Split (1)

train · 167k rows

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https://math.stackexchange.com/questions/3843410/understanding-the-definition-of-infinite-cartesian-product	1,638,309,982,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964359073.63/warc/CC-MAIN-20211130201935-20211130231935-00507.warc.gz	468,477,926	33,895	# Understanding the definition of infinite Cartesian product This is the first time I have come across the following definition for the infinite Cartesian product. I somewhat understand it, however, below I have pointed out where I am getting confused. Definition for the infinite Cartesian product: $$\prod_{i \in \mathbb{N}}\mathbb{X}_i = \{f: \operatorname{dom}(f) = \mathbb{N} \wedge \forall i, f(i) \in \mathbb{X}_i\}.$$ Here is where I am getting confused. Define the following: $$\mathbb{R}^2 = \mathbb{R} \times \mathbb{R} = \{(x,y): x \in \mathbb{R} \wedge y \in \mathbb{R}\} \quad (i)$$ $$\mathbb{R}^2 = \mathbb{R} \times \mathbb{R} = \prod_{i=1}^2\mathbb{R}_i = \{f: \operatorname{dom}(f) = \{1,2\} \wedge f(1) \in \mathbb{R}, f(2) \in \mathbb{R}\} \quad (ii)$$ Here is where I am getting confused. Lets say you want to write "express" the point where $$x = 5$$ and $$y = \pi$$, then using $$(i)$$ you would simply write $$(5,\pi)$$. How would you express the same for $$(ii)$$? In other words, I am just not sure how you would represent the same point using $$(ii)$$. I know that a function is, itself, a set. • Sep 28 '20 at 10:35 • Each element of the product is an $\infty$-tuple $(x_1, x_2, \ldots)$ where the $i$-th element belongs to set $X_i$. Sep 28 '20 at 10:36 • The case is similar to $\mathbb R \times \mathbb R$ where the elements a pairs $(r_1,r_2)$ where $r_i \in \mathbb R$. Sep 28 '20 at 10:37 • The sets $X^2$ and $X\times X$ are not identical but there's a natural bijection between them: a function $f:\{0,1\}\to X$ corresponds to the pair $(f(0),\,f(1))$. Sep 28 '20 at 10:44 • The "point" in (ii) is a function $f : \{ 1,2 \} \to \mathbb R$, where $f(1)=5$ and $f(2)= \pi$. Sep 28 '20 at 11:21 The only thing you are confused in here is notations. $$f$$ in this definition is a coordinate function: it maps coordinate index to the corresponding value, and so each function $$f$$ represents a single point in a product set. For example, point $$(x_1, x_2) \in \mathbb{R}^2$$ is exactly the function $$f : \{1,2\} \to \mathbb{R}$$ such that $$f(1) = x_1$$ and $$f(2) = x_2$$ (point $$(5,\pi)$$ may be represented as a such function $$f$$ with $$f(1) = 5$$ and $$f(2) = \pi$$). One of the reasons for using such notation is that it could be easily generalised to any number of coordinates (including infinity) – all you need is just expand the domain of $$f$$. Take a look to the definition of infinite Cartesian product written using coordinates (like in $$(i)$$): $$\prod_{i \in \mathbb{N}}\mathbb{X}_i = \{(x_1, x_2, \dots) \wedge \forall i, x_i \in \mathbb{X}_i\}.$$ It contains expression $$(x_1, x_2, \dots)$$ which is, in a sense, informal: it uses notation for points with finite count of coordinates (where coordinates are represented directly), for points with infinite number of coordinates (which are, actually, the infinite sequences). Using notation with coordinate functions gives you much more formal and explicit way to describe such objects; take a look at the formal definition of sequence, which uses the same construction. • Thank you! So, in essence, points in $\mathbb{R}^2$ can be represented as $(f(1), f(2))$ where $f(1) \in \mathbb{R}$ and $f(2) \in \mathbb{R}$, which is simply $(ii)$, correct? – user820163 Sep 28 '20 at 11:21 • @Dalton yes, sure, think about it as a just different notation for $(x_1, x_2)$ Sep 28 '20 at 11:22 • And, is $(ii)$ a set of a single function or a set of functions? – user820163 Sep 28 '20 at 11:28 • @Dalton $(ii)$ is a set of functions, where each function represents point $\mathbb{R}^2$ Sep 28 '20 at 11:31	1,195	3,612	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.21875	4	CC-MAIN-2021-49	latest	en	0.758755
https://www.allinterview.com/company/467-2/cmc/interview-questions/400/general-aptitude.html	1,547,691,272,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547583658681.7/warc/CC-MAIN-20190117020806-20190117042806-00403.warc.gz	728,471,379	10,753	Company Name Starts with ... # A B C D E F G H I J K L M N O P Q R S T U V W X Y Z • CMC aptitute test questions (29) • CMC interview questions (96) • CMC placement papers (4) • CMC technical test questions (2) CMC General Aptitude Interview Questions Each side of a rectangle is increased by 100% . How much the percentage of area will be increased 21099 Perimeter of the back wheel = 9 feet, front wheel = 7 feet on a certain distance the front wheel gets 10 revolutions more than back wheel . what is the distance? 5621 Perimeter of front wheel =30, back wheel = 20. If front wheel revolves 240 times. Howm many revolutions will the back wheel take? 7585 City A population is 68000, decreasing at a rate of 80 per year City B having population 42000 increasing at a rate of 120 per Year. In how many years both the cities will have same population 18822 Two cars, 15 km apart one is turning at a speed of 50kmph other at 40kmph . How will it take to two cars meet. 27925 A person wants to buy 3 paise and 5 paise stamps costing exactly one rupee. If he buys which of the following number of stamps. he wont able to buy 3 paise stamps 8562 There are 12 boys and 15 girls, How many different dancing groups can be formed. 14257 Which of the following fractions is less than 1/3 (1) 22/62 (2) 15/46 20070 Two circles , one circle is inscribed and another circle is outscribed over a square. What is the ratio of area of inner to outer circle. 7849 Perimeter of the back wheel = 9 feet, front wheel = 7 feet on a certain distance, the front wheel gets 10 revolutions more than the back wheel .What is the distance? 18078 There are 200 questions on a 3 hr examination.Among these questions are 50 mathematics problems.It is suggested that twice as much time be spent on each maths problem as for each other question.How many minutes should be spent on mathematics problems, plz give detail, how to solve? 19667 rajesh purchase 3 types of candies with 1 doller costin:\$0.05,\$0.02,\$0.01 he purchase 10 candies of \$.01 find quantity he buys for different candies 5362 Calculate how much oxygen is dissolved in 100 milliliters of blood if the blood is exposed to a partial pressure of oxygen of 120 torr? 1196 What is ISM / UNII bands? 602 ipls send me the type of interview questions asked in andhra bank clerical interview 1307 1) There is a singing competition for children going to be conducted at a local club. Parents have been asked to arrive at least an hour before and register their childrenâ€™s names with the Program Manager. Whenever a participant registers, the Program Manager has to position the name of the person in a list in alphabet order. Write a program to help the Program Manager do this by placing the name in the right place each time the Program Manger enters a name. The Logic should be written in Data Structures? 934 What control mechanisms might be most appropriate to ensure that action plans match targeted needs? 794 When will you multiply and devide with derating factors to the actual value while sizing the cable? 734 why genarator genarating voltage 11 KV? 458 How would you relate your key competencies to this position? 11186 How to find the max load of the engine? 842 problems and consequences of bpo ? 7653 5S training documents 1829 AFTER OVERHAULING KTA19G3 AND JUST ABOUT TO START ,BUT WE DIDNT FIX THE OIL SUMP AND TEST FOR -WATER LEAK FROM LINER O RING SIDE,THEN WE FILLED WATER AND FOUND WATER LEAK FROM INSIDE OF ENGINE -CAM FOLLOWER SIDE PLEASE GIVE ADVICE TO RECTIFY THE SAME 620 in a every intervew asking one common question why you are living current employer, which is the best answer 1758 what are the gbx,gdb and other tools in linux for core dump,and how to configure stdout,stderr..what is the difference between stdout ,stderr and coredump. how to create core dump and view that one? please can anybody explain me? 2975 how to turbine speed measurement? 432 CMC General Aptitude Interview Questions • C (4) • OOPS (1) • MFC (2) • VC++ AllOther (1) • PHP (2) • Programming Languages AllOther (1) • ASP (1) • JavaScript (1) • TCP IP (1) • Networking AllOther (1) • C Sharp (2) • ASP.NET (2) • VB.NET (2) • WCF (1) • OCI (1) • Forms Reports (1) • SQL Server (1) • MS Access (1) • Postgre (1) • SQL PLSQL (1) • Databases AllOther (1) • Linux Commands (1) • Data Structures (1) • Operating Systems AllOther (1) • Manual Testing (10) • Test Cases (1) • Testing AllOther (1) • Servers AllOther (1) • Core Java (9) • Struts (1) • Weblogic (1) • SAS (1) • Data Warehouse General (2) • HR Questions (11) • Clear Case (1) • SAP ABAP (1) • SAP SD (2) • SAP FI CO (1) • SAP B1 (1) • Embedded Systems AllOther (1) • Client Server Architecture (1) • Civil Engineering (1) • Electrical Engineering (1) • Electronics Communications (2) • Instrumentation (1) • Engineering AllOther (1) • Marketing Sales (1) • Taxation (1) • Accounting AllOther (1) • Call Centre AllOther (2) • General Knowledge_Current Affairs (1) • Everything Else AllOther (2) • General Aptitude (27) • Puzzles (1) • Placement Papers (4) • CCNA (1) • Oracle Certifications (1) • C C++ Errors (1)	1,400	5,193	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2019-04	longest	en	0.905017
https://www.chiefdelphi.com/t/autonomously-following-cubic-polynomial/387461/4	1,643,282,494,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320305260.61/warc/CC-MAIN-20220127103059-20220127133059-00161.warc.gz	739,043,827	6,671	# Autonomously Following Cubic Polynomial I have successfully generated a polynomial based off of two slopes and two points with thanks to @Ether and this thread. However, this does not get me as far as I’d like it to. For this to be complete, I need to steer along this path. There are several ways I could do this, and most involve taking the slope of a point along the path, and then adjusting my robot’s heading. In order to get this slope, I need an X value however. Or do I? Is there any better way to follow this polynomial? I really like the polynomial and would appreciate it if anyone had ideas. I am currently looking at my navX’s x-displacement, but I hear it is inaccurate, so I’d like to avoid it. Along with that, perhaps some calculus would allow me to find the slope with an arc length based off of my encoders? And lastly, how accurate is the Pigeon IMU in terms of displacement? Maybe I should look at replacing my navX with this. Thank you! Nowadays, there’s support built into WPILib for trajectory generation and tracking. We support clamped cubic splines and quintic splines as well as enforcement of arbitrary user-defined trajectory constraints. I suggest reading the following resources. Introduces trajectory generation: Talks about how our trajectory generation works internally (optional, but in case you’re curious): https://pietroglyph.github.io/trajectory-presentation/#/ Walks you through implementing trajectory generation/following on an actual robot: https://docs.wpilib.org/en/stable/docs/software/examples-tutorials/trajectory-tutorial/index.html 6 Likes It is not a good idea to use your IMU to figure out your displacement along the field. You should use a combination of your encoders and gyro in order to do this. Conveniently, WPILib has several odometry classes that utilize your encoder and gyro data to calculate your robot’s absolute position on the field. With the information about where you currently are on the field, and where you want to be, you can accurately follow a trajectory. With regards to pure angular displacement information from the NavX and the Pigeon IMU, I believe they use the same internal sensor so your values should not be too different (if at all). We haven’t had issues with either. 1 Like We used a similar idea to what I think you’re getting towards in 2018. The team has continued using it and you can find the code here. There is a whitepaper available here. Effectively what you will want to do is calculate the angle between your current heading and the slope of the tangent line at your current point in the polynomial. If your polynomial is defined on the domain [0, 1], and you have calculated the arc length of the polynomial in advance (we used this JS library), then you can find your desired tangent line by dividing your current distance traveled (average of your 2 encoders) by the total length of the polynomial. You’ll then use a simple P controller to constantly “chase” the next desired angle. This approach to auto is okay, and I think it’s one of the simpler ways to get it done; it’s relatively consistent, however it isn’t perfect and you’ll have to spend some time getting the autos perfect. However, if you are looking for long-term auto success, I recommend checking out the solutions currently provided by WPILib as mentioned by Tyler and Prateek above. They will be more accurate and better supported by the community in the long term. 1 Like The Pigeon and original NavX both use the same MPU-9250 IMU and will perform very similarly.	765	3,559	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.84375	3	CC-MAIN-2022-05	latest	en	0.948751
https://www.physicsforums.com/threads/photometry-determining-a-binary-star-system.545609/	1,660,414,395,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882571982.99/warc/CC-MAIN-20220813172349-20220813202349-00683.warc.gz	821,272,666	16,423	# Photometry: Determining a binary star system. Lavabug ## Homework Statement By performing CCD photometry on a pair of nearby stars A and B we obtain their relative magnitudes in the V filter and their colors: Star A: mV = 8.70 , (B − V )= 1.30 Star B: mV = 11.90 , (B − V )= 1.81 Star A is known to be a of a main sequence K0V type, while no other information on Star B is available. Argue if this consists of a visual binary system or if B is a background star. ## The Attempt at a Solution $$m_{\lambda}(N stars) = -2.5log(\sum10^{.0.4m_{\lambda}_i})$$ Not sure what to do. Using the expression above, I found the apparent magnitude for the system as a whole and it came out brighter than A (as expected), but by very little: 8.645. What do I need to look for to determine if its a binary or not? Staff Emeritus Gold Member It seems to me that, based on the colour of star B, you can figure out it's spectral type. At the very least, you know that it is redder than K0. Now, if it's a binary, then you also know that both stars are at the same distance from Earth. Therefore the difference in their apparent magnitudes is the same as their difference in absolute msgnitudes. So, under this assumption, you know how much dimmer (intrinsically) star B is than star A. Ask yourself whether it makes sense that a star on the main sequence that is 0.5 mag redder than a K0 star would also be that much dimmer. If it does make sense, then this can be a binary. If it doesn't, then the only explanation must be that star B is in fact much farther away. Lavabug It seems to me that, based on the colour of star B, you can figure out it's spectral type. At the very least, you know that it is redder than K0. Now, if it's a binary, then you also know that both stars are at the same distance from Earth. Therefore the difference in their apparent magnitudes is the same as their difference in absolute msgnitudes. So, under this assumption, you know how much dimmer (intrinsically) star B is than star A. Ask yourself whether it makes sense that a star on the main sequence that is 0.5 mag redder than a K0 star would also be that much dimmer. If it does make sense, then this can be a binary. If it doesn't, then the only explanation must be that star B is in fact much farther away. Thanks for the quick reply. How do I know that star B is redder than K0? Sorry if this sounds elementary, I'm just getting acquainted with calculating color indices, magnitudes etc. I'm attaching the table provided for problem-solving, it claims different values for a K0V star(I think everything is shifted by one unit), and according to the table, star B is a G0V type (comparing (B-V) indices), which would imply it is brighter than K0V on the absolute scale. (am I doing this right?) But on the apparent mag. scale star A is brighter, hence it is much closer than star B, am I on the right track? How far apart (or different in apparent magnitude) do they need to be in order to discard the possibility of it being a binary system? #### Attachments • table.pdf 728.7 KB · Views: 234 Staff Emeritus Gold Member Thanks for the quick reply. How do I know that star B is redder than K0? Sorry if this sounds elementary, I'm just getting acquainted with calculating color indices, magnitudes etc. I know star B is redder because its B-V colour index is larger. Here is how to interpret the colour index: B is the star's apparent magnitude in the B (blue) photometric band, which, if I recall correctly, uses filters centred on ~400 nm wavelength. V is the star's apparent magnitude in the V (visual) photometric band, which, if I recall correctly, uses filters centred on ~550 nm wavelength. If B-V is postiive, it means that B > V. Recall, that larger apparent magnitude = dimmer. Therefore, having B-V > 0 means that the star's observed B-band brightness is less than the its V-band brightness. The larger the colour index value is, the less emission is being received in the B-band relative to the V-band, and the (hence we infer) the redder the emission spectrum of the object must be. Since star B's colour index of 1.81 is larger than star A's colour index of 1.30, we conclude that star B has a redder spectrum (more emission at longer wavelengths, less at shorter wavelengths). I'm attaching the table provided for problem-solving, it claims different values for a K0V star(I think everything is shifted by one unit), and according to the table, star B is a G0V type (comparing (B-V) indices), which would imply it is brighter than K0V on the absolute scale. (am I doing this right?) No, I don't think you're doing this right. Even if everything were shifted by 1 unit, I have no idea how you'd get a spectral type G0V for star B, that just doesn't make any sense whatsoever. In any case, I don't think that everything is shifted by 1 unit. If it were, then the value of K0's B-V index in the table would be 0.30. It is not. The reason why the observed K0 colour index stated in the problem is greater than the theoretical one in the table is probably because of interstellar reddening. However, if both stars are in a binary, then they're at the same distance and hence you can assume they're both reddened by the same amount. So, although both of their colour indices will have changed, the difference between their colour indices will be the same as it was without the reddening. From what I just said in the previous sentence, since star B's colour index is about 0.5 mag larger than a K0 star, the table would seem to indicate that it is of type K7. So, under the assumptions we've made, star B must be of type K7. Now, compare the absolute magnitudes of K0 and K7 main sequence stars. According to the table, a K7 is only 2.2 mag dimmer than a K0. Yet, the observations show that star B is a whole 3.2 mag dimmer than star A. It doesn't fit where it should on the main sequence for a K7 star. The only thing we can conclude is that our initial assumption about the stars being at the same distance must have been wrong.	