practice Workbook 000i_ALG1HWPFM_890836. Solve linear and quadratic equations and inequalit ies, including systems of up to three linear equations with three. This will take you to the individual page of the worksheet. CHECK YOUR UNDERSTANDING Write your answers on notebook paper. The function fis a tensor. Graphing linear function: Type 1 - Level 2. The graph of any linear. A graph is linear if it is a straight line. Find an n-vector bfor which bTc= p0( ): This means that the derivative of the polynomial at a given point is a linear function of its coe cients. AP Calculus Topic: Analysis of Functions Materials: Student Activity pages Teacher Notes: This activity can be used early in the year after students have studied linear functions. Solving Equations Step 1. Linear Functions LINEAR FUNCTIONS REVIEW 7. Subsection LTLC Linear Transformations and Linear Combinations. Therefore, f of 1 is negative 2. 3Transforming Linear Functions. Several questions on functions are presented and their detailed solutions discussed. This was the unit that the city provided us last year, which is why we did not create a new one. Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. The piecewise linear approximation is implemented in a linear programming model by defining new variables xj1, xj2,…, xjr to represent. In this course, you will use x and y to write linear equations and n and u n to write recursive and explicit formulas for sequences of discrete points. Guess words about linear functions. Graphically, power functions can resemble exponential or logarithmic functions for some values of x. Linear Functions Vocabulary. 1 Introducing Functions Lesson Plan OBJECTIVES ˚Students will understand that for each input there can be only one output. the graph of the function f(x) = c. Chapter 2 Linear time series Prerequisites • Familarity with linear models. Homework: 1) Complete the Functions Story Problem Worksheet and Linear Function Word Problems if not finished in class. Graphing and Systems of Equations Packet 1 Intro. 2 Linear functions (EMA48) Functions of the form $$y=x$$ (EMA49) Functions of the form $$y=mx+c$$ are called straight line functions. Given the integers a;b > 0, we de ne greatest common divisor of a and b, as the largest number that divides both a and b. •Example: y = 3x + 2 •Solution: any ordered pair (x, y) that makes the equation true. Complete the table for and graph the resulting line. This final lesson in the unit culminates with the Go Public phase of the legacy cycle. Example Let be a uniform random variable on the interval , i. Graphing Linear Equations. Solving Systems of Linear Equations by Graphing HW : 4/13/2017: SOL Coach Book 1. Students should work in pairs or groups of three. This linear equation has m = 3 and b = -1. FUNDRAISING The Pep Club rented a shaved ice machine to sell shaved ice as a fundraiser. Some of the more important. Primary method for approaching these problems. Solving formulas is much like solving general linear equations. For instance, here is a function L from the set R2 to the set R3: L x1 x2. Lesson 3 Direct Variation. com You will need to understand how to project cash flow. Lesson 3 Solving Equations. Use Linear Combination to Solve Systems of Equations and Inequalities 1st: Rearrange the equations so terms line up as: Ax + By = C 2nd: Multiply none, or one, or both equations by constant(s) so that the coefficients of one of the variables are opposites. Transformations of Quadratic Functions Transformations of Functions Transformation: A change made to a figure or a relation such that the figure or the graph of the relation is shifted or changed in shape. Steps for Graphing a Linear Function (Slope-Intercept Form)! Identify and plot the y-intercept ! Use the slope to plot an additional point (Rise/Run) ! Draw a line through the two points EXAMPLES EXAMPLE 5: WRITING AND GRAPHING LINEAR EQUATIONS GIVEN A Y-INTERCEPT AND A SLOPE Write an equation of a line with the given slope and y-intercept. Review of Linear Functions (Lines) Find the slope of each line. > Topics include: Least-squares aproximations of over-determined equations and least-norm solutions of underdetermined equations. Linear equations are found throughout mathematics and the real world. Such a function can be used to describe variables that change at a constant rate. Find the inverse function for the following function: 3 a. except part (c). Chapter 3: Linear Functions. YOU WILL NOT BE USING A CALCULATOR FOR PART I MULTIPLE-CHOICE QUESTIONS, AND THEREFORE YOU SHOULD NOT USE ONE FOR THE REVIEW PACKETS. x 2 1 r(x) = 26 Compare Graphs of Linear Functions with the Graph of f(x) = x. In addition to using a table, equations can be graphed using the equations themselves Remember that Y=M(X)+B tells us the slope and the Y intercept. The y-intercept is where the graph crosses the y -axis. ) Modeling a Quadratic Function Using Various Word Problems Example 2: Complete each word problem using techniques learned in previous concepts. 3) Linear v Non Linear Functions 1 (8. (c) After how many. Because the fraction consists of the rise (the change in y, going up or down) divided by the run (the change in x, going from left to the right). Recently Modified. 3z+4=34 z=10 2. LESSON 3: LINEAR FUNCTIONS Study: Linear Functions Learn about slope and the three main forms of linear functions. the graphing and systems of linear equations reporting cluster The following four California content standards are included in the Graphing and Systems of Linear Equations reporting cluster and are represented in this booklet by 16 test questions. 5x – 3 = 2x – 27 8. 8K PDF/Acrobat 28 Jan 2016) Projector Resources. 4 - Linear Functions Writing equations for Horizontal and Vertical Lines. 1: Rate of Change and Slope Rate of Change - shows relationship between changing quantities. 2y+1=17 y=8 4. 21 Posts Related to Writing Linear Equations From Graphs Worksheet Pdf. The reason is that the domain and range of a linear function naturally span all real numbers unless the domain is restricted. org©2001 September 22, 2001 2 4. The function is defined by h(t) = -7t2 + 48t. Find the inverse function of f(x): State the domain of the inverse function f 1(x. Another special type of linear function is the Constant Function it is a horizontal line: f(x) = C. Luckily, this is not because function problems are inherently more difficult to solve than any other math problem, but because most students have simply not dealt with functions as much as they have other SAT math topics. Section 1: Quadratic Functions (Introduction) 3 1. the graph of the function f(x) = c. the graphing and systems of linear equations reporting cluster The following four California content standards are included in the Graphing and Systems of Linear Equations reporting cluster and are represented in this booklet by 16 test questions. linear algebra, the so-called pivot operation. Question #14: This question will assess if students understand the domain of a function as the inputs and the range as the outputs in an abstract scenario. H = 4T + 74 A company can make a total of 20 solar heaters for$13,900, while 10 heaters cost $7500. The table shows the cost per hour. x 2 1 r(x) = 26 Compare Graphs of Linear Functions with the Graph of f(x) = x. The size of the PDF file is 66677 bytes. Intersections between spheres, cylinders, or other quadrics can be found using quartic equations. Copy this to my account; E-mail to a friend; Find other activities. Linear Equations Worksheets: Linear Equations Worksheets Standard Form to Slope Intercept Form Worksheets Finding the Slope of an Equation of a Line Worksheets Find Slope From Two Points Worksheets Finding Slope Quizzes: Combining Like Terms Straight Line Graph Slope Formula - Finding slope of a line using point-point method System of Linear. Even though, mathematically speaking, these two points can be arbotrarily close, if you choose them to be too close, your line may deviate from what the graph. Note the ax ≡ b (mod n) iﬀ there is y ∈ Z such that ax+ ny = b (by equivalent formulation of equivalence mod n, ax ≡ b ( (mod n) iﬀ they diﬀer by a multiple of n). Can be inserted in interactive notebook, used as an anchor chart, or a quick reference for students. Multiple Choice Questions have been coming in Class 8 Linear Equations exams, thus do MCQs to test understanding of important topics in the chapters. The Next-Generation Advanced Algebra and Functions placement test is a computer adaptive assessment of test-takers’ ability for selected mathematics content. MULTIPLE CHOICE Plotting the points from each of the given tables as shown below, identify its correct. ferential equations by any discrete approximation method, construction of splines, and solution of systems of nonlinear algebraic equations represent just a few of the applications of numerical linear algebra. Mathematics (Linear) – 1MA0 STRAIGHT LINE GRAPHS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. A normal 12 point text isn’t the ideal solution for fast reading. Associate a given equation with a function whose zeros are the solutions of the equation. Introduction Diophantine equations are named for Diophantus of Alexandria who lived in the third century. In these cases, replace the function notation and solve rather than the x. Key Components Key Components 2. The slope is m = −0. Unit 2: Solve Linear Equations Instructor Notes The Mathematics of Writing and Solving Linear Equations Most students taking algebra already know the techniques for solving simple equations. Fluency in interpreting the parameters of linear functions is emphasized as well as setting up linear functions to model a variety of situations. Linear Functions Worksheets This collection of linear functions worksheets is a complete package and leaves no stone unturned. There are two methods to finding the slope of a linear relationship from a table. The coordinate plane has 4 quadrants. Free pdf on #248396. Population of Indiana '50 '60 '70 '80 '90 '00 3 2 4 5 Population (millions) 6 7 Year 0 3. com for more Free math videos and additional subscription based content!. Each section begins with a bite-sized introduction to a topic with an example, followed by practice exercises including word problems. These are to use the CDF, to trans-form the pdf directly or to use moment generating functions. Find solutions using the given "x" values. Then explore different ways to find the slope and y-intercept for a linear function. 2c+7=17 c=5 6. Find the b (y-intercept) and write the equation of the line in slope-intercept form. In order to use stochastic gradient descent with backpropagation of errors to train deep neural networks, an activation function is needed that looks and acts like a linear function, but is, in fact, a nonlinear function allowing complex relationships in the data to be learned. Chapter 5 - Linear Functions Name_____ Keller - Algebra 1 Notes 5. Linear functions are those that exhibit a constant rate of change, and their graphs form a straight line. Explain why this. Then, write the function formula. In the associated activities, students use linear models to depict Hooke's law as well as Ohm's law. ©K n2 v0r1 s45 SK Wupt 9a7 nS Xo uf htGwBaBrQeP nL1LOCR. For example, given ax+3=7, solve for x. The point is stated as an ordered pair (x,y). It is also important to know that any linear function can be written in the form f(x) mx -+- b, where m and b are constants. (a) f(x) = (x 4)2 (b) f(x) = p x+ 4 (c) f(x) = 1 x+ 1 (d) f(x) = x x+ 1 (e) f(x) = p x3 + 9 (f) f(x) = 1 p x2 + 1 3. Quadratic Functions-Worksheet Find the vertex and “a” and then use to sketch the graph of each function. Here is a small outline of some applications of linear equations. Give each student a card with one representation of a linear function from the LESSON 1 section of the INSTRUCTIONAL ACTIVITY SUPPLEMENT. Some of the worksheets below are Free Linear Equations Worksheets, Solving Systems of Linear Equations by Graphing, Solving equations by removing brackets & collecting terms, Solving a System of Two Linear Equations in Two Variables by Addition, …. mathematics content. a change in the size or position of a figure B. In its most basic form, a linear supply function looks as follows: y = mx + b. Grade 9 Math Solving Linear Equations. Systems of Linear Equations 0. Linear Algebra is one of the most important basic areas in Mathematics, having at least as great an impact as Calculus, and indeed it provides a signiﬂcant part of the machinery required to generalise Calculus to vector-valued functions of many variables. Pre-Algebra Chapter 8—Linear Functions and Graphing SOME NUMBERED QUESTIONS HAVE BEEN DELETED OR REMOVED. During a 45-minute lunch period, Albert (A) went running and Bill (B) walked for exercise. Linear function word problems Calculator tables. 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# Primes P such that ((P-1)/2)!=1 mod P

