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Dr. Carleen Eaton guides you through Algebra 1 with captivating lessons honed from teaching math and science for over 10 years. This course meets or exceeds all state standards and is essential to those having trouble with Algebra in high school or college. Carleen?s upbeat teaching style and real world examples will keep you engaged while learning. She covers everything in Algebra 1 from Linear Expressions to Systems of Equations and Rational Expressions. Along the way she has received multiple "Teacher of the Year" awards and rankings as one of the top instructors in California. Dr. Eaton received her M.D. from the UCLA School of Medicine.
II. Solving Linear Equations From Sentences to Equations 16:05 Addition and Subtraction Techniques 15:24 Multiplication and Division Techniques 15:41 Techniques for Multistep Equations 14:31 When the Variable is on Both Sides of the Equation 20:17 Ratios and Proportion 16:05 Applications of Percents 13:46 More Than One Variable 20:38
III. Functions Relations 16:58 Functions 19:27 Linear Functions 20:15
IV. Linear Functions and Their Graphs Slope and Rate of Change 19:46 Direct Variation 13:54 Slope Intercept Form of an Equation 12:06 Point Slope Form of an Equation 9:07 Parallel Lines and Perpendicular Lines 18:02 | 677.169 | 1 |
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Author Biography
Robert Wrede received his B.S. and M.A. degrees from Miami University, Oxford, Ohio. After teaching there for a year, he attended Indiana University and was awarded a Ph.D. in mathematics. He taught at San Jose State University from 1955 to 1994. He also consulted at IBM, the Naval Radiation Laboratory at Hunter's Point, and with several textbook companies. His primary interests have been in tensor analysis and relativity theory.
The late Murray R. Spiegel received the M.S. degree in Physics and the Ph.D. in Mathematics from Cornell University. He had positions at Harvard University, Columbia University, Oak Ridge, and Rensselaer Polytechnic Institute, and served as a mathematical consultant at several large companies. His last position was as a professor and chairman of Mathematics at the Rensselaer Polytechnic Institute, Hartford Graduate Center. He was interested in most branches of mathematics, especially those which involve applications to physics and engineering problems. He was the author of numerous journal articles and 14 books on various topics in mathematics. | 677.169 | 1 |
Abstract
There is widespread agreement within the scientific and education communities that undergraduate biology curricula fall short in providing students with the quantitative and interdisciplinary problem-solving skills they need to obtain a deep understanding of biological phenomena and be prepared fully to contribute to future scientific inquiry. MathBench Biology Modules were designed to address these needs through a series of interactive, Web-based modules that can be used to supplement existing course content across the biological sciences curriculum. The effect of the modules was assessed in an introductory biology course at the University of Maryland. Over the course of the semester, students showed significant increases in quantitative skills that were independent of previous math course work. Students also showed increased comfort with solving quantitative problems, whether or not they ultimately arrived at the correct answer. A survey of spring 2009 graduates indicated that those who had experienced MathBench in their course work had a greater appreciation for the role of mathematics in modern biology than those who had not used MathBench. MathBench modules allow students from diverse educational backgrounds to hone their quantitative skills, preparing them for more complex mathematical approaches in upper-division courses.
INTRODUCTION
Consensus is growing within the scientific and academic communities that undergraduate biological science curricula are in need of substantial reform (National Research Council [NRC], 2003, 2009; Steen, 2005; Project Kaleidoscope, 2006). Interdisciplinary and multidisciplinary approaches now dominate the forefront of biological sciences research at every level, from nanobiotechnology to ecosystem dynamics, but undergraduate curricula remain essentially disciplinary in nature. Mathematics and quantitative approaches in particular clearly constitute a key tool for modern life scientists, yet the undergraduate curriculum fails to help students appreciate the important relationships between mathematics and modern biology (Gross, 2000; Ewing, 2002; Hastings and Palmer, 2003; Cohen, 2004; Hoy, 2004; Jungck, 2005; Steen, 2005). Biology courses remain largely qualitative and descriptive rather than rigorously quantitative. Lack of quantitative training among our biology majors weakens students at a practical level, whether they plan to analyze environmental impacts (e.g., minimum viable population size) or pursue a career in medicine (e.g., nonlinear drug dose responses, Yuan, 2005).
Currently, biology undergraduates at many four-year institutions have few requirements relative to mathematics, far fewer than those in other sciences such as physics or chemistry. Although biology majors typically take math through calculus, these courses often use nonbiological examples or abstractions (Gross, 1994) and rarely teach other mathematical skills essential for modern biology (NRC, 2003). Many institutions have addressed this by developing special courses (capstone courses, advanced undergraduate seminars, or special research tracks) that introduce elite students to quantitative biology. Such courses undoubtedly produce excellent cutting-edge scholarship, but they are limited to only the strongest students and thus do not help to transform the culture of biology teaching and learning. A stronger approach, and one that is more likely to heal the damaging disjunction between mathematics and biology, is to infuse math throughout the entire undergraduate biology curriculum for all students (Gross, 2004; Labov et al., 2010).
Online modules (free-standing instructional elements, often Web-based, that can be used by multiple instructors and revisited throughout a course sequence) have become a key educational tool to integrate mathematics and biology. MathBench Biology Modules were designed to integrate quantitative approaches and mathematics more deeply into the introductory biology curriculum in a way that would reinforce biological concepts, increase math literacy, and prepare students to be receptive to more complicated mathematical approaches in upper-level courses (Nelson et al., 2009). The MathBench suite consists of 37 self-contained learning modules that use an informal tone to combat math anxiety and leverage the students' pre-existing knowledge of math. We developed the modules in alignment with generally accepted best practices in online instruction, including embedding concepts within storylines (Parker and Lepper, 1992; Mayer, 2008), using informal language (Ross, 1983; d'Ailly et al., 1997; Moreno and Mayer, 2000), proving opportunities for self-calibration (Hannafin and Sullivan, 1996), and striking a balance between demonstrating solutions to problems and allowing students to practice finding the solutions themselves (Hannafin and Sullivan, 1996). Each MathBench module consists of 8–30 interlinked, well-illustrated Web pages focusing on a single quantitative theme (e.g., diffusion, Punnett squares, data visualization, log transformations) plus interactive problems, interactive graphics, and quizzes to measure student comprehension. The quiz questions can be drawn randomly from a larger bank of test questions, allowing for individualized grading opportunities even in large-enrollment classes. MathBench modules are freely available online at mathbench.umd.edu.
MATERIALS AND METHODS
The impact of MathBench Biology Modules on undergraduate student quantitative skills was evaluated during the fall 2009 semester in a large introductory biology course at the University of Maryland, College Park. The course was BSCI 105 Principles of Biology 1: Cell and Molecular Biology, with a total enrollment of 614 students. The course was subdivided into five separate lecture sections of between 42 and 155 students, each taught by a different faculty member. Each lecture section was further subdivided into laboratory sections of approximately 20 students each, which met once each week for 3 h. Although BSCI 105 is designed for biology majors, it is also required in several other degree programs (e.g., chemistry, kinesiology, nutrition) and is taken by students from a wide variety of majors to fulfill medical school application requirements.
A series of nine MathBench modules was incorporated into the BSCI 105 laboratory curriculum. The modules were designed to introduce and reinforce specific quantitative skills and concepts that were identified by our faculty as being important for success in modern biology (Table 1). Modules covered topics such as basic lab techniques (metric system and use of the micropipetor), developing standard curves, logarithmic functions (particularly with respect to understanding pH), calculating molar weight, visualizing three-dimensional structures in two dimensions, probability calculations necessary for understanding BLAST searches, basic graphing (bar graphs, log), and simple statistical analyses. For example, the module entitled "Normal Distributions and the Scientific Method" begins with a brief overview of hypothesis testing, emphasizing the need for measuring outcomes, replicating measurements, and making comparisons. The module then uses a fanciful hypothetical scenario (the role of Fish2Whale fish food in increasing body length of tropical fish) to introduce students to the idea of characterizing variable traits within a population, such as organism size, using mean and SD. An interactive applet imbedded within the module allows students to manipulate these parameters and watch the effect of these manipulations directly on a virtual population of fish and graphically on a distribution of fish body lengths. Students are guided through their explorations with a systematic series of questions. At the conclusion of the module, students have the opportunity to manipulate means and standard deviations independently for two groups of virtual fish, representing control and treatment groups, allowing them to see the impact of these parameters on a scientist's ability to draw robust conclusions from an experiment (Figure 1).
MathBench modules used and specific quantitative skills and concepts emphasized in each
Students were assigned modules at predetermined intervals throughout the fall semester. Each of the nine modules was developed to complement a specific laboratory exercise, so that students made immediate practical use of what they had learned. Modules were completed as homework, outside of class time. Each module was followed by a quiz, which was administered online through the campus course management system. Student quiz scores in total constituted 16% of the laboratory course grade.
Quantitative Skill Survey Design and Administration
To measure changes in quantitative skills over the course of the semester, we developed an 18-question pretest instrument that touched upon nine of our 10 specific quantitative skills (Table 1). We did not create specific questions for the skill "Distill biology into math," as it represented the overarching goal of the project and would presumably be reflected in changes in total scores between the pretest and the posttest. The questions were crafted to allow each quantitative skill to be evaluated along a cognitive spectrum (Mayer, 2002), with one question designed to measure basic understanding and the other designed to reflect a more sophisticated understanding. For each question, students could select one of four multiple-choice options, or they could answer "I do not know how to approach this problem." The "I do not know…" option was included as a way of assessing the students' degree of comfort with solving quantitative problems, whether or not they ultimately arrived at the correct answer. Before its use in this study, the instrument was piloted in introductory biology and genetics classes to establish internal consistency, criterion validity, and content validity (Garson, 2007). A second version, with the same 18-question structure but differing only slightly in numerical details, was created to serve as a posttest. The posttest also asked students to reflect on whether their quantitative skills had improved as a result of the introductory biology course and, if so, which components of the course curriculum were responsible for the improvement. The posttest also contained two open-ended questions that allowed students to provide feedback on the modules and their implementation.
Students in BSCI 105 were asked to complete the pretest during the first two weeks of the semester, before the first MathBench module was assigned. At the end of the semester, following the completion of all MathBench modules and quizzes, they were asked to complete the posttest. Analyses of changes over the course of the semester included only those students who completed both the pretest and the posttest (n = 204). Analyses of self-reported assessments of improvement in quantitative skill used all students who completed the posttest (n = 396).
Survey of Attitudes in Graduates
In an effort to assess longer-term impacts of the MathBench initiative on student attitudes regarding the relationship between math and biology, we added several questions to an existing survey routinely administered to all graduating seniors in the College of Chemical and Life Sciences. Between 2005 and 2009, MathBench modules had been gradually introduced into selected sections of four introductory-level courses typically taken in the first two years of the biological sciences curriculum (Principles of Biology I, II, and III, and Principles of Genetics). Spring 2009 graduates (n = 307) were first asked whether they had used MathBench modules in any of their courses. Because MathBench had only been integrated into the curriculum fairly recently and had not yet been fully implemented, we expected that a large proportion of the graduating seniors would not have encountered it as part of their course work. They were then asked to select the statement that most closely represented their attitude toward math ("I hate math and try to avoid it," "I don't like math but I can cope with it," "I don't care about math one way or another," "I like math but I don't seek it out," or "I like math and enjoyed having course work that included math") and to describe their attitude regarding the relationship between math and biology ("Math is not relevant to biology," "Math can be useful in biology but it's not really necessary," "Math is helpful in biology," "Math is essential in biology if you want to do cutting-edge work," or "Math is essential for doing any biology, cutting-edge or not"). Attitudes of students who had encountered MathBench in their course work were then compared with those who had not.
Data Analysis
Pre- and posttest data were matched for all students completing both tests. We augmented the data set with the students' math eligibility level (determined by a math eligibility test administered by the university at orientation and completion of college mathematics course work) and whether or not they were concurrently enrolled in a math class. All statistical tests (multiple analysis of variance [MANOVA], Pearson product-moment correlation, Chi-square) were performed using JMP version 8.0.2.
RESULTS
Students showed significant improvement in quantitative skills over the course of the semester (pretest x = 7.3 correct out of 18, posttest x = 10.4 correct; repeated measures MANOVA F1196 = 125.4040, p < 0.0001). There was a significant effect of math eligibility on pre- and posttest scores, with students eligible for precalculus having the lowest scores and students eligible for calculus 3 or higher having the highest scores (F3196 = 11.0634, p < 0.0001; Figure 2). Nonetheless, the magnitude of improvement in scores between the pre- and posttest was independent of math eligibility (F3196 = 0.4764, NS). Students who were concurrently enrolled in a math class showed slightly, but significantly, greater improvements in scores than those not concurrently enrolled in math (F1196 = 6.2077, p < 0.05; Figure 3). Overall, the improvement in score from pre- to posttest was negatively correlated with pretest score, indicating that students with lower initial scores made more dramatic gains (Pearson product-moment correlation, r = −0.657, n = 204, p < 0.0001).
Changes in scores across the semester on an 18-item quantitative assessment for introductory biology students concurrently enrolled in a math class (n = 79) or not concurrently enrolled in a math class (n = 125).
Students did not show uniform improvement across all questions. When the questions were sorted by the proportion of students who answered them correctly, the largest gains were seen in the 12 questions that were most likely to be answered correctly on the pretest (i.e., the "easiest" questions; Figure 4). Little or no improvement was seen on the six questions that the students were least likely to have answered correctly on the pretest (the "hardest" questions).
Proportion of students answering correctly on the pre- and posttests for each of 18 questions on a quantitative skills assessment, with questions sorted by the proportion of students answering correctly on the pretest.
Math eligibility significantly affected level of risk aversion, as measured by the frequency with which students responded "I don't know how to approach this problem." Overall, 75% of students answered "I don't know…" to at least one pretest question, whereas 29% so answered on the posttest. Students having low math eligibility were most risk averse (repeated measures MANOVA F3196 = 5.6801, p < 0.01). As a whole, students were less risk averse after completing the MathBench modules, as evidenced by a significant decrease in the frequency of "I don't know…" responses between the pre- and the posttests (F1196 = 80.578, p < 0.0001). The greatest decreases in risk aversion were seen in the students with the lowest math eligibility (F3196 = 4.7257, p < 0.005) and those concurrently enrolled in a math course (F1196 = 4.0114, p < 0.05).
In the posttest, students were asked whether they had experienced an improvement in quantitative skills as a result of the course. Most students indicated that they had experienced "little" or "moderate" improvement (Figure 5). Of the students who reported improvement in quantitative skills (367 of 396, or 93% of students who completed the posttest), 71% attributed this improvement to the MathBench modules (Figure 6).
Student comments to the open-ended question, "What role did the MathBench modules have in the development of your scientific content knowledge and quantitative skills?" were overwhelmingly positive (83% of responses). Many students (31%) indicated that the modules helped them refresh their understanding of material learned in previous courses and high school. "The MathBench Biology Modules helped my content knowledge and quantitative skills by reviewing previous math skills I learned and elaborating on these skills to take my understanding of science-related math to another level." A small percentage of students (9%), largely those with more advanced mathematics backgrounds, found the modules too informal and simplistic. However, there were many positive comments from other students who appreciated those features (e.g., "its humorous style of writing and funny images make it easier to study without any pressure," "MathBench modules improved my quantitative skills because they made it very easy to see what the math was, they had good examples and explained all the math visually which really helped me"). Specific design features that students found helpful were its interactivity, opportunities for practice, immediate self-assessment feedback, and the ability of students to proceed at their own pace. "It helped to have interactive examples within the modules that allowed me to test my knowledge throughout the lesson." "I found it easier to go at my own pace with the modules rather than being in a classroom setting where I find it difficult to comprehend the material as quickly as the other students." Some indicated that the modules increased their level of engagement with the class (e.g., "I enjoyed the MathBench modules because during the week it kept me constantly active with the Bio class"). Many students specifically mentioned that they had a greater appreciation for the role of math in the biological sciences (e.g., "I now better understand the connection between math and Biology").
About half (51%) of the Spring 2009 graduates had encountered the modules in their course work. Regardless of their exposure to MathBench, a large majority of graduates (74%) either enjoyed or actively sought mathematical content in their course work. Graduates who had used MathBench were significantly more likely to describe the relationship between math and biology as "essential to cutting-edge biology" or "essential for all fields of biology" (χ2 = 13.7, df = 2, p < 0.001; Figure 7).
Responses of 2009 biological sciences graduates (n = 307) to a survey question regarding their attitude regarding the relationship between math and biology.
DISCUSSION
MathBench Biology Modules were effective in introducing and reinforcing quantitative concepts and skills into a large-enrollment introductory biology course. Over the course of the semester, students showed significant increases in quantitative skill that were independent of previous math course work. Specific attributes that students felt contributed to the development of their quantitative skills included interactivity, use of humor and informal language, opportunities for practice, immediate self-assessment feedback, and the ability of students to proceed at their own pace. This integration of mathematical approaches into biology courses appears to have lasting effects on student attitudes, as graduating seniors who encountered MathBench in their introductory course work reported a greater appreciation of the essential role of mathematics in the biological sciences than those who had not used MathBench.
These techniques are also thought to help the approximately one-third of college students who have trouble with quantitative work because of "math anxiety" (Betz, 1978). In our study, students with less-well-developed quantitative skills at the beginning of the semester (i.e., low math eligibility) were significantly more likely than those with stronger quantitative skills to avoid answering specific problems on the quantitative assessment. By the end of the semester, they showed increased comfort with solving quantitative problems, whether or not they ultimately arrived at the correct answer. Many of their comments on the open-ended assessment questions indicated that they felt more confidence in applying math to biological problems. This is heartening, especially in light of recent data that show that students with weak quantitative skills are wary of attempts to infuse mathematical approaches into introductory biology courses (Matthews et al., 2009).
Although students showed improvement across a range of quantitative skills and concepts, little to no improvement was seen in the "hardest" items on the assessment instrument. This result is not entirely surprising, given that the modules evaluated in this study were developed specifically for use in an introductory level course and therefore were designed to provide a basic level of conceptual understanding. It remains an open question whether this online modular approach is similarly effective in deepening quantitative understanding in more advanced course work. There is some evidence that the effectiveness of various pedagogical techniques differs depending on whether the intended audience is composed mainly of "novice" or "expert" learners, with many best practices shifting markedly for a more experienced audience (the so-called "expertise reversal effect," sensu Clark and Mayer, 2008).
Online modules are now widely used in quantitative biology. Some were written before technological tools for interactivity were widespread. Others, (e.g., ESTEEM and COMAP) include minimal explanatory text, couch explanations in formal language, and are best suited for upper- division students. MathBench modules differ in their approach, using simple, clear, and intuitive explanations of the math involved to help students bridge the gap between biology and mathematical formalism. As such, MathBench modules are complementary with many existing modules—indeed, other modules largely begin where MathBench leaves off. The modules' easy accessibility makes it possible for them to be disseminated widely in a variety of educational contexts, and their design is inclusive of diverse learning styles and educational backgrounds, making them a useful tool for preparing students for the complex mathematical approaches that have become essential to modern biology.
ACKNOWLEDGMENTS
We are grateful to all the faculty who helped develop ideas for the modules and used them in their classes. We also thank Leslie Ries, Katie Schneider, Tanya McLean, David Boothe, Aleksandra Ogurtsova, Awais Malik, Li Zhu, and Mike Landavere for their help in developing the modules and GoEun Na for assistance with pre- and posttest validation. Two anonymous reviewers provided feedback that greatly improved the manuscript. Support for this project was provided by a Howard Hughes Medical Institute Undergraduate Science Education Program grant to the University of Maryland; a National Science Foundation Course; Curriculum, and Laboratory Improvement grant (DUE–0736975); an instructional improvement grant from the University of Maryland Center for Teaching Excellence; and the College of Chemical and Life Sciences.
REFERENCES
American Association for the Advancement of Science, 1993. American Association for the Advancement of Science. New York: Oxford University Press; 1993. Benchmarks for Science Literacy. | 677.169 | 1 |
Mathematics for the people
Month: August 2011
Today I'm working on the first-semester calculus course I am teaching this fall. Changes to the course this fall include a new textbook and the use of Khan Academy. I have selected a new textbook from BVT publishing to keep the costs to the students down. I will be using Khan Academy to give the students a self-directed way to review algebra and pre-calculus material. Many students come into calculus unprepared to deal with the algebraic manipulations needed in calculus and lacking in their understanding of functions and function notation. Unfortunately, we don't have enough time in the semester to address all of the areas in which the students need help, so I hope this will help them to gain the competency they need to be successful.
As I did in the spring semester, I am requiring a significant amount of reading from the students, both of the textbook and of the trade paperback Calculus Diaries by Jennifer Ouelette. This semester, however, I am having students post their reading responses in a Facebook group, so we'll see how that experiment goes!
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Today I'm working on prepping the first week of group theory and abstract algebra. It's been interesting work. In game theory we're going to start with combinatorial game theory and we'll get to play a lot of games. That is fun and stimulating for the brain, but I also have to decide how much of the subject and methods of combinatorial game theory we should be learning together. What's the right balance between having interesting experiences and discussions together and learning formal theory?
In abstract algebra, my challenge this semester is to start at a place that the students can reach. Abstract algebra is unlike any other course most of my students have taken, and typically they feel very uncomfortable during the semester. This semester our first week will be spent on set theory with some interesting examples. This isn't the most exciting place to start the semester, but I hope to give the students enough tools to move forward without undue anxiety.
I'm also going to be doing "minute papers" at the end of each of these weekly classes this semester to get a sense of where the students are and what they need from me and from the class. I'll also do a more evaluation about three or four weeks in, and I'm going to supplement the traditional course evaluation with an end-of-term evaluation of my own.
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This week I am working on getting ready for my fall classes — creating syllabi to publish online, figuring out pacing and rough content, deciding how I will be grading and structuring classes. It's fun work and gets me excited about the semester.
I'm teaching a new course this semester in game theory. We're going to cover both combinatorial game theory and "regular" game theory, and the course is aimed at a beginning level. My biggest struggle right now is figuring out what I can reasonably accomplish in this topic with this audience. We're going to start the semester with combinatorial game theory, so I hope to get them playing lots of games from the start. About half-way through, we'll switch to the more traditional game theory topics. In this second part of the course, my aim is to have a lot of interesting discussions rather than to be too technical — I want to get them thinking about rational decision-making in a broad sense and give them an idea of how game theory addresses decision-making and strategies.
I'm also teaching abstract algebra and calculus I this semester, and those courses are in preparation as well. And I still have plenty of time to get distracted in making websites. | 677.169 | 1 |
CHIP AND NEW TECHNOLOGIES Visa Smart Debit/Credit Certificate Authority Public Keys Overview The EMV standard calls for the use of Public Key technology for offline authentication, for aspects of online
Phonics at a glance Phonics is Knowledge of the alphabetic code (26 letters, 44 phonemes, 140 different letter combinations) + Understanding of the skills of segmenting and blending Letters an d phonemes
DegreeWorks 4.09 Student Manual DegreeWorks is an online advising tool that helps Undergraduate students select courses from requirements that need to be fulfilled based on their academic plan. Accessing
Copthill School Information for Parents on Read Write Inc. What is the purpose of this handout? To inform parents of how reading is taught at Copthill To supply parents with clear information on the RWI
5 Page(s) withheld entirely at this location in the file. One or more of the following El(b)(1) (b)(7)(a) (05) (b)(2) (b)(7)(b) 0)(2 ) (b)(3) (b)(7)(c) (k)(1) (b)(7)(d) (k)(2) Eg(b)(7)(E) (k)(3) (b)(7)(f)Chapter 6 Eponential and Logarithmic Functions Section summaries Section 6.1 Composite Functions Some functions are constructed in several steps, where each of the individual steps is a function. For eample,Areas The area of ounded plane regions an e defined to the following rules: (A0) The area inside a square of side length 1 m is 1 m 2. (A1) An ongruent polgons enlose equal areas. (A2) The total area of
Math 31 Homework 1 Solutions Last modified: July 15, 01 This homework is due on Thursday, July 1th, 01 at 1:10pm Please turn it in during class, or in my mailbox in the main math office (next to 4W1) Please
Vehicle Identification Numbering System 00.03 IMPORTANT: See Subject 050 for the vehicle identification numbering system for vehicles built before May 1, 2000. Federal Motor Vehicle Safety Standard 115 | 677.169 | 1 |
5.2.1 Distance, Speed, and Time Relationships Using Simulation Software
Grade:
3-5
Standards:
Math Content:
Algebra
This example includes
a software simulation of two runners along a track. Students can control
the speeds and starting points of the runners, watch the race, and examine
a graph of the time-versus-distance relationship. The computer simulation
uses a context familiar to students, and the technology allows them to
analyze the relationships more deeply because of the ease of manipulating
the environment and observing the changes that occur. Activities like
this can help students in the upper elementary grades understand ideas
about functions and about representing change over time, as described
in the NCTM Algebra Standard.
This e-example contains the original applet which conforms more closely to the pointers in the book.
Follow-Up Questions and Tasks
Think about and discuss the following: What does the graph show? Did what happened match your prediction? If it did, how does the graph show what you predicted? If not, why do you think what happened was different from what you expected?
Click on Get Ready to position the boy and the girl to start a new race. Make a change in one of your settings (e.g., the length of the girl's stride or the boy's starting position). How will this change affect the graph? Run the simulation again and see what happens. Continue making changes and predicting the result. After each run of the simulation, think about what the graph shows and think about what happened and why.
Discussion
This example illustrates computer software that engages students in the upper elementary grades in ideas about functions and about representing change over time. The software and examples in this activity are based on the Trips software (Clements, Nemirovsky, and Sarama 1996). This software allows students to analyze change by setting the starting positions and length of stride (speed) for two runners. Students then observe the simulated races as they happen and relate the changing positions of the two runners to dynamic representations that change as the events occur. Students can predict the effects on the graph of changing the starting position or the length of the stride of either runner. They can observe and analyze how a change in one variable, such as length of stride, relates to a change in speed. This computer simulation uses a familiar context that students understand from daily life, and the technology allows them to analyze the relationships in this context deeply because of the ease of manipulating the environment and observing the changes that occur.
In this activity, students are working with functional relationships. As students work with this example, they need to be encouraged by the teacher to analyze how a change in the starting position or the length of the stride will affect the time needed to reach the finish lines. Acting out different stories about the "trips" can help students visualize the effect of, for example, increasing the length of the stride or having one runner start in a position ahead of the other runner. As students become familiar with the simulation, they can analyze each situation numerically by building a table showing the relationship between time and distance. By inspecting the track, the graph, and the table, students can become more precise in reasoning quantitatively about the relationships ("The length of the boy's stride is 2, so you know his distance by multiplying the time by 2"). Older elementary school students can relate the boy's and girl's trips proportionally ("The girl goes twice as far as the boy in the same amount of time"). Students can begin to describe rate of change informally by inspecting the slope of the line ("The girl's line is steeper because she is moving faster"). Interpreting two-variable graphs will be unfamiliar to many students in this age group. Part of the teacher's role is to help them connect what is happening on the graph to what is happening on the track: How long does it take for the boy to go the same distance as the girl has traveled in fifty "seconds"? How can you see this demonstrated on the track? On the graph? Where on the track does the girl catch up to the boy? Where is this point on the graph?
Additional Tasks and Questions
Set the starting position and length of stride for both runners. Run the simulation. Now write a story that describes the trip. For example, "The girl is going really fast. She catches up to and passes the boy, who is going slow," or "The girl started way behind the boy, who was already halfway to the tree by the time she got going. She went really fast and caught up to him more and more. Finally, at 75 she passed him and kept going really fast and got to the tree first."
Three motion stories are told below. Before the students use the simulation, have them physically simulate the motion stories (with their bodies). Then develop specific instructions (starting position and length of stride for each runner) to produce the action in the stories. Try out the instructions using the computer simulation above.
Motion Story 1. The boy and girl start from the same position. The girl gets to the tree ahead of the boy. Motion Story 2. The boy starts behind the girl. The boy gets to the tree before the girl. Motion Story 3. The boy starts at the tree and the girl starts at the house. The boy gets to the house before the girl gets to the tree.
Look at the two graphs below, which show the results of different motion stories. Develop a set of instructions to produce each trip.
Take Time to Reflect
Do you
think students would enjoy using this computer activity?
Why or why not? What are they likely to focus on?
How
can teachers help students become comfortable
moving among various techniques for organizing and
representing ideas about relationships and
functions?
What
important ideas about functions and representing
change over time can students learn while working on
this activity?
Acknowledgement
This activity and applet were adapted with permission from the Trips software, Clements, Nemirovsky, and Sarama (1996). The activity was adapted with permission from Tierney et al. (1998). | 677.169 | 1 |
Calculus: Worksheet (Study Guide) for Optimization Problems
Be sure that you have an application to open
this file type before downloading and/or purchasing.
69 KB|3 pages
Product Description
Here is a worksheet that lists the 8 steps needed to complete and optimization problem: constraints, domain, objective function, solving/substituting, taking the derivative, doing the test for max/min, and finally writing an answer statement for the problem.
Included is a step-by-step problem that utilizes the 8 steps. I encouraged my students to do every one of their problems on this worksheet so that they would have the steps clearly in their mind. When giving an exam, I did not provide the worksheet. | 677.169 | 1 |
Preparing students for college math
Giving students necessary math skills for jobs
Showing how math is connected/applied in the real world
To ensure students understand basics of algebra, geometry, and pre-calculus
Igniting passion in students for learning/understanding math | 677.169 | 1 |
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From sines and cosines to logarithms, conic sections, and polynomials, this
friendly guide takes the torture out of trigonometry, explaining basic concepts
in plain English, offering lots of easy-to-grasp example problems, and adding a
dash of humor and fun. It also explains the "why" of trigonometry, using
real-world examples that illustrate the value of trigonometry in a variety of
careers.
Mary Jane Sterling (Peoria, IL) has taught mathematics at Bradley University
in Peoria for more than 20 years. She is also the author of the highly
successful Algebra For Dummies (0-7645-5325-9). | 677.169 | 1 |
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In sixth grade mathematics, a variety of concepts are explored and reviewed. The course goal is to emphasize problem solving through the practical application of mathematical concepts. The curriculum covers number theory, statistical analysis, data organization, properties of polygons, operations with fractions, decimals and percents, ratios and proportions, probability, and area and perimeter, as well as lessons on number sense, divisibility rules, order of operations, exponents, integers, 2-dimensional geometric vocabulary, and the use of variables in solving basic algebraic equations.
Students in 7th grade mathematics focus on improving problem solving skills as well as working on their ability to explain and perform mathematical operations. This course is for students who are working at grade level in mathematics. The curriculum is problem-centered, and important mathematical concepts are embedded in engaging problems. The students will develop understanding and skills as they explore the problems individually, in a group, or with the class. The topics covered include patterns, an introduction to algebra with the use of variables, rational numbers, fractions and decimals, integers, ratios, proportions and percents, data analysis, probability, area, perimeter and volume.
Students do have the opportunity to take an accelerated course focuses on developing pre-Algebra skills. Important mathematical concepts are embedded in engaging problems. The students will develop understanding and skills as they explore the problems individually, in a group, or with the class. The topics covered include variables and equations, integers, ratios, proportions and percents, geometry, area, perimeter, and volume, data analysis, probability, linear equations and graphs, and multi-step equations.
The eighth grade pre-Algebra curriculum covers a wide variety of topics. Our goal is to prepare the students for a successful transition to high school mathematics courses. A variety of topics are covered in this course including operations with integers, evaluating algebraic expressions, solving one and two step equations and inequalities, graphing in the coordinate plane, understanding slope and y-intercept, operations with rational numbers, ratio, rates, proportion, percent, exponents and powers, surface area and volume, geometry, probability, and properties of parallel lines and related angles.
Students in 8th grade Algebra complete a comprehensive course in Algebra I. This course covers the same material that is covered in the high school curriculum. This course explores the use of variables; linear equations, slope and equations for lines, exponents and powers, quadratic equations and square roots, polynomials, solving systems of linear equations, and factoring, including the use of the quadratic formula. Additionally, the course will include a unit of geometry focused on parallel lines cut by a transversal, angle relationships, and constructing polygons using a ruler, compass and protractor. | 677.169 | 1 |
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Summary
Written for graduate students in mathematics or non-specialist mathematicians who wish to learn the basics about some of the most important current research in the field, this book provides an intensive, yet accessible, introduction to the subject of algebraic combinatorics. After recalling basic notions of combinatorics, representation theory, and some commutative algebra, the main material provides links between the study of coinvariant—or diagonally coinvariant—spaces and the study of Macdonald polynomials and related operators. This gives rise to a large number of combinatorial questions relating to objects counted by familiar numbers such as the factorials, Catalan numbers, and the number of Cayley trees or parking functions. The author offers ideas for extending the theory to other families of finite Coxeter groups, besides permutation groups | 677.169 | 1 |
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the common core mathematics companion the standards teachers in their understanding and learning of the common core mathematics standards. The common core mathematics companion the standards using the common core mathematics companion as core mathematics companion the standards . extends the popular common core mathematics companion the common core mathematics companion the standards decoded companion the standards . By linda m gojak and ruth harbin miles copublished with corwin in one short year since the release of k 2 and 3 5 companions to the common core state . Acknowledgments letter to grades 3 5 teachers letter to elementary school principals introduction a brief history of the common core | 677.169 | 1 |
Get the extra practice you need to succeed in your mathematics course with this hands-on Student Workbook. Designed to help you master the problem-solving skills and concepts presented in INTRODUCTORY ALGEBRA: AN APPLIED APPROACH, 9th Edition, this practical, easy-to-use workbook reinforces key concepts and promotes skill building.
Author Biography
Joanne Lockwood received a BA in English Literature from St. Lawrence University and both an MBA and a BA in mathematics from Plymouth State University. Ms. Lockwood taught at Plymouth State University and Nashua Community College in New Hampshire, and has over 20 years' experience teaching mathematics at the high school and college level. Ms. Lockwood has co-authored two bestselling developmental math series, as well as numerous derivative math texts and ancillaries. Ms. Lockwood's primary interest today is helping developmental math students overcome their challenges in learning math. Richard Aufmann is the lead author of two bestselling developmental math series and a bestselling college algebra and trigonometry series, as well as several derivative math texts. He received a BA in mathematics from the University of California, Irvine, and an MA in mathematics from California State University, Long Beach. Mr. Aufmann taught math, computer science, and physics at Palomar College in California, where he was on the faculty for 28 years. His textbooks are highly recognized and respected among college mathematics professors. Today, Mr. Aufmann's professional interests include quantitative literacy, the developmental math curriculum, and the impact of technology on curriculum development. | 677.169 | 1 |
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In this eBook, you will learn about the terms used in Polynomials and learn how to add and subtract Polynomials. You will also enjoy the ride of multiplying and dividing Polynomials. Grab your pen; it's going to be interesting! This guide is written with the love of math, detailing each step just like it will be explained in a personalized one-on-one teaching session.
Algebra 2 - Dividing Polynomials
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This book is a collection of essays centred around the subject of mathematical mechanization. It tries to deal with mathematics in a constructive and algorithmic manner so that reasoning becomes mechanical, automated and less laborious. The book is divided into three parts. Part I concerns historical developments of mathematics mechanization, especially in ancient China. Part II describes the underlying principles of polynomial equation-solving, with polynomial coefficients in fields restricted to the case of characteristic 0. Based on the general principle, some methods of solving such arbitrary polynomial systems may be found. This part also goes back to classical Chinese mathematics as well as treating modern works in this field. Finally, Part III contains applications and examples. Audience: This volume will be of interest to research and applied mathematicians, computer scientists and historians in mathematics.
A fixed-parameter is an algorithm that provides an optimal solution to a combinatorial problem. This research-level text is an application-oriented introduction to the growing and highly topical area of the development and analysis of efficient fixed-parameter algorithms for hard problems. The book is divided into three parts: a broad introduction that provides the general philosophy and motivation; followed by coverage of algorithmic methods developed over the years in fixed-parameter algorithmics forming the core of the book; and a discussion of the essential from parameterized hardness theory with a focus on W 1]-hardness, which parallels NP-hardness, then stating some relations to polynomial-time approximation algorithms, and finishing up with a list of selected case studies to show the wide range of applicability of the presented methodology. Aimed at graduate and research mathematicians, programmers, algorithm designers and computer scientists, the book introduces the basic techniques and results and provides a fresh view on this highly innovative field of algorithmic research. *Author: Niedermeier, Rolf *Series Title: Oxford Lecture Series in Mathematics and Its Applications *Series Number: NO. *Binding Type: Hardcover *Number of Pages: 312 *Publication Date: 2006/03/30 *Language: English *Dimensions: 9.48 x 6.34 x 0.86 inches
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Quantum weep: "Like the scent of burnt toast...."
Physics is hard. And that difficulty associated with physics leads to a general lack of understanding amongst a lot of people. And if we're talking about quantum physics you can amplify this as much of the theory is non-intuitive, even illogical. It isn't deterministic, its probabilistic meaning we can perform the same experiment twice in exactly the same conditions. Einstein, for one hated the idea that the universe maybe governed by probability. It led to one of his most famous quotes "God does not play dice". Add to this quantum theory isn't complete, there are varying interpretations, and where there is ambiguity, superstition can be squeezed in. To someone who has discovered the elegance and beauty of physics, the general lack of understanding among... Now on to something that ANGERS those who have actually put in the time and tremendous effort it takes to study physics. People exploiting the general public's ignorance of physics to support supernatural bullshit. They learnt advanced calculus, polynomials and quadratics and binomial theorem. They've stayed up until two or three in the morning agonising over a problem they just can't quite define. Let me be clear, and I say this in no uncertain terms. THERE IS NO AREA OF PHYSICS THAT IN ANYWAY SUPPORTS THE SUPERNATURAL. Let's look at a few common claims then a specific one. In the interest of fairness I won't say where or who posted it. It just happens to be quite a good example of some of the claims that are used to connect the paranormal and quantum physics. Wow, that's a lot of quantum related words thrown about with, unfortunately, very little understanding behind them. To be fair, Chris is just repeating common claims here, he has likely obtained them from another source and just accepted to be true, he's not really passing himself off as an expert, but he is circulating ignorance. First of all there is no "Law of observation". What I think Chris is referring to here is the observer effect. As in the very act of observing a physical process changes the outcome of that process. Now if you think of that in everyday terms, or the macroscopic world, that concept would be pretty stunning. Imagine if observing an oak tree grow caused to grow differently. Or change the outcome of a chemical reaction. One might reasonably conclude that something in the act of viewing, perhaps consciousness itself. has effected that process. That's what Chris has concluded, but here is the problem with that reasoning:. The observer effect as Chris seems to mean is seen only quantum and particle physics (the observer effect also applies in thermodynamics and electronics, in both cases its a direct effect of the instruments). Taking particle physics as an example, to observe an electron its necessary to bombard the electron with photons. The electron must interact with the photon, which naturally changes its state as the energy state of system depends on the energy of individual electrons, which changes as a result of. photon emission and absorption. Absorption of photons of particular energies moves an electron from its ground state to a corresponding excited state, emission moves it from an excited state to a lower excited state or a ground state,. Thus act of observing a quantum state, defined by observables such as energy, position and momentum, has changed that state, thus it changes all possible future states. One thing is clear:. Consciousness isn't a factor. To understand why see don't see these effects in the macroscopic world we have evolved to comprehend, consider this: One wouldn't expect the influx of photons to effect an oak tree because statistically we are talking about a lot of atoms, a lot... Another concept that Chris touches on is entanglement. Again this is a phenomena not seen on a macroscopic level, and one that Chris introduces with an immediate and fatal misunderstanding. Entanglement doesn't show that ALL THINGS are connected. Pairs or groups of particles are entangled when they. Source: Skeptic's Boot: The Rational Paranormal
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These figures were calculated by dividing the total union dues income by the total membership unions reported on their disclosure forms. Table 1 and most other tables in this report present the data in two ways: weighted and unweighted. The unweighted
Un-Common, Not Core
They also can't add fractions or do long division, which puts them at a severe disadvantage when they must add rational expressions or divide polynomials. Common Core exacerbates this problem. At every level, the problems are designed to be too hard to
The Golden Curve: Determining player value in freemium apps
The more complex your function is – i.e. combining three classes of functions and having tons of parameters or a polynomial of degree eight – the better it will fit to your data. However, this also means it will perform poorer for out-of-sample
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An approach to estimating prognosis using fractional polynomials in metastatic renal carcinoma
The MFP model allowed us to divide patients into four risk groups achieving median ... is extended to the first-order fractional polynomial or FP1 function Sauerbrei and Royston (1999) developed the MFP (multivariable fractional polynomial) approach ...
An approach to estimating prognosis using fractional polynomials in metastatic renal carcinoma
Hyperbolic homogeneous polynomials, oh my!
What do topology and combinatorics and n-dimensional space have to do with addition, subtraction, multiplication, and division? Yet there remains ... number solutions for several systems of homogeneous polynomial equations describing hyperbolic surfaces. | 677.169 | 1 |
Test ProblemsCan they? Many people take algebra for granted, whereas there is quite a lot of skill to manipulating an equation correctly, which only comes naturally after a lot of practice. And, if you manage to manipulate an equation correctly, then I'd say you were using theory and fundamental concepts!I'm sure you're capable. Maybe your math is just a little weak, but that will improve with time and practice. Everybody has their strengths and weaknesses. However, usually solving quadratics is not too big of a deal...it sounds like your math is just a bit weak and you need some practice. NMOS nmosYup. Did the same thing man. I actually got a C in calc I, II, and III . Then every math class after that I've had an A. My foundation (algebra) was horrible. It's better now, definitely not great. I just needed practice, and it sucks that the calc series had to be my algebra practice. Just hang in there... it will get better, but do yourself a favor and practice over winter and summer break. It will make your life easier.
i am grading calculus now and the problem that my students are having the hardets time with is fiunding the arc length of (3/4)x^4/3 - (3/8)x^(2/3).
basically because they cannot do the algebra to show that
[x^(1/3) - (1/4)x^(-1/3)]^2 + 1 = [x^(1/3) + (1/4)x^(-1/3)]^2
i explained over and over that (a-b)^2 + 4ab = (a+b)^2, so that if ab = 1/4, as it does in all these problems, then (a-b)^2 + 1 = (a+b)^2 is a perfect square. no use. more than half of a stronger than average class cannot get it. this is basic junior high algebra.
Now I think it's definitely the math part that discourages me. I can set up all the physics equations and can see how it should be solved theoretically. I think I lack the skill or manipulating equations as well as I would like. I just never thought it could be my math, since I do well in relatively advanced math classes, but I think you guys have a point about the importance of and how many people neglect algebra. | 677.169 | 1 |
Introduction
Welcome to Grade 11 - Functions, MCR3U.
This course has four strands; Characteristics of Functions, Exponential Functions, Trigonometric Functions, and Discrete Functions. There is a lot of new material and in order to do well, you must practice; that means, do your homework, and ask questions as soon as difficulties arise. | 677.169 | 1 |
Geometry, Relativity and the Fourth DimensionThis is a highly readable, popular exposition of the fourth dimension and the structure of the universe. A remarkable pictorial discussion of the curved space-time we call home,
About the Book: Self-Working Number Magic: 101 Foolproof TricksClear instructions for 101 tricks and problems, many based on important math principles. Calculations have been concealed; tricks are carefully streamlined for quick understan
About the Book: Introductory Graph TheoryGraph theory is used today in the physical sciences, social sciences, computer science, and other areas. Introductory Graph Theory presents a nontechnical introduction to this exciting field in a c
About the Book: Mathematics for the NonmathematicianIn this erudite, entertaining college-level text, Morris Kline, Professor Emeritus of Mathematics at New York University, provides the liberal arts student with a detailed treatment of m
About the Book: Ussr Olympiad Problem BookOver 300 challenging problems in algebra, arithmetic, elementary number theory and trigonometry, selected from the archives of the Mathematical Olympiads held at Moscow University. Most presuppose
About the Book: Concepts of Modern MathematicsSome years ago, -new math- took the countrys classrooms by storm. Based on the abstract, general style of mathematical exposition favored by research mathematicians, its goal was to teach stud
About the Book: Game Theory: A Nontechnical IntroductionA lucid and penetrating development of game theory that will appeal to the intuition . . . a most valuable contribution.- -- Douglas R. Hofstadter, author of Godel, Escher, Bach The
About the Book: Combinatorial Optimization: Algorithms and ComplexityThis book brings together in one volume the important ideas of computational complexity developed by computer scientists with the foundations of mathematical programming
About the Book: Calculus: An Intuitive and Physical ApproachApplication-oriented introduction relates the subject as closely as possible to science. In-depth explorations of the derivative, the differentiation and integration of the power
About the Book: Functions and GraphsThis volume presents students with problems and exercises designed to illuminate the properties of functions and graphs. The 1st part of the book employs simple functions to analyze the fundamental meth
About the Book: Counterexamples in AnalysisThese counterexamples, arranged according to difficulty or sophistication, deal mostly with the part of analysis known as -real variables, - starting at the level of calculus. The first half of t
About the Book: Mathematicians DelightThe main object of this book is to dispel the fear of mathematics. Many people regard mathematicians as a race apart, possessed of almost supernatural powers. While this is very flattering for success
About the Book: Basic Algebra Vol 1A classic text and standard reference for a generation, this volume and its companion are the work of an expert algebraist who taught at Yale for two decades. Nathan Jacobsons books possess a conceptual
About the Book: Topology and Geometry For PhysicistsDifferential geometry and topology are essential tools for many theoretical physicists, particularly in the study of condensed matter physics, gravity, and particle physics. Written by p
About the Book: What Is the Name of This Book?: The Riddle of Dracula and Other Logical PuzzlesIf youre intrigued by puzzles and paradoxes, these 200 mind-bending logic puzzles, riddles, and diversions will thrill you with challenges to y
About the Book: Concise History Of MathematicsThis compact, well-written history -- first published in 1948, and now in its fourth revised edition -- describes the main trends in the development of all fields of mathematics from the first | 677.169 | 1 |
Math
Mathematics Department Overview
The study of mathematics is an integral component of a variety of academic and vocational disciplines. All mathematics curricula were redesigned in 2016 to meet the various needs of Diman Regional Vocational Technical High School students to prepare them for MCAS 2.0, PSAT/SAT, and Accuplacer in accordance with NCTM and Common Core guidelines. Graduates who are continuing their education by enrolling in a two year or four year institution, enlisting in the military, or entering the workforce will be well prepared for the future after completing the comprehensive four year mathematics program.
In the News
Diman's nascent Math team multiplied their success at the vocational math meet, earning 2nd, 4th and 6th place out of 28 teams from 13 schools. At last year's event, the teams placed 9, 10, and 14 out of 23 teams. (4/11/17)
Two advisors, plus thirteen students, divided by merely one month of preparation equaled academic success for Diman on Monday, November 23. That's when three teams of Diman students competed in a vocational school math meet at Blue Hills Regional and earned 9th, 10th, and 14th place versus 23 other teams from 9 other schools. (12/2/2015)
Pi. It's the transcendental number "3.14…" used to calculate the area, perimeter, and volume of circles and cylinders. Every year, math departments and enthusiasts celebrate "Pi Day" on March 14 (3/14), but this year's celebration will be bigger, because the date will be 3/14/15, which corresponds to the next two digits of pi (3.1415) and only happens once a century. (3/5/2015)
A quadcopter drone, hovering 190 feet in the air and equipped with a digital camera, had perhaps the best view of the more than 1,500 Diman Regional Vocational Technical High School students and teachers who stood below holding square orange and white cards. (3/13/15) | 677.169 | 1 |
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Unformatted text preview: Math 216: Differential Equations Lab 3: Higher-Order Numerical Methods, Linearity and Superposition Goals In this lab we have two goals: we will first examine some new numerical methods to find approximations to the solutions to differential equation: that is, we will test-drive some new numerical solvers. We will implement an improved Euler's method, and will use Matlab's built in numerical solver which is an implementation of the Runge-Kutta method. Our second goal is to use these solvers to explore solutions to a linear and a nonlinear differential equation. This will allow us to see how superposition works (for linear equations) and doesn't (for nonlinear ones). Application: a pendulum The angular position of a simple pendulum can be described by differential equations. Con- sider the pendulum shown below. If θ measures the angular displacement of the pendulum bob from the vertical, then θ 00 ( t ) + g L sin( θ ( t )) = f ( t ) , (1) where g is the acceleration due to gravity, L is the length of the pendulum, and f ( t ) is any forcing imposed on the system. For this lab we will consider f ( t ) = 0. Note that this equation is conspicuously nonlinear ! As a result, we are unable to solve it in terms of elementary functions. Because of this, we often use its linearization, θ 00 ( t ) + g L θ ( t ) = f ( t ) , (2) which is valid for small angles θ and is obviously much easier (possible!) to solve. For this lab, we will take f ( t ) = 0, g = 9 . 81 m/s 2 , and L = 1 m (quite a big pendulum). 1 Prelab assignment Before arriving in the lab, answer the following questions. You will need your answers in lab to work the problems, and your recitation instructor may check that you have brought them. These problems are to be handed in as part of your lab report. First, consider the linearized equation modeling the pendulum's motion, (2), with the values indicated: θ 00 ( t ) + 9 . 81 θ ( t ) = 0 . 1. Suppose that we pull the pendulum back an angle of π 4 and then release it. Write the initial conditions that correspond to this physical situation, and then solve the linearized pendulum equation with these initial conditions. 2. Next, suppose that instead of pulling the pendulum back and releasing it, we give it a tap so that it has an initial velocity, say 2 m/s. Write out the initial conditions for this physical situation, and then solve the linearized pendulum equation with these initial conditions. 3. Now consider the combination of these initial conditions: we pull the bob of the pen- dulum back an angle of π 4 and also push it outwards with a speed of 2 m/s. Write down the initial conditions in this case. Then note that you can write down a solution to this initial problem using your work in the preceding two questions. Why are you able to do this? What is the solution to this initial value problem?...
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This note was uploaded on 02/08/2012 for the course MATH 216 taught by Professor Gavinlarose during the Winter '02 term at University of Michigan-Dearborn. | 677.169 | 1 |
Hi all, I just got started with my example of math trivia class. Boy! This thing is really tough! I just never seem to understand the point behind any topic . The result? My grades go down . Is there any expert who can lend me a helping hand?
That is true , there are programs that can assist you with homework. I think there are a few types that help you solve math problems, but I read that Algebrator stands out amongst them. I used the program when I was a student in College Algebra for helping me with example of math trivia, and it always helped me out since then. In time I understood all the topics, and after a while I was able to solve the most difficult of the tasks without the program. Don't worry; you won't have any problem using it. It was designed for students, so it's very easy to use. Basically you just have to type in the topic and that's it .Of course you should use it to learn math , not just copy the answers , because you won't learn that way.
Even I made use of Algebrator to understand the basic principles of Intermediate algebra a month back. It is worth putting the money in for the the purchase of Algebrator since it offers quality coaching in Remedial Algebra and is available at an affordable rate. | 677.169 | 1 |
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Get the confidence and math skills you need to get started with calculusAre you preparing for calculus? This hands-on workbook helps you master basic pre-calculus concepts and practice the types of problems you'll encounter in thecourse. You'll get hundreds of valuable exercises, problem-solving shortcuts, plenty of workspace, and step-by-step solutions to every problem. You'll also memorize the most frequently used equations, see how to avoid common mistakes, understand tricky trig proofs, and much more. Pre-Calculus Workbook For Dummies is the perfect tool for anyone who wants or needs more review before jumping into a calculus class. You'll get guidance and practical exercises designed to help you acquire the skills needed to excel in pre-calculus and conquer the next contender-calculus. Serves as a course guide to help you master pre-calculus concepts Covers the inside scoop on quadratic equations, graphing functions, polynomials, and more Covers the types of problems you'll encounter in your courseworkWith the help of Pre-Calculus Workbook For Dummies you'll learn how to solve a range of mathematical problems as well as sharpen your skills and improve your performance.
Yang Kuang, PhD, is a professor of mathematics at Arizona State University.
Michelle Rose Gilman is the co-author of Pre-Calculus For Dummies.
Introduction.
Part I: Setting the Foundation: The Nuts and Bolts of Pre-Calculus.
Chapter 1: Beginning at the Very Beginning: Pre-Pre-Calculus.
Chapter 2: Real Numbers Come Clean.
Chapter 3: Controlling Functions by Knowing Their Function.
Chapter 4: Searching Roots to Get the Degree.
Chapter 5: Exponential and Logarithmic Functions.
Part II: Trig Is the Key: Basic Review, the Unit Circle, and Graphs.
Chapter 6: Basic Trigonometry and the Unit Circle.
Chapter 7: Graphing and Transforming Trig Functions.
Part III: Digging into Advanced Trig: Identities, Theorems, and Applications. | 677.169 | 1 |
Geometry Course Overview:
This mastery-based geometry class integrates discussions, cartoons, anecdotes, examples, and plenty of exercises. With a greater emphasis on problems, rather than long introductory instructions, this is a great text for learning-through-doing.
Students will be challenged to reason mathematically and apply that reasoning to a wide range of real-world problems. This focus on mastery builds a solid foundation for future math, science, and engineering classes.
Curriculum Links:
Please note: I've chosen the "Honeybee" third edition of Jacobs Geometry. It is virtually the same as the pink "Math Tools" printing of the third edition; Master Books has simply removed a few "references to smoking, drinking, Martians, clairvoyance" et cetera. If you already own the pink "Math Tools" version (ISBN 9781619991095) and Answers (ISBN 9781619991163), they'll probably work fine.
Required Equipment:
A scientific (or graphing) calculator, protractor, and sturdy compass
Syllabus Outline: | 677.169 | 1 |
Algebra of calculus: Display Function in Textbook Style, Calculate Secant Line at an x-value of 'a' with a Distance of 'h, Calculate Tangent Line at an x-value of 'a', Plot all Three on Same Graph, Edit 'a' and/or 'h' or Function and 'a' and 'h'.
The scripts file in compiled form for MathStudio 6, MathStudio Express Apps and Web Version of 'mathstud.io'. Clicking the 'Use in Math Studio' button will open a blank untitled Math Studio (Web Version) file. In the MathStudio (Web Version) to use the last eleven global functions, shift enter to add a script function @f1(x) for field 1 and an expression in x for field 2, press enter. The first eight global functions do not require the @f1(x) script function. You can also use the source code to make your own scripts file. Note: scripts file will not work with older versions of standalone App. Older versions of standalone App must use clsVol1 (Source Code) file for making an 'Include' entry. Standalone process for writing the 'Include' entry's syntax varies with the platform. For example Android OS uses the source file's name in the Include(source filename) entry with the file being in the same directory. Window OS file must be compiled to a 'mathcode' extension and then a long pathname (available from its MathStudio menu) is used inside double quotes. Apple iOS8 MathStudio 6 clicks on the [Include] button found in the source code listing and the MathStudio Express use a Compile option from the Menu option followed by inserting an entry that says Include(Compiled filename) with no quotes. If you did the 'Include' correctly for the platform all the library functions will now be listed.
View a MathStudio Algebra Fundamental's Share directly by clicking the Share icon on a MathStudio web App screen and then clicking a MathStudio Algebra Fundamental Share or by clicking on one of our above Algebra Fundamental's Share links. Next, select the 'View in MathStudio' button. View the entries and related comments. Select an entry and toggle it between decimal and exact by pressing enter [return]. Modify individual entries and press enter [return] to see the results. Add entries, delete entries, toggles all entries between decimal and exact solutions by using the Add Entry, Delete Entry, Evaluate All commands found using the MathStudio menu (to the left of New) in the title bar. For any of our Shares, when you click the Math Studio menu item you will need to scroll the section of the screen up to see the Add Entry, Delete Entry, Evaluate All commands.
MathStudio 6 (Web Version) - Click in the screenshot of a Share to see all of the author's Shares.
MathStudio 6 (Web Version) - Click on the mathstud.io web App menu's 'MathStudio' (to the left of 'New') to see the Commands, 'Insert Entry', 'Delete Entry', and 'Evaluate All'. If the current file has lots of entries you will need to scroll to the top of this section. 'Insert Entry' adds an entry above the entry containing the cursor. 'Delete Entry' deletes the entry that contains the cursor. 'Evaluate All' recalculates all the entries in the current file. Using 'Evaluate All' toggles all entries answers between traditional math notation (fractions, square root, etc) and decimal form.
MathStudio (All Versions) - To see multiple answers on the same line use a pair of bracket, '[ ]', with MathStudio commands inside and separated by commas.
At our publisher's website, ComputerLearningService.com there are free two Chapter samples of the eBook, TI-Nspire Guide Algebra Fundamentals. The free samples are in Kindle, epub, and pdf formats. At YouTube there are free interactive comprehensive videos for all the eBook chapters. The sample chapters, YouTube videos, plus the complete eBook can provide you with additional practice screens and traditional manual techniques that are necessary for complete understanding and proper use of MathStudio's power.
The MathStudio illustration for each file in the Share are from version 6 of the iPad MathStudio App. The upload files for the were created with version 5 of MathStudio's Windows App. Therefore the downloads of the files should work in all version 5 and current versions of MathStudio standalone Apps.
See this thread's Opening Post 'clsVol1' file description to see how to create the 'Include' entry for the MathStudio Web Version and various MathStudio Apps platforms. The 'Include' statement allows us to use the source code global library functions in a MathStudio file.
Algebra Fundamentals Basic Rules of Algebra, Part 1 & 2 - Location xxxx refers to the location of the screen in our Kindle eBook, TI-Nspire Guide Algebra Fundamentals. You will find additional details for the TI-Nspire operations in the eBook plus manual solutions that can be used with either the TI-Nspire screens in the eBook or our MathStudio screens in our Shares.
Welcome to Math Studio's discussion forum. It is hoped that our Shares and this Discussion forum can serve as a vehicle to help you with learning math,
There is no one road to the learning of mathematics. Different approaches for different learners are needed to take learners to where they want to go. The goal of the Shares and discussion forum is give you the flexibility of various approaches aided by the use of MathStudio to reach your goals. | 677.169 | 1 |
-Fractional Evolution Equations and Inclusions: Analysis ....Fractional evolution inclusions are an important form of differential inclusions within nonlinear mathematical analysis. They are generalizations of the much more ...--Fractional Processes and Fractional-Order Signal ....Fractional processes are widely found in science, technology and engineering systems. In Fractional Processes and Fractional-order Signal Processing, some complex ...--Fractional Differential Equations: An Introduction to ....Fractional Differential Equations: An Introduction to Fractional Derivatives, Fractional Differential Equations, to Methods of Their Solution and Some of Their ...--The Fractional Calculus; Theory and Applications of ....The Fractional Calculus; Theory and Applications of Differentiation and Integration to Arbitrary Order: Amazon.it: Keith B. Oldham, Jerome Spanier: Libri in altre lingue- | 677.169 | 1 |
Also: Graph Paper; TI-83 or 84 graphing Calculator. The graphing calculator is a good investment for further math classes, and we will learn to use many of the features in this course.
Tuition: $490 per year Materials Fee: $15
Geometry (1 unit) - Jerry Jones Tuesdays & Thursdays 10:00-11:00
Prerequisite: Algebra 1.
Description: While continuing to review some Algebra 1 concepts, this course will develop familiarity with two- and-three dimensional objects and their properties, while making drawings and visualizing real-world applications. It includes the study of parallel lines, similar and congruent figures, triangles, polygons, circles, area and volume, an introduction to constructions, proofs, and graphed figures. Trigonometry is introduced. Deductive and inductive reasoning as well as investigative strategies in drawing conclusions are emphasized. May be taken after Algebra 2, but designed to be taken between Alg.1 & Alg.2.
Also: Protractor, Graph Paper; TI-83 or 84 Calculator. A compass is provided by the teacher.
Tuition: $495 per year Materials Fee: $15
Algebra 2 (1 unit) - Jerry Jones Tuesdays & Thursdays 11:00-12:00
Prerequisite: Algebra 1 and basic Geometry knowledge.
Description: Review of geometry and a further development of algebra is integrated with probability, statistics, graphing a wide array of functions (linear, polynomial, radical, exponential, and conic sections), matrices, logarithms, analytic geometry & trigonometry concepts. This course is powerful preparation for college and entrance exams.
Description: Advanced Algebra and trigonometry concepts are more fully developed (graphing non-linear systems, logarithms, and probability), and we introduce Calculus basics (limits, functions, differentiation, integration). This is a kinder, gentler calculus course with plenty of examples; (unlike most college and public school courses).
Grades:8-12 Description:This challenging 1.5 credit English, Ancient History, or Bible course glorifies God by comparing and contrasting the ideas of the Greek and Roman poets and philosophers with the teachings of the Bible. Students read and analyze classical texts from Homer, Job, Jonah, Esther, Plato, Aeschylus, Sophocles, Euripides, Aristophanes, Virgil, the Apostle Paul, Augustine, various early church fathers, Shakespeare, and others as literature and history, concerning their contribution to the culture of the Western world in the light of the Bible. Students will explore pagan-classical and Judeo-Christian worldviews of God, the gods, the human condition, and examine the resultant cultures. Classical mythology is examined as man's attempt to explain the order, structure, and origin of the universe. These explanations are contrasted with the truth as found in the Hebrew and Christian Scriptures. The Bible is presented in a literal, grammatico-historical sense. Writing assignments vary from study guides to short essays and quizzes.
This class utilizes the website extensively. There, students will find Thelma audio for almost every book and play. Students can read along in their text as Mrs. English reads aloud and explicates the text. Utilizing Thelma audio has made Classics available to many who were previously unprepared for this level of academic study.
Textbooks: Students purchase their own books. See booklist and bundle prices at
Tuition:$574 Materials Fee: $40
OLD TESTAMENT SURVEY (1 credit) - Thelma English
Wednesdays 12:30-2:30
Grades: 8-12 (grade 7 with permission: strong readers)
Description: In this one-credit English, Ancient History, or Bible course, students will read most of the Old Testament (20 chapters per week), exploring exegetical insights and archaeological evidence, to understand the Old Testament narrative as a whole. Students will learn a basic outline of ancient history, introductions to each book, and become familiar with the Hebrew and Greek alphabets to enable them to use scholarly resources. Short weekly writing assignments are designed to reinforce basic elements of the reading: map skills, mini-biographies, dating key figures and events, question sets, definitions, lists of judges and kings, short poems, Psalms explication, artistic expression options, and memorization.
Writing Fundamentals (1 unit) – Natalie Trust Tuesday 11:00-12:30 Grades: 8-12 Description: Students will learn different writing formats beginning with a basic paragraph & moving to five paragraph essays & then to research papers. MLA format will be taught & used to document research. Class will include a unit on creative writing as well. Textbooks: Provided in class Tuition: $400 Materials Fee: $30
INTERPERSONAL COMMUNICATION (1 unit)- Natalie Trust
Tuesday 9:30-11:00
Grades: 8-12 Description: With our ever-increasing dependence upon texting and various forms of social media to communicate with others, it is vital we develop and maintain excellent interpersonal skills. Students will be introduced to the importance of excellent communication skills within the settings of: family, workplace, & our society. The class will study the development of the self-concept, non-verbal
Description: Students in Acting and Understanding Shakespeare will be introduced to a variety of plays and sonnets written by the Bard himself. Through scene studies, monologues, and textual analysis, students will gain a deeper understanding of Shakespeare's contribution to the world of theatre and will learn how to master the challenging language of the Elizabethan era.
Students will:
· Act with other students and perform in groups and solo
· Learn how to analyze and embody Shakespeare's unique linguistic form
· Learn the basics of acting through vocal and movement work
· Gain a broader knowledge of Shakespeare's collection of works
· Present a selection of sonnets, scenes, and monologues at the end of the year in a showcase
Description: Dive into the human body as never before! This fun-filled class will look into the wonders of the human body, including anatomy, health and nutrition, physiology, diseases, survival skills and a unit on gender differences. The amusing "Blood and Guts" book will keep you enthralled and the hands-on experiments will keep you on your toes (like an exciting hands-on dissection!) Did I mention we're going to have a unit on Survival Skills? So many tangible life skills to learn in this class- you don't want to miss it!
Textbook: TBD
Tuition: $470 Materials Fee: $15
Physical Science (1 unit) - Jerry Jones
Wednesday 8:00-10:00
Prerequisites: none.
Description: Course begins with the basic building blocks of matter and measurement. We will introduce physics, work and energy, thermodynamics, electricity, magnetism, sound, light, optics, atoms, chemical compounds, mixtures and solutions. Every class meeting will include laboratory experiences, and reporting. This course should precede Chemistry and Physics.
Description: This challenging and exciting course is designed to give a broad overview of biology, covering topics such as zoology, botany, genetics, ecology, virology, and much more. There will also be a strong emphasis on human anatomy and physiology. The course is lecture/discussion oriented, with complimentary lab exercises, dissections, and potential field trips as well. From microscopic organisms to giant sequoias, this course will deepen a student's appreciation for and understand of this incredibly complex world that God created.
Description: All general topics of Chemistry will be covered, to give the student ample preparation for future college-level courses. The metric system will be used, and unit conversion will require some algebraic manipulation. Every class meeting will include laboratory experiences, and reporting.
Description: All general topics of Physics will be covered, to give the student ample preparation for future college-level courses. Five weeks will be devoted to basic electronics. The metric system will be used, and some problems will utilize basic trigonometry. Every class meeting will include laboratory experiences, and reporting.
Description: Starting in the Middle Ages and going through the 20th century, this course is a lively tour of the history of Western classical music. We will utilize extensive listening activities, art appreciation/history, and in-depth studies (including writing assignments) of key composers and their works. This course uses historical and cultural perspectives to trace how religious, social, and political changes influenced music (and vice versa), and how history and classical music/art are intertwined. We will also discover how many coffee beans Beethoven had to have per cup of coffee. From the deposition of the last Western Roman Emperor, to the Napoleonic Wars, to the birth of Jazz, this course covers over 1500 years of classical music and your ears will never be the same!
Textbook: two (2) textbooks:
1) Listen and You Will See: A Brief History of Classical Music by Nathaniel Rodrigues (available at Blurb.com)
2) A History of Western Music7th Edition (this edition is available for less than $20 online) by J. Peter Burkholder, Donald Jay Grout, Claude V. Palisca
Tuition: $325 Lab/Materials Fee: $40
World History & Geography for High School (I unit) - Randy Souders Tuesday, 11:00-1:00
Grades: 9-12 Description: History is fun again! Enjoy lectures, textbooks, films and discussions about our world of today and yesteryear. There will be papers, weekly quizzes, and a film. We will use the Bob Jones World History and Geography for High School. Study Early Civilizations, Romans, Vikings, Early Exploration, maps of the world and countries of the world.
Jr. High United States and World History & Geography Survey (I unit) - Randy Souders Tuesday, 1:00-2:30
Grades: 7-8 Description: History can be fun again! Enjoy lecture, book, films and discussions about of world of today and yesteryear. There will be papers, weekly quizzes and a final. We will have texts for each world history and geography using Bob Jones for Christian Schools. Study Early Civilizations, Romans, Vikings and early exploration, World Wars, Human Rights, and maps and countries of the world.
Description: This class is an Art Lab. The students may pursue a choice of curriculum including Jr High or High School Drawing, Principles of Design, Color in Art or Watercolor. We can also design a curriculum to suit your student. I do require Drawing or a strong knowledge of drawing as a prerequisite for the other classes. For more information please feel free to contact me. I will send a supply list after we have chosen a curriculum for your student and I have received your registration. | 677.169 | 1 |
Product details
AUTHOR
SUMMARY
This work teaches the basic principles of mathematics and applies them to cases that paramedics face in the field. Chapters cover maths rules and principles, ratios, proportions and conversion factors, fractions, decimals and percentages. Practice problems are scattered throughout.Mithriel Salmon is the author of 'Paramedic: Calculations For Medication Administration', published 2008 under ISBN 9780763746834 and ISBN 07637468 | 677.169 | 1 |
Relations and Functions
Be sure that you have an application to open
this file type before downloading and/or purchasing.
1 MB|4 pages
Product Description
Vocabulary on relations, functions, domain, range, and vertical line test. Examples of domain and range, vertical line test, and determining if a relation is a function. There is then practice on these topics. Relationships are represented as ordered pairs, tables, mapping diagrams, and graphs. A visual representation of a pop machine and the connection with functions.
I suggest using this as a note guide during the lesson. It could also be used as a worksheet for homework. | 677.169 | 1 |
Find a Pescadero look forward to hearing from you.In Algebra 1 students will learn how to identify and do the operations on real numbers. Also, they will learn to know how to solve and graph the linear equations, inequalities and system of two equations and inequalities. In Algebra 2 students will learn how to identify the mathematical concepts such as real and complex numbers as well as absolute value | 677.169 | 1 |
Introducing Systems of Equations Powerpoint
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this file type before downloading and/or purchasing.
448 KB|18 pages
Product Description
This powerpoint uses inspiration from the story the Tortoise and the Hare to model systems of equations. Students use this scenario to analyze when the Hare will pass the Tortoise. The scenario adjusts to model when the two will never pass and when the two will always be in the same location. Students then try modeling their own scenario using a Ferrari and a Minivan. This conceptually based powerpoint helps students to understand what is happening conceptually in a scenario, graphically, tabular and algebraically. This has been the basis for many great discussions in class. | 677.169 | 1 |
Maths
Maths is the only language spoken by everyone in the world and all young people who aspire to have a salaried job in the future need to master the subject of maths. Maths is used in nearly all aspects of life and at the Academy we aim to prepare students in the five branches of the National Curriculum:
· Functional Skills – mathematical processes and applications
· Number
· Algebra
· Geometry and measures
· Statistics
Key Stage 3- Years 7 and 8
The transition process between Key Stage 2 and Key Stage 3 is given high priority in the first two weeks when students enter the Academy. Students are given a diagnostic test to identify their strengths and weaknesses after which intervention programmes are put in place to help narrow any gaps in their learning. In Year 7, students are taught in mixed-ability groups with six sessions of Maths per week, covering:
· Number
· Algebra
· Shape, space and measure/ geometry
· Data handling
· Using and applying mathematics
During years 7 and 8 pupils are assessed once every term (with an end of year exam during the summer term) and this informs any set changes. Based on their end of year exam in year 7, pupils are placed into sets at the beginning of year 8.
Key Stage 4- Years 9 – 11
All pupils follow the Edexcel Linear programme in mathematics at key stage 4. Pupils are taught to develop both calculator and non-calculator skills and apply them to real-life scenarios to improve their functional skills. The linear programme is assessed in June with two examinations of equal weighting (one calculator, one non-calculator). Our students will sit two mock exams, the first in December and the second in March.
Key Stage 5- Years 12 and 13
In Year 12 the mathematics course is formed of three parts: two core modules and one applied module. In the two core modules, students build on their knowledge and understanding from GCSE topics including algebra, geometry and trigonometry but also explore new topics such as calculus. In the applied module, students use their mathematical skills in real-life scenarios: examples in 'statistics' include learning to analyse trends from data and the assessment of probabilities, whilst in 'mechanics' students learn to solve problems using concepts such as force, velocity, power and centres of gravity.
In Year 13 three further modules are taken: the core modules challenge and deepen understanding of topics met in year 12 mathematics. Students also have the opportunity to study any applied module they didn't take in year 12 (either statistics or mechanics). The course is solely examination-based: there is no coursework or controlled assessment. Each of the six modules will be assessed through a 90 minute examination and students sit three exams in each year.
Maths Gym
To help students who are struggling with maths we have established a 'Maths Gym' where pupils are given individual support, during one/ two periods a week, with the focus on the four key operations (addition, subtraction, multiplication and division). | 677.169 | 1 |
`[A] professor will sometimes prepare for a lecture t by writing some tes or browsing through the book but by lounging in the coffee room with his colleagues and bemoaning (a) the shortcomings of the students, (b) the shortcomings of the text, and (c) that professors are overqualified to teach calculus. Fortified by this yoga, the professor will then proceed to his class and give a lecture ranging from dreary to arrogant to boring to calamitous...The good news is that it requires more effort, more preparation, and more time to be a good teacher than to be a bad teacher. The proof is in this booklet.' ---from the Preface Lively and humorous, yet serious and sensible, this book is a practical guide to the teaching of mathematics. Eschewing generalities, Krantz emphasizes specifics---from how to deal with students who beg for extra points on an exam to mastering blackboard technique to how to use applications effectively. In addition, the book also deals with such sensitive subjects as cheating, bribery, and sexual harassment. Those teaching collegiate mathematics for the first time will find Krantz's advice especially helpful, and more experienced instructors will appreciate the book's elucidation of the fine points of excellent teaching. This book is intended for graduate students preparing for a career in college teaching of mathematics, mathematics instructors and professors.
Key Features
Author(s)
Steven G. Krantz
Publisher
American Mathematical Society
Date of Publication
28/10/1993
Language
English
Format
Paperback
ISBN-10
082180197X
ISBN-13
9780821801970
Subject
Education & Teaching
Publication Data
Place of Publication
Providence
Country of Publication
United States
Imprint
American Mathematical Society
Out-of-print date
02/10/2009
Dimensions
Weight
170 g
Height
230 | 677.169 | 1 |
Naive Lie Theory
ISBN-10: 0387782141
ISBN-13: 9780387782140 Until recently, lie theory has been reserved for practictioners, with no lie theory for mathematical beginners. This book aims to fill that gap and it covers all the basics at a level appropriate for junior/senior level undergraduates | 677.169 | 1 |
Mathematics
We believe in the importance of taking a rigorous approach to the teaching of pure maths.
We have learned from university academics that school-leavers with a good grounding in pure maths are better prepared for university – because they are ready to apply their knowledge in a variety of different disciplines and choose a preferred specialism.
As a result, our International GCSE Mathematics qualification has an emphasis on algebra, with the inclusion of calculus and matrices – making it an excellent preparation for A-level.
Our International AS / A-level Mathematics qualification also has an emphasis on pure maths, which forms 100% of Paper 1, as well as 50% of Paper 2 at AS and over 60% of the overall assessment at A-level.
Our International AS / A-level Further Mathematics qualification stretches the most able students and also has proportionally the same focus on pure maths as our International AS / A-level Mathematics qualification.
Each of these qualifications is rigorous, but also accessible. We put particular attention on achieving clarity in our assessments, so students succeed on the basis of their mathematical knowledge and not on their literacy | 677.169 | 1 |
in Life, Society, and contemporary approach to liberal arts math breaks away from traditional instruction and moves towards a more "modern" course that stresses rich ideas, little review, and more visualization. This reader-friendly book offers an accessible writing style and mathematical integrity. Its unique three-part organization (Life, Society, the World) presents readers with sound, relevant mathematics, leaving them with the (correct) impression that math is useful and affects their lives in many positive ways. | 677.169 | 1 |
Buy Algebra with a single home user licence NOW for just £25 + VAT (Click the orange 'Buy Now' button.)
If you like our Algebra CD-ROM, why not try its companion Trigonometry?
Algebra enables pupils to hone their maths skills with endless practice. The software offers guidance with on-screen help. The program then enables pupils to work step-by-step towards the correct answer. This program, written by teachers David Benjamin and Justin Dodd, is far and away the best available for developing algebra skills.
The sixteen sections of Algebra cover all the algebra on the curriculum.
Key Skills Removing Brackets Substitution Mensuration
Linear Functions One Stage Equations Two Stage Equations Equations with x on one side Equations with x on both sides | 677.169 | 1 |
Overview
If you need to know it, it's in this book!Math Workout for the SAT, 3rd Edition shows you exactly what to expect on the math portion of the SAT and includes:
• A thorough review of all SAT math topics from algebra to statistics • Key SAT math strategies and a breakdown of common SAT math mistakes • 50 quizzes and problem sets so you can practice your skills • Detailed answers and explanations for each practice problem
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Product Details
Meet the AuthorCustomer Reviews
Most Helpful Customer Reviews
Math Workout for the SAT, 3rd Edition 1 out of 5based on
0 ratings.
1 reviews.
Anonymous
More than 1 year ago
The first word that comes to mind when describing this book is inadequate. This book is the very soul and
definition of inadequate. Chapters are set up by attempting to explain a topic, but rarely provides a
comprehensive summary. Entire concepts are ignored completely. It is assumed that the student already know
everything about every topic.
The are only two or three practice problems in each section. The answer's explanations are perpetually
insufficient.
The one redeeming quality it might have is the practice problems in the back. If a student already knows
Algebra and Geometry very well, he or she might find such problems useful in familiarizing him or herself with
the types of problems found on the SAT.
However, I would not recommend anyone who is even slightly unsure about their own math skills to use this
book as a primary study guide. The sparse answer explanations and the poor summary of the topics are
supremely unhelpful. | 677.169 | 1 |
MATH ONLINE COURSES FOR COLLEGE CREDIT
Math online courses for college credit
TEST math online courses for college credit much interest will
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dimensional HarvardLinearAlgebra: Foundations to Frontiers (LAFF) is packed full of challenging, rewarding material that is essential for mathematicians, engineers, scientists, and anyone working with large datasets. Students appreciate our unique approach to teaching linearalgebra because:
In this course, you will learn all the standard topics that are taught in typical undergraduate linearalgebra courses all over the world, but using our unique method, you'll also get more! LAFF was developed following the syllabus of an introductory linearalgebra course at The University of Texas at Austin taught by Professor Robert van de Geijn, an expert on high performance linearalgebra libraries. Through short videos, exercises, visualizations, and programming assignments, you will study Vector and Matrix Operations, Linear Transformations, Solving Systems of Equations, Vector Spaces, Linear Least-Squares, and Eigenvalues and Eigenvectors. In addition, you will get a glimpse of cutting edge research on the development of linearalgebra libraries, which are used throughout computational science.
MATLAB licenses will be made available to the participants free of charge for the duration of the course.
Please note that the verified certificate option is not currently open for this course. Please enroll in the audit track and you will be emailed when the verified certificate option is open for enrollment.
The goal of this course, part of the Analytics: Essential Tools and Methods MicroMasters program, is for you to learn how to build these components and connect them using modern tools and techniques.
In the course, you'll see how computing and mathematics come together. For instance, "under the hood" of modern data analysis lies numerical linearalgebra, numerical optimization, and elementary data processing algorithms and data structures. Together, they form the foundations of numerical and data-intensive computing.
The hands-on component of this course will develop your proficiency with modern analytical tools. You will learn how to mash up Python, R, and SQL through Jupyter notebooks, among other tools. Furthermore, you will apply these tools to a variety of real-world datasets, thereby strengthening your ability to translate principles into practice | 677.169 | 1 |
A Level Mathematics
Did you know?
91%
achieved A*-B grade in Mathematics at A Level in 2016.
Why should I study Mathematics?
As an A Level subject, Mathematics combines very well with many areas of the curriculum and is often taken in conjunction with subjects from the Arts and Humanities, as well as the Sciences. Further Mathematics, as an additional A Level, is available for students who wish to take their study of the subject at school to a greater and more rigorous depth and is strongly recommended for particularly talented girls who are likely to want to study Engineering or Mathematics (alone or in combination with other subjects) at University. Universities often look favourably on Physics students who have studied double maths (see separate entry for Further Mathematics).
What will I need to study Mathematics?
An A or A* grade at GCSE is required.
Structure and Outline of the course:
The course structure is designed to develop understanding of mathematics and mathematical processes in a way that promotes confidence and fosters enjoyment. You will develop abilities to reason logically, generalise and extend your range of skills to be used in more difficult, less structured problems.
The specification is split into modules, in the following categories:
Pure Mathematics, Mechanics and Statistics:
• Pure Mathematics develops and expands on the work in algebra, geometry, graphs, coordinates and trigonometry first encountered in the GCSE course. Calculus is introduced. There is less numerical work; a genuine feel for algebra and a real familiarity with all its techniques are vital for success.
• Mechanics is concerned with modelling physical situations and using the techniques of pure mathematics to solve problems involving such concepts as force, velocity, power and centres of gravity, the underlying theme being Newton's Laws of Motion.
• Statistics is designed to encourage a mathematically analytical approach to practical situations and to develop mathematical models to assess probability and test hypotheses.
Method of Assessment:
All exams are 1½hrs long and contain a mixture of length of question. Each exam is specific to a given topic. There will be four examinations on Pure Mathematics, one on Statistics and one on Mechanics.
Student Comments:
"Maths was an automatic choice for A Level because I enjoy a challenge and the reward when you master a problem. The teachers in this department are confident and thorough and have helped me nurture any talent I have."
"Maths is very challenging but satisfying if you get to the end of two pages of working and find that you have got it right."
"What I like about Maths is that it is a challenge which isn't impossible, and it forces me to think and sharpens my brain." | 677.169 | 1 |
ISBN-10: 061848096X
ISBN-13: 9780618480968 In Review Notes next to examples and Prepare for the Next Section exercises provide point-of-use review. Extra support also comes from the Aufmann Interactive Method, featuring Try Exercises that allow students to practice math as it is presented and to more easily study for tests. Prepare for the Next Section Exercises appear at the end of exercises and are specifically written to review prerequisite skills the student will need in the next section. Answers at the end of the text and a reference section help students review if they get a wrong answer. Review Notes cover prerequisite skills to help students without the necessary knowledge to understand important concepts. These example-specific notes direct students to the appropriate pages where they can practice and review the skill, thus decreasing frustration and increasing success. Interactive Reading Support Questions engage students in learning mathematics and encourage them to think critically. Visualize the Solution graphics are often paired with Algebraic Solutions to assist visual learners in understanding concepts. Focus on Problem Solving features at the beginning of each chapter review then demonstrate various strategies used by successful problem solvers. This builds students' comfort level with problem solving and leaves them with a collection of tips and strategies to refer to throughout the course. Eduspace, Houghton Mifflin's online learning tool powered by Blackboard, is a customizable, powerful and interactive platform that provides instructors with text-specific online courses and content | 677.169 | 1 |
Rational Functions Unit
In addition to the final group project, each member of the group must also hand in a written explanation of the math behind your design.
You may also want to complete the American Foods, Inc. Adapted Knowledge Rating Scale Vocabulary Development, which is a long, ugly name for a simple concept. You don't have to fill it out, but it is probably a good way to reflect on your understanding of the language of math. It follows, then, that copying definitions out of the textbook or somewhere else probably isn't a great idea, the AFIAKRSVD will be much more helpful if you use your own words.
You may want to think about how to divide up the labor among the members of the group. The American Foods, Inc. Team Rolesmight help you think about the best way to do that.
bubbl.us can be a good way to organize your ideas, brainstorm, and create concept maps. Here is a concept map about functions that I created using it:
Good luck with this first project! Try not to get too frustrated or lost, while realizing that no one ever discovered anything meaningful without some frustration and wandering around in the dark first. | 677.169 | 1 |
Showing 1 to 5 of 5
MA1100 Lecture 2
Sets
Set Notations Set Relations Set Operations
1
Sets
A set is a well defined collection of objects Example (Standard sets) The set of all natural numbers The set of all integers Z The set of all rational numbers Q The set of all irrat
National University of Singapore
Department of Mathematics
Semester 2, 2014/2015
MA1101R Linear Algebra I
Homework 1
Instruction
(a) Do all the 10 problems and submit on Feb 2 (Monday) during lecture. This problem set is NOT the same as
the tutorial set.
National University of Singapore
Department of Mathematics
Semester 2, 2014/2015
MA1101R Linear Algebra I
Homework 4
Instruction
(a) Do all the 10 problems and submit on April 6 (Monday) during lecture. This problem set is NOT the same
as the tutorial set
National University of Singapore
Department of Mathematics
Semester 2, 2014/2015
MA1101R Linear Algebra I
Homework 2
Instruction
(a) Do all the 10 problems and submit on Feb 16 (Monday) during lecture. This problem set is NOT the same
as the tutorial set.
National University of Singapore
Department of Mathematics
Semester 2, 2014/2015
MA1101R Linear Algebra I
Homework 3
Instruction
(a) Do all the 7 problems and submit on March 16 (Monday) during lecture. This problem set is NOT the same
as the tutorial set | 677.169 | 1 |
Mathematics is the abstract science of number, quantity, and space. There are two types of Mathematics - first type studied in its own right - pure mathematics, second type applied to other disciplines such as physics and engineering - applied mathematics.
Here, we will focus only on pure mathematics. Students can use them to study for standardized tests. | 677.169 | 1 |
Product Description
▼▲
Get your student ready for college-level math with Switched-On Schoolhouse Trigonometry. A prep course designed for advanced math courses, this computer-based course will cover trigonometry in clear, step-by-step lessons. Meant for students who have passed Algebra II, topics taught include like-right angle trigonometry, trigonometric identities, graphing, the laws of sines and cosines, and polar coordinates through video clips, learning games, and engaging animation. Quizzes and tests are included for progress assessment. Easy for both parents and students to use, SOS features automatic grading and lesson planning, a built-in calendar, and message center! Algebra II is a prerequisite. 5 Units with Review and Final Exam.
This is a single-semester course. 5 units. Grades 9-12 | 677.169 | 1 |
Tehran UniversityAbout Henrik
This course is designed to build on algebraic and geometric concepts. It develops advanced algebra skills such as systems of equations, advanced polynomials, imaginary and complex numbers, quadratics, and concepts and includes the study of trigonometric functions. It also introduces matrices and their properties. The content of this course are important for students' success on both the ACT and college mathematics entrance exams.
Calculus
I am a 45 years old dad and have been teaching calculus for more than 20 years.
Tehran University
MathematicsAwesome!
Henrik was awesome! His knowledge base on math is above and beyond what I expected. You must get him if you every plan on getting a great understanding of math. | 677.169 | 1 |
Enabled for this site to function properly. CPM Home Textbook CCG. Chapter 1 Lessons.1 .
Math Homework Help and Answers. What s the answer to On Your Own on page 121 in Common Core Alg2 txtbk. Unable to access the question. Could you.
Enabled for this site to function properly. CPM Home Textbook CC2. Chapter 1 Lessons.7 .
Be enabled for this site to function properly. CPM Home Textbook CCA2. Chapter 1 Lessons.4. Chapter 2 Lessons. | 677.169 | 1 |
A Graphing Matter - Activities for Easing Into Algebra by Key Curriculum Press
Show your pre-algebra and algebra students fun and real-world applications of variables and relationships - and laugh while you graph! Activities in this book engage students with concrete problems and experiments in which variables represent quantities they can see and measure. These informal explorations of graphs and equations provide a bridge between the concrete and the abstract, building a foundation for more formal operations students will study in algebra. Published by Key Curriculum Press
THIS BOOK IS BRAND NEW AND IN EXCELLENT CONDITION | 677.169 | 1 |
9781576856864
1576856864
More Prices
Summary
The best way to master math is to practice, practice, practice-and 1001 Math Problems offers "mathophobes" and others who just need a little math tutoring the practice they need to succeed. Whether a new job suddenly requires figuring percentages or a student faces a standardized math test that could determine their future, the 1,001 math questions in this useful manual provide readers with the skill set that they need to master math, algebra, and geometry challenges. | 677.169 | 1 |
A Problem Book In Mathematics For Iit Jee
ISBN 9788188222254
ISBN-10
8188222259
Binding
Paperback
Number of Pages
920 Pages
Language
(English)
Subject
Entrance Exam Preparation
A Problem Book In Mathematics is a comprehensive book for students aspiring to study engineering from Indian Institute of Technology and other premier technical universities of India. The book is designed to help students build mathematics concepts by problem solving. The book comprises of chapter-wise questions. In addition, there several tips and tricks for each topic and practice questions for thorough revision. This book is essential standard XII students preparing for various engineering entrance examinations.
About Arihant Publications 4th Edition, etc. | 677.169 | 1 |
Math Links
The Mathematics and Computer Science Department has 7 permanent members with a wide variety of interests and areas of study. Small class sizes (5-15 for upper level courses) make interactions between students and professors quite easy. The professors take an interest in the progress of their students and are readily available for help with courses or advising. Students interested in mathematics may choose to major in mathematics, mathematics/physics, or economics/mathematics. Although the department does not directly support a major in Statistics or Computer Science, students interested in these disciplines should speak to a member of the department. If students are looking towards Engineering, we have a long standing tradition of excellent 3-2 programs; these are also described in the Whitman College Catalog.
Students majoring in mathematics have opportunities to work with faculty on research projects; some of these projects have resulted in presentations at professional meetings and/or the Whitman Undergraduate Conference and publication in mathematics journals. Examples of such projects include research on jet engines, Java programming for the game of Hex, curvature of plane curves, location modeling problems, a study of cubic reciprocity, and radial basis problems.
The department sponsors evening help sessions for calculus students, staffed by upper level mathematics students, and a mathematics club that provides a social-recreational mathematical opportunity for any interested student. In addition to weekly informal meetings and social activities, the club organizes a campus wide math contest. On occasion, outside speakers are invited to campus to give talks on subjects of current interest in mathematics. Senior mathematics majors are known by every member of the department and receive helpful advice concerning their future careers. Each spring, the department holds a banquet for all students majoring in mathematics, thus providing yet another opportunity for students and faculty to meet in a social setting.
Computing is an important aspect of a mathematics education and students are encouraged to become skilled in the use of computers. Students begin by taking introductory programming followed by several related courses, such as Calculus Lab, Engineering Mathematics and Mathematical Modeling. The department maintains its own cluster of GNU/Linux based PCs. During the evening hours, student consultants are available in the computer lab to help others with any problems that may arise. The consultants learn the basics of Unix system administration and are part of a healthy campus community of Linux users.
Mathematics majors and combined mathematics majors have many doors open to them as quantitative reasoning skills are in high demand. Our majors have gone into a number of different areas such as business, law, medicine, engineering, telecommunications, mathematics, teaching, actuarial science, and operations research. As far as graduate school is concerned, recent graduates have attended the University of Wisconsin, University of Illinois, Carnegie Mellon, Yale, and Oregon State. Other students have gone to work for companies such as Microsoft, Deloitte Consulting, Active Voice Communication, and Safeco. The Mathematical Association of America (MAA) has some short biographies of mathematics majors online, describing their post-graduation employment.
A student interested in majoring in mathematics is welcome to consult with any member of the Mathematics and Computer Science Department. | 677.169 | 1 |
Results in Maths Textbooks
1-25 of 104,814
It has been developed because of a need by students, parents, teachers and coaches for a comprehensive, well presented, easy to understand Maths summary book which covers the most important ideas in the recently developed National Year 7 Maths Curriculum throughout Australia.
It aims to improve mathematics learning for students with a range of abilities, needs and levels of interest. This new, full-colour series has been rewritten and developed for the Australian mathematics curriculum by AMSI, while retaining the structure, depth and approach of the original titlesFEATURES & BENEFITS Student Workbook. Complete year's work with full coverage of the Australian Curriculum. You're bound to find the product you want, at an awesome price! Books F, 1 or 2 (item300845204902 ). Decimals.
A mathematician who is known throughout the world as the "mathemagician," Arthur Benjamin mixes mathematics and magic to make the subject fun, attractive, and easy to understand. InThe Magic of Math, Benjamin does more than just teach skills: with a tip of his magic hat, he takes you on as his apprentice to teach you how to appreciate math the way he does decimals.
The Australian Curriculum Edition Targeting Maths Year 3 Student Book has been specifically written to meet the Australian Curriculum requirements of primary school Year 3. Features of the Targeting Maths program include.
Developed by highly experienced maths educators to reflect the NSW syllabus for the Australian Curriculum - Developed by highly experienced educators with years of knowledge and experience in maths teaching and in preparing exceptional maths resources, and who are committed to the principles of the NSW Syllabus and the Australian Curriculum.
This guide has been designed to meet all study needs, providing up-to-date information in an easy-to-use format. Sample HSC Exams have been updated for the new 2012 format. EXCEL HSC - MATHEMATICS STUDY GUIDE.
The Australian Curriculum Edition Targeting Maths Year 5 Student Book has been specifically written to meet the Australian Curriculum requirements of primary school Year 5. Features of the Targeting Maths program include.
The aim of the International Centre of Excellence for Education in Mathematics (ICE-EM) is to strengthen education in the mathematical sciences at all levels - from school to advanced research and contemporary applications in industry and commerce. | 677.169 | 1 |
1147.24
FREE
About the Book
Glencoe's Algebra 1 and Algebra 2 balance sound skill and concept development with applications, connections, problem solving, critical thinking, and technology. Whether your students are getting ready for college or the workplace, this program gives them the skills they need for success. | 677.169 | 1 |
Mathematics is a compulsory subjects from Years 7 through to 11. In our sixth form we also offer AS level, A Level, GSCE and Functional Skills in our Sixth Form.