1,492	6,034	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2022-33	latest	en	0.963213
http://www.jiskha.com/display.cgi?id=1305076018	1,496,031,811,000,000,000	text/html	crawl-data/CC-MAIN-2017-22/segments/1495463612013.52/warc/CC-MAIN-20170529034049-20170529054049-00391.warc.gz	692,473,608	3,734	# Physics posted by on . What will be the final temperature if 1.9 x 10^2 kJ of heat is added to 0.5 kg of ice at 0 deg. C? • Physics - , It takes 500g335 J/gC = 167,500 J to melt all the ice. If you add 190,000 J, you melt all the ice and have 22,500 J left to heat the water. Use the relation 22,500 J = M C (delta T) to determine the temperature rise, delta T. C = 4.18 J/gC M = 500 g	136	397	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.75	4	CC-MAIN-2017-22	latest	en	0.826941
https://www.scribd.com/document/278126154/Trigonometry	1,556,194,103,000,000,000	text/html	crawl-data/CC-MAIN-2019-18/segments/1555578721441.77/warc/CC-MAIN-20190425114058-20190425140058-00248.warc.gz	810,648,508	165,469	You are on page 1of 180 TRIGONOMETRY MICHAEL CORRAL Trigonometry Michael Corral Schoolcraft College The goal of this book is a bit different. It makes much of the material appear unmotivated. So this book presents material in a very different order than most books today. too. which is more intuitive for students to grasp. physics). I think that math instructors have a duty to prepare students for that. There are a few code samples in Chapter 6. as long as it produces accurate graphs. The prerequisites are high school algebra and geometry. Virtually no students will ever in their “everyday life” ﬁgure out the height of a tree with a protractor or determine the angular speed of a Ferris wheel. written in the Java and Python programming languages.g. though it could also be used in high schools. This text probably has a more geometric feel to it than most current trigonometry texts. general (oblique) triangles are presented. in fact. In Chapter 5 students are asked to use the free open-source software Gnuplot to graph some functions. with hints when I felt those were needed. It is suitable for a one-semester course at the college level. after starting with the right triangle deﬁnitions and some applications. Appendix B contains a brief tutorial on Gnuplot. any program can be used for those exercises. Students are far more likely to need trigonometry in other courses (e. However. In my experience. This book starts with the “old-fashioned” right triangle approach to the trigonometric functions. There are a few exercises that require the student to write his or her own computer program to solve some numerical computation problems. engineering. expanded with some exercises. For example. Instead of taking the (doomed) approach that students have to be shown that trigonometry is “relevant to their everyday lives” (which inevitably comes off as artiﬁcial). iii .Preface This book covers elementary trigonometry. Answers and hints to many of the odd-numbered and some of the even-numbered exercises are provided in Appendix A. I have tried to include some more challenging problems. this book has a different mindset: preparing students to use trigonometry as it is used in other courses. That was. one of the reasons I wanted to write this book. That seems like a more natural progression of topics. This book basically consists of my lecture notes from teaching trigonometry at Schoolcraft College over several years. hopefully sufﬁciently clear so that the reader can ﬁgure out what is being done even without knowing those languages. An average student should be able to do most of the exercises. instead of leaving general triangles until the end as is usually the case. presenting the deﬁnitions of the trigonometric functions and then immediately jumping into proving identities is too much of a detour from geometry to analysis for most students. I think that approaching the subject with too much of an analytic emphasis is a bit confusing to students. There are exercises at the end of each section. . . . .2 1. .2 4. . . . . . . . 65 71 78 82 87 87 90 95 100 103 Graphing the Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Law of Cosines . . . . . . 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Law of Tangents . . . . . . Area of a Sector . . . . . . . .5 3 Angles . . . . . . . . . . . . . . . . . . . . . . . . . . Circular Motion: Linear and Angular Speed . . Identities 3. 129 Numerical Methods in Trigonometry . . . . . .4 2.4 5 . . . . . . . . Basic Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . . . .3 4. . . . . . . . . . . . . . . .1 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 2. . . . . Rotations and Reﬂections of Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Circumscribed and Inscribed Circles . . . . . . . . Trigonometric Functions of an Acute Angle Applications and Solving Right Triangles . . . . . . . . . . . . . .4 4 . . . . . . . 103 Properties of Graphs of Trigonometric Functions . . . . . . Radian Measure 4. . . . . . . . . . .1 6. .3 2. . . . . . . . . .1 5. . . . . . . . . . . . . . . . Double-Angle and Half-Angle Formulas Other Identities . . . . . . . . . . . . . . . . . . Radians and Degrees . . . . . .3 6 . . Trigonometric Functions of Any Angle . . 14 . . . . . . . . . . . . . 109 Inverse Trigonometric Functions . . . . . . .4 1.3 3. . . .1 3.2 38 44 51 54 59 65 . . . . . .2 5. . . . . . . . . . . . . .5 2 The Law of Sines . . . 120 Additional Topics 6. . . . . . . . . . . . . . . . . . .1 1. .Contents Preface iii 1 1 Right Triangle Trigonometry 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 2. . . . . . . . .2 3.3 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sum and Difference Formulas . . . . . . . . 133 v . . . . . . . . . . . . . . . . . . . . 32 38 . . . . . . . . . 129 Solving Trigonometric Equations . . . . . . . . . . 24 . . . General Triangles 2. . . . . . . . . . . . . . . . The Area of a Triangle . . . Graphing and Inverse Functions 5. . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Polar Coordinates . . . . . . . . . . . . . .3 6. . . . . . . . . .4 Complex Numbers . . . . . . . . 146 Appendix A: Answers and Hints to Selected Exercises 152 Appendix B: Graphing with Gnuplot 155 GNU Free Documentation License 160 History 168 Index 169 . . . . . . . . . . . . . . . . . . . . . . . . .vi C ONTENTS 6. 1 . C.1.1 Angles Recall the following deﬁnitions from elementary geometry: (a) An angle is acute if it is between 0◦ and 90◦ . Before discussing those functions. meaning “measure”. such as Hipparchus and metro (μτρω and Ptolemy. we will review some basic terminology about angles. 2 Later in the text we will discuss negative angles and angles larger than 360◦ . the Egyptian scribe Ahmes recorded some rudimentary trigonometric calculations (concerning ratios of sides of pyramids) in the famous Rhind Papyrus sometime around 1650 B. known as the trigonometric functions. D. meaning “triangle”. 200. . The word “trigonometry” is derived from the Greek words trigono (τρ´ιγωνo). C.1 Right Triangle Trigonometry Trigonometry is the study of the relations between the sides and angles of triangles. used trigonometry in their study of astronomy between roughly 150 B.A .2 The following deﬁnitions will be used throughout the text: 1 Ahmes claimed that he copied the papyrus from a work that may date as far back as 3000 B. (a) acute angle (b) right angle Figure 1. For now we will only consider such angles. 1. ´ ). C.1 Trigonometry is distinguished from elementary geometry in part by its extensive use of certain functions of angles. (d) An angle is a straight angle if it equals 180◦ . For example.1 (c) obtuse angle (d) straight angle Types of angles In elementary geometry. (c) An angle is obtuse if it is between 90◦ and 180◦ . Though the ancient Greeks. angles are always considered to be positive and not larger than 360◦ . (b) An angle is a right angle if it equals 90◦ . its history is much older. we will sometimes use just a capital letter by itself (e. A. ∠ B ≤ 90◦ then ∠ A and ∠ B are complementary if ∠ A + ∠ B = 90◦ . if 0◦ ≤ ∠ A .1 (a) Two acute angles are complementary if their sum equals 90◦ .1. (b) Two angles between 0◦ and 180◦ are supplementary if their sum equals 180◦ . in a right triangle one of the angles is 90◦ and the other two angles are acute angles whose sum is 90◦ (i. It is also common to use letters (either uppercase or lowercase) from the Greek alphabet. shown in the table below. the other two angles are complementary angles).2 Types of pairs of angles Instead of using the angle notation ∠ A to denote an angle. In other words. ∠ B ≤ 180◦ then ∠ A and ∠ B are supplementary if ∠ A + ∠ B = 180◦ .1 Letters A B Γ Δ E Z H Θ α β γ δ ζ η θ The Greek alphabet Name Letters alpha beta gamma delta epsilon zeta eta theta I K Λ M N Ξ O Π ι κ λ μ ν ξ o π Name Letters iota kappa lambda mu nu xi omicron pi P Σ T Υ Φ X Ψ Ω ρ σ τ υ φ χ ψ ω Name rho sigma tau upsilon phi chi psi omega In elementary geometry you learned that the sum of the angles in a triangle equals 180◦ .e. y. to represent angles: Table 1. ∠ B ≤ 360◦ then ∠ A and ∠ B are conjugate if ∠ A +∠ B = 360◦ . (c) Two angles between 0◦ and 360◦ are conjugate (or explementary) if their sum equals 360◦ . B. x. In other words. ∠B ∠A ∠A (a) complementary ∠B ∠A ∠B (b) supplementary (c) conjugate Figure 1.2 Chapter 1 • Right Triangle Trigonometry §1. Thus. if 0◦ ≤ ∠ A . In other words.g. . t). C) or a lowercase variable name (e. if 0◦ ≤ ∠ A .g. Recall that in a right triangle one of the angles is a right angle. and that an isosceles triangle is a triangle with two sides of equal length. the circle’s radius).1 For each triangle below. and C are (distinct) points on a circle such that the line segment AB is a diameter of the circle. A = 35◦ and C = 20◦ . So we see that ∠ ACB = α + β. the angles are in terms of an unknown number α.1. Solution: For triangle ABC. Since AB is a diameter of the circle. which we can use to solve for α and then use that to solve for X . in the ﬁrst triangle above we will simply refer to the angle ∠ BAC as angle A. and Z: α + 3α + α = 180◦ ⇒ 5α = 180◦ ⇒ α = 36◦ ⇒ X = 36◦ . E = 53◦ and F = 90◦ . so 35◦ + B + 20◦ = 180◦ B = 180◦ − 35◦ − 20◦ ⇒ B = 125◦ . And since the angles of ABC must add up to 180◦ . This means that O AC is an isosceles triangle. B. For example.3 B A α β O B (b) Thales’ Theorem: ∠ ACB = 90◦ To prove this. the triangle ABC is a right triangle.3(b). as in Figure 1. For triangle X Y Z. Z = 36◦ Example 1. Y . determine the unknown angle(s): E B Y 3α ◦ 53 20◦ 35◦ α A D C F α X Z Note: We will sometimes refer to the angles of a triangle by their vertex points. we see that 180◦ = α + (α + β) + β = 2 (α + β). ∠ ACB = 90◦ . and we know that A + B + C = 180◦ . Thus. OBC is an isosceles triangle and ∠ OCB = ∠ OBC = β. Let α = ∠ BAC and β = ∠ ABC. let O be the center of the circle and draw the line segment OC. so D + E = 90◦ D = 90◦ − 53◦ ⇒ ⇒ D = 37◦ . Y = 3 × 36◦ = 108◦ . so α + β = 90◦ .3(a)).1 3 Example 1. ⇒ For the right triangle DEF. and we know that the two acute angles D and E are complementary. O A and OC have the same length (namely. and so ∠ OC A = ∠ O AC = α. QED . Likewise.1.1. then the angle ∠ ACB is a right angle (see Figure 1.2 Thales’ Theorem states that if A. In other words.Angles • Section 1. but we do know that X + Y + Z = 180◦ . C C αβ A O (a) Figure 1. Figure 1. then this divides ABC into two smaller triangles CBD and ACD.1. CD forms a right angle with AB).1.5 c−d A D (c) ACD Similar triangles ABC. which are both similar to ABC. Thus. Pythagorean Theorem: The square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of its legs. QED Note: The symbols ⊥ and ∼ denote perpendicularity and similarity. since ABC is similar to CBD. which has length c.1 In a right triangle.1. ACD Recall that triangles are similar if their corresponding angles are equal.4. Since ABC is similar to ACD. by proportionality of corresponding sides we see that AB is to CB (hypotenuses) as BC is to BD (vertical legs) ⇒ c a = a d ⇒ cd = a2 . respectively. .4 Chapter 1 • Right Triangle Trigonometry §1. In the right triangle ABC in Figure 1.1) Let us prove this. if a right triangle has a hypotenuse of length c and legs of lengths a and b. the length of the third side can be determined by using the Pythagorean Theorem: Theorem 1. as in Figure 1. comparing horizontal legs and hypotenuses gives c b = c−d b ⇒ b2 = c2 − cd = c2 − a2 ⇒ a2 + b 2 = c 2 . with lengths a and b. B c C d D c− A B a d b C (a) ABC a b d D C (b) CBD Figure 1. then the Pythagorean Theorem says: a2 + b 2 = c 2 (1. if we draw a line segment from the vertex C to the point D on the hypotenuse such that CD is perpendicular to AB (that is.4 the right angle is C. and the other two sides are called its legs. in c Figure 1. the hypotenuse is the line segment a AB. respectively. and BC and AC are the legs.1.1.5(a) below. For example. Thus. and that similarity implies that corresponding sides are proportional. the side opposite the right angle is called the hyB potenuse. The hypotenuse is always the longest side of a right A b C triangle (see Exercise 11). in the above proof we had CD ⊥ AB and ABC ∼ CBD ∼ ACD. CBD.1.4 By knowing the lengths of two sides of a right triangle. For example. Angles • Section 1. Find A and B if C = 24◦ . 3. the Pythagorean Theorem says that a2 + 42 = 52 a2 = 25 − 16 = 9 ⇒ ⇒ a = 3 . For Exercises 5-8. Find B if A = 15◦ and C = 50◦ . with C being the right angle. Find C if A = 110◦ and B = 31◦ . determine the length of the unknown side: B 5 A Y E a 4 2 D C z 1 1 e F 1 X Z Solution: For triangle ABC. we have 17 h 90◦ 8 2 2 2 h + 8 = 17 ⇒ 2 h = 289 − 64 = 225 ⇒ h = 15 ft . 2. Find A. 7. and B = 2α. and wall form a right triangle with a hypotenuse of length 17 ft (the length of the ladder) and legs with lengths 8 ft and h ft. B. A car goes 24 miles due north then 7 miles due east. the Pythagorean Theorem says that 12 + 12 = z2 ⇒ z2 = 2 ⇒ z = 2 . Find A and B if A = α and B = 2α. Example 1. ground. the Pythagorean Theorem says that e2 + 12 = 22 e2 = 4 − 1 = 3 ⇒ ⇒ e = 3 . Exercises For Exercises 1-4. ﬁnd the numeric value of the indicated angle(s) for the right triangle ABC. 8. and C if A = β and B = C = 4β. ﬁnd the numeric value of the indicated angle(s) for the triangle ABC.4 A 17 ft ladder leaning against a wall has its foot 8 ft from the base of the wall. Find A and B if A = θ and B = 1/θ . What is the straight distance between the car’s starting point and end point? . 6. 5. 