I was looking at Wilson's theorem: If $P$ is a prime then $(P-1)!\equiv -1\pmod P$. I realized this implies that for primes $P\equiv 3\pmod 4$, that $\left(\frac{P-1}{2}\right)!\equiv \pm1 \pmod P$.

Question: For which primes $P$ is $\left(\frac{P-1}{2}\right)!\equiv 1\pmod P$?

After convincing myself that it's not a congruence condition for $P,$ I found this sequence in OEIS. I'd appreciate any comments that shed light on the nature of such primes (for example, they appear to be of density 1/2 in all primes that are $3\bmod 4$).

Thanks,

Jacob

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Given that there are no comments of any note on the sequence in the OEIS, there's a fair chance that little is known about your question. –  Kevin Buzzard Feb 23 '10 at 10:28
For all p<=250000, p=3 mod 4, we have 5458 +1s and 5589 -1s. –  Kevin Buzzard Feb 23 '10 at 10:56

I am a newcomer here. If p >3 is congruent to 3 mod 4, there is an answer which involves only $p\pmod 8$ and $h\pmod 4$, where $h$ is the class number of $Q(\sqrt -p)$ . Namely one has $(\frac{p-1}{2})!\equiv 1 \pmod p$ if an only if either (i) $p\equiv 3 \pmod 8$ and $h\equiv 1 \pmod 4$ or (ii) $p\equiv 7\pmod 8$ and $h\equiv 3\pmod 4$.

The proof may not be original: since $p\equiv 3 \pmod 4$, one has to determine the Legendre symbol

$${{(\frac{p-1}{2})!}\overwithdelims (){p}} =\prod_{x=1}^{(p-1)/2}{x\overwithdelims (){p}}=\prod_{x=1}^{(p-1)/2}(({x\overwithdelims (){p}}-1)+1).$$ It is enough to know this modulo 4 since it is 1 or -1. By developping, one gets $(p+1)/2+S \pmod 4$, where $$S=\sum_{x=1}^{(p-1)/2}\Bigl({x\over p}\Bigr).$$ By the class number formula, one has $(2-(2/p))h=S$ (I just looked up Borevich-Shafarevich, Number Theory), hence the result, since $\Bigl({2\over p}\Bigr)$ depends only on $p \pmod 8$.

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That's very slick! –  David Speyer Feb 23 '10 at 13:02
Yes, very nice! My interpretation of the question is "do the primes for which the square root is 1 give a set of density 1/2?" and this at least gives some way of attacking the problem. –  Kevin Buzzard Feb 23 '10 at 13:09
+1. Salut, et bienvenu ! –  Chandan Singh Dalawat Feb 23 '10 at 13:58
In the paper emis.de/journals/EM/expmath/volumes/12/12.1/pp99_113.pdf they mention that Cohen proved modulo CL a conjecture of Hooley about a sum over $h(p)$. –  Victor Miller Feb 23 '10 at 15:14
Interesting. In particular, that paper claims that the odd part of h(p), as p runs through primes, seems to have the same distribution as the odd part of h(D), as D runs through square free integers. That's something I didn't know. I point out, however, that this paper deals with real quadratic fields. –  David Speyer Feb 23 '10 at 15:42

There is some history to this question. Dirichlet observed (see p. 275 of History of the Theory of Numbers,'' Vol. 1) that since we already know $(\frac{p-1}{2})! \equiv \pm 1 \bmod p$, computing modulo squares gives $(\frac{p-1}{2})! \equiv (-1)^{n} \bmod p$, where $n$ is the number of quadratic nonresidues mod $p$ which lie between 1 and $(p-1)/2$.