Subject Overview
Years 7 and 8
Throughout the first two years of mathematics we look to embed core skills, exploring the subject through investigations and problem solving. The key areas are number, geometry, algebra, statistics and probability.
Years 9 to 11
From Year 9 all students study for their GCSEs. We currently follow the EdExcel GCSE syllabus.
All exams for this course take place at the end of Year 11. There is no coursework in GCSE Mathematics.
Years 12 and 13
We offer AS and A Level Mathematics in the sixth form and follow the AQA syllabus. Due to the demands of the subject, we recommend that students achieve at least a Grade B in GCSE Mathematics if they would like to continue the subject at this level | 677.169 | 1 |
kingdomacademyhomeschool
A chronicle of our adventures in homeschooling.Thu, 22 Jun 2017 23:49:13 +0000enhourly1
TOS Review: UnLock Math
21 Jun 2017 17:44:01 +0000 we move through high school, the math gets harder and harder, at least for me. So I love using some type of computer-based program to teach those courses to take some of the pressure off. For this review, we got to try an online program from UnLock Math. My son is a rising junior and we checked out their UnLock Algebra2.
One of the things that I really like about this program is how complete it is. It features video leasson, testing, and automated grading, which just makes my life so much easier! The lessons are taught by real teachers and you get plenty of practice as well as access to complete solutions for the problem sets, which was helpful to me if I needed to go over something with my son.
Algebra2 has 15 units with topics ranging from polynomials, to quadratic functions, to statistics and matrices. It also includes mid-term and final exams. The program is very simple to use, especially for any kid familiar with computers. As a parent, you set up your account and add your student and they get their own log in info. When you log in you can check their progress and you can also view the program from the student's dashboard, in case you want to see what they're doing.
Units consist of multiple lessons which also include reviews of previous lessons. I like this method of teaching because even though students move on from a topic, they still practice the concept, which helps them retain it. Quizzes and tests are built in to each unit as well. Lessons begin with a short warm-up before students watch the video lesson. The videos are (thankfully) short, so this works well for kids with short attention spans. I also want to mention that the teachers in the videos are lively and engaged, something that is really important to my son. We have done online/computer-based classes before where the teacher drones on in a monotone voice and he quickly loses interest.
After watching the video, students complete a set of practice problems. There are about 10 problems per set and there are a few different sets, so they get plenty of practice options. Each lesson also includes a set of reference notes, which is a short summary of the lesson, complete with important vocabulary. These were very helpful and I printed them out to use them as study guides for my son for the exams.
The dashboard page also features handy graphics that allow students to track their own progress, which was a great motivator for my son. He could immediately see exactly how far he had come and how much he had left to do. It was also really easy for me to look at his login page and make sure he had done his work for the day!
On the whole, we really liked this program. I think it is a great option for parents who don't want to teach upper level math themselves. UnLock Math offers monthly and yearly pricing options, and you do get to choose your own start date, which is also nice.
If you would like to learn more, connect with UnLock Math on social media here:Filed under: Menu Planning, recipes]]> Review: Memoria Press (First Form Latin)
07 Jun 2017 14:26:02 +0000So as a kid I never liked brussels sprouts and as an adult I've avoided making them at home. However, a friend of mine made them as a side for a dinner party we went to and I decided to try some and I LOVED them! Maybe my tastebuds have changed now that I'm an adult, or perhaps it's just that my nana never made them that way, but this is now one of my favorite side dishes!
Directions:
1. Wash and trim the sprouts and place them in a large bowl.
2. Heat some oil in a skillet and cook the garlic until fragrant.
3. Add the bacon to the skillet and cook until cooked through, but not too crispy.
4. Pour the bacon and garlic over the brussels sprouts with about 1 tbsp of the bacon grease.
5. Add some olive oil to the bowl and season with salt and pepper to taste, then toss to coat.
6. Spread the sprouts over a baking pan sprayed with cooking spray or covered with foil.
7. Bake at 400 degrees for 20-30 minutes until sprouts begin to brown.
8. Sprinkle with Parmesan and cook for 5 minutes more.
This is so yummy, even my kids enjoy it! Thanks for stopping by, have a great week.My local grocery store recently had 10 pound bags of chicken leg quarters on sale for $5.00 so I grabbed a couple and put them in the freezer. I love chicken legs because they can be done so many ways and are easy to cook. This is my recipe for roasted chicken leg quarters.
Ingredients:
5 chicken leg quarters
1 cup buttermilk
1 tsp salt
1 tsp pepper
1 tsp paprika
1 tsp cayenne pepper
1 tsp onion powder
1 tsp garlic powder
Directions:
1. Pour the buttermilk into a bowl and add some salt and pepper to taste.
2. Place the chicken legs in a large resealable plastic bag and pour the buttermilk into the bag.
3. Marinate for 3-4 hours.
4. Mix the seasonings in a small bowl.
5. Remove the chicken legs from the buttermilk and shake off the excess. | 677.169 | 1 |
MATHSGEEKS
Overview
The MathsGeeks course is designed with the goal to improve conceptual
understanding, and increase analytical thinking and logical problem
solving skills. In addition to covering the requirement at school level,
it also covers advanced level questions, puzzles, MCQs and HOTS questions.
This course has been specially designed for students of classes VI to VIII
so as to create a strong base in Mathematics which will go on to help them
at many stages in life.
Course Highlights
Concept Clarity
Our course is structured in a way that all topics and
concepts are taught such that the students understand
everything with clarity and no doubts exist. Doubts, if
any, are clarified on various levels.
Comprehensive Material
Our integrated study material properly covers every aspect of each topic,
and explains it to the extent of being self sufficient.
We believe that our students should rely solely on the study material provided,
as no other reference book or guide will be needed.
Puzzles, MCQs, HOTS
In addition to sticking to the academic syllabus,
the course also enriches the learning of a student
through a variety of puzzles, MCQs and HOTS questions.
This imbibes an interest of the subject in the student,
and enhances his analytical and problem solving skills.
Competitive Edge
The course prepares a solid base for the student in the subject,
and makes it easier for him to do well in entrance exams
(Like JEE, SAT, CAT, GRE etc.) later. The course also helps the
students do well in Olympiads and other competitive exams held at various levels.
Level 1
For students in the 6th to 8th grades.
Course Contents
Algebra
Algebra
We will cover the following topics in Algebra - one of the most challenging
topics for students, but easy, once explained properly to them | 677.169 | 1 |
What is calculus used for
Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
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Calculus is the section in mathematics which is used in various fields and this branch of mathematics helps us analyze the behaviors of different functions. Calculus is a vast section and it includes various applications such as finding the area between the curves, maximum and minimum values possible for a function, structure of the graph and its continuity etc. These applications of calculus are used in different fields such as engineering and technology, business etc. and these concepts also help us find solutions to complex problems.
Example 1: Find the slope of the given straight line, f(x) = 2x + 11 using the derivative method.
In order to find the derivative, we can use the Power rule of the Derivatives: èd(xn)/dx = n * xn-1
We can distribute the derivative to both the terms:
èd(2x + 11)/dx è[ d(2x)/ dx ] + [ d(11)/ dx ]
Using the above formula, we get
è2(x1-1) + 0 = 2 Hence the slope or the derivative of the given straight line is 2 | 677.169 | 1 |
The students will follow the class syllabus throughout the year. Concepts for the week will be studied in class along with the practice problems. Students will then be ready to work assigned lessons and chapter tests at home. Homework is to be graded by the student/parent. Parents should monitor their student's homework each day making students accountable to complete homework assignments. All tests will be graded by the teacher.
Each student will need the Pre-Algebra Teaching Textbooks 2.0 – Pre-Algebra: Text and Answer Key/Test Bank or Pre-Algebra: CD Combo (You may buy the book/answer key only, or the books with the CD if you like to review lectures and see every problem worked out, but not the CD only set.) This class will be offered at the 1488 Campus on Wednesdays from 8:45-10:15 a.m. Tuition is $450 for the year, split into 10 payments of $45 (due July 2017-April 2018). A supply/printing fee of $5 is due in July
Algebra 1 - Teaching Textbooks - Brenda Hall
Expected time outside of class each week needed for preparation and homework will average between 4 – 5 hours.
Teaching Textbooks is a unique math program written specifically for homeschoolers in an easy-to-digest conversational tone. This course covers simple equations, negative numbers, fractions, powers, roots, second degree equations, graphing, solving systems of equations, inequalities, and irrational numbers. There are two options for parents to choose from: 1) purchase book, answer key/test bank only or 2) additionally purchase the audiovisual CD's to see solutions worked out for every problem. Please check the Teaching Textbooks web-site at teachingtextbooks.com.
How does this course work? First, students will follow the class syllabus throughout the year. Concepts for the week will be reviewed in class along with the practice problems for the week. Students will then be ready to work assigned lessons and chapter tests at home. Homework is to be graded by the student/parent. All tests will be graded by the teacher. Secondly, in-class quizzes will be given to monitor students' progress. Grades will be kept by the teacher and averaged grades will be sent each semester to the parents. Finally, a one-time $30 supply fee is due upon registration. This fee will cover the printing of chapter tests and handouts given to the parents at the beginning of the year.
The class will be one hour and 30 minutes. Pre-requisite: Pre-Algebra. Students should have a firm handle on addition, subtraction, multiplication, and division facts. They should be able to add, subtract, multiply and divide fractions with ease before taking Algebra I. Required: A TI-30XS calculator (at minimum). Those that plan to take Algebra II and Geometry should consider buying a TI-83 Plus or TI-84 calculator for all 3 classes. Supplies: Students will need a 2 or 3 inch 3 ring binder with tabs. Instructions on setting up folder will be sent via e-mail in August.
Tuition is $45/ month for 10 months and is due starting in July. Paymentsemester dated July 1, and 2nd semester dated Dec 1, 2017). NOTE: The first payment should include the supply fee.As we often turn students away once you have reserved your space, there will be no refunds given after July 1. Mail Class Registration Form and tuition payment to: Brenda Hall, 2005 Shasta Ridge, Conroe 77304. Mrs. Hall can be contacted at EBHall7@aol.com. | 677.169 | 1 |
Solving Math Problems
The following article presents a number of key concepts for successful math problem solving.
From the introduction:
"This paper describes a method for solving math problems.The basic idea is to combine two things:
First, a simple method for making handwritten notes while thinking about a problem.This method is aimed at supporting
– a step-by-step approach to problem solving and
– reflective thinking: Better understand and control what you do while solving a problem.
Second, a densely packed cheat sheet with broad advice on math problem solving. At present, this sheet focuses on general methods for problem solving. Later versions may contain material on specific domains like calculus or algebra.
Press the right button on the bar below the document to read it in full screen mode. | 677.169 | 1 |
Find a Framingham PrealgebraJust learn and follow the rules of this sensible math, and you will move forward in its complexity. From solving equations to multiplying polynomials to figuring out quadratics, it is helpful to walk through this subject knowing that it can assist you in the mathematics of digging a swimming poo | 677.169 | 1 |
Welcome to Math 25!
Our syllabus describes the class calendar, homework, grading, and other important information.
Math 25 builds off of Math 20 concepts and skills. There is not a lot of new math to learn. With lectures, discussions, and projects we review ratios, proportions, percents, and scale factors so a vague understanding of those topics grows into a solid understanding about how and when to use those ideas in real life.
As an analogy, most students leave Math 20 knowing what a "pliers" is. They understand how to use one. But in real life there are many kinds of pliers, each appropriate in different situations, and each used slightly differently.
Math 25 is about making your own "math toolbox". By the end of the term you will understand the different kinds of ratios, proportions, percents, and scale factors. You will know which are appropriate in various real-life situations, and how they are used slightly differently depending on the situation. | 677.169 | 1 |
Georgia High School Mathematics
Four New High School Math Resources Developed for Georgia
November 2010 Walch Education, a leading publisher of educational materials, is pleased to announce four new high school mathematics programs, completing its suite of Georgia high school math resources. Walch's materials, developed hand-in-hand with Georgia educators, have been built from the ground up to align with Georgia Performance Standards, Learning Tasks, and course descriptions.
"Our materials are developed to support consistent, coherent instruction that targets Georgia's Framework and Learning Tasks. They are designed to meet the needs of students across the spectrum and for user-friendly implementation" notes Walch Education's Vice President of Education, Jill Rosenblum.
The new programs include:
Math 1
Accelerated Math 1
Math 3 Support: GHSGT Review
Math 4
Math 1, Accelerated Math 1, and Math 4 include Teacher Resource Binders, Student Textbooks and Student Workbooks. Math 3 Support GHSGT Review is delivered as a student book containing a comprehensive set of materials for review and practice, and a packet of guidance and suggestion for teachers.
Walch's seven other Georgia High School math resources provide a consistent, complete and perfectly aligned suite of purpose-built mathematics support:
Together with Walch's middle school math materials, Walch Education's Georgia programs are being used in more than 100 schools with over 25,000 middle and high school students across the state.
"Our approach is distinctive in that we focus very specifically on the needs of a state and its educators" said Walch's President, Al Noyes, adding; "We have been working with Georgia curriculum leaders for about five years now, and have benefited greatly from their insights and our singular focus. These products are not only more effective than the "one size fits all" alternatives, they are less expensive. We're gratified by the support we have earned from educators across the state and look forward to continuing to help Georgia's teachers and students succeed."
For more information about these programs or tailoring materials to meet your school district's needs, please email Andrea Newman at: anewman@walch.com | 677.169 | 1 |
Here are my remaining office hours this semester. If you'd like to see me but can't make these times, please ask for an appointment. As always, you're
welcome to use my home number if you have a question.
Wed Apr 26, 8:30-10:50am
Thu Apr 27, 10:30am-4pm
Fri Apr 28, 8:30-11:30am
Sat Apr 29, 9:00-11:30am
Math Lab:
Have a problem and can't reach me to me for help? Try the
CofC Math Lab.
Course Objectives:
MATH 220 deepens the students' understanding of the definite integral
by studying
some of its
various applications and the basic techniques of integration.
We study sequences and series of real numbers, fundamental to any understanding of the real numbers and a stepping stone
to the study of power series, the students' introduction to the important topic of the approximation of functions.
Finally, we make our first steps towards a multivariate calculus with the study of calculus on curves given parametrically.
Student Learning Outcomes:
By the end of the course, students will be able to
Represent the following as definite integrals: area between curves, volume of a solid of revolution, average value of a function, arc length of a curve.
Identify properties of sequences (monotonicity, boundedness, convergence) and find the limits of sequences.
Determine whether an infinite series converges by choosing and applying a suitable convergence test.
Determine the radius of convergence of a power series.
Use Taylor Series to express functions as power series and to evaluate infinite series.
Represent plane curves as parametric equations, and recognize the plane curve that corresponds to given parametric equations.
Use derivatives and integrals to find slopes and lengths of parametric curves, and areas bounded by them.
Convert between Cartesian and polar coordinates, graph polar curves, and apply calculus to polar curves as for parametric curves.
Model mathematical questions with differential equations, and use basic methods for solving such equations.
General Education Student Learning Outcomes:
Students are expected to display a thorough understanding of the topics covered. In particular, upon completion of the course, students will be able to
model phenomena in mathematical terms,
solve problems using these models, and
demonstrate an understanding of the supporting theory behind the models apart from any particular application.
These outcomes will be assessed on the final exam.
Mathematics Program Student Learning Outcomes:
This course can be used to satisfy some requrements of the undergraduate mathematics degree program,
for which there are also some standard goals; students will:
use algebra, geometry, calculus and other track-appropriate sub-disciplines of mathematics to
model phenomena in mathematical terms;
use algebra, geometry, calculus and other track-appropriate sub-disciplines of mathematics to derive correct answers to challenging questions by applying the models from the previous Learning Outcome; and
You have two
choices, depending on which of our calculus courses you plan to take.
The cheapest option is probably to buy the book used online and then sell it yourself on Amazon when you're done with it.
Carefully save all receipts from the bookstore. If any of your books comes shrink-wrapped, DON'T unwrap it until you check with your instructor on the first day of class that you've bought the right book.
Students who are certain that they want to use WebAssign (see below)
might be able to
save money by buying the book bundled with a WebAssign access code at our bookstore.
Everyone else should try it for free first.
One of the options at our bookstore is a loose-leaf version of the entire book bundled with a WA code.
It's the cheapest way to obtain and book with WebAssign, but it tears very easily and will have almost no value as a used book after the semester.
It would be good to have your copy of the textbook by the first day of class,
but everyone will have free online (but not down-loadable) access to our book
for the first two weeks of class through WebAssign.
Students who purchase WebAssign will have online access to our book all semester.
I find flipping through the pages online to be pretty clumsy and an impediment
to learning from the book. I think even those students who buy WA will want
a hardcopy of the text, if they can afford it.
The Student Solutions Manual for Stewart's Calculus is optional.
It contains worked-out solutions to most of the odd-numbered
problems in the book. It might be helpful to look at the SSM once in a while, but, if
overused, the SSM will do you more harm than good.
WebAssign:
WebAssign is an online homework system that gives immediate
feedback and extra help on many of the problems in our text.
To set up your account, go to
click on "Enter Class Key" (or "Students/I Have a Class Key"),
and then enter our class key:
cofc
0370
2444
You're allowed to use WebAssign for free for about the first two weeks of the semester, starting from the first day of class.
You'll need to purchase a WebAssign access code if you want to use the system
after that. (If you purchased one for a course that used the same textbook in an earlier semester, you might not need to purchase another.) A required WebAssignment to be completed
during the free trial period will count as as your first quiz.
Later, optional WebAssign
problem sets will be available for students who find WA useful.
These optional problem sets will not be used in the calculation of your grade.
Exams and Grades:
We'll have four (4) 75-minute midterm exams, a 3-hour final exam,
and weekly one-question quizzes.
See Schedule below for dates.
All exams and quizes will be closed book: no notes, books, calculators, electronic devices, etc.
Although basic ideas we learn in this course can appear on several exams or
quizzes, each weekly quiz will be based primarily on material covered
since the time of the previous exam or quiz, and
each midterm exam will be based primarily on material covered since the
previous midterm.
Our
final exam in this course will be cumulative. Unless I
specifically tell you otherwise, you should assume that any topic of this
course could appear on the final.
Each of the midterm exams is worth
100 points, the final exam is worth 200 points, and the
weekly in-class quizzes are worth 50 points altogether.
I'll assign letter grades as follows:
Letter grade:
A
A-
B+
B
B-
C+
C
C-
D+
D
D-
Minimum required score:
90%
87%
83%
80%
77%
73%
70%
67%
63%
60%
57%
I won't drop any exams, but if you do better on the final exam than on
your worst midterm exam, I'll raise that (one) midterm exam score
by averaging it with your final exam (percentage) score.
Then,
at the end of the semester, I'll calculate your grade two ways--based on the
percent you earned of the 600 possible exam points, and again based on the
percent you earned of the 650 possible exam and quiz points--and
give you whichever letter grade comes out higher.
Attendance Policy:
Good attendance is a necessary first step towards a good grade.
I strongly recommend that you attend class every day.
If you're absent on a non-exam day, I'll assume that you have a
good reason for missing and will
not require an excuse. Read the text and try the homework
for the day you miss and then bring questions to me in my office. See
Make-up Policy for absences on exam days.
Note: College of Charleston policy requires me to take roll during the first two weeks after drop/add, until I determine that all of my students have attended
at least once, and report the results to the College. Any student who has not attended class at least once during those two weeks will be dropped from this class. These roll calls will not be used in my calculation of the remaining students' grades at the end of the semester.
Make-up Policy:
Exams:
If you must miss an exam, I expect you to contact me (using all
the numbers above)
and the
Absence Memo Office
as soon as possible. Do not delay.
I can allow you a make-up exam only if I determine that your absence at
exam time (and every reasonable time until the make-up) is excusable.
If you are not sick enough to see a doctor for your illness, then you are not sick enough to miss the exam.
An unexcused exam will be given
the grade zero, probably causing you to fail the course.
Quizzes:
At the end of the semester---starting from the date of the last in-class
quiz and ending on the last day of final exams---I'll allow
you to make up at most two (2)
quizzes that you've missed for any reason.
These makeups can only be used to replace quizzes that you've missed, not simply low scores.
No Absence Memo will be required for makeup quizzes.
The topic of the makeup quizzes can be from
anything we've covered during this semester and
will be taken outside of class at a mutually agreed upon time. Contact me after the last quiz to schedule a makeup.
If you think I've overlooked something when grading any of your work
and would like me to consider giving it a higher score, you must
write, sign, and date the following statement on the exam or quiz in question.
"Dear Professor Kunkle, Please regrade Problem(s) XYZ for a higher score.
I have not altered my work on this paper in any way since it was first graded."
After each class, do as many of the assigned problems as possible.
There will be a short time to ask questions about these at the beginning
of the next class.
If you run into dificulty, really try; don't flit from one
unsolved problem to the next.
Don't just do the homework until you get the right answer, but practice
homework problems until you can do them reliably on an exam.
Practice reading the instructions on homework problems.
If you are able to do the homework only after looking at some answers in
the back to figure out what the question is asking, then you're
not prepared for the exams.
Begin extra studying well in advance for the tests, at least a week.
Rework old problems that could appear on the test.
Write (and rewrite) a special
set of notes that summarize in your own words the important facts
for the test. Include in these notes the different types of
problems appearing in the homework and the steps you follow
to solve each type.
(For example, here are the
notes
written by an A student while studying for the first test in MATH 111 Precalculus.)
Calculators:
A calculator is of limited use in learning the material in this class,
so no specific model is required for this course.
Calculators will be excluded from all exams and quizzes.
Syllabus On Line:
If it becomes necessary for me to change any part of this syllabus,
you'll always find its most current version
at
.
Look for the last change date at the top of this document, and
the description of changes at the bottom.
Old Exams:
Here are the exams from my MATH 220, Spring 2016,
when I last taught this class.
Since course content and the order of topics
can change from one semester to the next,
these exams might not always cover the material you should be studying
for your exams.
Any student eligible for and needing accommodations because of a disability
is requested to contact Disability Services (953-1431) and speak privately with the professor during the first two weeks
of class or as soon as the student has been approved for services so that
reasonable accommodations can be arranged.
Assigned Problems:
This is a list of all the problems worth doing in each section we'll cover.
I won't collect these, but you should be doing them daily.
"5-25" means at least the odd numbered
problems between 5 and 25, inclusive, and preferably the even numbered problems as well.
* indicates a challenging but worthwhile problem.
** indicates a very challenging problem for your enjoyment only.
I won't put a ** problem on an exam,
and I probably won't have time to do one in class.
"12.rev" refers to the review exercises at the end of Chapter 12.
"App.D" refers to Appendix D in the back of our text.
[17] means to do problem 17 if time allows us to cover this topic in class.
Do the problems marked review in Chapter 3 to
review differentiation.
Do the problems marked review in Chapter 5 to
review the Fundamental Theorem and substition.
Do the problems marked review in Appendix D to
review trigonometry. | 677.169 | 1 |
Algebra II
Welcome
to Algebra II. In this course, we will cover the following topics:
Sequences and Series, Quadratics, Polynomials, Rational Functions,
Radical functions (sounds cool, I know), Exponential and Logarithmic
functions, Probability, Statistics, and Trigonometric functions.
Below,
I've attached useful links to documents like our syllabus. At the
bottom are the current assignments. Click on the assignment to find a
link to the book or the worksheet itself. Feel free to use the filter
to find previous assignments that you may have missed | 677.169 | 1 |
Help on maths coursework
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Is "math" or "maths" the correct word to use as the shortened or colloquial form of the word mathematics.
No My Assignment Help : Samples & Case Study Review Sample Assignment Samples & Case Study Review Sample: The biggest assignment sample database – Chat with. | 677.169 | 1 |
Calculus@Internet Homework problems, exercises and various VRML activities to assist in learning calculus. Site Map FAQ- Advertising. calculus@internet accessed 3684924 times since January 1999
ALVIRNE HIGH SCHOOL PROBLEM OF THE WEEK SITE Problems presented by Alvirne High School students and guests as they prepare for the Advanced Placement exam in calculus. Alvirne's AP calculus Class invites you to join them in their preparation
Calculus On The Web An internet tutoring utility for learning and practicing calculus. COW gives the student or interested Category Science Math calculusWelcome to calculus on the Web. The COW Library calculus on the Web ispartially supported by the National Science Foundation. A project of
E-Calculus Home Page calculus on the Web The COW Library Click on a button below to open a book General information desk. Contents of the COW library If you wish to log in for a recorded session, click on the Login button. calculus on the Web is
S.O.S. Math - Calculus Explains concepts in detail of limits, convergence of series, finding the derivative from the definition Category Science Math calculus Integral; More on the Area Problem; The Fundamental Theorem of calculus;Mean Value Theorems for Integrals. TECHNIQUES OF INTEGRATION
Welcome To The FHS AP Calculus Home Page An informal page with a general description of calculus. Included are an advanced placement practice Category Science Math calculus High School Math An Explanation of calculus and its Use What in the world is calculus? Haveyou ever asked this question? Actually, they get it using calculus.
AP Calculus On The Web AP calculus on the Web This site has been discontinued. Please visit our problemof the week site or the college board site for addtional information.
Calculus-Help.com Will Help You Survive Calculus! Features a new practice calculus problem every week with complete solutions. Includes an archive of Category Science Math calculus High School MathcalculusHelp.com offers calculus help to students of all ages, including freemultimedia tutorials and detailed calculus examples with complete solutions.
Calculus-Help.com Problem Of The Week Until then, take a shot at any of the problems in the extensive calculusHelp archiveby clicking the year numbers in the blue bar at the right edge of this
Springer LINK: Calculus Of Variations Journal with table of contents and article abstracts back to 1995. Full text available to subscribers only.Category Science Math calculus calculus of VariationsThe Springer Journal calculus of Variations and Partial Differential Equations publishestopquality contributions in the field of calculus of Variations and
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Math
At McAuliffe International School, our math department utilizes the Singapore math approach for mathematics. Our units and lessons are in complete alignment with the new Common Core State Standards that Colorado has adopted and is a mastery-based curriculum. Each lesson develops students' problem solving abilities by building their conceptual understandings, skills, mathematical processes, attitudes towards math, application of math in real-world contexts, and self-awareness. Additional resources are utilized to differentiate for below- and above-grade level students.
As an IB Middle Years Programme school, McAuliffe provides both Standard Mathematics and Accelerated Mathematics course sequences. The Standard Mathematics course sequence teaches Common Core Standards at the pace recommended by the state of Colorado. Upon successful completion of the standard pathway, students will be prepared for High School Algebra 1 during their 9th grade year and will be on track for completing Pre-Calculus math during high school.
The Accelerated course sequence concentrates 7th grade Pre-Algebra, 8th grade Algebra, and the basic components of High School Algebra I into two years. Accelerated courses include the same Common Core State Standards as the Standard courses, but the pacing of the courses is faster and the material is covered at a deeper conceptual level. Upon successful completion of the accelerated pathway, most students will be ready for Honors Geometry during their 9th grade year and will be prepared to take AP math or HL (Higher Level) IB Math during high school. Decisions to accelerate students into high school mathematics before ninth grade are based on solid evidence of student learning, work ethic, and commitment. Students who are not placed in the accelerated math sequence at McAuliffe may work with their designated high school to explore options for advanced coursework. | 677.169 | 1 |
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Mathematics for the Nonmathematician
Professor Kline begins with an overview, tracing the development of mathematics to the ancient Greeks, and following its evolution through the Middle Ages and the Renaissance to the present day. Subsequent chapters focus on specific subject areas, such as "Logic and Mathematics," "Number: The Fundamental Concept," "Parametric Equations and Curvilinear Motion," "The Differential Calculus," and "The Theory of Probability." Each of these sections offers a step-by-step explanation of concepts and then tests the student's understanding with exercises and problems. At the same time, these concepts are linked to pure and applied science, engineering, philosophy, the social sciences or even the arts. | 677.169 | 1 |
Choosing Among Alternative Upper-Level Courses
Choosing Among Alternative Upper-Level Courses
Below you will find brief summaries of the differences between alternative courses satisfying requirements for the math major.
You should talk to an adviser to help decide on the best courses for you.
Analysis: 500 or 765?
Both Math 500 and Math 765 are introductions to the theory of functions and calculus.
Here you will learn to prove some of the basic theorems of single-variable calculus.
Each course has Math 127 or 223 and Math 290 as its prerequisites.
The difference is in the level of sophistication or "mathematical maturity" required.
Before taking Math 765 you should have some experience in writing proofs, unless you have done exceptionally well in the calculus sequence and have strong skills in logic and writing.
Math courses at the 700 level generally require a greater time commitment than those at the 500 level.
Algebra: 558 or 791?
Both Math 558 and Math 791 are introductions to modern algebra.
Here you will learn about generalizations of familiar number systems, and prove basic facts about algebra.
A main difference is in the level of sophistication or "mathematical maturity" required.
Before taking Math 791 you should have some experience in writing proofs, unless you have done exceptionally well in the calculus sequence and have strong skills in logic and writing.
Math 558 often includes more on polynomials, complex numbers, modular arithmetic and concrete permutations groups.
Math 791 often goes further into the structure of abstract groups and rings.
Math courses at the 700 level generally require a greater time commitment than those at the 500 level.
Statistics: 526, 627 - 628 or 727 - 728?
Math 526 is a brief introduction to probability and some statistical methods useful in applications.
Its prerequisite is Math 116, 122 or 127.
Math 627-628 (with prerequisites Math 127 or 223 and Math 290) introduces the probability theory and theoretical statistics students need to begin graduate study in statistics.
The sequence includes some challenging multivariable calculus and some theorem-proof material.
A more advanced treatment of the 627-628 topics is given in courses 727-728, which prepare graduate students for the PhD qualifying exam in probability and statistics.
Math 727 is also a required course in the most commonly used KU program of study leading to a MA degree in mathematics. | 677.169 | 1 |
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Algebra Word Problems
No Problem!
By
Having a problem with word problems? Author Rebecca Wingard-Nelson introduces simple ways to tackle tricky word problems with algebra. Real world examples make the book easy to read and are great for students to use on their own, or with parents, teachers, or tutors. Free downloadable worksheets are available on
Other books in Math Busters Word Problems
Other books from the authors
Reviews
". . . offers a clear and lucid explanation of the basic tenets of solving word problems in algebra . . . This is an excellent clarification of a topic that many students find difficult to understand." | 677.169 | 1 |
Description:
The primary purpose of the book is to provide a convenient source of reference to those people who are appearing for engineering entrance examinations One has to refer a good number of books to understand the basics of graphs but this book will surely reduce the number of books that each of them needs to perform his job. This book has been designed for convenience.
In JEE Maths syllabus, graphs are not explicitly mentioned but For last 15-20 years questions based on graphs are being asked in IIT JEE question papers
We have tried our level best to present the graphs in easier way and good number of problems based on basic graphs and inequalities are discussed and solved in the book. All types of transformations are also covered up.
Please mail your suggestions at : admin@learnersplanet.com | 677.169 | 1 |
It is so important for students to understand that variables represent unknown numbers, and can be manipulated using the same Algebraic Properties to solve equations with one unknown. Knowing the reasoning behind the manipulations of equations is vital for complete understanding.
Some students may not get to this level of understanding until after they leave high school, and some will never get to that level of understanding. It is important as teachers, that we model real world equations, and how to apply these properties to literal equations.
Literal equations do bring relevance into the classroom. When students see scientist and engineers having to manipulate equations to solve for a certain variable, it makes it applicable for them. Verbalizing what properties that we are using as we work through these problems will make the properties second nature. That way students do not see solving equations and the Algebraic Properties as two different unlinked concepts. Instead, they see the reasoning behind the Math.
Importance of students understanding the Algebraic Properties
Vertical Alignment: Importance of students understanding the Algebraic Properties
I intend for today's Warm Up to take about 10 minutes for the students to complete and for me to review with the class. I allow students about three minutes to work on their own. Their task is to write down all of the algebraic properties they can name, then to define a literal equation. After giving them time to write, I take responses from the students and write the list of algebraic properties on the board. We then compare the responses to a sheet from the Virginia Department of Education that lists all of the algebraic properties. It is found at the following website:
As I review each property, I provide examples to demonstrate the properties. When we get to the definition of a literal equation, we discuss how this name applies to a formula in which all of the variables represent real numbers. Therefore, I say, "real number properties, such as these algebraic properties, apply to operations on literal equations as well as one-variable equations."
Teacher's Note: While reviewing the properties, I stress how all of the operations can be completed using addition or multiplication. I focus particular attention on the multiplicative inverse, taking an opportunity to re-teach this when reviewing problems that give my students difficulty. I demonstrate my approach in the video below.
Resources (1)
Resources
After reviewing the Warm Up, I hand each table a laminated copy of a PARCC High School Reference Sheet that I have created for this activity. My students are scheduled to take the PARCC Exam in Spring 2015. Based on the current schedule, the assessment will be held three-fourths through the curriculum, and, at the end of the year. I found the Reference Sheet at the following website:
The main objective of this activity is for my students to be able to solve one of the geometric formulas on the PARCC reference sheet for an indicated variable. I also want my students to be able to recognize different forms when solving equations. I have created a set of geometric formulas as individual cut-outs from the reference sheet. I place them in a cup for a random draw.
To start the activity, I draw a formula out of the cup. I ask the students to solve the formula for a specific variable (that I choose) on their individual white boards. As a student displays his/her work, I ask him/her to identify when an algebraic property is being used to solve the literal equation for the variable. I will probably probe for an explanation of why it works in this particular situation. As the class listens to explanations, I ask my students to write down different approaches, to help them recognize that flexibility can make things easier (or harder) based on the choice of method.
For example, if I draw the formula for the Volume of a Sphere first, I will ask my students to solve for r. Here are two examples of student work:
When comparing the work of these two students, I will point out that in Example 2 the student simplifies the formula further. I'll say, "Of course, that does not mean that Example 1's work is incorrect." Then we will discuss why both formulas are correct, and, equivalent.
Resources
I use today's Exit Slip as a formative assessment to check for student understanding of literal equations and how to apply the algebraic properties. I have used names for variables, rather than letters, to reinforce the theme that the properties apply to real numbers and variables that represent them. I plan to distribute the Exit Slip with about 10 minutes remaining in class.
The work on the Exit Slip should show the students ability to solve for Mass in the Density formula, and use that formula to solve an application problem. When reviewing the Exit Slip with students, I will discuss that requested calculations can be found by solving for Mass, first, and then making substitutions. However, I will also discuss how substitutions could be made first, and then solved for the Mass. If time allows, we may discuss when it makes sense to use one or the other approach. | 677.169 | 1 |
Amit Kathuria
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David Sanchez
Facts about AP Calculus AB, AP Calculus BC, AP statistics
Calculus AB and Calculus BC are designed to be college level calculus classes. With regards to the AP Calculus courses, you've 3 choices: you could take AB and BC calculation as a sequence, take AB Calculus only, or cut AB Calculus and go directly to BC Calculus. BC Calculus contains everything in AB Calculus, plus several additional issues. You may actually get an AB Calculus sub score whenever you take the BC test. So Calculus BC isn't necessarily more strenuous than Calculus AB. BC Calculus has to move faster since it covers increasingly material, which is what causes it to be more intense than AB.
Some schools teach AP Calculus BC in two class periods to easily fit into all the material, or have significantly more intensive summer projects. AP Calculus BC classes often cover everything in Calculus AB in the initial semester, while AB elongates that material out over the full year. If you do Choose to take Calculus AB and Calculus BC as a sequence - for instance take AB Calculus junior year and after that BC Calculus senior year - you do not need to worry about deciding between the two courses. Despite the fact you frequently get more school credit for BC.
Calculus, taking AB Calculus may be a great choice as well. Despite the fact that you aren't going to get to the additional issues that BC Calculus covers, you will still learn core calculus theories like limits, derivatives, and integrals. Taking AB Calculus may take back your schedule for one more course. AB Calculus will only be one class period, that could leave you room for another course. If you have not taken Pre Calculus yet, do not worry about deciding which calculus class you would like to take just yet.
AP statistics is also very important to get admission in good collage. One third portion of AP math exam AP statistics. AP statistics cover subjects such as dotplot, stemplot and histogram and inferences, etc. AP statistics can also enhance your score in exam. With shortcuts and tips that will save your time and give you ample time to revise your test during the actual exam. AP statistic cover following topics.
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There are some very valid reasons to choose to pursue this course with the TutorEye team and the very obvious one is the level of convenience that is attached to it. Yes, there is no commuting involved to get to those jam packed extra classes after school. You can access complete and well explained tutorials and videos in the convenience of your home on your very own personal computer. Just log in to the site and search through the list of the most appropriate videos and tutorials and start studying. Another major advantage of seeking online AP calculus AB, AP calculus BC and AP statistics tutors with the TutorEye team is the number of options that you get with a single concept. There are not one but many online tutors available that are well equipped to solve all your problems and clear every doubt that you might have lurking in your mind.
No matter how many AP calculus AB, AP calculus BC and AP statistics problems or doubts you have in your mind, you will get all the help you are going to need right on this one spot. All the topics are covered well right here and there is no need for you to seek any additional help once you have come here. TutorEye is your other school outside your school.
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CALCULUS
Basic Concepts
for High Schools
Translated from the Russian
by
V. KISIN and A. ZILBERMAN
MIR PUBLISHERS
Moscow
PREFACE
Many objects are obscure to us not
because our perceptions are poor,
but simply because these objects
are outside of the realm of our
conceptions.
Kosma Prutkov
CONFESSION OF THE AUTHOR. My first acquaintance with
calculus (or mathematical analysis) dates back to nearly a quarter of
a century. This happened in the Moscow Engineering Physics Institute during splendid lectures given at that time by Professor D. A. Vasilkov. Even now I remember that feeling of delight and almost happiness. In the discussions with my classmates I rather heatedly insisted
on a simile of higher mathematics to literature, which at that time
was to me the most admired subject. Sure enough, these comparisons
of mine lacked in objectivity. Nevertheless, my arguments were to
a certain extent justified. The presence of an inner logic, coherence,
dynamics, as well as the use of the most precise words to express a way
of thinking, these were the characteristics of the prominent pieces
of literature. They were present, in a different form of course, in
higher mathematics as well. I remember that all of a sudden elementary mathematics which until that moment had seemed to me very
dull and stagnant, turned to be brimming with life and inner motion
governed by an impeccable logic.
Years have passed. The elapsed period of time has inevitably
erased that highly emotional perception of calculus which has become
a working tool for me. However, my memory keeps intact that unusual
happy feeling which I experienced at the time of my initiation to this
extraordinarily beautiful world of ideas which we call higher mathematics.
CONFESSION OF THE READER. Recently our professor of
mathematics told us that we begin to study a new subject which
he called calculus. He said that this subject is a foundation of higher
mathematics and that it is going to be very difficult. We have already
studied real numbers, the real line, infinite numerical sequences, and
limits of sequences. The professor was indeed right saying that comprehension of the subject would present difficulties. I listen very
carefully to his explanations and during the same day study the
relevant pages of my textbook. I seem to understand everything, but
at the same time have a feeling of a certain dissatisfaction. It is difficult for me to construct a consistent picture out of the pieces obtained
in the classroom. It is equally difficult to remember exact wordings
and definitions, for example, the definition of the limit of sequence.
In other words, I fail to grasp something very important.
Perhaps, all things will become clearer in the future, but so far
calculus has not become an open book for me. Moreover, I do not
see any substantial difference between calculus and algebra. It seems
6
Preface
that everything has become rather difficult to perceive and even more
difficult to keep in my memory.
COMMENTS OF THE AUTHOR. These two confessions provide
an. opportunity to get acquainted with the two interlocutors in this
book: In fact,, the whole book is presented as a relatively free-flowing
dialogue between the AUTHOR and the READER. From one discussion to another the AUTHOR will lead the inquisitive and receptive
READER to different notions, ideas, and theorems of calculus,
emphasizing especially complicated or delicate aspects, stressing the
inner logic of proofs, and attracting the reader's attention to special
points. I hope that this form of presentation will help a reader of the
book in learning new definitions such as those of derivative, antiderivative, definite integral, differential equation, etc. I also expect that
it will lead the reader to better understanding of such concepts as
numerical sequence, limit of sequence, and function. Briefly, these
discussions are intended to assist pupils entering a novel world of
calculus. And if in the long run the reader of the book gets a feeling
of the intrinsic beauty and integrity of higher mathematics or even
is appealed to it, the author will consider his mission as successfully
completed.
Working on this book, the author consulted the existing manuals
and textbooks such as Algebra and Elements of Analysis edited by
A. N. Kolmogorov, as well as the specialized textbook by N. Ya.' Vilenkin and S. I. Shvartsburd Calculus. Appreciable help was given
to the author in the form of comments and recommendations by
N. Ya. Vilenkin, B. M. Ivlev, A. M. Kisin, S. N. Krachkovsky, and
N. Ch. Krutitskaya, who read the first version of the manuscript.
I wish to express gratitude for their advice and interest in my work.
I am especially grateful to A. N. Tarasova for her help in preparing
the manuscript.
.
CONTENTS
PREFACE
DIALOGUES
s
1. Infinite Numerical Sequence
9
2. Limit of Sequence
21
3. Convergent Sequence
30
4. Function
41
5. More on Function
53
6. Limit of Function
71
7. More on the Limit of Function
84
8. Velocity
94
9. Derivative
105
10. Differentiation
11. Antiderivative
117
-
134
12. Integral
146
13. Differential Equations
156
14. More on Differential Equations
PROBLEMS
178
168
DIALOGUE ONE
INFINITE NUMERICAL
SEQUENCE
AUTHOR. Let us start our discussions of calculus by
considering the definition of an infinite numerical sequence
or simply a sequence.
We shall consider the following examples of sequences:
1, 2, 4, 8, 16, 32, 64, 128, ...
(1).
5, 7, 9, 11, 13, 15, 17, 19, ...
(2)
1, 4, 9, 16, 25, 36, 49, 64, ...
(3)
y7, 2y2, ...
2' 3' 4, 5' 6, 7' 81 9'
1. Y2, V3, 2, y5, 1/
2
1
4
3
5
6
(4>
,
8
7
(5)
2, 0, -2, -4, -6, -8, - 10, - 12, ...
1,
1
1
1
2, 3, 4 , 5 ,
1,
2,
3,
1
4,
1, -1, 3,
2
1
4
6,
7 , 8 , .. .
(7)
...
-5, 7 , --7
7, 8,
1
3 , 3, 4 5 °
1
6
7
1
1
8
7 , 9 . .. .
1
(6)
1
1
1
5,
1
1
1,
1
1
(8)
.. .
(9)
(10)
Have a closer look at these examples. What do they have
in common?
READER. It is assumed that in each example there must
be an infinite number of terms in a sequence. But in general,
they are all different.
AUTHOR. In each example we have eight terms of
sequence. Could you write, say, the ninth term?
a..
READER. Sure, in the first example the ninth term must
be 256, while in the second example it must be 21.
Dialogue One
10
AUTHOR. Correct. It means that in all the examples
there is a certain law, which makes it possible to write down
the ninth, tenth, and other terms of the sequences. Note,
though, that if there is a finite number of terms in a sequence,
one may fail to discover the law which governs the infinite
:sequence.
READER. Yes, but in our case these laws are easily
recognizable. In example (1) we have the terms of an infinite
geometric progression with common ratio 2. In example (2)
we notice a sequence of odd numbers starting from 5. In
example (3) we recognize a sequence of squares of natural
numbers.
AUTHOR. Now let us look at the situation more rigorously. Let us enumerate all the terms of the sequence in
sequential order, i.e. 1, 2, 3, ..., . n, .... There is a certain
law (a rule) by which each of these natural numbers is
.assigned to a certain number (the corresponding term of
the sequence). In example (1) this arrangement is as follows:
1 2 4 8'16 32 ... 2n-1 ... (terms of the sequence)`
1234
5
6
...
4
...
(position numbers of the terms)
In order to describe a sequence it is sufficient to indicate
the term of the sequence corresponding to the number n,
i.e. to write down the term of the sequence occupying the
nth position. Thus, we can formulate the following definition
.of a sequence.
Definition:
We say that there is an infinite numerical sequence if every
natural number (position number) is unambiguously placed
in correspondence with a definite number (term of the sequence)
by a specific rule.
This relationship may be presented in the following
,general form
yi y2 y3 y4 y5 ... yn ...
TT T
1
T
T
T
2 3 4 5 .., n ...
The number yn is the nth term of the sequence, and the whole
:sequence is sometimes denoted by a symbol (yn).
Infinite Numerical Sequence
11
READER. We have been given a somewhat different
definition of a sequence: a sequence is a function defined on
a set of natural numbers (integers).
AUTHOR. Well, actually the two definitions are equivalent. However, I am not inclined to use the term "function"
too early. First, because the discussion of a function will
come later. Second, you will normally deal with somewhat
different functions, namely those defined not on a set of
integers but on the real line or within its segment. Anyway,
the above definition of a sequence is quite correct.
Getting back to our examples of sequences, let us look
in each case for an analytical expression (formula) for the
nth term. Go ahead.
READER. Oh, this is not difficult. In example (1) it is
yn = 2. In (2) it is yn = 2n + 3. In (3) it is yn = n2.
1
= n
In (4) it is yn = vn. In (5) it is yn = 1- n-{-1
n-}-1
In (6) it is yn = 4 - 2n. In (7) it is yn = n . In the remaining three examples I just do not know.
AUTHOR. Let us look at example (8). One can easily
see that if n is an even integer, then yn = n , but if n is
odd, then yn = n. It means that
yn =
n
ifn=2k
In ifn=2k-1
READER. Can I, in this particular case, find a single
analytical expression for yn?
AUTHOR. Yes, you can. Though I think you needn't.
Let us present yn in a different form:
yn=ann+bn n
and demand that the coefficient an be equal to unity if n is
odd, and to zero if n is even; the coefficient bn should behave
in quite an opposite manner. In this particular case these
coefficients can be determined as follows:
an = T[1-(-1)n]; bn =
[1 -F- (-1)nl
z
Dialogue One
12
Consequently,
Y. = 2 [1 - (- 1)n]
[1 + (- 1)n]
Tn1
Do in the same manner in the other two examples.
READER. For sequence (9) I can write
2(ni
Yn= 2.
1)
[1+(-1)n]
and for sequence (10)
[1-(-1)"]
yn
2n
} 2(nÂą1)[I
(-1)"]
AUTHOR. It is important to note that an analytical ex-
pression for the nth term of a given sequence is not necessarily a unique method of defining a sequence. A sequence can
be defined, for example, by recursion (or the recurrence
method) (Latin word recurrere means to run back). In this
case, in order to define a sequence one should describe the
first term (or the first several terms) of the sequence and
a recurrence (or a recursion) relation, which is an expression
for the nth term of the sequence via the preceding one (or
several preceding terms).
Using the recurrence method, let us present sequence (1),
as follows
2yn-1
Its first terms are
(11)
1, 1, 2, 3, 5, 8, 13, 21, ...
This sequence is known as the Fibonacci sequence (or
numbers).
READER. I understand, I have heard something about
the problem of Fibonacci rabbits.
Infinite Numerical Sequence
13
AUTHOR. Yes, it was this problem, formulated by Fibo-
iiacci, the 13th century Italian mathematician, that gave
the name to this sequence (11). The problem reads as follows.
A man places a pair of newly born rabbits into a warren and
wants to know how many rabbits he would have over a cer-
Symbol
denotes one pair of rabbits
Fig. 1.
lain period of time. A pair of rabbits will start producing
offspring two months after they were born and every following month one new pair of rabbits will appear. At the beginning (during the first month) the man will have in his warren
only one pair of rabbits (yl = 1); during the second month
lie will have the same pair of rabbits (y2 = 1); during the
third month the offspring will appear, and therefore the
number of the pairs of rabbits in the warren will grow to
two (y3 = 2); during the fourth month there will he one
wore reproduction of the first pair (y4 = 3); during the
fifth month there will be offspring both from the first and
second couples of rabbits (y5 = 5), etc. An increase of the
number of pairs in the warren from month to month is
plotted in Fig. 1. One can see that the numbers of pairs of
rabbits counted at the end of each month form sequence
(11), i.e. the Fibonacci sequence.
READER. But in reality the rabbits do not multiply in
accordance with such an idealized pattern. Furthermore, as
Dialogue One
14
time goes on, the first pairs of rabbits should obviously stop
proliferating.
AUTHOR. The Fibonacci sequence is interesting not
because it describes a simplified growth pattern of rabbits'
population. It so happens that this sequence appears, as if
by magic, in quite unexpected situations. For example, the
Fibonacci numbers are used to process information by computers and to optimize programming for computers. However,
this is a digression from our main topic.
Getting back to the ways of describing sequences,
I
would like to point out that the very method chosen to describe
a sequence is not of principal importance. One sequence may
be described, for the sake of convenience, by a formula for
the nth term, and another (as, for example, the Fibonacci
sequence), by the recurrence method. What is important,
however, is the method used to describe the law of correspondence, i.e. the law by which any natural number is placed in
correspondence with a certain term of the sequence. In a
Yn
f
I
Z -------T
o
I
I
I
I
I
I
!
(
I
i
1
l
2
1
3
4
5
1
6
1
7
8
1
l
9
10
n
Fig. 2
number of cases such a law can be formulated only by words.
The examples of such cases are shown below:
2, 3, 5, 7, 11, 13, 17, 19, 23, ...
3, 3.1, 3.14, 3.141, 3.1415, 3.14159,
(12
...
(13
Infinite Numerical Sequence
15
In both cases we cannot indicate either the formula for the,
ntli term or the recurrence relation. Nevertheless, you can
without great difficulties identify specific laws of corresponrlence and put them in words.
READER. Wait a minute. Sequence (12) is a sequence of
prime numbers arranged in an increasing order, while (13)
is, apparently, a sequence composed of decimal approximaI.ions, with deficit, for ac.
AUTHOR. You are absolutely right.
READER. It may seem that a numerical sequence differs.
from a random set of numbers by a presence of an intrinsic
degree of order that is reflected either by the formula for
1 he nth term or by the recurrence relation. However, the
Inst two examples show that such a degree of order needn't
ho present.
AUTHOR. Actually, a degree of order determined by
a formula (an analytical expression) is not mandatory. It
is important-, however, to have a law (a rule, a characteristic)
of correspondence, which enables one to relate any natural
lirimber to a certain term of a sequence. In examples (12)
rnd (13) such laws of correspondence are obvious. Therefore,
12) and (13) are not inferior (and not superior) to sequences
1
r
I
I
T
'
I
I
T
?
TF
I
I
I
I
I
I
I
I
Fig. 3
0)-(11) which permit an analytical description.
Later we shall talk about the geometric image (or map)
f a numerical sequence. Let us take two coordinate axes,
and y. We shall mark on the first axis integers 1, 2, 3, .. .
Definition:
A sequence (yn) is bounded if there are two numbers A and B,
labelling the range which encloses all the terms of a sequence
A < yn <, B (n = 1, 2, 3,
. .
.)
If it is impossible
to identify such two numbers
in particular, one can find only one of the two such
numbers, either the least or the greatest), such a sequence
(or,
is unbounded.
Do you find bounded sequences among our examples?
READER. Apparently, (5) is bounded.
AUTHOR. Find the numbers A and B for it.
READER. A=2, B = 1.
AUTHOR. Of course, but if there exists even one pair
of A and B, one may find any number of such pairs. You
could say, for example, that A = 0, B = 2, or A = -100,
B = 100, etc., and be equally right.
READER. Yes, but my numbers are more accurate.
Infinite Numerical Sequence
19
AUTHOR. From the viewpoint of the bounded sequence
definition, my numbers A and B are not better and not worse
than yours. However, your last sentence is peculiar. What
do you mean by saying "more accurate"?
READER. My A is apparently the greatest of all possible
lower bounds, while my B is the least of all possible upper
bounds.
AUTHOR. The first part of your statement is doubtlessly
correct, while the second part of it, concerning B, is not so
self-explanatory. It needs proof.
READER. But it seemed rathers,obvious. Because all
f.he terms of (5) increase gradually, and evidently tend to
unity, always remaining iless_than unity.
AUTHOR. Well, it is right. But it is not yet evident
that B = 1 is the least number for which y,, < B is valid
for all n. I stress the point again: your statement is not selfevident, it needs proof.
I shall note also that]"self-evidence" of your statement
about B = 1 is nothing but your subjective impression; it
is not a mathematically substantiated corollary.
READER. But how to prove that B =1 is, in this particular case, the least of all possible upper bounds?
AUTHOR. Yes, it can be proved. But let us not move
too fast and by all means beware of excessive reliance on
so-called self-evident impressions. The warning becomes
even more important in the light of the fact that the boundedness of a sequence does not imply at all that the greatest A
or the least B must be known explicitly.
Now, let us get back to our sequences and find other examples of bounded sequences.
READER. Sequence (7) is also bounded (one can easily
AUTHOR. You are quite right. Sequences (5), (7), (9),
(10), and (13) are bounded. Note that (5), (7), and (13) are
bounded and at the same time monotonic. Don't you feel
that this fact is somewhat puzzling?
READER. What's puzzling about it?
z
20
Dialogue One
AUTHOR. Consider, for example, sequence (5). Note
that each subsequent term is greater than the preceding
one. I repeat, each term! But the sequence contains an
infinite number of terms. Hence, if we follow the sequence
far enough, we shall see as many terms with increased magnitude (compared to the preceding term) as we wish. Nevertheless, these values will never go beyond a certain "boundary",
which in this case is unity. Doesn't it puzzle you?
READER. Well, generally speaking, it does. But I notice
that we add to each preceding term an increment which gradually becomes less and less.
AUTHOR. Yes, it is true. But this condition is obviously
insufficient to make such a sequence bounded. Take, for
example, sequence (4). Here again the "increments" added
to each term of the sequence gradually decrease; nevertheless,
the sequence is not bounded.
READER. We must conclude, therefore, that in (5) these
"increments" diminish faster than in (4).
AUTHOR. All the same, you have to agree that it is not
immediately clear that these "increments" may decrease
at a rate resulting in the boundedness of a sequence.
READER. Of course, I agree with that.
AUTHOR. The possibility of infinite but bounded sets
was not known, for example, to ancient Greeks. Suffice
it to recall the famous paradox about Achilles chasing
a turtle.
Let us assume that Achilles and the turtle are initially
separated by a distance of 1 km. Achilles moves 10 times
faster than the turtle. Ancient Greeks reasoned like this:
during the time Achilles covers 1 km the turtle covers 100 m.
By the time Achilles has covered these 100 m, the turtle
will have made another 10 m, and before Achilles has covered these 10 m, the turtle will have made 1 m more, and
so on. Out of these considerations a paradoxical conclusion
was derived that Achilles could never catch up with the
turtle.
This "paradox" shows that ancient Greeks failed to grasp
the fact that a monotonic sequence may be bounded.
READER. One has to agree that the presence of both the
monotonicity and boundedness is something not so simple
to understand.
Limit of Sequence
21
AUTHOR. Indeed, this is not so simple. It brings us
close to a discussion on the limit of sequence. The point
is that if a sequence is both monotonic and bounded, it
should necessarily have a limit.
Actually, this point can be considered as the "beginning"
of calculus.
DIALOGUE TWO
LIMIT OF SEQUENCE
AUTHOR. What mathematical operations do you know?
READER. Addition, subtraction, multiplication, division, involution (raising to a power), evolution (extracting
a root), and taking a logarithm or a modulus.
AUTHOR. In order to pass from elementary mathematics
to higher mathematics, this "list" should be supplemented
with one more mathematical operation, namely, that of
finding the limit of sequence; this operation is called sometimes the limit transition (or passage to the limit). By the
way, we shall clarify below the meaning of the last phrase
of the previous dialogue, stating that calculus "begins"
where the limit of sequence is introduced.
READER. I heard that higher mathematics uses the operations of differentiatton and integration.
AUTHOR. These operations, as we shall see, are in essence
nothing but the variations of the limit transition.
Now, let us get down to the concept of the limit of sequence.
Do you know what it is?
READER. I learned the definition of the limit of sequence
However, I doubt that I can reproduce it from memory.
AUTHOR. But you seem to "feel" this notion somehow?
Probably, you can indicate which of the sequences discussed
shove have limits and what the value of the limit is in each
rose.
READER. I think I can do this. The limit is 1 for sequence
(5), zero for (7) and (9), and n for (13).
AUTHOR. That's right. The remaining sequences have
no limits.
22
Dialogue Two
READER. By the way, sequence (9) is not monotonic ...
AUTHOR. Apparently, you have just remembered the end
of our previous dialogue wh( ro it was stated that if a sequence
.
is both monotonic and bounded, it has a limit.
READER. That's correct. But isn't this a contradiction?
AUTHOR. Where do you find the contradiction? Do you
think that from the statement "If a sequence is both monoton
is and bounded, it has a limit" one should necessarily draw
a reverse statement like "If a sequence has a limit, it must
be monotonic and bounded"? Later we shall see that a necessary condition for a limit is only the boundedness of a sequence. The monotonicity is not mandatory at all; consider,
for example, sequence (9).
Let us get back to the concept of the limit of sequence.
Since you have correctly indicated the sequences that have
limits, you obviously have some understanding of this
concept. Could you formulate it?
READER. A limit is a number to which a given sequence
tends (converges).
AUTHOR. What do you mean by saying "converges to a
number"?
READER. I mean that with an increase of the serial
number, the terms of a sequence converge very closely to
a certain value.
AUTHOR. What do you mean by saying "very closely"?
READER. Well, theWdifference'between the values of
the terms and the given number will become infinitely
small. Do you think any additional explanation is needed?
AUTHOR. The definition of the limit of sequence which
you have suggested can at best be classified as a subjective
impression. We have already discussed a similar situation in
the previous dialogue.
Let us see what is hidden behind the statement made
above. For this purpose, let us look at a rigorous definition
of the limit of sequence which we are going to examine in
detail.
Definition:
The number a is said to be the limit of sequence
if for
any positive number s there is a real number N such that for
all n > N the following inequality holds:
(i)
yx - a I < e
Ltmtt of Sequence
23
READER. I am afraid, it is beyond me to remember such
a definition.
AUTHOR. Don't hasten to remember. Try to comprehend
Ibis definition, to realize its structure and its inner logic.
You will see that every word in this phraso carries a definite
and necessary content, and that no other definition of the
limit of sequence could be more succinct (more delicate,
even).
First of all, let us note the logic of the sentence. A certain
number is the limit provided that for any s > 0 there is
n number N such that for all n > N inequality (1) holds.
f ri short, it is necessary that for any s a certain number N
xhou.ld exist.
Further, note two "delicate" aspects in this sentence.
I' first, the number N should exist for any positive number e.
Obviously, there is an infinite set of such e. Second, inequality (1) should hold always (i.e. for each e) for all n> N.
But there is an equally infinite set of numbers n!
READER. Now, the definition of the limit has become
more obscure.
AUTHOR. Well, it is natural. So far we have been examin-
ing the definition "piece by piece". It is very important
Ihat the "delicate" features, the "cream", so to say, are spotted from the very outset. Once you understand them, every1hing will fall into place.
In Fig. 7a there is a graphic image of a sequence. Strictly
speaking, the first 40 terms have been plotted on the graph.
Let us assume that if any regularity is noted in these 40
terms, we shall conclude that the regularity does exist
for n > 40.
Can we say that this sequence converges to the number a
(in other words, the number a is the limit of the sequence)?
READER. It seems plausible.
AUTHOR. Let" us, however, act not on the basis of our
impressions but on the basis of the definition of the limit
of sequence. So, we want to verify whether the number a is
the limit of The given sequence. What does our definition of
the limit prescribe us to do?
READER. We should take a positive number e.
AUTHOR. Which number?
READER. Probably, it must be small enough,
Limit of Sequence
25
AUTHOR. The words "small enough" are neither here
nor there. The number s must be arbitrary.
Thus, we take an arbitrary positive e. Let us have a look
at Fig. 7 and lay off on the y-axis an interval of length e,
both upward and downward from the same point a. Now,
let us draw through the points y = a + s and y = a - s
the horizontal straight lines that mark an "allowed" band
for our sequence. If for any term of the sequence inequality
(1) holds, the points on the graph corresponding to these
terms fall inside the "allowed" band. We see (Fig. 7b) that
starting from number 8, all the terms of the sequence stay
within the limits of the "allowed" band, proving the validity
of (1) for these terms. We, of course, assume that this situa-
tion will realize for all n > 40, i.e. for the whole infinite
"tail" of the sequence not shown in the diagram.
Thus, for the selected a the number N does exist. In
this particular case we found it to be 7.
READER. Hence, we can regard a as the limit of the
sequence.
AUTHOR. Don't you hurry. The definition clearly emphasizes: "for any positive s". So far we have analyzed only one
value of s. We should take another value of a and find N
not for a larger but for a smaller e. If for the second e the
search of N is a success, we should take a third, even smal-
ler e, and then a fourth, still smaller e, etc., repeating
each time the operation of finding N.
In Fig. 7c three situations are drawn up for el, P-2, and 63
(in this case s, > a > E3). Correspondingly, three "allowed"
hands
are
plotted on the graph. For a greater clarity,
each of these bands has its own starting N. We have chosen
N, = 7, N3 = 15, and N3 = 27.
Note that for each selected a we observe the same situation in Fig. 7c: up to a certain n, the sequence, speaking
figuratively, may be "indisciplined" (in other words, some
terms may fall out of the limits of the corresponding "allowed"
hand). However, after a certain n is reached, a very rigid
law sets in, namely, all the remaining terms of the sequence
(their number is infinite) do stay within the band.
READER. Do we really have to check it for an infinite
number of s values?
AUTHOR. Certainly not. Besides, it is impossible. We
25
Dialogue Two
must be sure that whichever value of a > 0 we take, there is
such N after which the whole infinite "tail" of the sequence
will get "locked up" within the limits of the corresponding
"allowed" band.
READER. And what if we are not so sure?
AUTHOR. If we are not and if one can find a value of a
such that it is impossible to "lock up" the infinite "tail" of
the sequence within the limits of its "allowed" hand, then a
is not the limit of our sequence.
READER. And when do we reach the certainty?
AUTHOR. We shall talk this matter over at a later stage
because it has nothing to do with the essence of the definition of the limit of sequence.
I suggest that you formulate this definition anew. Don't
try to reconstruct the wording given earlier, just try to put
it in your own words.
READER. I 11 try. The number a is the limit of a given
sequence if for any positive a there is (one can find) a serial
number n such that for all subsequent, numbers (i.e. for the
whole infinite "tail" of the sequence) the following inequality
holds: I y - a I < e.
AUTHOR. Excellent. You have almost repeated word by
word the definition that seemed to you impossible to remember.
READER. Yes, in reality it all has turned out to be quite
logical and rather easy.
AUTHOR. It is worthwhile to note that the dialectics of
thinking was clearly at work in this case: a concept becomes
"not difficult" because the "complexities" built into it were
clarified. First, we break up the concept into fragments,
the "complexities", then examine the "delicate"
points, thus trying to reach the "core" of the problem.
expose
Then we recompose the concept to make it integral, and, as
a result, this reintegrated concept becomes sufficiently
simple and comprehensible. In the future we shall try first
to find the internal structure and internal logic of the concepts and theorems.
I believe we can consider the concept. of the limit of se-
quence as thoroughly analyzed. I should like to add that,
as a result, the meaning of the sentence "the sequence converges to a" has been explained. I remind you that initially
Limit of Sequence
27
I his sentence seemed to you as requiring no additional explanations.
"READER. At the moment it does not seem so self-evident
any more. True, I see now quite clearly the idea behind it.
AUTHOR. Let us get back to examples (5), (7), and (9).
'I'hese are the sequences that we discussed at the beginning
of our talk. To begin with, we note that the fact that a
sequence (y,) converges to a certain number a is convention-
idly written as
lim yn =a
noo
(it, reads like this: "The limit of yn for n tending to infinity
is (e").
Using the definition of the limit, let us prove that
lim
n
n-.oo n+1
-1;
lim 1 = 0
n-.oo n
[1+(-1)nJ}=07
lim{ 2ni [1-(-1)"j-2(nt
1)
fn .o0
You will begin with the first of the above problems.
READER. 1 have to prove that
liim n
n- oo
I
1
n
choose an arbitrary value of a, for example, a = 0.1.
AUTHOR. I advise you to begin with finding the modulus
of Iyn - a I.
READER."In this case, the-modulus is
_1
n
n+1
I
i
n+1
=`
AUTHOR. Apparently a needn't be specified, at least at
the beginning.
READER. O.K. Therefore, for an arbitrary positive value
of a, I have to find N such that for all n > N the following
inequality holds
n+i
<e,
Dialogue Two
28
AUTHOR. Quite correct. Go on.
READER. The inequality can be rewritten in the form
It follows that the unknown N may be identified as an integral part of e - 1. Apparently, for all n > N the inequality
in question will hold.
AUTHOR. That's right. Let, for example, e = 0.01.
100-1 =99.
READER. Then N=-!, -1
AUTHOR. Let a = 0.001.
READER. Then N = e - 1 = 999.
AUTHOR. Let a = 0.00015.
READER. Then e - 1 = 6665.(6), so that N = 6665.
AUTHOR. It is quite evident that for any a (no matter
how small) we can find a corresponding N.
As to proving that the limits of sequences (7) and (9) are
zero, we shall leave it to the reader as an exercise.
READER. But couldn't the proof of the equality
lim n-IR1 = 1 be simplified?
n-oo
AUTHOR. Have a try.
READER. Well, first I rewrite the expression in the follown
= lim
ing way: lim n+1
1
Then I take into conn
sideration that with an increase in n, fraction
will
tend to zero, and, consequently, can be neglected against
n
unity. Hence, we may reject n and have: lim
AUTHOR. In practice this is the method generally used.
However one should note that in this case we have assumed,
first,
that lim
n-oo
0,
and, second, the validity of the
Limit of Sequence
20
ollowing rules
lim xn
limn = n-.oo
(2)
lim yn
yn
n-.oo
lira (xn + Zn) =1im xn + lira Zn
(3)
n-oo
n-oo
n-.oo
where Xn = 1, yn = 1 + 1n , and Zn = 1n . Later on we
shall discuss these rules, but at this juncture I suggest
that we simply use them to compute several limits. Let us
discuss two examples.
Example 1. Find lim 3n-1
5n-6
n-.oo
READER. It will be convenient to present the computation in the form
3-
1
n
lim 3n-1 = lim
n-0. 5n-6 n---00 5- n
"M
n_00
lim
(3- n )
(5-
AUTHOR. O.K. Example 2. Compute
-
3
5
)/
n
6n'-1
lim 5n'-t-2n-1
n-+oo
READER. We write
6na -1
lim 5n'+2n-1
_ lim 6n- 1n
5n+2--n
1
AUTHOR. Wait a moment! Did you think about the
reason for dividing both the numerator and denominator
of the fraction in the previous example by n? We did this
because sequences (3n - 1) and (5n - 6) obviously have
no limits, and therefore rule (2) fails. However, each of
sequences (3
n) and (5 n) has a limit.
-
-
READER. I have got your point. It means that in example
2 1 have to divide both the numerator and denominator
Ol
bialogue three
by n`2 to obtain the sequences with limits in both. According-
ly we obtain
6nz-1
6
lim 5n9-F2n-1 -gym 5+ 2
lim (6-1
n2
1
n-.oo
nz
_
1
n?
lit (5+_L
1
\
5
n+ao
AUTHOR. Well, we have examined the concept of the
limit of sequence. Moreover, we have learned a little how to
calculate limits. Now it is time to discuss some properties
of sequences with limits. Such sequences are called convergent.
DIALOGUE THREE
CONVERGENT SEQUENCE
AUTHOR. Let us prove the following
Theorem:
If a sequence has a limit, it is bounded.
We assume that a is the limit of a sequence (yn). Now
take an arbitrary value of E greater than 0. According to
the definition of the limit, the selected e can always be relat-
ed to N such that for all n > N, I yn - a I < e. Hence,
starting with n = N H- 1, all the subsequent terms of the
sequence satisfy the following inequalities
a-E<yn<a+a
As to the terms with serial numbers from 1 to N, it is always
possible to select both the greatest (denoted by B1) and the
least (denoted by A1) terms since the number of these terms
is finite.
Now we have to select the least value from a - e and Al
(denoted by A) and the greatest value from a + E and Bl
(denoted by B). It is obvious that A < yn < B for all the
terms of our sequence, which proves that the sequence (yn)
is bounded.
READER. I see.
Convergent Sequence
3i
AUTHOR. Not too well, it seems. Let us have a look at
the logical structure of the proof. We must verify that if the
sequence has a limit, there exist two numbers A and B such
that A < y, < B for each term of the sequence. Should
the sequence contain a finite number of terms, the existence
of such two numbers would be evident. However, the sequence
contains an infinite number of terms, the fact that complicates the situation.
READER. Now it is clear! The point is that if a sequence
has a limit a, one concludes that in the interval from a - e
Io a + a we have an infinite set of yn starting from n =
= N + 1 so that outside of this interval we shall find only
a finite number of terms (not larger than N).
AUTHOR. Quite correct. As you see, the limit "takes
care of" all the complications associated with the behaviour
of the infinite "tail" of a sequence. Indeed, I yn - a I < s
for all n > N, and this is the main "delicate" point of this
theorem. As to the first N terms of a sequence, it is essential
that their set is finite.
READER. Now it is all quite lucid. But what about e?
1 is value is not preset, we have to select it.
AUTHOR. A selection of a value for a affects only N.
If you take a smaller a, you will get, generally speaking,
a larger N. However, the number of the terms of a sequence
which do not satisfy
I
y,, - a I < s will remain finite.
And now try to answer the question about the validity of
the converse theorem: If a sequence is bounded, does it
imply it is convergent as well?
READER. The converse theorem is not true. For example,
sequence (10) which was discussed in the first dialogue is
bounded. However, it has no limit.
AUTHOR. Right you are. We thus come to a
Corollary:
The boundedness of a sequence is a necessary condition for
its convergence; however, it is not a sufficient condition. If
a sequence is convergent, it is bounded. If a sequence is unbounded, it is definitely nonconvergent.
READER. I wonder whether there is a sufficient condition
for the convergence of a sequence?
AUTHOR. We have already mentioned this condition
in the previous dialogue, namely, simultaneous validity
Dialogue Phree
32
of both the boundedness and monotonicity of a sequence.
The Weierstrass theorem states:
If a sequence is both bounded and monotonic, it has a limit.
Unfortunately, the proof of the theorem is beyond the
scope of this book; we shall not give it. I shall simply ask
you to look again at sequences
(5), (7), and (13) (see Dialogue
One), which satisfy the conditions of the Weierstrass theorem.
READER. As far as I understand, again the converse theorem is not true. Indeed, sequence
(9) (from Dialogue One) has a
limit but is not monotonic.
AUTHOR. That is correct. We
- bounded
sequences
- Monotonic
sequences
- convergent
thus come to the following
Conclusion:
If a sequence is both monotonic
and bounded, it is a sufficient
(but not necessary) condition for
its convergence .
READER. Well, one can easily get confused.
Fig. 8
AUTHOR. In order to avoid
confusion, let us have a look
at another illustration (Fig. 8). Let us assume that all bounded sequences are "collected" (as if we were picking marbles
sequences
scattered on the floor) in an area shaded by horizontal
lines, all monotonic sequences are collected in an area shaded
by tilted lines, and, finally, all convergent sequences are
collected in an area shaded by vertical lines. Figure 8 shows
how all these areas overlap, in accordance with the theorems
discussed above (the actual shape of all the areas is, of course,
absolutely arbitrary). As follows from the figure, the area
shaded vertically is completely included into the area shaded horizontally. It means that any convergent sequence must
be also bounded. The overlapping of the areas shaded horizon-
tally and by tilted lines occurs inside the area shaded vertically. It means that any sequence that is both bounded and
monotonic must be convergent as well. It is easy to deduce
that only five types of sequences are possible. In the figure
Convergent Sequence
33
Ilie points designated by A, B, C, D, and E identify five
sequences of different types. Try to name these sequences
uiid find the corresponding examples among the sequences
discussed in Dialogue One.
HEADER. Point A falls within the intersection of all the
Iliree areas. It represents a sequence which is at the same
lime bounded, monotonic, and convergent. Sequences (5),
(7), and (13) are examples of such sequences.
AUTHOR. Continue, please.
HEADER. Point B represents a bounded, convergent
1)IIL nonmonotonic sequence. One example is sequence (9).
Point C represents a bounded but neither convergent nor
ionotonic sequence. One example of such a sequence is
sequence (10).
Point D represents a monotonic but neither convergent
'liar bounded sequence. Examples of such sequences are (1),
(2), (3), (4), (6), (11), and (12).
Point E is outside of the shaded areas and thus represents
a sequence neither monotonic nor convergent nor bounded.
one example is sequence (8).
AUTHOR. What type of sequence is impossible then?
HEADER. There can be no bounded, monotonic, and
nonconvergent sequence. Moreover, it is impossible to have
Loth unboundedness and convergence in one sequence.
AUTHOR. As you see, Fig. 8 helps much to understand
f lie relationship between such properties of sequences as
l,uundedness, monotonicity, and convergence.
In what follows, we shall discuss only convergent sequences. We shall prove the following
Theorem:
A convergent sequence has only one limit.
This is the theorem of the uniqueness of the limit. It means
that a convergent sequence cannot have two or more limits.
Suppose the situation is contrary to the above statement.
Consider a convergent sequence with two limits a1 and a2
and select a value for 6 < la13 all . Now assume, for
example, that a = 1x13 all . Since a1 is a limit, then for
the selected value of a there is N1 such that for all n > N1
the terms of the sequence (its infinite "tail") must fall inside
3-01473
Dialogue Three
the interval 1 (Fig. 9). It means that we must have
y,, - al I < e. On the other hand, since a2 is a limit
there is N2 such that for all n> N2 the terms of the sequence
(again its infinite "Lail") must fall inside the interval 2,
It means that we must have I y - a2 I < e,. Hence, we
obtain that for all N greater than the largest among Nl
Fig. 9
and N2 the impossible must hold, namely, the terms of the
sequence must simultaneously belong to the intervals 1
and 2. This contradiction proves the theorem.
This proof contains at least two rather "delicate" points.
Can you identify them?
HEADER. I certainly notice one of them. If al and ay
are limits, no matter how the sequence behaves at the beginning, its terms in the long run have to concentrate simultaneously around al and a2i which is, of course, impossible,
AUTHOR. Correct. But there is one more "delicate"
point, namely, no matter how close al and a2 are, they
should inevitably be spaced by a segment (a gap) of a small
but definitely nonzero length.
READER. But it is self-evident.
AUTHOR. I agree. However, this "self-evidence" is connected to one more very fine aspect without which the very
calculus could not be developed. As you probably noted, one
cannot identify on the real line two neighbouring points.
If one point is chosen, it is impossible, in principle, to point
out its "neighbouring" point. In other words, no matter how
carefully you select a pair of points on the real line, it ie
always possible to find any number of points between the
two.
Take, for example, the interval [0, 11. Now, exclude the
point 1. You will have a half-open interval [0, fl. Can you
identify the largest number over this interval?
Convergent Sequence
35
READER. No, it is impossible.
AUTHOR. That's right. However, if there were a point
neighbouring 1, after the removal of the latter this "neighhour" would have become the largest number. I would like
to note here that many "delicate" points and many "secrets"
in the calculus theorems are ultimately associated with the
impossibility of identifying two neighbouring points on the
real line, or of specifying the greatest or least number on an
open interval of the real line.
But let us get back to the properties of convergent sequences and prove the following
Theorem:
If sequences (y,) and (Zn) are convergent (we denote their
limits by a and b, respectively), a sequence (Yn + Zn) is convergent too, its limit being a + b.
for the same s there is N2 such that for all the terms of the
second sequence with n > N2 we shall have I Zn - b I < s.
I f now we select the greatest among Nl and N2 (we denote
i I, by N), then for all n> N both
I Yn - a I < a and
Zn - b I < a. Well, this is as far as I can go.
AUTHOR. Thus, you have established that for an arbiI nary s there is N such that for all n> N both I yn - a I < e
and I zn -- b I < e simultaneously. And what can you say
a lout the modulus (yn + Zn) - (a + b) (for all n)?
I remind you that I A + B I<IA I+ IB
HEADER. Let us look at
I
I
I(yn+Zn)`(a+b) I= I(yn-a)+(Zn-b) i
< [ I yn - a I + I Z, - b II < (e + E) = 2E
AUTHOR. You have proved the theorem, haven't you?
HEADER. But we have only established that there is N
such that for all n > N we have (yn + Zn) - (a + b) I <
I
30
Dialogue Three
36
< 2e..
But we need to prove that
I(yn+zn)-(a+b) I < e
AUTHOR. Ah, that's peanuts, if you forgive the expression. In the case of the sequence (yn + zn) you select a value
of e, but for the sequences (yn) and (zn) you must select a
value of 2 and namely for this value find Nl and N2.
'T'hus, we have proved that if the sequences (yn) and (zn)
are convergent, the sequence (yn + zn) is convergent too.
We have even found a limit of the sum. And do you think
that the converse is equally valid?
READER. I believe it should be.
AUTHOR. You are wrong. Here is a simple illustration:
1
2
1
4
1
6
1
1
3
1
5
1
(Zn) = fit 3 s 4 t 5 t -61
1
g .. .
7
(yn+zn)=1,1,1, 1,
As you see, the sequences (yn) and (zn) are not convergent,
while the sequence (yn + zn) is convergent, its limit being
equal to unity.
'Thus, if a sequence (yn + zn) is convergent, two alternatives are possible:
sequences (yn) and (zn) are convergent as well, or
sequences (yn) and (zn) are divergent.
READER. But can it be that the sequence (yn) is convergent, while the sequence (zn) is divergent?
AUTHOR. It may be easily shown that this is impossible.
To begin with, let us note that if the sequence (yn) has a
limit a, the sequence (-yn) is also convergent and its limit
is -a. This follows from an easily proved equality
Tim (Cyn) = C lira yn
n-+w
n-w
where c is a constant.
Assume now that a sequence (yn + zn) is convergent to A,
and that (y,) is also convergent and its limit is a. Let us
apply the theorem on the sum of convergent sequences to
the sequences (Yn + zn) and (-y, ). As a result, we obtain
Convergent Sequence
37
that the sequence (yn, + zn - yn), i.e. (zn), is also convergent, with the limit A - a.
READER. Indeed (zn) cannot be divergent in this case.
AUTHOR. Very well. Let us discuss now one important
particular case of convergent sequences, namely, the socalled infinitesimal sequence, or simply, infinitesimal. This is
the name which is given to a convergent sequence with
a limit equal to zero. Sequences (7) and (9) from Dialogue
One are examples of infinitesimals.
Note that to any convergent sequence (yn) with a limit a
there corresponds an infinitesimal sequence (an), where
an = yn - a. That is why mathematical analysis. is also
called calculus of infinitesimals.
Now I invite you to prove the following
Theorem:
If (yn) is a bounded sequence and (an) is infinitesimal, then
(!Inan)
is infinitesimal as well.
READER. Let us select an arbitrary e > 0. We must
prove that there is N such that for all n > N the terms of
the sequence (ynan) satisfy the inequality I ynan I < e.
AUTHOR. Do you mind a hint? As the sequence (yn) is
hounded, one can find M such that I yn I < M for any n.
READER. Now all becomes very simple. We know that
the sequence (an) is infinitesimal. It means that for any
r' > 0 we can find N such that for all n > N I an I < E'.
For F,', I select M . Then, for n > N we have
Iynanl = IynIlanl<Mlanl<M M =e
This completes the proof.
AUTHOR. Excellent. Now, making use of this theorem,
it
is very easy to prove another
Theorem:
A sequence (ynzn) is convergent to ab if sequences (yn) and
(zn) are convergent to a and b, respectively.
Suppose y, = a + an and zn = b + Pn. Suppose also
That the sequences (an) and ((3n) are infinitesimal. Then we
can write:
Ynzn = ab + 'In, where 'Yn = ban + an + an Nn
Dialogue Three
38
Making use of the theorem we have just proved, we conclude
that the sequences (ban), (apn), and (a, 3n) are infinitesimal.
READER. But what justifies your conclusion about the
sequence (anF'n)?
AUTHOR. Because any convergent sequence (regardless
of whether it is infinitesimal or not) is bounded.
From the theorem on the sum of convergent sequences we
infer that the sequence (y,,) is infinitesimal, which immediate-
ly yields
lim (ynZn) = ab
n-0°
This completes the proof.
READER. Perhaps we should also analyze inverse variants in which the sequence (ynzn) is convergent. What can
be said in this case about the sequences (y,,) and (z,)?
AUTHOR. Nothing definite, in the general case. Obvious-
ly, one possibility is that (yn) and (zn) are convergent.
However, it is also possible, for example, for the sequence
(yn) to be convergent, while the sequence (zn) is divergent.
Here is a simple illustration:
(yn) = 1
1
1
4
9 ' 16
1
1
1
T5-,
n2 ,
(Zn) = 1, 2, 3, 4, 5, . . ., n, .. .
(ynZn) = 1,
1
1
1
2
3
4 ' 5
1
1
By the way, note that here we obtain an infinitesimal sequence by multiplying an infinitesimal sequence by an unbounded sequence. In the general case,
such multiplication needn't produce an infinitesimal.
Finally, there is a possibility when the sequence
is convergent, and the sequences (yn) and (z,,) are divergent..
Here is one example:
1
1
(yn) = 1, 4 ,
3' i6
1
(Zn) = 1, 2, 9 , 4,
1
7,
' 5, 36 '
1
25
.. .
1
6, 4o ,
(ynZn)=1, 2 e 3, 4 e 5, s e 7 e ..
1
1
1
1
1
1
Convergent Sequence
39
Now, let us formulate one more
Theorem:
(y,) and (zn) are sequences convergent to a and b
when b0, then a sequence (in) is also convergent, its
If
limit being L-° .
We shall omit the proof of this theorem.
READER. And what if the sequence (zn) contains zero
terms?
AUTHOR. Such terms are possible. Nevertheless, the
number of such terms can be only finite. Do you know why?
READER. I think, I can guess. The sequence (zn,) has
a nonzero limit b.
AUTHOR. Let us specify b > 0.
READER. Well, I select e = . There must be an integer N such that
znn-b
2 for all n > N. Obviously,
all Zn (the whole infinite "tail" of the sequence) will be pos-
itive. Consequently, the zero terms of the sequence (zn)
may only be encountered among a finite number of the
first N terms.
AUTHOR. Excellent. Thus, the number of zeros among
the terms of (zn) can only be finite. If such is the case, one
can surely drop these terms. Indeed, an elimination of any
finite number of terms of a sequence does not affect its properties.
For example, a convergent sequence still remains convergent,
with its limit unaltered. An elimination of a finite number
of terms may only change N (for a given E), which is certainly unimportant.
READER. It is quite evident to me that by eliminating
a finite number of terms one does not affect the convergence
of a sequence. But could an addition of a finite number of
terms affect the convergence of a sequence?
AUTHOR. A finite number of new terms does not affect
the convergence of a sequence either. No matter how many
new terms are added and what their new serial numbers are,
ones can always find the greatest number N after which the
whole infinite "tail" of the sequence is unchanged. No matter
how large the number of new terms may be and where you
40
Dialogue Three
insert them, the finite set of new terms cannot change the
infinite "tail" of the sequence. And it is the "tail" that determines the convergence (divergence) of a sequence.
Thus, we have arrived at the following
Conclusion:
Elimination, addition, and any other change of a finite
number of terms of a sequence do not affect either its convergence or its limit (if the sequence is convergent).
READER. I guess that an elimination of an infinite number of terms (for example, every other term) must not affect
the convergence of a sequence either.
AUTHOR. Here you must be very careful. If an initial
sequence is convergent, an elimination of an infinite number
of its terms (provided that the number of the remaining
terms is also infinite) does not affect either convergence or
the limit of the sequence. If, however, an initial sequence
is divergent, an elimination of an infinite number of its
terms may, in certain cases, convert the sequence into
a convergent one. For example, if you eliminate from diver-
gent sequence (10) (see Dialogue One) all the terms with
even serial numbers, you will get the convergent sequence
1,
1
3'
1
1
1
1
1
, 7' 9' 11' 13'
Suppose we form from a given convergent sequence two
new convergent sequences. The first new sequence will
consist of the terms of the initial sequence with odd serial
numbers, while the second will consists of the terms with
even serial numbers. What do you think are the limits of
these new sequences?
READER. It is easy to prove that the new sequences will
have the same limit as the initial sequence.
AUTHOR. You are right.
Note that from a given convergent sequence we can form
not only two but a finite number m of new sequences converg-
ing to the same limit. One way to do it is as follows. The
first new sequence will consist of the 1st, (m + 1)st,
(2m + 1)st, (3m + 1)st, etc., terms of the initial sequence.
The second sequence will consist of the 2nd, (m + 2)nd,
(2m + 2)nd, (3m + 2)nd, etc., terms of the initial sequence.
Function
41
Similarly we can form the third, the fourth, and other sequences.
In conclusion, let us see how one can "spoil" a convergent
sequence by turning it into divergent. Clearly, different
"spoiling" approaches are possible. Try to suggest something
simple.
READER. For example, we can replace all the terms with
even serial numbers by a constant that is not equal to the
limit of the initial sequence. For example, convergent
sequence (5) (see Dialogue One) can be "spoilt" in the following manner:
2, 2, 4, 2, s, 2, 8,2,...
AUTHOR. I see that you have mastered very well the essence of the concept of a convergent sequence. Now we are
ready for another substantial step, namely, consider one of
the most important concepts in calculus: the definition of
a function.
DIALOGUE FOUR
FUNCTION
READER. Functions are widely used in elementary
mathematics.
AUTHOR. Yes, of course. You are familiar with numerical
functions. Moreover, you have worked already with different
numerical functions. Nevertheless, it will be worthwhile to
dwell on the concept of the function. To begin with, what
is your idea of a function?
READER. As I understand it, a function is a certain correspondence between two variables, for example, between x
and y. Or rather, it is a dependence of a variable y on a
variable x.
AUTHOR. What do you mean by a "variable"?
READER. It is a quantity which may assume different
values.
42
Dialogue Four
AUTHOR. Can you explain what your understanding of
the expression "a quantity assumes a value" is? What does it
mean? And what are the reasons, in particular, that make
a quantity to assume this or that value? Don't you feel that
the very concept of a variable quantity (if you are going
to use this concept) needs a definition?
READER. O.K., what if I say: a function y = f (x)
symbolizes a dependence of y on x, where x and y are numbers.
AUTHOR. I see that you decided to avoid referring to the
concept of a variable quantity. Assume that x is a number
and y is also a number. But then explain, please, the meaning of the phrase "a dependence between two numbers".
READER. But look, the words "an independent variable"
and "a dependent variable" can be found in any textbook on
mathematics.
AUTHOR. The concept of a variable is given in textbooks
on mathematics after the definition of a function has been
introduced.
READER. It seems I have lost my way.
AUTHOR. Actually it is not all that difficult "to construct" an image of a numerical function. I mean image,
not mathematical definition which we shall discuss later.
In fact, a numerical function may be pictured as a "black
box" that generates a number at the output in response to a
number at the input. You put into this "black box" a number
(shown by x in Fig. 10) and the "black box" outputs a new
number (y in Fig. 10).
Consider, for example, the following function:
y=4x2-1
If the input is x = 2, the output is y = 15; if the input is
x = 3, the output is y = 35; if the input is x = 10, the
output is y = 399, etc.
READER. What does this "black box" look like? You
have stressed that Fig. 10 is only symbolic.
AUTHOR. In this particular case it makes no difference.
It does not influence the essence of the concept of a function.
But a function can also be "pictured" like this:
4 02-1
Function
43
The square in this picture is a "window" where you input,
the numbers. Note that there may be more than one "window". For example,
4 2-1
READER. Obviously, the function you have in mind is
4x2 -1
Y
Ixl+1
AUTHOR. Sure. In this case each specific value should
be input into both "windows" simultaneously.
EMM
"Black box"
working as a
function
0
Fig. 20
By the way, it is always important to see such a "window"
(or "windows") in a formula describing the function. Assume,
for example, that one needs to pass from a function y = f (x)
to a function y = / (x - 1) (on a graph of a function this
transition corresponds to a displacement of the curve in the
positive direction of the x-axis by 1). If you clearly understand the role of such a "window" ("windows"), you will
simply replace in this
(these "windows") x by
x - 1. Such an operation is illustrated by Fig. 11 which
represents the following function
4x2-1
Y
1=1+1
Dialogue Four
44
Obviously, as a result of substitution of x - 1 for x we
arrive at a new function (new "black box")
4(x-1)$-1
I
y = f( x ) to
see. If, for example, we wanted to pass from
i
,
y= f (x)
the function pictured in Fig. 19
Fig. 11
would be transformed as follows:
Y=
4
72
1
,zl-{-1
AUTHOR. Correct. Now. try to find y = f (x) if
2f ()-f(x)=3x
READER. I am at a loss.
AUTHOR. As a hint, I suggest replacing x by a .
READER. This yields
2f (x) - f (1) - 3
Now it is clear. Together with the initial equation, the
new equation forms a system of two equations for f (x)
Function
and f
45
1
2fls)-f(x)=3x
2f(x)_f(x)=
3
By multiplying all the terms of the second equation by 2
and then adding them to the first equation, we obtain
f (x) = x+ 2
AUTHOR. Perfectly true.
READER. In connection with your comment about the
numerical function as a "black box" generating a numerical
num
nu
funs
numerical
fanction
operator
numerical
function
functional
numerical
function
Fig. 12
output in response to a numerical input, I would like to ask
whether other types of "black boxes" are possible in calculus.
AUTHOR. Yes, they are. In addition to the numerical
function, we shall discuss the concepts of an operator and
a functional.
READER. I must confess I have never heard of such concepts.
AUTHOR. I can imagine. I think, however, that Fig. 12
will be helpful. Besides, it will elucidate the place and role
of the numerical function as a mathematical tool. Figure 12
shows that:
a numerical function is a "black box" that generates a number at the output in response to a number at the input;
an operator is a "black-box" that generates a numerical
46
Dialogue Pour
function at the output in response to a numerical function
at the input; it is said that an operator applied to a function
generates a new function;
a functional is a "black box" that generates a number at
the output in response to a numerical function at the input,
i.e. a concrete number is obtained "in response" to a concrete
function.
READER. Could you give examples of operators and functionals?
AUTHOR. Wait a minute. In the next dialogues we shall
analyze both the concepts of an operator and a functional.
So far, we shall confine ourselves to a general analysis of
both concepts. Now we get back to our main object, the
numerical function.
The question is: How to construct a "black box" that
generates a numerical function.
READER. Well, obviously, we should find a relationship,
or a law, according to which the number at the "output" of
the "black box" could be forecast for each specific number
introduced at the "input".
AUTHOR. You have put it quite clearly. Note that such
a law could be naturally referred to as the law of numerical
correspondence. However, the law of numerical correspondence
would not be a sufficient definition of a numerical function.
READER. What else do we need?
AUTHOR. Do you think that any number could be fed
into a specific "black box" (function)?
READER. I see. I have to define a set of numbers acceptable as inputs of the given function.
AUTHOR. That's right. This set is said to be the domain
of a function.
Thus, the definition of a numerical function is based on
two "cornerstones":
the domain of a function (a certain set of numbers), and
the law of numerical correspondence.
According to this law, every number from the domain of
a function is placed in correspondence with a certain number,
which is called the value of the function; the values form the
range of the function.
READER. Thus, we actually have to deal with two numer-
function
47
ical sets. On the one hand, we have a set called the domain
of a function and, on the other, we have a set called the
range of a function.
AUTHOR. At this juncture we have come closest to
a mathematical definition of a function which will enable
us to avoid the somewhat mysterious word "black box".
Look at Fig. 13. It shows the function y = V1 - x2.
Figure 13 pictures two numerical sets, namely, D (represented by the interval [-1, 1])
and E (the interval
[0, Q. For your convenience these
sets
are
shown on two different
real lines.