9. Then we see that the ladder.3 For each right triangle below. as in the picture on the right. A = α. For triangle DEF. So by the Pythagorean Theorem. Find B if A = 45◦ . Find A and B if A = φ and B = φ2 . 4. For triangle X Y Z. At what height is the top of the ladder touching the wall? Solution: Let h be the height at which the ladder touches the wall.1 5 Example 1. We can assume that the ground makes a right angle with the wall. 1. as in the picture on the right. (d) The triple in part(c) is known as Euclid’s formula for generating Pythagorean triples. then they form what is called a Pythagorean triple.e.12.10) is a Pythagorean triple.5 inches away from O. One end of a rope is attached to the top of a pole 100 ft high. and call the center of that circle O. For example. Does it match the physical measurement of P A? 15. and suppose that the vertex C lies on the circle. (Hint: Is a2 + b2 > a2 ?) 12.kc) for any integer k > 0. Suppose that ABC is a triangle with side AB a diameter of a circle with center O. tangent line A • nt O nge ta not P C Figure 1.1 10.4.1.6. 11. 3. (a) Show that (6. Can a right triangle have sides with lengths 2.m2 + n2 ) is a Pythagorean triple for all integers m > n > 0. The triple is normally written as (a. Let A be one of the points where this circle intersects the ﬁrst circle. with a2 + b2 = c2 . a notion which we will use throughout the text. (b) Show that if (a. what is the maximum distance along the ground from the base of the pole to where the other end can be attached? You may assume that the pole is perpendicular to the ground. and c of the sides of a right triangle are positive integers. Write down the ﬁrst ten Pythagorean triples generated by this formula. 2. as indicated by the dashed lines in the picture.1. Explain how this picture proves Thales’ Theorem. This exercise will describe how to draw a line through any point outside a circle such that the line intersects the circle at only one point.kb. so that ABC is in a new position.m2 − n2 . Prove that the hypotenuse is the longest side in every right triangle.6 Chapter 1 • Right Triangle Trigonometry §1. Now imagine that you rotate the circle 180◦ around its center. 5. 14. This is called a tangent line to the circle (see the picture on the left in Figure 1. m = 3 and n = 1.5) and (5. 2. Use this fact to explain why the line you drew is the tangent line through A and to calculate the length of P A. If the rope is 150 ft long. (3. i.b.6 On a sheet of paper draw a circle of radius 1 inch. m = 4 and n = 1. 2. C A O B . How would you interpret this geometrically? (c) Show that (2mn.c). Draw the line through P and A. b.6). If the lengths a. as in the picture on the right in Figure 1. and 6? Explain your answer.1. 4.13) are well-known Pythagorean triples. In general the tangent line through a point on a circle is perpendicular to the line joining that point to the center of the circle (why?).8. m = 5 and n = 1. Pick a point P which is 2. use: m = 2 and n = 1. 3.c) is a Pythagorean triple then so is (ka. Draw the circle which has OP as a diameter.b. 13. Trigonometric Functions of an Acute Angle • Section 1.2 7 1.2 Trigonometric Functions of an Acute Angle Table 1.2 Name of function h A nu te o yp c se b B a opposite Consider a right triangle ABC, with the right angle at C and with lengths a, b, and c, as in the ﬁgure on the right. For the acute angle A, call the leg BC its opposite side, and call the leg AC its adjacent side. Recall that the hypotenuse of the triangle is the side AB. The ratios of sides of a right triangle occur often enough in practical applications to warrant their own names, so we deﬁne the six trigonometric functions of A as follows: C The six trigonometric functions of A Abbreviation Deﬁnition sine A sin A = opposite side hypotenuse = a c cosine A cos A = hypotenuse = b c tangent A tan A = opposite side = a b cosecant A csc A = hypotenuse opposite side = c a secant A sec A = hypotenuse = c b cotangent A cot A = opposite side = b a We will usually use the abbreviated names of the functions. Notice from Table 1.2 that the pairs sin A and csc A, cos A and sec A, and tan A and cot A are reciprocals: csc A = 1 sin A sec A = 1 cos A cot A = 1 tan A sin A = 1 csc A cos A = 1 sec A tan A = 1 cot A 8 Chapter 1 • Right Triangle Trigonometry §1.2 Example 1.5 For the right triangle ABC shown on the right, ﬁnd the values of all six trigonometric functions of the acute angles A and B. B 5 Solution: The hypotenuse of ABC has length 5. For angle A, the opposite side BC has length 3 and the adjacent side AC has length 4. Thus: A 3 4 C sin A = opposite 3 = hypotenuse 5 cos A = 4 = hypotenuse 5 tan A = opposite 3 = 4 csc A = hypotenuse 5 = opposite 3 sec A = hypotenuse 5 = 4 cot A = 4 = opposite 3 For angle B, the opposite side AC has length 4 and the adjacent side BC has length 3. Thus: sin B = opposite 4 = hypotenuse 5 cos B = 3 = hypotenuse 5 tan B = opposite 4 = 3 csc B = hypotenuse 5 = opposite 4 sec B = hypotenuse 5 = 3 cot B = 3 = opposite 4 Notice in Example 1.5 that we did not specify the units for the lengths. This raises the possibility that our answers depended on a triangle of a speciﬁc physical size. For example, suppose that two different students are reading this textbook: one in the United States and one in Germany. The American student thinks that the lengths 3, 4, and 5 in Example 1.5 are measured in inches, while the German student thinks that they are measured in centimeters. Since 1 in ≈ 2.54 cm, the students are using triangles of different physical sizes (see Figure 1.2.1 below, not drawn to scale). B B 5 B 3 5 A 4 (a) Inches C A 3 4 C B A A (b) Centimeters Figure 1.2.1 C C (c) Similar triangles ABC ∼ A B C If the American triangle is ABC and the German triangle is A B C , then we see from Figure 1.2.1 that ABC is similar to A B C , and hence the corresponding angles Trigonometric Functions of an Acute Angle • Section 1.2 9 are equal and the ratios of the corresponding sides are equal. In fact, we know that common ratio: the sides of ABC are approximately 2.54 times longer than the corresponding sides of A B C . So when the American student calculates sin A and the German student calculates sin A , they get the same answer:3 ABC ∼ A B C BC AB = BC AB BC B C = AB AB sin A = sin A Likewise, the other values of the trigonometric functions of A and A are the same. In fact, our argument was general enough to work with any similar right triangles. This leads us to the following conclusion: When calculating the trigonometric functions of an acute angle A, you may use any right triangle which has A as one of the angles. Since we deﬁned the trigonometric functions in terms of ratios of sides, you can think of the units of measurement for those sides as canceling out in those ratios. This means that the values of the trigonometric functions are unitless numbers. So when the American student calculated 3/5 as the value of sin A in Example 1.5, that is the same as the 3/5 that the German student calculated, despite the different units for the lengths of the sides. Example 1.6 Find the values of all six trigonometric functions of 45◦ . 1 Solution: Since we may use any right triangle which has 45◦ as one of the angles, use the simplest one: take a square whose sides are all 1 unit long and divide it in half diagonally, as in the ﬁgure on the right. Since the two legs of the triangle ABC have the same length, ABC is an isosceles triangle, which means that the angles A and B are equal. So since A + B = 90◦ , this means that we must have A = B = 45◦ . By the Pythagorean Theorem, the length c of the hypotenuse is given by c2 = 12 + 12 = 2 ⇒ c = 2 . 1 2 B 1 45◦ A 1 C Thus, using the angle A we get: sin 45◦ = opposite 1 = hypotenuse 2 cos 45◦ = 1 = hypotenuse 2 tan 45◦ = hypotenuse = 2 opposite sec 45◦ = hypotenuse = 2 cot 45◦ = csc 45◦ = opposite 1 = = 1 1 1 = = 1 opposite 1 Note that we would have obtained the same answers if we hadused any right triangle similar to ABC. For example, if we multiply each side of ABC by 2, then we would have a similar triangle with legs of length 2 and hypotenuse of length 2. This would give us sin 45◦ = 22 , which equals 2 2· 2 = 1 2 as before. The same goes for the other functions. 3 We will use the notation AB to denote the length of a line segment AB. then the following relations hold: sin A = cos B sec A = csc B tan A = cot B sin B = cos A sec B = csc A tan B = cot A We say that the pairs of functions { sin.9 Write each of the following numbers as trigonometric functions of an angle less than 45◦ : (a) sin 65◦ . and 60◦ arise often in applications. Cofunction Theorem: If A and B are the complementary acute angles in a right triangle ABC.5 and 1. and cotangent got the “co” in their names. Example 1. secant and cosecant are cofunctions. secant and cosecant. { sec. 45◦ . That is how the functions cosine. (c) tan 59◦ . as in Figure 1. The Cofunction Theorem says that any trigonometric function of an acute angle is equal to its cofunction of the complementary angle.2 11 We now know the lengths of all sides of the triangle ABC. cot } are cofunctions.2 above. so by the Cofunction Theorem we know that sin 65◦ = cos 25◦ . cosecant. and tangent and cotangent of the complementary angles in Examples 1. (c) The complement of 59◦ is 90◦ − 59◦ = 31◦ and the cofunction of tan is cot. Generalizing those examples gives us the following theorem: Theorem 1. and tangent and cotangent are cofunctions. (b) cos 78◦ . csc }. so we have: adjacent 5 opposite 2 cos A = tan A = = = hypotenuse 3 adjacent 5 hypotenuse 3 csc A = = opposite 2 hypotenuse 3 sec A = = adjacent 5 adjacent 5 cot A = = opposite 2 You may have noticed the connections between the sine and cosine.Trigonometric Functions of an Acute Angle • Section 1. We can use the Pythagorean Theorem to generalize the right triangles in Examples 1. so tan 59◦ = cot 31◦ . (b) The complement of 78◦ is 90◦ − 78◦ = 12◦ and the cofunction of cos is sin. So sine and cosine are cofunctions.2. .7 and see what any 45 − 45 − 90 and 30 − 60 − 90 right triangles look like.2. so cos 78◦ = sin 12◦ . a 2 45◦ a 45◦ a (a) 45 − 45 − 90 Figure 1. Solution: (a) The complement of 65◦ is 90◦ − 65◦ = 25◦ and the cofunction of sin is cos.2. and { tan.2 2a 60◦ a 30◦ a 3 (b) 30 − 60 − 90 Two general right triangles (any a > 0) The angles 30◦ .7. cos }.6 and 1. and tan 75 = DE AE AE = AC − EC = and AE AD Hence. And ∠ ADB = 60◦ . B F 2 3 3 2 ◦ 30 A 45◦ E C 3 2 2 Now. So EC = FB = DE = DF + FE = ◦ sin 75 = DE AD = 3 +1 2 2 1 2 = + 6+ 2 4 1 2 3 2 3 2. From Figure 1. b = 21. place a 30−60−90 right triangle ADB with legs of length 3 and 1 on top of the hypotenuse − 45 − 90 right triangle ABC whose hypotenuse has of a 45 length 3. so we know that DF = FB = 1 . b = 24.2(a) we know that the length of each leg of ABC is the length of the 3 hypotenuse divided by 2. The hypotenuse BD of DFB has length 1 and DFB is a 45 − 45 − 90 right triangle. b = 5 9. Since 2 ∠ BAC = 45◦ and ∠ D AB = 30◦ . b = 15. Thus. a = 2. ﬁnd the values of the other ﬁve trigonometric functions of the acute angle A given the indicated value of one of the functions. Thus. 6+ 2 2 = 6− 2 4 sec 75◦ = − 1 2 . So AC = BC = = 32 . 6+ 2 Exercises For Exercises 1-10. so FE and BC are parallel. and tangent of ∠ D AE. b = 12. since it is the difference of ∠ ADB = 60◦ and ∠ ADE = 15◦ . c = 41 4. Thus. since it is the complement of ∠ D AB. a = 7. so that ADE is a right triangle. a = 9. we know that DE ⊥ AC and BC ⊥ AC.2 Example 1. we get csc 75◦ = = 3−1 2 4 . and FE = BC = = 3+ 1 2 ◦ . D ◦ Solution: Since 75◦ = 45◦ +30 . we see that ∠ D AE = 75◦ since it is the sum of those two angles. c = 6 A b C Figure 1.3 10. .2. 1. b = 40. a = 5. Then ∠ BDF = 45◦ . cosine. 6−2 . c = 8 For Exercises 11-18. FB and EC are both perpendicular to DE and hence FB is parallel to EC.12 Chapter 1 • Right Triangle Trigonometry §1. ∠ DBF = 45◦ since it is the complement of ∠ BDF. c = 5 7. cos 75 = Note: Taking reciprocals. b = 2. c = 29 6. b = 3 8. Notice that ∠ ADE = 15◦ . so that DFB is a right triangle. c = 17 5. as in the ﬁgure on the right. we need to ﬁnd the sine. a = 20. c = 25 B c a 2. a = 5. Also. 6+2 6− 2 . 6− 2 3 2 ◦ and cot 75◦ = = = 3− 1 2 3+1 2 3−1 2 = . 4 . Likewise. FBCE is a rectangle. a = 1. c = 13 3. a = 8. b = 7.10 Find the sine.3. Draw DE 1 perpendicular to AC.2. ﬁnd the values of all six trigonometric functions of angles A and B in the right triangle ABC in Figure 1. Draw BF perpendicular to DE. a = 1.2. and tangent of 75◦ . since it is the complement of ∠ D AE. since ∠ BCE is a right angle. cosine. engineering.11 A person stands 150 ft away from a ﬂagpole and measures an angle of elevation of 32◦ from his horizontal line of sight to the top of the ﬂagpole. we rounded off the answer for h to the nearest integer. h 32◦ 150 How did we know that tan 32◦ = 0. And 6 since none of the numbers we were given had decimal places.g. determining large distances or lengths by using measurements of angles and small. substitute x into the ﬁrst formula for h to get the height of the mountain: h = (1378 + 400) tan 25◦ = 1778 (0. where h = tan 32◦ 150 ⇒ ◦ h = 150 tan 32 = 150 (0. since they both equal h. surveying.6249 ? By using a calculator. Today. Use that equation to solve for x: x tan 25◦ − x tan 20◦ = 900 tan 20◦ − 400 tan 25◦ ⇒ x = 900 tan 20◦ − 400 tan 25◦ = 1378 ft tan 25◦ − tan 20◦ Finally. Assume that the person’s eyes are a vertical distance of 6 ft from the ground. Thus. Example 1. e. so ◦ h 20◦ 25◦ 500 400 x (x + 400) tan 25◦ = (x + 900) tan 20◦ .6249) = 94 . astronomy. and let x be the distance from the base of the mountain to the point directly beneath the top of the mountain. as in the picture on the right. known distances. trigonometry is widely used in physics.3 1. trigonometry was often used as a means of indirect measurement.4663) = 829 ft . Let h be the height of the mountain. What is the height of the ﬂagpole? Solution: The picture on the right describes the situation.12 A person standing 400 ft from the base of a mountain measures the angle of elevation from the ground to the top of the mountain to be 25◦ . and h = tan 20◦ x + 400 + 500 ⇒ h = (x + 900) tan 20◦ . Example 1. navigation. We see that the height of the ﬂagpole is h + 6 ft. Your calculator should be in degree mode for these examples. Then we see that h = tan 25◦ x + 400 ⇒ h = (x + 400) tan 25 .3 Applications and Solving Right Triangles Throughout its early development. The person then walks 500 ft straight back and measures the angle of elevation to now be 20◦ . and various ﬁelds of mathematics and other disciplines. the height of the ﬂagpole is h + 6 = 94 + 6 = 100 ft . How tall is the mountain? Solution: We will assume that the ground is ﬂat and not inclined relative to the base of the mountain. In this section we will see some of the ways in which trigonometry can be applied.14 Chapter 1 • Right Triangle Trigonometry §1. 1. 26-27 in W. M UNK AND G. 4280 = tan 24◦ x ⇒ x = 24◦ 4280 θ x 4280 = 9613 ft .14 An observer at the top of a mountain 3 miles above sea level measures an angle of depression of 2. Since the ground and the blimp’s horizontal line of sight are parallel.23◦ ⇒ r − r cos 2. Thus.3 15 Example 1.23◦ 3 H r θ r O Figure 1. . we have ∠ O AB = 90◦ . Since the angles in the triangle O AH add up to 180◦ . so we see that ∠ O AH = 90◦ − 2. Let θ be the angle ∠ AOH.23◦ 1 − cos 2. 1960. i.23◦ . Use this to estimate the radius of the earth.23◦ r = 3958. r+3 so solving for r we get r = (r + 3) cos 2.Applications and Solving Right Triangles • Section 1. and let H be the ocean horizon in the line of sight from A.H.4 Let r be the radius of the earth. and let B be a point on the horizontal line of sight from A (i.e. egg-shaped. Assuming the ground is ﬂat. we have θ = 180◦ − 90◦ − 87.23◦ = 87.23◦ to the ocean horizon. So by Exercise 14 in Section 1. as in Figure 1.1. how far away along the ground is the house from the blimp? Solution: Let x be the distance along the ground from the blimp to the house.J.6 miles. Solution: We will assume that the earth is a sphere. tan 24◦ Example 1.23◦ .e. Hence. The earth is an ellipsoid.e.77◦ . cos θ = OH r = OA r+3 ⇒ A B 2. we have O A = r + 3. London: Cambridge University Press.77◦ = 2. Let O be the center of the earth. Now since AB ⊥ O A. AH ⊥ OH and hence ∠ OH A = 90◦ .3. with an observed ellipticity of 1/297 (a sphere has ellipticity 0). 4 Of course it is not perfectly spherical. on the line perpendicular to O A). θ = 24◦ . Since A is 3 miles above sea level. OH = r. Note: This answer is very close to the earth’s actual (mean) radius of 3956. i.1 r = cos 2. we know from elementary geometry that the angle of elevation θ from the base of the house to the blimp is equal to the angle of depression from the blimp to the base of the house.23◦ ⇒ r = ⇒ 3 cos 2. Let the point A represent the top of the mountain. The Rotation of the Earth: A Geophysical Discussion. See pp. We see that the line through A and H is a tangent line to the surface of the earth (considering the surface as the circle of radius r through H as in the picture).3 miles . Also.13 A blimp 4280 ft above the ground measures an angle of depression of 24◦ from its horizontal line of sight to the base of a house on the ground.F M AC D ONALD.3. as in the picture to the right.23◦ = 3 cos 2. So ES = 92908394 − radius of earth = 92908394 − 3956. . Solution: Let B be the position of the sun.15 we used α = 0.00244◦ so the distance from the center of the earth to the sun is approximately 93 million miles .4) sin 0. For example. sin α sin 0. 