Jacobi (pp. 275-276 in Dickson's book) determined $n \bmod 2$ in terms of the class number $h_p$ of ${\mathbf Q}(\sqrt{-p})$, for $p \equiv 3 \bmod 4$ and $p \not= 3$. By the class number formula, $$\left(2-\left(\frac{2}{p}\right)\right)h_p = r-n,$$ where $r$ is the number of quadratic residues from 1 to $(p-1)/2$. Also $r + n = (p-1)/2$, so $$2n = \frac{p-1}{2} - \left(2 - \left(\frac{2}{p}\right)\right)h_p.$$ In particular, $h_p$ is odd when $p \equiv 3 \bmod 4$.

Taking cases if $p \equiv 3 \bmod 8$ and $p \equiv 7 \bmod 8$, we find both times that $n \equiv (h_p+1)/2 \bmod 2$, so $$\left(\frac{p-1}{2}\right)! \equiv (-1)^{(h_p+1)/2} \bmod p.$$

This shows why getting precise statistics on when the congruence has 1 on the right side will be hard.

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The following is a relevant classical paper:

Mordell, L. J. The congruence $(p-1/2)!\equiv ±1$ $({\rm mod}$ $p)$. Amer. Math. Monthly 68 1961 145--146.

http://www.math.uga.edu/~pete/Mordell61.pdf

Put $((p-1)/2)!\equiv(-1)^a\ (\text{mod}\,p)$, where $p$ is a prime $\equiv 3\ (\text{mod}\,4)$. The author proves the following result. If $p\equiv 3\ (\text{mod}\,4)$ and $p>3$, then $$a\equiv{\textstyle\frac 1{2}}\{1+h(-p)\}\quad(\text{mod}\,2), \tag1$$ where $h(-p)$ is the class number of the quadratic field $k(\surd-p)$ [$\mathbb{Q}(\sqrt{-p})$ must be meant here. --PLC]. The author points out that (1) follows easily from a result of Dirichlet; also that Jacobi had conjectured an equivalent result before the class number formula was known. (MathReview by L. Carlitz)

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The notation k(\sqrt{-p}) for our Q(\sqrt{-p}) is "classical" and was used e.g. by Hilbert in his Bericht. The idea was that k(\sqrt{-p}) is the field k you get by adjoining a square root of -p to the rationals. –  Franz Lemmermeyer Feb 23 '10 at 17:07
Hecke uses $K(\root l\of\mu;k)$ to denote $k(\root l\of\mu)$ in his Vorlesungen. –  Chandan Singh Dalawat Feb 24 '10 at 1:06

This is an attempt to justify the answer $1/2$ based on the Cohen-Lenstra heuristics. There will be a lot of nonsensical steps, and I am not an expert, so this should be viewed with caution.

As is observed above, this is equivalent to determining $h(p) \mod 4$, where $h(p)$ is the class number of $\mathbb{Q}(\sqrt{-p})$. Since $p$ is odd and $3 \mod 4$, the only ramified prime in $\mathbb{Q}(\sqrt{-p})$ is the principal ideal $(\sqrt{-p})$. Thus, there is no $2$-torsion in the class group and $h(p)$ is odd.

For any odd prime $q$, let $a(q,p)$ be the power of $q$ which divides $h(p)$. We want to compute the average value of $$\prod_{q \equiv 3 \mod 4} (-1)^{a(q,p)}.$$

First nonsensical step: Let's pretend that the CL-heuristics work the same way for the odd part of the class group of $\mathbb{Q}(\sqrt{-p})$, that they do for the odd part of the class group of $\mathbb{Q}(\sqrt{-D})$. We just saw above that the fact that $p$ is prime constrains the $2$-part of the class group; this claim says that it does not effect the distribution of anything else.

Then we are supposed to have: $$P(a(q,p)=0) = \prod_{i=1}^{\infty} (1-q^{-i}) = 1-1/q +O(1/q^2),$$ $$P(a(q,p)=1) = \frac{1}{q-1} \prod_{i=1}^{\infty} (1-q^{-i}) = 1/q +O(1/q^2),$$ and $$P(a(q,p) \geq 2) = O(1/q^2).$$

If you believe all of the above, then the average value of $(-1)^{a(p,q)}$ is $1-2/q+O(1/q^2)$.

Second nonsensical step: Let's pretend that $a(q,p)$ and $a(q',p)$ are uncorrelated. Furthermore, let's pretend that everything converges to its average value really fast, to justify the exchange of limits I'm about to do.

Then $$E \left( \prod_{q \equiv 3 \mod 4} (-1)^{a(q,p)} \right) = \prod_{q \equiv 3 \mod 4} \left( 1- 2/q + O(1/q^2) \right)$$.`

The right hand side is zero, just as if $h(p)$ were equally like to be $1$ or $3 \mod 4$.

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Most of the latex formulas don't parse on my browser (Firefox 3.5.8) (though some do). Does any one know why? –  Anonymous Feb 23 '10 at 17:33
Might you be missing the jsmath fonts? math.union.edu/~dpvc/jsMath/users/fonts.html –  David Speyer Feb 23 '10 at 17:50
It works! Great. Thanks! –  Anonymous Feb 23 '10 at 17:56


# multiply $2^{(a-1)b}$ by $2^b$ and get $2^{ab}$? How is this so?

I’m reading How To Prove It and in the following proof the author is doing some basic algebra with exponents that I just don’t understand. In Step 1.) listed below he is multiplying $2^b$ across each term in (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$) and gets the resulting set of terms in Step 2.) In particular I have no idea how he is getting $2^{ab}$ from multiplying $2^{(a-1)b}$ by $2^b$ again which is shown in the first sequence in Step 2.). When I do it I get $2^{(ab)(b) – (b)(b)}$ and assume this is as far as it can be taken. Can someone please help me understand what steps he is taking to to get his answer?

Theorem 3.7.1. Suppose n is an integer larger than 1 and n is not prime. Then $2^n$ − 1 is not prime. Proof. Since n is not prime, there are positive integers a and b such that a < n, b < n, and n = ab. Let x = $2^b$ − 1 and y = 1 + $2^b$ + $2^{2b}$ +· · ·+ $2^{(a−1)b}$. Then

xy = ($2^b$ − 1) · (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$)

Step 1.) = $2^b$ · (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$) − (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$)

Step 2.) = ($2^b$ + $2^{2b}$ + $2^{3b}$ +···+$2^{ab}$) − (1 + $2^b$ + $2^{2b}$ + ···+$2^{(a-1)b}$)

Step 3.) = $2^{ab}$ − 1

Step 4.) = $2^n$ − 1.