The set D is the domain
of the function, and E is
its range. Each number
in D corresponds to one
number in E (every input value is placed in
correspondence with one
output value). This correspondence is shown in
Fig. 13 by arrows point-
ing from D to E.
D
Fig. 13
READER. But Figure 13 shows that two different num-
bers in D correspond to one number in E.
AUTHOR. It does not contradict the statement "each
number in D corresponds to one number in E". I never said
that different numbers in D must correspond to different
numbers in E. Your remark (which actually stems from specific characteristics of the chosen function) is of no principal
significance. Several numbers in D may correspond to one
number in E. An inverse situation, however, is forbidden.
It is not allowed for one number in D to correspond to more
than one number in E. I emphasize that each number in D
must correspond to only one (not morel) number in E.
Now we can formulate a mathematical definition of the
numerical function.
Definition:
Take two numerical sets D and E in which each element x
Dialogue Four
48
of D (this is denoted by x ED) is placed in one-to-one correspond-
ence with one element y of E. Then we say that a function
y = f (x) is set in the domain D, the range of the function
being E. It is said that the argument x of the function y passes
through D and the values of y belong to E.
Sometimes it is mentioned (but more often omitted alto-
gether) that both D and E are subsets of the set of real
numbers R (by definition, R is the real line).
On the other hand, the definition of the function can be
reformulated using the term "mapping". Let us return again
to Fig. 13. Assume that the number of arrows from the points
of D to the points of E is infinite (just imagine that such
arrows have been drawn from each point of D). Would you
agree that such a picture brings about an idea that D is
mapped onto E?
READER. Really, it looks like mapping.
AUTHOR. Indeed, this mapping can be used to define
the f unction.
Definition:
A numerical function is a mapping of a numerical set D
(which is the domain of the function) onto another numerical
set E (the range of this function).
Thus, the numerical function is a mapping of one numerical
set onto another numerical set. The term "mapping" should be
understood as a kind of numerical correspondence discussed
above. In the notation y = f (x), symbol f means the function
itself (i.e. the mapping), with x E D and Y E E.
READER. If the numerical function is a mapping of one
numerical set onto another numerical set, then the operator
can be considered as a mapping of a set of numerical function
onto another set of functions, and the functional as a mapping of a set of functions onto a numerical set.
AUTHOR. You are quite right.
READER. I have noticed that you persistently use the
term "numerical function" (and I follow suit), but usually
one simply says "function". Just how necessary is the word
"numerical"?
AUTHOR. You have touched upon a very important
aspect. The point is that in modern mathematics the concept
of a function is substantially broader than the concept of a
numerical function. As a matter of fact, the concept of a
Function
49
function includes, as particular cases, a numerical function
as well as an operator and a functional, because the essence
in all the three is a mapping of one set onto another independently of the nature of the sets. You have noticed that
both operators and functionals are mappings of certain sets
onto certain sets. In a particular case of mapping of a numerical set onto a numerical set we come to a numerical function.
In a more general case, however, sets to be mapped can be
arbitrary. Consider a few examples.
Example 1. Let D be a set of working days in an academic
year, and E a set of students in a class. Using these sets,
we can define a function realizing a schedule for the students on duty in the classroom. In compiling the schedule,
each element of D (every working day in the year) is placed
in one-to-one correspondence with a certain element of E
(a certain student). This function is a mapping of the set
of working days onto the set of students. We may add that the
domain of the function consists of the working days and the
range is defined by the set of the students.
READER. It sounds a bit strange. Moreover, these sets
have finite numbers of elements.
AUTHOR. This last feature is not principal.
READER. The phrase "the values assumed on the set of
students" sounds somewhat awkward.
AUTHOR. Because you are used to interpret "value" as
"numerical value".
Let us consider some other examples.
Example 2. Let D be a set of all triangles, and E a set of
positive real numbers. Using these sets, we can define two
functions, namely, the area of a triangle and the perimeter
of a triangle. Both functions are mappings (certainly, of
different nature) of the set of the triangles onto the set of the
positive real numbers. It is said that the set of all the triangles is the domain of these functions and the set of the positive
real numbers is the range of these functions.
Example 3. Let D be a set of all triangles, and E a set of
all circles. The mapping of D onto E can be either a circle
inscribed in a triangle, or a circle circumscribed around
a.triangle. Both have the set of all the triangles as the domain
of the function and the set of all the circles as the range of
the function.
4-01473
Dialogue Four
50
By the way, do you think that it is possible to "construct"
an inverse function in a similar way, namely, to define
a function with all the circles as its domain and all the
triangles as its range?
READER. I see no objections.
AUTHOR. No, it is impossible. Because any number of
different triangles can be inscribed in or circumscribed
around a circle. In other words, each element of E (each
circle) corresponds to an infinite number of different elements
of D (i.e. an infinite number of triangles). It means that there
is no function since no mapping can be realized.
However, the situation can be improved if we restrict
the set of triangles.
READER. I guess I know how to do it. We must choose
the set of all the equilateral triangles as the set D. Then it
becomes possible to realize both a mapping of D onto E
(onto the set of all the circles) and an inverse mapping,
i.e. the mapping of E onto D, since only one equilateral
triangle could be inscribed in or circumscribed around a
given circle.
AUTHOR. Very good. I see that you have grasped the es-
sence of the concept of functional relationship. I should
emphasize that from the broadest point of view this concept
is based on the idea of mapping one set of objects onto
another set of objects. It means that a function can be realized
as a numerical function, an operator, or a functional. As we
have established above, a function may be represented by an
area or perimeter of a geometrical figure, such - as a circle
inscribed in a triangle or circumscribed around it, or it may
take the form of a schedule of students on duty in a classroom,
etc. It is obvious that a list of different functions may be
unlimited.
READER. I must admit that such a broad interpretation
of the concept of a function is very new to me.
AUTHOR. As a matter of fact, in a very diverse set of
possible functions (mappings), we shall use only numerical
functions, operators, and functionals. Consequently, we shall
refer to numerical functions as simply functions, while
operators and functionals will be pointed out specifically.
And now we shall examine the already familiar concept of
a numerical sequence as an example of mapping.
function
5t
READER. A numerical sequence is, apparently, a mapping of a set of natural numbers onto a different numerical
set. The elements of the second set are the terms of the
sequence. Hence, a numerical sequence is a particular case
of a numerical function. The domain of a function is repre-
sented by a set of natural numbers.
AUTHOR. This is correct. But you should bear in mind
that later on we shall deal with numerical functions whose
domain is represented by the real line, or by its interval (or
intervals), and whenever we mention a function, we shall
imply a numerical function.
In this connection it is worthwhile to remind you of the
classification of intervals. In the previous dialogue we have
already used this classification, if only partially.
First of all we should distinguish between the intervals
of finite length:
a closed interval that begins at a and ends at b is denoted
by [a, b]; the numbers x composing this interval meet the
inequalities a G x < b;
an open interval that begins at a and ends at b is denoted
by ]a, b[; the numbers x composing this interval meet the
inequalities a < x < b;
a half-open interval is denoted either by ]a, b] or [a, b[,
the former implies that a < x < b, and the latter that
a<x<b.
The intervals may also be infinite:
]-oo, oo[ (-oo<x<oo)-the real line
] a, oo [
(a<x<oo);
[a, oo [
(a<x<oo)
]-oo, b[ (-oo<x<b); ]-oo, b] (-oo<x<b)
Let us consider several specific examples of numerical
functions. Judging by the appearance of the formulas given
below, point out the intervals constituting the domains of
the following functions:
y=V1x2
Y=
k
Vx -1
Y2-x
(1)
(2)
(3)
btalogue pour
52
('' __
W
1
(
x-1
y-
1
(5)
V2-x
y=Vx-1+V
y
4)
1
= 7x -l +
(6)
1
2-x
(7)
(
y=Vx-1+
8)
1
y2-x
(9)
READER. It is not difficult. The domain of function (1)
is the interval (-1, 11; that of (2) is [1, oo[; that of (3)
is ]- oo, 21; that of (4) is 11, oo[; that of (5) is ]- oo, 2[;
that of (6) is [1, 2], etc.
AUTHOR. Yes, quite right, but may I interrupt you to
emphasize that if a function is a sum (a difference, or a
product) of two functions, its domain is represented by the
intersection of the sets which are the domains of the constituent functions. It is well illustrated by function (6). As
a matter of fact, the same rule must be applied to functions
(7)-(9). Please, continue.
READER. The domains of the remaining functions are (7)
]1, 21; (8) ]1, 2]; (9) 11, 21.
AUTHOR. And what can you say about the domain of
?
the function y = x - 2 -}READER. The domain of y = V x - 2 is [2, oo [, while
that of y = V 1- x is ] - oo, 1]. These intervals do not
intersect.
AUTHOR. It means that the formula y=Vx-2 +
+ 11r1 --x does not define any function.
More on Function
DIALOGUE FIVE
MORE ON FUNCTION
AUTHOR. Let us discuss the methods of defining functions. One of them has already been employed quite extensively. I mean the analytical description of a function by some
formula, that is, an analytical expression (for example,
expressions (f) through (9) examined at the end of the preceding dialogue).
READER. As a matter of fact, my concept of a function
was practically reduced to its representation by a-formula.
It was a formula that I had in mind whenever I spoke about
a dependence of a variable y on a variable x.
AUTHOR. Unfortunately, the concept of a function as a
formula relating x and y has long been rooted in the minds
of students. This is, of course, quite wrong. A function and
its formula are very different entities. It is one thing to
define a function as a mapping of one set (in our case it is
a numerical set) onto another, in other words, as a "black
box" that generates a number at the output in response
to a number at the input. It is quite another thing to have
just a formula, which represents only one of the ways of
defining a function. It is wrong to identify a function with
a formula giving its analytical description (unfortunately,
it happens sometimes).
READER. It seems that after the discussion in the previous dialogue about the function, such identification in
a general case is automatically invalidated. However, if we
confine ourselves only to numerical functions and if we bear
in mind that working with a function we always use a formula to describe it, a question arises: Why is it erroneous to
identify these two notions? Why should we always emphasize
the difference between the function and its formula?
AUTHOR. I'll tell you why. First, not every formula defines a function. Actually, at the end of the previous dialogue we already had such an example. I shall give you some
more: y =
i
+
1
,
y = log x + log (-x),
y
Dialogue Five
54
= Ysin _x- 2, y = log (sin x - 2), etc. These formulas
do not represent any functions.
Second (and this is more important), not all functions
can be written as formulas. One example is the so-called
Dirichlet function which is defined on the real line:
1
V-
0
if x is a rational number
if x is an irrational number
READER. You call this a function?
AUTHOR. It is certainly an unusual function, but still
a function. It is a mapping of a set of rational numbers to
unity and a set of irrational numbers to zero. The fact that
you cannot suggest any analytical expression for this function is of no consequence (unless you invent a special symbol
for the purpose and look at it as a formula).
However, there is one more, third and probably the most
important, reason why functions should not be identified
with their formulas. Let us look at the following expression:
cos x,
y
1-}- x2,
xC0
0<x<2
log(x-1), x>2
How many functions have I defined here?
READER. Three functions: a cosine, a quadratic function,
and a logarithmic function.
AUTHOR. You are wrong. The three formulas (y = cos x,
y = 1 + x2, and y = log (x - 1)) define in this case a
single function. It is defined on the real line, with the law of
numerical correspondence given as y = cos x over the inter-
val ] - oo, Of, as y = 1 + x2 over the interval [0, 2], and
as y = log (x - 1) over the interval 12, oo[.
READER. I've made a mistake because I did not, think
enough about the question.
AUTHOR. No, you have made the mistake because
subconsciously you identified a function with its analytical
expression, i.e. its formula. Later on, operating with functions, we shall use formulas rather extensively. However,
you should never forget that a formula is not all a function
is. It is only one way of defining it.
More on Function
55
The example above illustrates, by the way, that one should
not identify such notions as the domain of a function and
the range of x on which an analytical expression is defined
(i.e. the domain of an analytical expression). For example,
the expression 1 + x2 is defined on the real line. However,
in the example above this expression was used to define
the function only over the interval [0, 21.
It should be emphasized that the question about the domain of a function is of principal significance. It goes without
saying that the domain of a function cannot be wider than
the domain of an analytical expression used to define this
function. But it can be narrower.
READER. Does it mean that a cosine defined, for exam-
ple, over the interval [0, n] and a cosine defined over the
interval [at, 3n1 are two different functions?
AUTHOR. Strictly speaking, it does. A cosine defined,
for example, on the real line is yet another function. In
other words, using cosine we may, if we wish, define any
number of different functions by varying the domain of
these functions.
In the most frequent case, when the domain of a function
coincides with the domain of an analytical expression for
the function, we speak about a natural domain of the function. Note that in the examples in the previous dialogue we
dealt with the natural domains of the functions. A natural
domain is always meant if the domain of a function in
question is not specified (strictly speaking, the domain of
a function should be specified in every case).
READER. It turns out that one and the same function
can be described by different formulas and, vice versa, one
and the same formula can be used to "construct" different
functions.
AUTHOR. In the history of mathematics the realization
of this fact marked the final break between the concept of
a function and that of its analytical expression. This actually happened early in the 19th century when Fourier, the
French mathematician, very convincingly showed that it
is quite irrelevant whether one or many analytical expressions are used. to describe a function. Thereby an end was
put to the very long discussion among mathematicians about
identifying a function with its analytical expression.
Dialogue Five
56
It should be noted that similarly to other basic mathe-
matical concepts, the concept of a function went through a
long history of evolution. The term "function" was introduced by the German mathematician Leibnitz late in the
17th century. At that time this term had a rather narrow
meaning and expressed a relationship between geometrical
objects. The definition of a functional relationship, freed
from geometrical objects, was first formulated early in the
18th century by Bernoulli. The evolution of the concept of
a function can be conventionally broken up into three main
stages. During the first stage (the 18th century) a function
was practically identified with its analytical expression.
During the second stage (the 19th century) the modern concept of a function started to develop as a mapping of one
numerical set onto another. With the development of the
general theory of sets, the third stage began (the 20th century) when the concept of a function formerly defined only
for numerical sets was generalized over the sets of an arbitra-
ry nature.
READER. It appears that by overestimating the role
of a formula we inevitably slip back to the concepts of the
18th century.
AUTHOR. Let us discuss now one more way of defining
a function, namely, the graphical method. The graph of
a function y = f (x) is a set of points on the plane (x, y)
whose abscissas are equal to the values of the independent
variable (x), and whose ordinates are the corresponding
values of the dependent variable (y). The idea of the graphical method of defining a function is easily visualized. Figure
14a plots the graph of the function
x<0
0<x<2
log(x-1), x>2
f cosx,
Y=
1-x2,
discussed earlier. For a comparison, the graphs of the functions y = cos x, y = 1 --J- x2, and y = log (x - 1) are shown
within their natural domains of definition in the same figure
(cases (b), (c), and (d)).
READER. In Fig. 14a I notice an open circle. What dons
it wean?
More on Function
57
AUTHOR. This circle graphically represents a point ex-
cluded from the graph. In this particular case the point
(2, 0) does not belong to the graph of the function.
y- Lag(x-f)
im-f
0
f
x
Fig. 14
Figure 15 plots the graphs of the functions that were
discussed at the end of the previous dialogue. Let us have
a close look at them.
y=x
9A
0
y
i
y
f
x
I
t
0
11
(d)
y
f
0
(e)
A
9
I
-
f
f
I
0
1
fj
21
(9)
X
0
I
I
f1
21
(h)
Fig. 15
x
0
(i)
More on Function
59
READER. Obviously, in all the cases shown in Fig. 15
the domain of the function is supposed coinciding with
the domain of the corresponding analytical expression.
AUTHOR. Yes, you are right. In cases (b), (c), (d), and
(e) these domains are infinite intervals. Consequently, only
a part of each graph could be shown.
READER. In other cases, however, such as (g), (h), and
(i), the domains of the functions are intervals of finite
length. But here as well the figure has space for only a part
of each graph.
AUTHOR. That is right. The graph is presented in its
complete form only in cases (a) and (f). Nevertheless, the
behaviour of the graphs is quite clear for all the functions
in Fig.
15.
The cases which you noted, i.e. (g), (h), and (i), are very
interesting. Here we deal with the unbounded function
defined over the finite interval. The notion of boundedness
(unboundedness) has already been discussed with respect
to numerical sequences (see Dialogue One). Now we have
to extrapolate this notion to functions defined over intervals.
Definition:
A function y = f (x) is called bounded over an interval D
if one can indicate two numbers A and B such that
A < f (x)<B
for all x E D. If not, the function is called unbounded.
Note that within infinite intervals you may define both
bounded and unbounded functions. You are familiar with
examples of bounded functions: y = sin x and y = cos x.
Examples of unbounded functions are in Fig. 15 (cases (b),
(c), (d), and (e)).
READER. Over the intervals of finite length both bounded and unbounded functions may also be defined. Several
illustrations of such functions are also shown in Fig. 15:
the functions in cases (a) and (f) are bounded; the functions
in cases (g), (h), and (i) are unbounded.
AUTHOR. You are right.
READER. I note that in the cases that I have indicated
the bounded functions are defined over the closed intervals
Dialogue Five
60
([-1, 1] for (a) and [1, 2] for (f)), while the unbounded
functions are defined both over the open and half-open
intervals (]1, 2[ for (g), 11, 2] for (h), and If, 2[ for (i)).
AUTHOR. This is very much to the point. However, you
should bear in mind that it is possible to construct bounded
functions defined over open (half-open) intervals, and
The graphs, of these functions are shown in Fig. 16.
READER. It seems that the boundedness (unboundedness) of a function and the finiteness of the interval over
yvhicll it is defined are not interrelated. Am I right?
More on Function
61
AUTHOR. Not completely. There is, for example, the
following
L.Theorem:
k
If a function is defined over a closed interval and if it is
monotonic, the function is bounded.
READER. Obviously, the monotonicity of a function is
determined similarly to the monotonicity of a numerical
sequence.
AUTHOR. Yes, it is. Monotonic functions can be classified, as sequences, into nondecreasing and nonincreasing.
Definition:
A function y = f (x) is said to be nondecreasing over an
interval D if for any xl and x2 from this interval f (xl)
f (x2) if xl < x2 If, however, f (xl) >f (x2), the function
is said to be nonincreasing.
Can you prove the theorem formulated above?
READER. Let the function y = f (x) be defined over
the closed interval [a, b]. We denote f (a) = ya and f (b) _
= yb. To make the case more specific, let us assume that
the function is nondecreasing. It means that ya < yb.
I don't know how to proceed.
AUTHOR. Select an arbitrary point x over the interval [a, b].
READER. Since a < x and x < b, then, according to
the condition of the above theorem, Ya < f (x) and f (x)
yb. Thus, we get that Ya < f (x) < yb for all x in the
domain of the function. This completes the proof.
AUTHOR. Correct. So, if a monotonic function is defined
over a closed interval, it is bounded. As to a nonmonotonic
function defined over a closed interval, it may be either
bounded (Fig. 15a and f) or unbounded (Fig. 16b).
And now answer the following question: Is the function
y = sin x monotonic?
READER. No, it isn't.
AUTHOR. Well, your answer is as vague as my question.
First we should determine the domain of the function. If
we consider the function y = sin x as defined on the natural
domain (on the real line), then you are quite right. If,
however, the domain of the function is limited to the interval [- 2 , 2 ], the function becomes monotonic (nondecreasing).
btalogae lave
READER. I see that the question of the boundedness or
monotonicity of any function should be settled by taking
into account both the type of the analytical expression for
the function and the interval over which the
function is defined.
AUTHOR. This obser-
vation is valid not only
for the boundedness or
monotonicity but also
for other properties of
functions. For example,
is the function y = 1- x2
an even function?
READER. Evidently
the answer depends on
the domain of the func-
tion.
AU'1'HUH.
y=1-x? -1 <x< 2
Fig. 17
Yes,
of
course. If the function
is defined over an interval symmetric about the
origin of coordinates (for
example, on the real line or over the interval [-1, 1]),
the graph of the function will be symmetric about the
straight line x = 0. In this case y = 1 - x2 is an even
function. If, however, we assume that the domain of the
function is [-1, 2], the symmetry we have discussed above
is lost (Fig. 17) and, as a result, y = 1 - x2 is not even.
READER. It is obvious that your remark covers the
case of odd functions as well.
AUTHOR. Yes, it does. Here is a rigorous definition of
an even function.
Definition:
A function y = f (x) is said to be even if it is defined on
a set D symmetric about the origin and if f (-x) = f (x)
for all x ED.
By substituting f (-x) = -f (x) for f (-x) = f (x), we
obtain the definition of an odd function.
But let us return to monotonic functions.
If we drop the equality sign in the definition of a mono-
Strictly monotonic functions possess an interesting property: each has an inverse function.
READER. The concept of an inverse function has already
been used in the previous dialogue in conjunction with the
possibility of mapping a set of equilateral triangles onto
a set of circles. We saw that the inverse mapping, i.e. the
mapping of the set of circles onto the set of equilateral
triangles, was possible.
AUTHOR. That's right. Here we shall examine the
concept of an inverse function in greater detail (but for
numerical functions). Consider Fig. 18. Similarly to the
graphs presented in Fig. 13, it shows three functions:
(a)
y=jl1-xz,
-1<x<1
64
Dialogue pipe
(b)
y - sin x,
(C)
y= COS x,
- ! <x <-!
0<x<n
Here we have three mappings of one numerical set onto
another. In other words, we have three mappings of an
interval onto another interval. In case (a) the interval
[-1, 11 is mapped onto the interval [0, 1]; in (b) the
interval [ - 2 , 2 ] is mapped onto the interval [-1, 11;
and in (c) the interval [0, n] is mapped onto the interval
[-1, 11.
What is the difference between mappings (b) and (c), on
the one hand, and mapping (a), on the other?
READER. In cases (b) and (c) we have a one-to-one
correspondence, i.e. each point of the set D corresponds to
a single point of the set E and vice versa, i.e. each point
of E corresponds to only one point of D. In case (a), however, there is no one-to-one correspondence.
AUTHOR. Yes, you are right. Assume now that the
directions of all the arrows in the figure are reversed. Now,
will the mappings define a function in all the three cases?
READER. Obviously, in case (a) we will not have a function since then the reversal of the directions of the arrows
produces a forbidden situation, namely, one number corresponds to two numbers. In cases (b) and (c) no forbidden
situation occurs so that in these cases we shall have some
new functions.
AUTHOR. That is correct. In case (b) we shall arrive at
the function y = aresin x, which is the inverse function
with respect to y =sin x defined over the interval [- n2 a2 ] '
In case (c) we arrive at the function y = arccos x, which
is the inverse function with respect to y = cos x defined
over 10, n].
I would like to place more emphasis on the fact that in
order to obtain an inverse function from an initial function,
it is necessary to have a one-to-one correspondence between
the elements of the sets D and E. That is why the functions
y = sin x and y = cos x were defined not on their natural
domains but over such intervals where these functions are
More on Function
65
either increasing or decreasing. In other words, the initial
functions in cases (b) and (c) in Fig. 18 were defined as
strictly monotonic. A strict monotonicity is a sufficient
condition for the above-men4
tioned one-to-one correspondence between the elements of
D and E. No doubt you can
prove without my help the
following
Theorem:
If a function y = I (x)
is
strictly monotonic,
different x
are mapped onto different y.
READER. Thus, a sufficient
condition for the existence
of the inverse function is the
strict monotonicity of the
Fig. 19
initial function. Is this right?
AUTHOR. Yes, it is.
READER. But isn't the strict monotonicity of the initial
function also a necessary, condition for the existence of the
inverse function?
AUTHOR. No, it is not. A one-to-one correspondence
may also take place in the case of a nonmonotonic function.
For example,
1-x, 0<x<1
y
{x,
1<x<2
Have a look at the graph of this function shown in Fig. 10.
If a function is strictly monotonic, it has the inverse
function. However, the converse is not true.
READER. As I understand it, in order to obtain an
inverse function (when it exists), one should simply reverse
.the roles of x and y in the equation y = f (x) defining the
initial function. The inverse function will. then be given
by the equation x = F (y). As a result,. the range of the
initial function becomes the domain of the inverse function.
AUTHOR. That is correct. In practice a conversion of
the initial function to ' the inverse function can be easily
performed on a graph. The graph of the inverse function is
5-01673
Dialogue lave
always symmetric to the graph of the initial function about
a straight line y = x. It is illustrated in Fig. 20, which
shows several pairs of graphs of the initial and inverse
functions. A list of some pairs of functions with their domains is given below:
(a)
initial
- 00 < x < 00
y = x3,
- oo<x<oo
inverse y=Vx,
(b)
(c)
initial
y= x2,
inverse
y = 1/x,
y =10x,
initial
0 < x < 00
0 <x < 00
- oo < x < 00
0 < x < 00
inverse y = log x,
(d)
initial
(e)
initial y = cos x,
inverse y = arccos x,
initial y = tan x,
(f)
2<y<2
-1 < x < 1
0<x < it
-1 < x < 1
y = sin x,
inverse y = aresin x,
inverse
- 2 <x< 2
y=arctanx, -oo<x<oo
initial
y =cot x,
0<x<n
inverse y = arccot x, - oo < x < oo
All the domains of the inverse functions shown in the list
are the natural domains of the functions (however, in the
case of y = 3 x the natural domain is sometimes assumed
to be restricted to the interval [0, oo[ instead of the whole
(g)
real line). As to the initial functions, only two of them
(y = x3 and y = 10x) are considered in this case as defined
on their natural domains. The remaining functions are
defined over shorter intervals to ensure the strict mono-
tonicity of the functions.
Now we shall discuss the concept of a composite function.
Let us take as an example the function h (x) =V 1 + cos2x.
Consider
also the functions
V1+y2.
f (x) = cos x and g (y) _
READER. This f (x) notation is something new. So far
we used to write y = f (x).
.*
F+ /
1
2
-----arctaa x
0
Fig. 20
5
X
I
I
I
A
68
Dialogue Five
AUTHOR. You are right. However, it is expedient to
simplify the ,notation.
Consider the three functions: h (x), f (x), and g (y).
The function h (x) is a composite function composed of
f (x) and g (y):
h (x) = g [f (x)]
READER. I understand. Here, the values of f (x) are
used as the values of the independent variable (argument)
for g (y).
Fig. 21
Fig. 22
AUTHOR. Let us have a look at Fig. 21, which pictures
the mappings of sets in the case of our composite function,
h (x) _ 1/1 + cost x, with f (x) = cos x defined over the
interval [0, nI.
We see that the function f is a mapping of D (the interval [0, n1) onto G (the interval [-1, 11), that is, the mapping f. The function g (the function 1 1 + y2) is a mapping
of G onto E (the interval [1, V2]), that is, the mapping g.
Finally, the function h (the function Y1 + cost x defined
over the interval [0, n1) is a mapping of D onto E, that
is, the mapping h.
The mapping h is a result of the consecutive mappings f
and g, and is said to be the composition of mappings; the
following notation is used
h = g°f
More on Function
69
(the right-hand side of the equation should be read from
right to left: the mapping f is used first and then the mapping g).
READER. Obviously, for a composite function one can
also draw a diagram shown in Fig. 22.
AUTHOR. I have no objections. Although I feel that we
better proceed from the concept of a mapping of one set
onto another, as in Fig. 21.
READER. Probably, certain "difficulties" may arise
because the range of f is at the same time the domain of g?
AUTHOR. In any case, this observation must always be
kept in mind. One should not forget that the natural domain
of a composite function g If (x)1 is a portion (subset) of the
natural domain of f (x) for which the values of f belong to
the natural domain of g. This aspect was unimportant
in the example concerning g [f (x)1 = y1 + cos2 x because
all the values of / (even if cos x is defined on the whole real
line)
fall into the natural domain of g (y) = Y J + y2.
T can give you, however, a different example:
h(x)=j
-2, f(x)=11x-1, g(y)=1
The natural domain of f (x) is [1, oo[. Not any point in
this interval, however, belongs to the domain of the composite function h (x). Since the expression 1l y - 2 is mean-
ingful only if y > 2, and for y = 2 we have x = 5, the
natural domain of this composite function is represented
by [5, oo[, i.e. a subset smaller than the natural domain of
f (x)
Let us examine one more example of a composite function.
Consider the function y = sin (aresin x). You know that
aresin x can be regarded as an angle the sine of which is
equal to x. In other words, sin!(aresin x) = x. Can you
point out the difference between the composite function
y = sin (aresin x) and the function y = x?
READER. Yes, I can. The natural domain of the function
y = x is represented by the whole real line. As to the composite function y = sin (arcsin x), its natural domain coincides with the natural, domain of the function arcsin x,
i.e. with [-1, 1]. The graph of the function y = sin (arcsin x)
is shown in Fig. 23.
Dialogue Five
70
AUTHOR. Very good. In conclusion, let us get back to
the problem of the graphical definition of a function. Note
that there are functions whose graphs cannot be plotted in
principle, the whole curve or a
par t o f i t. F or exampl e, it i s
impossible to plot the graph of
the function y = sin 5 in the
j
i
if
-11
0
i
vicinity of x = 0 (Fig. 24).
It
is also im p ossible to have the
x
graph of the Dirichlet
function
mentioned above.
READER. It seemed to me
that the Dirichlet function had
Fig. 23
no graph at all.
AUTHOR. No, this is not
the case. Apparently, your idea of a graph of a function
is always a curve.
READER. But all the graphs that we have analyzed so
far were curves, and rather smooth curves, at that.
-1
u-s in(¢resin x)
f
2
y -Sinf
Fig. 24
AUTHOR. In the general case, such an image is not
obligatory. But it should be stressed that every function
has its graph, this graph being unique.
READER. Does this statement hold for functions that are
not numerical?
Limit of Function
71
AUTHOR. Yes, it does. In the most general case we can
give the following
Definition:
The graph of a function f defined on a set D with a range
on a set E is a set of all pairs (x, y) such that the first element
of the pair x belongs to D, while the second element of the pair
y belongs to E, y being a function of x (y = f (x)).
READER. So it turns out that the graph of a function
such as the area of a circle is actually a set of pairs each
consisting of a circle (an element x) and a positive number
(an element y) representing the area of a given circle.
AUTHOR. Precisely so. Similarly, the graph of. a function representing a schedule of students on duty in a classroom is a set of pairs each containing a date (an element x)
and the name of a student (an element y) who is on duty
on this date. Note also that in practice this function indeed
takes a graphic form.
If in a particular case both elements of the pair (both x
and y) are numbers, we arrive at the graph of the function
represented by a set of points on the coordinate plane. This
is the familiar graph of a numerical function.
DIALOGUE SIX
LIMIT OF FUNCTION
AUTHOR. Consider now the concept of the limit of
function.
READER. But we have already covered rather exten-
sively the concept of the limit of a numerical sequence. But
a sequence is nothing else but a function defined on a set
of natural numbers. Thus, having discussed the limit of
sequence, we become acquainted with the limit of function
as well. I wonder whether there is any point in a special
discussion of the concept of the limit of function.
AUTHOR. Undoubtedly, a further discussion will be very
much to the point. The functions we are concerned with
substantially differ from sequences (I have already emphasized this fact) because they are defined over intervals and
not on sets of natural numbers. This fact makes the concept
Dialogue Six
72
of the limit of function specific. Note, for example, that
every specific convergent sequence has only one limit. It
means that the words "the limit of a given sequence" are
self-explanatory. As for a function defined over an interval,
one can speak of an infinite number of "limits" because the
limit of function is found for each specific point x = a (or,
as we say, for x tending to a). Thus the phrase "the limit
of a given function" is meaningless because "the limit of
a given function must be considered only at each given
point a". Besides, this point a should either belong to the
domain of the function or coincide with one of the ends
of the domain.
READER. In this case the definition of the limit of
function should be very different from that of the limit of
sequence.
AUTHOR. Certainly, there is a difference.
Note, first of all, that we analyze a function y = f (x),
which is defined over a segment, and a point a in this segment (which may coincide with one of its ends when the
function is defined over an open or half-open interval).
READER. Do you mean to say that at the point x = a
the function f (x) may not be defined at all?
AUTHOR. That is quite correct. Now let us formulate
the definition of the limit of function.
Definition:
A number b is said to be the limit of a function f (x) at x
tending to a (the limit at point a) if for any positive value of s
there is a positive value of 6 such that for all x satisfying the
a and
conditions x belongs to the domain of the function; x
we have
Ix-aI<61
(1)
If(x) - b I <E
(2)
The standard notation is
lim f (x) = b
z-.a
READER. The definition of the limit of function is noticeably longer and more complicated than that of the limit
of sequence.
Limit of Function
73
AUTHOR. Note, first of all, that according to (1), point
x should belong to the interval ]a - 6, a + M. Point
x = a should be eliminated from this interval. The interval
]a - 6, a + S[ without point x = a is called a punctured
8-neighbourhood of point a.
We select an arbitrary positive number s. For s we want
to find another positive number S such that the value of the
function at any point x from the punctured 6-neighbourhood
of point a must be inside the interval lb - s, b + e[ (speak-
ing about any point x we imply only the points x in the
domain of the function). If there is such b for any s > 0, b
is said to be the limit of the function at point a. Otherwise,
b is not the limit of the function at point a.
READER. And what does your "otherwise" mean in
practice?
AUTHOR. Assume that the search for S has been success-
ful for n diminishing numbers F j, a2i ..., sn. But then
you notice that for a certain number e' it is impossible to
find the required number S, i.e. for any value of 6 no matter
how small) there is always at least one point x from the
punctured 6-neighbourhood of point a at which the value
of the function lies outside the interval ]b - s', b + e'[.
READER. But can it happen that we reduce the 8-neigh-
bourhood of point a so much that not a single point x, belong-
ing to the domain of the function, remains in the 6-neighbourhood?
AUTHOR. Obviously this is impossible. Because the
function is defined over an interval, and point a is taken
either from this interval or coincides with its end point.
READER. Everything seems clear. Apparently, in order
to root all this firmly in my mind we should discuss the
graph of a function.
AUTHOR. It is a good idea. Let us analyze, for the sake
of convenience, the graph of the function y = 1x (Fig. 25).
This figure illustrates only two situations. One of them
represents the selection of a1 (see the figure). It is easy to
infer that 6 is the value that we look for: the values of the
function at all points x from the S1-neighbourhood of point a
are inside the interval ]b - a1, b + a1[. These values are
represented by the portion of the graph between points A
And B. The second situation represents the selection of $j.
74
Dialogue Six
In this case the number that we seek for is 62: the values
of the function at points x from the d2-neighbourhood of
point a are represented by the portion of the graph between
points A' and B'.
Fig. 25
READER. Everything you have just described looks so
obvious that I see no "cream", to use your own words.
AUTHOR. "The cream" consists in the following. No
matter how small ]b - S, b + e [ is, one may always select
a 6-neighbourhood for point a such that for all points x in
this 6-neighbourhood (all points, with the exception of
point a itself and those at which the function is not defined)
the values of the function should by all means lie within
the indicated interval.
READER. Could you give an example of a function
violating this rule?
in the
AUTHOR. For instance, the function y = sin
vicinity of point x = 0. The graph of the function is plotted
in Fig. 24. Obviously, the smaller is I x I the greater is
the frequency with which the graph of the function oscillates
about the x-axis. For an infinitely small I x I the frequency
Limit of Function
75
of the oscillations tends to infinity. It is easy to prove that
the function y = sin z has no limit at x = 0.
READER. But this function is not defined at zero.
AUTHOR. You are right. However, this fact is irrelevant
from the viewpoint of the existence (or absence) of the
limit of the function at x = 0. This function is defined
over 1- oo, 0[ and 10, oo[. Point x = 0 is a common boun-
dary between the intervals over which the function sin X1
is defined.
But let us return to the concept of the limit. Can we,
for example, state that b = 0 is the limit of the, function
sin z1
at point x = 0?
READER. It seems that I get the point. As long as we
select e > 1, everything is O.K. But for any e < 1 it
becomes impossible to find a 6-neighbourhood of point
x = 0 such that at all points x ; 0 in this 6-neighbourhood
the values of the function sin 1 are inside the interval
1-e, e[. No matter how small the 6-neighbourhood of
point x = 0 is, it is the segment of finite length, so that
the graph of our function will oscillate infinitely many
times and thus will infinitely many times go beyond 1-e, e[.
AUTHOR. That's right. Note also that in order to be
convinced that a function has no limit, it is sufficient to
find a violation even more "modest". Namely, it is sufficient
that the graph of the function leave the interval 1-e, e[
at least once for any 6-neighbourhood.
b
READER. Apparently, not only b = 0 but no other
0 can be the limit of the function y = sin
at x = 0.
Because for any b
0 we can use the same arguments
as
x
for b = 0.
AUTHOR. Hence, we have proved that the function
y = sins has no limit at point x = 0.
READER. The reason for the absence of the limit at x = 0
lies in oscillations of the graph of the function. These oscillations become more and more frequent while approaching
x=0.
Dialogue Six
76
AUTHOR. But the reason is not confined only to the
infinitely increasing frequency of oscillations of the graph.
Another reason is the constancy of the amplitude of oscillations. Let us "slightly correct" our function by multiplying
sin x by x. The graph of the function y = x sin i is shown
Fig. 26
in Fig. 26. Do you think that b = 0 is the limit of this
function at x = 0?
READER. I am at a loss.
AUTHOR. I'll answer this question myself. Yes, it is.
The proof is within your reach if you use the definition of
the limit of function. You are welcome.
READER. We select an arbitrary s > 0. We should find
6> 0 such that x sin a- 0 I < a for all x (excluding
x = 0) satisfying the condition I x - 0 1 < 6. It seems
I
the definition of the limit of function) that b = 1 is the
limit of the function at point x = 1.
To begin with, consider the following inequality:
I yx-11 <E
Try to find a function g (e) such that I x - 1 1 < g (a)
for any x satisfying the condition I Yx - 1 1 < a.
READER. I understand that g (e) is actually the desired
6 corresponding to an arbitrary e.
AUTHOR. Yes, of course. We begin with some transformations. We shall proceed from the inequality:
ly-X-11<e
which can be rewritten in the form:
(3)
(1-a)<Vx<(s+1)
Since Vx > 0, the selection of e < 1 a fortiori (which,
of course, does not impair the generality of our proof)
allows us to square the last inequalities
(1 - e)2 < x < (1 + a)2
On removing the parentheses, we obtain
(-2e + e2) < (x - 1) < (2a + a2)
(4)
Note that inequalities (4) are equivalent to (3) (provided
that 0 < e < 1). Now let us proceed from (4) to a more
exacting inequality:
Ix-1 I<(2e-e2)
(5)
(since 0 < a < 1, we have (2a - a2) > 0). It is easy to
conclude that if (5) holds, inequalities (4) and, consequently,
biatogue Six
78
(3) will hold all the more. Thus, for an arbitrary e within
0 < e < 1, it is sufficient to take 8 = 2e - e2.
READER. What happens if e > 1?
AUTHOR. Then 8 determined for any s < 1 will be
AUTHOR. Yes, it is often the case. But not always.
Because the function f (x) may be undefined at point a.
Remember that the limit of the function x sin at point
x = 0 is zero, but the function itself is not defined
z at point
x=0.
READER. But perhaps the equality lim f (x) = f (a)
x,a
can be considered as valid in all the cases when f (x) is
defined at point a?
AUTHOR. This may not be correct either. Consider, for
example, a function which is called the "fractional part of
x". The standard notation for this function is {x}. The function is defined on the whole real line. We shall divide the
real line into half-intervals [n, n + 1[. For x in [n, n + 1[
we have {x} = x - n. The graph of the function y = {x}
is shown in Fig. 27.
Take, for example, x = 1. It is obvious that {x} is defined
at point x = 1 ({1} = 0). But does the function have the
limit at x = 1?
READER. It clearly has no limit. In any 6-neighbourhood of point x = 1 there may exist concurrently both the
points at which {x} assumes values greater than, for example,
2
, and the points at which {x} assumes values less than 3
Limit of Punction
90
It means that neither b = 1 nor b = 0 can be the limit of
the function at point x = 1, if only because it is impossible
to find an adequate 6 for s = 3 .
AUTHOR. I see that you have come to be rather fluent
in operating with limits of functions. My compliments.
i
Fig. 27
By the way, you have just proved the theorem on the uniqueness of the limit of function at a given point.
Theorem:
A function cannot have two (or more) limits at a given
point.
Now let us return to the equality
lim f (x) = f (a)
x-a
(6)
You already know that there are situations when lim f (x)
exists but f (a) does not exist and, vice versa, w'e'n f (a)
exists but lim f (x) does not exist. Finally, a situation is
possible when both lim f (x) and f (a) exist, but their
x-+a
values are not equal. I'll give you an example:
if xz 0
if x=0
The graph of this function is shown in Fig. 28. It is easy
to see that f (0) = 1, while lim f (x) = 0.
x- 0
You must be convinced by now that equality (6) is, not
always valid.
Dialogue Six
80
READER. But presumably, it is often true, isn't it?
AUTHOR. Yes, and if it is, the function f (x) is said to
be continuous at x = a.
Thus, we have arrived at a new important concept, name-
ly, that of the continuity of a function at a point. Let us
give the following
Definition:
A function f (x) is said to be continuous at a point x = a if
(1) it is defined at x = a,
(2) there is the limit of the function at x = a,
(3) this limit equals the value of the function at x = a;
or, in other words, the function f (x) is called continuous at
a point a if
lim f (x) = f (a)
x-a
I believe that the preceding discussion has brought us so
closely to this definition that it needs no additional explanation. I would only like to emphasize that the concept
of the continuity of a function is essentially local. Similarly
to the concept of the limit of function, it is related to a particular point x. A function may be either continuous at all
`points of an interval over which it is defined, or discontinuous at some of its points.
Taking the examples given above, can you single out
those functions that are discontinuous at particular points?
Limit of Function
81
READER. To begin with, I may refer to the function
whose graph is plotted in Fig. 28. This function is discontinuous at x = 0.
AUTHOR. Why?
READER. Because at this point the function assumes
the value y = 1, though the limit of the function at this
point is apparently zero.
AUTHOR. Very good. Can you give other examples?
READER. The function y = (x) (see Fig. 27) is discon-
tinuous at points x = 0, Âą1, t2, t3, ...
.
The function
(see Fig. 24) is discontinuous at x = 0 where
it is undefined and, moreover, has no limit. The ' function
y
sin
whose graph is shown in Fig. 14a (see the previous, dialogue)
is discontinuous at x = 2. The function y = tan x is discontinuous at points
x=
2
t Z n,
2 n, f 2
7E,
.. .
AUTHOR. That will do. Note that the points at which
the continuity of a function is violated are called disconti-
nuity points. We say that at these points a function has
a discontinuity. In passing through a discontinuity point
a graph of a function manifests a singularity. This fact is
well illustrated by the examples you have just indicated,
READER. The discontinuity points in all these examples
result in an interruption of the curve plotting the function.
One exception is the function y = sin 1 since it is simply
impossible to trace a graph of the function at x = 0.
AUTHOR. I may add that neither could you plot the
function y = tan x at its discontinuity points (since you
cannot draw a line which "goes into infinity").
READER. In any case, if a function is continuous every-
where in the domain (has no discontinuity points), its
graph is a continuous line: it can be drawn without lifting
the pencil from the paper.
AUTHOR. I agree. I would like to emphasize that the
continuity of a function at. a point x guarantees that a very
small displacement from this point will. result in a very small
change in the value of the function.
6-01473
82
Dialogue Six
Let us turn to Fig. 27 which is the graph of the function
y = {x}. Consider, for instance, x = 0.5. The function is
continuous at this point. It is quite evident that at a very
small displacement from the point (either to the left or to
the right) the value of the function will also change only
a little. Quite a different situation is observed if x = 1
(at one of the discontinuity points). At x = 1 the function
assumes the value y = 0. But an infinitesimal shift to the
left from the point x = 1 (take, for example, x = 0.999,
or x = 0.9999, or any other point no matter how close
to x = 1) will bring a sharp change in the value of the
function, from y = 0 to y x 1.
READER. Quite clear. I must admit, however, that the
local nature of the concept of a continuous function (i.e.
the fact that the continuity of a function is always related
to a specific point x) does not quite conform to the conven-
tional idea of continuity. Because continuity typically
implies a process and, consequently, a sort of an interval.
It seems that continuity should be related not to a specific
moment of time, but to an interval of time.
AUTHOR. It is an interesting observation. This local
character is a manifestation of one of the specific features
of calculus. When analyzing a function at a given point x,
you used to speak about its value only at this specific point;
but calculus operates not only with the value of a function
at a point but also with the limit of the function (or its
absence) at this point, with the continuity of the function
at the point. It means that on the basis of the information
about a function at a given point we may construct an image
of the behaviour of the function in the vicinity of this point.
Thus we can predict the behaviour of the function if the
point is slightly shifted from x.
So far we have made only the first step in this direction.
The next step will be the introduction of the concept of
a derivative. This will be the subject of discussion in Dialogues Eight and Nine.
READER. Nevertheless, I would like to note that in the
above examples a function was found to be either continuous everywhere over any interval of finite length or
discontinuous at a finite number of points. In this sense
the local nature of the concept of a discontinuity point is
Limit of Function
83
evident. The continuity of the function, however, is always
observed over a certain interval.
AUTHOR. First, the continuity of a function within an
interval does not interfere with the local nature of conti-
nuity. A function is continuous over an interval if it is continuous at all points of this interval.
Y
2
0
2
1
f
2
Fig. 29
Second, it is not difficult to construct an example in
which the number of discontinuity points over an interval
of finite length is infilzitely large. Let us look, for example,
at the following function:
2
Y-
for x=t1, t 2,Âą
4,
81 X16,
x2 for all the remaining points of the real
line,
including x = 0
The graph of this function is illustrated in Fig. 29. It is
easy to conclude that in any 6-neighbourhood of point
x = 0 the function has an infinite number of discontinuity
points.
Finally, I can give an example of a function which is
discontinuous at all points of an infinite interval. This is
a function you already know, the Dirichlet function (see
6*
S4
Dialogue Seven
the previous dialogue). Being defined on the whole real
line, the function has no limit at any point of the real
line; consequently, it is discontinuous at each point.
READER. This is the reason why we in principle cannot
plot the Dirichlet function by a graph.
AUTHOR. As to the most frequent functions, such as
power, exponential, logarithmic, trigonometric, and inverse
trigonometric, they are continuous at all points of the natural
domains of the corresponding analytical expressions. The
same can be said about composite functions obtained from
the above elementary functions. The continuity of all these
functions is proved in the more advanced courses of cal-
culus. We limit ourselves to a mere stating of the fact.
DIALOGUE SEVEN
MORE ON THE LIMIT
OF FUNCTION
READER. Comparing the definition of the limit of
a function at a point with the definition of the limit of
a numerical sequence, I come to the conclusion that these
two limits are of different nature.
AUTHOR. And I understand why. In fact, I did emphasize the difference myself in the previous dialogue, pointing
out, as you probably remember, that a sequence is a function defined on a set of integers, while the functions we are
discussing at the moment are defined over intervals. I doubt,
however, that you are justified in speaking about the difference in the nature of the limit of function and that of
sequence. In the final analysis (and this is essential) the
limit of a function at a point may be defined on the basis
of the limit of a numerical sequence.
READER. This is very interesting.
AUTHOR. Let us forget, for the time being, about the
definition of the limit of function given in the previous
dialogue. Consider a new definition.
We shall consider, as before, a function f (x) defined
over an interval, and a point x = a either taken within
the interval or coinciding with its end.
More on the Limit of Function
85
The definition of the limit of function may be formulated
as follows.
Definition:
A number b is said to be the limit of a function f (x) at
a point x = a if for any numerical sequence (xn) satisfying
the conditions:
(1) all x, belong to the domain of the function,
(2) x, 7A a for any n,
(3) the sequence (xn) is convergent and its limit is a; the
sequence of the corresponding values of the function (the sequence f (xn)) is convergent and its limit is b.
So we should "construct" a sequence convergent to a:
xi, x2, x3,
., xfi,
.
.
(we assume that all xn belong to the domain of the function
a for any n). This sequence automatically gene-
a
to a (all xn belong to the domain of the function; xn
for any n) there is the corresponding sequence of the values
of the function convergent to b, then b is said to be the
limit of the function f (x) at x a.
READER. The definition of the limit of function is quite
understandable in itself. But how does it relate to our
former definition of the limit of function?
Dialogue Seven
86
AUTHOR. The two are equivalent.
READER. But in form they are quite different)
AUTHOR. We can prove their equivalence. To begin
with, let the definition using a 8-neighbourhood of point a be
called "definition 1", and the definition using numerical
sequences, "definition 2".
Now, what two theorems must be proved to demonstrate
the equivalence of definitions 1 and 2? Can you formulate
these theorems?
READER. We have to prove two theorems, one direct
and the other converse. We want to prove that definition 2
follows from definition 1 and vice versa (i.e. definition 1
follows from definition 2).
AUTHOR. Correct. First, I shall prove the following
Theorem:
If a number b is the limit of a function f (x) at a point
a in terms of definition 1, it is the limit of the function f (x)
at a in terms of definition 2 as well.
Since b is the limit of the function f (x) at point a in
terms of definition 1 (this is given), consequently, for any
a > 0 there is 8>0 such that j f (x) - b I < e for all
a from a 8-neighbourhood of point a. Then we "construct" an arbitrary sequence (xn), requiring that it be
x
convergent to point a (any xn belong to the domain of the
a for any n). As a result we obtain a sefunction and x,
quence of the corresponding values of the function (the
sequence If (xn)]). We want to prove that the sequence
If (xn)] is convergent to b.
First, I select an arbitrary a > 0. I should find a number N such that I f (xn) - b I < e. for all n > N.
I cannot immediately find such N for an arbitrary e.
However, I can indicate for an arbitrary a such 8 that
I f (x) - b I < a if I x - a I< 8. Then I take this 6
and find a sequence (xn) convergent to a. Evidently, since
(xn) is convergent to a, 8 (as any other positive real number)
can be placed in correspondence with a number N such
that I xn - a I < 8 for all n > N. And, consequently, we
also have that I f (xn) - b I < e for all n > N. Hence,
we find that the thus found number N is actually the desired
number. It proves the convergence of the sequence If (xn)]
to b. Since the ' sequence (xn), which is convergent to a,
More on the Limit of Function
87
was chosen ("constructed") arbitrarily, we conclude that
the theorem's proof is completed.
If the line of reasoning is clear to you, try briefly to
recapitulate the logical structure of the proof.
READER. I shall try to present the structure of the
proof as a diagram (Fig. 30).
-al <d'
S>0
lf(X)-b <e
C
S>0
I
N
N
x,-a!<8
f(Xn)-bI <E
Fig. 30
AUTHOR. Your diagram is correct. Will you expand
on it.
READER. The first step of the proof: we find for an ar-
bitrary e> 0 a number S> 0 such that I f (x) - b I < a
if
Ix - aI<S.
The second step of the proof: we take S selected at the
convergent to a, and find
first step; choose a sequence
a number N such that x,, - a I < S for all n > N. Having
in mind the arguments used at the first step, we conclude
I
that If (x,,) - b I< a
if
I
x,ti - a I< S.
We have thus found for an arbitrary a > 0 a number N
such that I f (x,ti) - b I < a for all n > N. This completes
the proof.
AUTHOR. Correct. In conclusion I want to emphasize
several essential points on which the proof hinges. We
know that I f (x) - b I < a for any x from the 6-neighbourhood of a. Since a sequence (x,) is convergent to a, all x,,
(the whole infinite "tail" of the sequence (x,,) starting from
a certain number N + 1) are contained inside the 8-neigh-
Dialogue Seven
88
bourhood of point a. It then follows that all f
(the whole
infinite "tail" of the sequence If (x,,)] starting from the
same number N + 1) are contained inside the interval
I b - a, b + s[. This proves that the sequence If (x,,)]
converges to b.
READER. I understand.
AUTHOR. Now I am going to prove the following converse
-Theorem---,
If a number b is the limit of a function f (x) at a point
a in terms of definition 2, it is also the limit of the function
f (x) at a in terms of definition 1.
In this case I shall use the proof by contradiction. Assume
the contrary to what is to be proved, namely, assume that b,
the limit of f (x) in terms of definition 2, is not, however,
the limit of f (x) in terms of definition 1. Can you formulate
the last proposition (more exactly, the assumption)?
READER. As far as I remember, a similar formulation
has already been discussed in the previous dialogue. If b is
not the limit of the function f (x) at point a (in terms of
definition 1), it means that there is e' > 0 such that it is
impossible to find a necessary 6 > 0. Namely, no matter
what 6 we select, each time the function f (x) assumes a value
outside of ]b - s', b + s'[ for at least one point x from
the 6-neighbourhood of point a, i.e. the inequality I f (x) - b I < s' is violated.
AUTHOR. Correct. Assume that we have selected precisely this s' > 0. Next take an arbitrary 8 > 0, for instance, 6 = 1. As you have said, in any 6-neighbourhood
of point a and, hence, in the 61-neighbourhood of this
point there is at least one point x (denoted by x1) such that
If(x1)-bI>a'.
READER. What happens if the 61-neighbourhood con-
tains many such points x?
AUTHOR. It makes no difference. The important fact is
that there is at least one such point. If there are several
such points, take any one of them and denote it by x1.
Now we take a new 6, for instance, 6 = 2. According
to our assumption, the 62 neighbourhood of point a will
contain at least one point x (denoted by x2) such that
If(x2)-b f>s'.
More on the Limit of Function
89
Further we take 63 = 3. The 63 neighbourhood of point a
will also contain at least one point x (point x3) such that
If(x3)-bI>$'.
We can continue this process for a sequence of the 6-neigh-
bourhoods of point a
ss=1, 62=2, 63=g, ...
,
6n=n,...
Note that the 6-neighbourhoods are selected in such a way
that the sequence (Sn) converges to zero (is infinitesimal).
If each time we select from each 6-neighbourhood one
point x in which f (x) assumes a value outside of the inter-
inevitably converges to a. A sequence composed of the corresponding values of the function (the sequence If (x,)]) is
not convergent to b because for all n we have I f (xn) - b I > s'. It means that we obtained a sequence (xn)
convergent to a for which the sequence If (xn)] is divergent.
This contradicts the condition of the theorem which
states that b is the limit of the function at a in terms of
definition 2. It means that for any sequence (xn) convergent
to a the corresponding sequence If (xn)l must be convergent
to b. And the sequence (xn) that we have found contradicts
this condition.
Hence, the assumption that b, being the limit of the
function in terms of definition 2, is not at the same time
the limit of the function in terms of definition 1, is invalidated. This completes the proof of the theorem.
READER. I must admit of being wrong when I spoke
about different natures, of the limit of numerical sequence
and the limit of function at. . a point.
AUTHOR. These limits differ but their nature is the
same. The concept of the limit of function at a point is based,
as we have seen, on the concept of the limit of numerical
sequence.
That is why basic theorems about the limits of functions
are analogous to those about the limits of sequences.
Dialogue Seven
90
READER. We have already noted one of such theorems:
the theorem on the uniqueness of the limit of function at a
point.
AUTHOR. This theorem is analogous to that about the
uniqueness of the limit of numerical sequence.
I shall also give (without proof) the theorems on the limit
of the sum,. the product, and the ratio of functions.
Theorems:
If. functions f (x) and g (x) have limits at a point a, then
functions
If (x) + g (x)l,
It (x) g (x)],
((xx))
)
( gf
also have limits at this point. These limits equal the sum, product, and ratio, respectively, of the limits of the constituent
functions (in the last case it is necessary that the limit of the
function g (x) at a be different from zero).
Thus
lim It (x) + g (x)] = lim f (x) + lim g (x)
x-a
x-a
x-a
lim If (x) g (x)] = lim f (x) lim g (x)
x-a
x-a
x-a
lim f (x)
f (x)
x-.a ( g (x)
lim
)_
x-.a
lim g (x)
x- a
under an additional
condition: lim g (x) r 0
x-.a
READER. We have already discussed the similar theorems for numerical sequences.
AUTHOR. Next I wish to make two remarks, using for
the purpose specially selected examples.
Note 1. It is illustrated by the following example.
Obviously lim y1-x2= 0 and lim yx-1= 0. Does it
x-1
x-.1
mean that lim (y1-x2 +yx^ 1) = 0?
READER. The limit of the function y1 - x2 at x = 1
exists and is equal to zero. The limit of the function yx - 1
at x = 1 also exists and is also equal to zero. According
to the theorem on the limit of the sum, the limit of f (x) =
More on the Limit of Function
91
= y1 - x2 + yx - 1 must exist and be equal to the
sum of the two preceding limits, i.e. to zero.
AUTHOR. Nevertheless, f (x) = 1/1 - x2 + yx - 1 has
no limit at x = 1 for a simple reason that the expression
y1 - x2 + yx - 1 has meaning only at a single point
(point x = 1). Applying the theorem on the limit of the
sum, you have not taken into account the domains of the
functions y1 - x2 and yx - 1. The former has the natural domain over [-1, 11, while the latter over [1, oo[.
READER. Apparently your note also covers the cases
when the theorems on the limit of the product and the
limit of the ratio of functions are used.
AUTHOR. It goes without saying. Working with func-
tions, you must always consider their domains. The natural
domains of functions may intersect (or even coincide), but
sometimes they may not. This aspect must never be overlooked. Why do you think we never have such complications when working with sequences?
READER. Obviously because all numerical sequences
have one and the same domain, i.e. a set of natural numbers.
READER. What is this limit?
AUTHOR. It depends on the functions f (x) and g (x).
Let us show, for example, that
lim
a-.0
sin x
X
Diatogue_Seven
92
Note that the function sinx
is not defined at x = 0. This
x
fact, however, does not interfere with searching for the
limit of the function at x = 0.
We shall start with well-known inequalities:
sinx<x<tanx (0<x< Z )
An assumption that 0 < x < will not affect the generality of our results. Dividing sin x by each term of these
inequalities, we obtain
sinx
1>
x
>cosx
hence
sin x) < (1 - cos x)
0<(4
Next we take into account that
1-cos"x=2sinz 2 <2sin 2 <2 2 =x
Thus we have
0<(1-sinx)<
X
or
-x
sin X)
<0
X
whence
I1-
six I<Ixl
X
We thus arrive at the following inequality valid for
IxI<2
sinx
X
-1 I<IxI
(1)
By using this inequality, we can easily prove that the
sin
function
x has the limit at x = 0, and this limit is
More on the Limit of Function
93
unity. It will be convenient to use definition 1 for the
limit of function at a point.
Select an arbitrary e > 0, demanding for the sake of
simplicity that e < 11. For 6, it is sufficient to take 6 = e
since, according to (1), the condition I x - 0 1 < 6 immediately leads to
sin x
x
Thus, unity is indeed the limit of the function 9lu x at
x=0.
READER. Do we really have to resort to a similar line
of reasoning, based on the definition of the limit of function
at a point, each time we have to find the limit of
f (x)
lim
(x) = 0 ?
(
x-0
X-0
AUTHOR. No, of course not. The situation we are speaking about is known as an indeterminate form of the type
There are rules which enable one to analyze such a situation
5.
in a relatively straightforward manner and, so to say,
"resolve the indeterminacy". In practice it is usually not
difficult to solve the problem of existence of the limit of
a function (8((5))) at a specific point and find its value (if it
exists). A few rules of evaluation of indeterminate forms
of the type U (and other types as well) will be discussed
later. A systematic analysis of such rules, however, goes
beyond the scope of our dialogues.
It is important to stress here the following principle
(which is significant for further considerations): although
the theorem on the limit of the ratio is not valid in the
cases when lim g (x) .= 0, the limit of a function (f((x) )
8()
at a point a may exist if lim f (x) = 0. The example of
x_0
the limit of the function
sin x
illustration of this principle.
at x = 0 is a convincing
Dialogue Eight
94
READER. Presumably, a similar situation may take
place for numerical sequences as well?
AUTHOR. It certainly may. Here is a simple example:
( X " )=1 ,
1
g+
1
-i7-
1
64+
... ,
1
n9
,
.. .
(lim xn = 0)
n-.oo
(yn) = 1,
1
1
273,
1
4
,
... , n1 , .. .
(lim yn = 0)
It is readily apparent that the limit of the sequence
( 1--) is the limit of the sequence (n ) . This limit
does exist and is equal to zero.
READER. You mentioned that the existence of the
limit of a function (f (z)) at a, when both lim f (x) _
x-a
=0 and lim g (x) =0( the existence of the limit of the
x-a
type 0) ,
is very important for further considera-
tions. Why?
AUTHOR. The point is that one of the most important
concepts in calculus, namely, that of derivative, is based
on the limit of the type o. This will be clear in the subsequent dialogues.
DIALOGUE EIGHT
VELOCITY
AUTHOR. We are practically ready to tackle the concept
of a derivative. This concept, alongside with the concepts
of the limit of numerical sequence and the limit of function,
is one of the most important special concepts in calculus.
Velocity
95
We may approach the concept of a derivative by consider-
ing, for instance, a quantity widely used in physics: the
instantaneous velocity of nonuniform motion of a body.
READER. We have been familiarized with this notion
when studying kinematics in the course of physics, or, to be
precise, the kinematics of nonuniform motion in a straight
line.
AUTHOR. Exactly. What is your idea of the instanta-
neous velocity?
READER. The instantaneous velocity of a body is defined
as the velocity of a body at a given moment of time (at
a given point of its trajectory).
AUTHOR. And what is your idea of the velocity at a given
moment of time?
READER. Well, I see it as ... . If a body moves uni-
formly, at different moments of time its velocity remains
the same. If a body moves nonuniformly (accelerating or
decelerating), its velocity will, in the general case, vary
from moment to moment.
AUTHOR. Don't you feel that the phrase "velocity at
a given moment of time" is merely a paraphraze of the
"instantaneous velocity"? Six of one and half a dozen of the
other, eh? The term "velocity at a given moment of time"
calls for an explanation as much as the term "instantaneous
velocity".
To measure the velocity of a body, one should obviously
measure a certain distance (path) covered by the body, and
the time interval during which the distance is covered. But,
what path and period of time are meant when we refer
to the velocity at a given moment of time?
READER. Yes, in order to measure velocity, one must
actually know a certain path and time interval during which
the path is covered. But our subject is not the measurement,
it is a definition of the instantaneous velocity.
AUTHOR. For the time being we shall not bother about
a formal definition. It is more important to realize its
essential meaning. In order to do this, we cannot avoid the
aspect of measurements. Now, how would you find a way
to measure the velocity of a body at a given moment of time?
READER. I can take a short time interval At, that is,
the period from the given moment of time t to the moment
96
Dialogue Eight
t + At. During this time interval the body covers a distance As. If At is sufficiently small, the ratio At will give
the velocity of the body at the moment t.
AUTHOR. What do you mean by a sufficiently short time
interval? What do you compare it with? Is this interval
sufficiently small in comparison with a year, a month, an
hour, a minute, a second, or a millisecond?
READER. Perhaps, neither a year, a month, an hour
nor a minute will do in this case. I see now that the instantaneous velocity can only be measured with a certain degree
of accuracy. The smaller is At the smaller is the error with
which the instantaneous velocity is measured.
AUTHOR. In principle, the concept of the instantaneous
velocity (or, in other words, "velocity at a given moment
of time") must be independent of the measurement accuracy.
The velocity you are talking about, that is, the ratio AsI
Fig. 3t
is nothing more than the average velocity during At. Itlis
not the instantaneous velocity at all.. Of course, you are
right when you say that the smaller is At the closer is the
value of the average, velocity to the value of the instanta-
Velocity
97
neous velocity. However, no matter how small is At, the
ratio os is always only the average velocity during At.
READER. Then a better definition of the instantaneous
velocity is beyond me.
AUTHOR. Consider a graph of distance covered by a body
plotted as a function of time, that is, the graph of the
function s = s (t). This graph is shown in Fig. 31 by a solid
line. Note that in physics one typically uses the same symbol to denote both a function and its values (in this case
we use the symbol s).
READER. The figure also shows several thin lines.
AUTHOR. The thin lines (parabolas and straight lines)
are shown only to indicate how the graph of s = s (t) was
plotted. This graph is thus composed of "pieces" of parabolas and straight lines. For instance, for the time interval
from 0 to tl the graph is represented by a "piece" of the
extreme left-hand parabola (portion 0-1 of the graph).
Please recall the formula for the distance covered in a uniformly accelerated motion with zero initial velocity.
READER. This formula is
s (t) =
ate
(1)
2
where a is acceleration.
AUTHOR. And the extreme left-hand parabola is the
graph of the function represented by your formula.
READER. So for the time interval from 0 to tl the body
moves at a constant acceleration.
AUTHOR. Exactly.
READER. I see. For the time interval from .t1 to t2 the
body moves uniformly (portion 1-2 of the graph is a straight
line); from t2 to t3 the body moves at a constant decelera-
tion (the graph is an inverse parabola); from t3 to t4 the
body is not moving at all; from t4 to tb it moves at a constant
acceleration, and from tb to tg it moves at a constant deceleration.
AUTHOR. Precisely so. Now let us consider the graph of
the function s (t) shown in Fig. 31 from a purely mathema-
tical standpoint. Let us pose the following question: How
strongly do the values of the function change in response
7-01473
Dialogue tight
98
to the value of its argument t in different portions of the
graph?
READER. In portion 3-4 the values of the function
s (t) do not change at all, while in other portions they do.
A slower rate of change of the function is observed in the
vicinity of points 0, 3, 4, and 6; a faster rate of change
A
B
C
Fig. 32
is observed in the vicinity of points 1, 2, and 5. As a matter of fact, the rate of change is equally fast throughout
portion 1-2.
AUTHOR. You are a keen observer. And where do you
think the rate of change is faster, at point 2 or at point 5?
READER. Of course, at point 2. Here the graph of the
function has a much steeper slope than at point 5. '
AUTHOR. Let us turn to Fig. 32. Here in column A
two portions of the graph of the function s (t) are shown
separately, namely, those in the vicinity of points 2 and 5
(in Fig. 31 these portions are identified by dash circles).
In column B the portions of the graph close to points 2
and 5 are shown again, but this time with a two-fold increase
Velocity
99
in scale. Column C shows the result of another two-fold
scale increase. Obviously, as the scale increases, the curvature of the graph s (t) becomes less noticeable. We may say
that the graph has a property of "linearity on a small scale",
which enables us to consider the slope of the graph at a spe-
cific point. In Fig. 32 (in column C) it is shown that the
slope of the graph at point 2 is ai (the slope is measured
relative to the t-axis), while the slope at point 5 is a2, and
clearly a2 < al.
Denote the slope of the curve s (t) at the moment t by
a (t). Then tan a (t) is said to be the rate of change of the
function s (t) at the moment t, or simply the instantaneous
velocity.
READER. But why tangent?
AUTHOR. You immediately come to it by considering
portion 1-2 of the graph in Fig. 31. This portion represents
a uniform motion of the body, the rate of change of s (t)
being identical at all points. Obviously, it equals the
.average velocity during the time interval t2 - t1, which
is
Sg - Sl
t,-t,
= tan a.
READER. In Fig. 32 you have demonstrated a"straightening" of the graph by increasing its scale. But this straightening is only approximate. Why have you stopped at a mere
four-fold scale increase?
AUTHOR. We can get rid of this approximation and
formulate a more rigorous definition of a slope at a point.
To be more specific, we consider a segment of the graph
s (t) close to point 5. In this segment we select an arbitrary
point B and draw a secant through points 5 and B (Fig. 33).
Next, on the same graph between points 5 and B we select
an arbitrary point C and draw a new secant 5C. Further,
we select an arbitrary point D in the segment between 5
and C and draw a new secant 5D. We may continue this
process infinitely long and, as a result, we obtain a sequence
of secants which converges to a certain straight line (line 5A
in Fig. 33). This straight line is said to be tangent to the
curve at point 5. The slope of the tangent is said to be the
slope of the graph at a given point.
READER. If I understand you correctly we are now in
7*
100
Dialogue Bight
a position to formulate strictly the answer to the question
about the instantaneous velocity.
AUTHOR. Try to do it, then.
READER. The instantaneous velocity of a body at a moment of time t is the rate of change of s (t) at the moment t.
Fig. 33
Numerically it is equal to the tangent of the slope of (the tangent
line to the graph of the function s (t) at the moment t.
AUTHOR. Very good. But you should have mentioned
that s (t) expresses the distance covered by the body as
a function of time.
READER. This is true, my definition of the instantaneous
velocity is tied to the graph of s (t). What if the function
s (t) is not defined graphically?
AUTHOR. Anyway, a graph for s (t) always exists. The
only "inconvenience" in your definition is that it is necessary to take into account the scale of units on the coordinate
axes. If the unit of time (on the t-axis) and the unit of length
(on the s-axis) are represented by segments of identical
length, the instantaneous velocity at time t is
of length
tan a (t) unit
unit of time
If, however, the segment representing one unit of length
is n times greater than the segment representing one unit
Velocity
101
of time, the instantaneous velocity is
of length
n1 tan a (t) unit
unit of time
This "inconvenience", however, has no principal significance.
But it is also possible to formulate a definition of the
instantaneous velocity in a form free of graphic images.
5 _-+=--+- +-
I
t5
t
dt2
d ti
Fig. 34
Look at Fig. 34 which carries Fig. 33 one step further.
Figure 34 shows that the slope of the secant 5B is a ratio
asl
In other words, this is the average velocity for the
At,*
time interval from tb to t5 + At,. The slope of the secant
5C is Ot, that is, the average velocity for the time interval
from t6 to t6 + At2 (At2 < At,). The slope of the secant 5D
is ot3, that is, the average velocity for the time interval
from tb to tb + At, (At, < At2), etc. Thus, a sequence of
the secants converging to the tangent line (drawn at point 5
of the graph s (t)) corresponds to a sequence of the average
velocities converging to the slope as of the tangent line,
2
102
Dialogue Eight
that is, to the value of the instantaneous velocity at the
time moment t5.
READER. It comes out that the instantaneous velocity
is the limit of a sequence of average velocities.
AUTHOR. Precisely. The instantaneous velocity is in
fact the limit of a sequence of average velocities, provided
that the time interval over which the averaging is made
tends to zero converging to the moment of time t (viz., t5,
in Fig. 34).
Now let us formulate the definition in a more rigorous
manner. What we want to define is the instantaneous velocity of a body at a moment of time t. Consider an arbitrary
time interval from t to t + At,. The distance covered by
the body during this interval is Ask. The average velocity
of the body during this time interval is
Asl
vav (t+ At,) = At,
Next we select a shorter time interval At2, from t to t +
+ At2 (At2 < At,), during which a distance As2 is covered.
Consequently, the average velocity'Tover Ate is
vav (t, Ate) =
As2
Ate
We continue this process of selecting shorter and shorter
time intervals starting at the moment of time t. As a result,
we obtain a sequence of the average velocities
vav (t, Atl), vav (t, At2), vav (t, At3), .. .
The limit of this sequence for At --*-0 is the instantaneous
velocity at the moment of time t:
v (t) = lim vav (t, At)
(2)
At-0
Taking into account that
At)-s (t)
vav (t, At) = s (t-I- At
we rewrite expression (2) in the following form
(t)
v (t) = lim s (t+At)-s
At
At-0
I
(3)
103
Velocity
As a result, we can formulate the following definition
of the instantaneous velocity.
Definition:
The instantaneous velocity at a moment of time t is the
limit of a sequence of average velocities over time intervals
t + Ot for At -*O.
READER. Now I realize that instead of talking about
a sufficiently small time interval At (I am referring to our
talk about the ratio As at the beginning of the dialogue),
from t to
the argument should have been based on the limit transition
for At -*0. In other words, the instantaneous velocity
is not At
a sufficiently small At but lim AtS.
at at
At-0
AUTHOR. Exactly. The definition formulated above for
the instantaneous velocity not only exposes the gist of the
concept but gives a rule for its calculation, provided that
an analytical expression for s (t) is known. Let us make
such a calculation assuming that s (t) is given by expression (1).
READER. We should substitute (1) for (3). This gives
v (t) = lim
a (t -}- At)z
2
At-0
atz
2
At
AUTHOR. Go ahead. Remove the parentheses.
READER. This will give
-lima (t%+2t At+At2-tz)=
V (t)
At-0
2At
lim (at+ At)
= at
2
At-0
We have arrived at a familiar formula for the velocity of
uniformly accelerated motion with zero initial velocity:
v (t) = at
(4)
AUTHOR. You are absolutely right. I must congratulate
you: for the first time in your life you have carried out
104
Dialogue Eight
the so-called operation of differentiation. In other words,
you have determined for a given functions (t) its derivative,
that is, the function v (t).
READER. Does it mean that the instantaneous velocity
is a derivative?
AUTHOR. Note that a derivative exists only with respect
to a known initial function. If the initial function is s (t)
(path as a function of time), the derivative is the instantaneous velocity.
I
is
1
t5
4
t
Fig. 35
Let us return now to the graph s (t) shown in Fig. 31.
Our previous arguments and, in particular, relation (4),
allow us to transform the graph s (t) into a graph of the
derivative, that is, the function v (t). A comparison of
the two graphs is given in Fig. 35. I recommend that you
carefully analyze Fig. 35, interpreting it as a comparison
of the graph of a function s (t) and the graph of its rate
of change.
Derivative
105
DIALOGUE NINE
DERIVATIVE
AUTHOR. The previous dialogue gave us an opportunity
to introduce the concept of a derivative for a specific example
from physics (the instantaneous velocity of a body moving
nonuniformly along a straight line). Now let us examine
this concept from a purely mathematical viewpoint without
3,
Let us select an arbitrary point x = x0 from the domain of
the function. In the subsequent argument this point is considered as fixed. Now consider another point x from the
domain of the function and introduce a notation Ax = x - x0. The value Ax is called the increment of the independent
variable. The increment is considered with respect to the
fixed point x0. Depending on the point x, the value of Ax
may, be larger or smaller, positive or negative.
Now let us examine a difference between the values of
the function at points x = x0 + Ax and x = x0: Af (xo) _
= f (xo + Ax) - f (x0). The difference Af (xo) is said to be
the increment of a function f at a point x0. Since x0 is fixed,
Af (xo) should be considered as a function of a variable
increment Ax of the independent variable.
Dialogue Nine
106
READER. Then it is probably more logical to denote this
function by Af (Ax), and not by Af (xo), isn't it?
AUTHOR. Probably, you are right. However, the accepted notation is At (xo). Such a notation emphasized the fact
that the increment of f (in other words, the given function
of Ax) is referred to point xo.
With the concepts of the increment introduced, it is not
difficult to evaluate the rate of change of f close to xo.
xj
x1+Ax xo
xo+dx
Fig. 37
READER. This rate should be described by the ratio
For instance, if we compare A f (xo) with an increment of f at another point from the domain of the function
Af(xo)
Ax
.
(say, point x = x1), we may obtain an inequality
Af (zi)
Ax
>
Af (xo)
Ax
and therefore conclude that the rate of change of f close to
point xl is greater than that close to xo.
AUTHOR. Please, be careful. You have not said anything
about the value of the increment Ax. If Ax is too large,
the inequality you have just mentioned may lead to a wrong
conclusion. I shall make myself clearer by referring to
Fig. 37. As you see,
Af (x1)
Ax >
Af (xo)
Ax
Derivative
107
You must agree, however, that close to point xo the function
changes much faster (the graph of the function has a steeper
slope) than in the vicinity of x1.
READER. It is necessary that the value of the increment
Ax be sufficiently small. The smaller is Ax the more accurate
is the information about the rate of change of the function
close to the point under consideration.
AUTHOR. Well, we can do even better than this. We
may, for example, consider the limit of the ratio
for Ax 0 (remember the previous dialogue).
a limit similar to that discussed at the end of Dialogue
Seven, namely, a limit of the type
AUTHOR. Right. This limit, that is, the limit of the
type 0 is the main subject of this dialogue.
The primary requirement in this case is the existence
of the limit. It means that the function f should be such
that
lim F (Ox) = 0
AX-0
The necessary condition for satisfying this equality is
the continuity of f at x = x0. But we shall discuss this
problem later,
Dialogue Nine
108
If the limit of the type o (in other words, limit (1))
does exist, it is called "the derivative of the function f at
point x = xo and usually denoted by f' (xo).
Definition:
The derivative of a function f at a point xo (denoted by
f' (xo)) is the limit of the ratio of an increment of the function
f at the point xo (denoted by Of (xo)) to an increment Ax of
the independent variable for Ax -*-0:
f' (xo) = Ax-.0
lim
Af (x0)
Ax
or, in a more detailed notation,
f
`
(xo) = li ra f
(xo-I- Ax)-f (xo)
Ax
Ax-0
(2)
Note that you are already familiar with the right-hand
side of equation (2) (cf. expression (3) from the previous
dialogue).
READER. Actually the derivative of the function f at
point xo is the limit of the function
F
G
_
f (xo + Ax) - f (xo)
Ax
at Ax = 0. The independent variable of the function (F)
G
is the increment Ax.
AUTHOR. You are quite right. However, in what follows
you must use the definition of the derivative as formulated
above. This definition does not involve the function (F)
G
of Ax since this function plays, as you understand, only
an auxiliary role. We should simply bear in mind that the
phrase "the limit of the ratio of an increment A f (xo) to an
increment Ax for Ax --o-O" describes the limit of a function
of Ax, i.e. the function (G) , which is considered at Ax = 0.
The derivative can be also interpreted in terms of geometry.
READER. Shall we do it by using again the tangent
to the graph of a function?
bertvattve
ioo
AUTHOR. Yes, of course. Let us take the graph y = f (x)
(Fig. 38), fixing a point x = x0. Consider an increment
Ax1 of the argument; the corresponding increment of the
y-f(x)
Bi
Fig. 38
function at point x0 is Aft (xo). Denote the slope of the chord
AB1 by a1; it is readily apparent that
At, (xo)
Ax1
= tan ai
Next take an increment Ox2 (so that dx2 < 0x1). This
increment corresponds to the increment Oft (xo) of the
function f at point x0. Denote the slope of the chord AB2
by a2; it is similarly quite apparent that
Ate (xo)
Axa
= tan a2
Further, take an increment Axe (Ax. < Ax2), and so on.
As a result, we obtain an infinitesimal sequence of incre-
This leads to a new sequence of the values of the tangent
of the slopes of the chords AB1, AB2, AB3,
.
.
., ABn,
.. .
obtained as a sequence of the ratios of the two sequences
given above
tan a1, tan a2, tan a3,
. .
., tan an,
(3)
. . .
Both sequences (Axn) and (Afn (xo)) converge to zero. And
what can be said about the convergence of the sequence
n (x
n0) ) ?
(tan an) or, in other words, the sequence (L>Ax
)
n
READER. Obviously, the sequence ( Afn (:o))
n
con-
verges to f' (xo). In other words, the limit of (A Axn°) ) is
the derivative of f at xo.
AUTHOR. What are the grounds for this conclusion?
READER. Why, isn't it self-evident?
AUTHOR. Let me help you. Your conclusion is based
on definition 2 of the limit of function at a point. Don't
you think so?
READER. Yes, I agree. Indeed, a certain number (in
this case f' (xo)) is the limit of a function dJ (Ax) (in this
case (D = G) at Ax = 0 if for any sequence (Axn) convergent to zero the corresponding sequence (I (Axn)) converges
to this number. Sequence (3) is precisely the sequence
(I (Axn)) in our case.
AUTHOR. Correct. We have thus found that lim tan an=
n-oo
= f (xo). Now look at Fig. 38 and tell me which direction
is the limit for the sequence composed of the chords AB1,
AB2, AB3,
., AB,, ...?
READER. It is the direction of the tangent to the graph
.
.
f (x) at point x = x0.
berii.ative
iii
AUTHOR. Correct. Denote the slope of the tangent line
by ao. Thus
lim tan an = tan ao
n-,x
Consequently,
f' (xo) = tan ao
We thus obtain the following geometrical interpretation
of the derivative:
The derivative of a function f at a point xo is defined by the
slope of the tangent to the graph of the function f at the point
x=x0.
Fig. 39
Note that the slope of the tangent is measured relative
to the positive direction of the abscissa axis, so that the
derivative of f at point xo in Fig. 39 is positive (at this
point tan ao > 0), while at point xo the derivative of f
is negative (tan ao < 0).
But the geometrical interpretation of the derivative must
not upstage the basic idea that
The derivative of a function f at a point xo is the rate of
change of f at this point.
In the previous dialogue we analyzed the function s (t)
describing the dependence of the distance covered by a body
during the time t. In this case the derivative of s (t) at
a point t = to is the velocity of the body at the moment
of time t = to. If, however, we take v (t) as the initial function (the instantaneous velocity of a body as a function of
time), the derivative at t = to will have the meaning of
112
Dialogue Nine
the acceleration of the body at t = to. Indeed, acceleration
is the rate of change of the velocity of a body.
READER. Relation (2) seems to allow a very descriptive
(if.somewhat simplified) interpretation of the derivative.
We may say that
The derivative of a function y. = f (x) at a point x = xo
shows how much steeper the change in y is in comparison with
the change in x in the neighbourhood of x = x0.
AUTHOR. This interpretation of the derivative is quite
justified, and it may be useful at times.
Getting back to the geometrical interpretation of the
derivative, we should note that it immediately leads to the
following rather important
Conclusions:
The derivative of a function f = const (the derivative of
a constant) is zero at all the points.
The derivative of a function f = ax + b (where a and b
are constants) is constant at all the points and equals a.
The derivative of a function f = sin x is zero at the points
x = tnn (at these points the tangent to the graph of the
function is horizontal).
This "list" could, of course, be expanded.
Next I would like to attract your attention to the following: from the viewpoint of mathematics a derivative of
a function must also be considered as a certain function.
READER. But the derivative is a limit and, consequently, a numberl
AUTHOR. Let us clarify this. We have fixed a point
x = x0 and obtained for a function f (x) at this point the
number
Ii.
Af (x0)
Ax
0x-0
For each point x (from the domain of f) we have, in the
geenral case, its own number
lim
0x-.0
At (x)
Ax
This gives a mapping of a certain set of numbers x onto
.
a different set of numbers lim -The function which
Ax
Ax-+0
Derivative
113
represents this mapping of one numerical set onto another
is said to be the derivative and is denoted by f' (x).
READER. I see. So far we have considered only one
value of the function f' (x), namely, its value at the point
x=x0.
AUTHOR. I would like to remind you that in the previous dialogue we analyzed v (t) which was the derivative
of s (t). The graphs of the two functions (i.e. the initial
function s (t) and its derivative v (t)) were even compared
in Fig. 35.
READER. Now it is clear.
AUTHOR. I would like to make two remarks with regard
to f' (x).
Note 1. A function f' (x) is obtained only by using a function f (x). Indeed,
(x)
f' (x) = lim f (x+Ax)-/
Ax
0x-+0
(4)
It is as if there is a certain operator (recall Dialogue Four)
which generates f' (x) at the output in response to f (x) at
the input. In other words, this operator, applied to the
function f (x), "generates" f' (x). This operator is usually
denoted by ax. This notation should be interpreted as
a single entity and not as a ratio (it reads: "d over dx).
Consider an "image" T-_ II = I2I The squares in this
expression symbolize the familiar "windows". "Window" 1
is to input f (x), while "window" 2 outputs f' (x). Thus,
ax
f (x) = f' (x)
(5)
Definition:
The operation of obtaining f' (x) from f (x) is said to be
the differentiation of f (x).
The operator dx performs this operation over f (x) and
is said to be the operator of differentiation.
READER. But what exactly is dx doing with f (x)?
8-01473
Dialogue Nine
114
AUTHOR. It is exactly the operation prescribed by (4).
We may say that d "constructs" the ratio
TX_
f(x+Ax)-J(x)
Ax
from f (x) and determines the limit of this ratio (regarded
as a function of Ax) at Ax = 0.
READER. In other words, the operator dx performs
a certain limit transition operation, doesn't it?
AUTHOR. Certainly. The whole differential calculus
(and with it, integral calculus) can be formulated in terms
of certain limit transitions.
READER. Why should we introduce an operator d-dx if
it represents nothing else but the limit transition operation
described by (4)?
AUTHOR. You have posed a very important question.
The problem is that if we had formulated differential cal-
culus in terms of limits, using the relations of type (4),
all books on calculus should have been increased in their
volume several-fold and become hardly readable. The use
of the relations of type (5), instead of (4), makes it possible
to avoid this.
READER. But how can we use the relations of type (5)
without implicitly applying the relations of type (4)?
AUTHOR. What is done is this.
First, using (4), we find the result of applying the operator
dx to a sum, product, and ratio of functions, and to composite
or inverse functions provided that the result of applying the
operator to the initial function (or functions) is known. In
other words, the first step is to establish the rules for the
differentiation of functions.
Second, using (4), we find out the result of applying d
to some basic elementary functions (for instance, y = x",
y = sin x, and y = logo x).
After these two steps are completed you can practically
forget about the relations of type (4). In order to differentiate
a function, it is sufficient to express the function via basic
elementary functions (the derivatives of which were obtained
earlier) and apply the rules for differentiation.
READER. Does it mean that the relations of type (4)
berivative
115
could be put aside after they have been used, first, for
compiling a set of differentiation rules and, second, for
making a table of derivatives for basic elementary functions?
AUTHOR. Yes, this is the procedure. Using the differen-
tiation rules and the table of derivatives for some basic
elementary functions you are in a position to forget about
the relations of type (4) and are free to proceed further by
using the "language" of the relations of type (5). A formal
course of differential calculus could skip the analysis of
limit transition operations, that is, the relations of type
(4). It is quite sufficient for a student to learn a set of differen-
tiation rules and a table of derivatives of some functions.
READER. I certainly prefer to be given the foundation.
AUTHOR. Our next dialogue will be devoted to a discussion of the programme of actions as outlined above. At
the first step of the programme, the main rules for differentiation will be established on the basis of the relations of
(4) and, in addition, the derivatives of three functions
y = x2, y = sin x, and y = log,, x will be obtained. At the
second step, we shall obtain (without reference to the
relations of type (4)) the derivatives of the following func-
READER. I'll be looking forward to the next dialogue.
By the way, you wanted to make one more remark about
the derivative f' (x).
AUTHOR. Note 2 concerns the natural domain of a
derivative. Let a set D be the domain of f (x). The question
is whether D is also the domain of f' (x).
READER. In any case, the domain of f' (x) cannot be
wider than the domain of f (x) because in order to find
f' (x) we use f (x).
AUTHOR. A carefully balanced answer, to be sure.
The domain of f' (x) is in the general case a subset of D.
It is obtained from D as a result of elimination of those
points x for which lim AiOx(z) does not exist. By the way,
Ox-.o
this subset is called the domain of differentiability of f (x).
READER. What are the conditions of differentiability
of f (x) at any specific point x?
8*
116
Dialogue Nine'
AUTHOR. Obviously, these conditions are identical to
those of the existence of lim AtAx
(x) at point x. We have
Ax-.O
already observed that it is the limit of the type 0 , which
necessitates that both the numerator and denominator tend
Fig. 40
to zero. It means that f (x) must be continuous at x. The
following theorem could be proved rigorously.
Theorem:
The continuity of a function f (x) at a point x is a necessary
condition for the existence of f' (x) at x.
However, we shall not give the proof of this theorem
here. The simple qualitative arguments given above will
suffice.
READER. I wonder whether the continuity of a function
is also a sufficient condition for its differentiability.
AUTHOR. No, it is not. Consider, for example, the
function y = I log x 1. It is sufficient only to look at its
graph (Fig. 40) to conclude that at x = 1 the tangent to the
graph of the function is, strictly speaking, nonexistent (on
approaching x = I from the left we have one tangent, viz.,
the straight line AA, while on approaching x = 1 from the
right we have another tangent, viz., the straight line BB).
It means that y = I log x I does not have a derivative at
x = 1, although the function is continuous at this point.
Differentiation
117
In conclusion, let us turn to one interesting property of
differentiable functions. Let f (x) be a differentiable function, and its increment Af at x be related to the increment
Ax of the argument as follows:
Af = f' (x) Ax + q (Ax) Ax
(6)
where q (Ax) is a function of Ax. By dividing both parts
of (6) by Ax, we obtain
Af
Ax
= f' (x) +,q
(Ax)
Passing to a limit in both sides of the last equation for Ax
gives lim rj (Ax) = 0.
Ax-.0
Consequently, q (Ax) is an infinitesimal (we use the same
terminology as for numerical sequences, see Dialogue Three).
Conclusion:
An increment Af at a point x of a function f (x) differentiable
at this point can be represented by two summands, namely, a
summand proportional to the increment Ax of the argument
(this summand is f' (x) Ax) and a summand negligible in
comparison with the first for sufficiently small Ax (this summand is q (Ax) Ax, where r) (Ax) is infinitesimal).
READER.IIt seems that this is a formulation of the prop-
erty of "linearity on a small scale" that you mentioned in
the previous dialogue (see Fig. 32).
AUTHOR. Quite true. The main part of the increment of
a differentiable function (a summand linear with respect to
Ax) is called the differential of the function.
DIALOGUE TEN
DIFFERENTIATION
AUTHOR. Now our aim is a practical realization of the
programme outlined in the previous dialogue. This dialogue
could be considered as a drill on the calculation of deriva-
entiation of a function f (x) is defined as the operation of
obtaining f' (x) from f (x). This operation is performed by
using the operator of differentiation ax
-Ti- f (x) = f ' (x)
1. The Differenfiafion Rules
AUTHOR. Rule One. We shall prove the following
Theorem:
The derivative of the sum of two functions equals the sum
of their derivatives provided that they exist, i.e.
But I don't know what to do next.
AUTHOR. We shall repeat your writing but drop the
limit signs:
Au (r)
Ax
_
f(x+Ax)+g(--)-Ax)-f(x)-g(x)
_ f(x+Ax)-f(x)
Ax
Ax
+
g (x+Ax)-g (x)
Ax
Af (x)
Az
Ag (x)
T Ax
Differentiation
119
This gives
Au (x)
Yx
_
of (x)
1x
Og (x)
+ Lx
READER. I see. Next we use the well-known theorem on
the limit of the sum of functions, and the proof is complete.
AUTHOR. Quite correct. Rule Two. Let us prove the
next
Theorem:
A constant multiplier is factored out of the derivative, that is
dx
[a f (x)] = a d (x)
(2)
The theorem is immediately proved if we use the following
obvious equality`
A [a f (x)]
Ax
-a
Of ()
Ox
Rule Three. Now we shall consider the theorem on the
derivative of the product of two functions.
Theorem:
The derivative of a function u (x) = f (x) g (x) is calculated
by using the following formula:
(3)
u' (x) = f' (x) g (x) + f (x) g' (x)
provided that the derivatives f' (x) and g' (x) exist.
Formula (3) is called the Leibnitz formula. Another expression for the same formula is:
a (fg)=gdxf+f
g
READER. Apparently, as in the proof of the first
(x) and
theorem, we must express AuAx(x) through At Ax
og (x) . But how to do it?
Ax
AUTHOR. The simplest way is
u + Au = (f + Af) (g + Ag)
= fg + g Af + f Ag + Al Ag
Hence,
Au = g A/ + I Ag + Af Ag
Dialogue Ten
120
consequently,
Au (x)
Ax
= g (x)
Af (x)
Ax
+. f (x)
Ag (x)
Ax
Af
Ax(x)
+
Ag (x)
Now we find the limit for Ax --} 0. Notice that neither
g (x) nor f (x) depends on Ax, and Ag (x) tends to zero. As
a result,
1
im
Ax-.0
Au (x)
Ax
= g (x) 1im
Ax-O
A f (x)
Ax
-}- f (x) im
Ax-.0
Ag (x)
Ax
The theorem is thus proved.
Rule Four. The next theorem is related to the derivative
of the ratio of two functions.
Theorem:
READER. I shall proceed by analogy with the preceding
proof. I can write
u+Au f+Af
g+Ag
Hence,
f+Af
Au- g+Ag
f
g
_
gAf -fAg
g2+g Ag
This yields
Au
Ax
Af 9Ax_f Ag
Ax
g2+g Ag
Differentiation
121
Passing then to the limit for Ax -- - 0, I take into account
that neither g nor f depend on Ax, and that Ag also tends to
zero. Using the known theorems on the limit of the product
and the sum of functions, we obtain
Au
AX-0 Ax
lim
= lim I(
Ax-0
I
=
g
1
z
lim
g +gAg
Ax-0 \
lim Of
x-.0 Ax
- f lim
g
(g
Al
Ax
-f
Og
Ax
Ag
Ax-.0 Ax
This completes the proof.