908.8 . Example 1.14. Thus. The radius of the sun equals AS. the triangles E AS and EBS are similar. ◦ ∠ E AS = ∠ EBS = 90 . 293 miles .00244◦ .16 An observer on earth measures an angle of 32 4 from one visible edge of the sun to the other (opposite) edge. Hence. to get O A = 3956. which we mention only because some angle measurement devices do use minutes and seconds. a star) in space. Use this to estimate the radius of the sun. and a second is one-sixtieth of a minute.6 = 92904437. Use this to estimate the distance from the center of the earth to the sun. Since ∠ O AB = 90◦ .26722◦ = 433. let A be a point on the equator. 200 miles. The equatorial parallax of the sun has been observed to be approximately α = 0. as in the picture on the right. A degree can be divided into smaller units: a minute is one-sixtieth of a degree. then we say that the angle α = ∠ OBA is the equatorial parallax of the object.g. B E S 32 4 Solution: Let the point E be the earth and let S be the center of the sun. We want to ﬁnd the length of OB. Since we treated the sun in that example as a point. so the actual distance to the sun varies. Note: This answer is close to the sun’s actual (mean) radius of 432. The symbol for a minute is and the symbol for a second is .00244◦ ≈ 8.3 Example 1.4 miles.26722◦ = (92904437. If the earth is positioned in such a way that the angle ∠ O AB = 90◦ . Now. In Example 1.505◦ = 4◦ 30 18 : 30 18 degrees = 4. and let B represent an object (e. 4. ∠ AES = ∠ BES = 21 ∠ AEB = 12 (32 4 ) = 16 2 = (16/60) + (2/3600) = 0. And 4. we will ﬁnd the distance from the earth to the sun.15 we found the distance from the center of the earth to the sun to be 92. So since EB = E A (why?).6 miles. We will use the actual radius of the earth. Note: The earth’s orbit around the sun is an ellipse. Thus. 394 miles.00244◦ ).26722◦ . ES is the distance from the surface of the earth (where the observer stands) to the center of the sun. we have OA = sin α OB ⇒ OB = B α A O OA 3956.16 Chapter 1 • Right Triangle Trigonometry §1. sin (∠ AES) = AS ES ⇒ AS = ES sin 0. The observer’s lines of sight to the visible edges of the sun A are tangent lines to the sun’s surface at the points A and B. then we are justiﬁed in treating that distance as the distance between the centers of the earth and sun.6 = = 92908394 .5◦ = 4◦ 30 . Clearly AS = BS.505◦ + 4◦ 30 18 = 4 + 60 3600 In Example 1. mentioned at the end of Example 1. as in the picture on the right. Let O be the center of the earth. In the above example we used a very small angle (0.15 As another application of trigonometry to astronomy. so you will have to create one. Since the horizontal line segment BC is tangent to each roller. since OD A is a right triangle.69. Each roller touches both slanted sides of the V-block. Often no right triangle will be immediately evident. in which one circular roller sits on top of a smaller circular roller. ∠ OD A = E ∠ PEC = 90◦ . 5 This will often be worded as the line that is tangent to the circle. ∠ BOC = ∠ DOC. . OBC is a right triangle.17 The machine tool diagram on the right shows a symmetric V-block. Likewise.38 and so BC = 0. In applied problems it is not always obvious which right triangle to use. since BP = EP and ∠ PBC = ∠ PEC = 90◦ . We know that ∠ DOB = ∠ DO A = 53◦ . then ﬁnd the angle ∠ BOC. BC + BC = 1. so look for places where you can form perpendicular line segments. you can create right angles by using the perpendicularity of the tangent line to the circle at a point5 with the line that joins that point to the center of the circle.38. Thus. O d 1. BPC and EPC are congruent right triangles. When the problem contains a circle. their corresponding angles are equal.5◦ Hence.384) = 2. BC = EC. 53◦ = ∠ DOB = ∠ BOC + ∠ DOC = ∠ BOC + ∠ BOC = 2 ∠ BOC ⇒ ∠ BOC = 26. the diameter of the large roller is d = 2 × OB = 2 (1.69 = = 1. By symmetry. Find the diameter d of the large roller. OB = OD (since they each equal the radius of the large roller). given the information in the diagram. To do this. so we need to ﬁnd OB.14. Thus.5◦ . So ∠ DO A = 53 . ◦ ∠ DO A is the complement of ∠ O AD. 1.Applications and Solving Right Triangles • Section 1. so by the Pythagorean Theorem we have BC = DC: BC 2 = OC 2 − OB2 = OC 2 − OD 2 = DC 2 ⇒ BC = DC Thus.384 tan ∠ BOC tan 26. Example 1. Thus. we will show that OBC 37◦ B is a right triangle. But we know that BC = DC. There is no general strategy for this.3 8 Solution: The diameter d of the large roller is twice the radius D OB. and 1.15. Hence. we know that ODC is a right triangle. but remember that a right triangle requires a right angle. and then ﬁnd BC. ∠ OBC = ∠ PBC = 90◦ . P Since the slanted sides are tangent to each roller. which is why these sorts of problems can be difﬁcult. since their corresponding sides are equal. Now. since the vertical line through the centers of the rollers makes a 37◦ angle with each slanted side. A we have ∠ O AD = 37◦ . Thus. We did exactly that in Examples 1. OBC and ODC are congruent triangles (which we denote by OBC ∼ = ODC). C The length OB will then be simple to determine. And since ∠ OD A = 90◦ . So in particular. and we see from the diagram that EC + DC = 1. We now have all we need to ﬁnd OB: BC = tan ∠ BOC OB ⇒ OB = BC 0. Thus.16.768 .3 17 You may have noticed that the solutions to the examples we have shown required at least one right triangle. 18 Chapter 1 • Right Triangle Trigonometry §1.3 Example 1.18 A slider-crank mechanism is shown in Figure 1.3.2 below. As the piston moves downward the connecting rod rotates the crank in the clockwise direction, as indicated. piston a A c g rod ectin conn b C θ r cr an k B O Figure 1.3.2 Slider-crank mechanism The point A is the center of the connecting rod’s wrist pin and only moves vertically. The point B is the center of the crank pin and moves around a circle of radius r centered at the point O, which is directly below A and does not move. As the crank rotates it makes an angle θ with the line O A. The instantaneous center of rotation of the connecting rod at a given time is the point C where the horizontal line through A intersects the extended line through O and B. From Figure 1.3.2 we see that ∠ O AC = 90◦ , and we let a = AC, b = AB, and c = BC. In the exercises you will show that for 0◦ < θ < 90◦ , b2 − r 2 (sin θ )2 and a = r sin θ + b2 − r 2 (sin θ )2 tan θ . c = cos θ Applications and Solving Right Triangles • Section 1.3 19 For some problems it may help to remember that when a right trir angle has a hypotenuse of length r and an acute angle θ , as in the r sin θ picture on the right, the adjacent side will have length r cos θ and the θ opposite side will have length r sin θ . You can think of those lengths r cos θ as the horizontal and vertical “components” of the hypotenuse. Notice in the above right triangle that we were given two pieces of information: one of the acute angles and the length of the hypotenuse. From this we determined the lengths of the other two sides, and the other acute angle is just the complement of the known acute angle. In general, a triangle has six parts: three sides and three angles. Solving a triangle means ﬁnding the unknown parts based on the known parts. In the case of a right triangle, one part is always known: one of the angles is 90◦ . Example 1.19 Solve the right triangle in Figure 1.3.3 using the given information: B c (a) c = 10, A = 22◦ Solution: The unknown parts are a, b, and B. Solving yields: a = c sin A = 10 sin 22 = 3.75 b = c cos A = 10 cos 22◦ = 9.27 A a b C Figure 1.3.3 B = 90◦ − A = 90◦ − 22◦ = 68◦ (b) b = 8, A = 40◦ Solution: The unknown parts are a, c, and B. Solving yields: a = tan A b a = b tan A = 8 tan 40◦ = 6.71 b = cos A c c = b 8 = = 10.44 cos A cos 40◦ B = 90◦ − A = 90◦ − 40◦ = 50◦ (c) a = 3, b = 4 Solution: The unknown parts are c, A, and B. By the Pythagorean Theorem, c = a2 + b2 = 32 + 42 = 25 = 5 . Now, tan A = ab = 34 = 0.75. So how do we ﬁnd A? There should be a key labeled tan−1 on your calculator, which works like this: give it a number x and it will tell you the angle θ such that tan θ = x. In our case, we want the angle A such that tan A = 0.75: Enter: 0.75 This tells us that A =36.87 , approximately. Thus B = 90 − A = 90 − 36.87 = 53.13 . 1 1 Note: The sin and cos keys work similarly for sine and cosine, respectively. These keys use the inverse trigonometric functions, which we will discuss in Chapter 5. 20 Chapter 1 • Right Triangle Trigonometry §1.3 Exercises 1. From a position 150 ft above the ground, an observer in a building measures angles of depression of 12◦ and 34◦ to the top and bottom, respectively, of a smaller building, as in the picture on the right. Use this to ﬁnd the height h of the smaller building. 34◦ 12◦ 150 2. Generalize Example 1.12: A person standing a ft from the base of a mountain measures an angle of elevation α from the ground to the top of the mountain. The person then walks b ft straight back and measures an angle of elevation β to the top of the mountain, as in the picture on the right. Assuming the ground is level, ﬁnd a formula for the height h of the mountain in terms of a, b, α, and β. h h β b α a 3. As the angle of elevation from the top of a tower to the sun decreases from 64◦ to 49◦ during the day, the length of the shadow of the tower increases by 92 ft along the ground. Assuming the ground is level, ﬁnd the height of the tower. 4. Two banks of a river are parallel, and the distance between two points A and B along one bank is 500 ft. For a point C on the opposite bank, ∠ BAC = 56◦ and ∠ ABC = 41◦ , as in the picture on the right. What is the width w of the river? (Hint: Divide AB into two pieces.) 5. A tower on one side of a river is directly east and north of points A and B, respectively, on the other side of the river. The top of the tower has angles of elevation α and β from A and B, respectively, as in the picture on the right. Let d be the distance between A and B. Assuming that both sides of the river are at the same elevation, show that the height h of the tower is A 56◦ 500 41◦ B w C N E h β α d . h = 2 (cot α) + (cot β)2 A d B 6. The equatorial parallax of the moon has been observed to be approximately 57 of the earth to be 3956.6 miles, estimate the distance from the center of the earth to the moon. (Hint: See Example 1.15.) 7. An observer on earth measures an angle of 31 7 from one visible edge of the moon to the other (opposite) edge. Use this to estimate the radius of the moon. (Hint: Use Exercise 6 and see Example 1.16.) B = 26◦ 22.4 using the given information. A = 26◦ 20. b = 1. a = 2. Use this to ﬁnd AB. CE. A ball bearing sits between two metal grooves. a = 3. DC. a = 2.2 draw line segments from B perpendicular to O A and AC. c = 7 19. B c a 15. AC. ∠ BAC = θ and BC = a. solve the right triangle in Figure 1.3 8. What must the diameter of the ball bearing be for the distance between the vertexes of the grooves to be half an inch? You may assume that the top vertex is directly above the bottom vertex. AD. In this view. c = 2 21.7 9. b = 12 16. show that c = b2 − r 2 (sin θ )2 cos θ and a = r sin θ + b2 − r 2 (sin θ )2 tan θ . A = 8◦ 18. B = 35◦ 17.Applications and Solving Right Triangles • Section 1.18.3. 21 120◦ 1 2 90◦ 1. The machine tool diagram on the right shows a symmetric die punch.4 A b C . b = 3. as in the picture on the right. Find the amount d that the top of the roller rises above the top of the thread. (Hint: In Figure 1. (Hint: What is the angle ∠ ACD ?) 2 18 54◦ r 1 12 13. c = 2. in which a circular roller of diameter 1.5 10.8 inches as the distance across the top of the worm thread.3. A = 45◦ Figure 1.5 inches sits. (Hint: Extend the slanted sides of the thread until they meet at a point. In the ﬁgure on the right. In Exercise 9.) 14. what would the distance across the top of the worm thread have to be to make d equal to 0 inches? 30◦ 12.) d 1. Use the information in the diagram to ﬁnd the radius r. Repeat Exercise 9 using 1. and the slanted sides are tangent to that circle and form an angle of 54◦ . given the information in the diagram. the rounded tip is part of a circle of radius r. with the top groove having an angle of 120◦ and the bottom groove having an angle of 90◦ . a = 5. and DE in terms of θ and a. The machine tool diagram on the right shows a symmetric worm thread. The top and bottom sides of the die punch are horizontal. B = 8◦ 23. b = 2. B a θ A C D E For Exercises 15-23.3. For 0◦ < θ < 90◦ in the slider-crank mechanism in Example 1. 11. c = 6. ) (d) Use Figure 1. (c) Use Figure 1. In Example 1.5 to ﬁnd the exact values of sin 15◦ .3. P 7.5◦ .5◦ .22 Chapter 1 • Right Triangle Trigonometry §1.2. Now create a second semicircle as follows: Let A be the left endpoint of the ﬁrst semicircle. Then create a third semicircle in the same way: Let B be the left endpoint of the second semicircle. for any 0◦ < θ < 90◦ . So since tan 15◦ = cot 75◦ by the Cofunction 75 .5. you will need to use ∠ POQ = 60◦ and OP = 1 to ﬁnd the exact lengths of PQ and OQ. For example.3.3. cos 22. In fact.3.5 and is described below. it can be used to ﬁnd the exact values for the trigonometric functions of θ2 when those for θ are known. we found the exact values of all six trigonometric functions of 6−2 ◦ ◦ .5 to calculate the exact value of tan 7. cos 15◦ . and tan 22. then draw a new semicircle centered at A with radius equal to AP.3 24. .5◦ . This procedure can be continued indeﬁnitely to create more semicircles. we showed that cot 75 = 6+ 2 Theorem. (Hint: To start.3.5◦ . this means that tan 15◦ = 6−2 . and tan 15◦ . it can be shown that the line segment from the center of the new semicircle to P makes an angle with the horizontal line equal to half the angle from the previous semicircle’s center to P. as in Figure 1. (a) Explain why ∠ P AQ = 30◦ . The method is illustrated in Figure 1.5◦ C 15◦ B 30◦ A 60◦ O Q 1 Figure 1.5 Draw a semicircle of radius 1 centered at a point O on a horizontal line. In general. (Hint: What is the supplement of 60◦ ?) (b) Explain why ∠ PBQ = 15◦ and ∠ PCQ = 7. (e) Use the same method but with an initial angle of ∠ POQ = 45◦ to ﬁnd the exact values of sin 22. Let P be the point on the semicircle such that OP makes an angle of 60◦ with the horizontal line. Draw a line straight down from P to the horizontal line at the point Q. We will now describe another method for ﬁnding the 6+ 2 exact values of the trigonometric functions of 15◦ .10 in Section 1. then draw a new semicircle centered at B with radius equal to BP.5◦ . above sea level. θ (b) Find the angle θ that AB makes with the base of the cube. B B 28. That is. Calculate the area of this octagon. suppose that α. (b) The manufacturer needs to place a circular ring inside the container.3 25. A manufacturer needs to place ten identical ball bearings against the inner side of a circular container such that each ball bearing touches two other ball bearings. How many miles farther out is Person B’s horizon than Person A’s? (Note: 1 mile = 5280 ft) . would the area inside it increase or decrease? What number would the area approach. The picture on the right shows a cube whose sides are of length a > 0. 27. In Figure 1. What is the largest possible (outer) radius of the ring such that it is not on top of the ball bearings and its base is level with the base of the container? 26. (b) If you were to increase the number of sides of the polygon.6. 1 (a) Calculate the area of the octagon.3. called a regular octagon. The (inner) radius of the container is 4 cm.6 29.Applications and Solving Right Triangles • Section 1. That is. Show that: AD (a) BC = cot α − cot β (b) AC = AD · tan β tan β − tan α AD · sin α (c) BD = sin (β − α) (Hint: What is the measure of the angle ∠ ABD ?) α A β D C Figure 1.3. β. A circle of radius 1 is inscribed inside a polygon with eight sides of equal length. each of the eight sides is tangent to the circle. and AD are known. (c) Inscribe a regular octagon inside the same circle. draw a regular octagon such that each of its eight vertexes touches the circle. as in the picture on the right. if any? Explain. 