• Are you familiar with how $x^c\times x^d = x^{c+d}$? Are you familiar with how $(a-1)b + b = ab$? – JMoravitz Jun 10 '17 at 16:00
• My title should have read multiplying $2^{(a-1)b}$ by $2^b$ and get $2^{ab}$. Sorry about that. – maybedave Jun 10 '17 at 16:00
• @JMoravits. Not really and now I'm feeling a little dumb because from what you are saying, this is actually correct, no? – maybedave Jun 10 '17 at 16:01
• For integer values of $c$ and $d$, notice that $x^c\times x^d = \overbrace{\underbrace{x\times x\times x\times \cdots \times x}_{c~\text{times}}\times \underbrace{x\times x\times \cdots \times x}_{d~\text{times}}}^{c+d~~\text{times}}$. (The property can be extended to real values of $c$ and $d$ as well, look up a more in depth proof for that). As for why $(a-1)b+b=ab$, this is algebraic manipulation., $(a-1)b+b=ab-1b+b=ab-b+b=ab+(-b+b)=ab+0=ab$ – JMoravitz Jun 10 '17 at 16:07

$$2^{(a-1)b}2^b=2^{(a-1)b+b}=2^{ab-b+b}=2^{ab}.$$

• Ok, so what what you guys are saying is that I really should be adding the b to (a-1)b rather than multiplying it in. This is A BIG HELP! Thank you! – maybedave Jun 10 '17 at 16:04
• @maybedave It's $x^rx^s=x^{r+s}$... – Lord Shark the Unknown Jun 10 '17 at 16:05
• "rather than multiplying it in" You absolutely should not multiply them! Notice $64 =4*16=2^2*2^4=2^{2*4}=2^8=256$. Doesn't work. But $64= 2^2*2^4 = (2*2)*(2*2*2*2) = (2*2*2*2*2*2) = 2^6 = 64$. Addition has to work. Notice $2^4*2^4 = (2*2*2*2)*(2*2*2*2) = [(2*2*2*2)(2*2*2*2)(2*2*2*2)(2*2*2*2)] = 2^{4*4}$ just doesn't make any sense at all. – fleablood Jun 10 '17 at 16:18

Using the laws of exponents, $2^{(a-1)b}2^b = 2^{(a-1)b+b} = 2^{ab}$

Hint:

The correct rule is:

$$2^b 2^{(a-1)b}=2^{(a-1)b+b}$$

Basic rules $a^m a^n = (a*a*a*a*a........*a[n \text{ times}])*(a*a*...... *a[m \text{ times}]) = a*a*a*.......*a [n+m\text { times}] = a^{n+mm}$.

So $2^{(a-1)b}*2^b = (2*2*2*2*2........*2[(a-1)b \text{ times}])*(2*2*...... *2[b \text{ times}]) = 2*2*2*.......*2 [(a-1)b+b\text { times}] = 2^{(a-1)b + b}$

Meanwhile $(a-1)*b + b = a*b - 1*b + b = (ab) + (-b + b) = ab + 0 = ab$.

So $2^{(a-1)b}2^b = 2^{(a-1)b + b} = 2^{ab}$


# Converting a repeating decimal to ratio of integers

#### paulmdrdo

##### Active member
0.17777777777 convert into a ratio.

#### M R

##### Active member
Re: converting a repeating decimal to ratio of integers

Hi,
This is $$0.1 + 0.077777=\frac{1}{10}+\frac{7}{100}+\frac{7}{1000}+...$$ where you have a GP to sum.

Or $$\text{Let } x=0.0777..$$ so that $$10x=0.777..$$.

Subtracting gives $$9x=0.7$$ and so $$x=\frac{7}{90}$$. Now just add $$\frac{1}{10}+\frac{7}{90}$$ and simplify.

I should also say that we can write a decimal as a fraction but we can't write it as a ratio.

#### paulmdrdo

##### Active member
Re: converting a repeating decimal to ratio of integers

Hi,
This is $$0.1 + 0.077777=\frac{1}{10}+\frac{7}{100}+\frac{7}{1000}+...$$ where you have a GP to sum.

Or $$\text{Let } x=0.0777..$$ so that $$10x=0.777..$$.

Subtracting gives $$9x=0.7$$ and so $$x=\frac{7}{90}$$. Now just add $$\frac{1}{10}+\frac{7}{90}$$ and simplify.

I should also say that we can write a decimal as a fraction but we can't write it as a ratio.
what do you mean by "GP"?

#### M R

##### Active member
Re: converting a repeating decimal to ratio of integers

what do you mean by "GP"?
Sorry, I have to stop using abbreviations.

A GP is a geometric progression: $$a, ar, ar^2, ar^3...$$.

If you haven't met this then the second method I posted is fine.

#### soroban

##### Well-known member
Re: converting a repeating decimal to ratio of integers

Hello, paulmdrdo!

$$\text{Convert }\,0.1777\text{...}\,\text{ to a fraction.}$$

$$\begin{array}{ccc}\text{We have:} & x &=& 0.1777\cdots \\ \\ \text{Multiply by 100:} & 100x &=& 17.777\cdots \\ \text{Multiply by 10:} & 10x &=& \;\;1.777\cdots \\ \text{Subtract:} & 90x &=& 16\qquad\quad\; \end{array}$$

Therefore: .$$x \;=\;\frac{16}{90} \;=\;\frac{8}{45}$$

#### paulmdrdo

##### Active member
how would I decide what appropriate power of ten should i use?

for example i have 3.5474747474... how would you convert this one?

#### M R

##### Active member
Since two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.

If you use 1000 and 10 you will get

1000x=3547.474747...

10x=35.474747...

So 990x=3512 and x=3512/990=1756/495.

I'm adopting Soroban's approach as I prefer it to what I did earlier.

#### paulmdrdo

##### Active member
Since two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.

If you use 1000 and 10 you will get

1000x=3547.474747...

10x=35.474747...

So 990x=3512 and x=3512/990=1756/495.

I'm adopting Soroban's approach as I prefer it to what I did earlier.
"a difference of two in the powers of ten" -- what do you mean by this? sorry, english is not my mother tongue. bear with me.

Last edited:

#### M R

##### Active member
"a difference of two in the powers of ten" -- what do you me by this? sorry, english is not my mother tongue. bear with me.
No problem.

We have 10^3 and 10^1.

The difference between 3 and 1 is 3-1=2

#### Prove It

##### Well-known member
MHB Math Helper
how would I decide what appropriate power of ten should i use?

for example i have 3.5474747474... how would you convert this one?
You want to multiply by a power of 10 which enables you to only have the repeating digits shown, and then multiply by a higher power of ten to have exactly the same repeating digits. We require this so that when we subtract, the repeating digits are eliminated.

So in this case, since the 47 repeats, you want the first to read "something.4747474747..." and the second to read "something-else.4747474747..."

What powers of 10 will enable this?

#### MarkFL

Staff member
A quick method my dad taught me when I was little, is to put the repeating digits over an equal number of 9's.

1.) $$\displaystyle x=0.1\overline{7}$$

$$\displaystyle 10x=1.\overline{7}=1+\frac{7}{9}=\frac{16}{9}$$

$$\displaystyle x=\frac{16}{90}=\frac{8}{45}$$

2.) $$\displaystyle x=3.5\overline{47}$$

$$\displaystyle 10x=35.\overline{47}=35+\frac{47}{99}=\frac{3512}{99}$$

$$\displaystyle x=\frac{3512}{990}=\frac{1756}{495}$$


# Limit of $f(x)$ given that $f(x)/x$ is known

Given that $$\lim_{x \to 0} \dfrac{f(x)}{x}$$ exists as a real number, I am trying to show that $\lim_{x\to0}f(x) = 0$. There is a similar question here: Limit of f(x) knowing limit of f(x)/x.

But this question starts with the assumption of $$\lim_{x \to 0} \dfrac{f(x)}{x} = 0,$$ and all I am assuming is that the limit is some real number. So the product rule for limits doesn't really work here.

Or do I need to show that $$\lim_{x \to 0} \frac{f(x)}{x} = 0$$ and then apply the product rule?