AUTHOR. Very good. Now we shall discuss the, problem
of the differentiation of a composite function (for composite
functions, see Dialogue Five). Let w = h (x) be a composite
function, and h (x) = g If (x)]. This composite function is
the composition of two functions w = g (y) and y = f (x).
I remind you that the derivative f' (x) indicates how
faster y charges compared to x, and the derivative g' (y)
h' (x) = g' (y) f' (x)
READER. We have arrived at this rule using very simple
arguments. I wonder whether they can be regarded as a
proof of the rule.
AUTHOR. No, of course not. Therefore I am going to
give the proof of the differentiation rule for composite
functions.
Let the independent variable x have an increment Ax
such that x + Ax belongs to the domain of h (x). Then the
variable y will have an increment Ay = f (x + Ax) - f (x),
while the variable to will have an increment Aw =
= g (y + Ay) - g (y). Since the derivative g' (y) exists.
the increment Aw can be expressed as follows
Aw = g' (y) Ay + Tj Ay
Dialogue Ten
122
where q -*0 for Ay --),-0 (see expression (6) from the pre-
vious dialogue). Dividing both sides of the equation by
Ax, we obtain
Aw
x=
Ay
g,
Âąi
(y) Ax
Next we pass to the limit for Ax
Ay
Ax
0
lim oz = g' (y) lim o- + lim (, ox )
AX-0
Since
o ox
ex
Ax-O
Ax-.O
h' (x) and slim az = f' (x), we have
h' (x)=g'(y)f'(x)+f'(x) lim
x+0
And since Ay
0 for Ax 0,
lim 'q= lim q = 0
Ax-+0
Ay-+O
Hence we arrive at (5), namely, at the rule for the differentiation of composite functions.
Rule Six. Finally, I shall give (without proof) the rule
for the differentiation of inverse functions.
Theorem:
If a derivative y' (x) of an initial monotonic function y (x)
exists and is not equal to zero, the derivative of the inverse
function x (y) is calculated by the formula:
x' (y) =
y, (x)
(6)
READER. It seems that this formula can be easily obtained if we make use of the geometrical interpretation of
the derivative. Really, consider the graph of a monotonic
function y (x) (Fig. 41); its derivative at point x0 is tan a.
The same curve can, obviously, be regarded as the graph of
the inverse function x (y), with y considered as the independent variable instead of x, and x considered as the dependent
variable instead of y. But the derivative of the inverse
function at point yo is tan P (see the figure). Since a + f _
= 2 we have
tan =
1
tan a
Differentiation
123
This gives the above-cited differentiation rule for inverse
functions.
AUTHOR. I must admit that although your line of rea-
soning is not a rigorous mathematical proof, it is an example
of an effective application of geometrical concepts.
Fig. 41
2. The Differentiation of Functions y = x2, y = sin x,
and y = log,,x
AUTHOR. Using (4) from the previous dialogue, calculate
the derivatives of the three indicated functions. Start with
y = x2. Go ahead.
READER. I write
Hence,
This time, however, we should start with a discussion of the
transcendental number e (which is usually called the "base
of natural or Napieriari logarithms"). The number e may be
defined as the limit of a numerical sequence
e=lim(1+1)n
(9)
nyn
The approximate value of e is: e = 2.7182818284590...
Using (9), we can show that e is also the limit of y =
i
= (1 + x)x for x tending to zero
e=lim(1+x)
(10)
X-0
We shall omit the proof of (10).
READER. It seems that (10) follows logically from (9).
AUTHOR. Far from it. Don't forget that in (9) we deal
with the limit of a numerical sequence, while in (10) with the
limit of a function at a point. While n are integers, x belongs
to the real line (with the exception of x = 0). Therefore,
the transition from (9) to (10) requires a good deal of time
and space.
At this point I would have to find the limit for Ax --*0.
AUTHOR. I shall give you a hand here. We can rewrite
1
Ay
(x +Ax) ex = logo (1 +
= logo
1
= x logo
x
Ax
x
Ax lex
(1 -}
)
READER. I see. This gives
x
x
Ay
Ax
1
x
l0ga
1
1
+
Ax
x I)
To find the limit for Ax -* 0, we use (10). As a result
x
Ay
1
lim ox = x
Ax
Ax-limO logo (1 +-X-)
Ax
Ax -O
=
1
1
z
loge e = z
1
'
In a
(symbol In is the standard notation for the natural
logarithm).
AUTHOR. We have thus found that the operator dx
1
1
applied to the function y= 109a
ga X gives y- x' In a
x '
In a
Notice that the natural domain
1
1
x - lna in (11) is ]0, oo[.
of
dx 1ogo x
the function y =
We can sum up our conclusions now.
Using relation (4) from Dialogue Nine, first, we have
established the six differentiation rules and, second, we
have differentiated three functions. The results are summarized in Table 1, and Fig. 42 graphically represents the
d x2= 2x
x
f,-2x
f x2
d
d
.
Gn x=cos X
f -sin x
d leax
xIna
(a >0)
(a >0)
0
x
0
f=loyaX
f=
1
x ln.a
Fig. 42
Dialogue Ten
128
result of the action of the operator d on the three selected
functions. The left-hand column in the figure lists the
graphs of the three functions f (x), and the right-hand
column shows the graphs of the corresponding derivatives
f' W.
In what follows we shall not use formulas of type (4)
from Dialogue Nine, that is, we shall not operate in terms
of limit transitions. Using the results obtained above, we
shall find the derivatives for a number of elementary functions without calculating the relevant limits.
3. The Application of the Differentiation Rules
to Different Functions
READER. For n = 2 formula (12) holds and yields (7).
Assume now that (12) holds for n = m. We have to prove
that it is also true for n = m + 1. We write xm+i = xmx
and use the Leibnitz formula (Rule Three)
ax (xmx) -: x
dx xm + xm
tdx x
Since -Lx- x= 1 and according to the assumption ax xm
= mxm'1, we obtain
d
dx
xm+i = mxxm-1 + xm = (m + 1) xm
The proof is completed.
AUTHOR. The next example is the function
Differentiate this function using Rule Four and
y=x ".
(12).
Differentiation
129
Table 1
The Differentiation Rules
Rule i
(the differentiation of
d (f+g)
ax
the sum of functions)
Rule 2
dx (af)
=a
f+ ax g
f (a= const)
dx
a (fg)=ga fÂąf a
Rule 3
(the differentiation of
the product of functions)
d
x
d
y
d
x g
g-f-fdx
dxY
d
d
Rule 4
(the differentiation of
the ratio of functions)
d
dx
Rule 5
(the differentiation of
composite functions)
dx
(g) =
g If (x)] -
g2
(d
\df
d
x (y)
dy
=
Rule 6
(the differentiation of
inverse functions)
g
(f)) d f (x)
TX
1
d
ax
U (x)
READER. This is simple. Applying Rule Four, we obtain
d
/xx
l
r(
(
xn
By virtue of (12),
d
dx
9-01473
n
nx (n+1)
(13)
Dialogue Ten
ISO
AUTHOR. One particular result that follows from (13) is
d (1
1
x)
dx
(14)
x2
The next example is the function y = l/x.
READER. Here I shall use Rule Six (the differentiation
rule for inverse functions). The inverse function involved
is x = y2. Its derivative is given by (7). As a result,
d
dx V x
2y
d
dy
2 J/
Thus,
d vdx
x
1
2 V.x
(15)
AUTHOR. Now we can pass to the trigonometric functions. Consider the function y = cos x.
READER. I propose to use (8) and the identity sine x +
+ cos' x = 1. By differentiating both sides of the identity
and using Rule One, we obtain
d
x
sin2x-' dx cos2x=0
Next, by applying Rule Five (the differentiation rule for
composite functions) in conjunction with (7), we find
2sinxdx sinx+2cosxd cos x= 0
From (8), x sin x = cos x so that
d
dx
cos x = - sin x
AUTHOR. That is correct, although the result can be
obtained in a simpler way. Better use the identity cos x =
bi#erentiation
131
= sin(-- - x . Further, applying Rule Five, we obtain
2
sin (
- x)
y
here: y =
sin y
2
77 \ 2 - x/
-x
Making use of (8), we find
d
dx sin 2 - x)
sin y = -cosy= - sin x
Using now the suggested identity, we arrive at the final
result:
d
dx
cos x = - sin x
(16)
READER. The operation of differentiation thus "turns"
the sine into the cosine and, vice versa, that is, the cosine
into the sine.
Fig. 43
AUTHOR. Yes, it does. But in the last case the sign
changes too, that is, the cosine is transformed into the sine
with a negative sign. If you plot the graphs of sin x and
cos x in the same system of coordinates (Fig. 43), you will
find that at points x where one of the functions reaches
maximum or minimum (takes the value 1 or -1) the other
function vanishes. It is readily apparent that this fact has
direct relation to your remark. If, for example, at a certain
oint x the function sin x assumes its maximum value, the
9
13
Dialogue ?'en
tangent to its graph at the same point will, obviously, be
horizontal. Consequently, the derivative of the function
(i.e. cos x) must vanish at this point. I recommend that you
carefully analyze Fig. 43. In particular, follow the correspondence between the slope of the tangent to the graph of
the function drawn at different points and the sign of the
derivative at the same points..
Now turn to the next example, the function y = tan x.
Differentiate this function using the results of the differentiation of sin x and cos x and applying Rule Four.
READER. This will be easy:
d
cos x
sin x- sin x-dcos x
d r sin x l
dx
dx 1 cos x 1
co92 x
cosy x+sin2 x
coos
1
x
co92
x
Finally,
dx
tan x =
(17)
cosy x
AUTHOR. The result for y = cot x can be obtained
similarly:
d
dx
cot X= -
1
(18)
SW X
In order to differentiate y = arcsin x, we use Rule Six
dx aresin x =
d sin y
=
csy
cos (aresin x)
Since
cos (aresin x) = VT--x2
we obtain
d
dx
aresin x =
1/1-xz
(19)
Differentiation
133
In order to differentiate y = arccos x, it is sufficient to
use (19) and the identity
aresin x + arccos x = 2
Therefore,
d
dx
arccos x - -
1
V1-x-
(20)
Using Rule Six, we differentiate the function y = arctan x
dz arctan x =
I
d
7Y-
= cost y = [cos (arctan x)]2
tan ,y
Since
1
cos (arctan x) =
we obtain
arctan x =
1
(21)
1
And, finally, the differentiation of y = arccot x is carried
out by using the identity
arctan x + arccot x = 2
and yields
d arccot x = -
1
__
xa
(22)
We have thus performed the differentiation of all element-
ary trigonometric and inverse trigonometric functions.
Dialogue Eleven
134
In conclusion, let us examine the exponential function
y = ax. Using (11) and Rule Six, we obtain
dz ax=
1
d
d
=ylna=axlna
loge Y
This gives
dx
ax = ax In a
(23)
Result (23) is very interesting. We see that the differentiation of the exponential function y = ax again yields
the exponential function ax multiplied by the constant term
In a. In a particular case of a = e, we have In e = 1, and
therefore
d ex_ex
dx
(24)
The exponential function y = ex is simply called the exponential curve. From (24) it follows that differentiation
transforms this function into itself.
DIALOGUE ELEVEN
ANTIDERIVATIVE
READER. Differentiation is an operation of finding a
function f' (x) for a given function f (x). Presumably, an
inverse operation is possible as well, isn't it?
AUTHOR. An inverse operation indeed exists. It is called
integration. Integration of a function f (x) is an operation
by which the so-called antiderivativr is found for the given
function f (x).
seeking a derivative f' (x) for a given function f (x), and
now we deal with a situation in which the given function
f (x) is the derivative of a yet unknown function F (x).
AUTHOR. Absolutely right. Take, for example, a function
READER. But how did you find this antiderivative?
AUTHOR. This was simple. I resorted to the well-known
rules of differentiation but in a reverse order. In other words,
I mentally searched for a function that would yield our
function f (x) = 2x2 - 3x after differentiation. You can
easily verify that
F (x) = 3 3x2 - 3 2x = 2x2 - 3x
READER. But then why not take as this antiderivative,
example, a function F (x) = 3 x3 - 2 x2 + 2? It
will again yield F' (x) = 2x2 - 3x.
AUTHOR. You noticed a very important feature. Indeed,
an antiderivative found for a given function is not unique.
for
If F (x) is an antiderivative (for a function f), then any
function F (x) + C, where C is an arbitrary constant, is also
an antiderivative for the initial function because
d
AUTHOR. Precisely. Take a graph of one of the antiderivatives. By translating it along the y-axis, you will
obtain a family of the curves of antiderivatives for a given
function f. For example, let f (x) = sin x. The curves of
Fig. 44
antiderivatives for this function are plotted in Fig. 44.
These curves plot functions
F(x)=-cosx+C
(the dash curve is the graph of the function f (x) = sin x).
The constants C were taken with an increment of 0.5. By
reducing this increment, one can obviously obtain a pattern
of arbitrarily high density of F (x) curves.
The figure clearly shows that all the antiderivatives
belong to one family (in other words, correspond to the same
initial function f). This may not always be as clear if the
function is represented in an analytical form. Take, for
example, functions Fl = -cos x and F2 = 3 - 2 cost 2
It would be difficult to say at the first glance that these two
functions are the antiderivatives of one and the same function
(namely, f = sin x). However, since 2 cosy 2 = 1 + cos x,
we find
Fa(x)=3-I-cosx=-cosx+2
Antiderivative
137
READER. I guess it would be possible to find directly
that F' (x) = F' (x), wouldn't it?
AUTHiOR. Of course, it would:
dx F2 (x)
(see Table 2) which gives various functions f (x) in the
first column, the corresponding derivatives f (x) in the
Dialogue Eleven
138
Table 2
A List of Derivatives and Antiderivatives
for Selected Functions
f (x)
f' (x)
F (x)
0
ax+C
1
a
2
xn
nxn-1
3
ex
ex
ex+C
4
1
y
In x+C
n+1
1
x
2
1
5
2
x
3
sinx
coSx
7
cosx
- sinx
1
2 sinx
cosy x
1
-2 cosx
sin3 x
9i1u2 X
10
1
+ x2
- +
(1
sinx+C
tan x+C
CO92 x
9
x 1/x+C
-cos x+C
6
8
xn+1 + C
2)2
--cotx+C
arctanx+C
Antiderivative
139
second column, and the antiderivatives F (x) + C, corresponding to the functions f (x), in the third column. I want
to stress once more: the transformation f (x) -*f' (x) is the
operation of differentiation of the function f (x), and the
transformation f (x) --)- [F (x) + C] is the operation of
integration of the function f (x).
READER. Examples (8), (9), and (10) in Table 2 give
an impression that the transformation f (x) -- f' (x) is more
complicated than the transformation f (x) --> [F (x) -f- Cl.
AUTHOR. This impression stems from a special selection
of the functions f (x). Thus, it is easier to differentiate the
function tan x than the function core x . Indeed in the
latter case we have to use the rules for differentiation of a
ratio of two functions or of a composite function.
In general, it should be noted that the operation of inte-
gration is substantially more complicated than that of
differentiation. The differentiation of elementary functions
invariably gives elementary functions. By employing the
rules for differentiation discussed in the previous dialogue,
you will be able (and with no difficulties, as a rule) to differentiate practically any elementary function. But integration
is quite a different proposition. The rules for the integration
of elementary functions comprise numerous techniques, and
we would need several special dialogues to scan them. But
the main point is that not every elementary function has an
elementary function for its antiderivative. As one example,
shall mention the antiderivatives of such elementary
functions as log or l71 +xs . As a rule, in such cases
I
x
one is forced to resort to the methods of the so-called numerical integration.
be formed of fundamental elementary functions by a finite
number of the operations of addition, subtraction, multi-
140
Dialogue Eleven
plication, division, involution, evolution, and taking a
modulus, as well as by using the rules for obtaining inverse
and composite functions. All the functions used in the
previous dialogues are elementary (with an exception of the
Dirichlet function mentioned in Dialogue Five), and many
of them are fundamental elementary functions.
READER. My second question concerns the rules for
integration you refer to. Could you give at least some examples?
AUTHOR. I shall quote three simplest rules.
1. If F is an antiderivative for f, and G is an antiderivative
for g, an antiderivative for the sum of the functions f + g
is a function F + G.
2. If F is an antiderivative for f, an antiderivative for a
function af, where a is a constant, is a function aF.
3. If F (x) is an antiderivative for f (x), and a and b are
constants, an antiderivative for a function f (ax + b) is a
function a F (ax + b).
All the three rules are proved readily by using the rules
for differentiation (in the third rule one has to apply the
rule for the differentiation of composite functions). Indeed,
a rich collection of integration rules available in calculus.
But here these three rules will be sufficient since our goal
is quite modest: to give the fundamental idea of an antiderivative.
READER. Our discussion of a derivative covered its
geometrical interpretation as well. Is there a geometrical
interpretation of an antiderivatiye?
Antiderivative
!41
AUTHOR. Yes, there is. Let us find it (besides, we shall
need it later).
Consider a function f (x). For the sake of simplicity, assume that this function is monotonic (and even increasing).
Later we shall drop the
monotonicity of a func-
tion. The most important is that the function
be continuous over the
chosen interval (i.e. over
the interval on which it
is defined). Figure 45
shows a shaded area (the
so-called curvilinear tra-
pezoid) bounded by the
graph of the function
f (x), the interval [a, xl
of the x-axis, and two
perpendiculars
erected
from points a and x on
the x-axis. Let point a
be fixed; as for point
Fig. 45
x (the right-hand
end of the interval [a, x]), it is
not fixed and can assume values from a upward (within the
domain of definition of the function). Obviously, the area
of the curvilinear trapezoid shaded in the figure is a function
of x. We shall denote it by S (x).
Now turn to Fig. 46. Let us give an increment Ax to the
independent variable x. The interval [a, x + Ax] corre-
sponds to the area S (x + Ax). Denote AS (x) = S (x + Ax)-
- S (x). The increment AS (x) is, obviously, the area of
the shaded curvilinear trapezoid. The figure shows that
area ADEF < AS (x) < area ABCF
But the area ADEF is equal to f (x) Ax, and the area ABCF
is equal to f (x + Ax) Ax. Therefore,
f (x) Ax < AS (x) < f (x + Ax) Ax
or
f (x) <
As (X)
< f (x + Ax)
Dialogue Eleven
142
or
0
<°ix)-
f (x)) < [f (x) + Ax) - f (x)) = Af (x)
Now we find the limiting values of these inequalities for
Ax tending to zero. By virtue of the continuity of the
F
x x+Ax
a
Fig. 46
function f (x) we conclude that lim Af (x) = 0. ConseAxwo
quently,
lim
(AS(x)-f(x))=0
Ax
AX-0
As the function f (x) is independent of Ax, the last relation
yields
AS
(x) =
Ax o Ax
lim
f (x)
(2)
By the definition of derivative,
lim
AS (x)
Ax
= S' (x)
Consequently, relation (2) signifies that
f (x) = S' (x)
(3)
Antiderivative
143
Thus, in terms of geometry, the antiderivative of the function f, taken at point x, is the area of curvilinear trapezoid
bounded by the graph of the function f (x) over the interval
[a, x] of the x-axis.
READER. Presumably, it is one of the possible antiderivatives, isn't it?
AUTHOR. Definitely.
READER. But it is evident that the area S (x) also depends on the choice of point a.
AUTHOR. Absolutely correct. By choosing different
points a, we shall have different areas of curvilinear trapezoids and, correspondingly, different antiderivatives. But
all of them will be the antiderivatives of the function f
taken at point X. It is only important that in all cases
a < x.
READER. Then why is it that point a vanishes from
the final results?
AUTHOR. Your bewilderment is understandable. Let
us reformulate the results obtained above. Let F (x) be an
antiderivative of a function f (x) taken at point x. According
to (3), we can write
S (x) = F (x) + C
(here we have used the following theorem: if two functions
have equal derivatives, the functions will differ by a constant
If F (x) is an antiderivative of a function f (x), then the
area S (x) of a curvilinear trapezoid bounded by the graph
of the function f (x) over the interval [a, x] is given by the
difference F (x) - F (a).
You see now that point a is introduced explicitly.
READER. Now everything is clear.
AUTHOR. Relation (3) (and from it, (4)) can be obtained
Dialogue Eleven
144
for every continuous function; the monotonicity of a function
is not a necessary condition. Consider a function f (x) whose
graph is plotted in Fig. 47. We choose a point x and wish
to prove that for any e > 0 there is 6 > 0 such that
AS (x)
Ax -
f (x) I < e
for all Ax satisfying the condition I Ax I < 6.
I
(5)
READER. Shall we consider point x as fixed?
AUTHOR. Yes. Increments Ax and, correspondingly,
AS (x), are always considered for a definite point x.
So we take an arbitrary number e > 0 (shown in the
figure). As f (x) is a continuous function, there is a number
6 > 0 such that
If(x+Ax)-f(x)I<a
(6)
for all Ax satisfying the condition I Ax I < 6. This number 6
is the one we were to find.
Indeed, let us choose, for definiteness, that Ax > 0 but
specify that Ax < 6. The area of the curvilinear trapezoid
shaded in Fig. 47 will be denoted by AS (x) (this trapezoid
is bounded by the graph of the function f (x) over the interval
which is what we wanted to prove.
You see that a function/ needn't be monotonic: relation
(3) (and with it, (4)) is easily generalized to the case of an
arbitrary continuous function f.
Now let us turn again to Fig. 44 that gives a family gf
graphs of the antiderivative F (x) = -cos x + C for tli
Antiderivative
146
function f (x) = sin x. Indicate which of these graphs
(which antiderivative) stands for S (x) in each of the follow-
Fig. 47
and (c) a = n.
ing three cases: (a) a = 0, (b) a
READER. The question is clear. I denote the sought
functions by S1 (x), S2 (x), and S3 (x), respectively. These
AUTHOR. Correct. It is important to underline that in
each of the above three equalities the function F (x) is a
function chosen arbitrarily from the family of antiderivatives of f, shown in Fig. 44.
READER. It looks as if whatever the selected antiderivative of the function f is, the difference between its values
at two points depends only on the choice of these points but
not on the choice of a specific antiderivative.
AUTHOR. You have pointed out a property of principal
significance. It is so important that deserves a special dia-
logue.
DIALOGUE TWELVE
INTEGRAL
AUTHOR. We know already that the difference between
the values of an antiderivative at two arbitrary points
depends only on the choice of these points (and, evidently,
on the type of the initial function f (x)). As these two points
we choose points a and b, that is, consider an increment of
an antiderivative, F (b) - F (a). This increment plays a
very important role among the tools of calculus; it is called
the integral.
Definition:
The increment of an antiderivative F of a function f, i.e.
F (b) - F (a), is said to be the integral of f from a to b.
The notation of the integral is:
(it reads: "integral of f of x, dx, from a to b"). The numbers a
and b are the lower and upper limits of integration. The
function f is said to be integrand, and x the integration varP
able.
Integral
1li'7
Consequently, if F is one of the antiderivatives of the
function f, then the definition of an integral states that
b
f (x) dx = F (b) - F (a)
(1)
a
Formula (1) is known in the literature on mathematics as
the Newton-Leibnitz formula. Remember that F here is ar
arbitrary antiderivative of the function f.
V
f
Fig. 49
READER. As far as I understand, the integral of the
function f from a to b is precisely the area of the curvilinear
trapezoid bounded by the graph of the function f (x) over
the interval [a, b]. Is that right?
AUTHOR. Absolutely. The expression
is nothing less than the area of this geometrical figure.
Figure 49 shows three cases plotting different_ integrands:
(a) f (x) = 2x, (b) f (x) = x2,
100
(c) f (x)=yx
Dialogue Twelve
148
The limits of integration are chosen identical in all the
three cases: a = 1, b = 2. The corresponding areas of the
curvilinear trapezoids are shaded in the figure:
z
S,= 2xdx
i
2
S2 = f x2 dx
i
2
S3 = f vx dx
i
The numbers S1, S2i and S3 are different because the inte-grands f (x) are different.
We thus find that the expression
works as a functional (recall Dialogue Four). You "input" in
it a function f, and it "outputs" a number S.
By the way, you can easily find how this functional works.
To achieve this, use formula (1) and take into account that
at the "input", we obtain at the "output" the number 3
and with f x at the "input", we obtain at the "output" the
number
(2 2 - 1).
3
READER.
I see that we can rather easily find the areas
of various curvilinear trapezoids!
AUTHOR. More than only curvilinear trapezoids. For
instance, try to find the area of the figure shaded in Fig. 50.
READER. This area is the difference between the areas
of two curvilinear trapezoids:
0
Therefore,
S=-2
=3x
i
2_1_1
o^3o
rx -I'
I1
3
3
3
AUTHOR. Correct. Consider another example. Find the
area of *the shaded figure in Fig. 51.
READER. The graphs of the functions sin x and cos x
intersect at the point x = 4 Consequently, one has to
.
use the antiderivative of the function sin x over the"interval
Dialogue Twelve
150
CO,
4
C4
] , and that of the function cos x over the interval
2 ] Hence,
n
n
4
2
n
sin x dx + cos x dx = - cos x 14
S
n
-+- sin x
2
U
4
(cos 4 -cos o)
(sin
2
2 -sin 4 )
2
AUTHOR. Perfectly right. Now we shall discuss one
"fine point", returning to formula (1) and rewriting it in
the form
x
T f (t) dt = F (x) - F (a)
(2)
a
What has been changed by this rewriting?
READER. First, we have replaced the constant upper
limit of integration (the number b) by the variable limit of
integration (the variable x). Second, we have substituted the
integration variable t for the integration variable x.
AUTHOR. Only the first of these changes is significant.
The second (the substitution of the integration variable) is
of no consequence. It is easy to see that the formulas
b
b
b
f f (x) dx, 5 f (t) dt, S f (y) dy, S f (z) dz
a
a
a
a
are equivalent since all the four give F (b) - F (a). So it
does not matter what symbol is used for the integration
variable in each particular case.
READER. Why, then, did you have to substitute the
variable t for the integration variable x in (2)?
Integral
151
AUTHOR. Only not to confuse the integration variable
with the variable upper limit. These are different variables
and, of course, must be denoted by different symbols.
The expression
is called the integral with a variable upper limit. It is important that in contrast to the expression
this expression yields not a number but a function. According
to (2), this function is F (x) - F (a).
b
READER. But if the
x
then the
dt "black box" is a functional,
a
dt "black box" is an operator? (I have used
a
here our symbolic notation of "windows" into which the
function f must be input).
AUTHOR. Correct. This is immediately clear in the
following unusual table.
Table 3
f (x)
2x
3x2
4x3
5x4
6x5
3
7
15
31
63
x2-1
x4-1
a5- 1
x5-1
x3- 1
Dialogue Twelve
152
The second and third columns of this table show what the
x
2
"output" of the two "black boxes",
f (t) dt and
f (t) dt,
is when the "input" is a function f of the first column.
X
The integral J (...) dt is thus indeed an operator. Note
a
that its effect on a function is opposite to that of the operator
(we discussed this operator in Dialogue Nine).
T-
Indeed, take a function" f and first apply to it the opex
x
rator S!(...) dt and then the operator d : Fdz
f (t) dt) .
a
This gives
x(
a (I f (x) dt) = a [F (x) - F (a)] = a- F (x) = f (x)
a
i.e. we obtain the initial function f.
READER. We could apply these operators to the function
in the reverse order, couldn't we?
AUTHOR. Yes, we could. This means that the expression
x
fa (1-: f (t) ) dt
also gives the initial function f. At least, to within a constant term.
READER. Can it be verified?
AUTHOR. Yes, and very easily. What function is the
antiderivative for f' (x)?
READER. Obviously, the function f (x) + C.
AUTHOR. Therefore,
x
xx
a
(a f (t)) dt = J f' (t) dt = f (x) - f (a)
a
153
Integral
READER. Will it he correct to say that while the operator
d performs the operation of differentiation, the operator
x
J
a
(...) dt performs the operation of integration?
AUTHOR. Precisely. It might seem that the topic is
exhausted, but the discussion would be incomplete without
a clarification of one essential "subtlety". Throughout this
dialogue we operated with something we called "the area of
a curvilinear trapezoid" and found that this is the meaning
of the integral. But what is the "area
of
trapezoid"?
a curvilinear
READER. But surely this is self-evident. One glance at
the figures is enough.
AUTHOR. Look, for instance, at Fig. 45. It shows a shaded
geometrical figure called a curvilinear trapezoid. But it
says nothing about the area of the trapezoid.
READER. The area is a standard concept in geometry.
AUTHOR. No objections. But do not forget that in
geometry you normally apply this concept to a well-defined
set of figures: triangles, trapezoids, etc. And you remember
that difficulties arise when you try to determine the area
of a circle. By definition, the area of a circle is the limit of
the sequence of the areas of regular polygons inscribed in,
or circumscribed around, the circle, for an infinitely increasing number of the sides of the polygon.
READER. Presumably, the area of a curvilinear trapezoid
can also be defined as the limit of a specific sequence of areas?
AUTHOR. Yes, this is the normal approach. Consider a
curvilinear trapezoid bounded by the graph of a function
f (x) over the interval (a, b] (Fig. 52). Let us subdivide the
interval (a, b] into n subintervals of identical length Ax =
On each Ax-long interval, used as a base, we construct a
rectangle of altitude f (xk _1), where k is the subscript of
the right-hand end of this subinterval (this choice is arbitra-
Dialogue Twelve
154
ry; the left-hand end would do equally well). The area of
this rectangle is
f (xk-i) Ax
Consider now a sum of the areas of all such rectangles (this
total area is shaded in Fig. 52):
Sn (a, b) = f (xo) Ax + f (xi) Ax + ... + f (x,,-,) Ax
= If (x0) + J (x!) + ... + {J (xn-!) I
b
n a
As the function f (x) is continuous, the ensemble of all
these rectangles for sufficiently large n (sufficiently small Ax)
1'
Fig. 52
will be very close to the curvilinear trapezoid in question,
and, at any rate, the closer the larger n is (the smaller Ax).
It is, therefore, logical to assume the following
Definition:
The sequence of sums (Sn (a, b)) with n tending to infinity
has the limit said to be the area of the given curvilinear
trapezoid S (a, b):
S (a, b)
= lim Sn (a, b)
n-oo
(3)
READER. The area S (a, b) of a curvilinear trapezoid was
b
shown earlier to be the integral
f (x) dx; consequently,
Integral
155
definition (3) is a new definition of the integral:
b
f (x) dx = 1im Sn (a, b)
(4)
n-oo
a
Do you agree?
AUTHOR. Yes, certainly. And note that definition (4)
is independent, that is, it is not based on the concept of the
antiderivative.
Historically, by the way, the integral appeared as (4),
the fact that explains the origin of the standard notation.
Indeed, if definition (4) is rewritten in a slightly different
form
n
b
f (x) dx = lim (E f (xk_1) Ox)
a
n-.oo
(5)
k=1
you may notice a certain similarity in the form of the leftand right-hand sides of this equality. The very symbol
(the integral sign) originated from the letter S which was
often used to denote sums. The product f (xk_1) Ax evolved
to f (x) dx. In the 17th century mathematicians did not use
the concept of the limit. They treated integrals as "sums of
an infinitely large number of infinitely small addends",
with f (x) dx being these infinitesimal addends. In this sense,
1
the area of a curvilinear trapezoid S was defined as the
"sum of an infinitely large number of infinitely small areas
f (x) dx".
You realize, I hope, that such concepts were obviously
lacking mathematical rigorousness.
READER. This illustrates what you termed on many
occasions "subjective impressions".
AUTHOR. It must be clear to you by now that a strict
mathematical interpretation of the concept of the integral
is possible only if the limit transition is used. I have already
emphasized that the limit transition is the foundation of
calculus. If the concept of the limit is avoided ("limit of
156
Dialogue Thirteen
sequence" or "limit of function"), neither the derivative
nor integral can be treated rigorously.
READER. But the integral can be defined without resorting to (4). It is quite sufficient to use the Newton-Leibnitz
formula (1). And this formula does not involve any limit
transitions.
AUTHOR. But this formula involves the antiderivative.
And the antiderivative involves, in the long run, the concept of the derivative, that is, the unavoidable limit transition.
By the way, your last remark makes me touch the aspects
of introducing the integral in the literature. Two methodically distinct approaches are possible.
The first approach (the one used in these dialogues) assumes
that the operation of integration is directly introduced as
an operation inverse to differentiation. The Newton-Leibnitz formula (1) then serves, in fact, as the definition of the
integral: it is defined as an increment of the antiderivative.
The second approach assumes that the operation of integration is introduced as an independent operation, the integral being defined as the limit of a sequence formed of the
appropriate sums (see formula (4)). This approach corresponds to the historical progress in mathematics; indeed,
originally integral calculus was evolving independently of
differential calculus. The profound relationship between
the two branches of mathematics had been discovered only
by the end of the 17th century when the main problems of
the two were understood as mutually inverse. The NewtonLeibnitz formula (1) was precisely a reflection of this relation-
ship: it was demonstrated that the integral is none other
than an increment of the antiderivative.
DIALOGUE THIRTEEN
DIFFERENTIAL EQUATIONS
AUTHOR. You are, certainly, familiar with various types
of equations: algebraic, logarithmic, exponential, trigon
metric. They have a common feature: by solving these
Differential Equations
157
equations one arrives at numbers (these are the so-called
"roots" of equations). Now we are going to deal with a very
different type of equations, namely, equations whose solutions are functions. Among the equations subsumed into
this class are the so-called differential equations.
Consider a function f (x). We denote its first derivative
(the first-order derivative) by f' (x), its second derivative
by f" (x), its third derivative by f' (x), and so on.
Definition:
A differential equation is an equality relating x, f (x), f' (x),
f " (x), etc. A solution of a differential equation is a function
f (x)
READER. So far you have never mentioned the concepts
of second derivative or third derivative.
AUTHOR. True, and this is what we are going to do right
now.
READER. It is readily apparent that since a derivative
f' (x) is a function, it can be differentiated, thus yielding
a derivative of the derivative; I guess, this must be the second
derivative of the original function f (x).
AUTHOR. By differentiating the function f (x) n times
(of course, if this can be done with the given function), we
obtain a derivative of the nth order (in other words, "the
nth derivative"). Thus, the third derivative of f (x) is,
obviously,
f"
(x) - d a a f (x))
Note that we are, in fact, familiar with the second derivative. As the function f (x) is the first derivative of an anti-
derivative F (x) [I (x) = F (x)J, the function f' (x) can be
considered as the second derivative of the antiderivative
F (x):
f' (x) = F " (x)
READER. We know that the derivative of f (x) (to be
exact, its first derivative) is the rate of change of this function. Its magnitude is reflected in the slope of the graph of
the function f (x) at each point and is measured as the
tangent of the angle between the tangent line to the graph
and the abscissa axis. Could anything of this type be said
about the second derivative of f (x)?
i58
Dialogue Thirteen
AUTHOR. Evidently, the second derivative of f (x)
characterizes the rate at which the rate of change of the function changes with x, so that it is a finer characteristic of the
behaviour of the initial function. Look at Fig. 53. What
is the difference between functions f1 and f, at point x = xo?
READER. They have different first derivatives. I can
write:
! 1 (x0) = f2 (x0),
! 1 (x0)
f2 (x0)
AUTHOR. To complete the picture, note that at the point
in question the derivatives differ both in magnitude (the
figure clearly shows that I fz (xo) I < I f i (xo) I) and in sign:
fi (xo) > 0, fz (xo) < 0. We say, therefore, that the function
f1 increases (and rather rapidly) at point x = x0, while the
function f2 decreases (and comparatively slowly).
Now turn to Fig. 54. We observe that not only the values
of the functions f1 and f2 but also the values of their first
derivatives coincide at point x = xo:
f1 (xe) = f2 (Xe),
fi (xo) = f2 (xo)
However, the graph shows a difference in the behaviour of
the functions f1 and f2 in the vicinity of x0. Try to describe
this difference.
READER. In the vicinity of xo the graph of the function
fl is convex downward, while that of the function f2 is con-
Differential Equations
159
vex upward. Besides, the curvature is greater for the function f2 than for fl.
AUTHOR. These are precisely the finer features of the
behaviour of f (x) close to x = x0, and they can be identified
by finding the value of the second derivative at x0 (by calculating the value of f " (xo)). In the case shown in Fig. 54
we have
fi (x0
f2 (xo)
You will immediately see that f j (xo) > 0 and f j (xo) < 0.
Indeed, the slope of fl at x0 steadily increases; hence, the
slope of f i (x) is positive. On the contrary, the slope of f 2
at x0 steadily decreases; hence, the slope of fZ (x) is negative.
It is quite obvious (see the figure) that
I fi (xo) I < I fz (xo)
I
READER. In all likelihood, the third derivative of
f (x), i.e. f"' (xo), is a still finer characteristic of the behaviour of f (x) at x = x0. Am I right?
AUTHOR. Precisely. Unfortunately, it is virtually impossible to illustrate this simply enough on a graph of the
function f (x).
I think that it is enough for a discussion of derivatives of
different orders; let us move on to differential equations.
Note, first of all, that an equation of the type
f' (x) = (P (x)
(1)
where q (x) is a given function, can be considered as the
simplest particular case in the theory of differential equations; its solution is obtained by a straightforward integration.
growth (decay), and equation (3) is the differential equation
of harmonic oscillations.
Let us look at these equations more closely. We begin
with the differential equation of exponential growth (decay).
What conclusions can be drawn from the form of this equation?
READER. The form of equation (2) shows that the rate
of change of the function f (x) coincides with the value of the
function, to within a constant factor p, at each point x.
In other words, the function f (x) and its first derivative
f (x) coincide, to within the mentioned factor, at each
point x.
AUTHOR. Please, recall Dialogue Ten and tell me.what
functions could serve as solutions of this equation. What
are the functions for which the derivative coincides with the
function itself? In other words, what functions are transformed by differentiation into themselves?
READER. This property is typical of the exponential
function a" for a = e. It is called the exponential curve
and is often denoted by exp (x). We have found in Dialogue
Ten that
dx exp (x) = exp (x)
AUTHOR. Correct. This means that the function f (x) _
= exp (px) must be taken as a solution of the equation
f' (x) = pf (x). Indeed,
a eXP (Px) _ dy exp (y)) dx (Px) = p exp (y) =p exp (px)
For this reason equation (2) is called the differential
equation of exponential growth (decay). Obviously, we have
growth if p > 0, and decay if p < 0.
READER. Apparently, any function
/ (x) = C,expj(px)
where C is an arbitrary constant factor, is a solution of this
equation, because the constant C is factored out of the derivative.
Some of functions C exp (px) are plotted in Fig. 55 (we
have specified p > 0).
The formula f (x) = C exp (px), describing the whole
family of functions, is called the general solution of a given
differential equation. By fixing (i.e. specifying) a value of C,
one selects (singles out) a particular solution from the general
solution.
READER. How can it be done?
AUTHOR. Oh, this is elementary. It is sufficient to prescribe a specific value to the function f (x) at a certain point.
For example, let us prescribe
/ (x0) = yo
In this case we are interested in a single curve among the
curves of the whole family (see Fig. 55; the selected curve
is shown by a thicker solid line). This curve is a graph of the
function C exp (px) for which C exp (pxo) = yo, and, therefore, C = y0 exp (-pxo). Consequently, the particular solu1/2 i1-01473
Dialogue Thirteen
162
tion we are seeking for has the form
/ (x) = Yo exp [P (x - xo)1
(4)
READER. We thus obtain that in order to find a specific
(particular) solution of the differential equation f' (x) =
= p f (x), it is necessary to supplement the equation with an
additional condition: f (xo) = yo.
AUTHOR. Precisely. This condition is called the initial
condition.
Let us turn now to differential equation (3):
f" (x) = -qf (x) (q > 0)
READER. In this case the value of the function f (x)
coincides at each point not with the rate of change of the
function but with the rate of change of its rate of change,
with the sign reversed.
AUTHOR. In other words, the function f (x) is equal, to
within a constant factor, to its second derivative I" (x).
Recall what functions have this property.
READER. I,guess that the solutions of equation (3) are
functions sin x or cos x.
AUTHOR. To be precise: sin (vqx) or cos (Vq,x).
Indeed,
dx
or
dx sin (V q x)) = Yq dx cos (1/q x) = - q sin (Yq x)
d dx cos(vgx))-Vqd sin(yggx)_-gcos(/qx)
d
This is why the equation in question is called the differential
equation of harmonic oscillations.
It is easily seen that the general solution of equation (3)
can be written in the form
f (x) = C1 sin (yq x) + C2 cos (j/q x)
READER. But this gives us two integration constants
instead of one, as in the preceding case.
AUTHOR. Yes, and the reason is that differential equation
(3) contains the second derivative. Hence, it is necessary to
integrate twice in order to obtain the function f (x). And
we know that each integration leads to a family of antiderivatives, that is, generates an integration constant. In
the general case, the number of integration constants in the
general solution of a specific differential equation equals
the maximum order of derivative in this equation. The general
solution of equation (2) has a single integration constant
because it contains only the first derivative of the sought
function and does not involve derivatives of higher order.
The general solution of equation (3) has two integration
constants because the equation contains the second-order
derivative of the sought function and no derivatives of higher
order.
READER. And how do we write the initial condition for
equation (3)?
AUTHOR. One has to prescribe at a point x = xo a value
not only to the sought function but also to its first derivative.
In this case the initial conditions are written as follows:
f (xo) = to,
f' (xo) = fo
(6)
READER. And if a differential equation involved the
third derivative, and the general solution contained, as a
result, not two but three integration constants?
AUTHOR. In this case the initial conditions would prescribe values to the required function, its first derivative,
and its second derivative at a point x = xo:
f (xo) = to,
f' (xo) = fo,
f" (xo) = fo
But let us return to the general solution of equation (3).
It is usually written not in form (5) but in a somewhat
different form. Namely, either
f (x)=Asin (Vgx+a)
(7)
f (x) = A cos (V q x + 6)
(7a)
or
Formula (7a) is obtained from (7) if we set a = + n2
Dialogue Thirteen
164
In what follows we shall use notation (7). In this form
the role of the integration constants C1 and C2 in general
solution (5) is played by constants A and a. Formula (5)
is easily transformed by trigonometry to (7), by using the
formula for the sine of a sum. Indeed,
Asin (vgx+a) =Asin (ygx) cos a+Acos (V qx) sin m
so that
C1 = A cos a,
C2 = A sin a
Now try to obtain from general solution (7) a particular
solution satisfying initial conditions (6).
READER. We shall obtain it by expressing the constants
A and a via fo and fo. Equality (7) yields an expression for
the first derivative of f (x):
f (x) = A vq cos (yq x -{- a)
In this case initial conditions (6) take the form
sin (v g xo
I
a) =
cos(Vgxo+a)=
fo
A
(8)
fo
A vq
System (8) must be solved for the unknown constants A
and a. Squaring both equations of the system and summing
READER. Would it be correct to say that any differential
equation describes a process? I assume that / is a function
of time.
AUTHOR. Quite true. This is a point worthy of maximum
attention. In a sense, it reflects the principal essence of
Differential Equations
167
differential equations. Note: a differential equation relates
the values assumed by a function and some of its derivatives
at an arbitrary moment of time (at an arbitrary point in
space), so that a solution of the equation gives us a picture
of the process evolving in time (in space). In other words, a
differential equation embodies a local relation (a relation at
a point x, at a moment t) between f, f, f", . . ., thus yielding
a certain picture as a whole, a certain process, an evolution.
This is the principal idea behind the differential equations.
READER. And what is the role played by initial conditions?
AUTHOR. The role of initial (and boundary) conditions
is obvious. A differentia 1 equation per se can only describe
the character of evolution, of a given process. But a specific
pattern of evolution in a process is determined by concrete
initial conditions (for example, the coordinates and velocity
of a body at the initial moment of time).
READER. Can the character of the process "hidden" in a
differential equation be deduced simply from the form of
this equation?
AUTHOR. An experienced mathematician is normally
able to do it. One glance at equation (2a) is sufficient to
conclude that the process is an exponential growth (decay).
Equation (3a) is a clear message that the process involves
oscillations (to be precise, harmonic oscillations). Assume,
for example, that differential equation has the following
form
f"
(t) - pf' (t) + of (t) = 0
(p < 0, q > 0)
(18)
(compare it to equations (2a) and (3a)). We shall not analyze
this equation in detail. We only note that what it "hides"
is not a harmonic oscillatory process but a process of damped
oscillations. It can be shown (although we shall not do it)
that in this process the amplitude of oscillations will steadily
diminish with time by the exponential law exp (pt).
READER. Does it mean that equation (18) describes a
process which combines an oscillatory process and a process
of exponential decay?
AUTHOR. Precisely. It describes an oscillatory process,
but the amplitude of these oscillations decays with time.
Dialogue Fourteen
168
DIALOGUE FOURTEEN
MORE ON DIFFERENTIAL
EQUATIONS
AUTHOR. All the preceding dialogues (with an exception
of Dialogue Eight) left out, or very nearly so, any possible
physical content of the mathematical concepts and symbols
we were discussing. I wish to use this dialogue, which concludes the book, to "build a bridge" between higher mathematics and physics, with differential equations as a "building material". We shall analyze differential equations of
exponential decay and those of harmonic oscillations,
filling them with a specific physical content.
READER. In other words, you suggest discussing specific
physical processes?
AUTHOR. Yes, I do. I emphasize that differential equations play an outstanding role in physics. First, any more or
less real physical process cannot, as a rule, be described
without resorting to differential equations. Second, a typical
situation is that in which different physical processes are
described by one and the same differential equation. It is said
then that the physical processes are similar. Similar physical
processes lead to identical mathematical problems. Once
we know a solution of a specific differential equation, we
actually have the result for all similar physical processes
described by this particular differential equation.
Let us turn to the following specific problem in physics.
Imagine an ensemble of decaying radioactive atomic nuclei.
Denote by N (t) a function describing the number of atomic
nuclei per unit volume which have not decayed by the
moment of time t. We know that at the moment t = to
the number of nondecayed nuclei (per unit volume) is No,
and that the rate of decrease in the number of nondecayed
nuclei at the moment t is proportional to the number of
nondecayed nuclei at the given moment:
- N' (t) _ ti N (t)
(1)
More on Dtfterential Equations
169
is a proportionality factor; evidently, ti has the
T
dimension
of time; its physical meaning will be clarified
Here
later.
We are to find the function N (t).
This is our specific physical problem. Let us look at it
from the mathematical viewpoint.
N
r,
No
0
to
t
Fig. 56
READER. Equation (1) is a differential equation of type
(2a) from the preceding dialogue, in which p
The initial condition in this case is N (to) = No. By using
result (4) of the preceding dialogue, we immediately obtain
N (t) =No exp (- (t- to))
(2)
AUTHOR. Correct. The formula that you have written,
i.e. (2), describes the law of radioactive decay; we find that
this decay is exponential. The number of nondecayed nuclei
decreases with time exponentially (Fig. 56).
The constant i is, therefore, such a time interval during
which the number of nondecayed nuclei diminishes by a
factor of e (i.e. approximately by a factor of 2.7); indeed, in
this case In Nit) = In e = 1.
Let us turn now to a different physical problem. Let a
light wave with intensity 10 be incident perpendicularly at
a boundary (the so-called interface) of a medium; the wave
11
0
Fig. 57
propagates through the medium with gradually attenuating
intensity. We choose the x-axis as the wave propagation
direction and place the origin (point x = 0) on the interface
(Fig. 57). We want to find I (x), that is, the light intensity
as a function of the depth of penetration into the medium
(in other words, on the path traversed within this medium).
We also know that the rate of attenuation at a given point x
(i.e. the quantity -I' (x)) is proportional to the intensity
at this point:
- I' (x) = iI (x)
(3)
Y
Here rj is the proportionality factor whose dimension is,
obviously, that of inverse length; its physical meaning will
be clear somewhat later.
This, therefore, is the formulation of the physical problem.
More on Differential Equations
171
READER. It is readily apparent that, as in the preceding
case, we deal here with a differential equation of exponential
decay. The initial condition is 1 (0) = Io. By using result
(4) of the preceding dialogue, we obtain
I (x) = Io exp (-'ix)
(4)
AUTHOR. Formula (4) describes Bouguer's law, well
known in optics: as light penetrates the matter, its intensity
decays exponentially (see Fig. 57). We readily see that the
constant rj is a quantity inverse to the length along which
the light intensity diminishes by a factor of e. The constant
11 is called the linear absorption coefficient.
Note that results (2) and (4) describe two different physical
problems from different fields of physics. We describe here
two different physical processes. Nevertheless, the mathematical nature of these physical processes is the same: both
are described by the same differential equation.
Let us consider a different physical problem. Assume that
a ball with mass m, attached to fixed walls by elastic springs,
vibrates along the x-axis (Fig. 58). The origin x = 0 is
chosen in the position in which the ball is at equilibrium,
that is, half-way between the walls. The motion of the ball
is governed by Newton's second law:
ma = F
(5)
where a is acceleration, and F is the restoring force. We
assume that
F = -kx
(6)
where k is the elasticity factor characterizing the elasticity
of the spring.
We shall consider the displacement of the ball from the
equilibrium position (i.e. the quantity x) as a function of
time, x (t). This is the function we want to find.
12*
172
Dialogue Fourteen
We remind the reader that accelerati u is the second
derivative of a function which describes path as a function
of time: a = x' (t). Consequently, we can rewrite (5),
taking into account (6), in the form
mx'(t)+kx(t) = 0
or
x (t)+mx(t)=0
(7)
READER. This is a differential equation of type (3a) of
the preceding dialogue provided that q = kM
AUTHOR. This means that the general solution must be
of the form
x(t)=Asin(1/ tÂąa)
(8)
We thus find that the ball in the problem vibrates harmonically around its equilibrium position x = 0. The parameter A
is, obviously, the amplitude of vibrations. The parameter a
is called the initial phase of vibrations. Recalling relation
(17) of the previous dialogue, we conclude that the period
of vibrations is
By using (10), we rewrite general solution (8) in the form
(11)
x (t) = A sin (wt + a)
READER. And what about the initial conditions in this
case?
AUTHOR. Assume that the ball is at rest at t < 0. By
setting specific initial conditions at t = 0, we choose a
method by which vibrations are initiated at the moment
More on Differential Equations
173
t = 0. For example, let the initial conditions be given by
relations (11) of the previous dialogue:
x' (0) = vo
(12)
x (0) = 0,
This means that at the moment t = 0 the ball which is at
the equilibrium position (x'=0) starts moving at a velocity vo. According to relation (13) of the previous dialogue,
we obtain the following particular solution:'"
2(t) =
sin (wt)
(13)
Now try to discern the physical meaning of the initial
conditions of type (14) of the previous dialogue. 7
READER. These conditions have the form:
(14)
4(0) = xo, '- -x' (0) =10
This means that at the initial moment t = 0 the ball was
displaced from the equilibrium position by x = x0 and let
go. The corresponding particular solution, following from
relation (16) of the previous dialogue, takes the form
X '(t) = xo cos'(o t)
(15)
AUTHOR. In the first case we thus initiate vibrations by
imparting the initial velocity vo to the ball at the equilibrium position (in this case the amplitude A of vibrations is
V , and the initial phase a can be set equal to zero, in
accordance with (13)). In the second case the vibrations are
initiated by displacing the ball from the equilibrium position by x0 and then letting it go (in this case A = x0, and
the initial phase a can be set equal to 2 , in accordance
with (15)).
READER. Could we consider a case in which at t = 0
the ball is displaced from the equilibrium position by xl
and simultaneously given an initial velocity v1?
AUTHOR. Of course, this is one of the possible situations.
Figure 59 shows four vibration modes (four particular solutions) corresponding to four different initial conditions
(four different methods of starting the vibrations of the
Dialogue Fourteen
174
ball):
a=0.
(1) x (0) = 0,
x' (0) = vo;
in this case A =
(2) x (0) = xo,
x' (0) = 0;
in this case A=xo, a= -2n
-,
.
(3) x (0) = x1, x' (0) = v1 (the initial velocity imparted
to the ball has the same direction as the initial displacement); in this case A = A1, a = al (see the figure).
Fig. 59
(4) x (0) = x1, x' (0) = -v1 (the initial velocity impart-
ed to the ball has the direction opposite to that of the
initial displacement); in this case A = A1, a = n - a,
(see the figure).
As follows from relation (9) of the preceding dialogue,
A,
x;
(°-' )
2
(16)
and according to (10),
a, _- arctan (
VI
(17)
READER. I notice that by fixing specific initial conditions (in other words, by initiating the vibrations of the
ball by a specific method), we predetermine the amplitude
and initial phase of the vibrations.
More on Differential Equations
175
AUTHOR. Precisely. This is clearly shown in Fig. 59.
By the way, the same figure shows that the period of vibrations (their frequency) remains constant regardless of the
initial conditions.
To summarize, we note that a harmonic oscillation is
characterized by three parameters (see (11)): the amplitude
A, initial .phase a, and frequency
w. The first two parameters are
determined by the choice of initial conditions, and the last parameter is independent of them.
The above-described
process
of vibrations is one of the mechanical processes. Let us turnnow to a process of an essen-
tially
Fig. 60
physical naanalyze the
motion of electric charges in a circuit consisting of a capacitor with capacitance C and a coil with inductance L (Fig. 60).
different
ture. We shall
Let the capacitor plates have a charge Q (t) at a moment t;
correspondingly, the potential difference between the capacitor plates will be Q() . If the current in the circuit at the
moment t is i (t), then the potential difference generated in
the coil is -Li' (t). We know that it must be balanced out
by the potential difference across the capacitor plates:
- Li' (t) = QC()
(18)
Let us differentiate relation (18). This gives
- Li" (t) =
Q, (t,
(19)
Now we shall take into account that
Q' (t) = i (t)
(current intensity, or simply current, is the rate of change
of charge). As a result, equation (19) can be rewritten in
the form:
Dialogue Fourteen
176
or
i' (t) -f
LC
i(t)=O
(20)
The resultant differential equation is quite familiar, isn't it?
READER. This is a differential equation of type (3a)
of the preceding dialogue provided that q = - . We conclude, therefore, that the process in the circuit is harmonic.
AUTHOR. Note, however, that the process is not that of
mechanical vibrations of a ball attached to springs but the
process of electromagnetic oscillations in an electric circuit.
READER. As q = LC , and using relation (17) of the
previous dialogue, we obtain a relation for the period of
electromagnetic oscillations in the circuit:
(21)
T == 2n V -LC
The general solution of equation (20) is then
1 (t) = A sin ( PLC
z. w
t + a);
(22)
AUTHOR. Absolutely correct. The`two physical processes,
namely, the mechanical vibrations of a ball attached to
springs and the electromagnetic oscillations in a circuit, are
mathematically similar. They are described by the same
differential equation. Otherwise you couldn't write, nearly
automatically as you did, the period of oscillations (formula (21)) and the general solution (formula (22)).
In our dialogues we have discussed only two (and rather
simple) types of differential equations: those of exponential
growth (decay) and of harmonic oscillations. And we have
illustrated them with a number of physical processes of
very different kind.
More on Differential Equations
177
READER. I guess that the list of different differential
equations, and certainly the list of physical processes described by these equations, could be substantially enlarged.
AUTHOR. No doubt. This concludes our discussion of
differential equations. I want to note in conclusion that
differential equations are widely applied not only in physics
but in chemistry, biology, cybernetics, sociology, and
other fields of science as well.
PROBLEMS
1. Find a formula for the nth term from the first several
terms of the sequence:
10. Analyze the continuity and rdifferentiability of the
function f (x) = aresin (sin x) within the limits of the
natural domain of the function.
Answer. The natural domain of the function f (x) is
]-oo, oo[; the function is continuous everywhere; it is
differentiable at all points with the exception of points
x= t 2 Âą t, f 2 n, ....
It, Prove that the function f (x)
at point x = 0.
12. Prove that 3x6 - 5x3 - 30x < 40 if
has no derivative
Ix T < 2.
Hint. Find first that the maximum value of the polynomial f (x) = 3x6 - 5x3 - 30x over the interval
[-2, 2] is below 40. To do this, find the values of f (x)
at the end points of the interval [-2, 2] and at the
points at which the derivative of f (x) is zero (if these
points belong to the indicated interval).
13. Find the maximum and minimum values of the function
Note. The equation of the tangent to the graph of the
function f (x) at x = xo is: y = f (x0) + f' (x0) (x - xo),
where f' (x0) is the value of the derivative of the function at x0.
16. Find the derivatives of the following functions:
(a) f (x) _
(c)
solution for which the tangent to the graph at point xo
intersects the ordinate axis at point yl.
Answer. f (x) = 1 y so exp (x - xo).
Hint. Make use of the equation of tangent (see Note
to problem 15).
Problems
193
27. Consider the graphs of different particular solutions of
the equation f (x) = f (x). Verify that at the same
point xo the tangents to all these graphs intersect the
abscissa axis at a common point x = xo - 1.
28. Find the nth derivative of the following functions:
(a) f (x) = sin x; (b) f (x) - cos x. | 677.169 | 1 |
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Math journal prompts Algebra
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Students can exit this course and be placed in the next higher level math course if they score 90% in the Initial Assessment that will be administered during the first week of the course. Learning Outcomes:
By the end of this course, students should have achieved the following learning outcomes:
1. Students will demonstrate improvement of their pre-algebra skills by increasing their pretest scores by at least 25% during the post-test. 2. Students will demonstrate mastery level on each of the main concepts of the course (see below the Course Curriculum) by earning at least 70% on the assessments (i.e. quizzes) scheduled for each main concept. 3. Students will demonstrate college readiness to handle the rigor of the next mathematics course by attaining a score of at least 70% in the post-test.
4. Students will demonstrate higher confidence and motivation doing mathematics by
scoring on average of at least a point higher in the Likert scale administered at the start and end of the course.
Instructional Methods
Taking into consideration our diverse population of students and to ensure they are involved as much as possible in the learning process, a variety of teaching-learning methods will be used including:
Self-paced learning: Students complete the course content at their own pace based on their prior knowledge of the math concepts and skills covered in the course, and with the guidance of a suggested timeline. Students have the opportunity to complete course requirements before the end of the term or complete two courses in one academic term. Computer assisted instruction: A learning and assessment web-based system (i.e. ALEKS) is used to help students grasp and master the course content. ALEKS will assess students' prior knowledge of the course content and create a visual representation (i.e. pie chart) of what they know and need to learn. Based on this assessment, students work on the topics they are ready to learn. Students receive immediate feedback for their performance and are continuously assessed to guarantee mastery and retention of the course content. Individualized and small group instruction: The use of ALEKS in the classroom will allow the instructor to provide individualized and small group instruction based on the needs of the students. In addition, math tutors will be available during class time to assist students with their difficulties with the content. Online Learning Resources: ALEKS provides detailed explanations and demonstrations of the concepts and skills covered in the course. It also provides supplementary resources such as videos, animations, Power Point presentations, math dictionary, and the course textbook (i.e. e-book). In addition, students have access in Blackboard of additional instructor-made resources (i.e. handouts, Power Points, screencasts, etc.) and math links to other Internet sites that provide tutorials, virtual manipulatives, and multimedia materials. Video-lectures: The instructor will post short video lectures (5-10 minutes duration) about the topics that are covered in a particular slice of the pie in ALEKS. The video lectures will be posted in Blackboard. Graphic Organizers: Diagrammatic illustrations will be used in the classroom to help students organize and remember content area information. The graphic organizers will be posted in Blackboard. Math 2.0 Assignments: Students will use web2.0 tools to communicate mathematical ideas (i.e. concepts and skills) covered in this course. Students will have the opportunity to self-assess and reinforce what they have learned in class by participating and collaborating with peers using web2.0 tools in online discussions. In addition, students can take the opportunity to communicate their concerns or questions regarding course content. The instructor will use students' responses to provide the necessary assistance or enrichment activities that students need to succeed in the course. The discussions will take place in Blackboard, course wiki, or instructor's blog.
Supplies Students must always have the following supplies during class: 1. Notebook 2. Textbook 3. Pencils, erasers, sharpener, etc. The instructor and staff of the academic enhancement center will not supply papers and pencils for students who irresponsibly do not bring the basic materials to learn. A different way of learning math with ALEKS ALEKS is a web-based assessment and learning math system that uses artificial intelligent programming to provide an individualized learning experience for every student. The instructional model of this course will mainly consist on students actively learning at their own pace with the assistance of ALEKS, the online resources available in Blackboard course shell and the Internet. Students must take the initiative and responsibility to use all the available resources to actively learn the course content. Instructional time will be spent less on class lectures and more on learning by doing and reflecting. The instructor and tutors will be available during instructional time to answer their questions and provide the necessary assistance to help them succeed.
Course Curriculum
In this course, students will cover the following main concepts:
The main concepts are represented visually with a pie chart in ALEKS. Students meet the course objectives of each main concept when they filled the slice that represents the concept. In total, students must master 221 objectives (or topics) to pass this course. To see a detailed distribution of the 221 topics by main concepts, please visit the site and select the course Basic Math. A dark color in the pie represents how much students have mastered of a particular main topic and a light color represents how much of the main topic students still need to master. The course textbook is an important source of reference to help students master the course concepts. Each main concept (slice of the pie) consists of the following chapters in the course textbook: Course Concept Whole Numbers Fractions Decimals, Proportions, & Percents Measurement & Data Analysis Geometry Algebra Pie Slice Color Green Light Blue Dark Blue Orange Yellow Red Textbook Chapter(s) 1 2 3, 4, and 5 6 and 7 8 9 and 10
Grading Policy:
Letter Grades 70-100 0-69 Pass (P) Fail (F)
A student must score an average of at least 70% to pass this course.
Course Grading Criteria:
• Your grade for this course will be based on the following components: Grading Categories
ALEKS (Fill Pie Chart) Final Exam Quizzes Math2.0 Assignments TOTAL 25% 20% 35% 20% 100%
Important Notes • • • Students take a quiz only when they have filled a pie's slice. A quiz must be taken immediately after completing a slice of the pie. Students take quizzes in numerical order. See course outline below. Students must complete the whole pie; take all quizzes; and take Final Exam (Assessment) to have a chance passing the course. Quizzes must be taken on-site in presence of the instructor.
•
•
• •
•
Two scheduled assessments will take place during the term (Initial and Final Exam). If an assessment pops up in the system, contact the instructor immediately. Pie must be completed by the week prior finals. Incomplete grade is granted if students completed his/her pie and s/he took 3/4 of the quizzes by end of course.
Course Outline/Schedule:
This schedule is the suggested timeline (i.e. the slowest pace) that students should follow to complete successfully the course objectives. However, students can complete the course objectives at a faster pace. Students who fall behind the schedule are jeopardizing their chances to pass this course. ALEKS periodically prompts assessments that students must take every time.
COURSE POLICIES 1 Class Participation
Class participation and student interaction are important components of the learning experience. In order to achieve the maximum benefits from the course, all students are expected to read all assignments and related materials prior to each class and be prepared to actively be engaged in class discussions and doing math problems during class.
2 Practice Problems and Quizzes
All practice problems and quizzes are done in ALEKS. Students can do the practice problems at any time and any place since ALEKS is a web-based program. Quizzes must be taken only in the classroom. Students must earn at least 70% in the quiz.
3 Attendance
Educational research has proven there is a positive connection between attendance and academic success, so students are strongly urged to attend classes regularly. In the classroom, students can get the prompt assistance they may need to understand the course content. Onsite attendance is mandatory on Wednesdays. Online attendance is mandatory on Mondays and Fridays. Students must average 6 hours/week in ALEKS. 1 absence = 2 percent points deduction from overall average.
4 Use of Computers
"Computers and network systems offer powerful tools for communications among members of the St. Thomas community and of communities outside St. Thomas. When used appropriately, these tools can enhance dialogue, education, and communications. Unlawful or inappropriate use of these tools, however, can infringe on the rights of others. Activities that are expressively forbidden on St. Thomas' computers include but are not limited to the viewing, downloading or use of inappropriate materials, vandalism, virus propagation and installation of unauthorized materials. In addition, you are expected to act as a professional and use the equipment only when directed or appropriate to classroom activities. A lack of compliance with any of these directives could result in disciplinary action and dismissed of class or course.
5 Expected Classroom Behavior
Students have a responsibility to maintain both the academic and professional integrity of the school and to meet the highest standards of academic and professional conduct. Students are expected to do their own work on examinations, class preparation and assignments and to conduct themselves professionally when interacting with fellow students, faculty and staff. Academic and/or professional misconduct is subject to disciplinary action including course failure and/or probation of dismissal. No food allowed in the classroom. Dress appropriately to attend class. For additional clarification, please see Student Code of Conduct as stated in the Student Handbook.
6 Cell Phones and Calculators
Cell phones must be turned off or in vibrating mode. If a student must answer a phone call then the students must leave the classroom without disrupting the flow of the class. Students who spend a considerable amount of time attending a phone call outside the classroom will be considered absent. Calculators permitted during class only in certain circumstances the instructor considers appropriate.
7 Tutoring at the Academic Enhancement Center
Students who attend at least 10 hours of tutoring sessions will earn 5 percent points for their overall average. Please see front desk of the Academic Enhancement Center to find out the days and time of the tutoring sessions.
8 Incomplete Grade Students will be granted an incomplete grade if s/he completed the whole pie in ALEKS by week 15 and took ¾ of the scheduled quizzes. An incomplete grade grants the student another week to complete pending assignments. Rubric for Quizzes The following rubric (grading criteria) will be used to score quiz items. Points
1-point
Expectation
Correct answer. Work or process to support answer is logical and neatly organized. It reveals student understanding of concepts and skills. Incorrect answer. Work or process to support answer is logical and neatly organized. It reveals student understanding of concepts and skills. Minor computational or careless mistakes. Correct or Incorrect answer. Work or process to support answer is not logical or shown. It reveals student's misunderstanding of concepts and skills. Major computational mistakes.
1 1 2 3 , , , or - point 4 2 3 4
0-point
Math2.0 Assignments
Difficult Problems: The purpose of this writing assignment is for students to communicate the difficulties they are experiencing with a particular math problem. Students post the problem and a description of the difficulties they are experiencing in the discussion board of Blackboard (Bb). In addition, students must help two classmates by posting clear and concise explanations on how their classmates' problems are simplified or solved. Graphic Organizers: Students will work particular problems assigned by the instructor using visual aids. The graphic organizers will be located in Blackboard under the Assignment section. Students must print out the graphic organizer, work the problem, scan the assignment and submit to the instructor by using the drop box in Blackboard. Learning by Teaching: Students will create online presentations to teach their classmates about a concept/skill they learned in the course. These presentations will be created using web2.0 tools, productivity tools and/or open source software. The presentations will be posted in the course's wiki. Math Awareness: The purpose of this assignment is to help students answer the most frequent asked question in mathematics courses: "Why I need to learn this?" Students will read blog articles about mathematics awareness and post a comment in the blog.
Become an Active Learner
An active learner takes control and ownership of the learning process to meet the course's goals and expectations. Active learners decide why, what and how of their learning. They do not wait for learning to happen; instead, they make it happen. The instructional model of this course requires students to become active learners to meet successfully the course objectives. The following traits are typical of active learners:
1. 2. 3. 4. 5. 6. 7. 8. 9.
Identify personal goals and the steps necessary to achieve the goals. Use resources. Identify the people and tools available to aid in goal pursuit. Learn how to solve almost any problem they ever have to face. Look at situations objectively. Ask the right questions. Use time well. They organize and set priorities. Apply good reading, studying, and questioning skills to course materials. Apply good listening skills in the classroom. Find patterns and take effective notes to organize materials for studying.
10. Assess progress along the way and revise their plans.
Source:
English Second Language Learners
For students who do not speak English as their first language, the following suggestions may be helpful to succeed in this course:
1. Bring a dictionary that translates from the student's native language to English and vice
2. 3. 4. versa. If a student does not have a dictionary, the following website provides word and text translation: Find a classmate or group of students who speak English fluently to study for the class and to gain proficiency with the English language. If there is a classmate that speaks the same native language, students can ask for clarification or assistance using their native language as long it does not disrupt the classroom learning experience. The instructor of this course is bilingual (English-Spanish) and welcome students to speak Spanish during office hours or before-after class. In addition, there are many languages that have words which are pronounced and written similarly. Therefore, the instructor encourages students to sometimes use words in their native language to communicate ideas, concerns, or questions. If students learned different ways or methods for simplifying or solving math problems in their countries, the instructor encourages these students to share their methods with him.
5.
In addition, ALEKS offers the option of presenting course content in Spanish for the Spanish speaker students.
Students with Disabilities
Please note that students requiring accommodations as a result of a disability must contact Maritza Rivera (e-mail: mrivera@stu.edu and phone number: 305-628-6563) at the Academic Enhancement Center. Note for Changes: The instructor reserves the right to change this syllabus at any time during the term in order to better meet the needs of this particular class group. | 677.169 | 1 |
Program Components Introduction This guide describes the program components available for Pearson High School Math Common Core Edition. The new Common Core edition provides complete, comprehensive coverage
Pre-Algebra Program Organization Prentice Hall Mathematics supports student comprehension of the mathematics by providing well organized sequence of the content, structure of the daily lesson, systematic
Blended Instructional Design Introduction Before We Get Started This guide explains how to teach lesson with Prentice Hall Algebra 1, Geometry, and Algebra 2 the new Prentice Hall High School Math series.
Alabama Course of Study: Mathematics (Grades 9-12) NUMBER AND OPERATIONS 1. Simplify numerical expressions using properties of real numbers and order of operations, including those involving square roots,
7 Multiplying and Dividing Fractions Chapter Pacing Guide Please note that this pacing guide is based upon completing the entire text in 165 classes, approximately 50 minutes each. More time can be allotted
10 Geometry: Exploring Area Chapter Pacing Guide Please note that this pacing guide is based upon completing the entire text in 165 classes, approximately 50 minutes each. More time can be allotted for
13 Using Probability Chapter Pacing Guide Please note that this pacing guide is based upon completing the entire text in 165 classes, approximately 50 minutes each. More time can be allotted for this chapter
4 Multiplying and Dividing Decimals Chapter Pacing Guide Please note that this pacing guide is based upon completing the entire text in 165 classes, approximately 50 minutes each. More time can be allotted
11 Geometry: Using Area and Volume Chapter Pacing Guide Please note that this pacing guide is based upon completing the entire text in 165 classes, approximately 50 minutes each. More time can be allotted
7 Multiplying and Dividing Fractions Chapter Pacing Guide Please note that this pacing guide is based upon completing the entire text in 165 classes, approximately 50 minutes each. More time can be allotted
6 Adding and Subtracting Fractions Chapter Pacing Guide Please note that this pacing guide is based upon completing the entire text in 165 classes, approximately 50 minutes each. More time can be allotted
9 Algebra: Exploring Real Numbers Chapter Pacing Guide Please note that this pacing guide is based upon completing the entire text in 165 classes, approximately 50 minutes each. More time can be allotted
Assessment Support Introduction This guide introduces the diagnostic, formative, and summative assessment resources in the Pearson Algebra 1, Geometry, and Algebra 2 Common Core Edition. It explores both1 Texas High School Math Program Overview Introduction In this tutorial, you ll explore the Pearson Texas High School Math program: Algebra I, Geometry, and Algebra II. You ll review the program components | 677.169 | 1 |
A quick question about textbook quality
Why is that i always see and hear people complaining about textbooks? It is true that there are many bad textbooks but why do they buy them? Can't they read reviews or ask around like i do?
Also there are many good textbooks and as a self learner I think my opinion counts for something. A first course in general relativity, spacetime and geometry, introduction to topology, topology from a differentiable viewpoint etc.. are all very high quality.
Call me a cynic, but I'd have to say to most people whoa re constantly complaining must be lazy insoem sense(not to say that i'm a hard worker). What vexes me even more is that these people don't seem to care to go to their peers, or ask their profs for clarification. I'd KILL for that luxury.
Now i understand that some courses have bad textbooks but there are amny cheap reliable textbooks that you can find by asking professeurs... In many cases the good texts are cheaper than the expensive ones.
Anyway, I'm sorry for bothering you guys, but I just had to rant a little. | 677.169 | 1 |
ISBN-13: 9780471698043
Edition: 10 clear calculus content to help readers master the concepts and understand its relevance to the real world. It offers a balance of theory and applications to elevate their mathematical insights | 677.169 | 1 |
Course Overview: Reasoning based on probability and statistics gives modern society the ability to cope with uncertainty. It deals with the design of how data is collected, the analysis of the data, and the drawing of conclusions from the data.
Statistics has astonishing power to improve decision-making accuracy and test new ideas. It's a key analytical tool used in education, the social sciences, and business administration and is often a required college subject for majors in those areas. Statistics is frequently used for data analysis in the sciences and forms the mathematical basis for quality control in manufacturing.
AP Statistics is a college level class for students who have been highly successful in Algebra 2. It covers the topics needed for the College Board AP Statistics exam. Students passing this test may receive college credit.
In college, statistics is generally considered an intellectually challenging course. Correlations between the combined PSAT math and verbal scores with the passing rate in AP Statistics bears this out. However, AP Statistics has an advantage over the equivalent college course in that it takes an entire year to present what would be considered a semester of material in college.
Who Should Take This Class: Students with an interest in careers related to medicine, bioinformatics, genetics, business, mathematics, architecture, engineering, any of the sciences, or the social sciences including psychology, sociology, political science, and education. In college, statistics is required for most majors with the possible exception of the fine arts.
Credit: Although AP Statistics is considered an elective, it does count as 1 credit towards the graduation requirements of 4 math credits.
Summer Assignment: Some articles may be required summer reading, or often there is no summer assignment.
Prerequisites: Accelerated Geometry or above, or on-level Algebra 2 or above. A combined PSAT verbal and math score of 111 or higher is a good indication that you have the background needed to succeed in the course. Surprisingly, a score of 3 or higher on the AP World History exam is also a strong indication of future success in AP Statistics.
Calculators and Computers: For the purpose of understanding, students will learn all the traditional techniques of using statistics equations and tables for problem solving. However, in today's world, most statistical analysis is done with computer applications and so resources will be provided as follows:
TI-84 / TI-nSpireCalculator: virtually every traditional calculation will also be done on a TI-84.
Fathom Software: will be used by the teacher for demonstrating the basis behind several of the statistics techniques such as regression and correlation.
SAS Software: SAS is the market-leader in analytical software for businesses and universities. Students will utilize SAS throughout the course to reinforce statistical concepts when available. | 677.169 | 1 |
An Interesting Juxtaposition
Today's time in the classroom showed an interesting juxtaposition. The "low level'' class that is focusing on big, deep ideas is on track, but the more advanced class working on "straightforward material" spent a lot of time not going very far.
Math in Decision Making Success?
My liberal arts course students have made good progress on understanding the idea of a bijection. We haven't completely formalized it, but I think most of the class basically "gets it." In fact, today's four exercises proved to be not enough. They dispatched them quickly and we were done with ten minutes to spare! I'm afraid that means I might need to come up with something meatier to add to Wednesday's activities. This shouldn't be a problem—coming up with hard problems is easier than coming up with approachable ones.
Linear Algebra: The End of the Activity
It took a surprisingly long time to get through the "solve and sort" phase of the activity I started last time. But I am happy with the results. I asked the class to distill some lessons from the experience, and I got this list:
1. The size of the solution set depends on the number of pivots in the matrix.
2. The origin is a solution exactly when the system is homogeneous.
3. If you are working by hand, it can be helpful to look for clever tricks instead of just blindly following the algorithm.
4. If you had to do a lot of these, or even just one of any appreciable size, you want to use a computer. It is not difficult, just tedious.
So, it took a lot longer than I planned, but the main points came across. I used the end of our discussion to introduce the term "rank" and we talked a little about writing the solution set in the standard parametrized form using vector operations.
Dynamics
If you have bothered to look at these notes, and especially if you are doing something as odd as working through them, know that problem 29 was a total failure.
That is, I am sure the problem is fine, but it is totally misplaced. The students had no idea what to say about it. I'll have to find the right time to bring that point back up.
Also, one of the more advanced students wrote up a nice presentation of what it means for a sequence to converge along with his proof for task 15. I hope it helps out some for those students who haven't taken a real analysis course, yet. | 677.169 | 1 |
CHAPTER 1: BASIC ALGEBRA TECHNIQUES
1.1 SETS
Set Notation:
Set notation is used to describe membership in a class, collection or group.
A set is any collection of objects with a rule to determine if any arbitrary object belongs to the
collection or not. N
Translations of graphs
Shifting a graph vertically up or down.
Example 1:
The graph of the equation y = x2 is a parabola opening upwards with the y axis as an axis of symmetry and having its lowest point at the
origin. Given a table of values for the equa
Sets, Real Numbers and Intervals
Sets and set operations
A set is any well-defined collection of (mathematical) objects.
A set can be specified by listing its elements, or members, within curly brackets cfw_ . . . . .
A = cfw_ 1, 2, 3, 4 and B = cfw_ 4,
Reflections of graphs in the x and y axes
Introductory example.
Suppose that we have already constructed a table of values for the graph of the equation y = x + 4 and that we wish to draw the graph of
y = x + 4.
For each point ( x1, y1 ) on the graph of y
Rationalizing denominators and numerators
Introductory example.
Suppose that you are given the problem of obtaining an approximate decimal value for the expression
1
given that
2 1
rounded to 6 figures. Also suppose that you are not allowed to use a calcu
Rational expressions
Preliminaries.
Recall that a division of real numbers
a
1
( 1 )
=
a
a
b
, where b 0, can be performed by multiplying the real number a by the multiplicative inverse
of a. Thus
a
=a.
b
1
b
.
Consider a rational expression of the form
a
Rational exponents
Summary of formulas involving general integer exponents:
Before considering rational exponents it is appropriate to recall definitions and properties connected with integer exponents.
The following formulas define integer powers of a no
Solving quadratic inequalities
A sign chart of an expression is a number line that shows where the expression is positive, negative or 0.
For example, the following picture shows a sign chart for 2 x 3.
3
We can construct the sign chart by first observing
Quadratic equations and quadratic functions
Introduction.
A quadratic function is a function f of the form
f( x ) = a x2 + b x + c - (i),
where a 0.
A quadratic equation is an equation of the form
a x2 + b x + c = 0 - (ii),
where a 0.
The real number solu
CHAPTER 1: BASIC ALGEBRA TECHNIQUES
1.2 FUNDAMENTALS OF ARITHMETIC
Fractions:
A fraction a is made up of a numerator (the expression in the top) and a denominator (the
b
expression in the bottom). Note that: if numerator denominator, then the quantity has
CHAPTER 1: BASIC ALGEBRA TECHNIQUES
1.3 ALGEBRAIC EXPRESSIONS
Terminology:
Algebra uses symbols [ usually letters ] to represent quantities which are unknown or which can vary.
A term is made up of a coefficient a real number and/or variable(s), the unkno
CHAPTER 1: BASIC ALGEBRA TECHNIQUES
1.4 SOLVING EQUATIONS AND INEQUALITIES
Solving Polynomial Equations:
An equation is a statement of equality between two expressions, and this may or may not be true for
some values of the variable(s). Unless specified o
CHAPTER 5: INTRODUCTORY CALCULUS
5.3 Applications of Derivatives
Curve Sketching:
The value of the derivative of a function represents the slope of the tangent line to the curve of the
function at a particular point. Thus, if f x 0 on the interval a, b ,
CHAPTER 5: INTRODUCTORY CALCULUS
5.2 Differentiation and Derivatives
Determining an instantaneous rate of change is a common application. The expression discussed in
f ah f a
5.1, h , is often referred to as the Newton Quotient or difference quotient. If
CHAPTER 5: INTRODUCTORY CALCULUS
5.1 Limits
The Definition of a Limit at a Number:
One of the most fundamental concepts and tools in calculus is the limit, which is defined as follows.
Let f be a function defined on an open interval containing the number
CHAPTER 4: EXPONENTIAL AND LOGARITHMIC FUNCTIONS
4.3 Applications of Exponential and Logarithm Functions
Typical applications of these functions are to problems involving the rate of change of a quantity,
where that rate of change is proportional to the a
CHAPTER 4: EXPONENTIAL AND LOGARITHM FUNCTIONS
4.2 Logarithm Functions
Because an exponential function is always monotone, it has an inverse which is also a function. We
define the inverse function of fx exp a x to be the logarithm function base a, and de
CHAPTER 4: EXPONENTIAL AND LOGARITHMIC FUNCTIONS
4.1 Exponential Functions
We defined the expression a x in 1.2. The value a is called the base and x is referred to as the
exponent. Evaluating such expressions is called the exponentiation of x, base a. We
CHAPTER 3: ANALYTIC GEOMETRY
3.3 Trigonometric Functions
In order to describe trigonometric functions and derive some of their properties, we introduce the
unit circle. In the xy-plane, the unit circle is the set of all points one unit away from the origi
CHAPTER 3: ANALYTIC GEOMETRY
3.2 CONICS
There are special figures in the xy-plane called conic sections. These can be described as the
intersection of a right circular cone and a plane, and are called circles, ellipses, parabolas and
hyperbolas.
A right c
CHAPTER 3: ANALYTIC GEOMETRY
3.1 LINES
Two Points in a Plane:
We can compare two real numbers, a and b, by using the law of trichotomy: either a b, a b, or
a b. We can also compare the relative size of each value by comparing |a | to |b |, where |x | give
CHAPTER 2: FUNCTIONS
Some final notes on inverses:
S The graphs of f and f 1 are reflections of each other through the line y x.
S A relation/function is symmetric about the line y x if and only if the formulae for f and f 1
are equivalent.
S The formal d
CHAPTER 2: FUNCTIONS
2.1 THE BASICS
Cartesian Coordinates in the Plane:
In order to gain a better understanding of numbers and equations, we often use geometric
representations. We have already used the real number line to represent the value(s) of one va
Non-linear Systems of Equations in two variables
Example 1:
Question: Solve the following system of equations and interpret the solution graphically.
cfw_
y=x+1
x2 + y2 = 25
See [9.4, #7]
.
Solution:
Substituting y = x + 1 from the first equation into the
Multiplication of polynomials
Some terminology.
Consider the expression
a2 b + 2 a c 5 a3 d +
acd
3
+3
7,
where a, b, c and d are (unknown) real numbers. a, b, c and d are the variables in the expression.
This expression is the sum of the three terms: a2
Even and odd functions
A function f is called even provided that
f( x ) = f( x )
for all numbers x in the domain of f.
If a is a positive number, an even function has the same value at a negative number a as it has at the corresponding positive number a.
Solving Equations
Introductory example.
Solve the equation
x ( x + 2 ) = 15 - (i).
We need to find all the real numbers x such the statement x ( x + 2 ) = 15 is true.
Putting the problem into words, we need to find a number such that the product of this n | 677.169 | 1 |
Mathematics 11
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Description: This book is meant for the students studying in schools which follows CBSE or similar curriculum. It can also be referred for General Studies paper of UPSC's Civil Services Exam. All rights are res...
This book is meant for the students studying in schools which follows CBSE or similar curriculum. It can also be referred for General Studies paper of UPSC's Civil Services Exam. All rights are reserved by the publisher i.e. NCERT. The contents of the book has been downloaded from the site:
SETS
!In these days of conflict between ancient and modern studies; there
must surely be something to be said for a study which did not begin with Pythagoras and will not end with Einstein; but is the oldest and the youngest. — G.H. HARDY !
1.1 Introduction
The concept of set serves as a fundamental part of the present day mathematics. Today this concept is being used in almost every branch of mathematics. Sets are used to define the concepts of relations and functions. The study of geometry, sequences, probability, etc. requires the knowledge of sets. The theory of sets was developed by German mathematician Georg Cantor (1845-1918). He first encountered sets while working on "problems on trigonometric series". In this Chapter, we discuss some basic definitions and operations involving sets.
Georg Cantor (1845-1918)
1.2 Sets and their Representations
In everyday life, we often speak of collections of objects of a particular kind, such as, a pack of cards, a crowd of people, a cricket team, etc. In mathematics also, we come across collections, for example, of natural numbers, points, prime numbers, etc. More specially, we examine the following collections: (i) Odd natural numbers less than 10, i.e., 1, 3, 5, 7, 9 (ii) The rivers of India (iii) The vowels in the English alphabet, namely, a, e, i, o, u (iv) Various kinds of triangles (v) Prime factors of 210, namely, 2,3,5 and 7 (vi) The solution of the equation: x2 – 5x + 6 = 0, viz, 2 and 3. We note that each of the above example is a well-defined collection of objects in
2
MATHEMATICS
the sense that we can definitely decide whether a given particular object belongs to a given collection or not. For example, we can say that the river Nile does not belong to the collection of rivers of India. On the other hand, the river Ganga does belong to this colleciton. We give below a few more examples of sets used particularly in mathematics, viz. N Z Q R Z+ Q+ : : : : : : the set of all natural numbers the set of all integers the set of all rational numbers the set of real numbers the set of positive integers the set of positive rational numbers, and
R + : the set of positive real numbers. The symbols for the special sets given above will be referred to throughout this text. Again the collection of five most renowned mathematicians of the world is not well-defined, because the criterion for determining a mathematician as most renowned may vary from person to person. Thus, it is not a well-defined collection. We shall say that a set is a well-defined collection of objects. The following points may be noted : (i) Objects, elements and members of a set are synonymous terms. (ii) Sets are usually denoted by capital letters A, B, C, X, Y, Z, etc. (iii) The elements of a set are represented by small letters a, b, c, x, y, z, etc. If a is an element of a set A, we say that " a belongs to A" the Greek symbol ∈ (epsilon) is used to denote the phrase 'belongs to'. Thus, we write a ∈ A. If 'b' is not an element of a set A, we write b ∉ A and read "b does not belong to A". Thus, in the set V of vowels in the English alphabet, a ∈ V but b ∉ V. In the set P of prime factors of 30, 3 ∈ P but 15 ∉ P. There are two methods of representing a set : (i) Roster or tabular form (ii) Set-builder form. (i) In roster form, all the elements of a set are listed, the elements are being separated by commas and are enclosed within braces { }. For example, the set of all even positive integers less than 7 is described in roster form as {2, 4, 6}. Some more examples of representing a set in roster form are given below : (a) The set of all natural numbers which divide 42 is {1, 2, 3, 6, 7, 14, 21, 42}.
SETS
3
Thus, the above set can also be represented as {1, 3, 7, 21, 2, 6, 14, 42}. (b) (c)
!Note In roster form, the order in which the elements are listed is immaterial.
The set of all vowels in the English alphabet is {a, e, i, o, u}. The set of odd natural numbers is represented by {1, 3, 5, . . .}. The dots tell us that the list of odd numbers continue indefinitely.
It may be noted the set in roster form is !Note repeated, i.e., allthat while writing taken as distinct. For an elementthe not generally the elements are example, set of letters forming the word 'SCHOOL' is { S, C, H, O, L} or {H, O, L, C, S}. Here, the order of listing elements has no relevance. (ii) In set-builder form, all the elements of a set possess a single common property which is not possessed by any element outside the set. For example, in the set {a, e, i, o, u}, all the elements possess a common property, namely, each of them is a vowel in the English alphabet, and no other letter possess this property. Denoting this set by V, we write V = {x : x is a vowel in English alphabet} It may be observed that we describe the element of the set by using a symbol x (any other symbol like the letters y, z, etc. could be used) which is followed by a colon " : ". After the sign of colon, we write the characteristic property possessed by the elements of the set and then enclose the whole description within braces. The above description of the set V is read as "the set of all x such that x is a vowel of the English alphabet". In this description the braces stand for "the set of all", the colon stands for "such that". For example, the set A = {x : x is a natural number and 3 < x < 10} is read as "the set of all x such that x is a natural number and x lies between 3 and 10. Hence, the numbers 4, 5, 6, 7, 8 and 9 are the elements of the set A. If we denote the sets described in (a), (b) and (c) above in roster form by A, B, C, respectively, then A, B, C can also be represented in set-builder form as follows: A= {x : x is a natural number which divides 42} B= {y : y is a vowel in the English alphabet} C= {z : z is an odd natural number} Example 1 Write the solution set of the equation x2 + x – 2 = 0 in roster form. Solution The given equation can be written as (x – 1) (x + 2) = 0, i. e., x = 1, – 2 Therefore, the solution set of the given equation can be written in roster form as {1, – 2}. Example 2 Write the set {x : x is a positive integer and x2 < 40} in the roster form.
4
MATHEMATICS
Solution The required numbers are 1, 2, 3, 4, 5, 6. So, the given set in the roster form is {1, 2, 3, 4, 5, 6}. Example 3 Write the set A = {1, 4, 9, 16, 25, . . . }in set-builder form. Solution We may write the set A as A = {x : x is the square of a natural number} Alternatively, we can write A = {x : x = n2, where n ∈ N}
1 2 3 4 5 6 Example 4 Write the set { , , , , , } in the set-builder form. 2 3 4 5 6 7
Solution We see that each member in the given set has the numerator one less than the demominator. Also, the numerator begin from 1 and do not exceed 6. Hence, in the set-builder form the given set is
n ! " , where n is a natural number and 1 $ n $ 6& %x : x # n '1 ( )
Example 5 Match each of the set on the left described in the roster form with the same set on the right described in the set-builder form : (i) {P, R, I, N, C, A, L} (a) { x : x is a positive integer and is a divisor of 18} (ii) { 0 } (b) { x : x is an integer and x2 – 9 = 0} (iii) {1, 2, 3, 6, 9, 18} (c) {x : x is an integer and x + 1= 1} (iv) {3, –3} (d) {x : x is a letter of the word PRINCIPAL} Solution Since in (d), there are 9 letters in the word PRINCIPAL and two letters P and I are repeated, so (i) matches (d). Similarly, (ii) matches (c) as x + 1 = 1 implies x = 0. Also, 1, 2 ,3, 6, 9, 18 are all divisors of 18 and so (iii) matches (a). Finally, x2 – 9 = 0 implies x = 3, –3 and so (iv) matches (b).
EXERCISE 1.1
1. Which of the following are sets ? Justify your asnwer. (i) The collection of all the months of a year beginning with the letter J. (ii) The collection of ten most talented writers of India. (iii) A team of eleven best-cricket batsmen of the world. (iv) The collection of all boys in your class. (v) The collection of all natural numbers less than 100. (vi) A collection of novels written by the writer Munshi Prem Chand. (vii) The collection of all even integers.
SETS
5
2.
3.
4.
5.
(viii) The collection of questions in this Chapter. (ix) A collection of most dangerous animals of the world. Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol ∈ or ∉ in the blank spaces: (i) 5. . .A (ii) 8 . . . A (iii) 0. . .A (iv) 4. . . A (v) 2. . .A (vi) 10. . .A Write the following sets in roster form: (i) A = {x : x is an integer and –3 < x < 7} (ii) B = {x : x is a natural number less than 6} (iii) C = {x : x is a two-digit natural number such that the sum of its digits is 8} (iv) D = {x : x is a prime number which is divisor of 60} (v) E = The set of all letters in the word TRIGONOMETRY (vi) F = The set of all letters in the word BETTER Write the following sets in the set-builder form : (i) (3, 6, 9, 12} (ii) {2,4,8,16,32} (iii) {5, 25, 125, 625} (iv) {2, 4, 6, . . .} (v) {1,4,9, . . .,100} List all the elements of the following sets : (i) A = {x : x is an odd natural number} (ii) B = {x : x is an integer, – (iii) C = {x : x is an integer, x2 ≤ 4} (iv) D = {x : x is a letter in the word "LOYAL"} (v) E = {x : x is a month of a year not having 31 days} (vi) F = {x : x is a consonant in the English alphabet which precedes k }. Match each of the set on the left in the roster form with the same set on the right described in set-builder form: (i) {1, 2, 3, 6} (a) {x : x is a prime number and a divisor of 6} (ii) {2, 3} (b) {x : x is an odd natural number less than 10} (iii) {M,A,T,H,E,I,C,S} (c) {x : x is natural number and divisor of 6} (iv) {1, 3, 5, 7, 9} (d) {x : x is a letter of the word MATHEMATICS}.
1 9 <x< } 2 2
6.