23 r 4 (a) Find the common radius r of the ball bearings. respectively. Persons A and B are at the beach. as in the picture on the right. A (a) Find the length of the diagonal line segment AB. their eyes are 5 ft and 6 ft. ◦ 30◦ 390 Figure 1. explanation which says that in ancient times a person could travel 12 Babylonian miles in one day (i. The Babylonian mile was large enough (approximately 7 of our miles) to be divided into 30 equal parts for convenience. New York: Saunders College Publishing. notice that 30◦ and 390◦ have the same terminal side in Figure 1.4.4 Trigonometric Functions of Any Angle To deﬁne the trigonometric functions of any angle . If the rotation is counter-clockwise then we say that the angle is positive. thus giving 12 × 30 = 360 equal parts in a full rotation. For example. is one full revolution of the Earth around the Sun). is an angle of 360◦ . and OB is the terminal side of the angle (see Figure 1. It is often assumed that the number 360 was used because the Babylonians (supposedly) thought that there were 360 days in a year (a year. such that 360◦ is one revolution. or even just O.26 in H. there is another. or simply ∠ O. so that the terminal side coincides with the initial side.we need a more general deﬁnition of an angle.4. B B counter-clockwise direction (+) e id s al in m r te O clockwise direction (−) A initial side (a) angle ∠ AOB O A (b) positive and negative angles Figure 1. 5th ed.4. More than one full rotation creates an angle greater than 360◦ .1(b)).2. 1983. An Introduction to the History of Mathematics.1(a)).1 Deﬁnition of a general angle We denote the angle formed by this rotation as ∠ AOB. .including angles less than 0◦ or greater than 360◦ . denoted by the ray OB.. However. See p.4. since 30 + 360 = 390. so that the ray is in a new −−→ −−→ −−→ position. originated in ancient Babylonia. in the clockwise direction −−→ this would be −360◦ .4. We say that an angle is formed −−→ by rotating a ray O A about the endpoint O (called the vertex).e. one full rotation of the Earth about its axis). perhaps more likely.24 Chapter 1 • Right Triangle Trigonometry §1. −−→ One full counter-clockwise rotation of O A back onto itself (called a revolution).2 Angle greater than 360◦ 6 The system of measuring angles in degrees.4 1. The ray O A is called the initial side of the angle. and the angle is negative if the rotation is clockwise (see Figure 1.6 Not rotating O A constitutes an angle of 0◦ . E VES. of course. Trigonometric Functions of Any Angle • Section 1.4 25 We can now deﬁne the trigonometric functions of any angle in terms of Cartesian coordinates. Recall that the x y-coordinate plane consists of points denoted by pairs (x, y) of real numbers. The ﬁrst number, x, is the point’s x coordinate, and the second number, y, is its y coordinate. The x and y coordinates are measured by their positions along the x-axis and y-axis, respectively, which determine the point’s position in the plane. This divides the x y-coordinate plane into four quadrants (denoted by QI, QII, QIII, QIV), based on the signs of x and y (see Figure 1.4.3(a)-(b)). y (2, 3) QI x>0 y>0 QII x<0 y>0 QIII x<0 y<0 y 0 (−3, 2) x x 0 QIV x>0 y<0 (−2, −2) (3, −3) (b) Points in the plane y (x, y) r θ x 0 (c) Angle θ in the plane Figure 1.4.3 x y-coordinate plane Now let θ be any angle. We say that θ is in standard position if its initial side is the positive x-axis and its vertex is the origin (0, 0). Pick any point (x, y) onthe terminal side of θ a distance r > 0 from the origin (see Figure 1.4.3(c)). (Note that r = x2 + y2 . Why?) We then deﬁne the trigonometric functions of θ as follows: sin θ = y r cos θ = x r tan θ = y x (1.2) csc θ = r y sec θ = r x cot θ = x y (1.3) As in the acute case, by the use of similar triangles these deﬁnitions are well-deﬁned (i.e. they do not depend on which point (x, y) we choose on the terminal side of θ ). Also, notice that \| sin θ \| ≤ 1 and \| cos θ \| ≤ 1, since \| y \| ≤ r and \| x \| ≤ r in the above deﬁnitions. 26 Chapter 1 • Right Triangle Trigonometry §1.4 Notice that in the case of an acute angle these deﬁnitions are equivalent to our earlier deﬁnitions in terms of right triangles: draw a right triangle with angle θ such that x = adjacent side, y = opposite side, and r = hypotenuse. For example, this would give us sin θ = opposite y x r = hypotenuse and cos θ = r = hypotenuse , just as before (see Figure 1.4.4(a)). y y 90◦ se (x, y) hy po te r nu QII 90 < θ < 180◦ y opposite side θ 180◦ x 0 0◦ 0 QIII 180 < θ < 270◦ x QIV 270 < θ < 360◦ x QI 0 < θ < 90◦ 270◦ (a) Acute angle θ Figure 1.4.4 In Figure 1.4.4(b) we see in which quadrants or on which axes the terminal side of an angle 0 ≤ θ < 360◦ may fall. From Figure 1.4.3(a) and formulas (1.2) and (1.3), we see that we can get negative values for a trigonometric function. For example, sin θ < 0 when y < 0. Figure 1.4.5 summarizes the signs (positive or negative) for the trigonometric functions based on y QI sin + cos + tan + csc + sec + cot + QII sin + cos − tan − csc + sec − cot − QIII sin − cos − tan + csc − sec − cot + Figure 1.4.5 0 x QIV sin − cos + tan − csc − sec + cot − Signs of the trigonometric functions by quadrant Trigonometric Functions of Any Angle • Section 1.4 27 Example 1.20 (−1, 3) Find the exact values of all six trigonometric functions of 120◦ . y Solution: We know 120 = 180 − 60 . By Example 1.7 in Section 1.2, we see that we can use the point (−1, 3) on the terminal side of the 2 angle 120◦ in QII, since we saw in that example that a basic right 3 triangle with a 60 angle has adjacent side of length 1, opposite side of length 3, and hypotenuse of length 2, as in the ﬁgure on the right. 60◦ Drawing that triangle in QII so that the hypotenuse is on the terminal 0 side of 120◦ makes r = 2, x = −1, and y = 3. Hence: 1 −1 y 3 x y 3 sin 120 = cos 120 = tan 120 = = = = =− 3 r 2 r 2 x −1 csc 120◦ = r 2 = y 3 sec 120◦ = r 2 = = −2 x −1 cot 120◦ = 120◦ x −1 x = y 3 Example 1.21 Find the exact values of all six trigonometric functions of 225◦ . y Solution: We know that 225 = 180 + 45 . By Example 1.6 in Section 1.2, we see that we can use the point (−1, −1) on the terminal side of the angle 225◦ in QIII, since we saw in that example that a basic right triangle with a 45◦ angle has adjacent side of length 1, opposite side of length 1, and hypotenuse of length 2, as in the ﬁgure on the right. Drawing that triangle in QIII so that the hypotenuse is on the terminal side of 225◦ makes r = 2, x = −1, and y = −1. Hence: −1 y = sin 225 = r 2 −1 x cos 225 = = r 2 csc 225◦ = ◦ r =− 2 y sec 225◦ = 225◦ 1 45 1 x 0 ◦ 2 (−1, −1) −1 y tan 225 = = =1 x −1 r =− 2 x cot 225◦ = −1 x = =1 y −1 Example 1.22 Find the exact values of all six trigonometric functions of 330◦ . y Solution: We know that 330 = 360 − 30 . By Example 1.7 in Section 3 1.2, we see that we can use the point ( 3, −1) on the terminal side of a basic right the angle 225◦ in QIV, since we saw in that example that 0 30◦ triangle with a 30◦ angle has adjacent side of length 3, opposite 330◦ side of length 1, and hypotenuse of length 2, as in the ﬁgure on the 2 right. Drawing that triangle in QIV sothat the hypotenuse is on the ( 3, −1) terminal side of 330◦ makes r = 2, x = 3, and y = −1. Hence: −1 −1 y x 3 y cos 330 = tan 330◦ = = = = sin 330 = r 2 r 2 x 3 csc 330◦ = r = −2 y sec 330◦ = r 2 = x 3 cot 330◦ = x =− 3 y x 1 from Figure 1. Hence: y 0 = =0 r 1 cos 0◦ = x 1 = =1 r 1 tan 0◦ = y 0 = =0 x 1 r 1 = = undeﬁned y 0 sec 0◦ = r 1 = =1 x 1 cot 0◦ = x 1 = = undeﬁned y 0 sin 0◦ = csc 0◦ = Note that csc 0◦ and cot 0◦ are undeﬁned.6 we see that for 90◦ the terminal side is the positive y-axis. we do not have any 0◦ 180◦ right triangles to draw. −1) so that r = 1. and y = 0. 1) 90◦ Solution: These angles are different from the angles we have considered so far. However. in the formulas we would use r = 1.4. 0) functions are easy to calculate by picking the simplest points on their (−1. So unlike the previous examples. x = 1. (0. and y = −1.6 line segment from the origin to the point (1. and y = 0. for 180◦ use the point (−1. since division by 0 is not allowed. in that the terminal sides lie along either the x-axis or the y-axis. 0) terminal sides and then using the deﬁnitions in formulas (1. y (0. 180◦ . Again. and hence: sin 90◦ = y 1 = =1 r 1 cos 90◦ = x 0 = =0 r 1 csc 90◦ = r 1 = =1 y 1 sec 90◦ = r 1 = = undeﬁned x 0 tan 90◦ = y 1 = = undeﬁned x 0 cot 90◦ = x 0 = =0 y 1 Likewise. x = −1. and y = 1. Hence: sin 180◦ = csc 180◦ = y 0 = =0 r 1 r 1 = = undeﬁned y 0 cos 180◦ = −1 x = = −1 r 1 sec 180◦ = r 1 = = −1 x −1 tan 180◦ = cot 180◦ = y 0 = =0 x −1 −1 x = = undeﬁned y 0 Lastly. the values of the trigonometric x 0 (1. Similarly. −1) 270◦ For instance. as in Figure 1. so use the point (0. x = 0. 0) on its terminal side (the positive x-axis). for 270◦ use the point (0.2) and (1. 90◦ .4. Hence: sin 270◦ = −1 y = = −1 r 1 cos 270◦ = x 0 = =0 r 1 csc 270◦ = r 1 = = −1 y −1 sec 270◦ = r 1 = = undeﬁned x 0 tan 270◦ = −1 y = = undeﬁned x 0 cot 270◦ = x 0 = =0 y −1 . 0) as sort of a degenerate right triangle whose height is 0 and whose hypotenuse and base have the same length 1.6. for the angle 0◦ use the point (1. We have r = 1. you could think of the line segment from the origin to (0.4 Example 1. 1) as a degenerate right triangle whose base has length 0 and whose height equals the length of the hypotenuse. x = 0. 1).23 Find the exact values of all six trigonometric functions of 0◦ . and 270◦ . You could think of the Figure 1. Regardless.28 Chapter 1 • Right Triangle Trigonometry §1. 0) so that r = 1.3).4. 20-1. are called coterminal. we saw how the values of trigonometric functions of an angle θ larger than 90◦ were found by using a certain acute angle as part of a right triangle. sin (−45◦ ) = sin 315◦ . tan 540◦ = tan 180◦ .22. sin 360◦ = sin 0◦ .3 Angle 0◦ sin Table of trigonometric function values cos tan csc sec cot undeﬁned 3 0 1 0 undeﬁned 1 30◦ 1 2 3 2 1 3 2 3 45◦ 1 2 3 2 1 2 2 2 1 0 3 2 − 12 60◦ 90◦ 120◦ ◦ 150◦ 1 2 − 1 2 − 23 ◦ 0 −1 135 180 1 2 1 2 210◦ − 12 − 23 225◦ − 1 − 1 240◦ 2 − 23 270◦ −1 300◦ 315 ◦ 330◦ − 23 − 1 2 − 12 2 − 12 0 1 2 1 2 3 2 1 3 undeﬁned − 3 −1 2 1 2 3 2 1 3 1 undeﬁned 0 2 3 −2 − 2 − 1 2 3 −1 − 3 − 1 2 − 2 0 undeﬁned −1 1 3 −2 − 2 − 2 3 − 2 − 2 −2 1 3 −1 undeﬁned 0 2 2 − 1 −1 − 2 3 − 2 − 1 −2 2 3 3 1 3 undeﬁned − 3 3 3 3 undeﬁned 3 1 3 −1 − 3 Since 360◦ represents one full revolution. Angles such as these.Trigonometric Functions of Any Angle • Section 1. cos 390◦ = cos 30◦ . In Examples 1. For example. if two angles differ by an integer multiple of 360◦ then each trigonometric function will have equal values at both angles.4 29 The following table summarizes the values of the trigonometric functions of angles between 0◦ and 360◦ which are integer multiples of 30◦ or 45◦ : Table 1. That acute angle has a special name: if θ is a nonacute angle then we say that the reference angle for θ is the acute angle formed by the terminal side of θ and either the positive or . the trigonometric function values repeat every 360◦ . which have the same initial and terminal sides. In general. etc. 7. we see that 60◦ is the reference angle for the nonacute angle θ = 120◦ .4.8(a) that the point (−4.30 Chapter 1 • Right Triangle Trigonometry §1.4. Thus. Find the exact values of sin θ and tan θ . either sin θ = 3 5 and tan θ = − 43 or sin θ = − 53 and tan θ = 3 4 . and y y 3 3 so we have x = −4. y (a) Which angle between 0◦ and 360◦ has the same terminal side (and hence the same trigonometric function values) as θ ? 208◦ (b) What is the reference angle for θ ? ◦ ◦ ◦ Solution: (a) Since 928 = 2 × 360 + 208 .4.2. and r = 5. By the Pythagorean Theorem.5 that θ must be in either QII or QIII. Thus. 28◦ 0 x 928◦ Figure 1.24 Let θ = 928◦ . 30◦ is the reference angle for θ = 330◦ .4. the length of the opposite side must then be 3. as shown in Figure 1.8 (b) θ in QIII cos θ = − 45 When θ is in QII. and tan θ : . That is.8 in Section 1. we know from Figure 1. and cot θ and tan θ have the same sign. −3) is on the terminal side of θ . sin θ = r = 35 and tan θ = x = −34 .4. So in Example 1. So it sufﬁces to remember the signs of sin θ . as in Figure 1. we see from Figure 1. y = 3. −3) (a) θ in QII Figure 1.4. in Example 1.8 below: y y (−4. 45◦ is the reference angle for θ = 225◦ . so the reference angle for θ = 928◦ is 208◦ − 180◦ = 28◦ .21. (b) 928◦ and 208◦ have the same terminal side in QIII. we can use 4 as the length of the adjacent side and 5 as the length of the hypotenuse. Example 1. 3) θ 4 5 θ 3 4 0 x 0 x 3 5 (−4. sec θ and cos θ have the same sign. and r = 5. and in Example 1.20.7 Example 1. 3) is on the terminal side of θ .22. Since cos θ is negative.25 Suppose that cos θ = − 45 . Since cos θ = − 54 . we have two possibilities. Since reciprocals have the same sign. csc θ and sin θ have the same sign. y = −3. sin θ = r = −53 and tan θ = x = − −4 = 4 . Thus. and y y so we have x = −4.4. then θ has the same terminal side as 208◦ . When θ is in QIII. Thus. Solution: We can use a method similar to the one used to solve Example 1. we see from Figure 1.8(b) that the point (−4. cos θ . adjacent draw a right triangle and interpret cos θ as the ratio hypotenuse of two of its sides.4 negative x-axis. 22.Trigonometric Functions of Any Angle • Section 1. 1079◦ 10. sin θ = 1 2 28. 17. In which quadrant(s) do sine and cosine have the same sign? 12. tan θ = − 21 34. −514◦ 11. Expand Table 1.10 in Section 1. In which quadrant(s) do cosine and tangent have the same sign? 16. . cos θ = 2 5 26. 27. tan θ = 5 12 36. 2009◦ 9. tan θ = 0 35.3 to include all integer multiples of 15◦ . cos θ = 1 2 23. 317◦ 18. sin θ = − 21 29. tan θ = 1 For Exercises 37-40. cos θ = 0 25. 32. 127◦ 2. state in which quadrant or on which axis the given angle lies. In which quadrant(s) do sine and tangent have the same sign? 14. 230◦ 8. ﬁnd the exact values of sin θ and cos θ when tan θ has the indicated value. In which quadrant(s) do cosine and tangent have the opposite sign? For Exercises 17-21. −127◦ 3. sin θ = 0 30. Does sin 180◦ + sin 45◦ = sin 225◦ ? 38. See Example 1. −90◦ 6. 313◦ 4. 621◦ 7. Does cos 240◦ = (cos 120◦ )2 − (sin 120◦ )2 ? 41. sin θ = 1 For Exercises 32-36. Does cos 180◦ − cos 60◦ = cos 120◦ ? 40. −313◦ 5. 1. 275◦ For Exercises 22-26. 63◦ 19. cos θ = 1 For Exercises 27-31. −126◦ 20. Does tan 300◦ − tan 30◦ = tan 270◦ ? 39. ﬁnd the reference angle for the given angle.3 to answer the following questions. sin θ = − 23 31. ﬁnd the exact values of sin θ and tan θ when cos θ has the indicated value. use Table 1. tan θ = 1 2 33. 37.4 31 For an angle θ in standard position and a point (x. cos θ = − 12 24. 696◦ 21. y) on its terminal side: (a) sin θ has the same sign as y (b) cos θ has the same sign as x (c) tan θ is positive when x and y have the same sign (d) tan θ is negative when x and y have opposite signs Exercises For Exercises 1-10. In which quadrant(s) do sine and tangent have the opposite sign? 15.2. In which quadrant(s) do sine and cosine have the opposite sign? 13. ﬁnd the exact values of cos θ and tan θ when sin θ has the indicated value. 1 x x Rotation of an angle θ by 90◦ Thus. x) is on the terminal side of θ + 90◦ when (x. tan (θ + 90◦ ) = = − cot θ . we will take a look at how certain geometric operations can help simplify the use of trigonometric functions of any angle. y) is on the terminal side of θ . In general. and hence by similarity the remaining corresponding sides are also equal. To rotate an angle means to rotate its terminal side around the origin when the angle is in standard position. The two operations on which we will concentrate in this section are rotation and reﬂection. The rotation of θ by 90◦ does not change the length r of its terminal side. In Figure 1. This forces the other angle of the right triangle in QII to be θ .32 Chapter 1 • Right Triangle Trigonometry §1.5. cos (θ + 90◦ ) = = − sin θ . its complement.4) cos (θ + 90◦ ) = − sin θ (1.1 to match up those corresponding sides shows that the point (− y. y (x.1 we see an angle θ in QI which is rotated by 90◦ . x) θ r r θ + 90◦ x y 90◦ θ y Figure 1. the following relations hold for all angles θ : sin (θ + 90◦ ) = cos θ (1. and how some basic relations between those functions can be made.5 1.5) tan (θ + 90◦ ) = − cot θ (1. resulting in the angle θ + 90◦ in QII. so the hypotenuses of the similar right triangles are equal. Notice that the complement of θ in the right triangle in QI is the same as the supplement of the angle θ + 90◦ in QII. suppose we rotate an angle θ around the origin by 90◦ in the counterclockwise direction. Using Figure 1.5.5 Rotations and Reﬂections of Angles Now that we know how to deal with angles of any measure. by deﬁnition. and 90◦ equals 180◦ . the right triangle in QI is similar to the right triangle in QII. y) (− y.5. it is easy (see Exercise 4) to use similar arguments for the other quadrants. since the sum of θ . since the triangles have the same angles. Hence.6) . sin (θ + 90◦ ) = x x −y = cos θ . r r −y Though we showed this for θ in QI. For example. 