• Try proving it by contradiction: what happens if $\lim_{x\to 0}f(x)\neq 0$? – Ender Wiggins Aug 23 '18 at 9:00
• What prevents you from using product rule of limits? Perhaps you need to revisit the product rule in your text and then understand that it works fine here. – Paramanand Singh Aug 23 '18 at 14:23
• @FurryFerretMan Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… – gimusi Sep 17 '18 at 20:08

The product rule trick still works. If $\lim_{x \to 0} f(x)/x = R \in \mathbb R$, and obviously $\lim_{x \to 0} x = 0$, it follows that $$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{f(x)}{x} \times x = R \times 0 = 0.$$

• This is the most natural solution and I would not consider use of limit laws a trick rather it is the method. +1 – Paramanand Singh Aug 23 '18 at 14:19

We have that eventually

$$0\le \left|\frac{f(x)}{x}\right|\le M$$

therefore

$$0\le \left|f(x)\right|\le M|x| \to 0$$

Let $\lim_{x\to0}\dfrac{f(x)}{x}=l$ then $\bigg|\dfrac{f(x)}{x}-l\bigg|\leq M$ for some $M\in \mathbb{R}$. So $\bigg|\dfrac{f(x)}{x}\bigg|\leq |l|+M\Rightarrow |f(x)|\leq |x|(|l|+M) \Rightarrow \lim_{x\to 0} f(x)=0$

Let $x_n \rightarrow 0$.

$y_n:= f(x_n)/x_n$, we have

$y_n \rightarrow L.$

With

$f(x_n)=$

$(f(x_n)/x_n)(x_n)=(y_n)(x_n)$.

$\lim_{n \rightarrow \infty }f(x_n)=$

$\lim_{n \rightarrow \infty}((y_n)(x_n))=$

($\lim_{n \rightarrow \infty}(y_n))(\lim_{n \rightarrow \infty}(x_n))=$

$L \cdot 0=0.$


1. ## Divisibility by 6

Hey guys I need help with this question

$\text{prove}
(4^{n}-1)(3^{n}-1) \text{is divisible by 6 for all positive values of n, without Induction}$

and the answer given in the solution to the question is that since

$4^{n}-1=(1+3)^{n}-1 \therefore \text{Divisble by 3}$
$3^{n}-1=(1+2)^{n}-1 \therefore \text{Divisble by 2}$

$\therefore\text{Product is divisble of 2 and 3, therefore divisble by 6}$

I know by substituting in values for n in $(1+3)^{n}-1$ that they all appear divisible by 3, is this a definite rule, is there a proof of this without using induction?

The same goes for $(1+2)^{n}-1$, how do we know that this is divisble by 2 for all values of n without induction

If we can't use induction to prove those divisbility by 3 and 2, then isn't the question kinda invalid?

Thanks

2. 1.
I guess you don't know Newton's generalized binomial theorem...

$(a+b)^n=M_a+b^n=M_b+a^n$
I considered $M_k$ a multiple of k.

So:
$4^n-1=(1+3)^n-1=M_3+1-1=M_3$
$3^n-1=(1+2)^n-1=M_2+1-1=M_2$

2.
$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+ab^{n-2}+b^{n-1})$

So:
$4^n-1=(4-1)(4^{n-1}+4^{n-2}+...+4+1)=3(4^{n-1}+4^{n-2}+...+4+1)$
$3^n-1=(3-1)(3^{n-1}+3^{n-2}+...+3+1)=2(3^{n-1}+3^{n-2}+...+3+1)$

3. Originally Posted by aonin
Hey guys I need help with this question

$\text{prove}
(4^{n}-1)(3^{n}-1) \text{is divisible by 6 for all positive values of n, without Induction}$

and the answer given in the solution to the question is that since

$4^{n}-1=(1+3)^{n}-1 \therefore \text{Divisble by 3}$
$3^{n}-1=(1+2)^{n}-1 \therefore \text{Divisble by 2}$

$\therefore\text{Product is divisble of 2 and 3, therefore divisble by 6}$

I know by substituting in values for n in $(1+3)^{n}-1$ that they all appear divisible by 3, is this a definite rule, is there a proof of this without using induction?

The same goes for $(1+2)^{n}-1$, how do we know that this is divisble by 2 for all values of n without induction

If we can't use induction to prove those divisbility by 3 and 2, then isn't the question kinda invalid?

Thanks
The $\displaystle 3^n - 1$ is easy, without any tricks like below. $\displaystyle 3^n$ is an odd number, so subtracting 1 from it gives us an even number.

$(1 + 3)^n - 1$ is a bit trickier. Note that when we expand by the binomial theorem:

$(1 + 3)^n - 1 = (1^n + n \cdot 1^{n - 1}3^1 + ~...~ + n \cdot 1^13^{n - 1} + 3^n) - 1$

$\displaystyle = n \cdot 1^{n - 1}3^1 + ~...~ + n \cdot 1^13^{n - 1} + 3^n$
of which you can prove (using the combinatorial coefficients) that every coefficient is divisible by 3.

-Dan

4. $(1 + 3)^n - 1 = (1^n + 3 \cdot 1^{n - 1}3^1 + ~...~ + 3 \cdot 1^13^{n - 1} + 3^n) - 1$ - uhm, is not correct.

$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}\cdot a^{n-k} \cdot b^{k}$.

5. Originally Posted by veileen
$(1 + 3)^n - 1 = (1^n + 3 \cdot 1^{n - 1}3^1 + ~...~ + 3 \cdot 1^13^{n - 1} + 3^n) - 1$ - uhm, is not correct.

$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}\cdot a^{n-k} \cdot b^{k}$.
Whoops! I got carried away with the 3's. Thanks for the catch. I have fixed it in my post.

-Dan

6. Originally Posted by aonin
Hey guys I need help with this question

$\text{prove}
(4^{n}-1)(3^{n}-1) \text{is divisible by 6 for all positive values of n, without Induction}$

and the answer given in the solution to the question is that since
$4^{n}-1=(1+3)^{n}-1 \therefore \text{Divisble by 3}$
$3^{n}-1=(1+2)^{n}-1 \therefore \text{Divisble by 2}$
$\therefore\text{Product is divisble of 2 and 3, therefore divisble by 6}$
Note that $\displaystyle A=4^{n}-1=(1+3)^{n}-1=\sum\limits_{k = 1}^n {\binom{n}{k}\left( 3 \right)^k }$
$\displaystyle B=3^{n}-1=(1+2)^{n}-1=\sum\limits_{k = 1}^n {\binom{n}{k}\left( 2 \right)^k }$

Because the index begins with $k=1$ every term in $A$ is a multiple of 3. Similarly every term is B is even.

Hence every term in the product $AB$ is a multiple of 6.


# How to prove that $X$ is a completion of $Y$, for some metric spcaes $(X,d)$ and $(Y,d)$?

In our Real Analysis class we did a proof for existence of completion of metric space. However, I do not understand how do we use it practically. i.e How do I prove that $$X$$ is a completion of $$Y$$, for some metric spcaes $$(X,d)$$ and $$(Y,d)$$?

For example : How do I prove that $$(\mathbb{R},d_1)$$ is a completeion of $$(\mathbb{Q},d_1)$$? Here, $$d_1$$ is the standard Euclidean metric

$$\underline{\text{My attempt}}$$ -

I believe we should prove that $$\mathbb{Q}$$ is subset of $$\mathbb{R}$$ and $$\mathbb{R}$$ is complete. This would mean that all Cauchy sequence in $$\mathbb{Q}$$ converges in $$\mathbb{R}$$.