1.3 The Empty Set
Consider the set A = { x : x is a student of Class XI presently studying in a school } We can go to the school and count the number of students presently studying in Class XI in the school. Thus, the set A contains a finite number of elements. We now write another set B as follows:
6
MATHEMATICS
B = { x : x is a student presently studying in both Classes X and XI } We observe that a student cannot study simultaneously in both Classes X and XI. Thus, the set B contains no element at all. Definition 1 A set which does not contain any element is called the empty set or the null set or the void set. According to this definition, B is an empty set while A is not an empty set. The empty set is denoted by the symbol φ or { }. We give below a few examples of empty sets. (i) Let A = {x : 1 < x < 2, x is a natural number}. Then A is the empty set, because there is no natural number between 1 and 2. (ii) B = {x : x2 – 2 = 0 and x is rational number}. Then B is the empty set because the equation x2 – 2 = 0 is not satisfied by any rational value of x. (iii) C = {x : x is an even prime number greater than 2}.Then C is the empty set, because 2 is the only even prime number. (iv) D = { x : x2 = 4, x is odd }. Then D is the empty set, because the equation x2 = 4 is not satisfied by any odd value of x.
1.4 Finite and Infinite Sets
Let A = {1, 2, 3, 4, 5}, B = {a, b, c, d, e, g} and C = { men living presently in different parts of the world} We observe that A contains 5 elements and B contains 6 elements. How many elements does C contain? As it is, we do not know the number of elements in C, but it is some natural number which may be quite a big number. By number of elements of a set S, we mean the number of distinct elements of the set and we denote it by n (S). If n (S) is a natural number, then S is non-empty finite set. Consider the set of natural numbers. We see that the number of elements of this set is not finite since there are infinite number of natural numbers. We say that the set of natural numbers is an infinite set. The sets A, B and C given above are finite sets and n(A) = 5, n(B) = 6 and n(C) = some finite number. Definition 2 A set which is empty or consists of a definite number of elements is called finite otherwise, the set is called infinite. Consider some examples : (i) Let W be the set of the days of the week. Then W is finite. (ii) Let S be the set of solutions of the equation x2 –16 = 0. Then S is finite. (iii) Let G be the set of points on a line. Then G is infinite. When we represent a set in the roster form, we write all the elements of the set within braces { }. It is not possible to write all the elements of an infinite set within braces { } because the numbers of elements of such a set is not finite. So, we represent
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some infinite set in the roster form by writing a few elements which clearly indicate the structure of the set followed ( or preceded ) by three dots. For example, {1, 2, 3 . . .} is the set of natural numbers, {1, 3, 5, 7, . . .} is the set of odd natural numbers, {. . .,–3, –2, –1, 0,1, 2 ,3, . . .} is the set of integers. All these sets are infinite. the the !Note All infinite sets cannot be described inform,roster form. For example,this set of real numbers cannot be described in this because the elements of set do not follow any particular pattern. Example 6 State which of the following sets are finite or infinite : (i) {x : x ∈ N and (x – 1) (x –2) = 0} (ii) {x : x ∈ N and x2 = 4} (iii) {x : x ∈ N and 2x –1 = 0} (iv) {x : x ∈ N and x is prime} (v) {x : x ∈ N and x is odd} Solution (i) Given set = {1, 2}. Hence, it is finite. (ii) Given set = {2}. Hence, it is finite. (iii) Given set = φ. Hence, it is finite. (iv) The given set is the set of all prime numbers and since set of prime numbers is infinite. Hence the given set is infinite (v) Since there are infinite number of odd numbers, hence, the given set is infinite.
1.5 Equal Sets
Given two sets A and B, if every element of A is also an element of B and if every element of B is also an element of A, then the sets A and B are said to be equal. Clearly, the two sets have exactly the same elements. Definition 3 Two sets A and B are said to be equal if they have exactly the same elements and we write A = B. Otherwise, the sets are said to be unequal and we write A ≠ B. We consider the following examples : (i) Let A = {1, 2, 3, 4} and B = {3, 1, 4, 2}. Then A = B. (ii) Let A be the set of prime numbers less than 6 and P the set of prime factors of 30. Then A and P are equal, since 2, 3 and 5 are the only prime factors of 30 and also these are less than 6. Note A set does not change if one or more elements of the set are repeated. For example, the sets A = {1, 2, 3} and B = {2, 2, 1, 3, 3} are equal, since each
EXERCISE 1.2
1. Which of the following are examples of the null set (i) Set of odd natural numbers divisible by 2 (ii) Set of even prime numbers (iii) { x : x is a natural numbers, x < 5 and x > 7 } (iv) { y : y is a point common to any two parallel lines} Which of the following sets are finite or infinite (i) The set of months of a year (ii) {1, 2, 3, . . .} (iii) {1, 2, 3, . . .99, 100} (iv) The set of positive integers greater than 100 (v) The set of prime numbers less than 99 State whether each of the following set is finite or infinite: (i) The set of lines which are parallel to the x-axis (ii) The set of letters in the English alphabet (iii) The set of numbers which are multiple of 5
1.6 Subsets
Consider the sets : X = set of all students in your school, Y = set of all students in your class. We note that every element of Y is also an element of X; we say that Y is a subset of X. The fact that Y is subset of X is expressed in symbols as Y ⊂ X. The symbol ⊂ stands for 'is a subset of' or 'is contained in'. Definition 4 A set A is said to be a subset of a set B if every element of A is also an element of B. In other words, A ⊂ B if whenever a ∈ A, then a ∈ B. It is often convenient to use the symbol "⇒" which means implies. Using this symbol, we can write the definiton of subset as follows: A ⊂ B if a ∈ A ⇒ a ∈ B We read the above statement as "A is a subset of B if a is an element of A implies that a is also an element of B". If A is not a subset of B, we write A ⊄ B. We may note that for A to be a subset of B, all that is needed is that every element of A is in B. It is possible that every element of B may or may not be in A. If it so happens that every element of B is also in A, then we shall also have B ⊂ A. In this case, A and B are the same sets so that we have A ⊂ B and B ⊂ A ⇔ A = B, where "⇔" is a symbol for two way implications, and is usually read as if and only if (briefly written as "iff"). It follows from the above definition that every set A is a subset of itself, i.e., A ⊂ A. Since the empty set φ has no elements, we agree to say that φ is a subset of every set. We now consider some examples :
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MATHEMATICS
(i) The set Q of rational numbers is a subset of the set R of real numbes, and we write Q ⊂ R. (ii) If A is the set of all divisors of 56 and B the set of all prime divisors of 56, then B is a subset of A and we write B ⊂ A. (iii) Let A = {1, 3, 5} and B = {x : x is an odd natural number less than 6}. Then A ⊂ B and B ⊂ A and hence A = B. (iv) Let A = { a, e, i, o, u} and B = { a, b, c, d}. Then A is not a subset of B, also B is not a subset of A. Let A and B be two sets. If A ⊂ B and A ≠ B , then A is called a proper subset of B and B is called superset of A. For example, A = {1, 2, 3} is a proper subset of B = {1, 2, 3, 4}. If a set A has only one element, we call it a singleton set. Thus,{ a } is a singleton set. Example 9 Consider the sets φ, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. Insert the symbol ⊂ or ⊄ between each of the following pair of sets: (i) φ . . . B Solution (i) (ii) (iii) (iv) (ii) A . . . B (iii) A . . . C (iv) B . . . C φ ⊂ B as φ is a subset of every set. A ⊄ B as 3 ∈ A and 3 ∉ B A ⊂ C as 1, 3 ∈ A also belongs to C B ⊂ C as each element of B is also an element of C.
Example 10 Let A = { a, e, i, o, u} and B = { a, b, c, d}. Is A a subset of B ? No. (Why?). Is B a subset of A? No. (Why?) Example 11 Let A, B and C be three sets. If A ∈ B and B ⊂ C, is it true that A ⊂ C?. If not, give an example. Solution No. Let A = {1}, B = {{1}, 2} and C = {{1}, 2, 3}. Here A ∈ B as A = {1} and B ⊂ C. But A ⊄ C as 1 ∈ A and 1 ∉ C. Note that an element of a set can never be a subset of itself. 1.6.1 Subsets of set of real numbers As noted in Section 1.6, there are many important subsets of R. We give below the names of some of these subsets. The set of natural numbers N = {1, 2, 3, 4, 5, . . .} The set of integers Z = {. . ., –3, –2, –1, 0, 1, 2, 3, . . .} The set of rational numbers Q = { x : x =
p , p, q ∈ Z and q ≠ 0} q
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which is read " Q is the set of all numbers x such that x equals the quotient
p q , where
p and q are integers and q is not zero". Members of Q include –5 (which can be
rational. Members of T include 2 , 5 and ! . Some of the obvious relations among these subsets are: N ⊂ Z ⊂ Q, Q ⊂ R, T ⊂ R, N ⊄ T. 1.6.2 Intervals as subsets of R Let a, b ∈ R and a < b. Then the set of real numbers { y : a < y < b} is called an open interval and is denoted by (a, b). All the points between a and b belong to the open interval (a, b) but a, b themselves do not belong to this interval. The interval which contains the end points also is called closed interval and is denoted by [ a, b ]. Thus [ a, b ] = {x : a ≤ x ≤ b} We can also have intervals closed at one end and open at the other, i.e., [ a, b ) = {x : a ≤ x < b} is an open interval from a to b, including a but excluding b. ( a, b ] = { x : a < x ≤ b } is an open interval from a to b including b but excluding a. These notations provide an alternative way of designating the subsets of set of real numbers. For example , if A = (–3, 5) and B = [–7, 9], then A ⊂ B. The set [ 0, ∞) defines the set of non-negative real numbers, while set ( – ∞, 0 ) defines the set of negative real numbers. The set ( – ∞, ∞ ) describes the set of real numbers in relation to a line extending from – ∞ to ∞. On real number line, various types of intervals described above as subsets of R, are shown in the Fig 1.1.
Fig 1.1
Here, we note that an interval contains infinitely many points. For example, the set {x : x ∈ R, –5 < x ≤ 7}, written in set-builder form, can be written in the form of interval as (–5, 7] and the interval [–3, 5) can be written in setbuilder form as {x : –3 ≤ x < 5}.
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The number (b – a) is called the length of any of the intervals (a, b), [a, b], [a, b) or (a, b].
1.7 Power Set
Consider the set {1, 2}. Let us write down all the subsets of the set {1, 2}. We know that φ is a subset of every set . So, φ is a subset of {1, 2}. We see that {1} and { 2 }are also subsets of {1, 2}. Also, we know that every set is a subset of itself. So, { 1, 2 } is a subset of {1, 2}. Thus, the set { 1, 2 } has, in all, four subsets, viz. φ, { 1 }, { 2 } and { 1, 2 }. The set of all these subsets is called the power set of { 1, 2 }. Definition 5 The collection of all subsets of a set A is called the power set of A. It is denoted by P(A). In P(A), every element is a set. Thus, as in above, if A = { 1, 2 }, then P( A ) = { φ,{ 1 }, { 2 }, { 1,2 }} Also, note that n [ P (A) ] = 4 = 22 In general, if A is a set with n(A) = m, then it can be shown that n [ P(A)] = 2m.
1.8 Universal Set
Usually, in a particular context, we have to deal with the elements and subsets of a basic set which is relevant to that particular context. For example, while studying the system of numbers, we are interested in the set of natural numbers and its subsets such as the set of all prime numbers, the set of all even numbers, and so forth. This basic set is called the "Universal Set". The universal set is usually denoted by U, and all its subsets by the letters A, B, C, etc. For example, for the set of all integers, the universal set can be the set of rational numbers or, for that matter, the set R of real numbers. For another example, in human population studies, the universal set consists of all the people in the world.
1.9 Venn Diagrams
Most of the relationships between sets can be represented by means of diagrams which are known as Venn diagrams. Venn diagrams are named after the English logician, John Venn (1834-1883). These diagrams consist of rectangles and closed curves usually circles. The universal set is represented usually by a rectangle and its subsets by circles. In Venn diagrams, the elements of the sets are written in their respective circles (Figs 1.2 and 1.3)
Fig 1.2
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MATHEMATICS
Illustration 1 In Fig 1.2, U = {1,2,3, ..., 10} is the universal set of which A = {2,4,6,8,10} is a subset. Illustration 2 In Fig 1.3, U = {1,2,3, ..., 10} is the universal set of which A = {2,4,6,8,10} and B = {4, 6} are subsets, and also B ⊂ A. Fig 1.3 The reader will see an extensive use of the Venn diagrams when we discuss the union, intersection and difference of sets.
1.10 Operations on Sets
In earlier classes, we have learnt how to perform the operations of addition, subtraction, multiplication and division on numbers. Each one of these operations was performed on a pair of numbers to get another number. For example, when we perform the operation of addition on the pair of numbers 5 and 13, we get the number 18. Again, performing the operation of multiplication on the pair of numbers 5 and 13, we get 65. Similarly, there are some operations which when performed on two sets give rise to another set. We will now define certain operations on sets and examine their properties. Henceforth, we will refer all our sets as subsets of some universal set. 1.10.1 Union of sets Let A and B be any two sets. The union of A and B is the set which consists of all the elements of A and all the elements of B, the common elements being taken only once. The symbol '∪' is used to denote the union. Symbolically, we write A ∪ B and usually read as 'A union B'. Example 12 Let A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}. Find A ∪ B. Solution We have A ∪ B = { 2, 4, 6, 8, 10, 12} Note that the common elements 6 and 8 have been taken only once while writing A ∪ B. Example 13 Let A = { a, e, i, o, u } and B = { a, i, u }. Show that A ∪ B = A Solution We have, A ∪ B = { a, e, i, o, u } = A. This example illustrates that union of sets A and its subset B is the set A itself, i.e., if B ⊂ A, then A ∪ B = A. Example 14 Let X = {Ram, Geeta, Akbar} be the set of students of Class XI, who are in school hockey team. Let Y = {Geeta, David, Ashok} be the set of students from Class XI who are in the school football team. Find X ∪ Y and interpret the set. Solution We have, X ∪ Y = {Ram, Geeta, Akbar, David, Ashok}. This is the set of students from Class XI who are in the hockey team or the football team or both.
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Thus, we can define the union of two sets as follows: Definition 6 The union of two sets A and B is the set C which consists of all those elements which are either in A or in B (including those which are in both). In symbols, we write. A ∪ B = { x : x ∈A or x ∈B } The union of two sets can be represented by a Venn diagram as shown in Fig 1.4. The shaded portion in Fig 1.4 represents A ∪ B. Some Properties of the Operation of Union (i) A ∪ B = B ∪ A (Commutative law) Fig 1.4 (ii) ( A ∪ B ) ∪ C = A ∪ ( B ∪ C) (Associative law ) (iii) A ∪ φ = A (iv) A ∪ A = A (v) U ∪ A = U (Law of identity element, φ is the identity of ∪) (Idempotent law) (Law of U)
1.10.2 Intersection of sets The intersection of sets A and B is the set of all elements which are common to both A and B. The symbol '∩' is used to denote the intersection. The intersection of two sets A and B is the set of all those elements which belong to both A and B. Symbolically, we write A ∩ B = {x : x ∈ A and x ∈ B}. Example 15 Consider the sets A and B of Example 12. Find A ∩ B. Solution We see that 6, 8 are the only elements which are common to both A and B. Hence A ∩ B = { 6, 8 }. Example 16 Consider the sets X and Y of Example 14. Find X ∩ Y. Solution We see that element 'Geeta' is the only element common to both. Hence, X ∩ Y = {Geeta}. Example 17 Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = { 2, 3, 5, 7 }. Find A ∩ B and hence show that A ∩ B = B. Solution We have A ∩ B = { 2, 3, 5, 7 } = B. We note that B ⊂ A and that A ∩ B = B. Definition 7 The intersection of two sets A and B is the set of all those elements which belong to both A and B. Symbolically, we write A ∩ B = {x : x ∈ A and x ∈ B} The shaded portion in Fig 1.5 indicates the Fig 1.5 interseciton of A and B.
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MATHEMATICS
If A and B are two sets such that A ∩ B = φ, then U A and B are called disjoint sets. For example, let A = { 2, 4, 6, 8 } and B = { 1, 3, 5, 7 }. Then A and B are disjoint sets, A B because there are no elements which are common to A and B. The disjoint sets can be represented by means of Venn diagram as shown in the Fig 1.6 In the above diagram, A and B are disjoint sets. Fig 1.6 Some Properties of Operation of Intersection (i) A ∩ B = B ∩ A (Commutative law). (ii) ( A ∩ B ) ∩ C = A ∩ ( B ∩ C ) (Associative law). (iii) φ ∩ A = φ, U ∩ A = A (Law of φ and U). (iv) A ∩ A = A (Idempotent law) (v) A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) (Distributive law ) i. e., ∩ distributes over ∪ This can be seen easily from the following Venn diagrams [Figs 1.7 (i) to (v)].
(i)
(iii)
(ii)
(iv)
(v)
Figs 1.7 (i) to (v)
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1.10.3 Difference of sets The difference of the sets A and B in this order is the set of elements which belong to A but not to B. Symbolically, we write A – B and read as " A minus B". Example 18 Let A = { 1, 2, 3, 4, 5, 6}, B = { 2, 4, 6, 8 }. Find A – B and B – A. Solution We have, A – B = { 1, 3, 5 }, since the elements 1, 3, 5 belong to A but not to B and B – A = { 8 }, since the element 8 belongs to B and not to A. We note that A – B ≠ B – A. Example 19 Let V = { a, e, i, o, u } and B = { a, i, k, u}. Find V – B and B – V Solution We have, V – B = { e, o }, since the elements e, o belong to V but not to B and B – V = { k }, since the element k belongs to B but not to V. We note that V – B ≠ B – V. Using the setbuilder notation, we can rewrite the definition of difference as A – B = { x : x ∈ A and x ∉ B } The difference of two sets A and B can be represented by Venn diagram as shown in Fig 1.8. The shaded portion represents the difference of the two sets A and B. Remark The sets A – B, A ∩ B and B – A are mutually disjoint sets, i.e., the intersection of any of these two sets is the null set as shown in Fig 1.9.
1.11 Complement of a Set
Let U be the universal set which consists of all prime numbers and A be the subset of U which consists of all those prime numbers that are not divisors of 42. Thus, A = {x : x ∈ U and x is not a divisor of 42 }. We see that 2 ∈ U but 2 ∉ A, because 2 is divisor of 42. Similarly, 3 ∈ U but 3 ∉ A, and 7 ∈ U but 7 ∉ A. Now 2, 3 and 7 are the only elements of U which do not belong to A. The set of these three prime numbers, i.e., the set {2, 3, 7} is called the Complement of A with respect to U, and is denoted by
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A′. So we have A′ = {2, 3, 7}. Thus, we see that A′ = {x : x ∈ U and x ∉ A }. This leads to the following definition. Definition 8 Let U be the universal set and A a subset of U. Then the complement of A is the set of all elements of U which are not the elements of A. Symbolically, we write A′ to denote the complement of A with respect to U. Thus, A′ = {x : x ∈ U and x ∉ A }. Obviously A′ = U – A We note that the complement of a set A can be looked upon, alternatively, as the difference between a universal set U and the set A. Example 20 Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′. Solution We note that 2, 4, 6, 8, 10 are the only elements of U which do not belong to A. Hence A′ = { 2, 4, 6, 8,10 }. Example 21 Let U be universal set of all the students of Class XI of a coeducational school and A be the set of all girls in Class XI. Find A′. Solution Since A is the set of all girls, A′ is clearly the set of all boys in the class.
!Note
If A is a subset of the universal set U, then its complement A′ is also a subset of U. Again in Example 20 above, we have A′ = { 2, 4, 6, 8, 10 } (A′ ), = {x : x ∈ U and x ∉ A′} = {1, 3, 5, 7, 9} = A It is clear from the definition of the complement that for any subset of the universal Hence
The complement of the union of two sets is the intersection of their complements and the complement of the intersection of two sets is the union of their complements. These are called De Morgan's laws. These are named after the mathematician De Morgan. The complement A′ of a set A can be represented by a Venn diagram as shown in Fig 1.10. The shaded portion represents the complement of the set A. Some Properties of Complement Sets 1. Complement laws: (i) A ∪ A′ = U 2. De Morgan's law:
Fig 1.10
(ii) A ∩ A′ = φ
(i) (A - B)´ = A′ ∩ B′ (ii) (A . B ), = A′ ∪ B′
3. Law of double complementation : (A′ ), = A 4. Laws of empty set and universal set φ′ = U and U′ = φ. These laws can be verified by using Venn diagrams.
1.12 Practical Problems on Union and Intersection of Two Sets
In earlier Section, we have learnt union, intersection and difference of two sets. In this Section, we will go through some practical problems related to our daily life.The formulae derived in this Section will also be used in subsequent Chapter on Probability (Chapter 16).
EXERCISE 1.6
1. 2. 3. 4. 5. 6. 7. 8. If X and Y are two sets such that n ( X ) = 17, n ( Y ) = 23 and n ( X ∪ Y ) = 38, find n ( X ∩ Y ). If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and Y has 15 elements ; how many elements does X ∩ Y have? In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English? If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, how many elements does S ∪ T have? If X and Y are two sets such that X has 40 elements, X ∪ Y has 60 elements and X ∩ Y has 10 elements, how many elements does Y have? In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many people like both coffee and tea? In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis? In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?
Miscellaneous Examples
Example 28 Show that the set of letters needed to spell " CATARACT " and the set of letters needed to spell " TRACT" are equal. Solution Let X be the set of letters in "CATARACT". Then X = { C, A, T, R } Let Y be the set of letters in " TRACT". Then Y = { T, R, A, C, T } = { T, R, A, C } Since every element in X is in Y and every element in Y is in X. It follows that X = Y. Example 29 List all the subsets of the set { –1, 0, 1 }. Solution Let A = { –1, 0, 1 }. The subset of A having no element is the empty set φ. The subsets of A having one element are { –1 }, { 0 }, { 1 }. The subsets of A having two elements are {–1, 0}, {–1, 1} ,{0, 1}. The subset of A having three elements of A is A itself. So, all the subsets of A are φ, {–1}, {0}, {1}, {–1, 0}, {–1, 1}, {0, 1} and {–1, 0, 1}.
Miscellaneous Exercise on Chapter 1
1. Decide, among the following sets, which sets are subsets of one and another: A = { x : x ∈ R and x satisfy x2 – 8x + 12 = 0 }, B = { 2, 4, 6 }, C = { 2, 4, 6, 8, . . . }, D = { 6 }. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. (i) If x ∈ A and A ∈ B , then x ∈ B (ii) If A ⊂ B and B ∈ C , then A ∈ C (iii) If A ⊂ B and B ⊂ C , then A ⊂ C (iv) If A ⊄ B and B ⊄ C , then A ⊄ C (v) If x ∈ A and A ⊄ B , then x ∈ B (vi) If A ⊂ B and x ∉ B , then x ∉ A Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C. Show that the following four conditions are equivalent : (i) A ⊂ B(ii) A – B = φ (iii) A ∪ B = B (iv) A ∩ B = A Show that if A ⊂ B, then C – B ⊂ C – A. Assume that P ( A ) = P ( B ). Show that A = B Is it true that for any sets A and B, P ( A ) ∪ P ( B ) = P ( A ∪ B )? Justify your answer.
2.
3. 4. 5. 6. 7.
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8. 9. 10. 11.
12. 13.
14.
15.
16.
Show that for any sets A and B, A = ( A ∩ B ) ∪ ( A – B ) and A ∪ ( B – A ) = ( A ∪ B ) Using properties of sets, show that (i) A ∪ ( A ∩ B ) = A (ii) A ∩ ( A ∪ B ) = A. Show that A ∩ B = A ∩ C need not imply B = C. Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B. (Hints A = A ∩ ( A ∪ X ) , B = B ∩ ( B ∪ X ) and use Distributive law ) Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ. In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee? In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group? In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find: (i) the number of people who read at least one of the newspapers. (ii) the number of people who read exactly one newspaper. In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
Summary
This chapter deals with some basic definitions and operations involving sets. These are summarised below: " A set is a well-defined collection of objects. " A set which does not contain any element is called empty set. " A set which consists of a definite number of elements is called finite set, otherwise, the set is called infinite set. " Two sets A and B are said to be equal if they have exactly the same elements. " A set A is said to be subset of a set B, if every element of A is also an element of B. Intervals are subsets of R. " A power set of a set A is collection of all subsets of A. It is denoted by P(A).
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MATHEMATICS
" The union of two sets A and B is the set of all those elements which are either
in A or in B. " The intersection of two sets A and B is the set of all elements which are common. The difference of two sets A and B in this order is the set of elements which belong to A but not to B. " The complement of a subset A of universal set U is the set of all elements of U which are not the elements of A. " For any two sets A and B, (A ∪ B)′ = A′ ∩ B′ and ( A ∩ B )′ = A′ ∪ B′ " If A and B are finite sets such that A ∩ B = φ, then n (A ∪ B) = n (A) + n (B). If A ∩ B ≠ φ, then n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
Historical Note
The modern theory of sets is considered to have been originated largely by the German mathematician Georg Cantor (1845-1918 A.D.). His papers on set theory appeared sometimes during 1874 A.D. to 1897 A.D. His study of set theory came when he was studying trigonometric series of the form a1 sin x + a2 sin 2x + a3 sin 3x + ... He published in a paper in 1874 A.D. that the set of real numbers could not be put into one-to-one correspondence wih the integers. From 1879 onwards, he publishd several papers showing various properties of abstract sets. Cantor's work was well received by another famous mathematician Richard Dedekind (1831-1916 A.D.). But Kronecker (1810-1893 A.D.) castigated him for regarding infinite set the same way as finite sets. Another German mathematician Gottlob Frege, at the turn of the century, presented the set theory as principles of logic. Till then the entire set theory was based on the assumption of the existence of the set of all sets. It was the famous Englih Philosopher Bertand Russell (1872-1970 A.D.) who showed in 1902 A.D. that the assumption of existence of a set of all sets leads to a contradiction. This led to the famous Russell's Paradox. Paul R.Halmos writes about it in his book 'Naïve Set Theory' that "nothing contains everything". The Russell's Paradox was not the only one which arose in set theory. Many paradoxes were produced later by several mathematicians and logicians.
SETS
29
As a consequence of all these paradoxes, the first axiomatisation of set theory was published in 1908 A.D. by Ernst Zermelo. Another one was proposed by Abraham Fraenkel in 1922 A.D. John Von Neumann in 1925 A.D. introduced explicitly the axiom of regularity. Later in 1937 A.D. Paul Bernays gave a set of more satisfactory axiomatisation. A modification of these axioms was done by Kurt Gödel in his monograph in 1940 A.D. This was known as Von NeumannBernays (VNB) or Gödel-Bernays (GB) set theory. Despite all these difficulties, Cantor's set theory is used in present day mathematics. In fact, these days most of the concepts and results in mathematics are expressed in the set theoretic language.
—! — !
Chapter
2
RELATIONS AND FUNCTIONS
!Mathematics is the indispensable instrument of all physical research. – BERTHELOT !
2.1 Introduction
Much of mathematics is about finding a pattern – a recognisable link between quantities that change. In our daily life, we come across many patterns that characterise relations such as brother and sister, father and son, teacher and student. In mathematics also, we come across many relations such as number m is less than number n, line l is parallel to line m, set A is a subset of set B. In all these, we notice that a relation involves pairs of objects in certain order. In this Chapter, we will learn how to link pairs of objects from two sets and then introduce relations between the two objects in the pair. Finally, we will learn about G . W. Leibnitz (1646–1716) special relations which will qualify to be functions. The concept of function is very important in mathematics since it captures the idea of a mathematically precise correspondence between one quantity with the other.
2.2 Cartesian Products of Sets
Suppose A is a set of 2 colours and B is a set of 3 objects, i.e., A = {red, blue}and B = {b, c, s}, where b, c and s represent a particular bag, coat and shirt, respectively. How many pairs of coloured objects can be made from these two sets? Proceeding in a very orderly manner, we can see that there will be 6 distinct pairs as given below: (red, b), (red, c), (red, s), (blue, b), (blue, c), (blue, s). Thus, we get 6 distinct objects (Fig 2.1). Let us recall from our earlier classes that an ordered pair of elements taken from any two sets P and Q is a pair of elements written in small
Fig 2.1
RELATIONS AND FUNCTIONS
31
brackets and grouped together in a particular order, i.e., (p,q), p ∈ P and q ∈ Q . This leads to the following definition: Definition 1 Given two non-empty sets P and Q. The cartesian product P × Q is the set of all ordered pairs of elements from P and Q, i.e., P × Q = { (p,q) : p ∈ P, q ∈ Q } If either P or Q is the null set, then P × Q will also be empty set, i.e., P × Q = φ From the illustration given above we note that A × B = {(red,b), (red,c), (red,s), (blue,b), (blue,c), (blue,s)}. Again, consider the two sets: A = {DL, MP, KA}, where DL, MP, KA represent Delhi, Madhya Pradesh and Karnataka, respectively and B = {01,02, 03 03}representing codes for the licence plates of vehicles issued 02 by DL, MP and KA . 01 If the three states, Delhi, Madhya Pradesh and Karnataka were making codes for the licence plates of vehicles, with the DL MP KA restriction that the code begins with an element from set A, Fig 2.2 which are the pairs available from these sets and how many such pairs will there be (Fig 2.2)? The available pairs are:(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02), (KA,03) and the product of set A and set B is given by A × B = {(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02), (KA,03)}. It can easily be seen that there will be 9 such pairs in the Cartesian product, since there are 3 elements in each of the sets A and B. This gives us 9 possible codes. Also note that the order in which these elements are paired is crucial. For example, the code (DL, 01) will not be the same as the code (01, DL). As a final illustration, consider the two sets A= {a1, a2} and B = {b1, b2, b3, b4} (Fig 2.3). A × B = {( a1, b1), (a1, b2), (a1, b3), (a1, b4), (a2, b1), (a2, b2), (a2, b3), (a2, b4)}. The 8 ordered pairs thus formed can represent the position of points in the plane if A and B are subsets of the set of real numbers and it is obvious that the point in the position (a1, b2) will be distinct from the point in the position (b2, a1).
Fig 2.3
Remarks (i) Two ordered pairs are equal, if and only if the corresponding first elements are equal and the second elements are also equal.
EXERCISE 2.1
2" !5 1" !x 1. If % # 1, y – & $ % , & , find the values of x and y. 3( '3 3( '3 2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B). 3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G. 4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly. (i) If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}. (ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered
pairs (x, y) such that x ∈ A and y ∈ B. (iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ. If A = {–1, 1}, find A × A × A. If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that (i) A × (B ∩ C) = (A × B) ∩ (A × C). (ii) A × C is a subset of B × D. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.
5. 6. 7. 8. 9.
34
MATHEMATICS
10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of A × A.
2.3 Relations
Consider the two sets P = {a, b, c} and Q = {Ali, Bhanu, Binoy, Chandra, Divya}. The cartesian product of P and Q has 15 ordered pairs which can be listed as P × Q = {(a, Ali), (a,Bhanu), (a, Binoy), ..., (c, Divya)}. We can now obtain a subset of P × Q by introducing a relation R between the first element x and the Fig 2.4 second element y of each ordered pair (x, y) as R= { (x,y): x is the first letter of the name y, x ∈ P, y ∈ Q}. Then R = {(a, Ali), (b, Bhanu), (b, Binoy), (c, Chandra)} A visual representation of this relation R (called an arrow diagram) is shown in Fig 2.4. Definition 2 A relation R from a non-empty set A to a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B. The second element is called the image of the first element. Definition 3 The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain of the relation R. Definition 4 The set of all second elements in a relation R from a set A to a set B is called the range of the relation R. The whole set B is called the codomain of the relation R. Note that range ⊆ codomain. Remarks (i) A relation may be represented algebraically either by the Roster method or by the Set-builder method. (ii) An arrow diagram is a visual representation of a relation.
Example 8 The Fig 2.6 shows a relation between the sets P and Q. Write this relation (i) in set-builder form, (ii) in roster form.
Fig 2.5
What is its domain and range?
Solution It is obvious that the relation R is "x is the square of y". (i) In set-builder form, R = {(x, y): x is the square of y, x ∈ P, y ∈ Q} (ii) In roster form, R = {(9, 3), Fig 2.6 (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)} The domain of this relation is {4, 9, 25}. The range of this relation is {– 2, 2, –3, 3, –5, 5}. Note that the element 1 is not related to any element in set P. The set Q is the codomain of this relation. total number of relations set a !Note The of possible subsets of AthatB.can be definedpfrom an(B)A=toq, set B × If n(A ) = and then is the number
n (A × B) = pq and the total number of relations is 2pq.
Example 9 Let A = {1, 2} and B = {3, 4}. Find the number of relations from A to B. Solution We have, A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}. Since n (A×B ) = 4, the number of subsets of A×B is 24. Therefore, the number of relations from A into B will be 24. Remark A relation R from A to A is also stated as a relation on A.
Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈N}. Depict this relationship using roster form. Write down the domain and the range.
3.
A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form. The Fig2.7 shows a relationship between the sets P and Q. Write this relation (i) in set-builder form (ii) roster form. What is its domain and range? Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b): a , b ∈A, b is exactly divisible by a}. (i) Write R in roster form (ii) Find the domain of R (iii) Find the range of R. Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}. Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form. Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B. Let R be the relation on Z defined by R = {(a,b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R.
4.
5.
Fig 2.7
6. 7. 8. 9.
2.4 Functions In this Section, we study a special type of relation called function. It is one of the most important concepts in mathematics. We can, visualise a function as a rule, which produces new elements out of some given elements. There are many terms such as 'map' or 'mapping' used to denote a function. Definition 5 A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, a function f is a relation from a non-empty set A to a non-empty set B such that the domain of f is A and no two distinct ordered pairs in f have the same first element. If f is a function from A to B and (a, b) ∈ f, then f (a) = b, where b is called the image of a under f and a is called the preimage of b under f.
RELATIONS AND FUNCTIONS
37
The function f from A to B is denoted by f: A " B. Looking at the previous examples, we can easily see that the relation in Example 7 is not a function because the element 6 has no image. Again, the relation in Example 8 is not a function because the elements in the domain are connected to more than one images. Similarly, the relation in Example 9 is also not a function. (Why?) In the examples given below, we will see many more relations some of which are functions and others are not. Example 10 Let N be the set of natural numbers and the relation R be defined on N such that R = {(x, y) : y = 2x, x, y ∈ N}. What is the domain, codomain and range of R? Is this relation a function? Solution The domain of R is the set of natural numbers N. The codomain is also N. The range is the set of even natural numbers. Since every natural number n has one and only one image, this relation is a function. Example 11 Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not? (i) R = {(2,1),(3,1), (4,2)}, (ii) R = {(2,2),(2,4),(3,3), (4,4)} (iii) R = {(1,2),(2,3),(3,4), (4,5), (5,6), (6,7)} Solution (i) (ii) (iii) Since 2, 3, 4 are the elements of domain of R having their unique images, this relation R is a function. Since the same first element 2 corresponds to two different images 2 and 4, this relation is not a function. Since every element has one and only one image, this relation is a function.
2.4.1 Some functions and their graphs (i) Identity function Let R be the set of real numbers. Define the real valued function f : R → R by y = f(x) = x for each x ∈ R. Such a function is called the identity function. Here the domain and range of f are R. The graph is a straight line as shown in Fig 2.8. It passes through the origin.
Fig 2.8
(ii)
Constant function Define the function f: R → R by y = f (x) = c, x ∈ R where c is a constant and each x ∈ R. Here domain of f is R and its range is {c}.
Fig 2.9
RELATIONS AND FUNCTIONS
39
The graph is a line parallel to x-axis. For example, if f(x)=3 for each x∈R, then its graph will be a line as shown in the Fig 2.9. (iii) Polynomial function A function f : R → R is said to be polynomial function if for each x in R, y = f (x) = a0 + a1x + a2x2 + ...+ an xn, where n is a non-negative integer and a0, a1, a2,...,an∈R.
2 x are some examples
2
The functions defined by f(x) = x3 – x2 + 2, and g(x) = x4 +
of polynomial functions, whereas the function h defined by h(x) = x 3 + 2x is not a polynomial function.(Why?) Example 13 Define the function f: R → R by y = f(x) = x2, x ∈ R. Complete the Table given below by using this definition. What is the domain and range of this function? Draw the graph of f. x y = f(x) = x
2
The domain is all real numbers except 0 and its range is also all real numbers except 0. The graph of f is given in Fig 2.12.
FigFig 2.12 2.12
(v) The Modulus function The function f: R→R defined by f(x) = |x| for each x ∈R is called modulus function. For each non-negative value of x, f(x) is equal to x. But for negative values of x, the value of f(x) is the negative of the value of x, i.e.,
* x,x ) 0 f (x) $ + . , x,x - 0
The graph of the modulus function is given in Fig 2.13. (vi) Signum function The function f:R→R defined by
the set {–1, 0, 1}. The graph of the signum function is given by the Fig 2.14.
Fig 2.14
(vii) Greatest integer function The function f: R → R defined by f(x) = [x], x ∈R assumes the value of the greatest integer, less than or equal to x. Such a function is called the greatest integer function. From the definition of [x], we can see that [x] = –1 for –1 ≤ x < 0 [x] = 0 for 0 ≤ x < 1 [x] = 1 for 1 ≤ x < 2 [x] = 2 for 2 ≤ x < 3 and so on. The graph of the function is shown in Fig 2.15.
Fig 2.15
2.4.2 Algebra of real functions In this Section, we shall learn how to add two real functions, subtract a real function from another, multiply a real function by a scalar (here by a scalar we mean a real number), multiply two real functions and divide one real function by another. (i) Addition of two real functions Let f : X → R and g : X → R be any two real functions, where X ⊂ R. Then, we define (f + g): X → R by (f + g) (x) = f (x) + g (x), for all x ∈ X.
" If (a, b) = (x, y), then a = x and b = y. " If n(A) = p and n(B) = q, then n(A × B) = pq. "A×φ=φ " In general, A × B ≠ B × A. " Relation A relation R from a set A to a set B is a subset of the cartesian
product A × B obtained by describing a relationship between the first element x and the second element y of the ordered pairs in A × B.
" The image of an element x under a relation R is given by y, where (x, y) ∈ R, " The domain of R is the set of all first elements of the ordered pairs in a
relation R.
" The range of the relation R is the set of all second elements of the ordered
pairs in a relation R.
" Function A function f from a set A to a set B is a specific type of relation for
which every element x of set A has one and only one image y in set B. We write f: A→B, where f(x) = y.
" A is the domain and B is the codomain of f. " The range of the function is the set of images.
f f ( x) ( x) = g ( x) , x ∈ X, g(x) ≠ 0. g
Historical Note
The word FUNCTION first appears in a Latin manuscript "Methodus tangentium inversa, seu de fuctionibus" written by Gottfried Wilhelm Leibnitz (1646-1716) in 1673; Leibnitz used the word in the non-analytical sense. He considered a function in terms of "mathematical job" – the "employee" being just a curve. On July 5, 1698, Johan Bernoulli, in a letter to Leibnitz, for the first time deliberately assigned a specialised use of the term function in the analytical sense. At the end of that month, Leibnitz replied showing his approval. Function is found in English in 1779 in Chambers' Cylopaedia: "The term function is used in algebra, for an analytical expression any way compounded of a variable quantity, and of numbers, or constant quantities".
—! —
Chapter
3
TRIGONOMETRIC FUNCTIONS
!A mathematician knows how to solve a problem, he can not solve it. – MILNE !
3.1 Introduction
The word 'trigonometry' is derived from the Greek words 'trigon' and 'metron' and it means 'measuring the sides of a triangle'. The subject was originally developed to solve geometric problems involving triangles. It was studied by sea captains for navigation, surveyor to map out the new lands, by engineers and others. Currently, trigonometry is used in many areas such as the science of seismology, designing electric circuits, describing the state of an atom, predicting the heights of tides in the ocean, analysing a musical tone and in many other areas. In earlier classes, we have studied the trigonometric Arya Bhatt (476-550 B.C.) ratios of acute angles as the ratio of the sides of a right angled triangle. We have also studied the trigonometric identities and application of trigonometric ratios in solving the problems related to heights and distances. In this Chapter, we will generalise the concept of trigonometric ratios to trigonometric functions and study their properties.
3.2 Angles
Angle is a measure of rotation of a given ray about its initial point. The original ray is Vertex
Fig 3.1
50
MATHEMATICS
called the initial side and the final position of the ray after rotation is called the terminal side of the angle. The point of rotation is called the vertex. If the direction of rotation is anticlockwise, the angle is said to be positive and if the direction of rotation is clockwise, then the angle is negative (Fig 3.1). The measure of an angle is the amount of rotation performed to get the terminal side from the initial side. There are several units for Fig 3.2 measuring angles. The definition of an angle suggests a unit, viz. one complete revolution from the position of the initial side as indicated in Fig 3.2. This is often convenient for large angles. For example, we can say that a rapidly spinning wheel is making an angle of say 15 revolution per second. We shall describe two other units of measurement of an angle which are most commonly used, viz. degree measure and radian measure.
! 1 " 3.2.1 Degree measure If a rotation from the initial side to terminal side is # $ of % 360 & a revolution, the angle is said to have a measure of one degree, written as 1°. A degree is divided into 60 minutes, and a minute is divided into 60 seconds . One sixtieth of a degree is called a minute, written as 1′, and one sixtieth of a minute is called a second, written as 1″. Thus, 1° = 60′, 1′ = 60″ Some of the angles whose measures are 360°,180°, 270°, 420°, – 30°, – 420° are shown in Fig 3.3.
th
Fig 3.3
TRIGONOMETRIC FUNCTIONS
51
3.2.2 Radian measure There is another unit for measurement of an angle, called
the radian measure. Angle subtended at the centre by an arc of length 1 unit in a unit circle (circle of radius 1 unit) is said to have a measure of 1 radian. In the Fig 3.4(i) to (iv), OA is the initial side and OB is the terminal side. The figures show the angles whose measures are 1 radian, –1 radian, 1
1 1 radian and –1 radian. 2 2
(i)
(ii)
(iii)
(iv)
Fig 3.4 (i) to (iv)
We know that the circumference of a circle of radius 1 unit is 2π. Thus, one complete revolution of the initial side subtends an angle of 2π radian. More generally, in a circle of radius r, an arc of length r will subtend an angle of 1 radian. It is well-known that equal arcs of a circle subtend equal angle at the centre. Since in a circle of radius r, an arc of length r subtends an angle whose measure is 1 l radian, an arc of length l will subtend an angle whose measure is radian. Thus, if in r a circle of radius r, an arc of length l subtends an angle θ radian at the centre, we have θ =
l or l = r θ. r
52
MATHEMATICS
3.2.3 Relation between radian and real numbers
Consider the unit circle with centre O. Let A be any point on the circle. Consider OA as initial side of an angle. Then the length of an arc of the circle will give the radian measure of the angle which the arc will subtend at the centre of the circle. Consider the line PAQ which is tangent to the circle at A. Let the point A represent the real number zero, AP represents positive real number and AQ represents negative real numbers (Fig 3.5). If we rope the line AP in the anticlockwise direction along the circle, and AQ in the clockwise direction, then every real number will correspond to a radian measure and conversely. Thus, radian measures and real numbers can be considered as one and the same.
P 2
1 1 A 0 − −1 − −2 Q
O
Fig 3.5
3.2.4 Relation between degree and radian Since a circle subtends at the centre an angle whose radian measure is 2π and its degree measure is 360°, it follows that
2π radian = 360° or π radian = 180° The above relation enables us to express a radian measure in terms of degree measure and a degree measure in terms of radian measure. Using approximate value of π as
22 , we have 7
1 radian =
180' = 57° 16′ approximately. !
Also
1° =
! radian = 0.01746 radian approximately. 180
The relation between degree measures and radian measure of some common angles are given in the following table: Degree Radian 30° 45° 60° 90° 180°
!
270°
360°
2!
! 6
! 4
! 3
! 2
3! 2
TRIGONOMETRIC FUNCTIONS
53
Notational Convention
Since angles are measured either in degrees or in radians, we adopt the convention that whenever we write angle θ°, we mean the angle whose degree measure is θ and whenever we write angle β, we mean the angle whose radian measure is β. Note that when an angle is expressed in radians, the word 'radian' is frequently
! ! ( 45' are written with the understanding that π and 4 4 are radian measures. Thus, we can say that
omitted. Thus, ! ( 180' and Radian measure = Degree measure =
Example 3 Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm (use ! (
22 ). 7
54
MATHEMATICS
Solution Here l = 37.4 cm and θ = 60° = Hence, by r = r=
60! ! radian = 180 3
l , we have " 37.4×3 37.4×3×7 = = 35.7 cm ! 22
Example 4 The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (Use π = 3.14). Solution In 60 minutes, the minute hand of a watch completes one revolution. Therefore, in 40 minutes, the minute hand turns through or
Example 5 If the arcs of the same lengths in two circles subtend angles 65°and 110° at the centre, find the ratio of their radii. Solution Let r1 and r2 be the radii of the two circles. Given that θ1 = 65° = and
! 13! ) 65 = radian 180 36 ! 22! ) 110 = radian 180 36
θ2 = 110° =
Let l be the length of each of the arc. Then l = r1θ1 = r2θ2, which gives
5! 7! (iv) 3 6 A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? Find the degree measure of the angle subtended at the centre of a circle of 22 ). 7 In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii. Find the angle in radian through which a pendulum swings if its length is 75 cm and th e tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm
radius 100 cm by an arc of length 22 cm (Use ! (
11 16
5. 6. 7.
3.3 Trigonometric Functions
In earlier classes, we have studied trigonometric ratios for acute angles as the ratio of sides of a right angled triangle. We will now extend the definition of trigonometric ratios to any angle in terms of radian measure and study them as trigonometric functions. Consider a unit circle with centre at origin of the coordinate axes. Let P (a, b) be any point on the circle with angle AOP = x radian, i.e., length of arc AP = x (Fig 3.6). We define cos x = a and sin x = b Since ∆OMP is a right triangle, we have OM2 + MP2 = OP2 or a2 + b2 = 1 Thus, for every point on the unit circle, we have a2 + b2 = 1 or cos2 x + sin2 x = 1 Since one complete revolution subtends an angle of 2π radian at the centre of the circle, ∠AOB =
In earlier classes, we have discussed the values of trigonometric ratios for 0°, 30°, 45°, 60° and 90°. The values of trigonometric functions for these angles are same as that of trigonometric ratios studied in earlier classes. Thus, we have the following table: 0° sin cos tan 0 1 0
! 6
! 4
1 2 1 2
! 3
3 2
! 2
1 0 not defined
!
3! 2
–1 0 not defined
2!
1 2
3 2 1 3
0 –1 0
0 1 0
1 2
3
1
The values of cosec x, sec x and cot x are the reciprocal of the values of sin x, cos x and tan x, respectively. 3.3.1 Sign of trigonometric functions Let P (a, b) be a point on the unit circle with centre at the origin such that ∠AOP = x. If ∠AOQ = – x, then the coordinates of the point Q will be (a, –b) (Fig 3.7). Therefore cos (– x) = cos x and sin (– x) = – sin x Since for every point P (a, b) on the unit circle, – 1 ≤ a ≤ 1 and
Fig 3.7
3.3.2 Domain and range of trigonometric functions From the definition of sine
and cosine functions, we observe that they are defined for all real numbers. Further, we observe that for each real number x, – 1 ≤ sin x ≤ 1 and – 1 ≤ cos x ≤ 1 Thus, domain of y = sin x and y = cos x is the set of all real numbers and range is the interval [–1, 1], i.e., – 1 ≤ y ≤ 1.
! , n ∈ Z} and range is the set of all real numbers. The domain of 2 y = cot x is the set {x : x ∈ R and x ≠ n π, n ∈ Z} and the range is the set of all real numbers.
x ≠ (2n + 1) We further observe that in the first quadrant, as x increases from 0 to increases from 0 to 1, as x increases from third quadrant, as x increases from π to
decreases from 1 to 0 decreases from 0 to – 1 increases from –1 to 0 increases from 0 to 1 increases from 0 to ∞ increases from –∞to 0 increases from 0 to ∞ increases from –∞to 0
decreases from ∞ to 0 decreases from 0 to–∞ decreases from ∞ to 0 decreases from 0to –∞ increases from 1 to ∞ increases from –∞to–1 decreases from –1to–∞ decreases from ∞ to 1 increases from –∞to–1 decreases from–1to–∞
cosec decreases from ∞ to 1 increases from 1 to ∞
Similarly, we can discuss the behaviour of other trigonometric functions. In fact, we have the following table: Remark In the above table, the statement tan x increases from 0 to ∞ (infinity) for 0<x<
3! , 2π) and assumes arbitrarily large negative values as 2 x approaches to 2π. The symbols ∞ and – ∞ simply specify certain types of behaviour
of functions and variables. We have already seen that values of sin x and cos x repeats after an interval of 2π. Hence, values of cosec x and sec x will also repeat after an interval of 2π. We
Fig 3.8
Fig 3.9
Fig 3.10
Fig 3.11
TRIGONOMETRIC FUNCTIONS
61
Fig 3.12
Fig 3.13
shall see in the next section that tan (π + x) = tan x. Hence, values of tan x will repeat after an interval of π. Since cot x is reciprocal of tan x, its values will also repeat after an interval of π. Using this knowledge and behaviour of trigonometic functions, we can sketch the graph of these functions. The graph of these functions are given above: Example 6 If cos x = – 3 , x lies in the third quadrant, find the values of other five 5 trigonometric functions. Solution Since cos x = * Now or Hence
3.4 Trigonometric Functions of Sum and Difference of Two Angles
In this Section, we shall derive expressions for trigonometric functions of the sum and difference of two numbers (angles) and related expressions. The basic results in this connection are called trigonometric identities. We have seen that 1. sin (– x) = – sin x 2. cos (– x) = cos x We shall now prove some more results:
3.5 Trigonometric Equations
Equations involving trigonometric functions of a variable are called trigonometric equations. In this Section, we shall find the solutions of such equations. We have already learnt that the values of sin x and cos x repeat after an interval of 2π and the values of tan x repeat after an interval of π. The solutions of a trigonometric equation for which 0 ≤ x < 2π are called principal solutions. The expression involving integer 'n' which gives all solutions of a trigonometric equation is called the general solution. We shall use 'Z' to denote the set of integers. The following examples will be helpful in solving trigonometric equations: Example 18 Find the principal solutions of the equation sin x ( Solution We know that, sin
3 2
Historical Note
The. All this knowledge first went from India to middle-east and the main contribution of the siddhantas (Sanskrit astronomical works) to the history of mathematics. Bhaskara I (about 600 A.D.) gave formulae to find the values of sine functions for angles more than 90°. A sixteenth century Malayalam work Yuktibhasa (period) contains a proof for the expansion of sin (A + B). Exact expressin for sines or cosines of 18°, 36°, 54°, 72°, etc., are given by Bhaskara II. The symbols sin–1 x, cos–1 x, etc., for arc sin x, arc cos x, etc., were suggested by the astronomer Sir John F.W. Hersehel (1813 A.D.) The names height, and comparing the ratios:
H h ( = tan (sun's altitude) S s
Thales is also said to have calculated the distance of a ship at sea through the proportionality of sides of similar triangles. Problems on height and distance using the similarity property are also found in ancient Indian works.
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Chapter
4
PRINCIPLE OF MATHEMATICAL INDUCTION
!Analysis and natural philosopy owe their most important discoveries to
this fruitful means, which is called induction. Newton was indebted to it for his theorem of the binomial and the principle of universal gravity. – LAPLACE !
4.1 Introduction
One key basis for mathematical thinking is deductive reasoning. An informal, and example of deductive reasoning, borrowed from the study of logic, is an argument expressed in three statements: (a) Socrates is a man. (b) All men are mortal, therefore, (c) Socrates is mortal. If statements (a) and (b) are true, then the truth of (c) is established. To make this simple mathematical example, we could write: (i) Eight is divisible by two. G . Peano (ii) Any number divisible by two is an even number, (1858-1932) therefore, (iii) Eight is an even number. Thus, deduction in a nutshell is given a statement to be proven, often called a conjecture or a theorem in mathematics, valid deductive steps are derived and a proof may or may not be established, i.e., deduction is the application of a general case to a particular case. In contrast to deduction, inductive reasoning depends on working with each case, and developing a conjecture by observing incidences till we have observed each and every case. It is frequently used in mathematics and is a key aspect of scientific reasoning, where collecting and analysing data is the norm. Thus, in simple language, we can say the word induction means the generalisation from particular cases or facts.
PRINCIPLE OF MATHEMATICAL INDUCTION
87
In algebra or in other discipline of mathematics, there are certain results or statements that are formulated in terms of n, where n is a positive integer. To prove such statements the well-suited principle that is used–based on the specific technique, is known as the principle of mathematical induction.
4.2 Motivation
In mathematics, we use a form of complete induction called mathematical induction. To understand the basic principles of mathematical induction, suppose a set of thin rectangular tiles are placed on one end, as shown in Fig 4.1.
Fig 4.1
When the first tile is pushed in the indicated direction, all the tiles will fall. To be absolutely sure that all the tiles will fall, it is sufficient to know that (a) The first tile falls, and (b) In the event that any tile falls its successor necessarily falls. This is the underlying principle of mathematical induction. We know, the set of natural numbers N is a special ordered subset of the real numbers. In fact, N is the smallest subset of R with the following property: A set S is said to be an inductive set if 1∈ S and x + 1 ∈ S whenever x ∈ S. Since N is the smallest subset of R which is an inductive set, it follows that any subset of R that is an inductive set must contain N. Illustration Suppose we wish to find the formula for the sum of positive integers 1, 2, 3,...,n, that is, a formula which will give the value of 1 + 2 + 3 when n = 3, the value 1 + 2 + 3 + 4, when n = 4 and so on and suppose that in some manner we are led to believe that the formula 1 + 2 + 3+...+ n =
n ( n ! 1) is the correct one. 2
How can this formula actually be proved? We can, of course, verify the statement for as many positive integral values of n as we like, but this process will not prove the formula for all values of n. What is needed is some kind of chain reaction which will
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have the effect that once the formula is proved for a particular positive integer the formula will automatically follow for the next positive integer and the next indefinitely. Such a reaction may be considered as produced by the method of mathematical induction.
4.3 The Principle of Mathematical Induction
Suppose there is a given statement P(n) involving the natural number n such that (i) The statement is true for n = 1, i.e., P(1) is true, and (ii) If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., truth of P(k) implies the truth of P (k + 1). Then, P(n) is true for all natural numbers n. Property (i) is simply a statement of fact. There may be situations when a statement is true for all n ≥ 4. In this case, step 1 will start from n = 4 and we shall verify the result for n = 4, i.e., P(4). Property (ii) is a conditional property. It does not assert that the given statement is true for n = k, but only that if it is true for n = k, then it is also true for n = k +1. So, to prove that the property holds , only prove that conditional proposition: If the statement is true for n = k, then it is also true for n = k + 1. This is sometimes referred to as the inductive step. The assumption that the given statement is true for n = k in this inductive step is called the inductive hypothesis. For example, frequently in mathematics, a formula will be discovered that appears to fit a pattern like 1 = 12 =1 4 = 22 = 1 + 3 9 = 32 = 1 + 3 + 5 16 = 42 = 1 + 3 + 5 + 7, etc. It is worth to be noted that the sum of the first two odd natural numbers is the square of second natural number, sum of the first three odd natural numbers is the square of third natural number and so on.Thus, from this pattern it appears that 1 + 3 + 5 + 7 + ... + (2n – 1) = n2 , i.e, the sum of the first n odd natural numbers is the square of n. Let us write P(n): 1 + 3 + 5 + 7 + ... + (2n – 1) = n2. We wish to prove that P(n) is true for all n. The first step in a proof that uses mathematical induction is to prove that P (1) is true. This step is called the basic step. Obviously 1 = 12, i.e., P(1) is true. The next step is called the inductive step. Here, we suppose that P (k) is true for some
! One key basis for mathematical thinking is deductive reasoning. In contrast to
deduction, inductive reasoning depends on working with different cases and developing a conjective by observing incidences till we have observed each and every case. Thus, in simple language we can say the word 'induction' means the generalisation from particular cases or facts. ! The principle of mathematical induction is one such tool which can be used to prove a wide variety of mathematical statements. Each such statement is assumed as P(n) associated with positive integer n, for which the correctness
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for the case n = 1 is examined. Then assuming the truth of P(k) for some positive integer k, the truth of P (k+1) is established.
Historical Note
Unlike other concepts and methods, proof by mathematical induction is not the invention of a particular individual at a fixed moment. It is said that the principle of mathematical induction was known by the Phythagoreans. The French mathematician Blaise Pascal is credited with the origin of the principle of mathematical induction. The name induction was used by the English mathematician John Wallis. Later the principle was employed to provide a proof of the binomial theorem. De Morgan contributed many accomplishments in the field of mathematics on many different subjects. He was the first person to define and name "mathematical induction" and developed De Morgan's rule to determine the convergence of a mathematical series. G. Peano undertook the task of deducing the properties of natural numbers from a set of explicitly stated assumptions, now known as Peano's axioms.The principle of mathematical induction is a restatement of one of the Peano's axioms.
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Chapter
5
COMPLEX NUMBERS AND QUADRATIC EQUATIONS
!Mathematics is the Queen of Sciences and Arithmetic is the Queen of Mathematics. – GAUSS !
5.1 Introduction
In earlier classes, we have studied linear equations in one and two variables and quadratic equations in one variable. We have seen that the equation x2 + 1 = 0 has no real solution as x2 + 1 = 0 gives x2 = – 1 and square of every real number is non-negative. So, we need to extend the real number system to a larger system so that we can find the solution of the equation x2 = – 1. In fact, the main objective is to solve the equation ax2 + bx + c = 0, where D = b2 – 4ac < 0, which is not possible in the system of real numbers.
5.2 Complex Numbers
2
W. R. Hamilton (1805-1865)
Let us denote !1 by the symbol i. Then, we have i " !1 . This means that i is a solution of the equation x2 + 1 = 0. A number of the form a + ib, where a and b are real numbers, is defined to be a
# !1 $ complex number. For example, 2 + i3, (– 1) + i 3 , 4 % i & ' are complex numbers. ( 11 ) For the complex number z = a + ib, a is called the real part, denoted by Re z and b is called the imaginary part denoted by Im z of the complex number z. For example, if z = 2 + i5, then Re z = 2 and Im z = 5. Two complex numbers z1 = a + ib and z2 = c + id are equal if a = c and b = d.
5.3.3 Multiplication of two complex numbers Let z1 = a + ib and z2 = c + id be any two complex numbers. Then, the product z1 z2 is defined as follows: z1 z2 = (ac – bd) + i(ad + bc) For example, (3 + i5) (2 + i6) = (3 × 2 – 5 × 6) + i(3 × 6 + 5 × 2) = – 24 + i28 The multiplication of complex numbers possesses the following properties, which we state without proofs. (i) The closure law The product of two complex numbers is a complex number, the product z1 z2 is a complex number for all complex numbers z1 and z2. (ii) The commutative law For any two complex numbers z1 and z2, z1 z2 = z2 z1 . (iii) The associative law For any three complex numbers z 1 , z 2 , z 3 , (z1 z2) z3 = z1 (z2 z3). (iv) The existence of multiplicative identity There exists the complex number 1 + i 0 (denoted as 1), called the multiplicative identity such that z.1 = z, for every complex number z. (v) The existence of multiplicative inverse For every non-zero complex number z = a + ib or a + bi(a ≠ 0, b ≠ 0), we have the complex number
a –b 1 %i 2 or z–1 ), called the multiplicative inverse 2 2 (denoted by a %b a %b z of z such that
2
5.3.6 The square roots of a negative real number Note that i2 = –1 and ( – i)2 = i2 = – 1 Therefore, the square roots of – 1 are i, – i. However, by the symbol ! , we would mean i only. Now, we can see that i and –i both are the solutions of the equation x2 + 1 = 0 or x2 = –1. Similarly
Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13. 11. 4 – 3i 13. – i 5 % 3i 14. Express the following expression in the form of a + ib : 12.
,3 % i 5 - ,3 ! i 5 , 3 % 2 i- ! , 3 ! i 2 5.5 Argand Plane and Polar Representation
We already know that corresponding to each ordered pair of real numbers (x, y), we get a unique point in the XYplane and vice-versa with reference to a set of mutually perpendicular lines known as the x-axis and the y-axis. The complex number x + iy which corresponds to the ordered pair (x, y) can be represented geometrically as the unique point P(x, y) in the XY-plane and vice-versa. Some complex numbers such as 2 + 4i, – 2 + 3i, 0 + 1i, 2 + 0i, – 5 –2i and Fig 5.1 1 – 2i which correspond to the ordered pairs (2, 4), ( – 2, 3), (0, 1), (2, 0), ( –5, –2), and (1, – 2), respectively, have been represented geometrically by the points A, B, C, D, E, and F, respectively in the Fig 5.1. The plane having a complex number assigned to each of its point is called the complex plane or the Argand plane.
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105
Obviously, in the Argand plane, the modulus of the complex number x + iy =
x 2 % y 2 is the distance between the point P(x, y) to the origin O (0, 0)
(Fig 5.2). The points on the x-axis corresponds to the complex numbers of the form a + i 0 and the points on the y-axis corresponds to the complex numbers of the form
Fig 5.2
0 + i b. The x-axis and y-axis in the Argand plane are called, respectively, the real axis and the imaginary axis. The representation of a complex number z = x + iy and its conjugate z = x – iy in the Argand plane are, respectively, the points P (x, y) and Q (x, – y). Geometrically, the point (x, – y) is the mirror image of the point (x, y) on the real axis (Fig 5.3).
Fig 5.3
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5.5.1 Polar representation of a complex number Let the point P represent the nonzero complex number z = x + iy. Let the directed line segment OP be of length r and θ be the angle which OP makes with the positive direction of x-axis (Fig 5.4). We may note that the point P is uniquely determined by the ordered pair of real numbers (r, θ), called the polar coordinates of the point P. We consider the origin as the pole and the positive Fig 5.4 direction of the x axis as the initial line. We have, x = r cos θ, y = r sin θ and therefore, z = r (cos θ + i sin θ). The latter is said to be the polar form of the complex number. Here r " x 2 % y 2 " z is the modus of z and θ is called the argument (or amplitude) of z which is denoted by arg z. For any complex number z ≠ 0, there corresponds only one value of θ in 0 ≤ θ < 2π. However, any other interval of length 2π, for example – π < θ ≤ π, can be such an interval.We shall take the value of θ such that – π < θ ≤ π, called principal argument of z and is denoted by arg z, unless specified otherwise. (Figs. 5.5 and 5.6)
EXERCISE 5.2
Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2. 1. z = – 1 – i 2. z = – +i Convert each of the complex numbers given in Exercises 3 to 8 in the polar form: 3. 1 – i 4. – 1 + i 5. – 1 – i 6. – 3 7. +i 8. i
5.6 Quadratic Equations
We are already familiar with the quadratic equations and have solved them in the set of real numbers in the cases where discriminant is non-negative, i.e., ≥ 0, Let us consider the following quadratic equation:
ax 2 % bx % c " 0 with real coefficients a, b, c and a ≠ 0.
Also, let us assume that the b2 – 4ac < 0. Now, we know that we can find the square root of negative real numbers in the set of complex numbers. Therefore, the solutions to the above equation are available in the set of complex numbers which are given by x=
!b 2 b 2 ! 4ac !b 2 4ac ! b 2 i " 2a 2a
At this point of time, would know as to how many !Note an equation have? Insomeregard, be interested totheorem known as the roots does this the following Fundamental theorem of Algebra is stated below (without proof). "A polynomial equation has at least one root." As a consequence of this theorem, the following result, which is of immense importance, is arrived at: "A polynomial equation of degree n has n roots." Example 9 Solve x2 + 2 = 0 Solution We have, x2 + 2 = 0 or x2 = – 2 i.e., x = 2 !2 = 2 2 i
Historical Note The fact that square root of a negative number does not exist in the real number system was recognised by the Greeks. But the credit goes to the Indian mathematician Mahavira (850 A.D.) who first stated this difficulty clearly. "He mentions in his work 'Ganitasara Sangraha' as in the nature of things a negative (quantity) is not a square (quantity)', it has, therefore, no square root". Bhaskara, another Indian mathematician, also writes in his work Bijaganita, written in 1150. A.D. "There is no square root of a negative quantity, for it is not a square." Cardan (1545 A.D.) considered the problem of solving x + y = 10, xy = 40. He obtained x = 5 + !15 and y = 5 – !15 as the solution of it, which was discarded by him by saying that these numbers are 'useless'. Albert Girard (about 1625 A.D.) accepted square root of negative numbers and said that this will enable us to get as many roots as the degree of the polynomial equation. Euler was the first to introduce the symbol i for !1 and W.R. Hamilton (about 1830 A.D.) regarded the complex number a + ib as an ordered pair of real numbers (a, b) thus giving it a purely mathematical definition and avoiding use of the so called 'imaginary numbers'.
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Chapter
6
LINEAR INEQUALITIES
!Mathematics is the art of saying many things in many different ways. – MAXWELL!
6.1 Introduction
In earlier classes, we have studied equations in one variable and two variables and also solved some statement problems by translating them in the form of equations. Now a natural question arises: 'Is it always possible to translate a statement problem in the form of an equation? For example, the height of all the students in your class is less than 160 cm. Your classroom can occupy atmost 60 tables or chairs or both. Here we get certain statements involving a sign '<' (less than), '>' (greater than), '≤' (less than or equal) and ≥ (greater than or equal) which are known as inequalities. In this Chapter, we will study linear inequalities in one and two variables. The study of inequalities is very useful in solving problems in the field of science, mathematics, statistics, optimisation problems, economics, psychology, etc.
6.2 Inequalities
Let us consider the following situations: (i) Ravi goes to market with Rs 200 to buy rice, which is available in packets of 1kg. The price of one packet of rice is Rs 30. If x denotes the number of packets of rice, which he buys, then the total amount spent by him is Rs 30x. Since, he has to buy rice in packets only, he may not be able to spend the entire amount of Rs 200. (Why?) Hence 30x < 200 ... (1) Clearly the statement (i) is not an equation as it does not involve the sign of equality. (ii) Reshma has Rs 120 and wants to buy some registers and pens. The cost of one register is Rs 40 and that of a pen is Rs 20. In this case, if x denotes the number of registers and y, the number of pens which Reshma buys, then the total amount spent by her is Rs (40x + 20y) and we have 40x + 20y ≤ 120 ... (2)
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Since in this case the total amount spent may be upto Rs 120. Note that the statement (2) consists of two statements and 40x + 20y < 120 40x + 20y = 120 ... (3) ... (4)
Statement (3) is not an equation, i.e., it is an inequality while statement (4) is an equation.
Inequalities (5), (6), (9), (10) and (14) are strict inequalities while inequalities (7), (8), (11), (12), and (13) are slack inequalities. Inequalities from (5) to (8) are linear inequalities in one variable x when a ≠ 0, while inequalities from (9) to (12) are linear inequalities in two variables x and y when a ≠ 0, b ≠ 0. Inequalities (13) and (14) are not linear (in fact, these are quadratic inequalities in one variable x when a ≠ 0). In this Chapter, we shall confine ourselves to the study of linear inequalities in one and two variables only.
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6.3 Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation
Let us consider the inequality (1) of Section 6.2, viz, 30x < 200 Note that here x denotes the number of packets of rice. Obviously, x cannot be a negative integer or a fraction. Left hand side (L.H.S.) of this inequality is 30x and right hand side (RHS) is 200. Therefore, we have For x = 0, L.H.S. = 30 (0) = 0 < 200 (R.H.S.), which is true. For x = 1, L.H.S. = 30 (1) = 30 < 200 (R.H.S.), which is true. For x = 2, L.H.S. = 30 (2) = 60 < 200, which is true. For x = 3, L.H.S. = 30 (3) = 90 < 200, which is true. For x = 4, L.H.S. = 30 (4) = 120 < 200, which is true. For x = 5, L.H.S. = 30 (5) = 150 < 200, which is true. For x = 6, L.H.S. = 30 (6) = 180 < 200, which is true. For x = 7, L.H.S. = 30 (7) = 210 < 200, which is false. In the above situation, we find that the values of x, which makes the above inequality a true statement, are 0,1,2,3,4,5,6. These values of x, which make above inequality a true statement, are called solutions of inequality and the set {0,1,2,3,4,5,6} is called its solution set. Thus, any solution of an inequality in one variable is a value of the variable which makes it a true statement. We have found the solutions of the above inequality by trial and error method which is not very efficient. Obviously, this method is time consuming and sometimes not feasible. We must have some better or systematic techniques for solving inequalities. Before that we should go through some more properties of numerical inequalities and follow them as rules while solving the inequalities. You will recall that while solving linear equations, we followed the following rules: Rule 1 Equal numbers may be added to (or subtracted from) both sides of an equation. Rule 2 Both sides of an equation may be multiplied (or divided) by the same non-zero number. In the case of solving inequalities, we again follow the same rules except with a difference that in Rule 2, the sign of inequality is reversed (i.e., '<' becomes '>', ≤' becomes '≥' and so on) whenever we multiply (or divide) both sides of an inequality by a negative number. It is evident from the facts that 3 > 2 while – 3 < – 2, – 8 < – 7 while (– 8) (– 2) > (– 7) (– 2) , i.e., 16 > 14.
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119
Thus, we state the following rules for solving an inequality: Rule 1 Equal numbers may be added to (or subtracted from) both sides of an inequality without affecting the sign of inequality. Rule 2 Both sides of an inequality can be multiplied (or divided) by the same positive number. But when both sides are multiplied or divided by a negative number, then the sign of inequality is reversed. Now, let us consider some examples. Example 1 Solve 30 x < 200 when (i) x is a natural number, Solution We are given 30 x < 200 or (ii) x is an integer.
30 x 200 ! (Rule 2), i.e., x < 20 / 3. 30 30 (i) When x is a natural number, in this case the following values of x make the statement true. 1, 2, 3, 4, 5, 6. The solution set of the inequality is {1,2,3,4,5,6}. (ii) When x is an integer, the solutions of the given inequality are ..., – 3, –2, –1, 0, 1, 2, 3, 4, 5, 6 The solution set of the inequality is {...,–3, –2,–1, 0, 1, 2, 3, 4, 5, 6}
Example 2 Solve 5x – 3 < 3x +1 when (i) x is an integer, (ii) x is a real number. Solution We have, 5x –3 < 3x + 1 or 5x –3 + 3 < 3x +1 +3 (Rule 1) or 5x < 3x +4 or 5x – 3x < 3x + 4 – 3x (Rule 1) or 2x < 4 or x < 2 (Rule 2) (i) When x is an integer, the solutions of the given inequality are ..., – 4, – 3, – 2, – 1, 0, 1 (ii) When x is a real number, the solutions of the inequality are given by x < 2, i.e., all real numbers x which are less than 2. Therefore, the solution set of the inequality is x ∈ (– ∞, 2). We have considered solutions of inequalities in the set of natural numbers, set of integers and in the set of real numbers. Henceforth, unless stated otherwise, we shall solve the inequalities in this Chapter in the set of real numbers.
Example 7 The marks obtained by a student of Class XI in first and second terminal examination are 62 and 48, respectively. Find the number of minimum marks he should get in the annual examination to have an average of at least 60 marks. Solution Let x be the marks obtained by student in the annual examination. Then
62 $ 48 $ x % 60 3 or 110 + x ≥ 180 or x ≥ 70 Thus, the student must obtain a minimum of 70 marks to get an average of at least 60 marks.
Example 8 Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40. Solution Let x be the smaller of the two consecutive odd natural number, so that the other one is x +2. Then, we should have x > 10 and x + ( x + 2) < 40 Solving (2), we get 2x + 2 < 40 i.e., x < 19 From (1) and (3), we get 10 < x < 19 ... (1) ... (2)
... (3)
Since x is an odd number, x can take the values 11, 13, 15, and 17. So, the required possible pairs will be (11, 13), (13, 15), (15, 17), (17, 19)
x (5 x # 2) (7 x # 3) ! # 2 3 5 Ravi obtained 70 and 75 marks in first two unit test. Find the number if minimum marks he should get in the third test to have an average of at least 60 marks. To receive Grade 'A' in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita's marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade 'A' in the course. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
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25. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side. 26. A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second? [Hint: If x is the length of the shortest board, then x , (x + 3) and 2x are the lengths of the second and third piece, respectively. Thus, x + (x + 3) + 2x ≤ 91 and 2x ≥ (x + 3) + 5].
6.4 Graphical Solution of Linear Inequalities in Two Variables
In earlier section, we have seen that a graph of an inequality in one variable is a visual representation and is a convenient way to represent the solutions of the inequality. Now, we will discuss graph of a linear inequality in two variables. We know that a line divides the Cartesian plane into two parts. Each part is known as a half plane. A vertical line will divide the plane in left and right half planes and a non-vertical line will divide the plane into lower and upper half planes (Figs. 6.3 and 6.4).
Fig 6.3
Fig 6.4
A point in the Cartesian plane will either lie on a line or will lie in either of the half planes I or II. We shall now examine the relationship, if any, of the points in the plane and the inequalities ax + by < c or ax + by > c. Let us consider the line ax + by = c, a ≠ 0, b ≠ 0 ... (1)
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There are three possibilities namely: (i) ax + by = c (ii) ax + by > c (iii) ax + by < c. In case (i), clearly, all points (x, y) satisfying (i) lie on the line it represents and conversely. Consider case (ii), let us first assume that b > 0. Consider a point P (α,β) on the line ax + by = c, b > 0, so that aα + bβ = c.Take an arbitrary point Q (α , γ) in the half plane II (Fig 6.5). Now, from Fig 6.5, we interpret, γ > β (Why?) or b - > bβ or aα + b γ > aα + bβ (Why?) or aα + b γ > c i.e., Q(α, - ) satisfies the inequality ax + by > c. Fig 6.5 Thus, all the points lying in the half plane II above the line ax + by = c satisfies the inequality ax + by > c. Conversely, let (α, β) be a point on line ax + by = c and an arbitrary point Q(α, γ) satisfying ax + by > c so that aα + bγ > c aα + b γ > aα + bβ (Why?) . γ>β (as b > 0) . This means that the point (α, - ) lies in the half plane II. Thus, any point in the half plane II satisfies ax + by > c, and conversely any point satisfying the inequality ax + by > c lies in half plane II. In case b < 0, we can similarly prove that any point satisfying ax + by > c lies in the half plane I, and conversely. Hence, we deduce that all points satisfying ax + by > c lies in one of the half planes II or I according as b > 0 or b < 0, and conversely. Thus, graph of the inequality ax + by > c will be one of the half plane (called solution region) and represented by shading in the corresponding half plane. 1 The !Noteregion. region containing all the solutions of an inequality is called the solution 2. In order to identify the half plane represented by an inequality, it is just sufficient to take any point (a, b) (not online) and check whether it satisfies the inequality or not. If it satisfies, then the inequality represents the half plane and shade the region
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125
which contains the point, otherwise, the inequality represents that half plane which does not contain the point within it. For convenience, the point (0, 0) is preferred. 3. If an inequality is of the type ax + by ≥ c or ax + by ≤ c, then the points on the line ax + by = c are also included in the solution region. So draw a dark line in the solution region. 4. If an inequality is of the form ax + by > c or ax + by < c, then the points on the line ax + by = c are not to be included in the solution region. So draw a broken or dotted line in the solution region. In Section 6.2, we obtained the following linear inequalities in two variables x and y: 40x + 20y ≤ 120 ... (1) while translating the word problem of purchasing of registers and pens by Reshma. Let us now solve this inequality keeping in mind that x and y can be only whole numbers, since the number of articles cannot be a fraction or a negative number. In this case, we find the pairs of values of x and y, which make the statement (1) true. In fact, the set of such pairs will be the solution set of the inequality (1). To start with, let x = 0. Then L.H.S. of (1) is 40x + 20y = 40 (0) + 20y = 20y. Thus, we have 20y ≤ 120 or y ≤ 6 ... (2) For x = 0, the corresponding values of y can be 0, 1, 2, 3, 4, 5, 6 only. In this case, the solutions of (1) are (0, 0), (0, 1), (0,2), (0,3), (0,4), (0, 5) and (0, 6). Similarly, other solutions of (1), when x = 1, 2 and 3 are: (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (2, 0), (2, 1), (2, 2), (3, 0) This is shown in Fig 6.6. Let us now extend the domain of x and y from whole numbers to real numbers, and see what will be the solutions of (1) in this case. You will see that the graphical method of solution will be very convenient in this case. For this purpose, let us consider the (corresponding) equation and draw its graph. 40x + 20y = 120 ... (3) In order to draw the graph of the inequality (1), we take one point say (0, 0), in half plane I and check whether values of x and y satisfy the Fig 6.6 inequality or not.
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We observe that x = 0, y = 0 satisfy the inequality. Thus, we say that the half plane I is the graph (Fig 6.7) of the inequality. Since the points on the line also satisfy the inequality (1) above, the line is also a part of the graph. Thus, the graph of the given inequality is half plane I including the line itself. Clearly half plane II is not the part of the graph. Hence, solutions of inequality (1) will consist of all the points of its graph (half plane I including the line). We shall now consider some examples to explain the above procedure for solving a linear inequality involving two variables. Example 9 Solve 3x + 2y > 6 graphically.
Fig 6.7
Solution Graph of 3x + 2y = 6 is given as dotted line in the Fig 6.8. This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes (Fig 6.8) and determine if this point satisfies the given inequality, we note that 3 (0) + 2 (0) > 6 or 0 > 6 , which is false. Hence, half plane I is not the solution region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality. In other words, the shaded half plane II Fig 6.8 excluding the points on the line is the solution region of the inequality. Example 10 Solve 3x – 6 ≥ 0 graphically in two dimensional plane. Solution Graph of 3x – 6 = 0 is given in the Fig 6.9. We select a point, say (0, 0) and substituting it in given inequality, we see that: 3 (0) – 6 ≥ 0 or – 6 ≥ 0 which is false. Thus, the solution region is the shaded region on the right hand side of the line x = 2.
Fig 6.9
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Example 11 Solve y < 2 graphically. Solution Graph of y = 2 is given in the Fig 6.10. Let us select a point, (0, 0) in lower half plane I and putting y = 0 in the given inequality, we see that 1 × 0 < 2 or 0 < 2 which is true. Thus, the solution region is the shaded region below the line y = 2. Hence, every point below the line (excluding all the points on the line) determines the solution of the given inequality.
Fig 6.10
EXERCISE 6.2
Solve the following inequalities graphically in two-dimensional plane: 1. x + y < 5 2. 2x + y ≥ 6 3. 3x + 4y ≤ 12 4. y + 8 ≥ 2x 5. x – y ≤ 2 6. 2x – 3y > 6 7. – 3x + 2y ≥ – 6 8. 3y – 5x < 30 9. y < – 2 10. x > – 3. 6.5 Solution of System of Linear Inequalities in Two Variables In previous Section, you have learnt how to solve linear inequality in one or two variables graphically. We will now illustrate the method for solving a system of linear inequalities in two variables graphically through some examples. Example 12 Solve the following system of linear inequalities graphically. x+y≥5 ... (1) x–y≤3 ... (2) Solution The graph of linear equation x+y=5 is drawn in Fig 6.11. We note that solution of inequality (1) is represented by the shaded region above the line x + y = 5, including the points on the line. On the same set of axes, we draw the graph of the equation x – y = 3 as Fig 6.11 shown in Fig 6.11. Then we note that inequality (2) represents the shaded region above
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the line x – y = 3, including the points on the line. Clearly, the double shaded region, common to the above two shaded regions is the required solution region of the given system of inequalities. Example 13 Solve the following system of inequalities graphically 5x + 4y ≤ 40 ... (1) x≥2 ... (2) y≥3 ... (3) Solution We first draw the graph of the line 5x + 4y = 40, x = 2 and y = 3 Then we note that the inequality (1) represents shaded region below the line 5x + 4y = 40 and inequality (2) represents the shaded region right of line x = 2 but inequality (3) represents the shaded region above the line y = 3. Hence, shaded region (Fig 6.12) including all the point on the lines are also the solution of the given system of the linear inequalities. In many practical situations involving system of inequalities the variable x and y often represent quantities that cannot have negative values, for example, number of units produced, number of articles purchased, number of hours worked, etc. Clearly, in such cases, x ≥ 0, y ≥ 0 and the solution region lies only in the first quadrant. Example 14 Solve the following system of inequalities 8x + 3y ≤ 100 ... (1) x≥0 ... (2) y≥0 ... (3) Solution We draw the graph of the line 8x + 3y = 100 The inequality 8x + 3y ≤ 100 represents the shaded region below the line, including the points on the line 8x +3y =100 (Fig 6.13).
Fig 6.12
Fig 6.13
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Since x ≥ 0, y ≥ 0, every point in the shaded region in the first quadrant, including the points on the line and the axes, represents the solution of the given system of inequalities. Example 15 Solve the following system of inequalities graphically x + 2y ≤ 8 ... (1) 2x + y ≤ 8 ... (2) x>0 ... (3) y>0 ... (4) Solution We draw the graphs of the lines x + 2y = 8 and 2x + y = 8. The inequality (1) and (2) represent Fig 6.14 the region below the two lines, including the point on the respective lines. Since x ≥ 0, y ≥ 0, every point in the shaded region in the first quadrant represent a solution of the given system of inequalities (Fig 6.14).
–11 ≤ x ≤5 3 Example 18 Solve the system of inequalities: 3x – 7 < 5 + x 11 – 5 x ≤ 1 and represent the solutions on the number line.
which can be written as
... (1) ... (2)
Solution From inequality (1), we have 3x – 7 < 5 + x or x<6 ... (3) Also, from inequality (2), we have 11 – 5 x ≤ 1 or – 5 x ≤ – 10 i.e., x ≥ 2 ... (4) If we draw the graph of inequalities (3) and (4) on the number line, we see that the values of x, which are common to both, are shown by bold line in Fig 6.15.
Fig 6.15
Thus, solution of the system are real numbers x lying between 2 and 6 including 2, i.e., 2≤x<6
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Example 19 In an experiment, a solution of hydrochloric acid is to be kept between 30° and 35° Celsius. What is the range of temperature in degree Fahrenheit if conversion formula is given by C =
5 (F – 32), where C and F represent temperature in degree 9
Celsius and degree Fahrenheit, respectively. Solution It is given that 30 < C < 35. Putting C= 30 < or
5 (F – 32), we get 9 5 (F – 32) < 35, 9
9 9 × (30) < (F – 32) < × (35) 5 5
or 54 < (F – 32) < 63 or 86 < F < 95. Thus, the required range of temperature is between 86° F and 95° F. Example 20 A manufacturer has 600 litres of a 12% solution of acid. How many litres of a 30% acid solution must be added to it so that acid content in the resulting mixture will be more than 15% but less than 18%? Solution Let x litres of 30% acid solution is required to be added. Then Total mixture = (x + 600) litres Therefore 30% x + 12% of 600 > 15% of (x + 600) and 30% x + 12% of 600 < 18% of (x + 600) or and or and or or i.e.
11. A solution is to be kept between 68° F and 77° F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by F=
9 C + 32 ? 5
12. A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added? 13. How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content? 14. IQ of a person is given by the formula IQ =
MA × 100, CA
where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12 years old children, find the range of their mental age.
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Summary
" Two real numbers or two algebraic expressions related by the symbols <, >, ≤ " Equal numbers may be added to (or subtracted from ) both sides of an inequality. " Both sides of an inequality can be multiplied (or divided ) by the same positive
number. But when both sides are multiplied (or divided) by a negative number, then the inequality is reversed. " The values of x, which make an inequality a true statement, are called solutions of the inequality. " To represent x < a (or x > a) on a number line, put a circle on the number a and dark line to the left (or right) of the number a. " To represent x ≤ a (or x ≥ a) on a number line, put a dark circle on the number a and dark the line to the left (or right) of the number x. " If an inequality is having ≤ or ≥ symbol, then the points on the line are also included in the solutions of the inequality and the graph of the inequality lies left (below) or right (above) of the graph of the equality represented by dark line that satisfies an arbitrary point in that part. If an inequality is having < or > symbol, then the points on the line are not " included in the solutions of the inequality and the graph of the inequality lies to the left (below) or right (above) of the graph of the corresponding equality represented by dotted line that satisfies an arbitrary point in that part. " The solution region of a system of inequalities is the region which satisfies all the given inequalities in the system simultaneously. or ≥ form an inequality.
—! —
Chapter
7
PERMUTATIONS AND COMBINATIONS
!Every body of discovery is mathematical in form because there is no other guidance we can have – DARWIN!
7.1 Introduction
Suppose you have a suitcase with a number lock. The number lock has 4 wheels each labelled with 10 digits from 0 to 9. The lock can be opened if 4 specific digits are arranged in a particular sequence with no repetition. Some how, you have forgotten this specific sequence of digits. You remember only the first digit which is 7. In order to open the lock, how many sequences of 3-digits you may have to check with? To answer this question, you may, immediately, start listing all possible arrangements of 9 remaining digits taken 3 at a time. But, Jacob Bernoulli this method will be tedious, because the number of possible (1654-1705) sequences may be large. Here, in this Chapter, we shall learn some basic counting techniques which will enable us to answer this question without actually listing 3-digit arrangements. In fact, these techniques will be useful in determining the number of different ways of arranging and selecting objects without actually listing them. As a first step, we shall examine a principle which is most fundamental to the learning of these techniques.
7.2 Fundamental Principle of Counting
Let us consider the following problem. Mohan has 3 pants and 2 shirts. How many different pairs of a pant and a shirt, can he dress up with? There are 3 ways in which a pant can be chosen, because there are 3 pants available. Similarly, a shirt can be chosen in 2 ways. For every choice of a pant, there are 2 choices of a shirt. Therefore, there are 3 × 2 = 6 pairs of a pant and a shirt.
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Let us name the three pants as P1, P2 , P3 and the two shirts as S1, S2. Then, these six possibilities can be illustrated in the Fig. 7.1. Let us consider another problem of the same type. Sabnam has 2 school bags, 3 tiffin boxes and 2 water bottles. In how many ways can she carry these items (choosing one each). A school bag can be chosen in 2 different ways. After a school bag is chosen, a tiffin box can be chosen in 3 different ways. Hence, there are 2 × 3 = 6 pairs of school bag and a tiffin box. For each of these pairs a water Fig 7.1 bottle can be chosen in 2 differnt ways. Hence, there are 6 × 2 = 12 different ways in which, Sabnam can carry these items to school. If we name the 2 school bags as B1, B2, the three tiffin boxes as T1, T2, T3 and the two water bottles as W1, W2, these possibilities can be illustrated in the Fig. 7.2.
Fig 7.2
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In fact, the problems of the above types are solved by applying the following principle known as the fundamental principle of counting, or, simply, the multiplication principle, which states that "If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrence of the events in the given order is m×n." The above principle can be generalised for any finite number of events. For example, for 3 events, the principle is as follows: 'If an event can occur in m different ways, following which another event can occur in n different ways, following which a third event can occur in p different ways, then the total number of occurrence to 'the events in the given order is m × n × p." In the first problem, the required number of ways of wearing a pant and a shirt was the number of different ways of the occurence of the following events in succession: (i) (ii) the event of choosing a pant the event of choosing a shirt.
In the second problem, the required number of ways was the number of different ways of the occurence of the following events in succession: (i) (ii) (iii) the event of choosing a school bag the event of choosing a tiffin box the event of choosing a water bottle.
Here, in both the cases, the events in each problem could occur in various possible orders. But, we have to choose any one of the possible orders and count the number of different ways of the occurence of the events in this chosen order. Example 1 Find the number of 4 letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed. Solution There are as many words as there are ways of filling in 4 vacant places by the 4 letters, keeping in mind that the repetition is not allowed. The first place can be filled in 4 different ways by anyone of the 4 letters R,O,S,E. Following which, the second place can be filled in by anyone of the remaining 3 letters in 3 different ways, following which the third place can be filled in 2 different ways; following which, the fourth place can be filled in 1 way. Thus, the number of ways in which the 4 places can be filled, by the multiplication principle, is 4 × 3 × 2 × 1 = 24. Hence, the required number of words is 24.
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!Note If the repetition of the letters was allowed, how many words can be formed?
One can easily understand that each of the 4 vacant places can be filled in succession in 4 different ways. Hence, the required number of words = 4 × 4 × 4 × 4 = 256.
Example 2 Given 4 flags of different colours, how many different signals can be generated, if a signal requires the use of 2 flags one below the other? Solution There will be as many signals as there are ways of filling in 2 vacant places in succession by the 4 flags of different colours. The upper vacant place can be filled in 4 different ways by anyone of the 4 flags; following which, the lower vacant place can be filled in 3 different ways by anyone of the remaining 3 different flags. Hence, by the multiplication principle, the required number of signals = 4 × 3 = 12. Example 3 How many 2 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated? Solution There will be as many ways as there are ways of filling 2 vacant places in succession by the five given digits. Here, in this case, we start filling in unit's place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten's place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is 2 × 5, i.e., 10. Example 4 Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available. Solution A signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. Now, let us count the possible number of signals consisting of 2 flags, 3 flags, 4 flags and 5 flags separately and then add the respective numbers. There will be as many 2 flag signals as there are ways of filling in 2 vacant places in succession by the 5 flags available. By Multiplication rule, the number of ways is 5 × 4 = 20. Similarly, there will be as many 3 flag signals as there are ways of filling in 3 vacant places in succession by the 5 flags.
EXERCISE 7.1
1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) (ii) 2. 3. 4. 5. 6. repetition of the digits is allowed? repetition of the digits is not allowed?
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated? How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated? How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once? A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there? Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
7.3 Permutations
In Example 1 of the previous Section, we are actually counting the different possible arrangements of the letters such as ROSE, REOS, ..., etc. Here, in this list, each arrangement is different from other. In other words, the order of writing the letters is important. Each arrangement is called a permutation of 4 different letters taken all at a time. Now, if we have to determine the number of 3-letter words, with or without meaning, which can be formed out of the letters of the word NUMBER, where the repetition of the letters is not allowed, we need to count the arrangements NUM, NMU, MUN, NUB, ..., etc. Here, we are counting the permutations of 6 different letters taken 3 at a time. The required number of words = 6 × 5 × 4 = 120 (by using multiplication principle). If the repetition of the letters was allowed, the required number of words would be 6 × 6 × 6 = 216.
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Definition 1 A permutation is an arrangement in a definite order of a number of objects taken some or all at a time. In the following sub Section, we shall obtain the formula needed to answer these questions immediately. 7.3.1 Permutations when all the objects are distinct Theorem 1 The number of permutations of n different objects taken r at a time, where 0 < r ≤ n and the objects do not repeat is n ( n – 1) ( n – 2). . .( n – r + 1), which is denoted by nPr. Proof There will be as many permutations as there are ways of filling in r vacant places ...
! r vacant places "
This is a much more convenient expression for nPr than the previous one.
n! ' n! 0! Counting permutations is merely counting the number of ways in which some or all objects at a time are rearranged. Arranging no object at all is the same as leaving behind all the objects and we know that there is only one way of doing so. Thus, we can have
In particular, when r = n, n Pn '
n
P0 = 1 =
n! n! ' n ! ( n & 0)!
... (1)
Therefore, the formula (1) is applicable for r = 0 also. Thus
n
Pr '
n! ,0)r )n . # n & r $!
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Theorem 2 The number of permutations of n different objects taken r at a time, where repetition is allowed, is nr. Proof is very similar to that of Theorem 1 and is left for the reader to arrive at. Here, we are solving some of the problems of the pervious Section using the formula for nPr to illustrate its usefulness. In Example 1, the required number of words = 4P4 = 4! = 24. Here repetition is not allowed. If repeation is allowed, the required number of words would be 44 = 256. The number of 3-letter words which can be formed by the letters of the word
6! 6 NUMBER = P3 ' 3! = 4 × 5 × 6 = 120. Here, in this case also, the repetition is not
allowed. If the repetition is allowed,the required number of words would be 63 = 216. The number of ways in which a Chairman and a Vice-Chairman can be chosen from amongst a group of 12 persons assuming that one person can not hold more than one position, clearly
12
P2 '
12! ' 11 % 12 = 132. 10!