4 means that tan θ = − −run = run = m (just imagine in Figure 1. Since the run is always positive.2 First.3(b) the entire line being shifted horizontally to go through the origin. y y y = mx + b y = mx + b 0 b θ rise < 0 rise > 0 x θ run > 0 run > 0 (a) m > 0.5 33 Example 1.5. That is. where m is the slope of the line (deﬁned as m = rise run ) and b is the y-intercept. For m > 0 we see that θ is acute and tan θ = rise run = m. then m 2 = − m11 (see Figure 1. where the line crosses the yaxis (see Figure 1. We will show that the slopes of perpendicular lines are negative reciprocals.3.5. rise < 0 Figure 1. The line crosses the x-axis somewhere.3 If m < 0.5. i.26 Recall that any nonvertical line in the x y-coordinate plane can be written as y = mx + b. suppose that a line y = mx + b has nonzero slope. so that θ is unchanged and the point (−run.2(b)).5.2(a)). then we see that θ is obtuse and the rise is negative. so let θ be the angle that the positive x-axis makes with the part of the line above the x-axis. rise rise our deﬁnition of tan θ from Section 1.5. −rise) is on the terminal side of θ ). y = mx + b y y y = m2 x + b2 rise b m= run 0 rise run x 0 y = m1 x + b1 b2 x b1 (a) Slope of a line (b) Perpendicular lines Figure 1. as in Figure 1.5.e. θ obtuse. θ acute.Rotations and Reﬂections of Angles • Section 1. rise > 0 0 x b (b) m < 0. if y = m 1 x + b 1 and y = m 2 x + b 2 are nonvertical and nonhorizontal perpendicular lines. Hence: For a line y = mx + b with m . m 1 = tan θ and m 2 = tan (θ + 90◦ ).5. Hence. where θ is the angle formed by the positive x-axis and the part of the line above the x-axis. m 2 = − cot θ = − tan1 θ = − m11 .= 0.6) we know that tan (θ + 90◦ ) = − cot θ . the slope is given by m = tan θ . So if θ is the angle that the line y = m 1 x + b 1 makes with the positive x-axis.2(b) we see that if two lines y = m 1 x + b 1 and y = m 2 x + b 2 are perpendicular then rotating one line counterclockwise by 90◦ around the point of intersection gives us the second line. in Figure 1. QED . Now. But by formula (1. So by what we just showed. then θ + 90◦ is the angle that the line y = m 2 x + b 2 makes with the positive x-axis. and it is equivalent to a rotation of 180◦ around the origin. y) is on the terminal side of θ . That image is the reﬂection around the origin of the original object. this gives us: sin (θ − 90◦ ) = − cos θ (1.9) We now consider rotating an angle θ by 180◦ . Since (θ − 90◦ ) + 90◦ = θ .5. Notice from Figure 1. y y (x. which hold for all θ : (1. and are in the quadrant opposite θ . Notice that if we ﬁrst reﬂect the object in QI around the y-axis and then follow that with a reﬂection around the x-axis.5 Rotating an angle θ by 90◦ in the clockwise direction results in the angle θ − 90◦ .5.8) tan (θ − 90◦ ) = − cot θ (1. − y) is on the terminal side of θ ± 180◦ when (x.6) hold for any angle θ . We could use another geometric argument to derive trigonometric relations involving θ − 90◦ .12) A reﬂection is simply the mirror image of an object.5. − y) r θ + 180◦ 180◦ θ x x ◦ r (− x.34 Chapter 1 • Right Triangle Trigonometry §1.5 x . but it is easier to use a simple trick: since formulas (1.4 Rotation of θ by ±180◦ Since (− x.4 that the angles θ ± 180◦ have the same terminal side. y) θ − 180◦ (a) QI and QIII (b) QII and QIV Figure 1.5. Notice also that a reﬂection around the y-axis is equivalent to a reﬂection around the x-axis followed by a rotation of 180◦ around the origin.7) cos (θ − 90◦ ) = sin θ (1. y) θ + 180◦ 180◦ r (− x. − y) −180 θ −180◦ θ − 180◦ r (x. For example. in Figure 1. and its reﬂection around the x-axis is in QIV.11) tan (θ ± 180◦ ) = tan θ (1.10) cos (θ ± 180◦ ) = − cos θ (1. its reﬂection around the y-axis is in QII. y c c sin (θ ± 180◦ ) = − sin θ c c Figure 1.4)−(1.5 the original object is in QI. just replace θ by θ − 90◦ in each formula. we get the following relations. we get an image in QIII. 16) cos (90◦ − θ ) = sin θ (1.15).6 Reﬂection of θ around the x-axis So we see that reﬂecting a point (x. formulas (1. Hence: sin (−θ ) = − sin θ (1.10)−(1. Replacing θ by −θ in formulas (1.20) tan (180◦ − θ ) = − tan θ (1.5.2 to all θ . Thus. In general. − y) (a) QI and QIV r −θ (b) QII and QIII Figure 1. y) on the terminal side of θ .16)−(1.19) cos (180◦ − θ ) = − cos θ (1. gives: sin (90◦ − θ ) = cos θ (1. we see that the reﬂection of an angle θ around the x-axis is the angle −θ . then using formulas (1. y) r θ r θ x −θ r (x. for a point (x. and not on y. − y) x (x.21) .15) Notice that the cosine function does not change in formula (1. not just acute angles.15) give: sin (180◦ − θ ) = sin θ (1. the cosine function is even.14) because it depends on x.18) extend the Cofunction Theorem from Section 1.5 35 Applying this to angles.6.17) tan (90◦ − θ ) = cot θ (1. y) around the x-axis just replaces y by − y.13) cos (−θ ) = cos θ (1.6).4)−(1. Similarly. as in Figure 1.18) Note that formulas (1.14) tan (−θ ) = − tan θ (1.13)−(1.12) and (1. a function f (x) is an even function if f (− x) = f (x) for all x.5.13)−(1. y y (x. y) (x.Rotations and Reﬂections of Angles • Section 1. and it is called an odd function if f (− x) = − f (x) for all x. while the sine and tangent functions are odd. − y) Reﬂection of θ around the y-axis = 180◦ − θ It may seem that these geometrical operations and formulas are not necessary for evaluating the trigonometric functions. Also. y) θ = −43◦ r (x. as in Figure 1. which is not between 0◦ and 360◦ . so they are often used to prove general formulas in mathematics and other ﬁelds.8 Reﬂection around the y-axis: −43◦ and 223◦ 7 In Chapter 5 we will discuss why the sin−1 button returns that value. However.682 as the input. we get θ = −43◦ . we know that −43◦ and −43◦ + 360◦ = 317◦ have the same trigonometric function values.27 Find all angles 0◦ ≤ θ < 360◦ such that sin θ = −0. First.7 Since θ = −43◦ is in QIV. Solution: Using the sin−1 button on a calculator with −0.7 (x.36 Chapter 1 • Right Triangle Trigonometry §1.682. y) 180 − θ r r θ x −θ r Figure 1. Second.7.5. the formulas work for any angles. y) Figure 1.682 are θ = 223◦ and 317◦ . .5.5. y) ◦ (x. But 180◦ − θ = 180◦ − (−43◦ ) = 223◦ (see Figure 1.5. y (− x. as we will see later in the text. they can help in determining which angles have a given trigonometric function value.8). its reﬂection 180◦ − θ around the y-axis will be in QIII and have the same sine value. since we could just use a calculator. we see that the only angles between 0◦ and 360◦ with a sine of −0. So since angles in QI and QII have positive sine values.5 Notice that reﬂection around the y-axis is equivalent to reﬂection around the x-axis (θ → −θ ) followed by a rotation of 180◦ (−θ → −θ + 180◦ = 180◦ − θ ). there are two reasons for why they are useful. y x 180◦ − θ = 223◦ r (− x. Example 1. Prove formulas (1.12) and (1. 0) D A (a. . Find the angle between 0◦ and 360◦ which is the (a) reﬂection of θ around the x-axis (b) reﬂection of θ around the y-axis (c) reﬂection of θ around the origin 2. y) coordinates of the point D in terms of a and b.9816 14. sin θ = 0 11.Rotations and Reﬂections of Angles • Section 1.14. Mimic that proof to show that the formulas hold for θ in QII. sin θ = 0. Show that this result can also be proved using the formulas involving θ − 90◦ . b) C (0. 3. We proved formulas (1.1909 9. 5. Repeat Exercise 1 with θ = −248◦ . y2 ) (x3 .15). we claimed that in a right triangle ABC it was possible to draw a line segment CD from the right angle vertex C to a point D on the hypotenuse AB such that CD ⊥ AB. y1 ) x 17. cos θ = −0. for θ = 0◦ . as in the picture below.6) for any angle θ in QI.7813 12. (Hint: Only the last paragraph in that example needs to be modiﬁed. tan θ = 0.26 we used the formulas involving θ + 90◦ to prove that the slopes of perpendicular lines are negative reciprocals. What is the slope of the hypotenuse? What would be the slope of a line perpendicular to it?) Also.514 15. y1 ) (x1 . ﬁnd the (x. ﬁnd all angles 0◦ ≤ θ < 360◦ which satisfy the given equation: 7.e. It can be proved without using trigonometric functions that the slopes of perpendicular lines are negative reciprocals.) For Exercises 7 . 0) x 16. Let y = m 1 x + b 1 and y = m 2 x + b 2 be perpendicular lines (with nonzero slopes). i.5 37 Exercises 1.13)−(1.6294 13.19)-(1. Let θ = 32◦ . tan θ = −9. sin θ = 0.4)-(1.4226 10.6) for θ on the coordinate axes. Verify formulas (1. (Hint: Notice how ABC is placed on the x y-coordinate plane. 6. Repeat Exercise 1 with θ = 248◦ .) y y = m2 x + b2 y = m1 x + b1 (x2 . (Hint: Think of similar triangles and the deﬁnition of slope. cos θ = 0. Use the picture on the right to prove that claim. 180◦ . 270◦ .21) by using formulas (1. sin θ = −0.4226 8.4)-(1. y1 ) (x3 . Use the picture to show that m 2 = − m11 . 90◦ . In Example 1.2. In our proof of the Pythagorean Theorem in Section 1. 4. y B (0.10)−(1. y2 ) (x2 . In this chapter we will learn how to solve a general triangle in all four of the above cases. by similarity an inﬁnite number of triangles have the same angles. Though the methods described will work for right triangles. and c opposite the angles A. b sin B 38 sin A a = . or one side and one acute angle. and C. b. respectively. sin A sin B sin C (2.1 The Law of Sines Theorem 2.1) can be written as sin A sin B sin C . For a general triangle. and Case 4 can be solved using the law of cosines. we could ﬁnd the remaining sides and angles.3) .1) Note that by taking reciprocals. we will again need three pieces of information. then b c a = = . B. In each case we were actually given three pieces of information. c sin C sin B b = c sin C (2. Cases 1 and 2 can be solved using the law of sines. 2.2 General Triangles In Section 1. since we already knew one angle was 90◦ .3 we saw how to solve a right triangle: given two sides. triangles which do not have a right angle. As we will see. which may or may not have a right angle. equation (2. The four cases are: Case 1: Case 2: Case 3: Case 4: One side and two angles Two sides and one opposite angle Two sides and the angle between them Three sides Note that if we were given all three angles we could not determine the sides uniquely. that is. Case 3 can be solved using either the law of cosines or the law of tangents. = = a b c (2.2) and it can also be written as a collection of three equations: a sin A = . they are mostly used to solve oblique triangles. Law of Sines: If a triangle has sides of lengths a. and an obtuse triangle has one obtuse angle.1. There are two types of oblique triangles: an acute triangle has all acute angles. 1.1. In each case. For each triangle in Figure 2. (2. or it can be obtuse. draw the altitude1 from the vertex at C to the side AB.5) and (in Figure 2. sin A sin B (2. we see that h = sin A b (2.1. To prove the Law of Sines. as in Figure 2.1(b) the altitude lies outside the triangle.4) h = sin B a (2.8) sin B sin C so putting the last two equations together proves the theorem.6) and so putting a and A on the left side and b and B on the right side.1(a) the altitude lies inside the triangle.7) By a similar argument.1 Proof of the Law of Sines for an oblique triangle ABC Let h be the height of the altitude.5).1.1 39 Another way of stating the Law of Sines is: The sides of a triangle are proportional to the sines of their opposite angles. b (2. while in Figure 2. let ABC be an oblique triangle.19) in Section 1. 1 Recall from geometry that an altitude of a triangle is a perpendicular line segment from any vertex to the line containing the side opposite the vertex.1.5) and substituting that into equation (2.1(b). QED Note that we did not prove the Law of Sines for right triangles.1.1. .4) gives a sin B = sin A . Thus. C b h C a b a h 180◦ − B A c B A (a) Acute triangle c B (b) Obtuse triangle Figure 2. drawing the altitude from A to BC gives c b = .The Law of Sines • Section 2. as in Figure 2. Then ABC can be acute. we get b a = . ha = sin (180◦ − B) = sin B by formula (1.1(b). solving for h in equation (2.1.1(a). In Figure 2. since it turns out (see Exercise 12) to be trivially true for that case. and we know the side b. Solve the triangle ABC given a = 18.2 ◦ A = 25 c = 14.1 Example 2.8 (a) B = 44.2◦ . we have to solve twice for C and c : once for B = 44.8◦ .8◦ .2◦ : B = 44.5 that sin (180◦ − B) = sin B.2◦ C = 180◦ − A − B = 180◦ − 25◦ − 135. C = 19. So there is a second possible solution for B.2.2◦ b = 30 a = 18 ◦ A = 25 c = 40 C = 19. Using the sin−1 button on a calculator gives B = 44.8◦ .7 . and a c = sin C sin A ⇒ c = a sin C 10 sin 75◦ = sin A sin 41◦ ⇒ c = 14.1. C = 110.7 . namely 180◦ − 44. and b = 30. A = 41◦ .8◦ Figure 2.8◦ = ⇒ c= = sin C sin A sin A sin 25◦ ⇒ c = 14.2◦ (b) B = 135.2◦ = 19. We can use the Law of Sines to ﬁnd the other A = 25◦ opposite angle B.2◦ = ⇒ c= = sin C sin A sin A sin 25◦ ⇒ c = 40 c a a sin C 18 sin 19. then ﬁnd the third angle C by subtracting A and B from 180◦ . recall from Section 1. c = 14. we see that ◦ ◦ ◦ ◦ B = 180 − A − C = 180 − 41 − 75 A = 41◦ ◦ ⇒ a = 10 b B = 64 c B .4 a = 18 B = 135.2◦ . and c.8◦ b = 30 ◦ B = 44. Solve the triangle ABC given a = 10.1 C = 75◦ Case 1: One side and two angles. C = 110. Solution: We can ﬁnd the third angle by subtracting the other two angles from 180◦ .8◦ and once for B = 135.40 Chapter 2 • General Triangles §2.8◦ = 110.4 are the two possible sets of solutions.4 Hence. and C = 75◦ .8◦ = 135. then use the law of sines to ﬁnd the third side c. However. b. as shown in Figure 2. This means that there are two possible triangles. B = 44. we have sin B sin A = b a ⇒ sin B = b sin A 30 sin 25◦ = a 18 ⇒ a = 18 c B sin B = 0. So by the Law of Sines we have b a = sin B sin A ⇒ b = a sin B 10 sin 64◦ = sin A sin 41◦ ⇒ b = 13. A = 25◦ . then use the law of sines to ﬁnd the two unknown sides. In this example we need to ﬁnd B.2◦ Two possible solutions . Thus.7044 .8◦ c a a sin C 18 sin 110. First.2◦ C = 180◦ − A − B = 180◦ − 25◦ − 44. C b = 30 Solution: In this example we know the side a and its opposite angle A.2 Case 2: Two sides and one opposite angle.2◦ .1. c = 40 and B = 135. By the Law of Sines.8◦ B = 135. Example 2. there is no solution when a < h (this was the case in Example 2. A = 30◦ . For a triangle ABC.4).3(d) can not be used to determine B.3(d) that there may be two solutions. As we can see in Figure 2. A is obtuse or a right angle). That is. However. the point of intersection to the left of A in Figure 2. It is also possible for there to be exactly one solution or no solution at all.3 The ambiguous case when A is acute If A is acute. suppose that we know the sides a and b and the angle A.1. and b = 12. we have sin A sin B = b a ⇒ sin B = b sin A 12 sin 30◦ = a 5 ⇒ sin B = 1. a right triangle .1. then the altitude from C to AB has height h = b sin A.1 41 In Example 2. which is impossible since \| sin B \| ≤ 1 for any angle B. and imagine that the side a is attached at the vertex at C so that it can “swing” freely.3).1. There is a way to determine how many solutions a triangle has in Case 2. even though it appears from Figure 2.namely.1. If A is not acute (i. . since the dashed arc intersects the horizontal line at two points. Solution: By the Law of Sines.2). there may be more than one solution. C b A C a b h B (a) a < h: No solution b h a c A C a a B A (b) a = h: One solution B h B (c) h < a < b: Two solutions C b a b A c B (d) a ≥ b: One solution Figure 2. since that would make A an obtuse angle.when a = h.2 we saw what is known as the ambiguous case. and we assumed that A was acute. Thus.2 .1.3 Case 2: Two sides and one opposite angle. then the situation is simpler: there is no solution if a ≤ b. and there are two solutions when h < a < b (as was the case in Example 2. and there is exactly one solution if a > b (see Figure 2. Solve the triangle ABC given a = 5.The Law of Sines • Section 2. there is no solution .3(a)-(c). as indicated by the dashed arc in Figure 2. Draw the angle A and the side b.1. there is exactly one solution . Example 2. When a ≥ b there is only one solution.e.3 below. 1 C a a b b B A (a) a ≤ b: No solution Figure 2. . the largest side is opposite the largest angle. let C be the largest angle in a triangle ABC. this means that the largest side is opposite the largest angle. imagine that A is in standard position in the x y-coordinate plane and that we rotate the terminal side of A counterclockwise to the terminal side of the larger angle C. Of course. then A and B are also acute. y1 ) A x Figure 2. A.1.1.1 summarizes the ambiguous case of solving ABC when given a. y1 ) and (x2 . respectively. and b. So we just need to prove the result for when C is acute and for when C is obtuse. then we see from the picture that y1 ≤ y2 . Recall from Section 1. Since A ≤ C. In both cases. Since a right angle is the largest angle in a right triangle.1 that in a right triangle the hypotenuse is the largest side. we have A ≤ C and B ≤ C. To prove this. Table 2. If C is acute. y2 ) r C r (x1 . We will ﬁrst show that sin A ≤ sin C and sin B ≤ sin C. If C = 90◦ then we already know that its opposite side c is the largest side.5. r r y (x2 .g.1. sin A ≤ sin C and sin B ≤ sin C when C is acute. as in Figure 2. so that their distance to the origin is the same number r. B ≤ C implies that sin B ≤ sin C. e. the letters can be interchanged. and hence sin A = y1 y2 ≤ = sin C . etc. If we pick points (x1 . Thus.5 By a similar argument. We will now show that these inequalities hold when C is obtuse. What the Law of Sines does is generalize this to any triangle: In any triangle.4 A B (b) a > b: One solution The ambiguous case when A ≥ 90◦ Table 2.1 Summary of the ambiguous case 0◦ < A < 90◦ a < b sin A : No solution a = b sin A : One solution b sin A < a < b : Two solutions a ≥ b : One solution 90◦ ≤ A < 180◦ a ≤ b : No solution a > b : One solution There is an interesting geometric consequence of the Law of Sines. replace a and A by c and C.42 Chapter 2 • General Triangles C §2. y2 ) on the terminal sides of A and C. c and the angles A. and 10 seconds later measures an angle of inclination of 55◦ . b = 9 7. sin A ≤ sin C and sin B ≤ sin C when C is obtuse. c sin C sin C c By a similar argument. = c sin C a sin (B + C) 14. B. sinb B . c = 27 10. Prove the Law of Sines for right triangles. A = 94◦ . Use a ruler and a protractor to measure the sides a. a = 12. a = 15. b ≤ c. The Law of Sines says that the ratios sina A . C of the triangle. b = 40. Thus. a = 5. Thus. A = 42◦ . b = 12 9. Hence. a = 40. c = 30 6. 