However, consider this case -

Let $$(W,d)$$ be an incomplete Metric space and $$W \subset Y \subset X$$ and $$X,Y$$ are complete (with same metric).

Now, by my argument both $$X,Y$$ can be the completion of $$W$$. Is this alright? I read somewhere in Stack Exchange that completion is unique(I didn't really understand that answer)

$$\underline{\text{Approach to resolve this}}$$ - We choose the smallest, complete set which contains $$W$$ as its completion. We can prove that, since $$X$$ is complete any of it's closed subset is complete. We can also prove that $$\overline{W}$$ is the smallest closed set that contains $$W$$. Hence, $$\overline{W}$$ is the completion of $$W$$ (this would explain the notation in the completion proof).

Thus in the title - I just have to check if $$Y$$ is dense in $$X$$ to prove that $$X$$ is a completion of $$Y$$.

This would also solve my example question - We know that $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$ and $$\mathbb{R}$$ is complete. Thus $$\mathbb{R}$$ is completion of $$\mathbb{Q}$$.

I am fairly confident that this is correct. However, I am a Physics Master's student this is my first "Real" Math course ;) So I would appreciate if someone let's me know if this right and if not the corrections.

• What is $d_1$? ${}$ – José Carlos Santos Sep 27 '19 at 11:13
• $d_1$ is the standard Euclidean metric – Indigo1729 Sep 27 '19 at 11:29
• This follows by the universal property of completion of metric spaces. – ε--δ Sep 27 '19 at 11:59

To show that $$\mathbb{R}$$ is the Cauchy completion of $$\mathbb{Q}$$, it is not sufficient to show that $$\mathbb{Q}$$ is contained in $$\mathbb{R}$$ and $$\mathbb{R}$$ is complete. In fact, this just means that the completion of $$\mathbb{Q}$$ is contained within $$\mathbb{R}$$.

For example, the completion of $$\mathbb{Q}$$ is contained in $$\mathbb{C}$$ and the complex numbers are complete, but the completion of the rational numbers is not the complex numbers.

You should also verify that every element of $$\mathbb{R}$$ is the limit of a sequence of rational numbers (which follows directly from the fact that $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$ for example) in order to be sure that $$\mathbb{R}$$ is actually $$\mathbb{Q}$$'s completion. This resolves your worry about the uniqueness theorem for completions which you referred to.

One final point: one doesn't normally define the Cauchy completion in order to identify completions of metric spaces they already know (e.g. $$\mathbb{Q}$$ in your example). Instead, knowing that you can take the completion lets you construct new metric spaces which can be useful in their own right. Again for example, knowing that completions exist and taking the completion of $$\mathbb{Q}$$ is a way to construct the real numbers $$\mathbb{R}$$.

One nitpick: "I believe we should prove that $$\Bbb Q$$ is subset of $$\Bbb R$$ and $$\Bbb R$$ is complete. This would mean that all Cauchy sequence in $$\Bbb R$$ converge in $$\Bbb R$$."

To sum up, viewing $$\Bbb Q$$ as a subset of $$\Bbb R$$ (so we can avoid speaking of the embedding map and the image of $$\Bbb Q$$ under it), you need to show that

• the metric on $$\Bbb Q$$ is the restriction of the metric on $$\Bbb R$$
• $$\Bbb Q$$ is dense in $$\Bbb R$$ and of course
• $$\Bbb R$$ is complete under its metric

and that is essentially what you found out by yourself.

You're missing one... critical point!

It is not sufficient for $$X$$ to be a completion of $$W$$ that $$W \subseteq X$$ and $$X$$ is complete. You also need $$W$$ to be dense in $$X$$.

That being said, $$\mathbb R$$ is indeed a completion of $$\mathbb Q$$.


IntMath Home » Forum home » Methods of Integration » Integration Techniques

# Integration Techniques [Solved!]

### My question

You state in Section four that all angles are in radians and that the formulas do not work in degress.

Why do they only work in radians?

### Relevant page

4. Integration: Basic Trigonometric Forms

### What I've done so far

Just wanted to know if there are some integration techniques that work in degrees.

X

You state in Section four that all angles are in radians and that the formulas do not work in degress.

Why do they only work in radians?
Relevant page

4. Integration: Basic Trigonometric Forms

What I've done so far

Just wanted to know if there are some integration techniques that work in degrees.

## Re: Integration Techniques

Radians have the "magical" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus).

Degrees, on the other hand, can only be a measure of an angle (or of course, temperature).

Probably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.

Consider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):

We know the following for the green curve:

d/dx sin(x) = cos(x)

At x=0, the slope is cos(0) = 1

But now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:

At x=0^@, the slope is pi/180 cos(0^@) = pi/180

So if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.

In order for the following formula to "work", x needs to be in radians:

d/dx sin(x) = cos(x)

Hope it helps.

X

Radians have the "magical" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus).

Degrees, on the other hand, can only be a measure of an angle (or of course, temperature).

Probably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.

Consider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):

[graph]310,250;-1,10;-1.1,1.1,1.57,1;sin(x),sin(3.14x/180)[/graph]

We know the following for the green curve:

d/dx sin(x) = cos(x)

At x=0, the slope is cos(0) = 1

But now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:

At x=0^@, the slope is pi/180 cos(0^@) = pi/180

So if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.

In order for the following formula to "work", x needs to be in radians:

d/dx sin(x) = cos(x)

Hope it helps.

## Re: Integration Techniques

It does. A good example to use. Thank you.

X

It does.  A good example to use.  Thank you.


# How do I add the terms in the binomial expansion of $(100+2)^6$?

So, I stumbled upon the following question.

Using binomial theorem compute $102^6$.

Now, I broke the number into 100+2.

Then, applying binomial theorem

$\binom {6} {0}$$100^6(1)+\binom {6} {1}$$100^5(2)$+....

I stumbled upon this step. How did they add the humongous numbers? I am really confused.

Kindly help me clear my query.

• They're powers of ten.... And you don't mean "embezzled." Maybe you mean "confused." Mar 4 '17 at 4:39
• Correct! I need to work on my English skills, man! Thanks for the help tho! Mar 4 '17 at 4:46
• @symplectomorphic - actually, they probably meant "bamboozled", as it means something similar to "confused", but can be easily mixed up with "embezzled". Mar 4 '17 at 7:42

The long way:

\begin{align} & 10^{12} + 12 \cdot 10^{10} + 6 \cdot 10^9 + 16 \cdot 10^7 + 24 \cdot 10^5 +192 \cdot 10^2 + 64 \\ =\; & 10^{12} + (10+2) 10^{10} + 6 \cdot 10^9 + (10+6) 10^7 + (20+4) 10^5 +(100+90 +2) 10^2 + 60+4 \\ =\; & \color{red}1\cdot10^{12} + \color{red}1 \cdot 10^{11} + \color{red}2\cdot 10^{10} + \color{red}6 \cdot 10^9 + \color{red}1 \cdot 10^8 + \color{red}6 \cdot 10^7+\color{red}2 \cdot 10^6 + \color{red}4 \cdot 10^5+\color{red}1 \cdot 10^4+\color{red}9\cdot 10^3 + \color{red}2 \cdot 10^2 + \color{red}6 \cdot 10^1 + \color{red}4 \cdot 10^0 \end{align}

The latter is precisely the representation in base $\,10\,$ of $\;\color{red}{1126162419264}\,$.