7.3.4 Permutations when all the objects are not distinct objects Suppose we have to find the number of ways of rearranging the letters of the word ROOT. In this case, the letters of the word are not all different. There are 2 Os, which are of the same kind. Let us treat, temporarily, the 2 Os as different, say, O1 and O2. The number of permutations of 4-different letters, in this case, taken all at a time is 4!. Consider one of these permutations say, RO 1O2T. Corresponding to this permutation,we have 2 ! permutations RO1O2T and RO2O1T which will be exactly the same permutation if O1 and O2 are not treated as different, i.e., if O1 and O2 are the same O at both places.
Let us now find the number of ways of rearranging the letters of the word INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears 3 times. Temporarily, let us treat these letters different and name them as I1, I2, T1, T2, T3. The number of permutations of 9 different letters, in this case, taken all at a time is 9 !. Consider one such permutation, say, I1 NT1 SI2 T2 U E T3. Here if I1, I2 are not same
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MATHEMATICS
and T1, T2, T3 are not same, then I1, I2 can be arranged in 2! ways and T1, T2, T3 can be arranged in 3! ways. Therefore, 2! × 3! permutations will be just the same permutation corresponding to this chosen permutation I1NT1SI2T2UET3. Hence, total number of different permutations will be
9! 2! 3!
We can state (without proof) the following theorems: Theorem 3 The number of permutations of n objects, where p objects are of the same kind and rest are all different =
n! p! .
In fact, we have a more general theorem. Theorem 4 The number of permutations of n objects, where p1 objects are of one kind, p2 are of second kind, ..., pk are of kth kind and the rest, if any, are of different kind is
n! . p1! p2! ... pk!
Example 9 Find the number of permutations of the letters of the word ALLAHABAD. Solution Here, there are 9 objects (letters) of which there are 4A's, 2 L's and rest are all different. Therefore, the required number of arrangements =
9! 5% 6% 7 %8% 9 ' = 7560 4! 2! 2
Example 10 How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed? Solution Here order matters for example 1234 and 1324 are two different numbers. Therefore, there will be as many 4 digit numbers as there are permutations of 9 different digits taken 4 at a time.
9 Therefore, the required 4 digit numbers = P4 =
9! 9! = = 9 × 8 × 7 × 6 = 3024. # 9 – 4 $! 5!
Example 11 How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed? Solution Every number between 100 and 1000 is a 3-digit number. We, first, have to
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count the permutations of 6 digits taken 3 at a time. This number would be 6P3. But, these permutations will include those also where 0 is at the 100's place. For example, 092, 042, . . ., etc are such numbers which are actually 2-digit numbers and hence the number of such numbers has to be subtracted from 6P3 to get the required number. To get the number of such numbers, we fix 0 at the 100's place and rearrange the remaining 5 digits taking 2 at a time. This number is 5P2. So The required number
Example 14 Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that (i) all vowels occur together (ii) all vowels do not occur together. Solution (i) There are 8 different letters in the word DAUGHTER, in which there are 3 vowels, namely, A, U and E. Since the vowels have to occur together, we can for the time being, assume them as a single object (AUE). This single object together with 5 remaining letters (objects) will be counted as 6 objects. Then we count permutations of these 6 objects taken all at a time. This number would be 6P6 = 6!. Corresponding to each of these permutations, we shall have 3! permutations of the three vowels A, U, E taken all at a time . Hence, by the multiplication principle the required number of permutations = 6 ! × 3 ! = 4320. (ii) If we have to count those permutations in which all vowels are never together, we first have to find all possible arrangments of 8 letters taken all at a time, which can be done in 8! ways. Then, we have to subtract from this number, the number of permutations in which the vowels are always together.
8 ! – 6 ! × 3 ! = 6 ! (7×8 – 6) = 2 × 6 ! (28 – 3) = 50 × 6 ! = 50 × 720 = 36000 Example 15 In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable ? Therefore, the required number
Solution Total number of discs are 4 + 3 + 2 = 9. Out of 9 discs, 4 are of the first kind
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(red), 3 are of the second kind (yellow) and 2 are of the third kind (green). Therefore, the number of arrangements
9! = 1260 . 4! 3! 2!
Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, (i) do the words start with P (ii) do all the vowels always occur together (iii) do the vowels never occur together (iv) do the words begin with I and end in P? Solution There are 12 letters, of which N appears 3 times, E appears 4 times and D appears 2 times and the rest are all different. Therefore
12! The required number of arrangements ' 3! 4! 2! ' 1663200
(i) Let us fix P at the extreme left position, we, then, count the arrangements of the remaining 11 letters. Therefore, the required of words starting with P are
'
(ii)
11! ' 138600 . 3! 2! 4!
for the time
There are 5 vowels in the given word, which are 4 Es and 1 I. Since, they have to always occur together, we treat them as a single object being. This single object together with 7 remaining objects will account for 8 objects. These 8 objects, in which there are 3Ns and 2 Ds, can be rearranged in
8! 3! 2! ways. Corresponding to each of these arrangements, the 5 vowels E, E, E,
5! ways. Therefore, by multiplication principle the 4! required number of arrangements
E and I can be rearranged in
=
(iii)
8! 5! % ' 16800 3! 2! 4!
The required number of arrangements = the total number of arrangements (without any restriction) – the number of arrangements where all the vowels occur together.
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MATHEMATICS
(iv)
= 1663200 – 16800 = 1646400 Let us fix I and P at the extreme ends (I at the left end and P at the right end). We are left with 10 letters. Hence, the required number of arrangements =
10! = 12600 3! 2! 4!
EXERCISE 7.3
1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated? 2. How many 4-digit numbers are there with no digit repeated? 3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated? 4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even? 5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position? 6. Find n if n – 1P3 : nP4 = 1 : 9. 7. Find r if (i) 5 Pr ' 2 6 Pr&1 (ii) 5 Pr ' 6 Pr &1 . 8. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once? 9. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. (i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but first letter is a vowel? 10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together? 11. In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?
7.4 Combinations
Let us now assume that there is a group of 3 lawn tennis players X, Y, Z. A team consisting of 2 players is to be formed. In how many ways can we do so? Is the team of X and Y different from the team of Y and X ? Here, order is not important. In fact, there are only 3 possible ways in which the team could be constructed.
PERMUTATIONS AND COMBINATIONS
149
Fig. 7.3
These are XY, YZ and ZX (Fig 7.3). Here, each selection is called a combination of 3 different objects taken 2 at a time. In a combination, the order is not important. Now consider some more illustrations. Twelve persons meet in a room and each shakes hand with all the others. How do we determine the number of hand shakes. X shaking hands with Y and Y with X will not be two different hand shakes. Here, order is not important. There will be as many hand shakes as there are combinations of 12 different things taken 2 at a time. Seven points lie on a circle. How many chords can be drawn by joining these points pairwise? There will be as many chords as there are combinations of 7 different things taken 2 at a time. Now, we obtain the formula for finding the number of combinations of n different objects taken r at a time, denoted by nCr.. Suppose we have 4 different objects A, B, C and D. Taking 2 at a time, if we have to make combinations, these will be AB, AC, AD, BC, BD, CD. Here, AB and BA are the same combination as order does not alter the combination. This is why we have not included BA, CA, DA, CB, DB and DC in this list. There are as many as 6 combinations of 4 different objects taken 2 at a time, i.e., 4C2 = 6. Corresponding to each combination in the list, we can arrive at 2! permutations as 2 objects in each combination can be rearranged in 2! ways. Hence, the number of permutations = 4C2 × 2!. On the other hand, the number of permutations of 4 different things taken 2 at a time = 4P2. Therefore
4
P2 = 4C2 × 2! or
4! ' 4C2 # 4 & 2 $! 2!
Now, let us suppose that we have 5 different objects A, B, C, D, E. Taking 3 at a time, if we have to make combinations, these will be ABC, ABD, ABE, BCD, BCE, CDE, ACE, ACD, ADE, BDE. Corresponding to each of these 5C3 combinations, there are 3! permutations, because, the three objects in each combination can be
These examples suggest the following theorem showing relationship between permutaion and combination: Theorem 5 n Pr ' n C r r! , 0 < r ≤ n. Proof Corresponding to each combination of nCr we have r ! permutations, because r objects in every combination can be rearranged in r ! ways. Hence, the total number of permutations of n different things taken r at a time is nCr × r!. On the other hand, it is
n
n r
We define nC0 = 1, i.e., the number of combinations of n different things taken nothing at all is considered to be 1. Counting combinations is merely counting the number of ways in which some or all objects at a time are selected. Selecting nothing at all is the same as leaving behind all the objects and we know that there is only one way of doing so. This way we define nC0 = 1. As
Example 18 A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women? Solution Here, order does not matter. Therefore, we need to count combinations. There will be as many committees as there are combinations of 5 different persons
5 taken 3 at a time. Hence, the required number of ways = C3 '
5! 4 % 5 ' ' 10 . 3! 2! 2
Now, 1 man can be selected from 2 men in 2C1 ways and 2 women can be selected from 3 women in 3C2 ways. Therefore, the required number of committees
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MATHEMATICS
2 3 = C1 % C2 '
2! 3! % ' 6. 1! 1! 2! 1!
Example 19 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these (i) (ii) (iii) (iv) (v) four cards are of the same suit, four cards belong to four different suits, are face cards, two are red cards and two are black cards, cards are of the same colour?
Solution There will be as many ways of choosing 4 cards from 52 cards as there are combinations of 52 different things, taken 4 at a time. Therefore The required number of ways =
52
C4 '
52! 49 % 50 % 51% 52 ' 4! 48! 2% 3% 4
= 270725
(i) There are four suits: diamond, club, spade, heart and there are 13 cards of each suit. Therefore, there are 13C4 ways of choosing 4 diamonds. Similarly, there are 13 C4 ways of choosing 4 clubs, 13C4 ways of choosing 4 spades and 13C4 ways of choosing 4 hearts. Therefore The required number of ways = 13C4 + 13C4 + 13C4 + 13C4. = 4% (ii) There are13 cards in each suit. Therefore, there are 13C1 ways of choosing 1 card from 13 cards of diamond, 13 C1 ways of choosing 1 card from 13 cards of hearts, 13C1 ways of choosing 1 card from 13 cards of clubs, 13C1 ways of choosing 1 card from 13 cards of spades. Hence, by multiplication principle, the required number of ways = 13C1 × 13C1 × 13C1× 13C1 = 134
(iii) There are 12 face cards and 4 are to be slected out of these 12 cards. This can be done in 12C4 ways. Therefore, the required number of ways =
13! ' 2860 4! 9!
12! ' 495 . 4! 8!
PERMUTATIONS AND COMBINATIONS
153
(iv) There are 26 red cards and 26 black cards. Therefore, the required number of ways = 26C2 × 26C2
EXERCISE 7.4
1. 2. 3. 4. 5. 6. 7. If nC8 = nC2, find nC2. Determine n if (ii) 2nC3 : nC3 = 11 : 1 (i) 2nC2 : nC2 = 12 : 1 How many chords can be drawn through 21 points on a circle? In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls? Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour. Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination. In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers? A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected. In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
8. 9.
Miscellaneous Examples
Example 20 How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE ? Solution In the word INVOLUTE, there are 4 vowels, namely, I,O,E,Uand 4 consonants, namely, N, V, L and T.
154
MATHEMATICS
The number of ways of selecting 3 vowels out of 4 = 4C3 = 4. The number of ways of selecting 2 consonants out of 4 = 4C2 = 6.
Therefore, the number of combinations of 3 vowels and 2 consonants is 4 × 6 = 24.
Now, each of these 24 combinations has 5 letters which can be arranged among themselves in 5 ! ways. Therefore, the required number of different words is 24 × 5 ! = 2880. Example 21 A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girl ? (ii) at least one boy and one girl ? (iii) at least 3 girls ? Solution (i) Since, the team will not include any girl, therefore, only boys are to be selected. 5 boys out of 7 boys can be selected in 7C5 ways. Therefore, the required number of ways (ii)
7! 6% 7 7 = C5 ' 5! 2! ' 2 ' 21
Since, at least one boy and one girl are to be there in every team. Therefore, the team can consist of (a) 1 boy and 4 girls (c) 3 boys and 2 girls (b) 2 boys and 3 girls (d) 4 boys and 1 girl.
Example 22 Find the number of words with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary, what will be the 50th word? Solution There are 5 letters in the word AGAIN, in which A appears 2 times. Therefore, the required number of words =
5! ' 60 . 2!
To get the number of words starting with A, we fix the letter A at the extreme left position, we then rearrange the remaining 4 letters taken all at a time. There will be as many arrangements of these 4 letters taken 4 at a time as there are permutations of 4 different things taken 4 at a time. Hence, the number of words starting with
4! = 12 as after placing G 2! at the extreme left position, we are left with the letters A, A, I and N. Similarly, there are 12 words starting with the next letter I. Total number of words so far obtained = 24 + 12 + 12 =48.
A = 4! = 24. Then, starting with G, the number of words ' The 49th word is NAAGI. The 50th word is NAAIG. Example 23 How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4? Solution Since, 1000000 is a 7-digit number and the number of digits to be used is also 7. Therefore, the numbers to be counted will be 7-digit only. Also, the numbers have to be greater than 1000000, so they can begin either with 1, 2 or 4. The number of numbers beginning with 1 =
6! 4 % 5 % 6 ' = 60, as when 1 is 3! 2! 2
fixed at the extreme left position, the remaining digits to be rearranged will be 0, 2, 2, 2, 4, 4, in which there are 3, 2s and 2, 4s. Total numbers begining with 2 =
6! 3% 4 % 5 % 6 ' = 180 2! 2! 2
6! ' 4 % 5 % 6 = 120 3!
and total numbers begining with 4 '
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MATHEMATICS
Therefore, the required number of numbers = 60 + 180 + 120 = 360. Alternative Method
7! The number of 7-digit arrangements, clearly, 3! 2! ' 420 . But, this will include those
numbers also, which have 0 at the extreme left position. The number of such
6! arrangements 3! 2! (by fixing 0 at the extreme left position) = 60.
Therefore, the required number of numbers = 420 – 60 = 360. If one one digits given in list is repeated, will !Note that in or more thanthe digits can be usedthe many times as isitgiven be understood any number, as in the list, e.g., in the above example 1 and 0 can be used only once whereas 2 and 4 can be used 3 times and 2 times, respectively. Example 24 In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together? Solution Let us first seat the 5 girls. This can be done in 5! ways. For each such arrangement, the three boys can be seated only at the cross marked places. × G × G × G × G × G ×. There are 6 cross marked places and the three boys can be seated in 6P3 ways. Hence, by multiplication principle, the total number of ways
6! 3! = 4 × 5 × 2 × 3 × 4 × 5 × 6 = 14400.
= 5! × 6P3 = 5!×
Miscellaneous Exercise on Chapter 7
1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER ? 2. How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together? 3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (i) exactly 3 girls ? (ii) atleast 3 girls ? (iii) atmost 3 girls ? 4. If the different permutations of all the letter of the word EXAMINATION are
PERMUTATIONS AND COMBINATIONS
157
5. 6.
7.
8. 9. 10.
11.
listed as in a dictionary, how many words are there in this list before the first word starting with E ? How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated ? The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet ? In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions ? Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible ? From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen ? In how many ways can the letters of the word ASSASSINATION be arranged so that all the S's are together ?
Summary
" Fundamental principle of counting If an event can occur in m
different ways, following which another event can occur in n different ways, then the total number of occurrence of the events in the given order is m × n. " The number of permutations of n different things taken r at a time, where repetition is not allowed, is denoted by nPr and is given by nPr =
n! (n & r )! ,
where 0 ≤ r ≤ n. " n! = 1 × 2 × 3 × ...×n " n! = n × (n – 1) ! " The number of permutations of n different things, taken r at a time, where repeatition is allowed, is nr. " The number of permutations of n objects taken all at a time, where p1 objects
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MATHEMATICS
are of first kind, p2 objects are of the second kind, ..., pk objects are of the kth
n! kind and rest, if any, are all different is p ! p !... p ! . 1 2 k
" The number of combinations of n different things taken r at a time, denoted by
n
Cr , is given by nCr = '
n! r!( n & r )! , 0 ≤ r ≤ n.
Historical Note
The concepts of permutations and combinations can be traced back to the advent of Jainism in India and perhaps even earlier. The credit, however, goes to the Jains who treated its subject matter as a self-contained topic in mathematics, under the name Vikalpa. Among the Jains, Mahavira, (around 850 A.D.) is perhaps the world's first mathematician credited with providing the general formulae for permutations and combinations. In the 6th century B.C., Sushruta, in his medicinal work, Sushruta Samhita, asserts that 63 combinations can be made out of 6 different tastes, taken one at a time, two at a time, etc. Pingala, a Sanskrit scholar around third century B.C., gives the method of determining the number of combinations of a given number of letters, taken one at a time, two at a time, etc. in his work Chhanda Sutra. Bhaskaracharya (born 1114 A.D.) treated the subject matter of permutations and combinations under the name Anka Pasha in his famous work Lilavati. In addition to the general formulae for nCr and nPr already provided by Mahavira, Bhaskaracharya gives several important theorems and results concerning the subject. Outside India, the subject matter of permutations and combinations had its humble beginnings in China in the famous book I–King (Book of changes). It is difficult to give the approximate time of this work, since in 213 B.C., the emperor had ordered all books and manuscripts in the country to be burnt which fortunately was not completely carried out. Greeks and later Latin writers also did some scattered work on the theory of permutations and combinations. Some Arabic and Hebrew writers used the concepts of permutations and combinations in studying astronomy. Rabbi ben Ezra, for instance, determined the number of combinations of known planets taken two at a time, three at a time and so on. This was around 1140 A.D. It appears that Rabbi ben Ezra did not
PERMUTATIONS AND COMBINATIONS
159
know the formula for nCr. However, he was aware that nCr = nCn–r for specific values n and r. In 1321 A.D., Levi Ben Gerson, another Hebrew writer came up with the formulae for nPr , nPn and the general formula for nCr. The first book which gives a complete treatment of the subject matter of permutations and combinations is Ars Conjectandi written by a Swiss, Jacob Bernoulli (1654 – 1705 A.D.), posthumously published in 1713 A.D. This book contains essentially the theory of permutations and combinations as is known today.
—! —
Chapter
8
BINOMIAL THEOREM
!Mathematics is a most exact science and its conclusions are capable of absolute proofs. – C.P. STEINMETZ!
8.1 Introduction
In earlier classes, we have learnt how to find the squares and cubes of binomials like a + b and a – b. Using them, we could evaluate the numerical values of numbers like (98)2 = (100 – 2)2, (999)3 = (1000 – 1)3, etc. However, for higher powers like (98)5, (101)6, etc., the calculations become difficult by using repeated multiplication. This difficulty was overcome by a theorem known as binomial theorem. It gives an easier way to expand (a + b)n, where n is an integer or a rational number. In this Chapter, we study binomial theorem for positive integral indices only.
8.2 Binomial Theorem for Positive Integral Indices
Blaise Pascal (1623-1662)
Let us have a look at the following identities done earlier: (a+ b)0 = 1 a+b≠0 (a+ b)1 = a + b (a+ b)2 = a2 + 2ab + b2 (a+ b)3 = a3 + 3a2b + 3ab2 + b3 (a+ b)4 = (a + b)3 (a + b) = a4 + 4a3b + 6a2b2 + 4ab3 + b4 In these expansions, we observe that (i) The total number of terms in the expansion is one more than the index. For example, in the expansion of (a + b)2 , number of terms is 3 whereas the index of (a + b)2 is 2. (ii) Powers of the first quantity 'a' go on decreasing by 1 whereas the powers of the second quantity 'b' increase by 1, in the successive terms. (iii) In each term of the expansion, the sum of the indices of a and b is the same and is equal to the index of a + b.
BINOMIAL THEOREM
161
We now arrange the coefficients in these expansions as follows (Fig 8.1):
Fig 8.1
Do we observe any pattern in this table that will help us to write the next row? Yes we do. It can be seen that the addition of 1's in the row for index 1 gives rise to 2 in the row for index 2. The addition of 1, 2 and 2, 1 in the row for index 2, gives rise to 3 and 3 in the row for index 3 and so on. Also, 1 is present at the beginning and at the end of each row. This can be continued till any index of our interest. We can extend the pattern given in Fig 8.2 by writing a few more rows.
Fig 8.2
Pascal's Triangle
The structure given in Fig 8.2 looks like a triangle with 1 at the top vertex and running down the two slanting sides. This array of numbers is known as Pascal's triangle, after the name of French mathematician Blaise Pascal. It is also known as Meru Prastara by Pingla. Expansions for the higher powers of a binomial are also possible by using Pascal's triangle. Let us expand (2x + 3y)5 by using Pascal's triangle. The row for index 5 is 1 5 10 10 5 1 Using this row and our observations (i), (ii) and (iii), we get (2x + 3y)5 = (2x)5 + 5(2x)4 (3y) + 10(2x)3 (3y)2 +10 (2x)2 (3y)3 + 5(2x)(3y)4 +(3y)5 = 32x5 + 240x4y + 720x3y2 + 1080x2y3 + 810xy4 + 243y5.
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MATHEMATICS
Now, if we want to find the expansion of (2x + 3y)12, we are first required to get the row for index 12. This can be done by writing all the rows of the Pascal's triangle till index 12. This is a slightly lengthy process. The process, as you observe, will become more difficult, if we need the expansions involving still larger powers. We thus try to find a rule that will help us to find the expansion of the binomial for any power without writing all the rows of the Pascal's triangle, that come before the row of the desired index. For this, we make use of the concept of combinations studied earlier to rewrite the numbers in the Pascal's triangle. We know that
n
Cr !
n! r!(n – r )! , 0 ≤ r ≤ n and
n is a non-negative integer. Also, nC0 = 1 = nCn The Pascal's triangle can now be rewritten as (Fig 8.3)
Fig 8.3
Pascal's triangle
Observing this pattern, we can now write the row of the Pascal's triangle for any index without writing the earlier rows. For example, for the index 7 the row would be
7
C0 7C1 7C2 7C3 7C4 7C5 7C6 7C7.
Thus, using this row and the observations (i), (ii) and (iii), we have (a + b)7 = 7C0 a7 + 7C1a6b + 7C2a5b2 + 7C3a4b3 + 7C4a3b4 + 7C5a2b5 + 7C6ab6 + 7C7b7 An expansion of a binomial to any positive integral index say n can now be visualised using these observations. We are now in a position to write the expansion of a binomial to any positive integral index.
The coefficients nCr occuring in the binomial theorem are known as binomial coefficients. There are (n+1) terms in the expansion of (a+b)n, i.e., one more than the index. In the successive terms of the expansion the index of a goes on decreasing by unity. It is n in the first term, (n–1) in the second term, and so on ending with zero in the last term. At the same time the index of b increases by unity, starting with zero in the first term, 1 in the second and so on ending with n in the last term. In the expansion of (a+b)n, the sum of the indices of a and b is n + 0 = n in the first term, (n – 1) + 1 = n in the second term and so on 0 + n = n in the last term. Thus, it can be seen that the sum of the indices of a and b is n in every term of the expansion. Taking a = x and b = – y, we obtain (x – y)n = [x + (–y)]n = nC0xn + nC1xn – 1(–y) + nC2xn–2(–y)2 + nC3xn–3(–y)3 + ... + nCn (–y)n = nC0xn – nC1xn – 1y + nC2xn – 2y2 – nC3xn – 3y3 + ... + (–1)n nCn yn
General and Middle Terms
In the binomial expansion for (a + b)n, we observe that the first term is n C0an, the second term is nC1an–1b, the third term is nC2an–2b2, and so on. Looking at the pattern of the successive terms we can say that the (r + 1)th term is n Cran–rbr. The (r + 1)th term is also called the general term of the expansion (a + b)n. It is denoted by Tr+1. Thus Tr+1 = nCr an–rbr. Regarding the middle term in the expansion (a + b)n, we have (i) If n is even, then the number of terms in the expansion will be n + 1. Since
1& % In the expansion of ' x $ ( , where x ≠ 0, the middle term is x* )
i.e., (n + 1)th term, as 2n is even.
2n
% 2 n $1 $ 1 & ' ( , 2 ) *
th
%1& It is given by Cnx ' ( = 2nCn (constant). )x*
2n n
n
This term is called the term independent of x or the constant term. Example 5 Find a if the 17th and 18th terms of the expansion (2 + a)50 are equal. Solution The (r + 1)th term of the expansion (x + y)n is given by Tr + 1 = nCrxn–ryr. For the 17th term, we have, r + 1 = 17, i.e., r = 16 Therefore, Similarly, Given that So
50
Example 7 Find the coefficient of x6y3 in the expansion of (x + 2y)9. Solution Suppose x6y3 occurs in the (r + 1)th term of the expansion (x + 2y)9. Now Tr+1 = 9Cr x9 – r (2y)r = 9Cr 2 r . x9 – r . y r . Comparing the indices of x as well as y in x6y3 and in Tr + 1 , we get r = 3. Thus, the coefficient of x6y3 is
9
C3 2 3 =
9! 3 9.8.7 3 .2 = . 2 = 672. 3! 6! 3.2
Example 8 The second, third and fourth terms in the binomial expansion (x + a)n are 240, 720 and 1080, respectively. Find x, a and n. Solution Given that second term T2 = 240
170
MATHEMATICS
We have So Similarly
T2 = nC1xn – 1 . a
n n n
C1xn–1 . a = 240 C2xn–2 a2 = 720
n–3 3
... (1) ... (2) ... (3)
and C3x a = 1080 Dividing (2) by (1), we get
n
C 2 x n# 2 a 2 720 ! i.e., n 240 C1 x n #1a
(n # 1)! a . !6 (n # 2)! x
... (4)
or
a 6 ! x ( n # 1)
a 9 ! x 2( n # 2)
Dividing (3) by (2), we have ... (5)
From (4) and (5),
6 9 ! n #1 2 (n # 2) .
Hence, from (1), 5x4a = 240, and from (4),
Thus, n = 5
a 3 ! x 2
Solving these equations for a and x, we get x = 2 and a = 3. Example 9 The coefficients of three consecutive terms in the expansion of (1 + a)n are in the ratio1: 7 : 42. Find n. Solution Suppose the three consecutive terms in the expansion of (1 + a)n are (r – 1)th, rth and (r + 1)th terms. The (r – 1)th term is nCr – 2 ar – 2, and its coefficient is nCr – 2. Similarly, the coefficients of rth and (r + 1)th terms are nCr – 1 and nCr , respectively. Since the coefficients are in the ratio 1 : 7 : 42, so we have,
n n n
In the expansion of (1 + a)m+n, prove that coefficients of am and an are equal.
10. The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r. 11. Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n – 1. 12. Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.
Example 11 If the coefficients of ar – 1, ar and ar + 1 in the expansion of (1 + a)n are in arithmetic progression, prove that n2 – n(4r + 1) + 4r2 – 2 = 0. Solution The (r + 1)th term in the expansion is nCrar. Thus it can be seen that ar occurs in the (r + 1)th term, and its coefficient is nCr. Hence the coefficients of ar – 1, ar and ar + 1 are nCr – 1, nCr and nCr + 1, respectively. Since these coefficients are in arithmetic progression, so we have, nCr – 1+ nCr + 1 = 2.nCr. This gives
Example 12 Show that the coefficient of the middle term in the expansion of (1 + x)2n is equal to the sum of the coefficients of two middle terms in the expansion of (1 + x)2n – 1. Solution As 2n is even so the expansion (1 + x)2n has only one middle term which is
% 2n & $ 1( ' 2 ) *
th
Thus, the coefficient of a4 in the given product is – 438. Example 14 Find the rth term from the end in the expansion of (x + a)n. Solution There are (n + 1) terms in the expansion of (x + a)n. Observing the terms we can say that the first term from the end is the last term, i.e., (n + 1)th term of the expansion and n + 1 = (n + 1) – (1 – 1). The second term from the end is the nth term of the expansion, and n = (n + 1) – (2 – 1). The third term from the end is the (n – 1)th term of the expansion and n – 1 = (n + 1) – (3 – 1) and so on. Thus rth term from the end will be term number (n + 1) – (r – 1) = (n – r + 2) of the expansion. And the (n – r + 2)th term is nCn – r + 1 xr – 1 an – r + 1.
%3 1 & Example 15 Find the term independent of x in the expansion of ' x $ 3 ( , x > 0. 2 x* )
18
Solution We have Tr + 1 =
18
Cr
- x.
3
18 # r
% 1 & ' 3 ( )2 x *
r
!
18
Cr x
18 # r 3
. 2
r
1
r .x 3
=
18
C2
1 .x 2r
18 # 2 r 3
Since we have to find a term independent of x, i.e., term not having x, so take We get r = 9. The required term is 18C9
18 # 2r !0. 3
1 . 29
Example 16 The sum of the coefficients of the first three terms in the expansion of
3 & % ' x # 2 ( , x ≠ 0, m being a natural number, is 559. Find the term of the expansion x * ) containing x3. 3 & % Solution The coefficients of the first three terms of ' x # 2 ( are mC0, (–3) mC1 x * ) m and 9 C2. Therefore, by the given condition, we have
m
Miscellaneous Exercise on Chapter 8
1. Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively. 2. Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal. 3. Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem. 4. If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer. [Hint write an = (a – b + b)n and expand] 5. Evaluate
-
3$ 2
. #6
3# 2
4
.
6
.
2
2 2 6. Find the value of a $ a # 1
-
. $ -a
n
# a2 # 1 .
.
4
7. Find an approximation of (0.99)5 using the first three terms of its expansion. 8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the
%4 1 & end in the expansion of ' 2 $ 4 ( is 3* )
Historical Note
The ancient Indian mathematicians knew about the coefficients in the expansions of (x + y)n, 0 ≤ n ≤ 7. The arrangement of these coefficients was in the form of a diagram called Meru-Prastara, provided by Pingla in his book Chhanda shastra (200B.C.). This triangular arrangement is also found in the work of Chinese mathematician Chu-shi-kie in 1303 A.D. The term binomial coefficients was first introduced by the German mathematician, Michael Stipel (1486-1567A.D.) in approximately 1544 A.D. Bombelli (1572 A.D.) also gave the coefficients in the expansion of (a + b)n, for n = 1,2 ...,7 and Oughtred (1631 A.D.) gave them for n = 1, 2,..., 10. The arithmetic triangle, popularly known as Pascal's triangle and similar to the Meru-Prastara of Pingla was constructed by the French mathematician Blaise Pascal (1623-1662 A.D.) in 1665. The present form of the binomial theorem for integral values of n appeared in Trate du triange arithmetic, written by Pascal and published posthumously in 1665 A.D.
—! —
Chapter
9
SEQUENCES AND SERIES
!Natural numbers are the product of human spirit. – DEDEKIND !
9.1 Introduction
In mathematics, the word, "sequence" is used in much the same way as it is in ordinary English. When we say that a collection of objects is listed in a sequence, we usually mean that the collection is ordered in such a way that it has an identified first member, second member, third member and so on. For example, population of human beings or bacteria at different times form a sequence. The amount of money deposited in a bank, over a number of years form a sequence. Depreciated values of certain commodity occur in a sequence. Sequences have important applications in several Fibonacci (1175-1250) spheres of human activities. Sequences, following specific patterns are called progressions. In previous class, we have studied about arithmetic progression (A.P). In this Chapter, besides discussing more about A.P.; arithmetic mean, geometric mean, relationship between A.M. and G.M., special series in forms of sum to n terms of consecutive natural numbers, sum to n terms of squares of natural numbers and sum to n terms of cubes of natural numbers will also be studied.
9.2 Sequences
Let us consider the following examples: Assume that there is a generation gap of 30 years, we are asked to find the number of ancestors, i.e., parents, grandparents, great grandparents, etc. that a person might have over 300 years. Here, the total number of generations = 300 ! 10 30
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MATHEMATICS
The number of person's ancestors for the first, second, third, …, tenth generations are 2, 4, 8, 16, 32, …, 1024. These numbers form what we call a sequence. Consider the successive quotients that we obtain in the division of 10 by 3 at different steps of division. In this process we get 3,3.3,3.33,3.333, ... and so on. These quotients also form a sequence. The various numbers occurring in a sequence are called its terms. We denote the terms of a sequence by a1, a2, a3, …, an, …, etc., the subscripts denote the position of the term. The nth term is the number at the nth position of the sequence and is denoted by an. The nth term is also called the general term of the sequence. Thus, the terms of the sequence of person's ancestors mentioned above are: a1 = 2, a2 = 4, a3 = 8, …, a10 = 1024. Similarly, in the example of successive quotients a1 = 3, a2 = 3.3, a3 = 3.33, …, a6 = 3.33333, etc. A sequence containing finite number of terms is called a finite sequence. For example, sequence of ancestors is a finite sequence since it contains 10 terms (a fixed number). A sequence is called infinite, if it is not a finite sequence. For example, the sequence of successive quotients mentioned above is an infinite sequence, infinite in the sense that it never ends. Often, it is possible to express the rule, which yields the various terms of a sequence in terms of algebraic formula. Consider for instance, the sequence of even natural numbers 2, 4, 6, … a2 = 4 = 2 × 2 Here a1 = 2 = 2 × 1 a3 = 6 = 2 × 3 .... .... .... .... .... .... a4 = 8 = 2 × 4 .... .... .... .... .... ....
a23 = 46 = 2 × 23, a24 = 48 = 2 × 24, and so on. In fact, we see that the nth term of this sequence can be written as an = 2n, where n is a natural number. Similarly, in the sequence of odd natural numbers 1,3,5, …, the nth term is given by the formula, an = 2n – 1, where n is a natural number. In some cases, an arrangement of numbers such as 1, 1, 2, 3, 5, 8,.. has no visible pattern, but the sequence is generated by the recurrence relation given by a1 = a2 = 1 a3 = a1 + a2 an = an – 2 + an – 1, n > 2 This sequence is called Fibonacci sequence.
SEQUENCES AND SERIES
179
In the sequence of primes 2,3,5,7,…, we find that there is no formula for the nth prime. Such sequence can only be described by verbal description. In every sequence, we should not expect that its terms will necessarily be given by a specific formula. However, we expect a theoretical scheme or a rule for generating the terms a1, a2, a3,…,an,… in succession. In view of the above, a sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it of the type {1, 2, 3...k}. Sometimes, we use the functional notation a(n) for an.
9.3 Series
Let a1, a2, a3,…,an, be a given sequence. Then, the expression a1 + a2 + a3 +,…+ an + ... is called the series associated with the given sequence .The series is finite or infinite according as the given sequence is finite or infinite. Series are often represented in compact form, called sigma notation, using the Greek letter " (sigma) as means of indicating the summation involved. Thus, the series a1 + a2 + a3 + ... + an is abbreviated as
" ak .
k !1
n
Remark When the series is used, it refers to the indicated sum not to the sum itself. For example, 1 + 3 + 5 + 7 is a finite series with four terms. When we use the phrase "sum of a series," we will mean the number that results from adding the terms, the sum of the series is 16. We now consider some examples. Example 1 Write the first three terms in each of the following sequences defined by the following: (i) an = 2n + 5, (ii) an =
9.4 Arithmetic Progression (A.P.)
Let us recall some formulae and properties studied earlier. A sequence a1, a2, a3,…, an ,… is called arithmetic sequence or arithmetic progression if an + 1 = an + d, n ∈ N, where a1 is called the first term and the constant term d is called the common difference of the A.P. Let us consider an A.P. (in its standard form) with first term a and common difference d, i.e., a, a + d, a + 2d, ... Then the nth term (general term) of the A.P. is an = a + (n – 1) d. We can verify the following simple properties of an A.P. : (i) If a constant is added to each term of an A.P., the resulting sequence is also an A.P. (ii) If a constant is subtracted from each term of an A.P., the resulting sequence is also an A.P. (iii) If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P. (iv) If each term of an A.P. is divided by a non-zero constant then the resulting sequence is also an A.P. Here, we shall use the following notations for an arithmetic progression: a = the first term, l = the last term, d = common difference, n = the number of terms. Sn= the sum to n terms of A.P. Let a, a + d, a + 2d, …, a + (n – 1) d be an A.P. Then l = a + (n – 1) d
Therefore S1 = a1 = P, S2 = a1 + a2 = 2P + Q So that a2 = S2 – S1 = P + Q Hence, the common difference is given by d = a2 – a1 = (P + Q) – P = Q. Example 6 The sum of n terms of two arithmetic progressions are in the ratio (3n + 8) : (7n + 15). Find the ratio of their 12th terms. Solution Let a1, a2 and d1, d2 be the first terms and common difference of the first and second arithmetic progression, respectively. According to the given condition, we have
Hence, the required ratio is 7 : 16. Example 7 The income of a person is Rs. 3,00,000, in the first year and he receives an increase of Rs.10,000 to his income per year for the next 19 years. Find the total amount, he received in 20 years. Solution Here, we have an A.P. with a = 3,00,000, d = 10,000, and n = 20. Using the sum formula, we get,
20 [600000 $ 19 ( 10000] = 10 (790000) = 79,00,000. 2 Hence, the person received Rs. 79,00,000 as the total amount at the end of 20 years. S20 !
9.4.1 Arithmetic mean Given two numbers a and b. We can insert a number A between them so that a, A, b is an A.P. Such a number A is called the arithmetic mean (A.M.) of the numbers a and b. Note that, in this case, we have A – a = b – A, i.e., A =
a$b 2
We may also interpret the A.M. between two numbers a and b as their average
a $b . For example, the A.M. of two numbers 4 and 16 is 10. We have, thus 2
constructed an A.P. 4, 10, 16 by inserting a number 10 between 4 and 16. The natural
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MATHEMATICS
question now arises : Can we insert two or more numbers between given two numbers so that the resulting sequence comes out to be an A.P. ? Observe that two numbers 8 and 12 can be inserted between 4 and 16 so that the resulting sequence 4, 8, 12, 16 becomes an A.P. More generally, given any two numbers a and b, we can insert as many numbers as we like between them such that the resulting sequence is an A.P. Let A1, A2, A3, …, An be n numbers between a and b such that a, A1, A2, A3, …, An, b is an A.P. Here, b is the (n + 2) th term, i.e., b = a + [(n + 2) – 1]d = a + (n + 1) d. This gives
EXERCISE 9.2
1. Find the sum of odd integers from 1 to 2001. 2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5. 3. In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112. 4. How many terms of the A.P. – 6, #
11 , – 5, … are needed to give the sum –25? 2
5. In an A.P., if pth term is
1 1 and qth term is , prove that the sum of first pq p q
6. 7. 8. 9. 10. 11.
1 (pq +1), where p ≠ q. 2 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term. Find the sum to n terms of the A.P., whose kth term is 5k + 1. If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference. The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms. If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms. Sum of the first p, q and r terms of an A.P are. a, b and c, respectively.
terms is Prove that
a b c ( q # r ) $ ( r # p ) $ ( p # q) ! 0 p q r
12. The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of mth and nth term is (2m – 1) : (2n – 1). 13. If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m. 14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P. 15. If
16. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m.
a n $ bn is the A.M. between a and b, then find the value of n. a n#1 $ b n#1
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MATHEMATICS
17. A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount he will pay in the 30th instalment? 18. The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120° , find the number of the sides of the polygon.
In each of these sequences, how their terms progress? We note that each term, except the first progresses in a definite order. In (i), we have and so on.
1 a –1 a –1 a –1 and so on. In (ii), we observe, a1 ! , 2 ! , 3 ! , 4 ! 9 a1 3 a2 3 a3 3
Similarly, state how do the terms in (iii) progress? It is observed that in each case, a a a a1 ! 2, to the 3 ! 2, 4 ! 2 every term except the first term bears a constant ratio 2 ! 2, term immediately preceding a1 a2 a3
1 it. In (i), this constant ratio is 2; in (ii), it is – and in (iii), the constant ratio is 0.01. 3 Such sequences are called geometric sequence or geometric progression abbreviated as G.P. A sequence a1, a2, a3, …, an, … is called geometric progression, if each term is
non-zero and
ak $ 1 ak
= r (constant), for k ≥ 1.
By letting a1 = a, we obtain a geometric progression, a, ar, ar2, ar3,…., where a is called the first term and r is called the common ratio of the G.P. Common ratio in
1 and 0.01, respectively. 3 As in case of arithmetic progression, the problem of finding the nth term or sum of n terms of a geometric progression containing a large number of terms would be difficult without the use of the formulae which we shall develop in the next Section. We shall use the following notations with these formulae: a = the first term, r = the common ratio, l = the last term,
geometric progression (i), (ii) and (iii) above are 2, –
SEQUENCES AND SERIES
187
n = the numbers of terms, Sn = the sum of n terms. 9.5.1 General term of a G .P. Let us consider a G.P. with first non-zero term 'a' and common ratio 'r'. Write a few terms of it. The second term is obtained by multiplying a by r, thus a2 = ar. Similarly, third term is obtained by multiplying a2 by r. Thus, a3 = a2r = ar2, and so on. We write below these and few more terms. st 1 term = a1 = a = ar1–1, 2nd term = a2 = ar = ar2–1, 3rd term = a3 = ar2 = ar3–1 4th term = a4 = ar3 = ar4–1, 5th term = a5 = ar4 = ar5–1 Do you see a pattern? What will be 16th term? a16 = ar16–1 = ar15 Therefore, the pattern suggests that the nth term of a G.P. is given by
Example10 Which term of the G.P., 2,8,32, ... up to n terms is 131072? Solution Let 131072 be the nth term of the given G.P. Here a = 2 and r = 4. Therefore 131072 = an = 2(4)n – 1 or 65536 = 4n – 1 This gives 48 = 4n – 1. So that n – 1 = 8, i.e., n = 9. Hence, 131072 is the 9th term of the G.P.
188
MATHEMATICS
Example11 In a G.P., the 3rd term is 24 and the 6th term is 192.Find the 10th term. Solution Here, a3 ! ar ! 24
2 5
... (1)
and a6 ! ar ! 192 ... (2) Dividing (2) by (1), we get r = 2. Substituting r = 2 in (1), we get a = 6. Hence a10 = 6 (2)9 = 3072. Example12 Find the sum of first n terms and the sum of first 5 terms of the geometric
We have S10 = 2(210 – 1) = 2046 Hence, the number of ancestors preceding the person is 2046. 9.5.3 Geometric Mean (G .M.) The geometric mean of two positive numbers a and b is the number ab . Therefore, the geometric mean of 2 and 8 is 4. We observe that the three numbers 2,4,8 are consecutive terms of a G.P. This leads to a generalisation of the concept of geometric means of two numbers. Given any two positive numbers a and b, we can insert as many numbers as we like between them to make the resulting sequence in a G.P. Let G1, G2,…, Gn be n numbers between positive numbers a and b such that a,G1,G2,G3,…,Gn,b is a G.P. Thus, b being the (n + 2)th term,we have
b ! ar n $ 1 ,
or
1
+ b ,n $ 1 . r!/ 0 1a2 + b , n$1 G 3 ! ar ! a / 0 , 1a2
3 3
1
Hence
2 + b , n$1 G1 ! ar ! a / 0 , + b , n$1 2 G 2 ! ar ! a / 0 , 1a2 1a2
+ b , n $1 G n ! ar ! a / 0 1a2
n
n
Example17 Insert three numbers between 1 and 256 so that the resulting sequence is a G.P. Solution Let G1, G2,G3 be three numbers between 1 and 256 such that 1, G1,G2,G3 ,256 is a G.P.
9.6 Relationship Between A.M. and G.M.
Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively. Then
A!
Thus, we have
a$b and G ! ab 2
A–G=
a $b a $ b # 2 ab # ab = 2 2
70 2 From (1), we obtain the relationship A ≥ G.
=
5
a# b
6
2
... (1)
Example 18 If A.M. and G.M. of two positive numbers a and b are 10 and 8, respectively, find the numbers. Solution Given that and From (1) and (2), we get a + b = 20 ... (3) ab = 64 ... (4) 2 Putting the value of a and b from (3), (4) in the identity (a – b) = (a + b)2 – 4ab, we get (a – b)2 = 400 – 256 = 144 or a – b = ± 12 ... (5) Solving (3) and (5), we obtain a = 4, b = 16 or a = 16, b = 4 Thus, the numbers a and b are 4, 16 or 16, 4 respectively.
A.M. !
a$b !10 2
... (1) ... (2)
G.M. ! ab ! 8
192
MATHEMATICS
EXERCISE 9.3
1. Find the 20th and nth terms of the G.P. 2. 3. 4. 5.
5 5 5 , , , ... 2 4 8 Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2. The 5 th, 8 th and 11 th terms of a G.P. are p, q and s, respectively. Show that q2 = ps. The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7th term. Which term of the following sequences:
2, 2 2 , 4,... is 128 ?
39 12. The sum of first three terms of a G.P. is and their product is 1. Find the 10 common ratio and the terms. 13. How many terms of G.P. 3, 32, 33, … are needed to give the sum 120? 14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P. 15. Given a G.P. with a = 729 and 7th term 64, determine S7. 16. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term. 17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
SEQUENCES AND SERIES
193
18. Find the sum to n terms of the sequence, 8, 88, 888, 8888… . 19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8,
20. 21. 22. 23. 24.
1 . 2 Show that the products of the corresponding terms of the sequences a, ar, ar2, …arn – 1 and A, AR, AR2, … ARn – 1 form a G.P, and find the common ratio. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18. If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq – r br – pcP – q = 1. If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from
16, 32 and 128, 32, 8, 2, (n + 1)th to (2n)th term is
a and b. 28. The sum of two numbers is 6 times their geometric means, show that numbers are in the ratio 3 $ 2 2 : 3 # 2 2 . 29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A 8 ( A $ G )( A # G ) .
a n$1 $ b n$1 may be the geometric mean between a n $ bn
5
65
6
30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour ? 31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually? 32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
194
MATHEMATICS
9.7 Sum to n Terms of Special Series
We shall now find the sum of first n terms of some special series, namely; (i) 1 + 2 + 3 +… + n (sum of first n natural numbers) (ii) 12 + 22 + 32 +… + n2(sum of squares of the first n natural numbers) (iii) 13 + 23 + 33 +… + n3(sum of cubes of the first n natural numbers). Let us take them one by one. (i) Sn=1 + 2 + 3 + … + n, then Sn =
Miscellaneous Exercise On Chapter 9
1. Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term. 2. If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers. 3. Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3(S2 – S1) 4. Find the sum of all numbers between 200 and 400 which are divisible by 7. 5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5. 6. Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder. 7. If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and
x !1
" f (x) ! 120 , find the value of n.
n
8. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms. 9. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P. 10. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers. 11. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. 12. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms. 13. If
1 ( 22 $ 2 ( 32 $ ... $ n ( (n $ 1) 2 3n $ 5 ! 12 ( 2 $ 2 2 ( 3 $ ... $ n2 ( (n $ 1) 3n $ 1 . A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him? Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him? A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years. A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on.
SEQUENCES AND SERIES
201
It took 8 more days to finish the work. Find the number of days in which the work was completed.
Summary
! By a sequence, we mean an arrangement of a number in a definite order
according to some rule. Also, we define a sequence as a function whose domain is the set of natural numbers or some subsets of the type {1, 2, 3....k). A sequence containing a finite number of terms is called a finite sequence. A sequence is called infinite if it is not a finite sequence.
! Let a1, a2, a3, ... be the sequence, then the sum expressed as a1 + a2 + a3 + ...
is called series. A series is called finite series if it has got finite number of terms. ! An arithmetic progression (A.P.) is a sequence in which terms increase or decrease regularly by the same constant. This constant is called common difference of the A.P. Usually, we denote the first terms of A.P. by a, the common difference by d and the last term by l. The general term or the nth term of the A.P. is given by an = a + (n – 1) d. The sum Sn of the first n terms of an A.P. is given by
n n Sn = )2a + 5 n – 16 d * = 5 a + l 6 . 3 4 2 2 a+b i.e., the 2
! The arithmetic mean A of any two numbers a and b is given by
sequence a, A, b is in A.P.
! A sequence is said to be a geometric progression or G.P., if the ratio of any
term to its preceding term is same throughout. This constant factor is called the common ratio. Usually, we denote the first term of a G.P. by a and its common ratio by r. The general or the nth term of G.P. is given by an= arn – 1. The sum Sn of the first n terms of G.P. is given by
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MATHEMATICS
Sn =
a rn – 1 r –1
5
6 or a 51– r 6 , if r ' 1
n
1–r
! The geometric mean (G.M.) of any two positive numbers a and b is given by
ab i.e., the sequence a, G, b is G.P.
Historical Note
Evidence is found that Babylonians, some 4000 years ago, knew of arithmetic and geometric sequences. According to Boethius (510 A.D.), arithmetic and geometric sequences were known to early Greek writers. Among the Indian mathematician, Aryabhatta (476 A.D.) was the first to give the formula for the sum of squares and cubes of natural numbers in his famous work Aryabhatiyam, written around 499 A.D. He also gave the formula for finding the sum to n terms of an arithmetic sequence starting with p th term. Noted Indian mathematicians Brahmgupta (598 A.D.), Mahavira (850 A.D.) and Bhaskara (1114-1185 A.D.) also considered the sum of squares and cubes. Another specific type of sequence having important applications in mathematics, called Fibonacci sequence, was discovered by Italian mathematician Leonardo Fibonacci (1170-1250 A.D.). Seventeenth century witnessed the classification of series into specific forms. In 1671 A.D. James Gregory used the term infinite series in connection with infinite sequence. It was only through the rigorous development of algebraic and set theoretic tools that the concepts related to sequence and series could be formulated suitably.
—! —
Chapter
10
STRAIGHT LINES
!G eometry, as a logical system, is a means and even the most powerful
means to make children feel the strength of the human spirit that is of their own spirit. – H. FREUDENTHAL!
10.1 Introduction
We are familiar with two-dimensional coordinate geometry from earlier classes. Mainly, it is a combination of algebra and geometry. A systematic study of geometry by the use of algebra was first carried out by celebrated French philosopher and mathematician René Descartes, in his book 'La Géométry, published in 1637. This book introduced the notion of the equation of a curve and related analytical methods into the study of geometry. The resulting combination of analysis and geometry is referred now as analytical geometry. In the earlier classes, we initiated René Descartes the study of coordinate geometry, where we studied about (1596 -1650) coordinate axes, coordinate plane, plotting of points in a plane, distance between two points, section formule, etc. All these concepts are the basics of coordinate geometry. Let us have a brief recall of coordinate geometry done in earlier classes. To recapitulate, the location of the points (6, – 4) and (3, 0) in the XY-plane is shown in Fig 10.1. We may note that the point (6, – 4) is at 6 units distance from the y-axis measured along the positive x-axis and at 4 units distance from the x-axis measured along the negative y-axis. Similarly, the point (3, 0) is at 3 units distance from the y-axis measured along the positive x-axis and has zero distance from the x-axis. Fig 10.1 We also studied there following important formulae:
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I. Distance between the points P (x1, y1) and Q (x2, y2) is
PQ #
!x
2
– x1 " $ ! y2 – y1 "
2
2
For example, distance between the points (6, – 4) and (3, 0) is
! 3 % 6 "2 $ ! 0 $ 4 "2 #
9 $ 16 # 5 units.
II. The coordinates of a point dividing the line segment joining the points (x1, y1) and (x2, y2) internally, in the ratio m: n are ) )
Remark If the area of the triangle ABC is zero, then three points A, B and C lie on a line, i.e., they are collinear. In the this Chapter, we shall continue the study of coordinate geometry to study properties of the simplest geometric figure – straight line. Despite its simplicity, the line is a vital concept of geometry and enters into our daily experiences in numerous interesting and useful ways. Main focus is on representing the line algebraically, for which slope is most essential.
10.2 Slope of a Line
A line in a coordinate plane forms two angles with the x-axis, which are supplementary.
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The angle (say) θ made by the line l with positive direction of x-axis and measured anti clockwise is called the inclination of the line. Obviously 0° ≤ θ ≤ 180° (Fig 10.2). We observe that lines parallel to x-axis, or coinciding with x-axis, have inclination of 0°. The inclination of a vertical line (parallel to or coinciding with y-axis) is 90°. Definition 1 If θ is the inclination of a line l, then tan θ is called the slope or gradient of Fig 10.2 the line l. The slope of a line whose inclination is 90° is not defined. The slope of a line is denoted by m. Thus, m = tan θ, θ ≠ 90° It may be observed that the slope of x-axis is zero and slope of y-axis is not defined. 10.2.1 Slope of a line when coordinates of any two points on the line are given We know that a line is completely determined when we are given two points on it. Hence, we proceed to find the slope of a line in terms of the coordinates of two points on the line. Let P(x 1, y 1 ) and Q(x 2, y 2 ) be two points on non-vertical line l whose inclination is θ. Obviously, x1 ≠ x2, otherwise the line will become perpendicular to x-axis and its slope will not be defined. The inclination of the line l may be acute or obtuse. Let us take these two cases. Draw perpendicular QR to x-axis and PM perpendicular to RQ as shown in Figs. 10.3 (i) and (ii). Case 1 When angle θ is acute: In Fig 10.3 (i), ∠MPQ = θ. Therefore, slope of line l = m = tan θ. But in ∆MPQ, we have tan ! #
Fig 10. 3 (i)
Consequently, we see that in both the cases the slope m of the line through the points (x1, y1) and (x2, y2) is given by m #
y2 % y1 . x2 % x1
10.2.2 Conditions for parallelism and perpendicularity of lines in terms of their slopes In a coordinate plane, suppose that non-vertical lines l1 and l2 have slopes m1 and m2, respectively. Let their inclinations be α and β, respectively. If the line l1 is parallel to l2 (Fig 10.4), then their inclinations are equal, i.e., α = β, and hence, tan α = tan β Therefore m1 = m2, i.e., their slopes are equal. Conversely, if the slope of two lines l1 and l2 is same, i.e., m1 = m2. Then tan α = tan β.
Fig 10. 4
By the property of tangent function (between 0° and 180°), α = β. Therefore, the lines are parallel.
Let us consider the following example.
Example 1 Find the slope of the lines: (a) (b) (c) (d) Passing through the points (3, – 2) and (–1, 4), Passing through the points (3, – 2) and (7, – 2), Passing through the points (3, – 2) and (3, 4), Making inclination of 60° with the positive direction of x-axis.
Solution (a) The slope of the line through (3, – 2) and (– 1, 4) is
m#
4 % ( %2) 6 3 # #% . %1 % 3 %4 2
(b) The slope of the line through the points (3, – 2) and (7, – 2) is
m#
% 2 % ( % 2) 0 # # 0. 7%3 4
(c) The slope of the line through the points (3, – 2) and (3, 4) is
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4 % ( % 2) 6 # , which is not defined. 3%3 0 (d) Here inclination of the line α = 60°. Therefore, slope of the line is m#
m = tan 60° =
3.
10.2.3 Angle between two lines When we think about more than one line in a plane, then we find that these lines are either intersecting or parallel. Here we will discuss the angle between two lines in terms of their slopes. Let L1 and L2 be two non-vertical lines with slopes m1 and m2, respectively. If α1 and α2 are the inclinations of lines L1 and L2, respectively. Then
m 2 % m1 is positive, then tan θ will be positive and tan φ will be negative, 1 $ m1 m 2 m 2 % m1 is negative, then tan θ will be negative and tan φ will be positive, 1 $ m1 m 2
which means θ will be acute and φ will be obtuse. Case II If
which means that θ will be obtuse and φ will be acute. Thus, the acute angle (say θ) between lines L1 and L2 with slopes m1 and m2, respectively, is given by
tan ! # m 2 % m1 , as 1 $ m1m 2 - 0 1 $ m1m 2
... (1)
The obtuse angle (say φ) can be found by using φ =1800 – θ. Example 2 If the angle between two lines is
# 1 and slope of one of the lines is , find 4 2
the slope of the other line. Solution We know that the acute angle θ between two lines with slopes m1 and m2 is given by Let m1 =
tan ! # m2 % m1 1 $ m1m 2
... (1)
1 # , m2 = m and θ = . 4 2
1 2 1 2
Now, putting these values in (1), we get
# tan # 4
m%
1 1$ m 2
or 1 #
m%
1 1$ m 2
,
which gives
1 1 m% 2 # 1 or % 2 # –1. 1 1 1$ 1$ m m 2 2 m%
1 Therefore m # 3 or m # % . 3
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Hence, slope of the other line is
1 3 or % . Fig 10.7 explains the 3
reason of two answers.
Fig 10. 7
Example 3 Line through the points (–2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x. Solution Slope of the line through the points (– 2, 6) and (4, 8) is
m1 #
8%6 2 1 # # 4 % ! %2 " 6 3
Slope of the line through the points (8, 12) and (x, 24) is
24 % 12 12 # x%8 x %8 Since two lines are perpendicular, m1 m2 = –1, which gives m2 # 1 12 . # %1 or x = 4 . 3 x %8
10.2.4 Collinearity of three points We know that slopes of two parallel lines are equal. If two lines having the same slope pass through a common point, then two lines will coincide. Hence, if A, B and C are three points in the XY-plane, then they will lie on a line, i.e., three points are collinear (Fig 10.8) if and only if slope of AB = slope of BC.
Example 5 In Fig 10.9, time and distance graph of a linear motion is given. Two positions of time and distance are recorded as, when T = 0, D = 2 and when T = 3, D = 8. Using the concept of slope, find law of motion, i.e., how distance depends upon time. Solution Let (T, D) be any point on the line, where D denotes the distance at time T. Therefore, points (0, 2), (3, 8) and (T, D) are collinear so that
EXERCISE 10.1
1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area. 2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle. 3. Find the distance between P (x1, y1) and Q (x2, y2) when : (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis. 4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4). 5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).
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6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle. 7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise. 8. Find the value of x for which the points (x, – 1), (2,1) and (4, 5) are collinear. 9. Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram. 10. Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2). 11. The slope of a line is double of the slope of another line. If tangent of the angle between them is
1 , find the slopes of the lines. 3
12. A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k – y1 = m (h – x1). 13. If three points (h, 0), (a, b) and (0, k) lie on a line, show that
a b $ # 1. h k
14. Consider the following population and year graph (Fig 10.10), find the slope of the line AB and using it, find what will be the population in the year 2010?
Fig 10.10
10.3 Various Forms of the Equation of a Line
We know that every line in a plane contains infinitely many points on it. This relationship between line and points leads us to find the solution of the following problem:
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How can we say that a given point lies on the given line? Its answer may be that for a given line we should have a definite condition on the points lying on the line. Suppose P (x, y) is an arbitrary point in the XY-plane and L is the given line. For the equation of L, we wish to construct a statement or condition for the point P that is true, when P is on L, otherwise false. Of course the statement is merely an algebraic equation involving the variables x and y. Now, we will discuss the equation of a line under different conditions. 10.3.1 Horizontal and vertical lines If a horizontal line L is at a distance a from the x-axis then ordinate of every point lying on the line is either a or – a [Fig 10.11 (a)]. Therefore, equation of the line L is either y = a or y = – a. Choice of sign will depend upon the position of the line according as the line is above or below the y-axis. Similarly, the equation of a vertical line at a distance b from the x-axis is either x = b or x = – b [Fig 10.11(b)].
Fig 10.11
Example 6 Find the equations of the lines parallel to axes and passing through (– 2, 3). Solution Position of the lines is shown in the Fig 10.12. The y-coordinate of every point on the line parallel to x-axis is 3, therefore, equation of the line parallel tox-axis and passing through (– 2, 3) is y = 3. Similarly, equation of the line parallel to y-axis and passing through (– 2, 3) is x = – 2.
Fig 10.12
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10.3.2 Point-slope form Suppose that P0 (x0, y0) is a fixed point on a non-vertical line L, whose slope is m. Let P (x, y) be an arbitrary point on L (Fig 10.13). Then, by the definition, the slope of L is given by
m#
y % y0 , i.e., y % y 0 # m ! x % x 0 " x % x0
...(1)
Since the point P0 (x0 , y0) along with all points (x, y) on L satisfies (1) and no other point in the plane satisfies (1). Equation (1) is indeed the equation for the given line L.
Fig 10.13
Thus, the point (x, y) lies on the line with slope m through the fixed point (x0, y0), if and only if, its coordinates satisfy the equation y – y0 = m (x – x0) Example 7 Find the equation of the line through (– 2, 3) with slope – 4. Solution Here m = – 4 and given point (x0 , y0) is (– 2, 3). By slope-intercept form formula (1) above, equation of the given line is y – 3 = – 4 (x + 2) or 4x + y + 5 = 0, which is the required equation. 10.3.3 Two-point form Let the line L passes through two given points P1 (x1, y1) and P2 (x2, y2). Let P (x, y) be a general point on L (Fig 10.14). The three points P1, P2 and P are collinear, therefore, we have slope of P1P = slope of P1P2
Fig 10.14
i.e.,
y % y1 y 2 % y1 # , x % x1 x 2 % x1
or y % y 1 #
y 2 % y1 ( x % x1 ). x 2 % x1
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Thus, equation of the line passing through the points (x1, y1) and (x2, y2) is given by
y % y1 #
y 2 % y1 ( x % x1) x 2 % x1
... (2)
Example 8 Write the equation of the line through the points (1, –1) and (3, 5). Solution Here x1 = 1, y1 = – 1, x2 = 3 and y2 = 5. Using two-point form (2) above for the equation of the line, we have
y % !% 1" #
or
5 % !% 1" !x % 1" 3 %1
% 3 x $ y $ 4 # 0 , which is the required equation.
10.3.4 Slope-intercept form Sometimes a line is known to us with its slope and an intercept on one of the axes. We will now find equations of such lines. Case I Suppose a line L with slope m cuts the y-axis at a distance c from the origin (Fig10.15). The distance c is called the yintercept of the line L. Obviously, coordinates of the point where the line meet the y-axis are (0, c). Thus, L has slope m and passes through a fixed point (0, c). Therefore, by point-slope form, the equation of L is y % c # m ( x % 0 ) or y # mx $ c Thus, the point (x, y) on the line with slope Fig 10.15 m and y-intercept c lies on the line if and only if y # mx $ c ...(3) Note that the value of c will be positive or negative according as the intercept is made on the positive or negative side of the y-axis, respectively. Case II Suppose line L with slope m makes x-intercept d. Then equation of L is ... (4) Students may derive this equation themselves by the same method as in Case I. Example 9 Write the equation of the lines for which tan θ = inclination of the line and (i) y-intercept is –
y # m( x % d )
1 , where θ is the 2
3 (ii) x-intercept is 4. 2
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Solution (i) Here, slope of the line is m = tan θ =
1 3 and y - intercept c = – . 2 2
Therefore, by slope-intercept form (3) above, the equation of the line is
y#
which is the required equation.
1 3 x % or 2 y % x $ 3 # 0 , 2 2 1 and d = 4. 2
(ii) Here, we have m = tan θ =
Therefore, by slope-intercept form (4) above, the equation of the line is
y#
1 ( x % 4) or 2 y % x $ 4 # 0 , 2
which is the required equation. 10.3.5 Intercept - form Suppose a line L makes x-intercept a and y-intercept b on the axes. Obviously L meets x-axis at the point (a, 0) and y-axis at the point (0, b) (Fig .10.16). By two-point form of the equation of the line, we have
y%0#
i.e.,
b%0 (x % a) or ay # %bx $ ab , 0%a
x y $ # 1. a b
Fig 10.16
Thus, equation of the line making intercepts a and b on x-and y-axis, respectively, is
x y $ #1 a b
... (5)
Example 10 Find the equation of the line, which makes intercepts –3 and 2 on the x- and y-axes respectively. Solution Here a = –3 and b = 2. By intercept form (5) above, equation of the line is
x y $ #1 %3 2
or
2x % 3 y $ 6 # 0 .
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10.3.6 Normal form Suppose a non-vertical line is known to us with following data: (i) Length of the perpendicular (normal) from origin to the line. (ii) Angle which normal makes with the positive direction of x-axis. Let L be the line, whose perpendicular distance from origin O be OA = p and the angle between the positive x-axis and OA be ∠XOA = ω. The possible positions of line L in the Cartesian plane are shown in the Fig 10.17. Now, our purpose is to find slope of L and a point on it. Draw perpendicular AM on the x-axis in each case.
Fig 10.17
In each case, we have OM = p cos ω and MA = p sin ω, so that the coordinates of the point A are (p cos ω, p sin ω). Further, line L is perpendicular to OA. Therefore The slope of the line L = % Thus, the line L has slope %
1 1 cos $ #% #% . slope of OA tan $ sin $
cos $ and point A ! p cos $, p sin $" on it. Therefore, by sin $
point-slope form, the equation of the line L is
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y % p sin $ # %
cos $ ! x % p cos $ " or sin $
x cos $ $ y sin $ # p (sin 2$ $ cos 2$)
or x cos ω + y sin ω = p. Hence, the equation of the line having normal distance p from the origin and angle ω which the normal makes with the positive direction of x-axis is given by x cos ω + y sin ω = p ... (6) Example 11 Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with positive direction of x-axis is 15°. Solution Here, we are given p = 4 and ω = 150 (Fig10.18).
Example 12 The Fahrenheit temperature F and absolute temperature K satisfy a linear equation. Given that K = 273 when F = 32 and that K = 373 when F = 212. Express K in terms of F and find the value of F, when K = 0. Solution Assuming F along x-axis and K along y-axis, we have two points (32, 273) and (212, 373) in XY-plane. By two-point form, the point (F, K) satisfies the equation
K % 273 #
or
373 % 273 100 ! F % 32 " or K % 273 # ! F % 32 " 212 % 32 180
... (1)
5 ! F % 32 " $ 273 9 which is the required relation. K#
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When K = 0, Equation (1) gives
0#
5 ! F % 32 " $ 273 9
or F % 32 # %
273 . 9 # % 491.4 5
or F= % 459.4 .
Alternate method We know that simplest form of the equation of a line is y = mx + c. Again assuming Falong x-axis and K along y-axis, we can take equation in the form K = mF + c ... (1) Equation (1) is satisfied by (32, 273) and (212, 373). Therefore 273 = 32m + c ... (2) and 373 = 212m + c Solving (2) and (3), we get m= ... (3)
We know, that mx + c, !NoteFor finding thesethe equation y = we needcontains two constants, namely, m and c. two constants, two conditions satisfied by the equation of line. In all the examples above, we are given two conditions to determine the equation of the line.
EXERCISE 10.2
In Exercises 1 to 8, find the equation of the line which satisfy the given conditions: 1. Write the equations for the x-and y-axes. 2. Passing through the point (– 4, 3) with slope 3. Passing through (0, 0) with slope m. 4. Passing through 2, 2 3 and inclined with the x-axis at an angle of 75o. 5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2. 6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30o with positive direction of the x-axis. 7. Passing through the points (–1, 1) and (2, – 4).
1 . 2
!
"
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8. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 300. 9. The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R. 10. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6). 11. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line. 12. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3). 13. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9. 14.
15. 16.
17.
18.
2# with the 3 positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin. The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line. The length L (in centimetrs) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134 when C = 110, express L in terms of C. The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre? P (a, b) is the mid-point of a line segment between axes. Show that equation
Find equation of the line through the point (0, 2) making an angle of the line is
x y $ # 2. a b
19. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line. 20. By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear.
10.4 General Equation of a Line
In earlier classes, we have studied general equation of first degree in two variables, Ax + By + C = 0, where A, B and C are real constants such that A and B are not zero simultaneously. Graph of the equation Ax + By + C = 0 is always a straight line.
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Therefore, any equation of the form Ax + By + C = 0, where A and B are not zero simultaneously is called general linear equation or general equation of a line. 10.4.1 Different forms of Ax + By + C = 0 The general equation of a line can be reduced into various forms of the equation of a line, by the following procedures: (a) Slope-intercept form If B ≠ 0, then Ax + By + C = 0 can be written as
y#%
where
A C x % or y # mx $ c B B
... (1)
A C and c # % . B B We know that Equation (1) is the slope-intercept form of the equation of a line m#% A C , and y-intercept is % . B B
We know that equation (1) is intercept form of the equation of a line whose x-intercept is %
C C and y-intercept is % . A B
If C = 0, then Ax + By + C = 0 can be written as Ax + By = 0, which is a line passing through the origin and, therefore, has zero intercepts on the axes. (c) Normal form Let x cos ω + y sin ω = p be the normal form of the line represented by the equation Ax + By + C = 0 or Ax + By = – C. Thus, both the equations are same and therefore,
Proper choice of signs is made so that p should be positive. Example 13 Equation of a line is 3x – 4y + 10 = 0. Find its (i) slope, (ii) x - and y-intercepts. Solution (i) Given equation 3x – 4y + 10 = 0 can be written as
10.5 Distance of a Point From a Line
The distance of a point from a line is the length of the perpendicular drawn from the point to the line. Let L : Ax + By + C = 0 be a line, whose distance from the point P (x1, y1) is d. Draw a perpendicular PM from the point P to the line L (Fig10.19). If
Fig10.19
the line meets the x-and y-axes at the points Q and R, respectively. Then, coordinates of the points are is given by
1 area (1PQR) # PM.QR , which gives PM = 2
. 2 2 A $B Thus, the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1, y1) is given by
d# A x1 $ B y1 $ C
2 2 A $B
or
d#
A x1 $ B y1 $ C
.
10.5.1 Distance between two parallel lines We know that slopes of two parallel lines are equal. Therefore, two parallel lines can be taken in the form ... (1) y = mx + c1 and y = mx + c2 ... (2) Line (1) will intersect x-axis at the point
+ c1 ( A ) % m , 0 & as shown in Fig10.20. * '
Fig10.20
Distance between two lines is equal to the length of the perpendicular from point A to line (2). Therefore, distance between the lines (1) and (2) is
If lines are given in general form, i.e., Ax + By + C1 = 0 and Ax + By + C2 = 0,
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then above formula will take the form d #
C1 % C 2 A 2 $ B2
Students can derive it themselves. Example 18 Find the distance of the point (3, – 5) from the line 3x – 4y –26 = 0. Solution Given line is 3x – 4y –26 = 0 ... (1) Comparing (1) with general equation of line Ax + By + C = 0, we get A = 3, B = – 4 and C = – 26. Given point is (x1, y1) = (3, –5). The distance of the given point from given line is
d# Ax1 $ By1 $ C A $B
2 2
7. Find equation of the line parallel to the line 3 x % 4 y $ 2 # 0 and passing through the point (–2, 3). 8. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3. 9. Find angles between the lines
3 x $ y # 1and x $ 3 y # 1.
10. The line through the points (h, 3) and (4, 1) intersects the line 7 x % 9 y % 19 # 0. at right angle. Find the value of h. 11. Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x –x1) + B (y – y1) = 0. 12. Two lines passing through the point (2, 3) intersects each other at an angle of 60o. If slope of one line is 2, find equation of the other line. 13. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2). 14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0. 15. The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c. 16. If p and q are the lengths of perpendiculars from the origin to the lines x cos ! % y sin ! # k cos 2! and x sec θ + y cosec θ = k, respectively, prove that p2 + 4q2 = k2. 17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A. 18. If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that
1 x y # # or x # 1, y # 1 . %2%3 %9$ 4 %2%3
Therefore, the point of intersection of two lines is (1, 1). Since above three lines are concurrent, the point (1, 1) will satisfy equation (3) so that 5.1 + k .1 – 3 = 0 or k = – 2. Example 21 Find the distance of the line 4x – y = 0 from the point P (4, 1) measured along the line making an angle of 135° with the positive x-axis. Solution Given line is 4x – y = 0 In order to find the distance of the line (1) from the point P (4, 1) along another line, we have to find the point of intersection of both the lines. For this purpose, we will first find the equation of the second line (Fig 10.21). Slope of second line is tan 135° = –1. Equation of the line with slope – 1 through the point P (4, 1) is ... (1)
Fig 10.21
y – 1 = – 1 (x – 4) or x + y – 5 = 0
... (2)
Solving (1) and (2), we get x = 1 and y = 4 so that point of intersection of the two lines is Q (1, 4). Now, distance of line (1) from the point P (4, 1) along the line (2) = The distance between the points P (4, 1) and Q (1, 4). =
!1 % 4 "2 $ ! 4 % 1"2 # 3
2 units .
Example 22 Assuming that straight lines work as the plane mirror for a point, find the image of the point (1, 2) in the line x % 3 y $ 4 # 0 . Solution Let Q (h, k) is the image of the point P (1, 2) in the line x – 3y + 4 = 0 ... (1)
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MATHEMATICS
Fig10.22
Therefore, the line (1) is the perpendicular bisector of line segment PQ (Fig 10.22). Hence so that Slope of line PQ =
k % 2 %1 # h %1 1 3
Solving these two relations we get k = 0 or h = equations y = 0 or x =
5 . Thus, the point (h, k) satisfy the 3
5 , which represent straight lines. Hence, path of the point 3
equidistant from the lines (1) and (2) is a straight line.
Miscellaneous Exercise on Chapter 10
1. Find the values of k for which the line (k–3) x – (4 – k2) y + k2 –7k + 6 = 0 is (a) Parallel to the x-axis, (b) Parallel to the y-axis, (c) Passing through the origin. 2. Find the values of θ and p, if the equation x cos θ + y sinθ = p is the normal form of the line 3 x + y + 2 = 0. 3. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and – 6, respectively. 4. What are the points on the y-axis whose distance from the line
x y $ # 1 is 3 4
4 units. 5. Find perpendicular distance from the origin of the line joining the points (cosθ, sin θ) and (cos φ, sin φ). 6. Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0. 7. Find the equation of a line drawn perpendicular to the line 8. 9. 10. 11. 12.
x y $ # 1 through the 4 6
point, where it meets the y-axis. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point. If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1(c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0. Find the equation of the lines through the point (3, 2) which make an angle of 45o with the line x – 2y = 3. Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
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13. Show that the equation of the line passing through the origin and making an angle θ with the line y # mx $ c is
y m $ tan ! # / . x 1 % m tan !
14. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4? 15. Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0. 16. Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point. 17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find the equation of the legs (perpendicular sides) of the triangle. 18. Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror. 19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m. 20. If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y +7 = 0 is always 10. Show that P must move on a line. 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0. 22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A. 23. Prove that the product of the lengths of the perpendiculars drawn from the
" Slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2)
y 2 % y1 y1 % y 2 # , x 1 - x 2. x 2 % x1 x1 % x 2 " If a line makes an angle á with the positive direction of x-axis, then the slope of the line is given by m = tan α, α ≠ 90°. " Slope of horizontal line is zero and slope of vertical line is undefined.
" Two lines are parallel if and only if their slopes are equal. " Two lines are perpendicular if and only if product of their slopes is –1. " Three points A, B and C are collinear, if and only if slope of AB = slope of BC. " Equation of the horizontal line having distance a from the x-axis is either
y = a or y = – a. " Equation of the vertical line having distance b from the y-axis is either x = b or x = – b. " The point (x, y) lies on the line with slope m and through the fixed point (xo, yo), if and only if its coordinates satisfy the equation y – y o = m (x – xo). " Equation of the line passing through the points (x1, y1) and (x2, y2) is given by
y % y1 #
y 2 % y1 ( x % x1). x 2 % x1
" The point (x, y) on the line with slope m and y-intercept c lies on the line if and
only if y # mx $ c . " If a line with slope m makes x-intercept d. Then equation of the line is y = m (x – d). " Equation of a line making intercepts a and b on the x-and y-axis, respectively, is
x y $ # 1. a b
" The equation of the line having normal distance from origin p and angle between
normal and the positive x-axis ω is given by x cos $ $ y sin $ # p . " Any equation of the form Ax + By + C = 0, with A and B are not zero, simultaneously, is called the general linear equation or general equation of a line. " The perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1, y1) . 2 2 A $B " Distance between the parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0, is given by d # is given by d #
Ax1 $ B y1 $ C
C1 % C 2
2 2 A $B
.
Chapter
11
CONIC SECTIONS
!Let the relation of knowledge to real life be very visible to your pupils
and let them understand how by knowledge the world could be transformed. – BERTRAND RUSSELL !
11.1 Introduction
In the preceding Chapter 10, we have studied various forms of the equations of a line. In this Chapter, we shall study about some other curves, viz., circles, ellipses, parabolas and hyperbolas. The names parabola and hyperbola are given by Apollonius. These curves are in fact, known as conic sections or more commonly conics because they can be obtained as intersections of a plane with a double napped right circular cone. These curves have a very wide range of applications in fields such as planetary motion, Apollonius (262 B.C. -190 B.C.) design of telescopes and antennas, reflectors in flashlights and automobile headlights, etc. Now, in the subsequent sections we will see how the intersection of a plane with a double napped right circular cone results in different types of curves.
11.2 Sections of a Cone
Let l be a fixed vertical line and m be another line intersecting it at a fixed point V and inclined to it at an angle α (Fig11.1). Suppose we rotate the line m around the line l in such a way that the angle α remains constant. Then the surface generated is a double-napped right circular hollow cone herein after referred as
Fig 11. 1
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237
Fig 11. 2
Fig 11. 3
cone and extending indefinitely far in both directions (Fig11.2). The point V is called the vertex; the line l is the axis of the cone. The rotating line m is called a generator of the cone. The vertex separates the cone into two parts called nappes. If we take the intersection of a plane with a cone, the section so obtained is called a conic section. Thus, conic sections are the curves obtained by intersecting a right circular cone by a plane. We obtain different kinds of conic sections depending on the position of the intersecting plane with respect to the cone and by the angle made by it with the vertical axis of the cone. Let β be the angle made by the intersecting plane with the vertical axis of the cone (Fig11.3). The intersection of the plane with the cone can take place either at the vertex of the cone or at any other part of the nappe either below or above the vertex. 11.2.1 Circle, ellipse, parabola and hyperbola When the plane cuts the nappe (other than the vertex) of the cone, we have the following situations: (a) When β = 90o, the section is a circle (Fig11.4). (b) When α < β < 90o, the section is an ellipse (Fig11.5). (c) When β = α; the section is a parabola (Fig11.6). (In each of the above three situations, the plane cuts entirely across one nappe of the cone). (d) When 0 ≤ β < α; the plane cuts through both the nappes and the curves of intersection is a hyperbola (Fig11.7).
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MATHEMATICS
Fig 11. 4
Fig 11. 5
Fig 11. 6
Fig 11. 7
11.2.2 Degenerated conic sections When the plane cuts at the vertex of the cone, we have the following different cases: (a) When α < β ≤ 90o, then the section is a point (Fig11.8). (b) When β = α, the plane contains a generator of the cone and the section is a straight line (Fig11.9). It is the degenerated case of a parabola. (c) When 0 ≤ β < α, the section is a pair of intersecting straight lines (Fig11.10). It is the degenerated case of a hyperbola.
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239
In the following sections, we shall obtain the equations of each of these conic sections in standard form by defining them based on geometric properties.
Fig 11. 8
Fig 11. 9
11.3 Circle
Fig 11. 10
Definition 1 A circle is the set of all points in a plane that are equidistant from a fixed point in the plane. The fixed point is called the centre of the circle and the distance from the centre to a point on the circle is called the radius of the circle (Fig 11.11).
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MATHEMATICS
Fig 11. 11
Fig 11. 12
The equation of the circle is simplest if the centre of the circle is at the origin. However, we derive below the equation of the circle with a given centre and radius (Fig 11.12). Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on the circle (Fig11.12). Then, by the definition, | CP | = r . By the distance formula, we have
(x – h)2 ! (y – k )2 " r
In each of the following Exercises 6 to 9, find the centre and radius of the circles.
10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16. 11. Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0. 12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3). 13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes. 14. Find the equation of a circle with centre (2,2) and passes through the point (4,5). 15. Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?
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MATHEMATICS
11.4 Parabola
Definition 2 A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the line) in the plane. The fixed line is called the directrix of the parabola and the fixed point F is called the focus (Fig 11.13). ('Para' means 'for' and 'bola' means 'throwing', i.e., the shape described when you throw a ball in the air). If the fixed point lies on the fixed !Notethe set of points in the plane, which line, then are equidistant from the fixed point and the fixed line is the straight line through the fixed point and perpendicular to the fixed line. We call this straight line as degenerate case of the parabola. A line through the focus and perpendicular to the directrix is called the axis of the parabola. The point of intersection of parabola with the axis is called the vertex of the parabola (Fig11.14).
Fig 11. 13
11.4.1 Standard equations of parabola The
equation of a parabola is simplest if the vertex is at the origin and the axis of symmetry is along the x-axis or y-axis. The four possible such orientations of parabola are shown below in Fig11.15 (a) to (d).
Fig 11.14
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243
Fig 11.15 (a) to (d)
We will derive the equation for the parabola shown above in Fig 11.15 (a) with focus at (a, 0) a > 0; and directricx x = – a as below: Let F be the focus and l the directrix. Let FM be perpendicular to the directrix and bisect FM at the point O. Produce MO to X. By the definition of parabola, the mid-point O is on the parabola and is called the vertex of the parabola. Take O as origin, OX the x-axis and OY perpendicular to it as the y-axis. Let the distance from the directrix to the focus be 2a. Then, the coordinates of the focus are (a, 0), and the equation of the directrix is x + a = 0 as in Fig11.16. Fig 11.16 Let P(x, y) be any point on the parabola such that PF = PB, ... (1) where PB is perpendicular to l. The coordinates of B are (– a, y). By the distance formula, we have
and so P(x,y) lies on the parabola. Thus, from (2) and (3) we have proved that the equation to the parabola with vertex at the origin, focus at (a,0) and directrix x = – a is y2 = 4ax. Discussion In equation (2), since a > 0, x can assume any positive value or zero but no negative value and the curve extends indefinitely far into the first and the fourth quadrants. The axis of the parabola is the positive x-axis. Similarly, we can derive the equations of the parabolas in: Fig 11.15 (b) as y2 = – 4ax, Fig 11.15 (c) as x2 = 4ay, Fig 11.15 (d) as x2 = – 4ay, These four equations are known as standard equations of parabolas. standard parabolas have focus coordinate !Note Thethe originequations of the directrix is parallelon one of the coordinate axis; vertex at and thereby to the other axis. However, the study of the equations of parabolas with focus at any point and any line as directrix is beyond the scope here. From the standard equations of the parabolas, Fig11.15, we have the following observations: 1. Parabola is symmetric with respect to the axis of the parabola.If the equation has a y2 term, then the axis of symmetry is along the x-axis and if the equation has an x2 term, then the axis of symmetry is along the y-axis. When the axis of symmetry is along the x-axis the parabola opens to the (a) right if the coefficient of x is positive, (b) left if the coefficient of x is negative. When the axis of symmetry is along the y-axis the parabola opens (c) upwards if the coefficient of y is positive. (d) downwards if the coefficient of y is negative.
2.
3.
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11.4.2 Latus rectum
Definition 3 Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose end points lie on the parabola (Fig11.17). To find the Length of the latus rectum of the parabola y2 = 4ax (Fig 11.18). By the definition of the parabola, AF = AC. But Hence AC = FM = 2a AF = 2a.
And since the parabola is symmetric with respect to x-axis AF = FB and so AB = Length of the latus rectum = 4a.
Fig 11.17
Fig 11.18
Example 5 Find the coordinates of the focus, axis, the equation of the directrix and latus rectum of the parabola y2 = 8x. Solution The given equation involves y2, so the axis of symmetry is along the x-axis. The coefficient of x is positive so the parabola opens to the right. Comparing with the given equation y2 = 4ax, we find that a = 2.
Fig 11.19
Thus, the focus of the parabola is (2, 0) and the equation of the directrix of the parabola is x = – 2 (Fig 11.19). Length of the latus rectum is 4a = 4 × 2 = 8.
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MATHEMATICS
Example 6 Find the equation of the parabola with focus (2,0) and directrix x = – 2. Solution Since the focus (2,0) lies on the x-axis, the x-axis itself is the axis of the parabola. Hence the equation of the parabola is of the form either y2 = 4ax or y2 = – 4ax. Since the directrix is x = – 2 and the focus is (2,0), the parabola is to be of the form y 2 = 4ax with a = 2. Hence the required equation is y2 = 4(2)x = 8x Example 7 Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2). Solution Since the vertex is at (0,0) and the focus is at (0,2) which lies on y-axis, the y-axis is the axis of the parabola. Therefore, equation of the parabola is of the form x2 = 4ay. thus, we have x2 = 4(2)y, i.e., x2 = 8y. Example 8 Find the equation of the parabola which is symmetric about the y-axis, and passes through the point (2,–3). Solution Since the parabola is symmetric about y-axis and has its vertex at the origin, the equation is of the form x2 = 4ay or x2 = – 4ay, where the sign depends on whether the parabola opens upwards or downwards. But the parabola passes through (2,–3) which lies in the fourth quadrant, it must open downwards. Thus the equation is of the form x2 = – 4ay. Since the parabola passes through ( 2,–3), we have 22 = – 4a (–3), i.e., a = Therefore, the equation of the parabola is
1 3
$1% x2 = # 4 & ' y, i.e., 3x2 = – 4y. ( 3)
EXERCISE 11.2
In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. 1. y2 = 12x 4. x2 = – 16y 2. x2 = 6y 5. y2 = 10x 3. 6. y2 = – 8x x2 = – 9y
In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:
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247
7. Focus (6,0); directrix x = – 6 9. Vertex (0,0); focus (3,0)
8. Focus (0,–3); directrix y = 3 10. Vertex (0,0); focus (–2,0)
11. Vertex (0,0) passing through (2,3) and axis is along x-axis. 12. Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis.
11. 5 Ellipse
Definition 4 An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. The two fixed points are called the foci (plural of 'focus') of the ellipse (Fig11.20). Note The constant sum of !distances of a point onwhich is thefrom the the the ellipse two fixed points is always greater than the distance between the two fixed points.
Fig 11.20
The mid point of the line segment joining the foci is called the centre of the ellipse. The line segment through the foci of the ellipse is called the major axis and the line segment through the centre and perpendicular to the major axis is called the minor axis. The end points of the major axis are called the vertices of the ellipse(Fig 11.21).
Fig 11.21
Fig 11.22
We denote the length of the major axis by 2a, the length of the minor axis by 2b and the distance between the foci by 2c. Thus, the length of the semi major axis is a and semi-minor axis is b (Fig11.22).
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MATHEMATICS
11.5.1 Relationship between semi-major axis, semi-minor axis and the distance of the focus from the centre of the ellipse (Fig 11.23). Take a point P at one end of the major axis. Sum of the distances of the point P to the foci is F1 P + F2P = F1O + OP + F2P (Since, F1P = F1O + OP) = c + a + a – c = 2a Take a point Q at one end of the minor axis. Sum of the distances from the point Q to the foci is F1Q + F2Q =
Fig 11.23
b2 ! c2
!
b2 ! c2 = 2 b2 ! c 2
Since both P and Q lies on the ellipse. By the definition of ellipse, we have 2 b 2 ! c 2 = 2a, i.e., or a 2 = b2 + c2 , i.e., a = b2 ! c 2 c=
11.5.2 Special cases of an ellipse In the equation c2 = a2 – b2 obtained above, if we keep a fixed and vary c from 0 to a, the resulting ellipses will vary in shape.
a 2 # b2 .
Case (i) When c = 0, both foci merge together with the centre of the ellipse and a2 = b2, i.e., a = b, and so the ellipse becomes circle (Fig11.24). Thus, circle is a special case of an ellipse which is dealt in Section 11.3. Case (ii) When c = a, then b = 0. The ellipse reduces to the line segment F1F2 joining the two foci (Fig11.25).
Fig 11.24
11.5.3 Eccentricity
Fig 11.25
Definition 5 The eccentricity of an ellipse is the ratio of the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse (eccentricity is denoted by e) i.e., e "
c . a
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249
Then since the focus is at a distance of c from the centre, in terms of the eccentricity the focus is at a distance of ae from the centre. 11.5.4 Standard equations of an ellipse The equation of an ellipse is simplest if the centre of the ellipse is at the origin and the foci are
(a) Fig 11.26
on the x-axis or y-axis. The two such possible orientations are shown in Fig 11.26. We will derive the equation for the ellipse shown above in Fig 11.26 (a) with foci on the x-axis. Let F1 and F2 be the foci and O be the midpoint of the line segment F1F2. Let O be the origin and the line from O through F2 be the positive x-axis and that through F1as the negative x-axis. Let, the line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F1 be (– c, 0) and F2 be (c, 0) (Fig 11.27). Let P(x, y) be any point on the ellipse such that the sum of the distances from P to the two x2 y 2 ! "1 foci be 2a so given a 2 b2 PF1 + PF2 = 2a. ... (1) Fig 11.27 Using the distance formula, we have
P(x, y) lies on the ellipse. Hence from (2) and (3), we proved that the equation of an ellipse with centre of the origin and major axis along the x-axis is
x2 y 2 ! = 1. a 2 b2 Discussion From the equation of the ellipse obtained above, it follows that for every point P (x, y) on the ellipse, we have
x2 y2 " 1 # 2 ≤ 1, i.e., x2 ≤ a2, so – a ≤ x ≤ a. a2 b
Therefore, the ellipse lies between the lines x = – a and x = a and touches these lines. Similarly, the ellipse lies between the lines y = – b and y = b and touches these lines. Similarly, we can derive the equation of the ellipse in Fig 11.26 (b) as These two equations are known as standard equations of the ellipses. Note The standard equations of ellipses have centre at the origin and the major and minor axis are coordinate axes. However, the study of the ellipses with centre at any other point, and any line through the centre as major and the minor axes passing through the centre and perpendicular to major axis are beyond the scope here. From the standard equations of the ellipses (Fig11.26), we have the following observations: 1. Ellipse is symmetric with respect to both the coordinate axes since if (x, y) is a point on the ellipse, then (– x, y), (x, –y) and (– x, –y) are also points on the ellipse. 2. The foci always lie on the major axis. The major axis can be determined by finding the intercepts on the axes of symmetry. That is, major axis is along the x-axis if the coefficient of x2 has the larger denominator and it is along the y-axis if the coefficient of y2 has the larger denominator.
x2 y2 ! "1 . b2 a 2
!
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MATHEMATICS
11.5.5 Latus rectum
Definition 6 Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose end points lie on the ellipse (Fig 11.28). To find the length of the latus rectum of the ellipse
x2 y 2 + =1 a 2 b2
Fig 11. 28
Let the length of AF2 be l. Then the coordinates of A are (c, l ),i.e., (ae, l )
Since the ellipse is symmetric with respect to y-axis (of course, it is symmetric w.r.t. both the coordinate axes), AF2 = F2B and so length of the latus rectum is
Example 9 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse
x2 y 2 ! "1 25 9
Solution Since denominator of
x2 y2 is larger than the denominator of , the major 25 9
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253
axis is along the x-axis. Comparing the given equation with a = 5 and b = 3. Also
x2 y 2 ! " 1 , we get a 2 b2
c " a 2 – b 2 " 25 – 9 " 4
Therefore, the coordinates of the foci are (– 4,0) and (4,0), vertices are (– 5, 0) and (5, 0). Length of the major axis is 10 units length of the minor axis 2b is 6 units and the
2b 2 18 4 " and latus rectum is . 5 a 5 Example 10 Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse 9x2 + 4y2 = 36.
eccentricity is Solution The given equation of the ellipse can be written in standard form as
x2 y2 ! "1 4 9 y2 x2 Since the denominator of is larger than the denominator of , the major axis is 9 4 along the y-axis. Comparing the given equation with the standard equation
x2 y2 ! " 1 , we have b = 2 and a = 3. b2 a 2
Also and c=
e"
a2 – b2
c 5 " a 3
=
9–4" 5
Hence the foci are (0, 5 ) and (0, – 5 ), vertices are (0,3) and (0, –3), length of the major axis is 6 units, the length of the minor axis is 4 units and the eccentricity of the
5 . 3 Example 11 Find the equation of the ellipse whose vertices are (± 13, 0) and foci are (± 5, 0). Solution Since the vertices are on x-axis, the equation will be of the form
Example 12 Find the equation of the ellipse, whose length of the major axis is 20 and foci are (0, ± 5). Solution Since the foci are on y-axis, the major axis is along the y-axis. So, equation of the ellipse is of the form Given that a = semi-major axis " and the relation
Example 13 Find the equation of the ellipse, with major axis along the x-axis and passing through the points (4, 3) and (– 1,4). Solution The standard form of the ellipse is and (–1, 4) lie on the ellipse, we have
x2 y2 ! 2 = 1. Since the points (4, 3) a2 b
... (1) ….(2)
16 9 ! 2 "1 2 a b
and
1 16 ! 2 =1 2 a b
247 247 2 and b " . 7 15
2 Solving equations (1) and (2), we find that a "
Hence the required equation is
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255
x2 y2 ! " 1 , i.e., 7x2 + 15y2 = 247. $ 247 % 247 & ' ( 7 ) 15
EXERCISE 11.3
In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse. 1. 4.
x2 y2 ! "1 36 16 x2 y2 ! "1 25 100
Ends of major axis (0, ± 5 ), ends of minor axis (± 1, 0) Length of major axis 26, foci (± 5, 0) Length of minor axis 16, foci (0, ± 6). Foci (± 3, 0), a = 4 b = 3, c = 4, centre at the origin; foci on a x axis. Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6). 20. Major axis on the x-axis and passes through the points (4,3) and (6,2). 11.6 Hyperbola Definition 7 A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant.