13. as are A and B. So by what we showed above for acute angles. Thus. c is the largest side. But we know from Section 1. B = 75◦ . A = 40◦ . so by the Law of Sines we have sin A sin C a a = ≤ = 1 ⇒ ≤ 1 ⇒ a ≤ c. For a triangle ABC. Explain why in Case 1 (one side and two angles) there is always exactly one solution. B ≤ 180◦ − C. B = 45◦ . If the airplane is ﬂying at a constant speed and at a steady altitude of 6000 ft in a straight line directly over the observer. show that .1 43 If C is obtuse. . solve the triangle ABC.e. What relation does that common ratio have to the diameter of your circle? 11. A = 25◦ . c = 35 3. we must have A ≤ 180◦ − C. (Note: 1 mile = 5280 ft) 10 seconds pass ✈ ✈ 55◦ 6000 ft 30◦ 12. (Hint: One of the angles is known. b = 7 5. a ≤ c and b ≤ c. sin A ≤ sin C if C is acute or obtuse. If A > 180◦ − C then A + C > 180◦ . a = 10. B = 25◦ 2.The Law of Sines • Section 2. A = 47◦ .5 that sin C = sin (180◦ − C). One diagonal of a parallelogram is 17 cm long and makes angles of 36◦ and 15◦ with the sides. For a triangle ABC. QED Exercises For Exercises 1-9. b. which is impossible. Verify this for your triangle. we know that sin A ≤ sin (180◦ − C) and sin B ≤ sin (180◦ − C). a = 22. i. Find the lengths of the sides. = c sin C 15. then 180◦ − C is acute. c = 15 4. b = 15 8.) sin A ± sin B a±b . a = 5. A = 35◦ . 1. An observer on the ground measures an angle of inclination of 30◦ to an approaching airplane. show that 16. sinc C are equal. Draw a circle with a radius of 2 inches and inscribe a triangle inside the circle. A = 94◦ . Likewise. ﬁnd the speed of the airplane in miles per hour. A = 50◦ . 2.10) 2 2 2 (2. Solve the triangle ABC given A = 30◦ .2. and c opposite the angles A. C b=4 a Solution: Using the Law of Sines. we can use the Law of Cosines: Theorem 2. In Figure 2.2.44 Chapter 2 • General Triangles §2. making the problem impossible to solve. respectively.2. b. let ABC be an oblique triangle.2.1(a) the altitude divides AB into two line segments with lengths x and c − x. c = a + b − 2ab cos C .4 Case 3: Two sides and the angle between them. Let h be the height of the altitude. In each case.1(b) the altitude extends the side AB by a distance x. (2. Example 2.2 The Law of Cosines We will now discuss how to solve a triangle in Case 3: two sides and the angle between them. then a2 = b2 + c2 − 2bc cos A .1(b). Law of Cosines: If a triangle has sides of lengths a. while in Figure 2. But that would just result in the equation sina30◦ = 5 sin C . B. and C. we have A = 30◦ 4 5 a = = . as in Figure 2.9) 2 2 2 (2.2.1 Proof of the Law of Cosines for an oblique triangle ABC h . as in Figure 2. which we already knew and which still has two unknowns! Thus. First.2 2.11) b = c + a − 2ca cos B . For example. this problem can not be solved using the Law of Sines. draw the altitude from the vertex at C to the side AB. let us see what happens when we try to use the Law of Sines for this case. To solve the triangle in the above example. To prove the Law of Cosines. or it can be obtuse. C C b c−x A a h b a x c (a) Acute triangle B A c 180◦ − B x B (b) Obtuse triangle Figure 2. Then ABC can be acute. b = 4. sin 30◦ sin B sin C c=5 B where each of the equations has two unknown parts. and c = 5.1(a). to solve for a we could use the equation sin4 B = sin5 C to solve for sin B in terms of sin C and substitute that into the equation sina30◦ = sin4 B . One cycle will give you the second formula. we see by the Pythagorean Theorem that h2 = a2 − x2 (2. since it turns out (see Exercise 15) that all three formulas reduce to the Pythagorean Theorem for that case.2. and replace c by a (likewise for the capital letters). (2. and another cycle will give you the third. since the other two can be obtained by “cycling” through the letters a. notice that it sufﬁces to remember just one of the three formulas (2. .9)-(2. substituting the expression for h2 in equation (2. QED Note that we did not prove the Law of Cosines for right triangles. replace b by c.5 that cos (180◦ − B) = − cos B. But we see from Figure 2.13) Thus. x = −a cos B and so b2 = a2 + c2 − 2ca cos B .11).1(a) that x = a cos (180◦ − B).1.2.2 45 For each triangle in Figure 2.2. Thus. But we see from Figure 2.1(a) that x = a cos B.12) into equation (2. so b2 = a2 + c2 − 2ca cos B .12) into equation (2. By similar arguments for A and C we get the other two formulas.16) So for both acute and obtuse triangles we have proved formula (2.The Law of Cosines • Section 2. Notice that the proof was for B acute and obtuse.14) And for the obtuse triangle in Figure 2.1(b) we see that b2 = h2 + (c + x)2 .1(a) we see that b2 = h2 + (c − x)2 .2. and we know from Section 1. b. (2. That is. and c.2. (2. (2. Also.10) in the Law of Cosines.13) gives b2 = a2 − x2 + (c − x)2 = a2 − x2 + c2 − 2cx + x2 = a2 + c2 − 2cx .15) Thus.15) gives b2 = a2 − x2 + (c + x)2 = a2 − x2 + c2 + 2cx + x2 = a2 + c2 + 2cx . The Law of Cosines can be viewed as a generalization of the Pythagorean Theorem. replace a by b. substituting the expression for h2 in equation (2.12) and likewise for the acute triangle in Figure 2. 5 Case 3: Two sides and the angle between them. and c = 5. to ﬁnd the angle B. b = 4. regardless of the size of the angle. we have 2 2 A = 30◦ a c=5 B 2 a = b + c − 2bc cos A = 42 + 52 − 2(4)(5) cos 30◦ = 6.6091 2(5)(2. and if the cosine is negative then the angle is obtuse.5◦ a 2.5 that there was only one solution. as we saw in Section 2. since it would make A + B = 157. then we have a formula for the square of the third side. which (as we learned in Section 2. But C = 22.52 .5.5: b sin A 4 sin 30◦ = = 0. both an acute angle and its obtuse supplement have the same positive sine. Why not use the Law of Sines.1) means that C has to be the largest angle.5◦ . To see this.52) cos B = B = 52.4.5◦ would not be the largest angle in this solution.52 How would we know which answer is correct? We could not immediately rule out B = 127. However. C b=4 Solution: We will use the Law of Cosines to ﬁnd a. This is in contrast to using the sine function.36 ⇒ a = 2. c. sin B = .5◦ or 127.5◦ Thus.5◦ ⇒ C = 97.46 Chapter 2 • General Triangles §2. which has a simpler formula? The reason is that using the cosine function eliminates any ambiguity: if the cosine is positive then the angle is acute. The reason is simple: when joining two line segments at a common vertex to form an angle. b. First. which seems like it could be a valid solution.g. We will now solve the triangle from Example 2. then use C = 180◦ − A − B.5◦ as too large. 2 Now we use the formula for b to ﬁnd B: b2 = c2 + a2 − 2ca cos B ⇒ ⇒ ⇒ c 2 + a2 − b 2 2ca 52 + (2. this solution is impossible. Solve the triangle ABC given A = 30◦ . use it again to ﬁnd B. and A). Notice in Example 2.1. there is exactly one way to connect their free endpoints with a third line segment. For Case 3 this will always be true: when given two sides and their included angle. the triangle will have exactly one solution. Notice in the Law of Cosines that if two sides and their included angle are known (e. Why? Because the largest side in the triangle is c = 5. C = 180◦ − A − B = 180◦ − 30◦ − 52.5◦ < 180◦ and so C = 22. suppose that we had used the Law of Sines to ﬁnd B in Example 2.2 The angle between two sides of a triangle is often called the included angle. You may be wondering why we used the Law of Cosines a second time in Example 2.7937 ⇒ B = 52. and hence we have a contradiction. Example 2.52)2 − 42 cos B = = 0.5◦ . as the following example shows. The following example gives an idea of how to do this. We will now see how to use the Law of Cosines for that case.5◦ Thus. and c = 4. Solve the triangle ABC given a = 2.9◦ . there is no solution . It may seem that there is always a solution in Case 4 (given all three sides). Example 2. 2bc 2(3)(6) c=6 B which is impossible since \| cos A \| ≤ 1. though it is less commonly used for that purpose than the Law of Sines. b=3 a=2 c=6 The Law of Cosines can also be used to solve triangles in Case 2 (two sides and one opposite angle).7 Case 4: Three sides.6875 2(4)(2) cos B = ⇒ B = 46.6◦ ⇒ cos C = Now we use the formula for c2 to ﬁnd C: c2 = a2 + b2 − 2ab cos C ⇒ ⇒ a2 + b 2 − c 2 2ab 22 + 32 − 42 cos C = = −0. C b=3 a=2 Solution: If we blindly try to use the Law of Cosines to ﬁnd A. A = 180◦ − B − C = 180◦ − 46. i.2 47 It remains to solve a triangle in Case 4.6 Case 4: Three sides. C Solution: We will use the Law of Cosines to ﬁnd B and C.5◦ ⇒ A = 28. Example 2. First. b = 3.25 2(2)(3) C = 104.The Law of Cosines • Section 2. Thus. given three sides.139 . but that is not true. then use A = 180◦ − B − C.6◦ − 104. we get 2 2 2 a = b + c − 2bc cos A ⇒ A b 2 + c 2 − a2 32 + 62 − 22 cos A = = = 1. Solve the triangle ABC given a = 2. . b = 3. and c = 6.e. We could have saved ourselves some effort by recognizing that the length of one of the sides (c = 6) is greater than the sums of the lengths of the remaining sides (a = 2 and b = 3). we use the formula for b2 to ﬁnd B: b2 = c2 + a2 − 2ca cos B 2 ⇒ ⇒ 2 b=3 A a=2 c=4 B 2 c + a − b 2ca 42 + 22 − 32 cos B = = 0. which (as the picture on the right shows) is impossible in a triangle. c2 + d 2 = a2 + b2 − 2ab cos C + a2 + b2 + 2ab cos C QED c a c2 + d 2 = a2 + b2 + a2 + b2 = 2 (a2 + b2 ) . or two real roots (corresponding to the number of solutions in Case 2).8 Case 2: Two sides and one opposite angle.8◦ . and let the diagonals opposite the angles C and D have lengths c and d. Thus.2◦ .2. c = 14.7111 2ca 2(40)(18) ⇒ B = 44. Like the Law of Sines. we have 54. which is close to what we found before (the small difference being due to different rounding). we know that c2 = a2 + b2 − 2ab cos C .9 Use the Law of Cosines to prove that the sum of the squares of the diagonals of any parallelogram equals the sum of the squares of the sides. and d 2 = a2 + b2 − 2ab cos D . A = 25◦ .2. b D b Figure 2. For c = 40 we get cos B = c 2 + a2 − b 2 402 + 182 − 302 = = 0. To solve this using the Law of Cosines.48 Chapter 2 • General Triangles §2. By properties of parallelograms. C = 19.8◦ . = 2 (a2 + b2 ) .2 Example 2.1 we used the Law of Sines to show that there are two sets of solutions for this triangle: B = 44. Then we need to show that a d C By the Law of Cosines. c = 40 and B = 135. Solution: In Example 2.2◦ . C = 110. and b = 30. ﬁrst ﬁnd c by using the formula for a2 : a2 = b2 + c2 − 2bc cos A ⇒ 182 = 302 + c2 − 2(30)c cos 25◦ ⇒ c2 − 54. c = 2(1) Note that these are the same values for c that we found before. the Law of Cosines can be used to prove some geometric facts.38 ± (54. Solve the triangle ABC given a = 18. we know that D = 180◦ − C. Example 2.38)2 − 4(1)(576) = 40 or 14. so d 2 = a2 + b2 − 2ab cos (180◦ − C) = a2 + b2 + 2ab cos C . as in Figure 2. respectively. Solution: Let a and b be the lengths of the sides. which is a quadratic equation in c. so we know that it can have either zero. The other solution set can be obtained similarly.2 . since cos (180◦ − C) = − cos C. one. as in the following example.4.4 .2.2 from Section 2.38 c + 576 = 0 . By the quadratic formula.7◦ ⇒ C = 110.3◦ . cos A + cos B + cos C = a2 (b + c − a) + b2 (a + c − b) + c2 (a + b − c) .) 8. solve the triangle ABC. . Find the distance between the centers of the circles.9)-(2. At either point of intersection. Show that the sum of the squares of the three medians of a triangle is 3⁄4 the sum of the squares of the sides.11)) for right triangles. a = 6. respectively. Recall from elementary geometry that a median of a triangle is a line segment from any vertex to the midpoint of the opposite side. 1. 10. 16. 60◦ 3. a b c 2abc 14. a = 11.4 Exercise 11 11. even if one of the cosines is negative. b = 8. B = 60◦ . Find the angles of the triangle whose vertexes are the centers of the circles. Two trains leave the same train station at the same time.2 15. If one train travels at an average speed of 100 mi/hr and the other at an average speed of 90 mi/hr. b = 3. respectively. c = 16 7. b = 4. Two circles of radii 5 and 3 cm.2 49 Exercises For Exercises 1-6. how far apart are the trains after half an hour? 9. a = 7. 5. b = 13.3 below. Show that for any triangle ABC. c2 < a2 + b2 if C is acute.The Law of Cosines • Section 2. (Hint: The diagonals bisect each other. moving along straight tracks that form a 35◦ angle. 12.4 above. c = 1 6. c = 6 3. c2 > a2 + b2 if C is obtuse. A = 30◦ . a = 7. 2 It turns out that 1 < cos A + cos B + cos C ≤ 3/2 for any triangle.2. c = 9 4. Use the Law of Cosines to show that for any triangle ABC. c = 9 5.2.3 Exercise 10 Figure 2.5 Figure 2. the tangent lines to the circles form a 60◦ angle.5 4 x 5 6 2 3 5. Show that for any triangle ABC. Find the length x of the diagonal of the quadrilateral in Figure 2. c = 12 2. b = 4. as we will see later. 13. Prove the Law of Cosines (i.2.2. Find the lengths of the sides of the parallelogram. The diagonals of a parallelogram intersect at a 42◦ angle and have lengths of 12 and 7 cm. cos A cos B cos C a2 + b 2 + c 2 + + = . intersect at two points.e. Three circles with radii of 4. are tangent to each other externally. 2abc What do the terms in parentheses represent geometrically? Use your answer to explain why cos A + cos B + cos C > 0 for any triangle. and c2 = a2 + b2 if C is a right angle. A = 60◦ . formulas (2. and 6 cm. as in Figure 2. 5. Let E be the angle of elevation from the earth station to the satellite. 18. Satellite Communications. an earth station.2. 1986.5 Use the Law of Cosines to show that d = rs re 1+ rs 2 re −2 cos γ .2. 22-25 in T. and let d be the distance from the earth station to the satellite. B OSTIAN. Suppose that a satellite in space.2 17. 2 re re 1+ −2 cos γ rs rs cos E = Note: This formula allows the angle of elevation E to be calculated from the coordinates of the earth station and the subsatellite point (where the line from the satellite to the center of the earth crosses the surface of the earth).11) in our statement of the Law of Cosines. and the center of the earth all lie in the same plane. .3 3 See pp. P RATT AND C. New York: John Wiley & Sons. The Dutch astronomer and mathematician Willebrord Snell (1580-1626) wrote the Law of Cosines as 2ab 1 = 2 2 1 − cos C c − (a − b) in his trigonometry text Doctrina triangulorum (published a year after his death). let r s be the distance from the center of the earth to the satellite (called the orbital radius of the satellite). rs and then use E = ψ − 90◦ and the Law of Sines to show that sin γ . Let r e be the radius of the earth. and let γ and ψ be the angles shown in Figure 2. Show that this formula is equivalent to formula (2. satellite ✸ local horizontal d rs E ❚ earth station ψ γ re center of earth Figure 2.50 Chapter 2 • General Triangles §2.W. while formula (2. and c opposite the angles A.7◦ We can ﬁnd the remaining side c by using the Law of Sines: c = a sin C 5 sin 96◦ = sin A sin 54.20) is more convenient when b > a. we can switch the order of the letters in each of the above formulas. a−b = a+b tan tan 1 2 (A − B) 1 2 (A + B) ⇒ ⇒ ⇒ b=3 a=5 c A tan 12 (A − B) 5−3 = 5+3 tan 12 (84◦ ) tan 12 (A − B) = 28 tan 42◦ = 0.3 51 2.2251 ◦ 1 ⇒ A − B = 25. B. We now have two equations involving A and B. Thus.17) tan 12 (B − C) b−c .4◦ ⇒ A = 54.17) as tan 12 (B − A) b−a .09 ⇒ B = 29. and C = 96◦ . Example 2. = c+a tan 12 (C + A) (2.3. we can rewrite formula (2. = a+b tan 12 (A + B) (2. then tan 12 (A − B) a−b .7 .4◦ A + B = 84◦ −− − − − − − − 2A = 109.3 The Law of Tangents We have shown how to solve a triangle in all four cases discussed at the beginning of this chapter.The Law of Tangents • Section 2. For example. = b+c tan 12 (B + C) (2. = b+a tan 12 (B + A) (2. An alternative to the Law of Cosines for Case 3 (two sides and the included angle) is the Law of Tangents: Theorem 2.10 C = 96◦ Case 3: Two sides and the included angle.20) and similarly for the other formulas. Solution: A + B + C = 180◦ .7◦ ⇒ c = 6.3◦ B . so A + B = 180◦ − C = 180◦ − 96◦ = 84◦ .17). respectively. which we can solve by adding the equations: A − B = 25. Law of Tangents: If a triangle has sides of lengths a.18) tan 12 (C − A) c−a . b = 3.19) Note that since tan (−θ ) = − tan θ for any angle θ . If a > b.7◦ ⇒ B = 84◦ − 54. by the Law of Tangents. b. and C. then it is usually more convenient to use formula (2. Solve the triangle ABC given a = 5.4◦ 2 (A − B) = 12. 