• Thanks for the help! Perhaps the most convincing and noteworthy answer. Mar 4 '17 at 10:52

That number is $1\; 12 \; 6\; 16 \; 24\;192\; 64$ (space added for emphasis). Notice a relationship with the coefficients of the powers of $10$?

$10^{12} + 12\times10^{10} + 6\times10^9 + 16*10^7 + 24\times10^5 + 192\times10^2 + 64$ $= 10^{12} + 10^{11} + 2\times10^{10} + 6\times10^9 + 10^8 + 6\times10^7 + 2\times10^6 + 4\times10^5 + 10^4 + 9\times10^3 + 2\times10^2 + 6\times10^1 + 4\times10^0= 1126162419264$

• Thank you very much, sir! I really appreciate it. Mar 4 '17 at 10:51

Since it's all powers of ten you could add those quite easily. If it were something else you could have needed to use some calculator. But here you have just powers of tens, which basically keep everything the same. Here's the number with a few spaces to make you understand it.

$\text{1 12 6 16 24 192 64}$ Those powers just end up just putting those digits in the right order because we use a number system with ten digits.

• Ok! Why not 10112...? Mar 4 '17 at 4:50
• You see say you're adding something like a $12 \times 10^{10}$. You can see that those powers are all spaced evenly. By that I mean to say whenever you have say n digits to be added, over there they had decreased the exponent of 10 by n
– SBM
Mar 4 '17 at 4:53
• Moreover, saying a number equals something like $123.xyz$ is the same thing as $1 \times 10^2 + 2 \times 10^1 + 3 \times 10^0 + x \times 10^{-1} \ldots$ and so on. As long as you keep those evenly spaced, it remains same. By x, y and z I meant arbitrary digits not variables
– SBM
Mar 4 '17 at 4:56
• I got your point. Mar 4 '17 at 5:04

Start adding from the rightmost term($64$); you will notice that the number of trailing zeros on the preceding term is exactly the same as the number of digits in following term. $$...2400000\quad +\\ .....19200\quad +\\ .......64\quad$$ ...and so on till the first.


# Prove by induction that $5^n - 1$ is divisible by $4$.

Prove by induction that $5^n - 1$ is divisible by $4$.

How should I use induction in this problem. Do you have any hints for solving this problem?

Thank you so much.

• What do you know about the question or what have you tried? Dec 6 '13 at 7:10
• I have a vague feeling that this question has already been covered on the site.
Dec 6 '13 at 12:37

We prove that for all $n \in \mathbb{N}$, $4 \mid \left( 5^n-1 \right)$. (Notationally, this says $4$ divides $5^n-1$ with a zero remainder).

1. For a basis, let $n=1$. Then $$5^1-1=4,$$ and clearly $4\mid4$.

2. Assume that $5^n-1$ is is divisible by $4$ for $n=k, \, k \in\mathbb{N}$. Then by this assumption, $$4 \mid \left( 5^k-1 \right) \Rightarrow 5^k-1=4m, \, m \in \mathbb{Z}.$$ (This notationally means that $5^k-1$ is an integer multiple of $4$.)

3. Let $n=k+1$. Then \begin{align*} 5^{k+1}-1 &= 5^k \cdot 5-1 \\ &=5^k(4+1)-1 \\ &=4\cdot 5^k+5^k -1 \\ &=4\cdot5^k+4m\\ &=4\left( 5^k+m \right). \end{align*} Since $4\mid4\left( 5^k+m \right)$, we may conclude, by the axiom of induction, that the property holds for all $n \in \mathbb{N}$.

• Nice, clear, clean... a good answer indeed! Dec 6 '13 at 16:53
• this should be the accepted answer Jan 15 '17 at 5:43

First prove the base case $n=1$. Then induct and make use of the fact that $$(5^{n+1}-1) - (5^n-1) = 4 \cdot 5^n$$to conclude what you want.

Without induction, you can use the identity

$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+...+a+1)$$

Of course you would still need induction or something to prove this identity.

• Yes. I removed my comment once I saw your updated answer. To be really pedantic, even in my answer, we still need induction first to show that $5^n$ is an integer, to then claim that $4$ divides $4 \cdot 5^n$.
– user17762
Dec 6 '13 at 7:24

without induction $$5^n-1=(4+1)^n-1$$ $$=4^n+n4^{n-2}+...+1-1$$ the only term in $(1+4)^n$ not being multiplied by a power of $4$ is $1$ but it disappears due to the $-1$.

Why induction? $5^n$ ends in $\dots25$ for $n>1$, so $5^n-1$ ends in $\dots24$.

• Technically, you still need induction to show that $5^n$ ends in $25$ for $n>1$.
– user17762
Dec 6 '13 at 7:17
• @user17762 $25 * 5 = 25 \mod 100$ Dec 6 '13 at 13:48
• @ratchetfreak Technically I think you still need induction to prove the statement about an arbitrary $n$. Of course, using the fact you mention, the induction becomes trivial (like proving by induction that $1^n = 1$ for all $n$ using the fact that 1 is the multiplicative identity.) Dec 6 '13 at 18:48

$\displaystyle{5^{n + 1} - 1 = \left(5^{n} - 1\right)5 + 4}$

it's even more general:

$k$ divides $(k+1)^n-1$ with $k,n \in \mathbb{N}$

simply by modular arithmetic:

$$k+1 = 1 \mod {k} \\ \Downarrow \\(k+1)^n=1 \mod {k}$$

To prove by induction you:

1. Assume the proposition is true for n

2. Show that if it is true for n, then it is also true for n+1

3. Show that it is true for n=1

Then you know that it will be true for all natural numbers.

In this case:

1. Assume $5^n-1$ is divisible by 4

2. Say $m=5^n$, so $m-1$ is divisible by 4

• $5^{n+1}-1$ = $5m - 1$
• $5m - 1$ = $5(m-1) + 4$
• Since $m-1$ is divisible by 4 and 4 is divisible by 4, then this expression is divisible by 4.
3. For $n=1$, $5^1 - 1 = 4$ which is divisible by 4

And there you are...

This isn't by induction, but I think it's a nice proof nonetheless, certainly more enlightening: $\displaystyle 5^n-1=(1+4)^n-1=\sum_{k=0}^n {n\choose k}4^k-1=1+\sum_{k=1}^n {n\choose k}4^k-1=\sum_{k=1}^n {n\choose k}4^k$ which is clearly divisible by $4$.


1. ## Partial Fractions

2. I believe you forgot (to list in your final answer) the first term of one of the original lines in the problem.

3. My answer is the part in red (Part 3).

4. Originally Posted by r_maths
My answer is the part in red (Part 3).
Hmm, I get $A = 1, B = 2$, which leads to a final answer of: $x+\frac{1}{x+1}+\frac{2}{x+2}$. I do not see how the x term is not supposed to be there.

5. I did include the x

I got the same answer as you, but the answer from the book gives a different answer...

6. Originally Posted by r_maths
I did include the x

I got the same answer as you, but the answer from the book gives a different answer...

I do not see how B equals one, so the book is probably wrong unless both of us missed something.

7. Hello, r_maths!

Express $\frac{x^3+3x^2+5x+4}{(x+1)(x+2)}$ in partial fractions.

$\frac{x^3+3x^2+5x+4}{(x+1)(x+2)} \;=\;x + \frac{3x+4}{(x+1)(x+2)} \;=\;x + \frac{1}{x+1} + \frac{2}{x+2}\quad{\color{red}Right!}$

Book answer: . $x + \frac{1}{x+1} + \frac{1}{x+2}\quad{\color{red}Wrong!}$
$\text{Their answer adds up to: }\;\frac{x^3+3x^2+4x+3}{(x+1)(x+2)}$ . . . . obviously not the original fraction.