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MATHEMATICS
Fig 11.29
The term "difference" that is used in the definition means the distance to the further point minus the distance to the closer point. The two fixed points are called the foci of the hyperbola. The mid-point of the line segment joining the foci is called the centre of the hyperbola. The line through the foci is called the transverse axis and the line through the centre and perpendicular to the transverse axis is called the conjugate axis. The points at which the hyperbola intersects the transverse axis are called the vertices of the hyperbola (Fig 11.29). We denote the distance between the two foci by 2c, the distance between two vertices (the length of the transverse axis) by 2a and we define the quantity b as b =
c2 – a2
Fig 11.30
Also 2b is the length of the conjugate axis (Fig 11.30). To find the constant P1F2 – P1F1 :
11.6.1 Eccentricity
c is called the eccentricity of the a hyperbola. Since c ≥ a, the eccentricity is never less than one. In terms of the eccentricity, the foci are at a distance of ae from the centre.
Definition 8 Just like an ellipse, the ratio e = 11.6.2 Standard equation of Hyperbola The equation of a hyperbola is simplest if the centre of the hyperbola is at the origin and the foci are on the x-axis or y-axis. The two such possible orientations are shown in Fig11.31.
(a) Fig 11.31
(b)
We will derive the equation for the hyperbola shown in Fig 11.31(a) with foci on the x-axis. Let F1 and F2 be the foci and O be the mid-point of the line segment F1F2. Let O be the origin and the line through O through F2 be the positive x-axis and that through F 1 as the negative x-axis. The line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F1 be (– c,0) and F2 be (c,0) (Fig 11.32). Let P(x, y) be any point on the hyperbola such that the difference of the distances from P to the farther point minus the closer point be 2a. So given, PF1 – PF2 = 2a Fig 11.32
x ≥ 1, i.e., x ≤ – a or x ≥ a. Therefore, no portion of the curve lies between the a
y 2 x2 Similarly, we can derive the equation of the hyperbola in Fig 11.31 (b) as 2 # 2 = 1 a b
These two equations are known as the standard equations of hyperbolas. equations of have transverse and !NotetheThe standard axes and the hyperbolasthe origin. However, conjugate axes as coordinate centre at there are hyperbolas with any two perpendicular lines as transverse and conjugate axes, but the study of such cases will be dealt in higher classes. From the standard equations of hyperbolas (Fig11.29), we have the following observations: 1. Hyperbola is symmetric with respect to both the axes, since if (x, y) is a point on the hyperbola, then (– x, y), (x, – y) and (– x, – y) are also points on the hyperbola.
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2.
The foci are always on the transverse axis. It is the positive term whose denominator gives the transverse axis. For example, has transverse axis along x-axis of length 6, while has transverse axis along y-axis of length 10.
x2 y 2 – "1 9 16
y 2 x2 – "1 25 16
11.6.3 Latus rectum Definition 9 Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola. As in ellipse, it is easy to show that the length of the latus rectum in hyperbola is
2b 2 . a Example 14 Find the coordinates of the foci and the vertices, the eccentricity,the length of the latus rectum of the hyperbolas:
Miscellaneous Examples
Example 17 The focus of a parabolic mirror as shown in Fig 11.33 is at a distance of 5 cm from its vertex. If the mirror is 45 cm deep, find the distance AB (Fig 11.33). Solution Since the distance from the focus to the vertex is 5 cm. We have, a = 5. If the origin is taken at the vertex and the axis of the mirror lies along the positive x-axis, the equation of the parabolic section is y2 = 4 (5) x = 20 x Note that x = 45. Thus y2 = 900 Therefore y = * 30 Hence AB = 2y = 2 × 30 = 60 cm.
Fig 11.33 Example 18 A beam is supported at its ends by supports which are 12 metres apart. Since the load is concentrated at its centre, there
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263
is a deflection of 3 cm at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection 1 cm? Solution Let the vertex be at the lowest point and the axis vertical. Let the coordinate axis be chosen as shown in Fig 11.34.
Fig 11.34
The equation of the parabola takes the form x2 = 4ay. Since it passes through
Example 19 A rod AB of length 15 cm rests in between two coordinate axes in such a way that the end point A lies on x-axis and end point B lies on y-axis. A point P(x, y) is taken on the rod in such a way that AP = 6 cm. Show that the locus of P is an ellipse. Solution Let AB be the rod making an angle θ with OX as shown in Fig 11.35 and P (x, y) the point on it such that AP = 6 cm. Since AB = 15 cm, we have PB = 9 cm. From P draw PQ and PR perpendicular on y-axis and x-axis, respectively.
Miscellaneous Exercise on Chapter 11
1. 2. 3. If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola? The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle. An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end. A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis. Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum. A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man. An equilateral triangle is inscribed in the parabola y2 = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
4. 5.
6. 7.
8.
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Summary
In this Chapter the following concepts and generalisations are studied.
" A circle is the set of all points in a plane that are equidistant from a fixed point
in the plane.
" A parabola is the set of all points in a plane that are equidistant from a fixed
line and a fixed point in the plane.
" The equation of the parabola with focus at (a, 0) a > 0 and directrix x = – a is
y2 = 4ax. " Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose end points lie on the hyperbola. " Length of the latus rectum of the parabola y2 = 4ax is 4a. " An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant.
x2 y 2 " The equations of an ellipse with foci on the x-axis is a 2 + b 2 = 1 .
" Latus rectum of an ellipse is a line segment perpendicular to the major axis
through any of the foci and whose end points lie on the ellipse.
" The eccentricity of an ellipse is the ratio between the distances from the centre
of the ellipse to one of the foci and to one of the vertices of the ellipse. " A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant.
" The equation of a hyperbola with foci on the x-axis is :
x2 y 2 # "1 a 2 b2
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" Latus rectum of hyperbola is a line segment perpendicular to the transverse
axis through any of the foci and whose end points lie on the hyperbola.
" Length of the latus rectum of the hyperbola :
x2 y 2 2b2 # 2 " 1 is : . a a2 b
" The eccentricity of a hyperbola is the ratio of the distances from the centre of
the hyperbola to one of the foci and to one of the vertices of the hyperbola.
Historical Note
Geometry is one of the most ancient branches of mathematics. The Greek geometers investigated the properties of many curves that have theoretical and practical importance. Euclid wrote his treatise on geometry around 300 B.C. He was the first who organised the geometric figures based on certain axioms suggested by physical considerations. Geometry as initially studied by the ancient Indians and Greeks, who made essentially no use of the process of algebra. The synthetic approach to the subject of geometry as given by Euclid and in Sulbasutras, etc., was continued for some 1300 years. In the 200 B.C., Apollonius wrote a book called 'The Conic' which was all about conic sections with many important discoveries that have remained unsurpassed for eighteen centuries. Modern analytic geometry is called 'Cartesian' after the name of Rene Descartes (1596-1650 A.D.) whose relevant 'La Geometrie' was published in 1637. But the fundamental principle and method of analytical geometry were already discovered by Pierre de Fermat (1601-1665 A.D.). Unfortunately, Fermats treatise on the subject, entitled Ad Locus Planos et So LIDOS Isagoge (Introduction to Plane and Solid Loci) was published only posthumously in 1679 A.D. So, Descartes came to be regarded as the unique inventor of the analytical geometry. Isaac Barrow avoided using cartesian method. Newton used method of undetermined coefficients to find equations of curves. He used several types of coordinates including polar and bipolar. Leibnitz used the terms 'abscissa', 'ordinate' and 'coordinate'. L' Hospital (about 1700 A.D.) wrote an important textbook on analytical geometry. Clairaut (1729 A.D.) was the first to give the distance formula although in clumsy form. He also gave the intercept form of the linear equation. Cramer
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267
(1750 A.D.) made formal use of the two axes and gave the equation of a circle as ( y – a)2 + (b – x)2 = r He gave the best exposition of the analytical geometry of his time. Monge (1781 A.D.) gave the modern 'point-slope' form of equation of a line as y – y′ = a (x – x′) and the condition of perpendicularity of two lines as aa′ + 1 = 0. S.F. Lacroix (1765–1843 A.D.) was a prolific textbook writer, but his contributions to analytical geometry are found scattered. He gave the 'two-point' form of equation of a line as
y –!=
!, – ! (x – ") ", – "
(! – ax – b) 1 ! a2
.
and the length of the perpendicular from (α, β) on y = ax + b as
$ a, – a % His formula for finding angle between two lines was tan θ " & 1 ! aa, ' . It is, of ( ) course, surprising that one has to wait for more than 150 years after the invention of analytical geometry before finding such essential basic formula. In 1818, C. Lame, a civil engineer, gave mE + m′E′ = 0 as the curve passing through the points of intersection of two loci E = 0 and E′ = 0. Many important discoveries, both in Mathematics and Science, have been linked to the conic sections. The Greeks particularly Archimedes (287–212 B.C.) and Apollonius (200 B.C.) studied conic sections for their own beauty. These curves are important tools for present day exploration of outer space and also for research into behaviour of atomic particles.
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Chapter
12
INTRODUCTION TO THREE DIMENSIONAL GEOMETRY
!Mathematics is both the queen and the hand-maiden of all sciences – E.T. BELL!
12.1 Introduction
You may recall that to locate the position of a point in a plane, we need two intersecting mutually perpendicular lines in the plane. These lines are called the coordinate axes and the two numbers are called the coordinates of the point with respect to the axes. In actual life, we do not have to deal with points lying in a plane only. For example, consider the position of a ball thrown in space at different points of time or the position of an aeroplane as it flies from one place to another at different times during its flight. Similarly, if we were to locate the position of the Leonhard Euler (1707-1783) lowest tip of an electric bulb hanging from the ceiling of a room or the position of the central tip of the ceiling fan in a room, we will not only require the perpendicular distances of the point to be located from two perpendicular walls of the room but also the height of the point from the floor of the room. Therefore, we need not only two but three numbers representing the perpendicular distances of the point from three mutually perpendicular planes, namely the floor of the room and two adjacent walls of the room. The three numbers representing the three distances are called the coordinates of the point with reference to the three coordinate planes. So, a point in space has three coordinates. In this Chapter, we shall study the basic concepts of geometry in three dimensional space.* * For various activities in three dimensional geometry one may refer to the Book, "A Hand Book for
designing Mathematics Laboratory in Schools", NCERT, 2005.
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269
12.2 Coordinate Axes and Coordinate Planes in Three Dimensional Space
Consider three planes intersecting at a point O such that these three planes are mutually perpendicular to each other (Fig 12.1). These three planes intersect along the lines X′OX, Y′OY and Z′OZ, called the x, y and z-axes, respectively. We may note that these lines are mutually perpendicular to each other. These lines constitute the rectangular coordinate system. The planes XOY, YOZ and ZOX, called, respectively the XY-plane, YZ-plane and the ZX-plane, are known as the three coordinate planes. We take Fig 12.1 the XOY plane as the plane of the paper and the line Z′OZ as perpendicular to the plane XOY. If the plane of the paper is considered as horizontal, then the line Z′OZ will be vertical. The distances measured from XY-plane upwards in the direction of OZ are taken as positive and those measured downwards in the direction of OZ′ are taken as negative. Similarly, the distance measured to the right of ZX-plane along OY are taken as positive, to the left of ZX-plane and along OY′ as negative, in front of the YZ-plane along OX as positive and to the back of it along OX′ as negative. The point O is called the origin of the coordinate system. The three coordinate planes divide the space into eight parts known as octants. These octants could be named as XOYZ, X′OYZ, X′OY′Z, XOY′Z, XOYZ′, X′OYZ′, X′OY′Z′ and XOY′Z′. and denoted by I, II, III, ..., VIII , respectively.
12.3 Coordinates of a Point in Space
Having chosen a fixed coordinate system in the space, consisting of coordinate axes, coordinate planes and the origin, we now explain, as to how, given a point in the space, we associate with it three coordinates (x,y,z) and conversely, given a triplet of three numbers (x, y, z), how, we locate a point in the space. Given a point P in space, we drop a Fig 12.2 perpendicular PM on the XY-plane with M as the foot of this perpendicular (Fig 12.2). Then, from the point M, we draw a perpendicular ML to the x-axis, meeting it at L. Let OL be x, LM be y and MP be z. Then x,y and z are called the x, y and z coordinates, respectively, of the point P in the space. In Fig 12.2, we may note that the point P (x, y, z) lies in the octant XOYZ and so all x, y, z are positive. If P was in any other octant, the signs of x, y and z would change
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MATHEMATICS
accordingly. Thus, to each point P in the space there corresponds an ordered triplet (x, y, z) of real numbers. Conversely, given any triplet (x, y, z), we would first fix the point L on the x-axis corresponding to x, then locate the point M in the XY-plane such that (x, y) are the coordinates of the point M in the XY-plane. Note that LM is perpendicular to the x-axis or is parallel to the y-axis. Having reached the point M, we draw a perpendicular MP to the XY-plane and locate on it the point P corresponding to z. The point P so obtained has then the coordinates (x, y, z). Thus, there is a one to one correspondence between the points in space and ordered triplet (x, y, z) of real numbers. Alternatively, through the point P in the space, we draw three planes parallel to the coordinate planes, meeting the x-axis, y-axis and z-axis in the points A, B and C, respectively (Fig 12.3). Let OA = x, OB = y and OC = z. Then, the point P will have the coordinates x, y and z and we write P (x, y, z). Conversely, given x, y and z, we locate the three points A, B and C on the three coordinate axes. Through the points A, B and C we draw planes parallel to Fig 12.3 the YZ-plane, ZX-plane and XY-plane, respectively. The point of interesection of these three planes, namely, ADPF, BDPE and CEPF is obviously the point P, corresponding to the ordered triplet (x, y, z). We observe that if P (x, y, z) is any point in the space, then x, y and z are perpendicular distances from YZ, ZX and XY planes, respectively. on the x-axis will be as (x,0,0) and the coordinates of any point in the YZ-plane will be as (0, y, z). Remark The sign of the coordinates of a point determine the octant in which the point lies. The following table shows the signs of the coordinates in eight octants. Table 12.1
s ant Oct nates rdi Coo
!Note The coordinates of the origin O are (0,0,0). The coordinates of any point
I + + +
II – + +
III – – +
IV + – +
V + + –
VI – + –
VII – – –
VIII + – –
x y z
INTRODUCTION TO THREE DIMENSIONAL GEOMETRY
271
Example 1 In Fig 12.3, if P is (2,4,5), find the coordinates of F. Solution For the point F, the distance measured along OY is zero. Therefore, the coordinates of F are (2,0,5). Example 2 Find the octant in which the points (–3,1,2) and (–3,1,– 2) lie. Solution From the Table 12.1, the point (–3,1, 2) lies in second octant and the point (–3, 1, – 2) lies in octant VI.
EXERCISE 12.1
1. 2. 3. A point is on the x -axis. What are its y-coordinate and z-coordinates? A point is in the XZ-plane. What can you say about its y-coordinate? Name the octants in which the following points lie: (1,. 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (– 4, 2, –5), (– 4, 2, 5), (–3, –1, 6) (2, – 4, –7). 4. Fill in the blanks: (i) The x-axis and y-axis taken together determine a plane known as_______. (ii) The coordinates of points in the XY-plane are of the form _______. (iii) Coordinate planes divide the space into ______ octants.
12.4 Distance between Two Points
We have studied about the distance between two points in two-dimensional coordinate system. Let us now extend this study to three-dimensional system. Let P(x1, y1, z1) and Q ( x2, y2, z2) be two points referred to a system of rectangular axes OX, OY and OZ. Through the points P and Q draw planes parallel to the coordinate planes so as to form a rectangular parallelopiped with one diagonal PQ (Fig 12.4). Now, since ∠PAQ is a right angle, it follows that, in triangle PAQ, PQ2 = PA 2 + AQ2
12.5 Section Formula
In two dimensional geometry, we have learnt how to find the coordinates of a point dividing a line segment in a given ratio internally. Now, we extend this to three dimensional geometry as follows: Let the two given points be P(x1, y1, z1) and Q (x2, y2, z2). Let the point R (x, y, z)
274
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MATHEMATICS
divide PQ in the given ratio m : n internally. Draw PL, QM and RN perpendicular to the XY-plane. Obviously PL || RN || QM and feet of these perpendiculars lie in a XY-plane. The points L, M and N will lie on a line which is the intersection of the plane containing PL, RN and QM with the XY-plane. Through the point R draw a line ST parallel to the line LM. Line ST will intersect the line LP externally at the point S and the line MQ at T, as shown in Fig 12.5. Also note that quadrilaterals LNRS and NMTR are parallelograms. The triangles PSR and QTR are similar. Therefore,
These are the coordinates of the mid point of the segment joining P (x1, y1, z1) and Q (x2, y2, z2). Case 2 The coordinates of the point R which divides PQ in the ratio k : 1 are obtained by taking k #
Generally, this result is used in solving problems involving a general point on the line passing through two given points. Example 7 Find the coordinates of the point which divides the line segment joining the points (1, –2, 3) and (3, 4, –5) in the ratio 2 : 3 (i) internally, and (ii) externally. Solution (i) Let P (x, y, z) be the point which divides line segment joining A(1, – 2, 3) and B (3, 4, –5) internally in the ratio 2 : 3. Therefore
Therefore, the required point is (–3, –14, 19). Example 8 Using section formula, prove that the three points (– 4, 6, 10), (2, 4, 6) and (14, 0, –2) are collinear. Solution Let A (– 4, 6, 10), B (2, 4, 6) and C(14, 0, – 2) be the given points. Let the point P divides AB in the ratio k : 1. Then coordinates of the point P are
Therefore, C (14, 0, –2) is a point which divides AB externally in the ratio 3 : 2 and is same as P.Hence A, B, C are collinear. Example 9 Find the coordinates of the centroid of the triangle whose vertices are (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3). Solution Let ABC be the triangle. Let the coordinates of the vertices A, B,C be (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3), respectively. Let D be the mid-point of BC. Hence coordinates of D are
Example 10 Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, – 8) is divided by the YZ-plane. Solution Let YZ-plane divides the line segment joining A (4, 8, 10) and B (6, 10, – 8) at P (x, y, z) in the ratio k : 1. Then the coordinates of P are
Example 13 The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates of A and B are (3, –5, 7) and (–1, 7, – 6), respectively, find the coordinates of the point C. Solution Let the coordinates of C be (x, y, z) and the coordinates of the centroid G be (1, 1, 1). Then
Miscellaneous Exercise on Chapter 12
1. Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex. 2. Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) and (6, 0, 0). 3. If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10) and R(8, 14, 2c), then find the values of a, b and c.
INTRODUCTION TO THREE DIMENSIONAL GEOMETRY
279
4. Find the coordinates of a point on y-axis which are at a distance of 5 2 from the point P (3, –2, 5). 5. A point R with x-coordinate 4 lies on the line segment joining the points P(2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R. [Hint Suppose R divides PQ in the ratio k : 1. The coordinates of the point R are given
" In three dimensions, the coordinate axes of a rectangular Cartesian coordinate
system are three mutually perpendicular lines. The axes are called the x, y and z-axes. " The three planes determined by the pair of axes are the coordinate planes, called XY, YZ and ZX-planes.
" The three coordinate planes divide the space into eight parts known as octants. " The coordinates of a point P in three dimensional geometry is always written
in the form of triplet like (x, y, z). Here x, y and z are the distances from the YZ, ZX and XY-planes.
" (i) "
Any point on x-axis is of the form (x, 0, 0)
(ii) Any point on y-axis is of the form (0, y, 0) (iii) Any point on z-axis is of the form (0, 0, z). Distance between two points P(x1, y1, z1) and Q (x2, y2, z2) is given by
PQ # ( x2 ! x1 )2 " ( y2 ! y1 )2 " ( z2 ! z1 )2
" The coordinates of the point R which divides the line segment joining two
points P (x1 y1 z1) and Q (x2, y2, z2) internally and externally in the ratio m : n are given by
Historical Note
Rene' Descartes (1596–1650 A.D.), the father of analytical geometry, essentially dealt with plane geometry only in 1637 A.D. The same is true of his co-inventor Pierre Fermat (1601-1665 A.D.) and La Hire (1640-1718 A.D.). Although suggestions for the three dimensional coordinate geometry can be found in their works but no details. Descartes had the idea of coordinates in three dimensions but did not develop it. J.Bernoulli (1667-1748 A.D.) in a letter of 1715 A.D. to Leibnitz introduced the three coordinate planes which we use today. It was Antoinne Parent (1666-1716 A.D.), who gave a systematic development of analytical solid geometry for the first time in a paper presented to the French Academy in 1700 A.D. L.Euler (1707-1783 A.D.) took up systematically the three dimensional coordinate geometry, in Chapter 5 of the appendix to the second volume of his "Introduction to Geometry" in 1748 A.D. It was not until the middle of the nineteenth century that geometry was extended to more than three dimensions, the well-known application of which is in the Space-Time Continuum of Einstein's Theory of Relativity.
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Chapter
13
LIMITS AND DERIVATIVES
!With the Calculus as a key, Mathematics can be successfully applied to the explanation of the course of Nature – WHITEHEAD !
13.1 Introduction
This chapter is an introduction to Calculus. Calculus is that branch of mathematics which mainly deals with the study of change in the value of a function as the points in the domain change. First, we give an intuitive idea of derivative (without actually defining it). Then we give a naive definition of limit and study some algebra of limits. Then we come back to a definition of derivative and study some algebra of derivatives. We also obtain derivatives of certain standard functions.
13.2 Intuitive Idea of Derivatives
Physical experiments have confirmed that the body dropped from a tall cliff covers a distance of 4.9t2 metres in t seconds, i.e., distance s in metres covered by the body as a function of time t in seconds is given by s = 4.9t2. The adjoining Table 13.1 gives the distance travelled in metres at various intervals of time in seconds of a body dropped from a tall cliff. The objective is to find the veloctiy of the body at time t = 2 seconds from this data. One way to approach this problem is to find the average velocity for various intervals of time ending at t = 2 seconds and hope that these throw some light on the velocity at t = 2 seconds. Average velocity between t = t1 and t = t2 equals distance travelled between t = t1 and t = t2 seconds divided by (t2 – t1). Hence the average velocity in the first two seconds
From Table 13.2, we observe that the average velocity is gradually increasing. As we make the time intervals ending at t = 2 smaller, we see that we get a better idea of the velocity at t = 2. Hoping that nothing really dramatic happens between 1.99 seconds and 2 seconds, we conclude that the average velocity at t = 2 seconds is just above 19.551m/s. This conclusion is somewhat strengthened by the following set of computation. Compute the average velocities for various time intervals starting at t = 2 seconds. As before the average velocity v between t = 2 seconds and t = t2 seconds is =
The following Table 13.3 gives the average velocity v in metres per second between t = 2 seconds and t2 seconds.
Table 13.3
t2 v
4
3
2.5
2.2
2.1
2.05
2.01
29.4 24.5 22.05 20.58 20.09 19.845 19.649
Here again we note that if we take smaller time intervals starting at t = 2, we get better idea of the velocity at t = 2. In the first set of computations, what we have done is to find average velocities in increasing time intervals ending at t = 2 and then hope that nothing dramatic happens just before t = 2. In the second set of computations, we have found the average velocities decreasing in time intervals ending at t = 2 and then hope that nothing dramatic happens just after t = 2. Purely on the physical grounds, both these sequences of average velocities must approach a common limit. We can safely conclude that the velocity of the body at t = 2 is between 19.551m/s and 19.649 m/s. Technically, we say that the instantaneous velocity at t = 2 is between 19.551 m/s and 19.649 m/s. As is well-known, velocity is the rate of change of distance. Hence what we have accomplished is the following. From the given data of distance covered at various time instants we have estimated the rate of change of the distance at a given instant of time. We say that the derivative of the distance function s = 4.9t2 at t = 2 is between 19.551 and 19.649. An alternate way of viewing this limiting process is shown in Fig 13.1. This is a plot of distance s of the body from the top of the cliff versus the time t elapsed. In the limit as the sequence of time intervals h1, h2, ..., approaches zero, the sequence of average velocities approaches the same limit as does the Fig 13.1 sequence of ratios
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MATHEMATICS
C1B1 C2 B2 C3 B3 , ... , , AC1 AC2 AC3
where C1B1 = s1 – s0 is the distance travelled by the body in the time interval h1 = AC1, etc. From the Fig 13.1 it is safe to conclude that this latter sequence approaches the slope of the tangent to the curve at point A. In other words, the instantaneous velocity v(t) of a body at time t = 2 is equal to the slope of the tangent of the curve s = 4.9t2 at t = 2.
13.3 Limits
The above discussion clearly points towards the fact that we need to understand limiting process in greater clarity. We study a few illustrative examples to gain some familiarity with the concept of limits. Consider the function f(x) = x2. Observe that as x takes values very close to 0, the value of f(x) also moves towards 0 (See Fig 2.10 Chapter 2). We say
lim f # x $ ! 0
x %0
(to be read as limit of f (x) as x tends to zero equals zero). The limit of f (x) as x tends to zero is to be thought of as the value f (x) should assume at x = 0. In general as x → a, f (x) → l, then l is called limit of the function f (x) which is symbolically written as lim f # x $ ! l . x %a Consider the following function g(x) = |x|, x & 0. Observe that g(0) is not defined. Computing the value of g(x) for values of x very near to 0, we see that the value of g(x) moves towards 0. So, lim g(x) = 0. This is intuitively x%0 clear from the graph of y = |x| for x & 0. (See Fig 2.13, Chapter 2). Consider the following function.
x2 " 4 , x&2. x"2 Compute the value of h(x) for values of x very near to 2 (but not at 2). Convince yourself that all these values are near to 4. This is somewhat strengthened by considering the graph of the function y = h(x) given here (Fig 13.2). h# x$ !
Fig 13.2
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285
In all these illustrations the value which the function should assume at a given point x = a did not really depend on how is x tending to a. Note that there are essentially two ways x could approach a number a either from left or from right, i.e., all the values of x near a could be less than a or could be greater than a. This naturally leads to two limits – the right hand limit and the left hand limit. Right hand limit of a function f(x) is that value of f(x) which is dictated by the values of f(x) when x tends to a from the right. Similarly, the left hand limit. To illustrate this, consider the function
( 1, x ' 0 f # x$ ! ) + 2, x * 0
Graph of this function is shown in the Fig 13.3. It is clear that the value of f at 0 dictated by values of f(x) with x ≤ 0 equals 1, i.e., the left hand limit of f (x) at 0 is
lim f ( x) !1 .
x%0
Similarly, the value of f at 0 dictated by values of f (x) with x > 0 equals 2., i.e., the right hand limit of f (x) at 0 is
x %0 ,
lim f ( x ) ! 2 .
Fig 13.3
In this case the right and left hand limits are different, and hence we say that the limit of f (x) as x tends to zero does not exist (even though the function is defined at 0). Summary We say xlim– f(x) is the expected value of f at x = a given the values of f near %a x to the left of a. This value is called the left hand limit of f at a. We say lim, f ( x ) is the expected value f at x = a given the values of
x%a
f near x to the right of a. This value is called the right hand limit of f(x) at a. If the right and left hand limits coincide, we call that common value as the limit of f(x) at x = a and denote it by lim f(x). x %a Illustration 1 Consider the function f(x) = x + 10. We want to find the limit of this function at x = 5. Let us compute the value of the function f(x) for x very near to 5. Some of the points near and to the left of 5 are 4.9, 4.95, 4.99, 4.995. . ., etc. Values of the function at these points are tabulated below. Similarly, the real number 5.001,
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MATHEMATICS
5.01, 5.1 are also points near and to the right of 5. Value of the function at these points are also given in the Table 13.4.
Table 13.4
x f(x)
4.9 14.9
4.95 14.95
4.99 14.99
4.995 14.995
5.001 15.001
5.01 15.01
5.1 15.1
From the Table 13.4, we deduce that value of f(x) at x = 5 should be greater than 14.995 and less than 15.001 assuming nothing dramatic happens between x = 4.995 and 5.001. It is reasonable to assume that the value of the f(x) at x = 5 as dictated by the numbers to the left of 5 is 15, i.e.,
x %5 –
Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are both equal to 15. Thus,
x %5 "
lim f # x $ ! lim, f # x $ ! lim f # x $ ! 15 .
x %5 x %5
This conclusion about the limit being equal to 15 is somewhat strengthened by seeing the graph of this function which is given in Fig 2.16, Chapter 2. In this figure, we note that as x appraches 5 from either right or left, the graph of the function f(x) = x +10 approaches the point (5, 15). We observe that the value of the function at x = 2 also happens to be equal to 12. Illustration 2 Consider the function f(x) = x3. Let us try to find the limit of this function at x = 1. Proceeding as in the previous case, we tabulate the value of f(x) at x near 1. This is given in the Table 13.5.
Table 13.5
x f(x)
0.9 0.729
0.99 0.970299
0.999 0.997002999
1.001 1.003003001
1.01 1.030301
1.1 1.331
From this table, we deduce that value of f(x) at x = 1 should be greater than 0.997002999 and less than 1.003003001 assuming nothing dramatic happens between
LIMITS AND DERIVATIVES
287
x = 0.999 and 1.001. It is reasonable to assume that the value of the f(x) at x = 1 as dictated by the numbers to the left of 1 is 1, i.e.,
x %1"
Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are both equal to 1. Thus,
x %1"
lim f # x $ ! lim, f # x $ ! lim f # x $ ! 1 .
x %1 x %1
This conclusion about the limit being equal to 1 is somewhat strengthened by seeing the graph of this function which is given in Fig 2.11, Chapter 2. In this figure, we note that as x approaches 1 from either right or left, the graph of the function f(x) = x3 approaches the point (1, 1). We observe, again, that the value of the function at x = 1 also happens to be equal to 1. Illustration 3 Consider the function f(x) = 3x. Let us try to find the limit of this function at x = 2. The following Table 13.6 is now self-explanatory.
Table 13.6
x f(x)
1.9 5.7
1.95 5.85
1.99 5.97
1.999 5.997
2.001 6.003
2.01 6.03
2.1 6.3
As before we observe that as x approaches 2 from either left or right, the value of f(x) seem to approach 6. We record this as
x %2 "
lim f # x $ ! lim, f # x $ ! lim f # x $ ! 6
x %2 x %2
Its graph shown in Fig 13.4 strengthens this fact. Here again we note that the value of the function at x = 2 coincides with the limit at x = 2. Illustration 4 Consider the constant function f(x) = 3. Let us try to find its limit at x = 2. This function being the constant function takes the same
Graph of f(x) = 3 is anyway the line parallel to x-axis passing through (0, 3) and is shown in Fig 2.9, Chapter 2. From this also it is clear that the required limit is 3. In fact, it is easily observed that lim f # x $ ! 3 for any real number a. x %a Illustration 5 Consider the function f(x) = x2 + x. We want to find lim f # x $ . We x %1 tabulate the values of f(x) near x = 1 in Table 13.7.
Table 13.7
x f(x)
0.9 1.71
0.99 1.9701
0.999 1.997001
1.01 2.0301
1.1 2.31
1.2 2.64
From this it is reasonable to deduce that
x %1"
lim f # x $ ! lim f # x $ ! lim f # x $ ! 2 . ,
x %1 x %1
From the graph of f(x) = x2 + x shown in the Fig 13.5, it is clear that as x approaches 1, the graph approaches (1, 2). Here, again we observe that the
lim f (x) = f (1) x%1
1 for x * 0 . We want to know lim f (x). x%0 x2 Here, observe that the domain of the function is given to be all positive real numbers. Hence, when we tabulate the values of f(x), it does not make sense to talk of x approaching 0 from the left. Below we tabulate the values of the function for positive x close to 0 (in this table n denotes any positive integer). From the Table 13.10 given below, we see that as x tends to 0, f(x) becomes larger and larger. What we mean here is that the value of f(x) may be made larger than any given number.
As usual we make a table of x near 0 with f(x). Observe that for negative values of x we need to evaluate x – 2 and for positive values, we need to evaluate x + 2.
Table 13.11
x f(x)
– 0.1 – 2.1
– 0.01 – 2.01
– 0.001 – 2.001
0.001 2.001
0.01 2.01
0.1 2.1
From the first three entries of the Table 13.11, we deduce that the value of the function is decreasing to –2 and hence.
x %0 "
lim f # x $ ! "2
LIMITS AND DERIVATIVES
291
From the last three entires of the table we deduce that the value of the function is increasing from 2 and hence
x % 0,
lim f # x $ ! 2
Since the left and right hand limits at 0 do not coincide, we say that the limit of the function at 0 does not exist. Graph of this function is given in the Fig13.6. Here, we remark that the value of the function at x = 0 is well defined and is, indeed, equal to 0, but the limit of the function at x = 0 is not even defined. Illustration 10 As a final illustration, we find lim f # x $ , x %1 where
(x , 2 x & 1 f # x$ ! ) x !1 + 0
Table 13.12 Fig 13.6
x f(x)
0.9 2.9
0.99 2.99
0.999 2.999
1.001 3.001
1.01 3.01
1.1 3.1
As usual we tabulate the values of f(x) for x near 1. From the values of f(x) for x less than 1, it seems that the function should take value 3 at x = 1., i.e.,
x %1"
lim f # x $ ! 3 .
Similarly, the value of f(x) should be 3 as dictated by values of f(x) at x greater than 1. i.e.
x %1,
Graph of function given in Fig 13.7 strengthens our deduction about the limit. Here, we
Fig 13.7
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MATHEMATICS
note that in general, at a given point the value of the function and its limit may be different (even when both are defined). 13.3.1 Algebra of limits In the above illustrations, we have observed that the limiting process respects addition, subtraction, multiplication and division as long as the limits and functions under consideration are well defined. This is not a coincidence. In fact, below we formalise these as a theorem without proof. Theorem 1 Let f and g be two functions such that both lim f (x) and lim g(x) exist.
x %a x %a
(ii) Limit of difference of two functions is difference of the limits of the functions, i.e.,
lim [f(x) – g(x)] = lim f(x) – lim g(x).
x %a x %a x %a
(iii) Limit of product of two functions is product of the limits of the functions, i.e.,
lim [f(x) . g(x)] = lim f(x). lim g(x).
x %a x %a x %a
(iv) Limit of quotient of two functions is quotient of the limits of the functions (whenever the denominator is non zero), i.e.,
lim
x %a
g # x$
f # x$
!
lim f # x $
x %a
lim g # x $
x% a
! Note
In particular as a special case of (iii), when g is the constant function
such that g(x) = 7 , for some real number 7 , we have
lim -# 7. f $ # x $ . ! 7.lim f # x $ . / 0
x%a x %a
In the next two subsections, we illustrate how to exploit this theorem to evaluate limits of special types of functions. 13.3.2 Limits of polynomials and rational functions A function f is said to be a polynomial function if f(x) is zero function or if f(x) = a0 + a1x + a2x2 +. . . + anxn, where ais are real numbers such that an ≠ 0 for some natural number n. We know that lim x = a. Hence
x %a
= a0 , a1a , a2 a 2 , ... , an a n = f #a$ (Make sure that you understand the justification for each step in the above!)
g # x$ A function f is said to be a rational function, if f(x) = h x , where g(x) and h(x) # $
are polynomials such that h(x) ≠ 0. Then
lim f # x $ ! lim
x %a x% a
g # x$ h# x$
!
lim g # x $
x %a
lim h # x $
x% a
!
g #a$ h # a$
However, if h(a) = 0, there are two scenarios – (i) when g(a) ≠ 0 and (ii) when g(a) = 0. In the former case we say that the limit does not exist. In the latter case we can write g(x) = (x – a)k g1 (x), where k is the maximum of powers of (x – a) in g(x) Similarly, h(x) = (x – a)l h1 (x) as h (a) = 0. Now, if k ≥ l, we have
lim f # x $ !
x%a
We remark that we could cancel the term (x – 1) in the above evaluation because x & 1. Evaluation of an important limit which will be used in the sequel is given as a theorem below. Theorem 2 For any positive integer n,
lim xn " an ! na n"1 . x %a x " a
Remark The expression in the above theorem for the limit is true even if n is any rational number and a is positive.
13.4 Limits of Trigonometric Functions The following facts (stated as theorems) about functions in general come in handy in calculating limits of some trigonometric functions. Theorem 3 Let f and g be two real valued functions with the same domain such that f (x) ≤ g( x) for all x in the domain of definition, For some a, if both lim f(x) and x %a
lim g(x) exist, then lim f(x) ≤ lim g(x). This is illustrated in Fig 13.8. x %a x %a x %a
Fig 13.8
Theorem 4 (Sandwich Theorem) Let f, g and h be real functions such that f (x) ≤ g( x) ≤ h(x) for all x in the common domain of definition. For some real number a, if lim f(x) = l = lim h(x), then lim g(x) = l. This is illustrated in Fig 13.9. x %a x %a
x% a
Fig 13.9
Given below is a beautiful geometric proof of the following important inequality relating trigonometric functions.
A general rule that needs to be kept in mind while evaluating limits is the following.
f # x$ Say, given that the limit lim g x exists and we want to evaluate this. First we check x%a # $
the value of f (a) and g(a). If both are 0, then we see if we can get the factor which is causing the terms to vanish, i.e., see if we can write f(x) = f1 (x) f2(x) so that f1 (a) = 0 and f2 (a) ≠ 0. Similarly, we write g(x) = g1 (x) g2(x), where g1(a) = 0 and g2(a) ≠ 0. Cancel out the common factors from f(x) and g(x) (if possible) and write
13.5 Derivatives
We have seen in the Section 13.2, that by knowing the position of a body at various time intervals it is possible to find the rate at which the position of the body is changing. It is of very general interest to know a certain parameter at various instants of time and try to finding the rate at which it is changing. There are several real life situations where such a process needs to be carried out. For instance, people maintaining a reservoir need to know when will a reservoir overflow knowing the depth of the water at several instances of time, Rocket Scientists need to compute the precise velocity with which the satellite needs to be shot out from the rocket knowing the height of the rocket at various times. Financial institutions need to predict the changes in the value of a particular stock knowing its present value. In these, and many such cases it is desirable to know how a particular parameter is changing with respect to some other parameter. The heart of the matter is derivative of a function at a given point in its domain of definition.
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MATHEMATICS
Definition 1 Suppose f is a real valued function and a is a point in its domain of definition. The derivative of f at a is defined by
h provided this limit exists. Derivative of f (x) at a is denoted by f′(a). lim f # a , h$ " f #a $
h %0
Observe that f′ (a) quantifies the change in f(x) at a with respect to x.
We now present a geometric interpretation of derivative of a function at a point. Let y = f(x) be a function and let P = (a, f(a)) and Q = (a + h, f(a + h) be two points close to each other on the graph of this function. The Fig 13.11 is now self explanatory.
Fig 13.11
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MATHEMATICS
We know that f B # a $ ! lim
f #a , h$ " f #a$
h From the triangle PQR, it is clear that the ratio whose limit we are taking is precisely equal to tan(QPR) which is the slope of the chord PQ. In the limiting process, as h tends to 0, the point Q tends to P and we have QR ! lim Q % P PR h This is equivalent to the fact that the chord PQ tends to the tangent at P of the curve y = f(x). Thus the limit turns out to be equal to the slope of the tangent. Hence lim
h %0
h %0
f #a , h$ " f #a$
f B # a $ ! tan " .
For a given function f we can find the derivative at every point. If the derivative exists at every point, it defines a new function called the derivative of f . Formally, we define derivative of a function as follows. Definition 2 Suppose f is a real valued function, the function defined by
h wherever the limit exists is defined to be the derivative of f at x and is denoted by f′(x). This definition of derivative is also called the first principle of derivative. h Clearly the domain of definition of f′ (x) is wherever the above limit exists. There are different notations for derivative of a function. Sometimes f′ (x) is denoted by lim f # x , h$ " f # x$
h %0
Thus
f ' # x $ ! lim
f # x , h$ " f # x$
h %0
d # f # x $$ or if y = f(x), it is denoted by dy . This is referred to as derivative of f(x) dx dx or y with respect to x. It is also denoted by D (f (x) ). Further, derivative of f at x = a
is also denoted by
13.5.1 Algebra of derivative of functions Since the very definition of derivatives involve limits in a rather direct fashion, we expect the rules for derivatives to follow closely that of limits. We collect these in the following theorem. Theorem 5 Let f and g be two functions such that their derivatives are defined in a common domain. Then
The proofs of these follow essentially from the analogous theorem for limits. We will not prove these here. As in the case of limits this theorem tells us how to compute derivatives of special types of functions. The last two statements in the theorem may be restated in the following fashion which aids in recalling them easily: Let u ! f # x $ and v = g (x). Then
# uv $B
= u Bv , uvB
This is referred to a Leibnitz rule for differentiating product of functions or the product rule. Similarly, the quotient rule is
LIMITS AND DERIVATIVES
309
< u =B u Bv " uvB > ? = v2 @vA
Now, let us tackle derivatives of some standard functions. It is easy to see that the derivative of the function f(x) = x is the constant function 1. This is because f'(x) = lim
f # x , h$ " f # x$ h
h %0
= lim
h %0
x,h"x h
= lim1 ! 1 . h%0 We use this and the above theorem to compute the derivative of f(x) = 10x = x + .... + x (ten terms). By (1) of the above theorem
Alternatively, we may also prove this by induction on n and the product rule as follows. The result is true for n = 1, which has been proved earlier. We have
d n d x.x n"1 x = dx dx
=
# $
#
$ # $
$
d d # x $ . xn "1 , x. x n"1 (by product rule) dx dx
n "1
# $
#
= 1.x
, x. # n " 1$ x n " 2 (by induction hypothesis)
= x n "1 , # n " 1$ x n"1 ! nx n"1 . Remark The above theorem is true for all powers of x, i.e., n can be any real number (but we will not prove it here). 13.4.2 Derivative of polynomials and trigonometric functions We start with the
following theorem which tells us the derivative of a polynomial function.
Theorem 7 Let f(x) = an x n , an"1 x n"1 , .... , a1 x , a0 be a polynomial function, where ai s are all real numbers and an & 0. Then, the derivative function is given by
1 . Here, we use the fact tan x that the derivative of tan x is sec2 x which we saw in Example 17 and also that the derivative of the constant function is 0.
Alternatively, this may be computed by noting that cot x !
" The expected value of the function as dictated by the points to the left of a
point defines the left hand limit of the function at that point. Similarly the right hand limit. " Limit of a function at a point is the common value of the left and right hand limits, if they coincide.
Historical Note
In the history of mathematics two names are prominent to share the credit for inventing calculus, Issac Newton (1642 – 1727) and G.W. Leibnitz (1646 – 1717). Both of them independently invented calculus around the seventeenth century. After the advent of calculus many mathematicians contributed for further development of calculus. The rigorous concept is mainly attributed to the great
320
MATHEMATICS
mathematicians, A.L. Cauchy, J.L.Lagrange and Karl Weierstrass. Cauchy gave the foundation of calculus as we have now generally accepted in our textbooks. Cauchy used D' Alembert's limit concept to define the derivative of a function. Starting with definition of a limit, Cauchy gave examples such as the limit of
sin D
D
; y f ( x , i ) " f ( x) , and called the limit for for D = 0. He wrote ; x ! i
i % 0, the "function derive'e, y′ for f ′ (x)".
Before 1900, it was thought that calculus is quite difficult to teach. So calculus became beyond the reach of youngsters. But just in 1900, John Perry and others in England started propagating the view that essential ideas and methods of calculus were simple and could be taught even in schools. F.L. Griffin, pioneered the teaching of calculus to first year students. This was regarded as one of the most daring act in those days. Today not only the mathematics but many other subjects such as Physics, Chemistry, Economics and Biological Sciences are enjoying the fruits of calculus.
—! —
Chapter
14
MATHEMATICAL REASONING
!There are few things which we know which are not capable of
mathematical reasoning and when these can not, it is a sign that our knowledge of them is very small and confused and where a mathematical reasoning can be had, it is as great a folly to make use of another, as to grope for a thing in the dark when you have a candle stick standing by you. – ARTHENBOT !
14.1 Introduction
In this Chapter, we shall discuss about some basic ideas of Mathematical Reasoning. All of us know that human beings evolved from the lower species over many millennia. The main asset that made humans "superior" to other species was the ability to reason. How well this ability can be used depends on each person's power of reasoning. How to develop this power? Here, we shall discuss the process of reasoning especially in the context of mathematics. In mathematical language, there are two kinds of reasoning – inductive and deductive. We have already discussed the inductive reasoning in the context of mathematical induction. In this Chapter, we shall discuss some fundamentals of deductive reasoning.
George Boole (1815 - 1864)
14.2 Statements
The basic unit involved in mathematical reasoning is a mathematical statement. Let us start with two sentences: In 2003, the president of India was a woman. An elephant weighs more than a human being.
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MATHEMATICS
When we read these sentences, we immediately decide that the first sentence is false and the second is correct. There is no confusion regarding these. In mathematics such sentences are called statements. On the other hand, consider the sentence: Women are more intelligent than men. Some people may think it is true while others may disagree. Regarding this sentence we cannot say whether it is always true or false . That means this sentence is ambiguous. Such a sentence is not acceptable as a statement in mathematics. A sentence is called a mathematically acceptable statement if it is either true or false but not both. Whenever we mention a statement here, it is a "mathematically acceptable" statement. While studying mathematics, we come across many such sentences. Some examples are: Two plus two equals four. The sum of two positive numbers is positive. All prime numbers are odd numbers. Of these sentences, the first two are true and the third one is false. There is no ambiguity regarding these sentences. Therefore, they are statements. Can you think of an example of a sentence which is vague or ambiguous? Consider the sentence: The sum of x and y is greater than 0 Here, we are not in a position to determine whether it is true or false, unless we know what x and y are. For example, it is false where x = 1, y = –3 and true when x = 1 and y = 0. Therefore, this sentence is not a statement. But the sentence: For any natural numbers x and y, the sum of x and y is greater than 0 is a statement. Now, consider the following sentences : How beautiful! Open the door.
Where are you going? Are they statements? No, because the first one is an exclamation, the second an order and the third a question. None of these is considered as a statement in mathematical language. Sentences involving variable time such as "today", "tomorrow" or "yesterday" are not statements. This is because it is not known what time is referred here. For example, the sentence Tomorrow is Friday
MATHEMATICAL REASONING
323
is not a statement. The sentence is correct (true) on a Thursday but not on other days. The same argument holds for sentences with pronouns unless a particular person is referred to and for variable places such as "here", "there" etc., For example, the sentences
She is a mathematics graduate. Kashmir is far from here. are not statements. Here is another sentence There are 40 days in a month. Would you call this a statement? Note that the period mentioned in the sentence above is a "variable time" that is any of 12 months. But we know that the sentence is always false (irrespective of the month) since the maximum number of days in a month can never exceed 31. Therefore, this sentence is a statement. So, what makes a sentence a statement is the fact that the sentence is either true or false but not both. While dealing with statements, we usually denote them by small letters p, q, r,... For example, we denote the statement "Fire is always hot" by p. This is also written as p: Fire is always hot. Example 1 Check whether the following sentences are statements. Give reasons for your answer. (i) 8 is less than 6. (iii) The sun is a star. (v) There is no rain without clouds. (ii) Every set is a finite set. (iv) Mathematics is fun. (vi) How far is Chennai from here?
Solution (i) This sentence is false because 8 is greater than 6. Hence it is a statement. (ii) This sentence is also false since there are sets which are not finite. Hence it is a statement. (iii) It is a scientifically established fact that sun is a star and, therefore, this sentence is always true. Hence it is a statement. (iv) This sentence is subjective in the sense that for those who like mathematics, it may be fun but for others it may not be. This means that this sentence is not always true. Hence it is not a statement.
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MATHEMATICS
(v) It is a scientifically established natural phenomenon that cloud is formed before it rains. Therefore, this sentence is always true. Hence it is a statement. (vi) This is a question which also contains the word "Here". Hence it is not a statement. The above examples show that whenever we say that a sentence is a statement we should always say why it is so. This "why" of it is more important than the answer.
EXERCISE 14.1
1. Which of the following sentences are statements? Give reasons for your answer. (i) There are 35 days in a month. (ii) Mathematics is difficult. (iii) The sum of 5 and 7 is greater than 10. (iv) The square of a number is an even number. (v) The sides of a quadrilateral have equal length. (vi) Answer this question. (vii) The product of (–1) and 8 is 8. (viii) The sum of all interior angles of a triangle is 180°. (ix) Today is a windy day. (x) All real numbers are complex numbers. 2. Give three examples of sentences which are not statements. Give reasons for the answers.
14.3 New Statements from Old
We now look into method for producing new statements from those that we already have. An English mathematician, "George Boole" discussed these methods in his book "The laws of Thought" in 1854. Here, we shall discuss two techniques. As a first step in our study of statements, we look at an important technique that we may use in order to deepen our understanding of mathematical statements. This technique is to ask not only what it means to say that a given statement is true but also what it would mean to say that the given statement is not true.
14.3.1 Negation of a statement The denial of a statement is called the negation of
the statement. Let us consider the statement: p: New Delhi is a city The negation of this statement is
MATHEMATICAL REASONING
325
It is not the case that New Delhi is a city This can also be written as It is false that New Delhi is a city. This can simply be expressed as New Delhi is not a city. Definition 1 If p is a statement, then the negation of p is also a statement and is denoted by ∼ p, and read as 'not p'. the negation !Note"ItWhile formingare also used. of a statement, phrases like, "It is not the case" or is false that" Here is an example to illustrate how, by looking at the negation of a statement, we may improve our understanding of it. Let us consider the statement p: Everyone in Germany speaks German. The denial of this sentence tells us that not everyone in Germany speaks German. This does not mean that no person in Germany speaks German. It says merely that at least one person in Germany does not speak German. We shall consider more examples. Example 2 Write the negation of the following statements. (i) Both the diagonals of a rectangle have the same length. (ii)
7 is rational.
Solution (i) This statement says that in a rectangle, both the diagonals have the same length. This means that if you take any rectangle, then both the diagonals have the same length. The negation of this statement is It is false that both the diagonals in a rectangle have the same length This means the statement There is atleast one rectangle whose both diagonals do not have the same length. (ii) The negation of the statement in (ii) may also be written as It is not the case that 7 is rational. This can also be rewritten as
7 is not rational.
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MATHEMATICS
Example 3 Write the negation of the following statements and check whether the resulting statements are true, (i) Australia is a continent. (ii) There does not exist a quadrilateral which has all its sides equal. (iii) Every natural number is greater than 0. (iv) The sum of 3 and 4 is 9. Solution (i) The negation of the statement is It is false that Australia is a continent. This can also be rewritten as Australia is not a continent. We know that this statement is false. (ii) The negation of the statement is It is not the case that there does not exist a quadrilateral which has all its sides equal. This also means the following: There exists a quadrilateral which has all its sides equal. This statement is true because we know that square is a quadrilaterial such that its four sides are equal. (iii) The negation of the statement is It is false that every natural number is greater than 0. This can be rewritten as There exists a natural number which is not greater than 0. This is a false statement. (iv) The negation is It is false that the sum of 3 and 4 is 9. This can be written as The sum of 3 and 4 is not equal to 9. This statement is true. 14.3.2 Compound statements Many mathematical statements are obtained by combining one or more statements using some connecting words like "and", "or", etc. Consider the following statement p: There is something wrong with the bulb or with the wiring. This statement tells us that there is something wrong with the bulb or there is
MATHEMATICAL REASONING
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something wrong with the wiring. That means the given statement is actually made up of two smaller statements: q: There is something wrong with the bulb. r: There is something wrong with the wiring. connected by "or" Now, suppose two statements are given as below: p: 7 is an odd number. q: 7 is a prime number. These two statements can be combined with "and" r: 7 is both odd and prime number. This is a compound statement. This leads us to the following definition: Definition 2 A Compound Statement is a statement which is made up of two or more statements. In this case, each statement is called a component statement. Let us consider some examples. Example 4 Find the component statements of the following compound statements. (i) The sky is blue and the grass is green. (ii) It is raining and it is cold. (iii) All rational numbers are real and all real numbers are complex. (iv) 0 is a positive number or a negative number. Solution Let us consider one by one (i) The component statements are p: The sky is blue. q: The grass is green. The connecting word is 'and'. (ii) The component statements are p: It is raining. q: It is cold.
The connecting word is 'and'.
(iii)The component statements are p: All rational numbers are real. q: All real numbers are complex.
The connecting word is 'and'.
(iv)The component statements are
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p: 0 is a positive number. q: 0 is a negative number.
The connecting word is 'or'.
Example 5 Find the component statements of the following and check whether they are true or not. (i) A square is a quadrilateral and its four sides equal. (ii) All prime numbers are either even or odd. (iii) A person who has taken Mathematics or Computer Science can go for MCA. (iv) Chandigarh is the capital of Haryana and UP. (v)
2 is a rational number or an irrational number.
(vi) 24 is a multiple of 2, 4 and 8. Solution (i) The component statements are p: A square is a quadrilateral. q: A square has all its sides equal. We know that both these statements are true. Here the connecting word is 'and'. (ii) The component statements are p: All prime numbers are odd number. q: All prime numbers are even number. Both these statements are false and the connecting word is 'or'. (iii) The component statements are p: A person who has taken Mathematics can go for MCA. q: A person who has taken computer science can go for MCA. Both these statements are true. Here the connecting word is 'or'. (iv) The component statements are p: Chandigarh is the capital of Haryana. q: Chandigarh is the capital of UP. The first statement is true but the second is false. Here the connecting word is 'and'. (v) The component statements are
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p: q:
2 is a rational number. 2 is an irrational number.
The first statement is false and second is true. Here the connecting word is 'or'. (vi) The component statements are p: 24 is a multiple of 2. q: 24 is a multiple of 4. r: 24 is a multiple of 8. All the three statements are true. Here the connecting words are 'and'. Thus, we observe that compound statements are actually made-up of two or more statements connected by the words like "and", "or", etc. These words have special meaning in mathematics. We shall discuss this mattter in the following section.
EXERCISE 14.2
1. Write the negation of the following statements: (i) Chennai is the capital of Tamil Nadu. (ii) (iii) (iv)
2 is not a complex number
All triangles are not equilateral triangle. The number 2 is greater than 7.
(v) Every natural number is an integer. 2. Are the following pairs of statements negations of each other: (i) The number x is not a rational number. The number x is not an irrational number. The number x is a rational number. The number x is an irrational number. 3. Find the component statements of the following compound statements and check whether they are true or false. (i) Number 3 is prime or it is odd. (ii) (iii) All integers are positive or negative. 100 is divisible by 3, 11 and 5. (ii)
14.4 Special Words/Phrases
Some of the connecting words which are found in compound statements like "And",
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"Or", etc. are often used in Mathematical Statements. These are called connectives. When we use these compound statements, it is necessary to understand the role of these words. We discuss this below.
14.4.1 The word "And" Let us look at a compound statement with "And".
p: A point occupies a position and its location can be determined. The statement can be broken into two component statements as q: A point occupies a position. r: Its location can be determined. Here, we observe that both statements are true. Let us look at another statement. p: 42 is divisible by 5, 6 and 7. This statement has following component statements q: 42 is divisible by 5. r: 42 is divisible by 6. s: 42 is divisible by 7. Here, we know that the first is false while the other two are true. We have the following rules regarding the connective "And" 1. 2. The compound statement with 'And' is true if all its component statements are true. The component statement with 'And' is false if any of its component statements is false (this includes the case that some of its component statements are false or all of its component statements are false). Example 6 Write the component statements of the following compound statements and check whether the compound statement is true or false. (i) A line is straight and extends indefinitely in both directions. (ii) 0 is less than every positive integer and every negative integer. (iii) All living things have two legs and two eyes. Solution (i) The component statements are p: A line is straight. q: A line extends indefinitely in both directions.
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Both these statements are true, therefore, the compound statement is true. (ii) The component statements are p: 0 is less than every positive integer. q: 0 is less than every negative integer. The second statement is false. Therefore, the compound statement is false. (iii) The two component statements are p: All living things have two legs. q: All living things have two eyes. Both these statements are false. Therefore, the compound statement is false. Now, consider the following statement. p: A mixture of alcohol and water can be separated by chemical methods. This sentence cannot be considered as a compound statement with "And". Here the word "And" refers to two things – alcohol and water. This leads us to an important note. "And" is always compound statement !NoteinDo not think that a statement with word "And" is notaused as a conjunction. as shown the above example. Therefore, the 14.4.2 The word "Or" Let us look at the following statement. p: Two lines in a plane either intersect at one point or they are parallel. We know that this is a true statement. What does this mean? This means that if two lines in a plane intersect, then they are not parallel. Alternatively, if the two lines are not parallel, then they intersect at a point. That is this statement is true in both the situations. In order to understand statements with "Or" we first notice that the word "Or" is used in two ways in English language. Let us first look at the following statement. p: An ice cream or pepsi is available with a Thali in a restaurant. This means that a person who does not want ice cream can have a pepsi along with Thali or one does not want pepsi can have an ice cream along with Thali. That is, who do not want a pepsi can have an ice cream. A person cannot have both ice cream and pepsi. This is called an exclusive "Or". Here is another statement. A student who has taken biology or chemistry can apply for M.Sc. microbiology programme. Here we mean that the students who have taken both biology and chemistry can apply for the microbiology programme, as well as the students who have taken only one of these subjects. In this case, we are using inclusive "Or". It is important to note the difference between these two ways because we require this when we check whether the statement is true or not.
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Let us look at an example. Example 7 For each of the following statements, determine whether an inclusive "Or" or exclusive "Or" is used. Give reasons for your answer. (i) To enter a country, you need a passport or a voter registration card. (ii) The school is closed if it is a holiday or a Sunday. (iii) Two lines intersect at a point or are parallel. (iv) Students can take French or Sanskrit as their third language. Solution (i) Here "Or" is inclusive since a person can have both a passport and a voter registration card to enter a country. (ii) Here also "Or" is inclusive since school is closed on holiday as well as on Sunday. (iii) Here "Or" is exclusive because it is not possible for two lines to intersect and parallel together. (iv) Here also "Or" is exclusive because a student cannot take both French and Sanskrit. Rule for the compound statement with 'Or' 1. 2. A compound statement with an 'Or' is true when one component statement is true or both the component statements are true.
A compound statement with an 'Or' is false when both the component statements are false. For example, consider the following statement. p: Two lines intersect at a point or they are parallel The component statements are q: Two lines intersect at a point. r: Two lines are parallel. Then, when q is true r is false and when r is true q is false. Therefore, the compound statement p is true. Consider another statement. p: 125 is a multiple of 7 or 8. Its component statements are q: 125 is a multiple of 7. r: 125 is a multiple of 8. Both q and r are false. Therefore, the compound statement p is false.
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Again, consider the following statement: p: The school is closed, if there is a holiday or Sunday. The component statements are q: School is closed if there is a holiday. r: School is closed if there is a Sunday. Both q and r are true, therefore, the compound statement is true. Consider another statement. p: Mumbai is the capital of Kolkata or Karnataka. The component statements are q: Mumbai is the capital of Kolkata. r: Mumbai is the capital of Karnataka. Both these statements are false. Therefore, the compound statement is false. Let us consider some examples. Example 8 Identify the type of "Or" used in the following statements and check whether the statements are true or false: (i)
2 is a rational number or an irrational number.
(ii) To enter into a public library children need an identity card from the school or a letter from the school authorities. (iii) A rectangle is a quadrilateral or a 5-sided polygon. Solution (i) The component statements are p: 2 is a rational number. q: 2 is an irrational number. Here, we know that the first statement is false and the second is true and "Or" is exclusive. Therefore, the compound statement is true. (ii) The component statements are p: To get into a public library children need an identity card. q: To get into a public library children need a letter from the school authorities. Children can enter the library if they have either of the two, an identity card or the letter, as well as when they have both. Therefore, it is inclusive "Or" the compound statement is also true when children have both the card and the letter. (iii) Here "Or" is exclusive. When we look at the component statements, we get that the statement is true.
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14.4.3 Quantifiers Quantifiers are phrases like, "There exists" and "For all". Another phrase which appears in mathematical statements is "there exists". For example, consider the statement. p: There exists a rectangle whose all sides are equal. This means that there is atleast one rectangle whose all sides are equal. A word closely connected with "there exists" is "for every" (or for all). Consider a statement. p: For every prime number p,
p is an irrational number.
This means that if S denotes the set of all prime numbers, then for all the members p of the set S,
p is an irrational number.
In general, a mathematical statement that says "for every" can be interpreted as saying that all the members of the given set S where the property applies must satisfy that property. We should also observe that it is important to know precisely where in the sentence a given connecting word is introduced. For example, compare the following two sentences: 1. For every positive number x there exists a positive number y such that y < x. 2. There exists a positive number y such that for every positive number x, we have y < x. Although these statements may look similar, they do not say the same thing. As a matter of fact, (1) is true and (2) is false. Thus, in order for a piece of mathematical writing to make sense, all of the symbols must be carefully introduced and each symbol must be introduced at precisely the right place – not too early and not too late. The words "And" and "Or" are called connectives and "There exists" and "For all" are called quantifiers. Thus, we have seen that many mathematical statements contain some special words and it is important to know the meaning attached to them, especially when we have to check the validity of different statements.
EXERCISE 14.3
1. For each of the following compound statements first identify the connecting words and then break it into component statements. (i) All rational numbers are real and all real numbers are not complex. (ii) Square of an integer is positive or negative. (iii) The sand heats up quickly in the Sun and does not cool down fast at night. (iv) x = 2 and x = 3 are the roots of the equation 3x2 – x – 10 = 0.
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2. Identify the quantifier in the following statements and write the negation of the statements. (i) There exists a number which is equal to its square. (ii) For every real number x, x is less than x + 1. (iii) There exists a capital for every state in India. 3. Check whether the following pair of statements are negation of each other. Give reasons for your answer. (i) x + y = y + x is true for every real numbers x and y. (ii) There exists real numbers x and y for which x + y = y + x. 4. State whether the "Or" used in the following statements is "exclusive "or" inclusive. Give reasons for your answer. (i) Sun rises or Moon sets. (ii) (iii) To apply for a driving licence, you should have a ration card or a passport. All integers are positive or negative.
14.5 Implications
In this Section, we shall discuss the implications of "if-then", "only if" and "if and only if ". The statements with "if-then" are very common in mathematics. For example, consider the statement. r: If you are born in some country, then you are a citizen of that country. When we look at this statement, we observe that it corresponds to two statements p and q given by p : you are born in some country. q : you are citizen of that country. Then the sentence "if p then q" says that in the event if p is true, then q must be true. One of the most important facts about the sentence "if p then q" is that it does not say any thing (or places no demand) on q when p is false. For example, if you are not born in the country, then you cannot say anything about q. To put it in other words" not happening of p has no effect on happening of q. Another point to be noted for the statement "if p then q" is that the statement does not imply that p happens. There are several ways of understanding "if p then q" statements. We shall illustrate these ways in the context of the following statement. r: If a number is a multiple of 9, then it is a multiple of 3. Let p and q denote the statements p : a number is a multiple of 9. q: a number is a multiple of 3.
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Then, if p then q is the same as the following: 1. p implies q is denoted by p ⇒ q. The symbol ⇒ stands for implies. This says that a number is a multiple of 9 implies that it is a multiple of 3. 2. p is a sufficient condition for q. This says that knowing that a number as a multiple of 9 is sufficient to conclude that it is a multiple of 3. 3. p only if q. This says that a number is a multiple of 9 only if it is a multiple of 3. 4. q is a necessary condition for p. This says that when a number is a multiple of 9, it is necessarily a multiple of 3. 5. ∼q implies ∼p. This says that if a number is not a multiple of 3, then it is not a multiple of 9.
14.5.1 Contrapositive and converse Contrapositive and converse are certain
other statements which can be formed from a given statement with "if-then". For example, let us consider the following "if-then" statement. If the physical environment changes, then the biological environment changes. Then the contrapositive of this statement is If the biological environment does not change, then the physical environment does not change. Note that both these statements convey the same meaning. To understand this, let us consider more examples. Example 9 Write the contrapositive of the following statement: (i) If a number is divisible by 9, then it is divisible by 3. (ii) If you are born in India, then you are a citizen of India. (iii) If a triangle is equilateral, it is isosceles. Solution The contrapositive of the these statements are (i) If a number is not divisible by 3, it is not divisible by 9. (ii) If you are not a citizen of India, then you were not born in India. (iii) If a triangle is not isosceles, then it is not equilateral. The above examples show the contrapositive of the statement if p, then q is "if ∼q, then ∼p". Next, we shall consider another term called converse. The converse of a given statement "if p, then q" is if q, then p.
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For example, the converse of the statement p: If a number is divisible by 10, it is divisible by 5 is q: If a number is divisible by 5, then it is divisible by 10. Example 10 Write the converse of the following statements. (i) If a number n is even, then n2 is even. (ii) If you do all the exercises in the book, you get an A grade in the class. (iii) If two integers a and b are such that a > b, then a – b is always a positive integer. Solution The converse of these statements are (i) If a number n2 is even, then n is even. (ii) If you get an A grade in the class, then you have done all the exercises of the book. (iii) If two integers a and b are such that a – b is always a positive integer, then a > b. Let us consider some more examples. Example 11 For each of the following compound statements, first identify the corresponding component statements. Then check whether the statements are true or not. (i) If a triangle ABC is equilateral, then it is isosceles. (ii) If a and b are integers, then ab is a rational number. Solution (i) The component statements are given by p : Triangle ABC is equilateral. q : Triangle ABC is Isosceles. Since an equilateral triangle is isosceles, we infer that the given compound statement is true. (ii) The component statements are given by p : a and b are integers. q : ab is a rational number. since the product of two integers is an integer and therefore a rational number, the compound statement is true. 'If and only if', represented by the symbol '⇔' means the following equivalent forms for the given statements p and q. (i) p if and only if q (ii) q if and only if p
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(iii) p is necessary and sufficient condition for q and vice-versa (iv) p ⇔ q Consider an example. Example 12 Given below are two pairs of statements. Combine these two statements using "if and only if ". (i) p: If a rectangle is a square, then all its four sides are equal. q: If all the four sides of a rectangle are equal, then the rectangle is a square. (ii) p: If the sum of digits of a number is divisible by 3, then the number is divisible by 3. q: If a number is divisible by 3, then the sum of its digits is divisible by 3. Solution (i) A rectangle is a square if and only if all its four sides are equal. (ii) A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
EXERCISE 14.4
1. Rewrite the following statement with "if-then" in five different ways conveying the same meaning. If a natural number is odd, then its square is also odd. 2. Write the contrapositive and converse of the following statements. (i) (ii) (iii) (iv) (v) (i) (ii) (iii) (iv) If x is a prime number, then x is odd. If the two lines are parallel, then they do not intersect in the same plane. Something is cold implies that it has low temperature. You cannot comprehend geometry if you do not know how to reason deductively. x is an even number implies that x is divisible by 4. You get a job implies that your credentials are good. The Bannana trees will bloom if it stays warm for a month. A quadrilateral is a parallelogram if its diagonals bisect each other. To get an A+ in the class, it is necessary that you do all the exercises of the book.
3. Write each of the following statements in the form "if-then"
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4. Given statements in (a) and (b). Identify the statements given below as contrapositive or converse of each other. (a) (i) (ii) (b) (i) (ii) If you live in Delhi, then you have winter clothes. If you do not have winter clothes, then you do not live in Delhi. If you have winter clothes, then you live in Delhi. If a quadrilateral is a parallelogram, then its diagonals bisect each other. If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
14.6 Validating Statements
In this Section, we will discuss when a statement is true. To answer this question, one must answer all the following questions. What does the statement mean? What would it mean to say that this statement is true and when this statement is not true? The answer to these questions depend upon which of the special words and phrases "and", "or", and which of the implications "if and only", "if-then", and which of the quantifiers "for every", "there exists", appear in the given statement. Here, we shall discuss some techniques to find when a statement is valid. We shall list some general rules for checking whether a statement is true or not. Rule 1 If p and q are mathematical statements, then in order to show that the statement "p and q" is true, the following steps are followed. Step-1 Show that the statement p is true. Step-2 Show that the statement q is true. Rule 2 Statements with "Or" If p and q are mathematical statements , then in order to show that the statement "p or q" is true, one must consider the following. Case 1 By assuming that p is false, show that q must be true. Case 2 By assuming that q is false, show that p must be true. Rule 3 Statements with "If-then"
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In order to prove the statement "if p then q" we need to show that any one of the following case is true. Case 1 By assuming that p is true, prove that q must be true.(Direct method) Case 2 By assuming that q is false, prove that p must be false.(Contrapositive method) Rule 4 Statements with "if and only if " In order to prove the statement "p if and only if q", we need to show. (i) If p is true, then q is true and (ii) If q is true, then p is true Now we consider some examples. Example 13 Check whether the following statement is true or not. If x, y ∈ Z are such that x and y are odd, then xy is odd. Solution Let p : x, y ∈ Z such that x and y are odd q : xy is odd To check the validity of the given statement, we apply Case 1 of Rule 3. That is assume that if p is true, then q is true. p is true means that x and y are odd integers. Then x = 2m + 1, for some integer m. y = 2n + 1, for some integer n. Thus xy = (2m + 1) (2n + 1) = 2(2mn + m + n) + 1 This shows that xy is odd. Therefore, the given statement is true. Suppose we want to check this by using Case 2 of Rule 3, then we will proceed as follows. We assume that q is not true. This implies that we need to consider the negation of the statement q. This gives the statement ∼q : Product xy is even. This is possible only if either x or y is even. This shows that p is not true. Thus we have shown that ∼q ⇒ ∼p
!Note
The above example illustrates that to prove p ⇒ q, it is enough to show ∼q ⇒ ∼p which is the contrapositive of the statement p ⇒ q. Example 14 Check whether the following statement is true or false by proving its contrapositive. If x, y ∈ Ζ such that xy is odd, then both x and y are odd. Solution Let us name the statements as below
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p : xy is odd. q : both x and y are odd. We have to check whether the statement p ⇒ q is true or not, that is, by checking its contrapositive statement i.e., ∼q ⇒ ∼p Now ∼q : It is false that both x and y are odd. This implies that x (or y) is even. Then x = 2n for some integer n. Therefore, xy = 2ny for some integer n. This shows that xy is even. That is ∼p is true. Thus, we have shown that ∼q ⇒ ∼p and hence the given statement is true. Now what happens when we combine an implication and its converse? Next, we shall discuss this. Let us consider the following statements. p : A tumbler is half empty. q : A tumbler is half full. We know that if the first statement happens, then the second happens and also if the second happens, then the first happens. We can express this fact as If a tumbler is half empty, then it is half full. If a tumbler is half full, then it is half empty. We combine these two statements and get the following: A tumbler is half empty if and only if it is half full. Now, we discuss another method. 14.5.1 By Contradiction Here to check whether a statement p is true, we assume that p is not true i.e. ∼p is true. Then, we arrive at some result which contradicts our assumption. Therefore, we conclude that p is true. Example 15 Verify by the method of contradiction. p:
7 is irrational
Solution In this method, we assume that the given statement is false. That is we assume that such that
7 is rational. This means that there exists positive integers a and b
a 7 ! , where a and b have no common factors. Squaring the equation, b
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a2 ⇒ a2 = 7b2 ⇒ 7 divides a. Therefore, there exists an integer c such b2 that a = 7c. Then a2 = 49c2 and a2 = 7b2 Hence, 7b2 = 49c2 ⇒ b2 = 7c2 ⇒ 7 divides b. But we have already shown that 7 divides a. This implies that 7 is a common factor of both of a and b which contradicts our earlier assumption that a and b have no common factors. This shows that the
we get 7 ! assumption 7 is rational is wrong. Hence, the statement 7 is irrational is true. Next, we shall discuss a method by which we may show that a statement is false. The method involves giving an example of a situation where the statement is not valid. Such an example is called a counter example. The name itself suggests that this is an example to counter the given statement. Example 16 By giving a counter example, show that the following statement is false. If n is an odd integer, then n is prime. Solution The given statement is in the form "if p then q" we have to show that this is false. For this purpose we need to show that if p then ∼q. To show this we look for an odd integer n which is not a prime number. 9 is one such number. So n = 9 is a counter example. Thus, we conclude that the given statement is false. In the above, we have discussed some techniques for checking whether a statement is true or not. In mathematics, counter examples are used statement. !Note generating examples in favour of a statement to disprove the validity of However, do not provide the statement.
EXERCISE 14.5
1. Show that the statement p: "If x is a real number such that x3 + 4x = 0, then x is 0" is true by (i) direct method, (ii) method of contradiction, (iii) method of contrapositive 2. Show that the statement "For any real numbers a and b, a2 = b2 implies that a = b" is not true by giving a counter-example. 3. Show that the following statement is true by the method of contrapositive. p: If x is an integer and x2 is even, then x is also even. 4. By giving a counter example, show that the following statements are not true. (i) p: If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle. (ii) q: The equation x2 – 1 = 0 does not have a root lying between 0 and 2.
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5.
Which of the following statements are true and which are false? In each case give a valid reason for saying so. (i) p: Each radius of a circle is a chord of the circle. (ii) q: The centre of a circle bisects each chord of the circle. (iii) r: Circle is a particular case of an ellipse. (iv) s: If x and y are integers such that x > y, then –x < – y. (v) t : 11 is a rational number.
Miscellaneous Examples
Example 17 Check whether "Or" used in the following compound statement is exclusive or inclusive? Write the component statements of the compound statements and use them to check whether the compound statement is true or not. Justify your answer. t: you are wet when it rains or you are in a river. Solution "Or" used in the given statement is inclusive because it is possible that it rains and you are in the river. The component statements of the given statement are p : you are wet when it rains. q : You are wet when you are in a river. Here both the component statements are true and therefore, the compound statement is true. Example 18 Write the negation of the following statements: (i) p: For every real number x, x2 > x. (ii) q: There exists a rational number x such that x2 = 2. (iii) r: All birds have wings. (iv) s: All students study mathematics at the elementary level. Solution (i) The negation of p is "It is false that p is" which means that the condition x2 > x does not hold for all real numbers. This can be expressed as
∼p: There exists a real number x such that x2 < x. (ii) Negation of q is "it is false that q", Thus ∼q is the statement. ∼q: There does not exist a rational number x such that x2 = 2.
This statement can be rewritten as ∼q: For all real numbers x, x2 " 2
(iii) The negation of the statement is ∼r: There exists a bird which have no wings.
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(iv) The negation of the given statement is ∼s: There exists a student who does not study mathematics at the elementary level. Example 19 Using the words "necessary and sufficient" rewrite the statement "The integer n is odd if and only if n2 is odd". Also check whether the statement is true. Solution The necessary and sufficient condition that the integer n be odd is n2 must be odd. Let p and q denote the statements p : the integer n is odd. q : n2 is odd. To check the validity of "p if q", we have to check whether "if p then q" and "if q then p" is true. Case 1 If p, then q If p, then q is the statement: If the integer n is odd, then n2 is odd. We have to check whether this statement is true. Let us assume that n is odd. Then n = 2k + 1 when k is an integer. Thus n2 = (2k + 1)2 = 4k2 + 4k + 1 Therefore, n2 is one more than an even number and hence is odd. Case 2 If q, then p If q, then p is the statement If n is an integer and n2 is odd, then n is odd. We have to check whether this statement is true. We check this by contrapositive method. The contrapositive of the given statement is: If n is an even integer, then n2 is an even integer n is even implies that n = 2k for some k. Then n2 = 4k2. Therefore, n2 is even. Example 20 For the given statements identify the necessary and sufficient conditions. t: If you drive over 80 km per hour, then you will get a fine. Solution Let p and q denote the statements: p : you drive over 80 km per hour. q : you will get a fine. The implication if p, then q indicates that p is sufficient for q. That is driving over 80 km per hour is sufficient to get a fine. Here the sufficient condition is "driving over 80 km per hour": Similarly, if p, then q also indicates that q is necessary for p. That is
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When you drive over 80 km per hour, you will necessarily get a fine. Here the necessary condition is "getting a fine".
Miscellaneous Exercise on Chapter 14
Write the negation of the following statements: (i) p: For every positive real number x, the number x – 1 is also positive. (ii) q: All cats scratch. (iii) r: For every real number x, either x > 1 or x < 1. (iv) s: There exists a number x such that 0 < x < 1. 2. State the converse and contrapositive of each of the following statements: (i) p: A positive integer is prime only if it has no divisors other than 1 and itself. (ii) q: I go to a beach whenever it is a sunny day. (iii) r: If it is hot outside, then you feel thirsty. 3. Write each of the statements in the form "if p, then q" (i) p: It is necessary to have a password to log on to the server. (ii) q: There is traffic jam whenever it rains. (iii) r: You can access the website only if you pay a subsciption fee. 4. Rewrite each of the following statements in the form "p if and only if q" (i) p: If you watch television, then your mind is free and if your mind is free, then you watch television. (ii) q: For you to get an A grade, it is necessary and sufficient that you do all the homework regularly. (iii) r: If a quadrilateral is equiangular, then it is a rectangle and if a quadrilateral is a rectangle, then it is equiangular. 5. Given below are two statements p : 25 is a multiple of 5. q : 25 is a multiple of 8. Write the compound statements connecting these two statements with "And" and "Or". In both cases check the validity of the compound statement. 6. Check the validity of the statements given below by the method given against it. (i) p: The sum of an irrational number and a rational number is irrational (by contradiction method). (ii) q: If n is a real number with n > 3, then n2 > 9 (by contradiction method). Write the following statement in five different ways, conveying the same meaning. p: If a triangle is equiangular, then it is an obtuse angled triangle. 1.
7.
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Summary
" A mathematically acceptable statement is a sentence which is either true or " Explained the terms:
– Negation of a statement p: If p denote a statement, then the negation of p is denoted by ∼p. – Compound statements and their related component statements: A statement is a compound statement if it is made up of two or more smaller statements. The smaller statements are called component statements of the compound statement. – The role of "And", "Or", "There exists" and "For every" in compound statements. – The meaning of implications "If ", "only if ", " if and only if ". A sentence with if p, then q can be written in the following ways. – p implies q (denoted by p # q) – p is a sufficient condition for q – q is a necessary condition for p – p only if q – ∼q implies ∼p – The contrapositive of a statement p ⇒ q is the statement ∼ q ⇒ ∼p . The converse of a statement p ⇒ q is the statement q ⇒ p. p ⇒ q together with its converse, gives p if and only if q. " The following methods are used to check the validity of statements: (i) direct method (ii) contrapositive method (iii) method of contradiction (iv) using a counter example. false.
Historical Note
The first treatise on logic was written by Aristotle (384 B.C.-322 B.C.). It was a collection of rules for deductive reasoning which would serve as a basis for the study of every branch of knowledge. Later, in the seventeenth century, German mathematician G. W. Leibnitz (1646 – 1716 A.D.) conceived the idea of using symbols in logic to mechanise the process of deductive reasoning. His idea was realised in the nineteenth century by the English mathematician George Boole (1815–1864 A.D.) and Augustus De Morgan (1806–1871 A.D.) , who founded the modern subject of symbolic logic.
Chapter
15
STATISTICS
!"Statistics may be rightly called the science of averages and their estimates." – A.L.BOWLEY & A.L. BODDINGTON !
15.1 Introduction
We know that statistics deals with data collected for specific purposes. We can make decisions about the data by analysing and interpreting it. In earlier classes, we have studied methods of representing data graphically and in tabular form. This representation reveals certain salient features or characteristics of the data. We have also studied the methods of finding a representative value for the given data. This value is called the measure of central tendency. Recall mean (arithmetic mean), median and mode are three measures of central tendency. A measure of central Karl Pearson tendency gives us a rough idea where data points are (1857-1936) centred. But, in order to make better interpretation from the data, we should also have an idea how the data are scattered or how much they are bunched around a measure of central tendency. Consider now the runs scored by two batsmen in their last ten matches as follows: Batsman A : 30, 91, 0, 64, 42, 80, 30, 5, 117, 71 Batsman B : 53, 46, 48, 50, 53, 53, 58, 60, 57, 52 Clearly, the mean and median of the data are Batsman A Batsman B Mean 53 53 Median 53 53 Recall that, we calculate the mean of a data (denoted by x ) by dividing the sum of the observations by the number of observations, i.e.,
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MATHEMATICS
x!
1 n
i !1
" xi
n
Also, the median is obtained by first arranging the data in ascending or descending order and applying the following rule.
$ n # 1% If the number of observations is odd, then the median is & ' ( 2 )
th
observation.
th
$n% If the number of observations is even, then median is the mean of & ' (2)
th
and
$n % & # 1' observations. (2 ) We find that the mean and median of the runs scored by both the batsmen A and B are same i.e., 53. Can we say that the performance of two players is same? Clearly No, because the variability in the scores of batsman A is from 0 (minimum) to 117 (maximum). Whereas, the range of the runs scored by batsman B is from 46 to 60. Let us now plot the above scores as dots on a number line. We find the following diagrams:
For batsman A
For batsman B
Fig 15.1
Fig 15.2
We can see that the dots corresponding to batsman B are close to each other and are clustering around the measure of central tendency (mean and median), while those corresponding to batsman A are scattered or more spread out. Thus, the measures of central tendency are not sufficient to give complete information about a given data. Variability is another factor which is required to be studied under statistics. Like 'measures of central tendency' we want to have a single number to describe variability. This single number is called a 'measure of dispersion'. In this Chapter, we shall learn some of the important measures of dispersion and their methods of calculation for ungrouped and grouped data.
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349
15.2 Measures of Dispersion
The dispersion or scatter in a data is measured on the basis of the observations and the types of the measure of central tendency, used there. There are following measures of dispersion: (i) Range, (ii) Quartile deviation, (iii) Mean deviation, (iv) Standard deviation. In this Chapter, we shall study all of these measures of dispersion except the quartile deviation.
15.3 Range
Recall that, in the example of runs scored by two batsmen A and B, we had some idea of variability in the scores on the basis of minimum and maximum runs in each series. To obtain a single number for this, we find the difference of maximum and minimum values of each series. This difference is called the 'Range' of the data. In case of batsman A, Range = 117 – 0 = 117 and for batsman B, Range = 60 – 46 = 14. Clearly, Range of A > Range of B. Therefore, the scores are scattered or dispersed in case of A while for B these are close to each other. Thus, Range of a series = Maximum value – Minimum value. The range of data gives us a rough idea of variability or scatter but does not tell about the dispersion of the data from a measure of central tendency. For this purpose, we need some other measure of variability. Clearly, such measure must depend upon the difference (or deviation) of the values from the central tendency. The important measures of dispersion, which depend upon the deviations of the observations from a central tendency are mean deviation and standard deviation. Let us discuss them in detail.
15.4 Mean Deviation
Recall that the deviation of an observation x from a fixed value 'a' is the difference x – a. In order to find the dispersion of values of x from a central value 'a' , we find the deviations about a. An absolute measure of dispersion is the mean of these deviations. To find the mean, we must obtain the sum of the deviations. But, we know that a measure of central tendency lies between the maximum and the minimum values of the set of observations. Therefore, some of the deviations will be negative and some positive. Thus, the sum of deviations may vanish. Moreover, the sum of the deviations from mean ( x ) is zero. Also Mean of deviations !
Sum of deviations 0 ! !0 Number of observations n
Thus, finding the mean of deviations about mean is not of any use for us, as far as the measure of dispersion is concerned.
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MATHEMATICS
Remember that, in finding a suitable measure of dispersion, we require the distance of each value from a central tendency or a fixed number 'a'. Recall, that the absolute value of the difference of two numbers gives the distance between the numbers when represented on a number line. Thus, to find the measure of dispersion from a fixed number 'a' we may take the mean of the absolute values of the deviations from the central value. This mean is called the 'mean deviation'. Thus mean deviation about a central value 'a' is the mean of the absolute values of the deviations of the observations from 'a'. The mean deviation from 'a' is denoted as M.D. (a). Therefore, M.D.(a) =
Sum of absolute values of deviations from 'a' . Number of observations
Remark Mean deviation may be obtained from any measure of central tendency. However, mean deviation from mean and median are commonly used in statistical studies. Let us now learn how to calculate mean deviation about mean and mean deviation about median for various types of data 15.4.1 Mean deviation for ungrouped data Let n observations be x1, x2, x3, ...., xn. The following steps are involved in the calculation of mean deviation about mean or median: Step 1 Calculate the measure of central tendency about which we are to find the mean deviation. Let it be 'a'. Step 2 Find the deviation of each xi from a, i.e., x1 – a, x2 – a, x3 – a,. . . , xn– a Step 3 Find the absolute values of the deviations, i.e., drop the minus sign (–), if it is there, i.e., x1 * a , x2 * a , x3 * a , ...., xn * a Step 4 Find the mean of the absolute values of the deviations. This mean is the mean deviation about a, i.e.,
M.D.(a) !
Thus and
" xi * a
i !1
n
n
1 M.D. ( x ) = n
M.D. (M) =
" xi * x
i !1 n i !1
n
, where x = Mean
1 n
" xi * M , where M = Median
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351
Note In this Chapter, we shall use the symbol M to denote median unless stated otherwise.Let us now illustrate the steps of the above method in following examples. Example 1 Find the mean deviation about the mean for the following data: 6, 7, 10, 12, 13, 4, 8, 12 Solution We proceed step-wise and get the following: Step 1 Mean of the given data is
Note Instead of carrying out the steps every time, we can carry on calculation, step-wise without referring to steps. Example 2 Find the mean deviation about the mean for the following data : 12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5 Solution We have to first find the mean ( x ) of the given data
15.4.2 Mean deviation for grouped data We know that data can be grouped into two ways : (a) Discrete frequency distribution, (b) Continuous frequency distribution. Let us discuss the method of finding mean deviation for both types of the data. (a) Discrete frequency distribution Let the given data consist of n distinct values x1, x2, ..., xn occurring with frequencies f1, f2 , ..., fn respectively. This data can be represented in the tabular form as given below, and is called discrete frequency distribution: x : x1 x2 x3 ... xn f : f1 f2 f3 ... fn (i) Mean deviation about mean First of all we find the mean x of the given data by using the formula
STATISTICS
353
x!
" xi fi
i !1 n
n
" fi
i !1
!
1 N
" xi fi ,
i !1
n
where
"x f
i !1 i
n
i
denotes the sum of the products of observations xi with their respective
frequencies fi and N !
"f
i !1
n
i
is the sum of the frequencies.
Then, we find the deviations of observations xi from the mean x and take their absolute values, i.e., xi * x for all i =1, 2,..., n. After this, find the mean of the absolute values of the deviations, which is the required mean deviation about the mean. Thus
M.D. ( x ) !
"f
i !1
n
i
xi * x
"f
i !1
n
=
i
1 n " f i xi * x N i !1
(ii) Mean deviation about median To find mean deviation about median, we find the median of the given discrete frequency distribution. For this the observations are arranged in ascending order. After this the cumulative frequencies are obtained. Then, we identify , where 2 N is the sum of frequencies. This value of the observation lies in the middle of the data, therefore, it is the required median. After finding median, we obtain the mean of the absolute values of the deviations from median.Thus, the observation whose cumulative frequency is equal to or just greater than
N
M.D.(M) !
1 N
"f
i !1
n
i
xi * M
Example 4 Find mean deviation about the mean for the following data : xi 2 5 6 8 10 12 fi 2 8 10 7 8 5 Solution Let us make a Table 15.1 of the given data and append other columns after calculations.
(b) Continuous frequency distribution A continuous frequency distribution is a series in which the data are classified into different class-intervals without gaps alongwith their respective frequencies. For example, marks obtained by 100 students are presented in a continuous frequency distribution as follows : Marks obtained 0-10 10-20 20-30 30-40 Number of Students 12 18 27 20 40-50 17 50-60 6
(i) Mean deviation about mean While calculating the mean of a continuous frequency distribution, we had made the assumption that the frequency in each class is centred at its mid-point. Here also, we write the mid-point of each given class and proceed further as for a discrete frequency distribution to find the mean deviation. Let us take the following example.
Solution We make the following Table 15.4 from the given data :
Table 15.4
Marks obtained 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Number of students fi 2 3 8 14 8 3 2 40
N !
Mid-points xi 15 25 35 45 55 65 75
f ix i
xi * x
fi xi * x
30 75 280 630 440 195 150 1800
30 20 10 0 10 20 30
60 60 80 0 80 60 60 400
Here
" fi
i !1
7
! 40, " fi xi ! 1800,
i !1
7
" fi
i !1
7
xi * x ! 400
Therefore
x!
1 7 1800 " fi xi ! 40 ! 45 N i !1
and
M.D. + x , !
1 7 1 " fi xi * x ! 40 - 400 ! 10 N i !1
Shortcut method for calculating mean deviation about mean We can avoid the tedious calculations of computing x by following step-deviation method. Recall that in this method, we take an assumed mean which is in the middle or just close to it in the data. Then deviations of the observations (or mid-points of classes) are taken from the
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357
assumed mean. This is nothing but the shifting of origin from zero to the assumed mean on the number line, as shown in Fig 15.3
Fig 15.3
If there is a common factor of all the deviations, we divide them by this common factor to further simplify the deviations. These are known as step-deviations. The process of taking step-deviations is the change of scale on the number line as shown in Fig 15.4
Fig 15.4
The deviations and step-deviations reduce the size of the observations, so that the computations viz. multiplication, etc., become simpler. Let, the new variable be denoted by d i !
xi * a , where 'a' is the assumed mean and h is the common factor. Then, the h
n
mean x by step-deviation method is given by
" f i di x ! a # i !1 -h N
Let us take the data of Example 6 and find the mean deviation by using stepdeviation method.
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MATHEMATICS
Take the assumed mean a = 45 and h = 10, and form the following Table 15.5.
Table 15.5
!Note The step deviation method is applied to compute x . Rest of the procedure is same.
(ii) Mean deviation about median The process of finding the mean deviation about median for a continuous frequency distribution is similar as we did for mean deviation about the mean. The only difference lies in the replacement of the mean by median while taking deviations. Let us recall the process of finding median for a continuous frequency distribution. The data is first arranged in ascending order. Then, the median of continuous frequency distribution is obtained by first identifying the class in which median lies (median class) and then applying the formula
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359
N *C Median ! l # 2 -h f
where median class is the class interval whose cumulative frequency is just greater than or equal to
N , N is the sum of frequencies, l, f, h and C are, respectively the lower 2 limit , the frequency, the width of the median class and C the cumulative frequency of the class just preceding the median class. After finding the median, the absolute values
of the deviations of mid-point xi of each class from the median i.e., xi * M are obtained.
n " fi xi * M N i !1 The process is illustrated in the following example:
Find the mean deviation about the median for the data in Exercises 7 and 8.
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361
Find the mean deviation about the mean for the data in Exercises 9 and 10. 9. Income 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 per day Number 4 8 9 10 7 5 4 3 of persons 10. Height 95-105 105-115 115-125 125-135 135-145 145-155 in cms Number of 9 13 26 30 12 10 boys 11. Find the mean deviation about median for the following data : Marks 0-10 10-20 20-30 30-40 40-50 50-60 Number of 6 8 14 16 4 2 Girls 12. Calculate the mean deviation about median age for the age distribution of 100 persons given below: Age 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55 Number 5 6 12 14 26 12 16 9 [Hint Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval] 15.4.3 Limitations of mean deviation In a series, where the degree of variability is very high, the median is not a representative central tendency. Thus, the mean deviation about median calculated for such series can not be fully relied. The sum of the deviations from the mean (minus signs ignored) is more than the sum of the deviations from median. Therefore, the mean deviation about the mean is not very scientific.Thus, in many cases, mean deviation may give unsatisfactory results. Also mean deviation is calculated on the basis of absolute values of the deviations and therefore, cannot be subjected to further algebraic treatment. This implies that we must have some other measure of dispersion. Standard deviation is such a measure of dispersion.