$\frac{x}{1}\cdot{\color{blue}\frac{(x+1)(x+2)}{(x+ 1)(x+2)}} + \frac{1}{x+1}\cdot{\color{blue}\frac{x+2}{x+2}} + \frac{2}{x+2}\cdot{\color{blue}\frac{x+1}{x+1}}$

. . $= \;\frac{x(x+1)(x+2) + (x+2) + 2(x+1)}{(x+1)(x+2)}$

. . $= \;\frac{x^3+3x^2+2x + x + 2 + 2x + 2}{(x+1)(x+2)}$

. . $= \;\frac{x^3+3x^2+5x+4}{(x+1)(x+2)}$ . . . . See?

8. $\frac{{x^3 + 3x^2 + 5x + 4}}
{{(x + 1)(x + 2)}} = \frac{{x^3 + 3x^2 + 2x + 3x + 4}}
{{(x + 1)(x + 2)}} = \frac{{x(x + 1)(x + 2) + (3x + 4)}}
{{(x + 1)(x + 2)}}$
.

Now

$x + \frac{{3x + 4}}
{{(x + 1)(x + 2)}} = x + \frac{{(x + 2) + 2(x + 1)}}
{{(x + 1)(x + 2)}} = x + \frac{1}
{{x + 1}} + \frac{2}
{{x + 2}}$
.

Hence $\frac{{x^3 + 3x^2 + 5x + 4}}
{{(x + 1)(x + 2)}} = x + \frac{1}
{{x + 1}} + \frac{2}
{{x + 2}}$
.


# Can we determine summands from their partial sums?

Assume there are non-negative numbers $$\lambda_1\le \ldots\le \lambda_n\in[0,\infty)$$. You are given the (ordered) list $$s_1\le\ldots\le s_{2^n}\in[0,\infty)$$ of all partial sums, i.e. every $$s_i$$ is of the form $$s_i=\sum_{k\in K_i}\lambda_k$$ for some unique but unknown $$K_i\subseteq\{1,\ldots,n\}$$.

Question: Can we determine the $$\lambda_1,\ldots,\lambda_n$$ from knowing the $$s_1,\ldots,s_{2^n}$$?

Some obvious facts are

1. Since there is some $$s_i$$ corresponding to $$K_i=\emptyset$$, $$s_1=\cdots=s_i=0$$.
2. $$\lambda_1=s_2$$, since no non-trivial partial sum can be smaller as the smallest possible summand.
3. $$s_{2^n}=\sum_{k=1}^n\lambda_k$$, since no partial sum can be bigger.
4. For $$n=2$$ the answer is yes, since $$\lambda_1=s_2$$ and $$\lambda_2=s_4-s_2$$.

Note: For my use case it would be sufficient to know if for any given $$s_1\le\cdots\le s_{2^n}$$ there is at most one possibility for $$\lambda_1\le\cdots\le\lambda_n$$.

1. By calculating how many $$s_i$$ are zero, you can determine how many $$\lambda_i$$ are zero. If there are any, they will be creating repeated entries $$s_i$$ throughout, but in a systematic manner where one can eliminate their effect; indeed, if there are $$k$$ zeroes, then there will be $$k$$ extra duplicates of every entry. For simplicity, suppose $$\lambda_1 > 0$$.

2. Now, indeed, $$\lambda_1 = s_2$$.

3. We proceed by induction. Suppose we have identified $$\lambda_1,\ldots,\lambda_j$$ and calculated the partial sums consisting of only $$\lambda_1,\ldots,\lambda_j$$, such as $$\lambda_1+\lambda_3$$. Then the smallest remaining partial sum must be $$\lambda_{j+1}$$. Proof: Otherwise the smallest remaining partial sum would have to be a sum with at least one unknown $$\lambda_m$$ that is (by definition) not yet in the list of known partial sums, which would imply that $$\lambda_m$$ is smaller than the smallest remaining partial sum; a contradiction.

4. Now that $$\lambda_{j+1}$$ is also known, consider the known list of partial sums to include all sums of $$\lambda_1,\ldots,\lambda_{j+1}$$.

Actually, separate treatment of steps 1 and 2 is not necessary; one only has to initialize the list of known partial sums with the empty sum $$s_1=0$$.

This method probably works even if you are missing entires from the list of partial sums, as long as all the missing entries are strictly greater than $$\lambda_n$$.

• One has to be careful: Assume we have $\lambda_1=1,\lambda_2=2,\lambda_3=3,\lambda_4=4,...$ and already computed $\lambda_1,\lambda_2$. The set of known partial sums is $K=\{0,1,2,3\}$, so the smallest partial sum not in $K$ would be $4\ne\lambda_3$. The problem here is that $\lambda_3=\lambda_1+\lambda_2$ is itself a partial sum. I believe that this can be fixed by using multisets of partial sums: In the above situation the multiset of all partial sums would be $\{0,1,2,3,3,4,...\}$, so after removing the multiset $K$ the smallest remaining element would be $3=\lambda_3$ as required. – Robert Rauch Apr 25 at 7:47
• @RobertRauch Indeed, but since you are using an ordered list of partial sums with $2^{n}$ elements, it already contains the information about multiplicity. – Tommi Brander Apr 25 at 7:55

Here is a streamlined version of Tommi Brander's solution: $$\newcommand{\IN}{\mathbb{N}}$$

Lemma: Let $$0\le\lambda_1\le\cdots\le\lambda_n$$ and $$\mathcal{P}\doteq\left\{\lambda_K\doteq\sum_{k\in K}\lambda_k\left|K\subseteq\mathbb{N}_n\right.\right\}$$ the set of their partial sums, where $$\IN_n\doteq\{1,2,\ldots,n\}$$. For $$0\le l\le n$$ consider the function $$m_l:\mathcal{P}\to\IN_0$$ with $$m_l(\lambda)=\#\{K\subseteq\IN_l\mid\lambda_K=\lambda\}$$. Then for all $$1\le l\le n$$, $$$$\lambda_l=\min\left\{\lambda\in\mathcal{P}\mid m_{l-1}(\lambda) In particular, noting that $$m_0(\lambda)=\mathbf{1}(\lambda=0)$$, the $$\lambda_1,\ldots,\lambda_n$$ can be recovered iteratively only from $$\mathcal{P}$$ and $$m_n$$.

Proof: Let $$\mathcal{P}_l\doteq\left\{\lambda\in\mathcal{P}\mid m_{l-1}(\lambda). By construction, $$\lambda_l\in\mathcal{P}_l$$ and therefore $$\lambda_*=\min\mathcal{P}_l$$ exists and satisfies $$\lambda_*\le\lambda_l$$. Since $$m_{l-1}(\lambda_*), there is some $$K\subseteq\IN_n$$ with $$K\not\subseteq\IN_{l-1}$$ such that $$\lambda_*=\lambda_K$$. In particular there is $$r\in K$$ with $$r\ge l$$ and, hence, $$\lambda_r\in\mathcal{P}_l$$. By positivity of the $$\lambda_i$$ we find that $$\lambda_r\le\lambda_K=\lambda_*$$ and therefore $$\lambda_*=\lambda_r$$ by minimality of $$\lambda_*$$. Since $$r\ge l$$ we conclude $$\lambda_l\le \lambda_r=\lambda_*\le\lambda_l$$, thus $$\lambda_*=\lambda_l$$.