15.5 Variance and Standard Deviation
Recall that while calculating mean deviation about mean or median, the absolute values of the deviations were taken. The absolute values were taken to give meaning to the mean deviation, otherwise the deviations may cancel among themselves. Another way to overcome this difficulty which arose due to the signs of deviations, is to take squares of all the deviations. Obviously all these squares of deviations are
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MATHEMATICS
non-negative. Let x1, x2, x3, ..., xn be n observations and x be their mean. Then
(x1 * x)2 # (x2 *x)2 # .......# (xn * x )2 !
"(x * x) .
2 i !1 i
n
If this sum is zero, then each ( xi * x ) has to be zero. This implies that there is no dispersion at all as all observations are equal to the mean x . If
" (x * x)
i !1 i
n
2
is small , this indicates that the observations x1, x2, x3,...,xn are
close to the mean x and therefore, there is a lower degree of dispersion. On the contrary, if this sum is large, there is a higher degree of dispersion of the observations from the mean x . Can we thus say that the sum
" (x * x)
i !1 i
n
2
is a reasonable indicator
of the degree of dispersion or scatter? Let us take the set A of six observations 5, 15, 25, 35, 45, 55. The mean of the observations is x = 30. The sum of squares of deviations from x for this set is
" (x * x)
i !1 i
6
2
= (5–30)2 + (15–30)2 + (25–30)2 + (35–30)2 + (45–30)2 +(55–30)2
= 625 + 225 + 25 + 25 + 225 + 625 = 1750 Let us now take another set B of 31 observations 15, 16, 17,. The mean of these observations is y = 30 Note that both the sets A and B of observations have a mean of 30. Now, the sum of squares of deviations of observations for set B from the mean y is given by
will tend to say that the set A of six observations has a lesser dispersion about the mean than the set B of 31 observations, even though the observations in set A are more scattered from the mean (the range of deviations being from –25 to 25) than in the set B (where the range of deviations is from –15 to 15). This is also clear from the following diagrams. For the set A, we have
Fig 15.5
For the set B, we have
Fig 15.6
Thus, we can say that the sum of squares of deviations from the mean is not a proper measure of dispersion. To overcome this difficulty we take the mean of the squares of
1 n the deviations, i.e., we take " ( xi * x ) 2 . In case of the set A, we have n i !1
1 1 × 1750 = 291.67 and in case of the set B, it is × 2480 = 80. 6 31 This indicates that the scatter or dispersion is more in set A than the scatter or dispersion in set B, which confirms with the geometrical representation of the two sets. Mean !
Thus, we can take
1 " ( xi * x ) 2 as a quantity which leads to a proper measure n
of dispersion. This number, i.e., mean of the squares of the deviations from mean is called the variance and is denoted by . 2 (read as sigma square). Therefore, the variance of n observations x1, x2,..., xn is given by
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MATHEMATICS
1 n . ! " ( xi * x ) 2 n i !1
2
15.5.1 Standard Deviation In the calculation of variance, we find that the units of individual observations xi and the unit of their mean x are different from that of variance, since variance involves the sum of squares of (xi– x ). For this reason, the proper measure of dispersion about the mean of a set of observations is expressed as positive square-root of the variance and is called standard deviation. Therefore, the standard deviation, usually denoted by . , is given by
.!
1 n " ( xi * x ) 2 n i !1
... (1)
Let us take the following example to illustrate the calculation of variance and hence, standard deviation of ungrouped data. Example 8 Find the Variance of the following data: 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 Solution From the given data we can form the following Table 15.7. The mean is calculated by step-deviation method taking 14 as assumed mean. The number of observations is n = 10
Table 15.7
Solution Presenting the data in tabular form (Table 15.8), we get
Table 15.8
xi 4 8 11 17 20 24 32
fi 3 5 9 5 4 3 1 30
fi x i 12 40 99 85 80 72 32 420
xi – x –10 –6 –3 3 6 10 18
( xi * x ) 2
100 36 9 9 36 100 324
f i ( xi * x ) 2 300 180 81 45 144 300 324 1374
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MATHEMATICS
N = 30,
7
" fi xi ! 420, " fi + xi * x ,
i !1 i !1
7
7
2
! 1374
Therefore
x!
" fi xi
i !1
N
2
!
1 - 420 ! 14 30
Hence
1 variance (. ) = N
=
" fi (xi *
i !1
7
x )2
1 × 1374 = 45.8 30
and
Standard deviation (. ) ! 45 .8 = 6.77
15.5.3 Standard deviation of a continuous frequency distribution The given continuous frequency distribution can be represented as a discrete frequency distribution by replacing each class by its mid-point. Then, the standard deviation is calculated by the technique adopted in the case of a discrete frequency distribution. If there is a frequency distribution of n classes each class defined by its mid-point xi with frequency fi, the standard deviation will be obtained by the formula
1 - 293.77 = 6.12 48 Therefore, Standard deviation ( . ) = 6.12
= 15.5.4. Shortcut method to find variance and standard deviation Sometimes the values of xi in a discrete distribution or the mid points xi of different classes in a continuous distribution are large and so the calculation of mean and variance becomes tedious and time consuming. By using step-deviation method, it is possible to simplify the procedure. Let the assumed mean be 'A' and the scale be reduced to
1 times (h being the h
... (1)
width of class-intervals). Let the step-deviations or the new values be yi. i.e.
yi ! xi * A h
Calculate the standard deviation and mean diameter of the circles. [ Hint First make the data continuous by making the classes as 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5 - 48.5, 48.5 - 52.5 and then proceed.]
15.6 Analysis of Frequency Distributions
In earlier sections, we have studied about some types of measures of dispersion. The mean deviation and the standard deviation have the same units in which the data are given. Whenever we want to compare the variability of two series with same mean, which are measured in different units, we do not merely calculate the measures of dispersion but we require such measures which are independent of the units. The measure of variability which is independent of units is called coefficient of variation (denoted as C.V.) The coefficient of variation is defined as
C.V. !
.
x
- 100 , x 5 0 ,
where σ and x are the standard deviation and mean of the data. For comparing the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.
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373
15.6.1 Comparison of two frequency distributions with same mean Let x1 and σ1 be the mean and standard deviation of the first distribution, and x2 and σ2 be the mean and standard deviation of the second distribution. Then and
1 C.V. (1st distribution) = x - 100 1
- 100 ... (2) x It is clear from (1) and (2) that the two C.Vs. can be compared on the basis of values
of . 1 and . 2 only. Thus, we say that for two series with equal means, the series with greater standard deviation (or variance) is called more variable or dispersed than the other. Also, the series with lesser value of standard deviation (or variance) is said to be more consistent than the other. Let us now take following examples: Example 13 Two plants A and B of a factory show following results about the number of workers and the wages paid to them. A B No. of workers Average monthly wages Variance of distribution of wages 5000 Rs 2500 81 6000 Rs 2500 100
.2
In which plant, A or B is there greater variability in individual wages? Solution The variance of the distribution of wages in plant A ( . 12 ) = 81 Therefore, standard deviation of the distribution of wages in plant A ( . 1 ) = 9
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MATHEMATICS
Also, the variance of the distribution of wages in plant B ( . 2 2 ) = 100 Therefore, standard deviation of the distribution of wages in plant B ( .2 ) = 10 Since the average monthly wages in both the plants is same, i.e., Rs.2500, therefore, the plant with greater standard deviation will have more variability. Thus, the plant B has greater variability in the individual wages. Example 14 Coefficient of variation of two distributions are 60 and 70, and their standard deviations are 21 and 16, respectively. What are their arithmetic means. Solution Given C.V. (1st distribution) = 60, . 1 = 21 C.V. (2nd distribution) = 70, . 2 = 16 Let x1 and x2 be the means of 1st and 2nd distribution, respectively. Then C.V. (1st distribution) = Therefore 60 =
.1
x1 × 100
21 21 -100 or x1 ! - 100 ! 35 x1 60
and i.e.
C.V. (2nd distribution) = 70 =
.2
x2 ×100
16 16 - 100 or x2 ! - 100 ! 22.85 x2 70
Example 15 The following values are calculated in respect of heights and weights of the students of a section of Class XI : Height Weight Mean Variance 162.6 cm 127.69 cm2 52.36 kg 23.1361 kg2
Can we say that the weights show greater variation than the heights? Solution To compare the variability, we have to calculate their coefficients of variation. Given Therefore Also Variance of height = 127.69cm2 Standard deviation of height = Variance of weight = 23.1361 kg
2
An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results: Firm A No. of wage earners Mean of monthly wages Variance of the distribution of wages (i) Which firm A or B pays larger amount as monthly wages? (ii) Which firm, A or B, shows greater variability in individual wages? 586 Rs 5253 100 Firm B 648 Rs 5253 121
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4.
The following is the record of goals scored by team A in a football session: No. of goals scored No. of matches 0 1 1 9 2 7 3 5 4 3
5.
For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent? The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
" xi ! 212 ,
i !1
50
" xi2 ! 902.8 ,
i !1
50
" yi ! 261 ,
i !1
50
" yi2 ! 1457.6
i !1
50
Which is more varying, the length or weight?
Miscellaneous Examples
Example 16 The variance of 20 observations is 5. If each observation is multiplied by 2, find the new variance of the resulting observations. Solution Let the observations be x1, x2, ..., x20 and x be their mean. Given that variance = 5 and n = 20. We know that Variance . 2 !
20
+ ,
1 n
" (xi * x )2 , i.e., 5 !
i !1
20
1 20 " (xi * x )2 20 i !1
... (1)
or
" (xi * x )2 = 100
i !1
If each observation is multiplied by 2, and the new resulting observations are yi , then yi = 2xi i.e., xi = Therefore i.e.
y!
reader may note that if each observation is multiplied by constant !Note Theof the resulting observations becomes k times the originalavariance. k, the variance Example17 The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations. Solution Let the other two observations be x and y. Therefore, the series is 1, 2, 6, x, y. Now or Therefore Also i.e. 8.24 = Mean x = 4.4 =
Solution Let x be the mean of x1, x2, ...,xn . Then the variance is given by
. 12 =
1 n
" (xi * x )
i !1
n
2
If 'a is added to each observation, the new observations will be yi = xi + a Let the mean of the new observations be y . Then
y=
... (1)
1 n
" yi ! n " (xi # a)
i !1 i !1
n
1
n
n 0 1/n 1 xi # " a 2 = = n 1" n i !1 4 3 i !1
"x #
i !1 i
n
na !x#a n
... (2)
i.e. y = x+a Thus, the variance of the new observations
.2
2
1 = n
1 = n
" (yi * y ) 2 =
i !1
n i !1
n
1 n " ( xi # a * x * a) 2 n i !1
[Using (1) and (2)]
" (xi * x )2 = . 12
Thus, the variance of the new observations is same as that of the original observations. Note We may note that adding (or subtracting) a positive number to (or from) each observation of a group does not affect the variance. Example 19 The mean and standard deviation of 100 observations were calculated as 40 and 5.1, respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation? Solution Given that number of observations (n) = 100 Incorrect mean ( x ) = 40, Incorrect standard deviation (σ) = 5.1 We know that
Miscellaneous Exercise On Chapter 15
1. 2. 3. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations. Given that x is the mean and σ2 is the variance of n observations x1, x2, ...,xn. Prove that the mean and variance of the observations ax1, ax2, ax3, ...., axn are a x and a2 σ2, respectively, (a ≠ 0). The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by 12. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below: Subject Mean Standard deviation Mathematics 42 12 Physics 32 15 Chemistry 40.9 20
4.
5.
6.
7.
which of the three subjects shows the highest variability in marks and which shows the lowest? The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.
- 100, x 5 0. x For series with equal means, the series with lesser standard deviation is more consistent or less scattered.
" Coefficient of variation (C.V.)
.
Historical Note
'Statistics' is derived from the Latin word 'status' which means a political state. This suggests that statistics is as old as human civilisation. In the year 3050 B.C., perhaps the first census was held in Egypt. In India also, about 2000 years ago, we had an efficient system of collecting administrative statistics, particularly, during the regime of Chandra Gupta Maurya (324-300 B.C.). The system of collecting data related to births and deaths is mentioned in Kautilya's Arthshastra (around 300 B.C.) A detailed account of administrative surveys conducted during Akbar's regime is given in Ain-I-Akbari written by Abul Fazl.
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MATHEMATICS
Captain John Graunt of London (1620-1674) is known as father of vital statistics due to his studies on statistics of births and deaths. Jacob Bernoulli (1654-1705) stated the Law of Large numbers in his book "Ars Conjectandi', published in 1713. The theoretical development of statistics came during the mid seventeenth century and continued after that with the introduction of theory of games and chance (i.e., probability). Francis Galton (1822-1921), an Englishman, pioneered the use of statistical methods, in the field of Biometry. Karl Pearson (1857-1936) contributed a lot to the development of statistical studies with his discovery of Chi square test and foundation of statistical laboratory in England (1911). Sir Ronald A. Fisher (1890-1962), known as the Father of modern statistics, applied it to various diversified fields such as Genetics, Biometry, Education, Agriculture, etc.
—! —
Chapter
16
PROBABILITY
!Where a mathematical reasoning can be had, it is as great a folly to
make use of any other, as to grope for a thing in the dark, when you have a candle in your hand. – JOHN ARBUTHNOT !
16.1 Introduction
In earlier classes, we studied about the concept of probability as a measure of uncertainty of various phenomenon. We have obtained the probability of getting an even number in throwing a die as
3 1 i.e., . Here the 6 2
total possible outcomes are 1,2,3,4,5 and 6 (six in number). The outcomes in favour of the event of 'getting an even number' are 2,4,6 (i.e., three in number). In general, to obtain the probability of an event, we find the ratio of the number of outcomes favourable to the event, to the total Kolmogorove number of equally likely outcomes. This theory of probability (1903-1987) is known as classical theory of probability. In Class IX, we learnt to find the probability on the basis of observations and collected data. This is called statistical approach of probability. Both the theories have some serious difficulties. For instance, these theories can not be applied to the activities/experiments which have infinite number of outcomes. In classical theory we assume all the outcomes to be equally likely. Recall that the outcomes are called equally likely when we have no reason to believe that one is more likely to occur than the other. In other words, we assume that all outcomes have equal chance (probability) to occur. Thus, to define probability, we used equally likely or equally probable outcomes. This is logically not a correct definition. Thus, another theory of probability was developed by A.N. Kolmogorov, a Russian mathematician, in 1933. He
384
MATHEMATICS
laid down some axioms to interpret probability, in his book 'Foundation of Probability' published in 1933. In this Chapter, we will study about this approach called axiomatic approach of probability. To understand this approach we must know about few basic terms viz. random experiment, sample space, events, etc. Let us learn about these all, in what follows next.
16.2 Random Experiments
In our day to day life, we perform many activities which have a fixed result no matter any number of times they are repeated. For example given any triangle, without knowing the three angles, we can definitely say that the sum of measure of angles is 180°. We also perform many experimental activities, where the result may not be same, when they are repeated under identical conditions. For example, when a coin is tossed it may turn up a head or a tail, but we are not sure which one of these results will actually be obtained. Such experiments are called random experiments. An experiment is called random experiment if it satisfies the following two conditions: (i) It has more than one possible outcome. (ii) It is not possible to predict the outcome in advance. Check whether the experiment of tossing a die is random or not? In this chapter, we shall refer the random experiment by experiment only unless stated otherwise. 16.2.1 Outcomes and sample space A possible result of a random experiment is called its outcome. Consider the experiment of rolling a die. The outcomes of this experiment are 1, 2, 3, 4, 5, or 6, if we are interested in the number of dots on the upper face of the die. The set of outcomes {1, 2, 3, 4, 5, 6} is called the sample space of the experiment. Thus, the set of all possible outcomes of a random experiment is called the sample space associated with the experiment. Sample space is denoted by the symbol S. Each element of the sample space is called a sample point. In other words, each outcome of the random experiment is also called sample point. Let us now consider some examples. Example 1 Two coins (a one rupee coin and a two rupee coin) are tossed once. Find a sample space. Solution Clearly the coins are distinguishable in the sense that we can speak of the first coin and the second coin. Since either coin can turn up Head (H) or Tail(T), the possible outcomes may be
PROBABILITY
385
Heads on both coins = (H,H) = HH Head on first coin and Tail on the other = (H,T) = HT Tail on first coin and Head on the other = (T,H) = TH Tail on both coins = (T,T) = TT Thus, the sample space is S = {HH, HT, TH, TT} ordered pairs of !Note The outcomes of this experiment arethe ordered pairs.H and T. For the sake of simplicity the commas are omitted from Example 2 Find the sample space associated with the experiment of rolling a pair of dice (one is blue and the other red) once. Also, find the number of elements of this sample space. Solution Suppose 1 appears on blue die and 2 on the red die. We denote this outcome by an ordered pair (1,2). Simlarly, if '3' appears on blue die and '5' on red, the outcome is denoted by the ordered pair (3,5). In general each outcome can be denoted by the ordered pair (x, y), where x is the number appeared on the blue die and y is the number appeared on the red die. Therefore, this sample space is given by S = {(x, y): x is the number on the blue die and y is the number on the red die}. The number of elements of this sample space is 6 × 6 = 36 and the sample space is given below: {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} Example 3 In each of the following experiments specify appropriate sample space (i) (ii) A boy has a 1 rupee coin, a 2 rupee coin and a 5 rupee coin in his pocket. He takes out two coins out of his pocket, one after the other. A person is noting down the number of accidents along a busy highway during a year.
Solution (i) Let Q denote a 1 rupee coin, H denotes a 2 rupee coin and R denotes a 5 rupee coin. The first coin he takes out of his pocket may be any one of the three coins Q, H or R. Corresponding to Q, the second draw may be H or R. So the result of two draws may be QH or QR. Similarly, corresponding to H, the second draw may be Q or R. Therefore, the outcomes may be HQ or HR. Lastly, corresponding to R, the second draw may be H or Q. So, the outcomes may be RH or RQ.
386
MATHEMATICS
Thus, the sample space is S={QH, QR, HQ, HR, RH, RQ} (ii) The number of accidents along a busy highway during the year of observation can be either 0 (for no accident ) or 1 or 2, or some other positive integer. Thus, a sample space associated with this experiment is S= {0,1,2,...} Example 4 A coin is tossed. If it shows head, we draw a ball from a bag consisting of 3 blue and 4 white balls; if it shows tail we throw a die. Describe the sample space of this experiment. Solution Let us denote blue balls by B1, B2, B3 and the white balls by W1, W2, W3, W4. Then a sample space of the experiment is S = { HB1, HB2, HB3, HW1, HW2, HW3, HW4, T1, T2, T3, T4, T5, T6}. Here HBi means head on the coin and ball Bi is drawn, HWi means head on the coin and ball Wi is drawn. Similarly, Ti means tail on the coin and the number i on the die. Example 5 Consider the experiment in which a coin is tossed repeatedly until a head comes up. Describe the sample space. Solution In the experiment head may come up on the first toss, or the 2nd toss, or the 3rd toss and so on till head is obtained. Hence, the desired sample space is S= {H, TH, TTH, TTTH, TTTTH,...}
EXERCISE 16.1
In each of the following Exercises 1 to 7, describe the sample space for the indicated experiment. 1. A coin is tossed three times. 2. A die is thrown two times. 3. A coin is tossed four times. 4. A coin is tossed and a die is thrown. 5. A coin is tossed and then a die is rolled only in case a head is shown on the coin. 6. 2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the sample space for the experiment in which a room is selected and then a person. 7. One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and the number on its uppermost face is noted. Describe the sample space. 8. An experiment consists of recording boy–girl composition of families with 2 children. (i) What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their births?
PROBABILITY
387
(ii) 9. 10.
11.
12.
13.
14.
15.
16.
What is the sample space if we are interested in the number of girls in the family? A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment. An experiment consists of tossing a coin and then throwing it second time if a head occurs. If a tail occurs on the first toss, then a die is rolled once. Find the sample space. Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or non – defective(N). Write the sample space of this experiment. A coin is tossed. If the out come is a head, a die is thrown. If the die shows up an even number, the die is thrown again. What is the sample space for the experiment? The numbers 1, 2, 3 and 4 are written separatly on four slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the sample space for the experiment. An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this experiment. A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black balls. If it shows head, we throw a die. Find the sample space for this experiment. A die is thrown repeatedly untill a six comes up. What is the sample space for this experiment?
16.3 Event We have studied about random experiment and sample space associated with an experiment. The sample space serves as an universal set for all questions concerned with the experiment. Consider the experiment of tossing a coin two times. An associated sample space is S = {HH, HT, TH, TT}. Now suppose that we are interested in those outcomes which correspond to the occurrence of exactly one head. We find that HT and TH are the only elements of S corresponding to the occurrence of this happening (event). These two elements form the set E = { HT, TH} We know that the set E is a subset of the sample space S . Similarly, we find the following correspondence between events and subsets of S.
388
MATHEMATICS
Description of events Number of tails is exactly 2 Number of tails is atleast one Number of heads is atmost one Second toss is not head Number of tails is atmost two Number of tails is more than two
The above discussion suggests that a subset of sample space is associated with an event and an event is associated with a subset of sample space. In the light of this we define an event as follows. Definition Any subset E of a sample space S is called an event. 16.3.1 Occurrence of an event Consider the experiment of throwing a die. Let E denotes the event " a number less than 4 appears". If actually '1' had appeared on the die then we say that event E has occurred. As a matter of fact if outcomes are 2 or 3, we say that event E has occurred Thus, the event E of a sample space S is said to have occurred if the outcome ! of the experiment is such that ! ∈ E. If the outcome ! is such that ! ∉ E, we say that the event E has not occurred. 16.3.2 Types of events Events can be classified into various types on the basis of the elements they have. 1. Impossible and Sure Events The empty set φ and the sample space S describe events. In fact φ is called an impossible event and S, i.e., the whole sample space is
called the sure event.
To understand these let us consider the experiment of rolling a die. The associated sample space is S = {1, 2, 3, 4, 5, 6} Let E be the event " the number appears on the die is a multiple of 7". Can you write the subset asociated with the event E? Clearly no outcome satisfies the condition given in the event, i.e., no element of the sample space ensures the occurrence of the event E. Thus, we say that the empty set only correspond to the event E. In other words we can say that it is impossible to have a multiple of 7 on the upper face of the die. Thus, the event E = ! is an impossible event. Now let us take up another event F "the number turns up is odd or even". Clearly
PROBABILITY
389
F = {1, 2, 3, 4, 5, 6,} = S, i.e., all outcomes of the experiment ensure the occurrence of the event F. Thus, the event F = S is a sure event. 2. Simple Event If an event E has only one sample point of a sample space, it is called a simple (or elementary) event. In a sample space containing n distinct elements, there are exactly n simple events. For example in the experiment of tossing two coins, a sample space is S={HH, HT, TH, TT} There are four simple events corresponding to this sample space. These are E1= {HH}, E2={HT}, E3= { TH} and E4={TT}. 3. Compound Event If an event has more than one sample point, it is called a Compound event. For example, in the experiment of "tossing a coin thrice" the events E: 'Exactly one head appeared' F: 'Atleast one head appeared' G: 'Atmost one head appeared' etc. are all compound events. The subsets of S associated with these events are E={HTT,THT,TTH} F={HTT,THT, TTH, HHT, HTH, THH, HHH} G= {TTT, THT, HTT, TTH} Each of the above subsets contain more than one sample point, hence they are all compound events. 16.3.3 Algebra of events In the Chapter on Sets, we have studied about different ways of combining two or more sets, viz, union, intersection, difference, complement of a set etc. Like-wise we can combine two or more events by using the analogous set notations. Let A, B, C be events associated with an experiment whose sample space is S. 1. Complementary Event For every event A, there corresponds another event A " called the complementary event to A. It is also called the event 'not A'. For example, take the experiment 'of tossing three coins'. An associated sample space is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} Let A={HTH, HHT, THH} be the event 'only one tail appears' Clearly for the outcome HTT, the event A has not occurred. But we may say that the event 'not A' has occurred. Thus, with every outcome which is not in A, we say that 'not A' occurs.
2. The Event 'A or B' Recall that union of two sets A and B denoted by A ∪ B contains all those elements which are either in A or in B or in both. When the sets A and B are two events associated with a sample space, then 'A ∪ B' is the event 'either A or B or both'. This event 'A ∪ B' is also called 'A or B'. Therefore Event 'A or B' = A ∪ B = {ω : ω ∈ A or # ∈ B} 3. The Event 'A and B' We know that intersection of two sets A ∩ B is the set of those elements which are common to both A and B. i.e., which belong to both 'A and B'. If A and B are two events, then the set A ∩ B denotes the event 'A and B'. A ∩ B = {ω : ω ∈ A and ω ∈ B} Thus, For example, in the experiment of 'throwing a die twice' Let A be the event 'score on the first throw is six' and B is the event 'sum of two scores is atleast 11' then A = {(6,1), (6,2}, (6,3), (6,4), (6,5), (6,6)}, and B = {(5,6), (6,5), (6,6)} so A ∩ B = {(6,5), (6,6)} Note that the set A ∩ B = {(6,5), (6,6)} may represent the event 'the score on the first throw is six and the sum of the scores is atleast 11'.
4. The Event 'A but not B' We know that A–B is the set of all those elements which are in A but not in B. Therefore, the set A–B may denote the event 'A but not B'.We know that A – B = A ∩ B´ Example 6 Consider the experiment of rolling a die. Let A be the event 'getting a prime number', B be the event 'getting an odd number'. Write the sets representing the events (i) Aor B (ii) A and B (iii) A but not B (iv) 'not A'. Solution Here S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5} Obviously (i) 'A or B' = A ∪ B = {1, 2, 3, 5} (ii) 'A and B' = A ∩ B = {3,5} (iii) 'A but not B' = A – B = {2} (iv) 'not A' = A′ = {1,4,6}
PROBABILITY
391
16.3.4 Mutually exclusive events In the experiment of rolling a die, a sample space is S = {1, 2, 3, 4, 5, 6}. Consider events, A 'an odd number appears' and B 'an even number appears' Clearly the event A excludes the event B and vice versa. In other words, there is no outcome which ensures the occurrence of events A and B simultaneously. Here A = {1, 3, 5} and B = {2, 4, 6} Clearly A ∩ B = φ, i.e., A and B are disjoint sets. In general, two events A and B are called mutually exclusive events if the occurrence of any one of them excludes the occurrence of the other event, i.e., if they can not occur simultaneously. In this case the sets A and B are disjoint. Again in the experiment of rolling a die, consider the events A 'an odd number appears' and event B 'a number less than 4 appears' Obviously A = {1, 3, 5} and B = {1, 2, 3} Now 3 ∈ A as well as 3 ∈ B Therefore, A and B are not mutually exclusive events. Remark Simple events of a sample space are always mutually exclusive. 16.3.5 Exhaustive events Consider the experiment of throwing a die. We have S = {1, 2, 3, 4, 5, 6}. Let us define the following events A: 'a number less than 4 appears', B: 'a number greater than 2 but less than 5 appears' and C: 'a number greater than 4 appears'. Then A = {1, 2, 3}, B = {3,4} and C = {5, 6}. We observe that A ∪ B ∪ C = {1, 2, 3} ∪ {3, 4} ∪ {5, 6} = S. Such events A, B and C are called exhaustive events. In general, if E1, E2, ..., En are n events of a sample space S and if
E1 % E 2 % E 3 % ... % E n $ % E i $ S
i $1 n
then E1, E2, ...., En are called exhaustive events.In other words, events E1, E2, ..., En are said to be exhaustive if atleast one of them necessarily occurs whenever the experiment is performed. Further, if Ei ∩ Ej = φ for i ≠ j i.e., events Ei and Ej are pairwise disjoint and
i $1
EXERCISE 16.2
1. A die is rolled. Let E be the event "die shows 4" and F be the event "die shows even number". Are E and F mutually exclusive? 2. A die is thrown. Describe the following events: (i) A: a number less than 7 (iii) C: a multiple of 3 (v) E: an even number greater than 4 (ii) (iv) (vi) B: a number greater than 7 D: a number less than 4 F: a number not less than 3
Also find A ∪ B, A ∩ B, E ∪ F, D ∩ E, A – C, D – E, F′, E ∩ F′, 3. An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events: A: the sum is greater than 8, B: 2 occurs on either die C: the sum is at least 7 and a multiple of 3. Which pairs of these events are mutually exclusive? 4. Three coins are tossed once. Let A denote the event 'three heads show", B denote the event "two heads and one tail show", C denote the event" three tails show and D denote the event 'a head shows on the first coin". Which events are (i) mutually exclusive? (ii) simple? (iii) Compound? 5. Three coins are tossed. Describe (i) Two events which are mutually exclusive. (ii) Three events which are mutually exclusive and exhaustive. (iii) Two events, which are not mutually exclusive. (iv) Two events which are mutually exclusive but not exhaustive. (v) Three events which are mutually exclusive but not exhaustive. 6. Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice & 5. Describe the events (i) A ′ (ii) not B (iii) A or B (iv) A and B (v) A but not C (vi) B or C (vii) B and C (viii) A ∩ B′ ∩ C′ 7. Refer to question 6 above, state true or false: (give reason for your answer) (i) A and B are mutually exclusive (ii) A and B are mutually exclusive and exhaustive (iii) A = B′
394
MATHEMATICS
(iv) A and C are mutually exclusive (v) A and B′ are mutually exclusive. (vi) A′, B′, C are mutually exclusive and exhaustive.
16.4 Axiomatic Approach to Probability
In earlier sections, we have considered random experiments, sample space and events associated with these experiments. In our day to day life we use many words about the chances of occurrence of events. Probability theory attempts to quantify these chances of occurrence or non occurrence of events. In earlier classes, we have studied some methods of assigning probability to an event associated with an experiment having known the number of total outcomes. Axiomatic approach is another way of describing probability of an event. In this approach some axioms or rules are depicted to assign probabilities. Let S be the sample space of a random experiment. The probability P is a real valued function whose domain is the power set of S and range is the interval [0,1] satisfying the following axioms (ii) P (S) = 1 (i) For any event E, P (E) ' 0 (iii) If E and F are mutually exclusive events, then P(E ∪ F) = P(E) + P(F). It follows from (iii) that P(φ) = 0. To prove this, we take F = φ and note that E and φ are disjoint events. Therefore, from axiom (iii), we get P (E ∪ φ) = P (E) + P (φ) or P(E) = P(E) + P (φ) i.e. P (φ) = 0. Let S be a sample space containing outcomes #1 , #2 ,..., #n , i.e., S = {ω1, ω2, ..., ωn} It follows from the axiomatic definition of probability that (i) 0 ≤ P (ωi) ≤ 1 for each ωi ∈ S (ii) P (ω1) + P (ω2) + ... + P (ωn) = 1 (iii) For any event A, P(A) = ∑ P(ωi ), ωi ∈ A. that the called ! Note It may be noted we writesingleton {ω } is}). elementary event and for notational convenience, P(ω ) for P({ω
i i i
For example, in 'a coin tossing' experiment we can assign the number of the outcomes H and T. i.e. P(H) = and P(T) = (1)
to each
Clearly this assignment satisfies both the conditions i.e., each number is neither less than zero nor greater than 1 and
PROBABILITY
395
P(H) + P(T) =
1 1 + =1 2 2 1 1 , and probability of T = 2 2
Therefore, in this case we can say that probability of H = If we take P(H) =
3 . 4 We find that both the assignments (1) and (2) are valid for probability of H and T. In fact, we can assign the numbers p and (1 – p) to both the outcomes such that 0 ≤ p ≤ 1 and P(H) + P(T) = p + (1 – p) = 1 This assignment, too, satisfies both conditions of the axiomatic approach of probability. Hence, we can say that there are many ways (rather infinite) to assign probabilities to outcomes of an experiment. We now consider some examples. Example 9 Let a sample space be S = {ω1, ω2,..., ω6}.Which of the following assignments of probabilities to each outcome are valid? Outcomes # 1 #2 #3 #4 #5 #6
Yes, in this case, probability of H = and probability of T = (a) (b) (c) (d) (e)
1 6
1
1 6
0
1 6
0
1 6
0
1 6
0
1 6
0
1 8
1 12
2 3
1 3
1 3
(
1 4
(
1 3
1 12
0.2
1 6
0.3
1 6
0.4
1 6
0.5
3 2
0.6
0.1
Solution (a) Condition (i): Each of the number p(ωi) is positive and less than one. Condition (ii): Sum of probabilities =
1 1 1 1 1 1 ) ) ) ) ) $1 6 6 6 6 6 6
396
MATHEMATICS
Therefore, the assignment is valid (b) Condition (i): Each of the number p(ωi) is either 0 or 1. Condition (ii) Sum of the probabilities = 1 + 0 + 0 + 0 + 0 + 0 = 1 Therefore, the assignment is valid (c) Condition (i) Two of the probabilities p(ω5) and p(ω6) are negative, the assignment is not valid (d) (e)
16.4.1 Probability of an event Let S be a sample space associated with the experiment 'examining three consecutive pens produced by a machine and classified as Good (non-defective) and bad (defective)'. We may get 0, 1, 2 or 3 defective pens as result of this examination. A sample space associated with this experiment is S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG}, where B stands for a defective or bad pen and G for a non – defective or good pen. Let the probabilities assigned to the outcomes be as follows Sample point: Probability: BBB BBG BGB GBB BGG GBG GGB GGG
16.4.2 Probabilities of equally likely outcomes Let a sample space of an experiment be S = {ω1, ω2,..., ωn}. Let all the outcomes are equally likely to occur, i.e., the chance of occurrence of each simple event must be same. i.e. P(ωi) = p, for all ωi ∈ S where 0 ≤ p ≤ 1 Since
* P(!i ) $1 i.e., p + p + ... + p (n times) = 1
i$1
n
or
np = 1 i.e., p =
1 n
Let S be a sample space and E be an event, such that n(S) = n and n(E) = m. If each out come is equally likely, then it follows that
The points HTH and THH are common to both A and B . In the computation of P(A) + P(B) the probabilities of points HTH and THH, i.e., the elements of A ∩B are included twice. Thus to get the probability P(A % B) we have to subtract the probabilities of the sample points in A ∩ B from P(A) + P(B) i.e.
Thus we observe that, P( A % B) $ P( A) ) P( B) ( P (A / B) In general, if A and B are any two events associated with a random experiment, then by the definition of probability of an event, we have
P - A % B . $ * p - !i . , +!i , A % B .
Thus, for mutually exclusive events A and B, we have P( A % B) $ P ( A) ) P (B) , which is Axiom (iii) of probability. 16.4.4 Probability of event 'not A' Consider the event A = {2, 4, 6, 8} associated with the experiment of drawing a card from a deck of ten cards numbered from 1 to 10. Clearly the sample space is S = {1, 2, 3, ...,10} If all the outcomes 1, 2, ...,10 are considered to be equally likely, then the probability of each outcome is
3 2 = 1 ( $ 1 ( P ( A) 5 5 Also, we know that A′ and A are mutually exclusive and exhaustive events i.e., A ∩ A′ = φ and A ∪ A′ = S or P(A ∪ A′) = P(S) Now P(A) + P(A′) = 1, by using axioms (ii) and (iii). or P( A " ) = P(not A) = 1 – P(A) We now consider some examples and exercises having equally likely outcomes unless stated otherwise.
Example 10 One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be (i) a diamond (ii) not an ace (iii) a black card (i.e., a club or, a spade) (iv) not a diamond (v) not a black card. Solution When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes is 52. (i) Let A be the event 'the card drawn is a diamond' Clearly the number of elements in set A is 13. Therefore, P(A) =
13 1 $ 52 4
i.e. Probability of a diamond card = (ii) We assume that the event 'Card drawn is an ace' is B Therefore 'Card drawn is not an ace' should be B′. We know that P(B′) = 1 – P(B) = 1 (
(iv) We assumed in (i) above that A is the event 'card drawn is a diamond', so the event 'card drawn is not a diamond' may be denoted as A' or 'not A' Now P(not A) = 1 – P(A) = 1 ( (v)
1 3 $ 4 4 The event 'card drawn is not a black card' may be denoted as C′ or 'not C'.
We know that P(not C) = 1 – P(C) = 1 (
1 1 $ 2 2 1 2
Therefore, Probability of not a black card =
Example 11 A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be (i) red, (ii) yellow, (iii) blue, (iv) not blue, (v) either red or yellow. Solution There are 9 discs in all so the total number of possible outcomes is 9. Let the events A, B, C be defined as A: 'the disc drawn is red' B: 'the disc drawn is yellow' C: 'the disc drawn is blue'. (i) The number of red discs = 4, i.e., n (A) = 4 Hence
1 2 $ 3 3 (v) The event 'either red or yellow' may be described by the set 'A or C' Since, A and C are mutually exclusive events, we have
Therefore P(not C) = 1 ( P(A or C) = P (A ∪ C) = P(A) + P(C) =
4 1 7 ) $ 9 3 9
Example 12 Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that (a) Both Anil and Ashima will not qualify the examination. (b) (c) Atleast one of them will not qualify the examination and Only one of them will qualify the examination.
Solution Let E and F denote the events that Anil and Ashima will qualify the examination, respectively. Given that P(E) = 0.05, P(F) = 0.10 and P(E ∩ F) = 0.02. Then (a) The event 'both Anil and Ashima will not qualify the examination' may be expressed as E´ ∩ F´.
(c) The event only one of them will qualify the examination is same as the event either (Anil will qualify, and Ashima will not qualify) or (Anil will not qualify and Ashima will qualify) i.e., E ∩ F´ or E´ ∩ F, where E ∩ F´ and E´ ∩ F are mutually exclusive. Therefore, P(only one of them will qualify) = P(E ∩ F´ or E´ ∩ F) = P(E ∩ F´) + P(E´ ∩ F) = P (E) – P(E ∩ F) + P(F) – P (E ∩ F) = 0.05 – 0.02 + 0.10 – 0.02 = 0.11 Example 13 A committee of two persons is selected from two men and two women. What is the probability that the committee will have (a) no man? (b) one man? (c) two men? Solution The total number of persons = 2 + 2 = 4. Out of these four person, two can be selected in 4 C 2 ways. (a) No men in the committee of two means there will be two women in the committee. Out of two women, two can be selected in 2 C 2 $ 1 way. Therefore (b)
P - no man . $
2 4
C 2 12 2 2 1 1 $ $ 423 6 C2
One man in the committee means that there is one woman. One man out of 2
can be selected in 2 C1 ways and one woman out of 2 can be selected in 2 C1 ways. Together they can be selected in 2 C1 2 2 C1 ways. Therefore (c)
P - One man . $
2
1 2 3 4 5 6 15 14 14 14 14 14 14 14 2. A coin is tossed twice, what is the probability that atleast one tail occurs? 3. A die is thrown, find the probability of following events: (i) A prime number will appear, (ii) A number greater than or equal to 3 will appear, (iii) A number less than or equal to one will appear, (iv) A number more than 6 will appear, (v) A number less than 6 will appear. 4. A card is selected from a pack of 52 cards. (a) How many points are there in the sample space? (b) Calculate the probability that the card is an ace of spades. (c) Calculate the probability that the card is (i) an ace (ii) black card.
5. A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. find the probability that the sum of numbers that turn up is (i) 3 (ii) 12 6. There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman? 7. A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts. 8. Three coins are tossed once. Find the probability of getting (i) 3 heads (ii) 2 heads (iii) atleast 2 heads (iv) atmost 2 heads (v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) atmost two tails
2 is the probability of an event, what is the probability of the event 'not A'. 11 10. A letter is chosen at random from the word 'ASSASSINATION'. Find the probability that letter is (i) a vowel (ii) a consonant
9. If
PROBABILITY
405
11. In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of Winning the prize in the game. [Hint order of the numbers is not important.] 12. Check whether the following probabilities P(A) and P(B) are consistently defined (i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6 (ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8 13. Fill in the blanks in following table: P(A) P(B) P(A / B) P(A % B) (i) (ii) (iii)
1 3 1 5 1 15
(i) P(E or F), (ii) P(not E and not F). Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive. A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B) In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology. In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing atleast one of them is 0.95. What is the probability of passing both? The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?
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MATHEMATICS
21. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that (i) (ii) (iii) The student opted for NCC or NSS. The student has opted neither NCC nor NSS. The student has opted NSS but not NCC.
Miscellaneous Examples
Example 14 On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability that she visits (i) (iii) (v) A before B? A first and B last? A just before B? (ii) (iv) A before B and B before C? A either first or second?
Solution The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e., 24.Therefore, n (S) = 24. Since the number of elements in the sample space of the experiment is 24 all of these outcomes are considered to be equally likely. A sample space for the experiment is S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB BACD, BADC, BDAC, BDCA, BCAD, BCDA CABD, CADB, CBDA, CBAD, CDAB, CDBA DABC, DACB, DBCA, DBAC, DCAB, DCBA} (i) Let the event 'she visits A before B' be denoted by E Therefore, E = {ABCD, CABD, DABC, ABDC, CADB, DACB ACBD, ACDB, ADBC, CDAB, DCAB, ADCB} Thus (ii)
P-E. $
n-E. n - S.
$
12 1 $ 24 2
Let the event 'Veena visits A before B and B before C' be denoted by F. Here F = {ABCD, DABC, ABDC, ADBC}
Example 15 Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains (i) all Kings (ii) 3 Kings (iii) atleast 3 Kings. Solution Total number of possible hands =
52
– P - A / B. ( P - A / C . ) P - A / B / C . Example 17 In a relay race there are five teams A, B, C, D and E. (a) (b) What is the probability that A, B and C finish first, second and third, respectively. What is the probability that A, B and C are first three to finish (in any order) (Assume that all finishing orders are equally likely)
Solution If we consider the sample space consisting of all finishing orders in the first three places, we will have 5 P3 , i.e., a probability of (a)
5! = 5 × 4 × 3 = 60 sample points, each with - 5 ( 3 .!
1 . 60 A, B and C finish first, second and third, respectively. There is only one finishing order for this, i.e., ABC. 1 . 60 A, B and C are the first three finishers. There will be 3! arrangements for A, B and C. Therefore, the sample points corresponding to this event will be 3! in number.
P (A, B and C are first three to finish) $
Thus P(A, B and C finish first, second and third respectively) = (b)
So
3! 6 1 $ $ 60 60 10
Miscellaneous Exercise on Chapter 16
1. A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that (i) all will be blue? (ii) atleast one will be green? 2. 4 cards are drawn from a well – shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?
PROBABILITY
409
3. A die has two faces each with number '1', three faces each with number '2' and one face with number '3'. If die is rolled once, determine (i) P(2) (ii) P(1 or 3) (iii) P(not 3) 4. In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets. 5. Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that (a) you both enter the same section? (b) you both enter the different sections? 6. Three letters are dictated to three persons and an evelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope. 7. A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (i) P(A ∪ B) (ii) P(A´ ∩ B´) (iii) P(A ∩ B´) (iv) P(B ∩ A´) 8. From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows: S. No. 1. 2. 3. 4. 5. Name Harish Rohan Sheetal Alis Salim Sex M M F F M Age in years 30 33 46 28 41
A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years? 9. If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when, (i) the digits are repeated? (ii) the repetition of digits is not allowed? 10. The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?
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MATHEMATICS
Summary
In this Chapter, we studied about the axiomatic approach of probability. The main features of this Chapter are as follows:
" Sample space: The set of all possible outcomes " Sample points: Elements of sample space " Event: A subset of the sample space " Impossible event : The empty set " Sure event: The whole sample space " Complementary event or 'not event' : The set A′ or S – A " Event A or B: The set A ∪ B " Event A and B: The set A ∩ B " Event A and not B: The set A – B " Mutually exclusive event: A and B are mutually exclusive if A ∩ B = φ " Exhaustive and mutually exclusive events: Events E1, E2,..., En are mutually " Probability: Number P (ωi) associated with sample point # i such that
(i) 0 ≤ P (ωi) ≤ 1 (iii) P(A) =
i
i
" If A and B are mutually exclusive, then P(A or B) = P(A) + P(B) " If A is any event, then
P(not A) = 1 – P(A)
PROBABILITY
411
Historical Note
Probability theory like many other branches of mathematics, evolved out of practical consideration. It had its origin in the 16th century when an Italian physician and mathematician Jerome Cardan (1501–1576) wrote the first book on the subject "Book on Games of Chance" (Biber de Ludo Aleae). It was published in 1663 after his death. In 1654, a gambler Chevalier de Metre approached the well known French Philosoher and Mathematician Blaise Pascal (1623–1662) for certain dice problem. Pascal became interested in these problems and discussed with famous French Mathematician Pierre de Fermat (1601–1665). Both Pascal and Fermat solved the problem independently. Besides, Pascal and Fermat, outstanding contributions to probability theory were also made by Christian Huygenes (1629–1665), a Dutchman, J. Bernoulli (1654–1705), De Moivre (1667–1754), a Frenchman Pierre Laplace (1749–1827), A Frenchman and the Russian P.L Chebyshev (1821–1897), A. A Markov (1856–1922) and A. N Kolmogorove (1903–1987). Kolmogorove is credited with the axiomatic theory of probability. His book 'Foundations of Probability' published in 1933, introduces probability as a set function and is considered a classic.
—! —
Appendix
1
INFINITE SERIES
A.1.1 Introduction
As discussed in the Chapter 9 on Sequences and Series, a sequence a1, a2, ..., an, ... having infinite number of terms is called infinite sequence and its indicated sum, i.e., a1 + a2 + a3 + ... + an + ... is called an infinte series associated with infinite sequence. This series can also be expressed in abbreviated form using the sigma notation, i.e., a1 + a2 + a3 + . . . + an + . . . =
#a
k "1
!
k
In this Chapter, we shall study about some special types of series which may be required in different problem situations.
A.1.2 Binomial Theorem for any Index
In Chapter 8, we discussed the Binomial Theorem in which the index was a positive integer. In this Section, we state a more general form of the theorem in which the index is not necessarily a whole number. It gives us a particular type of infinite series, called Binomial Series. We illustrate few applications, by examples. We know the formula (1 + x)n = n C0 + n C1 x + . . . + n Cn xn Here, n is non-negative integer. Observe that if we replace index n by negative integer or a fraction, then the combinations n C r do not make any sense. We now state (without proof), the Binomial Theorem, giving an infinite series in which the index is negative or a fraction and not a whole number. Theorem The formula
m $ m & 1% 2 m $ m & 1%$ m & 2 % 3 x ' x ' ... 1.2 1.2.3
A.1.3 Infinite Geometric Series
From Chapter 9, Section 9.3, a sequence a 1 , a 2, a 3, ..., a n is called G.P., if
ak '1 ak = r (constant) for k = 1, 2, 3, . . ., n–1. Particularly, if we take a1 = a, then the
resulting sequence a, ar, ar2, . . ., arn–1 is taken as the standard form of G.P., where a is first term and r, the common ratio of G.P. Earlier, we have discussed the formula to find the sum of finite series a + ar + ar2 + ... + arn – 1 which is given by
Sn "
a 1& rn 1& r
$
%.
In this section, we state the formula to find the sum of infinite geometric series a + ar + ar2 + .. . . .+ arn – 1 +. . . . and illustrate the same by examples. Let us consider the G.P. 1, Here a = 1, r =
A.1.4 Exponential Series
Leonhard Euler (1707 – 1783), the great Swiss mathematician introduced the number e in his calculus text in 1748. The number e is useful in calculus as 6 in the study of the circle. Consider the following infinite series of numbers
1 1 1 1 ' ' ' ' ... ... (1) 1! 2! 3! 4! The sum of the series given in (1) is denoted by the number e Let us estimate the value of the number e. Since every term of the series (1) is positive, it is clear that its sum is also positive. Consider the two sums 1'
Thus, e2 lies between 7.355 and 7.4. Therefore, the value of e2, rounded off to one decimal place, is 7.4.
A.1.5 Logarithmic Series
Another very important series is logarithmic series which is also in the form of infinite series. We state the following result without proof and illustrate its application with an example. Theorem If | x | < 1, then
x 2 x3 ' & ... 2 3 The series on the right hand side of the above is called the logarithmic series. log e $1 ' x % " x &
Here, we have used the facts ? ' @ = p and ?@ " q . We know this from the given roots of the quadratic equation. We have also assumed that both | ? x | < 1 and
| @ x | < 1.
—! —
Appendix
2
MATHEMATICAL MODELLING
A.2.1 Introduction
Much of our progress in the last few centuries has made it necessary to apply mathematical methods to real-life problems arising from different fields – be it Science, Finance, Management etc. The use of Mathematics in solving real-world problems has become widespread especially due to the increasing computational power of digital computers and computing methods, both of which have facilitated the handling of lengthy and complicated problems. The process of translation of a real-life problem into a mathematical form can give a better representation and solution of certain problems. The process of translation is called Mathematical Modelling. Here we shall familiaries you with the steps involved in this process through examples. We shall first talk about what a mathematical model is, then we discuss the steps involved in the process of modelling.
A.2.2 Preliminaries
Mathematical modelling is an essential tool for understanding the world. In olden days the Chinese, Egyptians, Indians, Babylonians and Greeks indulged in understanding and predicting the natural phenomena through their knowledge of mathematics. The architects, artisans and craftsmen based many of their works of art on geometric prinicples. Suppose a surveyor wants to measure the height of a tower. It is physically very difficult to measure the height using the measuring tape. So, the other option is to find out the factors that are useful to find the height. From his knowledge of trigonometry, he knows that if he has an angle of elevation and the distance of the foot of the tower to the point where he is standing, then he can calculate the height of the tower. So, his job is now simplified to find the angle of elevation to the top of the tower and the distance from the foot of the tower to the point where he is standing. Both of which are easily measurable. Thus, if he measures the angle of elevation as 40° and the distance as 450m, then the problem can be solved as given in Example 1.
422
MATHEMATICS
Example 1 The angle of elevation of the top of a tower from a point O on the ground, which is 450 m away from the foot of the tower, is 40°. Find the height of the tower. Solution We shall solve this in different steps. Step 1 We first try to understand the real problem. In the problem a tower is given and its height is to be measured. Let h denote the height. It is given that the horizontal distance of the foot of the tower from a particular point O on the ground is 450 m. Let d denotes this distance. Then d = 450m. We also know that the angle of elevation, denoted by θ, is 40°. The real problem is to find the height h of the tower using the known distance d and the angle of elevation θ. Step 2 The three quantities mentioned in the problem are height, distance and angle of elevation. So we look for a relation connecting these three quantities. This is obtained by expressing it geometrically in the following way (Fig 1). AB denotes the tower. OA gives the horizontal distance from the point O to foot of the tower. ∠AOB is the angle of elevation. Then we have tan θ =
h or h = d tan θ d
... (1)
Fig 1
This is an equation connecting θ, h and d.
Step 3 We use Equation (1) to solve h. We have θ = 40°. and d = 450m. Then we get h = tan 40° × 450 = 450 × 0.839 = 377.6m Step 4 Thus we got that the height of the tower approximately 378m. Let us now look at the different steps used in solving the problem. In step 1, we have studied the real problem and found that the problem involves three parameters height, distance and angle of elevation. That means in this step we have studied the real-life problem and identified the parameters. In the Step 2, we used some geometry and found that the problem can be represented geometrically as given in Fig 1. Then we used the trigonometric ratio for the "tangent" function and found the relation as h = d tan θ So, in this step we formulated the problem mathematically. That means we found an equation representing the real problem.
MATHEMATICAL MODELLING
423
In Step 3, we solved the mathematical problem and got that h = 377.6m. That is we found Solution of the problem. In the last step, we interpreted the solution of the problem and stated that the height of the tower is approximately 378m. We call this as Interpreting the mathematical solution to the real situation In fact these are the steps mathematicians and others use to study various reallife situations. We shall consider the question, "why is it necessary to use mathematics to solve different situations." Here are some of the examples where mathematics is used effectively to study various situations. 1. Proper flow of blood is essential to transmit oxygen and other nutrients to various parts of the body in humanbeings as well as in all other animals. Any constriction in the blood vessel or any change in the characteristics of blood vessels can change the flow and cause damages ranging from minor discomfort to sudden death. The problem is to find the relationship between blood flow and physiological characteristics of blood vessel. 2. In cricket a third umpire takes decision of a LBW by looking at the trajectory of a ball, simulated, assuming that the batsman is not there. Mathematical equations are arrived at, based on the known paths of balls before it hits the batsman's leg. This simulated model is used to take decision of LBW. 3. Meteorology department makes weather predictions based on mathematical models. Some of the parameters which affect change in weather conditions are temperature, air pressure, humidity, wind speed, etc. The instruments are used to measure these parameters which include thermometers to measure temperature, barometers to measure airpressure, hygrometers to measure humidity, anemometers to measure wind speed. Once data are received from many stations around the country and feed into computers for further analysis and interpretation. 4. Department of Agriculture wants to estimate the yield of rice in India from the standing crops. Scientists identify areas of rice cultivation and find the average yield per acre by cutting and weighing crops from some representative fields. Based on some statistical techniques decisions are made on the average yield of rice. How do mathematicians help in solving such problems? They sit with experts in the area, for example, a physiologist in the first problem and work out a mathematical equivalent of the problem. This equivalent consists of one or more equations or inequalities etc. which are called the mathematical models. Then
424
MATHEMATICS
solve the model and interpret the solution in terms of the original problem. Before we explain the process, we shall discuss what a mathematical model is. A mathematical model is a representation which comprehends a situation. An interesting geometric model is illustrated in the following example. Example 2 (Bridge Problem) Konigsberg is a town on the Pregel River, which in the 18th century was a German town, but now is Russian. Within the town are two river islands that are connected to the banks with seven bridges as shown in (Fig 2). People tried to walk around the town in a way that only crossed each bridge once, but it proved to be difficult problem. Leonhard Euler, a Swiss Fig 2 mathematician in the service of the Russian empire Catherine the Great, heard about the problem. In 1736 Euler proved that the walk was not possible to do. He proved this by inventing a kind of diagram called a network, that is made up of vertices (dots where lines meet) and arcs (lines) (Fig3). He used four dots (vertices) for the two river banks and the two islands. These have been marked A, B and C, D. The seven lines (arcs) are the seven bridges. You can see that 3 bridges (arcs) join to riverbank, A, and 3 join to riverbank B. 5 bridges (arcs) join to island C, and 3 join to island D. This means that all the vertices have an odd number of arcs, so Fig 3 they are called odd vertices (An even vertex would have to have an even number of arcs joining to it). Remember that the problem was to travel around town crossing each bridge only once. On Euler's network this meant tracing over each arc only once, visiting all the vertices. Euler proved it could not be done because he worked out that, to have an odd vertex you would have to begin or end the trip at that vertex. (Think about it). Since there can only be one beginning and one end, there can only be two odd vertices if you are to trace over each arc only once. Since the bridge problem has 4 odd vertices, it just not possible to do!
MATHEMATICAL MODELLING
425
After Euler proved his Theorem, much water has flown under the bridges in Konigsberg. In 1875, an extra bridge was built in Konigsberg, joining the land areas A and D (Fig 4). Is it possible now for the Konigsbergians to go round the city, using each bridge only once? Here the situation will be as in Fig 4. After the addition of the new edge, both the vertices A and D have become even degree vertices. Fig 4 However, B and C still have odd degree. So, it is possible for the Konigsbergians to go around the city using each bridge exactly once. The invention of networks began a new theory called graph theory which is now used in many ways, including planning and mapping railway networks (Fig 4).
A.2.3 What is Mathematical Modelling?
Here, we shall define what mathematical modelling is and illustrate the different processes involved in this through examples. Definition Mathematical modelling is an attempt to study some part (or form) of the real-life problem in mathematical terms. Conversion of physical situation into mathematics with some suitable conditions is known as mathematical modelling. Mathematical modelling is nothing but a technique and the pedagogy taken from fine arts and not from the basic sciences. Let us now understand the different processes involved in Mathematical Modelling. Four steps are involved in this process. As an illustrative example, we consider the modelling done to study the motion of a simple pendulum. Understanding the problem This involves, for example, understanding the process involved in the motion of simple pendulum. All of us are familiar with the simple pendulum. This pendulum is simply a mass (known as bob) attached to one end of a string whose other end is fixed at a point. We have studied that the motion of the simple pendulum is periodic. The period depends upon the length of the string and acceleration due to gravity. So, what we need to find is the period of oscillation. Based on this, we give a precise statement of the problem as Statement How do we find the period of oscillation of the simple pendulum? The next step is formulation. Formulation Consists of two main steps. 1. Identifying the relevant factors In this, we find out what are the factors/
426
MATHEMATICS
parameters involved in the problem. For example, in the case of pendulum, the factors are period of oscillation (T), the mass of the bob (m), effective length (l ) of the pendulum which is the distance between the point of suspension to the centre of mass of the bob. Here, we consider the length of string as effective length of the pendulum and acceleration due to gravity (g), which is assumed to be constant at a place. So, we have identified four parameters for studying the problem. Now, our purpose is to find T. For this we need to understand what are the parameters that affect the period which can be done by performing a simple experiment. We take two metal balls of two different masses and conduct experiment with each of them attached to two strings of equal lengths. We measure the period of oscillation. We make the observation that there is no appreciable change of the period with mass. Now, we perform the same experiment on equal mass of balls but take strings of different lengths and observe that there is clear dependence of the period on the length of the pendulum. This indicates that the mass m is not an essential parameter for finding period whereas the length l is an essential parameter. This process of searching the essential parameters is necessary before we go to the next step. 2. Mathematical description This involves finding an equation, inequality or a geometric figure using the parameters already identified. In the case of simple pendulum, experiments were conducted in which the values of period T were measured for different values of l. These values were plotted on a graph which resulted in a curve that resembled a parabola. It implies that the relation between T and l could be expressed T2 = kl It was found that k !
4! 2 . This gives the equation g
... (1)
T ! 2!
l g
... (2)
Equation (2) gives the mathematical formulation of the problem.
Finding the solution The mathematical formulation rarely gives the answer directly. Usually we have to do some operation which involves solving an equation, calculation or applying a theorem etc. In the case of simple pendulums the solution involves applying the formula given in Equation (2).
MATHEMATICAL MODELLING
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The period of oscillation calculated for two different pendulums having different lengths is given in Table 1
Table 1
Interpretation/Validation
A mathematical model is an attempt to study, the essential characteristic of a real life problem. Many times model equations are obtained by assuming the situation in an idealised context. The model will be useful only if it explains all the facts that we would like it to explain. Otherwise, we will reject it, or else, improve it, then test it again. In other words, we measure the effectiveness of the model by comparing the results obtained from the mathematical model, with the known facts about the real problem. This process is called validation of the model. In the case of simple pendulum, we conduct some experiments on the pendulum and find out period of oscillation. The results of the experiment are given in Table 2.
Table 2
Now, we compare the measured values in Table 2 with the calculated values given in Table 1. The difference in the observed values and calculated values gives the error. For example, for l = 275 cm, and mass m = 385 gm, error = 3.371 – 3.36 = 0.011 which is small and the model is accepted. Once we accept the model, we have to interpret the model. The process of describing the solution in the context of the real situation is called interpretation of the model. In this case, we can interpret the solution in the following way: (a) The period is directly proportional to the square root of the length of the pendulum.
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(b) It is inversely proportional to the square root of the acceleration due to gravity. Our validation and interpretation of this model shows that the mathematical model is in good agreement with the practical (or observed) values. But we found that there is some error in the calculated result and measured result. This is because we have neglected the mass of the string and resistance of the medium. So, in such situation we look for a better model and this process continues. This leads us to an important observation. The real world is far too complex to understand and describe completely. We just pick one or two main factors to be completely accurate that may influence the situation. Then try to obtain a simplified model which gives some information about the situation. We study the simple situation with this model expecting that we can obtain a better model of the situation. Now, we summarise the main process involved in the modelling as (a) Formulation (b) Solution (c) Interpretation/Validation The next example shows how modelling can be done using the techniques of finding graphical solution of inequality. Example 3 A farm house uses atleast 800 kg of special food daily. The special food is a mixture of corn and soyabean with the following compositions
Table 3
Material Corn Soyabean
Nutrients present per Kg Protein .09 .60
Nutrients present per Kg Fibre .02 .06
Cost per Kg Rs 10 Rs 20
The dietary requirements of the special food stipulate atleast 30% protein and at most 5% fibre. Determine the daily minimum cost of the food mix. Solution Step 1 Here the objective is to minimise the total daily cost of the food which is made up of corn and soyabean. So the variables (factors) that are to be considered are x = the amount of corn y = the amount of soyabean z = the cost Step 2 The last column in Table 3 indicates that z, x, y are related by the equation z = 10x + 20y ... (1) The problem is to minimise z with the following constraints:
MATHEMATICAL MODELLING
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The farm used atleast 800 kg food consisting of corn and soyabean i.e., x + y ≥ 800 ... (2) (b) The food should have atleast 30% protein dietary requirement in the proportion as given in the first column of Table 3. This gives 0.09x + 0.6y ≥ 0.3 (x + y) ... (3) (c) Similarly the food should have atmost 5% fibre in the proportion given in 2nd column of Table 3. This gives 0.02x + 0.06 y ≤ 0.05 (x + y) ... (4) We simplify the constraints given in (2), (3) and (4) by grouping all the coefficients of x, y. Then the problem can be restated in the following mathematical form. Statement Minimise z subject to x + y ≥ 800 0.21x – .30y ≤ 0 0.03x – .01y ≥ 0 This gives the formulation of the model. Step 3 This can be solved graphically. The shaded region in Fig 5 gives the possible solution of the equations. From the graph it is clear that the minimum value is got at the point (470.6,329.4) i.e., x = 470.6 and y = 329.4.
(a)
Fig 5
This gives the value of z as z = 10 × 470.6 + 20 × 329.4 = 11294 This is the mathematical solution.
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MATHEMATICS
Step 4 The solution can be interpreted as saying that, "The minimum cost of the special food with corn and soyabean having the required portion of nutrient contents, protein and fibre is Rs 11294 and we obtain this minimum cost if we use 470.6 kg of corn and 329.4 kg of soyabean." In the next example, we shall discuss how modelling is used to study the population of a country at a particular time. Example 4 Suppose a population control unit wants to find out "how many people will be there in a certain country after 10 years" Step 1 Formulation We first observe that the population changes with time and it increases with birth and decreases with deaths. We want to find the population at a particular time. Let t denote the time in years. Then t takes values 0, 1, 2, ..., t = 0 stands for the present time, t = 1 stands for the next year etc. For any time t, let p (t) denote the population in that particular year. Suppose we want to find the population in a particular year, say t0 = 2006. How will we do that. We find the population by Jan. 1st, 2005. Add the number of births in that year and subtract the number of deaths in that year. Let B(t) denote the number of births in the one year between t and t + 1 and D(t) denote the number of deaths between t and t + 1. Then we get the relation P (t + 1) = P (t) + B (t) – D (t) Now we make some assumptions and definitions 1. 2.
B (t ) P (t ) is called the birth rate for the time interval t to t + 1. D (t) P (t) is called the death rate for the time interval t to t + 1.
Assumptions 1. The birth rate is the same for all intervals. Likewise, the death rate is the same for all intervals. This means that there is a constant b, called the birth rate, and a constant d, called the death rate so that, for all t ≥ 0,
b!
B (t ) P (t )
and
d!
D (t ) P (t )
... (1)
2. There is no migration into or out of the population; i.e., the only source of population change is birth and death. As a result of assumptions 1 and 2, we deduce that, for t ≥ 0,
for t = 0, 1, 2, ... The constant 1 + b – d is often abbreviated by r and called the growth rate or, in more high-flown language, the Malthusian parameter, in honor of Robert Malthus who first brought this model to popular attention. In terms of r, Equation (4) becomes , t = 0, 1, 2, ... ... (5) P(t) = P(0)r t P(t) is an example of an exponential function. Any function of the form cr t, where c and r are constants, is an exponential function. Equation (5) gives the mathematical formulation of the problem. Step 2 – Solution Suppose the current population is 250,000,000 and the rates are b = 0.02 and d = 0.01. What will the population be in 10 years? Using the formula, we calculate P(10). P(10) = (1.01)10 (250,000,000) = (1.104622125) (250,000,000) = 276,155,531.25 Step 3 Interpretation and Validation Naturally, this result is absurd, since one can't have 0.25 of a person. So, we do some approximation and conclude that the population is 276,155,531 (approximately). Here, we are not getting the exact answer because of the assumptions that we have made in our mathematical model. The above examples show how modelling is done in variety of situations using different mathematical techniques. Since a mathematical model is a simplified representation of a real problem, by its very nature, has built-in assumptions and approximations. Obviously, the most important
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MATHEMATICS
question is to decide whether our model is a good one or not i.e., when the obtained results are interpreted physically whether or not the model gives reasonable answers. If a model is not accurate enough, we try to identify the sources of the shortcomings. It may happen that we need a new formulation, new mathematical manipulation and hence a new evaluation. Thus mathematical modelling can be a cycle of the modelling process as shown in the flowchart given below:
EXERCISE 14.1
(i) This sentence is always false because the maximum number of days in a month is 31. Therefore, it is a statement. (ii) This is not a statement because for some people mathematics can be easy and for some others it can be difficult. (iii) This sentence is always true because the sum is 12 and it is greater than 10. Therefore, it is a statement. (iv) This sentence is sometimes true and sometimes not true. For example the square of 2 is even number and the square of 3 is an odd number. Therefore, it is not a statement. (v) This sentence is sometimes true and sometimes false. For example, squares and rhombus have equal length whereas rectangles and trapezium have unequal length. Therefore, it is not a statement. (vi) It is an order and therefore, is not a statement. (vii) This sentence is false as the product is (–8). Therefore, it is a statement. (viii) This sentence is always true and therefore, it is a statement. (ix) It is not clear from the context which day is referred and therefore, it is not a statement. (x) This is a true statement because all real numbers can be written in the form a + i × 0. 2. The three examples can be: (i) Everyone in this room is bold. This is not a statement because from the context it is not clear which room is reffered here and the term bold is not precisely defined. (ii) She is an engineering student. This is also not a statement because who 'she' is. (iii) "cos2θ is always greater than 1/2". Unless, we know what θ is, we cannot say whether the sentence is true or not. 1.
ANSWERS
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EXERCISES 14.2
1. (i) Chennai is not the capital of Tamil Nadu.
2 is a complex number. All triangles are equilateral tringles. The number 2 is not greater than 7. Every natural number is not an integer. The negation of the first statement is "the number x is a rational number." which is the same as the second statement" This is because when a number is not irrational, it is a rational. Therefore, the given pairs are negations of each other. The negation of the first statement is "x is an irrational number" which is the same as the second statement. Therefore, the pairs are negations of each other. Number 3 is prime; number 3 is odd (True). All integers are positive; all integers are negative (False). 100 is divisible by 3,100 is divisible by 11 and 100 is divisible by 5 (False).
(ii) (iii) (iv) (v) 2. (i)
(ii)
3.
(i) (ii) (iii)
EXERCISE 14.3
(i) "And". The component statements are: All rational numbers are real. All real numbers are not complex. (ii) "Or". The component statements are: Square of an integer is positive. Square of an integer is negative. (iii) "And". the component statements are: The sand heats up quickily in the sun. The sand does not cool down fast at night. (iv) "And". The component statements are: 2 x = 2 is a root of the equation 3x – x – 10 = 0 2 x = 3 is a root of the equation 3x – x – 10 = 0 2. (i) "There exists". The negation is There does not exist a number which is equal to its square. (ii) "For every". The negation is There exists a real number x such that x is not less than x + 1. (iii) "There exists". The negation is There exists a state in India which does not have a capital. 1.
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MATHEMATICS
No. The negation of the statement in (i) is "There exists real number x and y for which x + y ≠ y + x", instead of the statement given in (ii). 4. (i) Exclusive (ii) Inclusive (iii) Exclusive 3.
EXERCISE 14.4
1. (i) (ii) (iii) (iv) (v) 2. (i) A natural number is odd implies that its square is odd. A natural number is odd only if its square is odd. For a natural number to be odd it is necessary that its square is odd. For the square of a natural number to be odd, it is sufficient that the number is odd If the square of a natural number is not odd, then the natural number is not odd. The contrapositive is If a number x is not odd, then x is not a prime number. The converse is If a number x in odd, then it is a prime number. The contrapositive is If two lines intersect in the same plane, then they are not parallel The converse is If two lines do not interesect in the same plane, then they are parallel The contrapositive is If something is not at low temperature, then it is not cold The converse is If something is at low temperature, then it is cold The contrapositive is If you know how to reason deductively, then you can comprehend geometry. The converse is If you do not know how to reason deductively, then you can not comprehend geometry. This statement can be written as "If x is an even number, then x is divisible by 4". The contrapositive is, If x is not divisible by 4, then x is not an even number. The converse is, If x is divisible by 4, then x is an even number. If you get a job, then your credentials are good. If the banana tree stays warm for a month, then it will bloom.
(ii)
(iii)
(iv)
(v)
3.
(i) (ii)
ANSWERS
461
(iii) (iv) 4. a (i) (ii) b (i) (ii)
If diagonals of a quadrilateral bisect each other, then it is a parallelogram. If you get A+ in the class, then you do all the exercises in the book. Contrapositive Converse Contrapositive Converse
EXERCISE 14.5
5. (i) False. By definition of the chord, it should intersect the circle in two points. (ii) False. This can be shown by giving a counter example. A chord which is not a dimaeter gives the counter example. (iii) True. In the equation of an ellipse if we put a = b, then it is a circle (Direct Method) (iv) True, by the rule of inequality (v) False. Since 11 is a prime number, therefore 11 is irrational.
Miscellaneous Exercise on Chapter 14
(i) (ii) (iii) (iv) 2. (i) 1. There exists a positive real number x such that x–1 is not positive. There exists a cat which does not scratch. There exists a real number x such that neither x > 1 nor x < 1. There does not exist a number x such that 0 < x < 1. The statement can be written as "If a positive integer is prime, then it has no divisors other than 1 and itself. The converse of the statement is If a positive integer has no divisors other than 1 and itself, then it is a prime. The contrapositive of the statement is If positive integer has divisors other than 1 and itself then it is not prime. (ii) The given statement can be written as "If it is a sunny day, then I go to a beach. The converse of the statement is If I go to beach, then it is a sunny day. The contrapositive is If I do not go to a beach, then it is not a sunny day. (iii) The converse is If you feel thirsty, then it is hot outside. The contrapositive is If you do not feel thirsty, then it is not hot outside.
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MATHEMATICS
(i) If there is log on to the server, then you have a password. (ii) If it rains, then there is traffic jam. (iii) If you can access the website, then you pay a subscription fee. 4. (i) You watch television if and only if your mind in free. (ii) You get an A grade if and only you do all the homework regularly. (iii) A quadrilateral is equiangular if and only if it is a rectangle. 5. The compound statement with "And" is 25 is a multiple of 5 and 8 This is a false statement. The compound statement with "Or" is 25 is a multiple of 5 or 8 This is true statement. 7. Same as Q1 in Exercise 14.4 3.
(iii) "Getting at most two tails", and "getting exactly two tails" (iv) "Getting exactly one head" and "getting exactly two heads" (v) "Getting exactly one tail", "getting exactly two tails", and getting exactly three tails" | 677.169 | 1 |
Math X
This is a course of independent study aimed at preparing very motivated students for a university program in pure mathematics. The basic component consists of the academic stream of high-school math, including AP Calculus AB. The enrichment components are mainly based on Paul Zeitz's The Art and Craft of Problem-Solving. Gilbert Strang's Introduction to Linear Algebra provides a capstone in Grade 12.
Current Assignment
September 2016 – January 2016: Complete Precalculus 11 and Zeitz chapters 3 and 6. I've added enrichment problem sets to chapters 1 and 2 of the Precalculus book. If there is time left over at the end of the semester, do Zeitz chapter 7. | 677.169 | 1 |
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Math UMA007 : Numerical Analysis
Course Description Numerical Analysis (MATH UMA007) is a one semester, upper-level
module that emphasizes the mathematics used to
design numerical methods, and to analyse
their properties. Students also experiment with implementing algorithms in Matlab/Octave. Course credit are 4.5. The details are given in the Course Policies. | 677.169 | 1 |
Unizor's Shop
Unizor's Shop
Welcome to Unizor - an Advanced Math course for high school students and teachers (
Here we prove theorems, solve problems, take exams, enroll into different subjects.
We discourage memorization and encourage derivation.
We develop creativity, logic and analytical thinking.
We believe, this course is an excellent tool to develop intelligent critical mind.
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A series of lectures introducing a concept of vectors, their representations, geometrical and algebraic meaning and operations with vectors - addition, subtraction, multiplication by a constant, scalar product and vector product.
A set of lectures dedicated to advanced geometry for high school students.
Special attention is given to proofs of theorem and solving problems.
This set of lectures is a part of the Advanced Math 4 Teenagers course on the Web site
This document describes the composition of the Math Concepts subject of the advanced course of Mathematics for high school students and teachers presented on the Internet at
Click "Math Concepts" on the Unizor main page to get started.
All you want to know about Trigonometry, but were afraid to ask.
Rigorous definitions of trigonometric functions for any angles (not necessarily acute), their graphs and numerous problems related to trigonometric identities, equations and inequalities.
The subject of Probability is introduced on the elementary level with discussions about its meaning (measure of frequency of past event occurrences or amount of knowledge about event, or measure of prediction about its future occurrence).
Concepts of Conditional probability, random variables, distribution of probabilities and characteristics of these distributions (expected value and variance) are introduced and supplemented with numerous examples and problems.
I'd like to introduce you to my educational site Unizor, which is created with an idea to redirect studying mathematics from simple memorization of facts, usually forgotten immediately after passing an exam, towards development of intelligence, creativity and analytical thinking, which I consider the real purpose to study mathematics and important throughout the whole life, regardless of profession.
Texas A&M University has published a short article about Unizor at
Combinatorics is extremely important subject of Mathematics.
Studying it and, especially, solving problems presented and analyzed in this course would prove to be extremely important to develop creativity, logic and analytical thinking in students.
This is one of the most useful math subjects in everyday life filled with necessities of making choices.
A relatively superficial introduction to matrices as linear transformations.
Most of the attention is dedicated to 2x2 and 3x3 matrices.
Rules of operation with matrices are discussed as well as their role in solving the systems of linear equations.
This document describes the composition of the Algebra subject of the advanced course of Mathematics for high school students and teachers presented on the Internet at
Click "Algebra" on the Unizor main page to get started.
Unizor features a curriculum for advanced course of mathematics for high schools. The word "advanced" we mean in terms of depth of learning, rather than in terms of number of topics covered.
Most important characteristic that distinguishes this course from many other on-line math courses is its emphasis on problem solving, rigorousness of presentation and proof of theorems as the main methodology.
Unizor emphasizes "Knowledge" and "Why?" rather than "Skill" and "How?".
Unizor is a mind development tool inasmuch as gym is a tool to develop muscles and stamina. | 677.169 | 1 |
ractical Algebra: A Self-Teaching Guide
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Practical Algebra If you studied algebra years ago and now need a refresher course in order to use algebraic principles on the job, or if you're a student who needs an introduction to the subject, here's the perfect book for you. Practical Algebra is an easy and fun-to-use workout program that quickly puts you in command of all the basic concepts and tools of algebra. With the aid of practical, real-life examples and applications, you'll learn: * The basic approach and application of algebra to problem solving* The number system (in a much broader way than you have known it from arithmetic)* Monomials and polynomials; factoring algebraic expressions; how to handle algebraic fractions; exponents, roots, and radicals; linear and fractional equations* Functions and graphs; quadratic equations; inequalities; ratio, proportion, and variation; how to solve word problems, and more Authors Peter Selby and Steve Slavin emphasize practical algebra throughout by providing you with techniques for solving problems in a wide range of disciplines - from engineering, biology, chemistry, and the physical sciences, to psychology and even sociology and business administration.Step by step, Practical Algebra shows you how to solve algebraic problems in each of these areas, then allows you to tackle similar problems on your own, at your own pace. Self-tests are provided at the end of each chapter so you can measure your mastery | 677.169 | 1 |
PROBLEM SOLVING FOR THE ELEMENTARY TEACHER
PROBLEM SOLVING FOR THE ELEMENTARY TEACHER
2013 Fall Term
3 Units
Mathematics 370
This course is primarily for pre-service elementary and middle school teachers. Students will learn a variety of problem solving strategies applicable in elementary and middle school. The applications will cover many different areas of mathematics.
Other Requirements: PREREQ: MATH 149
Class Schedule
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Sections may be full or not open for registration. Please use WINS if you wish to register for a course. | 677.169 | 1 |
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PowerPoint Slideshow about 'Differentiation' - finneaCalculus was developed in the 17th century by Sir Issac Newton and Gottfried Leibniz who disagreed fiercely over who originated it.
Calculus provides a tool for solving problems involving motion. For example, methods obtained from calculus are used to study the orbits of planets, to calculate the flight path of a rocket and to predict the path of a particle through a magnetic field.
In fact calculus can be used to study any situation where a rate of increase or decrease is involved. (rate of change)
If we remember from standard grade distance time graphs. The speed was obtained from the gradient of the graph. | 677.169 | 1 |
Category Archives: 15. Formulae
Overview This builds on the Basic Algebra section. It covers; how to write your own algebraic formulae, how to evaluate expressions and formula by substituting values for variables and how to change the subject of a formula. Formulae Questions 1. … Continue reading → | 677.169 | 1 |
Maths - Course Description
Maths is one of the core subjects items that are covered are: Analysing and displaying data, Number skills, Equations, functions and formulae, Fractions
Spring Term, Angles and shapes, Decimals, Equations Summer Term, Multiplicative reasoning, Perimeter, area and volume, Sequences and graphs.
Creativity is just connecting things. When you ask creative people how they did something, they feel a little guilty because they didn't really do it, they just saw something. It seemed obvious to them after a while. | 677.169 | 1 |
Algebra 2 Curriculum
This bundle contains warm-ups, notes, homework assignments, quizzes, unit tests, a midterm test, end of year review materials, and a final exam for Algebra 2. This bundle does not contain activities. is designed to review solving systems of equations by graphing (or writing systems of equation by the graph). This activity also gets students up and about.
Place the 10 cards on the wall around your room. Students pick any card to begin with. They should graph the function on the bott combines the skill of writing algebraic expressions with the classic board game CLUE. Students will use the clues they gather from correctly writing algebraic expressions to solve the mystery of Who Killed Mr. X. Pression.
This kit offers two ways to use this activity.
Use as a Sca
This activity will get your students out of their seats and working cooperatively in small groups. They will use their knowledge of solving systems of equations in order to solve problems. This maze requires students to know how to solve using substitution and elimination.
This is one of my favoriSimplifying Exponent Expression Task Cards
Included in this product is a set of 30 task cards on simplifying exponent expressions, a student recording sheet and an answer key. The task cards are available with a colored border or with just a black and white border. All of the task ca
These task cards meet common core state standard CCSS.Math.Content.8.F.B.4 Determine the rate of change from two (x, y) values, including reading these from a table or from a graph.
There are 36 total slope cards; 30 cards are for finding the slope given two points (using the slope formula) and 6 a
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Writing linear equations can be tricky. Students need to be able to manipulate various forms of information to create the equation of the line.
Included in this activity are 36 task cards featuring:
(8) slope and a point,
(8) function tables,
(6) two points,
(6) x- & y-i
Bring to life the traditional practice class or homework assignment with some global competency and diversity! While your students practice algebra, they can learn about important and inspiration international figures!
Person Puzzles are designed to highlight individuals with diverse backgrounds who
Students match 20 cards of graphs, scatter plots, sets of coordinates, mappings and tables to cutouts of each relation's correct domain and range in this activity. Once all cards are matched, students can also sort their cards into "function" or "not a function" and record their answers on the inclu
Students solve systems of equations word problems on 10 task cards in this activity. All systems can be solved with elimination (one or both equations may need multiplication first).
Also included is one bonus card that asks students to find the solution to a system of three equations in 3 variabl
Multi Step Equation Word Problem Task Cards:
Included in this set are 28 multi-step equations word problem task cards, a student answer sheet, and an answer key. These cards are all short answer questions. Students are given a multi-step equation word problem and are asked to find
Absolute Value Inequalities Coloring Activity (includes Interval Notation)
This is a fun way for students to practice solving absolute value inequalities and representing solutions in interval notation. There are 10 inequalities total, including many in which the absolute value must be isolated. S
Ahoy! Thar's different ways to teach Systems 'o Equations 'n Inequalities. Take a voyage with me! (NOTE: This project is not written in English it is written in PIRATE).
Students work develop their graphing linear equations and inequalities skills through three different pirate adventures. There is
This activity is designed for students to complete after learning about solving systems of linear equations. The activity gives students a real-life perspective on systems of equations by using the motivation that only candy can provide. Students will receive a paper bag (stapled closed) filled wit
Students match linear and nonlinear function graphs to short motion stories that describe the graphs.
Each story describes a walk between home and school at a mixture of constant and nonconstant rates. Students will need to think about the shapes of graphs when speed is increasing, decreasing or
This activity will get your students out of their seats and working cooperatively in small groups. They will use their knowledge of graphing systems of equations & inequalities in order to solve problems.
This is one of my favorite activities to do as a review and as a way to deepen students' u
This activity is designed to help students with graphing systems of linear inequalities (or for writing the inequalities from graphs). This activity also gets students up and about.
Purpose of activity is to allow students to practice while giving them the ability to self-check, thus ensuringAngle of Elevation and Depression Math Lib
In this activity, students will generate pieces to a story as they move throughout ten stations. They will practice solving angle of elevation and depression word problems. Their answer to each station will give them a piece of their story (who, doing what,
Students must work their way through this "choose your own adventure" zombie apocalypse flip book by correctly solving 14 polynomials. Correct answers will take students through the story while incorrect answers will take them to death by zombies.
Most of the polynomials require a GCF to be factore
This maze is designed to give students practice simplifying radical expressions without using a plain old worksheet. When I give my students activities like this, I usually ask that they turn in a sheet of paper that has all of their work on it so that I can see that they actually took the time to fi
This activity combines the skill of solving quadratics by factoring when a=1 with the classic board game CLUE. Students will use the clues they gather from correctly solving quadratic equations by factoring to solve the mystery of Who Killed Mr. Quad.
This kit offers two ways to use this activity.
Students sort relations {tables, graphs, equations, mappings, coordinate pairs} into the correct category of "function" or "not a function". This download also includes a "Why?" sheet that asks students to explain their reasoning behind their decisions. Alternate answer sheets with QR codes are inc self-checking maze has 23 systems of equations designed to be solved by substitution.
Not all boxes are used in this maze to prevent students from just guessing the correct route. In order to complete this maze students will have to solve 11 systems.
Answer key is included.
Please view the pThese two sorting activities will help your students practice identifying end behaviors for polynomial functions. They will classify each function according to its end behavior using cards with a mix of equations, explanations, and graphs.
** This resource is also available as part of a DISCOUNTED
Classroom Tested - Student Approved { Just Print and Go }
TpT purchaser feedback:
"I really like the idea of using this type of manipulative with older kids. I think it adds a level of engagement that can be elusive with pen and paper. Nice job!"
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Solving Two Step Equations Partner Activity:
This is an activity that can be used to practice solving 2-step equations. It is meant to be completed in partner groups. Each page has two lists of problems. These lists are labeled "partner 1" and "partner 2." Each equation is differen
Students work with tables, graphs, coordinate pairs, equations, scatter plots, and word problems to identify which relations are functions, which between 2 functions shows a greater rate of change, and if a function is linear or nonlinear.
You may choose to have your students simply identify the fLaw of Sines and Law of Cosines - Math Lib Bundle
Math Lib activities are a class favorite! With the following two math lib activities, students will practice using the Law of Sines and Law of Cosines to find missing side and angle measures in triangles. Students must be aware of the ambig
This is a set of 30 cards that will strengthen students' skills in matching a system of linear equations to a word problem. There are 15 systems and 15 matching word problems.
Due to the fact that these systems all use the same numbers they are not intended to be solved. Attempting to solv
***** Click here for a video demonstration of Algebra Clue. *****
This activity is ideal for end-of-the-year exam review for the equations and inequalities units. It will get your students out of their seats and working cooperatively in small gr
This activity consists of 18 easy trinomials (no coefficient on x^2) that student must factor using the 36 binomials. Students cut out the 36 binomials and glue them onto the corresponding trinomial. This activity is self checking because they only have a certain set of binomials that could be the f
Graphing Linear Inequalities Task Cards:
Included in this set are 24 task cards on graphing linear inequalities, a student answer sheet, and an answer key. These cards are all questions that require students to graph. There are 2 types of problems. The first 12 cards ask students t
Systems of Equations CARD MATCH & AROUND THE ROOM Activities with fun Superhero vs. Villains Theme
This is a resource for reviewing Solving Systems of Equations using all 3 methods: Graphing, Elimination and Substitution. 7 problems are set up
This activity is designed to help students practice compound interest type problems. All problems use the compound interest formula. Student find the future value of items, IRAs and Saving Accounts assume a constant rate is increase. Problems compound annually, semiannually, quarterly, and monthl
One Step Inequalities Scavenger Hunt:
***This game includes: 7 pages of solving one step inequality questions, student recording sheet, and an answer key.***
To play:
1) Cut each question page in half to separate the question boxes. You will have 14 questions. Each question asksParent Functions and Transformations Reference Book
This reference book was created to use as a review of transformations and the following function families: linear, absolute value, quadratic, cubic, square root, cube root, exponential, logarithmic, and reciprocal
Students will graph the both th
Complex Numbers Scavenger Hunt (All Operations)
This scavenger hunt activity consists of 24 problems in which students practice simplifying, adding, subtracting, multiplying, and dividing complex numbers. For division, students must be able to rationalize the denominator, which includes multiplying
In this activity students work in pairs and sort cards into two groups; functions and non-functions. Students must discuss and decide if the relation on each card represents a function or not.
Relations are represented as graphs, tables, sets of ordered pairs and mapping diagrams.
A graphic orga | 677.169 | 1 |
Systems of Equations (Bundled): Real World Scenarios and More!
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this file type before downloading and/or purchasing.
16 MB|N/A
Product Description
This bundle contains 6-lessons covering all the major concepts of systems of linear equations. The topics are covered in a fun, and engaging way! If you buy all of these products separately, you will spend $16.00. Bundled together you will spend only $10.00. This is a 38% savings!! The bundle contains all of the following:
Lesson 6: Systems of Equations Special Cases: Trees and More Trees (11 pages)
NOTE: When writing my lessons I try to keep the spirit of the common core in mind. In other words, all of my lessons contain real-world scenarios that are rigorous but not impossible. At the same time, students are guided through the lessons with hints to help them be successful. You will notice this in all of my lessons. No matter how my lessons are designed, I always encourage students to work together and collaborate. I hope this bundle engages and helps your students to be successful. Thanks, and Let's Algebra!
REMEMBER: If you would like to see a preview of each lesson, simply view the preview for each individual product in my store. | 677.169 | 1 |
Keep Those Algebra Skills Sharp Part 2 Spring 2014 (Editable)
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69 KB|2 pages
Product Description
This worksheet is designed as a homework assignment on the night of an assessment, so students keep their understanding up-to-date. It includes the topics of analyzing a graph, and explaining in precise terms the meaning of the slope and y-intercept within the context of the problem, determining unknown terms of an arithmetic and geometric sequence, solving quadratic equations by graphing, and statistics (mean, median, mode, variance, and standard deviation). | 677.169 | 1 |
ISBN-10: 0471433314
ISBN-13: 9780471433316 fundamental concepts and techniques of real analysis for readers in all of these areas. It helps one develop the ability to think deductively, analyze mathematical situations and extend ideas to a new context. Like the first three editions, this edition maintains the same spirit and user-friendly approach with addition examples and expansion on Logical Operations and Set Theory. There is also content revision in the following areas: introducing point-set topology before discussing continuity, including a more thorough discussion of limsup and limimf, covering series directly following sequences, adding coverage of Lebesgue Integral and the construction of the reals, and drawing reader attention to possible applications wherever possible | 677.169 | 1 |
Course Title: Apply mathematical techniques to scientific contexts
Part A: Course Overview
Program: C4386 Certificate IV in Tertiary Preparation
Course Title: Apply mathematical techniques to scientific contexts
Portfolio: SEH Portfolio Office
Nominal Hours: 70Assessment may consist of written tests, in class activities, presentations & written reports.
Assessment Tasks
Assessment 1:
Assignment 1 (20%)
Assessment 2:
Quiz 1 (15%)
Assessment 3:
Assignment 2 (20%)
Assessment 4:
Quiz 2 (15%)
Assessment 5:
Exam (30%)
To be deemed competent you must demonstrate an understanding of all aspects required of the competency. Assessment methods have been designed to measure your achievement of each competency in a flexible manner over multiple tasks. | 677.169 | 1 |
For ages: 8-10, 11-13
Date Added: 2016-08-08 The supportive materials for Math Mammoth Grade 6 curriculum include:
- the answer keys for the two student worktexts - the tests - additional cumulative review lessons - a worksheet maker (Internet access required) - bonus software Soft-Pak
NOTE: This product does NOT include the actual student books. It is intended to be used in conjunction with the printed student books, which you need... [click here for more]
For ages: 11-13, High School
Date Added: 2016-08-08 The supportive materials for Math Mammoth Grade 7 curriculum includes:
- the answer keys for the two student worktexts - the tests - additional cumulative review lessons - a worksheet maker (Internet access required) - probability simulations - bonus software Soft-Pak
NOTE: This product does NOT include the actual student books. It is intended to be used in conjunction with the printed... [click here for more] | 677.169 | 1 |
This is quite simply an outstanding resource for students (and their teachers!). Many exercises are provided and the answers are all at the back. OCR have very helpfully provided the document as a Word document.
From the University of Cambridge comes Underground Mathematics which started in 2012 as the Cambridge Mathematics Education Project (CMEP). The site provides a library of rich resources for age 16+ students with the aim of "Enabling all students to explore the connections that underpin mathematics". Underground Mathematics is being developed by the University of Cambridge, funded by a grant from the UK Department for Education. The resources are free for all users; you can read more about the team and their philosophy here. Follow Underground Mathematics on Twitteror Facebook.
Underground Maths have given brilliant resources suggestions clearly mapped to the subject content for the new A Level. This is an Excel spreadsheet; for each content statement, Underground Maths have suggested up to three rich resources and up to three Review questions. Each suggestion is hyperlinked to take you directly to the resource on the Underground Mathematics site. Resources that are particularly good at supporting the overarching theme of Mathematical modelling have been highlighted.
These of course are suggestions. There are so many outstanding resources on Underground Maths. In our department, like many others I am sure we will be exploring the resources and bookmarking our own favourites.
TheA level mathematics working group reportis a very valuable document I do believe we should always look both backwards and forwards to inform our teaching; where have our students been and where are they going?
Educas state: "The objective of this qualification is to assist the understanding of the problem-solving cycle of planning, collecting, processing and discussing in meaningful contexts and to use statistical software to process real data sets. It has been specifically designed to be taught in schools and colleges to equip learners aged 16-19 with a broad range of skills empowering them to successfully negotiate statistical problems in Higher Education or the world of work."
This discovery led me to a treasure trove of resources from Educas for Mathematics at all levels, including GCSE – this definitely needs further exploration! | 677.169 | 1 |
BUYER OFFERS: We are a retail store with set pricing and unfortunately we can't fulfil any requests to sell items for less than the listed price.
Description: New Century Maths for the Australian Curriculum Years 7 ' 10 is designed to meet the requirements of the new Australian Curriculum. Written by the original NSW author team, these new editions retain all of the successful features of New Century Maths: Chapter outline, Wordbank, Investigation, Technology, Mental skills, Language of maths, Topic overview with mind map and Glossary. supplement to the printed text or as a standalone option for schools seeking a digital-only resource solution. *Complimentary access to NelsonNet is available to teachers who use the accompanying student book as a core resource in their classroom. Contact your education consultant for access codes and conditions | 677.169 | 1 |
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