3285 ✓ The small difference (≈ 0. which works even when the sides are equal.17) would be 0 (since tan 0◦ = 0). 5 Named after the German astronomer and mathematician Karl Mollweide (1774-1825).21): sin 12 (A − B) a−b = c cos 12 C sin 12 (54.09 cos 48◦ 0. B = 29. making computation easier. For this reason. This means that the Law of Tangents is of no help in Case 3 when the two known sides are equal. sin 12 (A − B) a−b . c sin 12 C (2.7◦ = 6.09 cos 12 (96◦ ) 2 sin 12.21) cos 12 (A − B) a+b = . Example 2.3◦ ) 5−3 = 6. and so both sides of formula (2. so we can conclude that both sides of the equation agree. the Law of Tangents was more popular than it is today since it lent itself better than the Law of Cosines to what was known as logarithmic computation. either equation can be used to check a solution of a triangle. and hence the solution is correct.3284 = 0.10.09. Solution: Recall that the full solution was a = 5. the Law of Tangents seems to have fallen out of favor in trigonometry books lately. requires slightly fewer steps. and is perhaps more straightforward. then we know that the solution is correct.7◦ . For this reason.22) Note that all six parts of a triangle appear in both of Mollweide’s equations. b = 3. 4 Before the advent of electronic calculators. and perhaps also because of the somewhat unusual way in which it is used. computations with large numbers were handled by taking logarithms and looking up values in a logarithm table.11 Use one of Mollweide’s equations to check the solution of the triangle from Example 2.7◦ − 29.3 Note that in any triangle ABC. If both sides of the equation agree (more or less).52 Chapter 2 • General Triangles §2.3◦ . A = 54. and = c cos 12 C (2.4 Related to the Law of Tangents are Mollweide’s equations:5 Mollweide’s equations: For any triangle ABC.0001) is due to rounding errors from the original solution. It does not seem to have any advantages over the Law of Cosines. We will check this with equation (2. Ratios (such as in the Law of Tangents and the Law of Sines) could be replaced by differences of logarithms. if a = b then A = B (why?). c = 6. and C = 96◦ . . In those days. 2◦ .3 53 Example 2. We will prove the Law of Tangents and Mollweide’s equations in Chapter 3. C = 111. A = 30◦ . use the Law of Tangents to solve the triangle ABC. a = 12. c = 6 3. The quadrilateral has eight parts: four sides and four angles. c = 9. b = 8. show that the triangle ABC is isosceles. c = 9.The Law of Tangents • Section 2. 1. A = 39◦ .5◦ 7. A = 19. and C = 65◦ ? Solution: Before using Mollweide’s equations. For any triangle ABC. c = 10. c = 9.6◦ 6. B = 68. R IDER. For any triangle ABC. where we will be able to supply brief analytic proofs.5◦ . show that c = b cos A + a cos B .2◦ . respectively. B = 60◦ .6 Exercises For Exercises 1-3. C = 70. Let ABCD be a quadrilateral which completely contains its two diagonals. A = 27. If b cos A = a cos B . check if it is possible for a triangle to have the given parts. 1 2C. a = 6. So check with Mollweide’s equation (2. b = 7. C = 60◦ 2. show that tan 1 a− b 2 (A − B) = a+ b cot a− b a+ b .86 ✗ Here the difference is far too large. b = 4. What is the smallest number of parts that you would need to know to solve the quadrilateral? Explain your answer. a = 7. (Hint: Draw the altitude from the vertex c − a cos B C to AB.) Notice that this formula provides another way of solving a triangle in Case 3 (two sides and the included angle). a sin B . In this case all those conditions hold. B = 70.44 = 1. 9.5◦ . Show that tan 12 (A − B) = 8. New York: The Macmillan Company. show that tan A = 10. C = 92. 6 There are (complex) geometric proofs of the Law of Tangents and Mollweide’s equations. B = 40.R. Plane and Spherical Trigonometry. so we conclude that there is no triangle with these parts.5◦ 1.7◦ .5◦ ) = 9 sin 32. This is another check of a triangle. simpler checks are that the angles add up to 180◦ and that the smallest and largest sides are opposite the smallest and largest angles. b = 9. 1942. a = 5. 11. . See pp.22): cos 12 (A − B) a+b = c sin 12 C cos 12 (55◦ − 60◦ ) 6+7 = 9 sin 12 (65◦ ) 13 cos (−2. Let ABC be a right triangle with C = 90◦ . b = 7. b = 7.8◦ 5. c = 9 For Exercises 4-6. 12. For any triangle ABC. B = 60◦ . a = 3. A = 55◦ .12 Can a triangle have the parts a = 6. 96-98 in P. 4. 1 a b c A B (c) A obtuse Area of ABC In each case we draw an altitude of height h from the vertex at C to AB.1(b). b = 5. in which A can be either acute. Case 1: Two sides and the included angle.25) Notice that the height h does not appear explicitly in these formulas. Example 2.54 Chapter 2 • General Triangles §2. or obtuse.4 The Area of a Triangle In elementary geometry you learned that the area of a triangle is one-half the base times the height. Suppose that we have a triangle ABC. C b=5 a Solution: Using formula (2. We will now use that.4. C b h c A C C a a b h B h c A B ◦ (b) A = 90 (a) A acute Figure 2. These formulas have the advantage of being in terms of parts of the triangle. so that the area (which we will denote by the letter K) is given by K = 12 hc.1. combined with some trigonometry. without having to ﬁnd h separately. b. We thus get the following formula: Area = K = 1 2 bc sin A (2. But we see that h = b sin A in each of the triangles (since h = b and sin A = sin 90◦ = 1 in Figure 2.13 Find the area of the triangle ABC given A = 33◦ .23). and h = b sin (180◦ − A) = b sin A in Figure 2. b. to derive more formulas for the area when given various parts of the triangle. and c are known.53 A = 33◦ c=7 B . a right angle.23) The above formula for the area of ABC is in terms of the known parts A. Assume that A.4.4. and c.4. and c = 7. the area K is given by: K = = 1 2 1 2 bc sin A (5)(7) sin 33◦ K = 9. Similar arguments for the angles B and C give us: Area = K = Area = K = 1 2 1 2 ac sin B (2. as in Figure 2.1(c)).4 2. although it is implicitly there.24) ab sin C (2. Solution: Using formula (2.). Then the area K of the triangle is Area = K = s (s − a) (s − b) (s − c) . (2.7 By the Law of Sines we know that c = a sin C .29) To prove this.D. (2.58 K = C = 40◦ a = 12 b A = 115◦ c B = 25◦ Case 3: Three sides. 2s = a + b + c is the perimeter of the triangle).4.26) Similar arguments for the sides b and c give us: b2 sin A sin C 2 sin B 2 c sin A sin B Area = K = 2 sin C Area = K = (2. and all three angles are known.e. ﬁrst remember that the area K is one-half the base times the height.30) 7 Note that this is equivalent to knowing just two angles and a side (why?). . B = 25◦ . say. sin A so substituting this into formula (2. 8 Due to the ancient Greek engineer and mathematician Heron of Alexandria (c.24) we get: Area = K = a2 sin B sin C 2 sin A (2. as before in Figure 2. Using c as the base and the altitude h as the height. C = 40◦ . Suppose that we have a triangle ABC in which all three sides are known. and c. Suppose that we have a triangle ABC in which one side.28) Example 2.1. b.14 Find the area of the triangle ABC given A = 115◦ . Then Heron’s formula8 gives us the area: Heron’s formula: For a triangle ABC with sides a. and a = 12. 10-70 A. Squaring both sides gives us K 2 = 14 h2 c2 .27) (2. let s = 12 (a + b + c) (i. the area K is given by: a2 sin B sin C 2 sin A 122 sin 25◦ sin 40◦ = 2 sin 115◦ K = 21. a.4 55 Case 2: Three angles and any side.26).The Area of a Triangle • Section 2. we have K = 21 hc. 32) By the Law of Cosines we know that b 2 + c 2 − a2 2bc + b2 + c2 − a2 (b + c)2 − a2 ((b + c) + a) ((b + c) − a) = = = 2bc 2bc 2bc 2bc (a + b + c) (b + c − a) = .2. And in Figure 2. C C b h a c D A h D B (a) A acute Figure 2.4. in either case we have (AD)2 = b2 (cos A)2 . 2bc 1 + cos A = 1 + and similarly b 2 + c 2 − a2 2bc − b2 − c2 + a2 a2 − (b − c)2 (a − (b − c)) (a + (b − c)) = = = 2bc 2bc 2bc 2bc (a − b + c) (a + b − c) = .32). (2.2(a). we have K 2 = 14 b2 c2 (1 + cos A) (1 − cos A) . Thus. so upon taking square roots we get K = s (s − a) (s − b) (s − c) .4 In Figure 2. (2. we see that K 2 = s (s − a) (s − b) (s − c) .2(b) we see that AD = b cos (180◦ − A) = − b cos A.31) (Note that the above equation also holds when A = 90◦ since cos 90◦ = 0 and h = b). QED . Hence.4.56 Chapter 2 • General Triangles §2. and so h2 = b2 − b2 (cos A)2 = b2 (1 − (cos A)2 ) = b2 (1 + cos A) (1 − cos A) .4.4. we see that AD = b cos A. substituting equation (2. In Figure 2.30). let D be the point where the altitude touches AB (or its extension).2 a b A c B (b) A obtuse Proof of Heron’s formula By the Pythagorean Theorem. = · · · 2 2 2 2 K2 = 1 4 b2 c2 and since we deﬁned s = 12 (a + b + c).31) into equation (2. we see that h2 = b2 − (AD)2 . 2bc 1 − cos A = 1 − Thus. substituting these expressions into equation (2. we have (a + b + c) (b + c − a) (a − b + c) (a + b − c) · 2bc 2bc a+b+ c b+ c−a a−b+ c a+b− c . However.The Area of a Triangle • Section 2.pdf . which has 14 digits.is K = 0. Solution: To use Heron’s formula. This is a problem because a = 1000000.berkeley. some calculators (e.33) To use this formula.15 Find the area of the triangle ABC given a = 5.g.16 Find the area of the triangle ABC given a = 1000000. The actual area . For the triangle in Example 2. it is basically 1. as in the following example.9 Luckily there is a better formula10 for the area of a triangle when the three sides are known: For a triangle ABC with sides a ≥ b ≥ c.com/cd/E19957-01/806-3568/ncg_goldberg. Then perform the operations inside the square root in the exact order in which they appear in the formula.4 57 Example 2. and c = 7.g. Most calculators can store 12-14 digits internally (even if they display less). and c = 0. b = 999999. There is an excellent overview of this important subject in the article What Every Computer Scientist Should Know About Floating-Point Arithmetic by D.8 .16.html 10 Due to W.accurate to 15 decimal places . What is amazing about this formula is that it is just Heron’s formula rewritten! The use of parentheses is what forces the correct order of operations for numerical stability.0000029. depending on how they store values internally.eecs. Example 2. and so we would get s − a = 0.0000008. Then take the square root and divide by 4.99999999999895. Solution: Using Heron’s formula with s = 12 (a + b + c) = 12 (5 + 4 + 7) = 8. Other calculators may give some other inaccurate answer. and hence may round off that value of a + b + c to 2000000. b = 4.edu/~wkahan/Triangle. the TI-83 Plus) will give a rounded down value of 1000000. available at http://docs. exhibiting what is called numerical instability for “extreme” triangles. Notice that the actual value of a + b + c is 2000000. i. the area K is given by: K = s (s − a) (s − b) (s − c) = 8 (8 − 5) (8 − 4) (8 − 7) = 96 ⇒ K = 4 6 ≈ 9. When we then divide that rounded value for a + b + c by 2 to get s. C b=4 A a=5 c=7 B Heron’s formula is useful for theoretical purposes (e. in deriving other formulas). 9 This is an issue even on modern computers. it is not well-suited for calculator use. we need to calculate s = 21 (a + b + c). causing Heron’s formula to give us an area of 0 for the triangle! And this is indeed the incorrect answer that the TI-83 Plus returns. Goldberg.9999979.e. sort the names of the sides so that a ≥ b ≥ c. the area is: Area = K = 14 (a + (b + c)) (c − (a − b)) (c + (a − b)) (a + (b − c)) (2. Kahan: http://www. The above example shows how problematic ﬂoating-point arithmetic can be. including the use of parentheses. the above formula gives an answer of exactly K = 1 on the same TI-83 Plus calculator that failed with Heron’s formula.oracle. c = 35 3.3 θ B 6 D A Exercise 7 Figure 2. Let ABCD be a quadrilateral which completely contains its two diagonals. K = 21 AC · BD sin θ . Find the area of the quadrilateral in Figure 2.) 11. b = 6. i. A = 70◦ .16. 9. From formula (2.4 Exercise 8 8.3 below. B = 48◦ . (Hint: Factor the expression inside the square root. (Hint: In Heron’s formula replace s by 21 (a + b + c).16.4. a = 10. 1202-1261). then use formula (2. b = 3.4. B = 1◦ . Show that the triangle area formula (2.34) is equivalent to Heron’s formula. ﬁnd the area of the triangle ABC.34) For the triangle in Example 2. a = 2.4. c = 4 6. C = 8◦ . c = 11 4. c = 5 7.26) derive the following formula for the area of a triangle ABC: Area = K = a2 sin B sin C 2 sin (B + C) 10.23) to ﬁnd the area.5 4 2 5. the area is: Area = K = 1 2 a2 c 2 − a2 + c 2 − b 2 2 2 (2. A = 171◦ . as in Figure 2. Show that the area K of ABCD is equal to half the product of its diagonals and the sine of the angle they form. A = 10◦ .4 above. Exercises For Exercises 1-6.4.5 Figure 2.Chapter 2 • General Triangles 58 §2. Show that the triangle area formula Area = K = 14 (a + (b + c)) (c − (a − b)) (c + (a − b)) (a + (b − c)) is equivalent to Heron’s formula. C 3. the above formula gives an answer of exactly K = 1 on the same TI-83 Plus calculator that failed with Heron’s formula. c = 12 2.4 Another formula11 for the area of a triangle given its three sides is given below: For a triangle ABC with sides a ≥ b ≥ c.e. 1. C = 122◦ .) 12. Find the angle A in Example 2. a = 5. b = 2 5. B = 95◦ . b = 4. . Did it work? 11 Due to the Chinese mathematician Qiu Jiushao (ca. sin A sin B sin C This common ratio has a geometric meaning: it is the diameter (i. which intercept the same arc BC 1 as the central angle ∠ O. we need to review some elementary geometry. 1950. ∠ A is an inscribed angle that intercepts the arc BC.A.5. called the circumscribed circle of the triangle.5. and hence ∠ A = ∠ D = 2 ∠ O (so ∠ O = 2 ∠ A = 2 ∠ D ).5.1(c) shows two inscribed angles. this means a = sin A. A central angle of a circle is an angle whose vertex is the center O of the circle and whose sides (called radii) are line segments from O to two points on the circle.Circumscribed and Inscribed Circles • Section 2.4. ∠ O is a central angle and we say that it intercepts the arc BC. AVERY. If an inscribed angle ∠ A and a central angle ∠ O intercept the same arc. namely a b c = = .5. then ∠ A = 12 ∠ O .1(b).5 59 2. Before proving this.) 12 For a proof. twice the radius) of the unique circle in which ABC can be inscribed. 210-211 in R. In We state here without Figure 2. We will now prove our assertion about the common ratio in the Law of Sines: Theorem 2.5 Circumscribed and Inscribed Circles Recall from the Law of Sines that any triangle ABC has a common ratio of sides to sines of opposite angles. inscribed angles which intercept the same arc are equal. see pp. and c = sin C. the radius R of its circumscribed circle is given by: 2R = a b c = = sin A sin B sin C (Note: For a circle of diameter 1.5. In Figure 2. b = sin B. A A O B D O C (a) Central angle ∠ O B C (b) Inscribed angle ∠ A Figure 2. (2. Figure 2.1 B (c) ∠ A = ∠ D = C 1 2 ∠O Types of angles in a circle An inscribed angle of a circle is an angle whose vertex is a point A on the circle and whose sides are line segments (called chords) from A to two other points on the circle. Plane Geometry. Thus.e. 12 proof a useful relation between inscribed and central angles: Theorem 2.35) . ∠ A and ∠ D. For any triangle ABC.1(a). Boston: Allyn & Bacon. and so 2 R = AB = c = 1c = sinc C . Then AB is a diameter of the circle.Chapter 2 • General Triangles 60 §2.6. as in Figure 2. draw a perpendicular line segment from O to AB at the point D. i.5. B 5 Solution: We know that ABC is a right triangle. the hypotenuse is a diameter of the circumscribed circle. sin C = 1.5. so C = 90◦ by Thales’ Theorem.17 Find the radius R of the circumscribed circle for the triangle ABC whose sides are a = 3. say. 2R = a 3 = 3 = 5 sin A ⇒ R = 2.5.5.5 . 5 Note that since R = 2.e. OA R 2R sin C so by the Law of Sines the result follows if O is inside or outside ABC. and c = 5. For any right triangle. ∠ ACB = ∠ AOD.2 below. or on the triangle. Hence. Thus.5. So as we see from Figure 2.5 To prove this. which is the same as AB. Then O can be either inside. b = 4. C C C a b b R A c 2 A a O c 2 R c 2 a b B R O R D D c 2 c A R O B R B (a) O inside ABC (b) O outside ABC Figure 2. so AOB is an isosceles triangle. and so the center O of the circle is the midpoint of AB. .3 Corollary 2. So since C = ∠ ACB. on the side AB.5. O A 3 4 C Figure 2. and the result again follows by the Law of Sines. from elementary geometry we know that OD bisects both the angle ∠ AOB and the side AB. as in Figure 2.2(c). Hence.4 that ∠ ACB = 12 ∠ AOB. the center of the circle is the midpoint of the hypotenuse. sin A = 3/5. Thus. AB must be a diameter of the circle. let O be the center of the circumscribed circle for a triangle ABC. Now suppose that O is on ABC. But since the inscribed angle ∠ ACB and the central angle ∠ AOB intercept the same arc AB. the diameter of the circle is 5. outside. we know from Theorem 2.3. QED Example 2. So ∠ AOD = 12 ∠ AOB and AD = 2c .2 (c) O on ABC Circumscribed circle for ABC The radii O A and OB have the same length R. In the ﬁrst two cases. we have c AD c c sin C = sin ∠ AOD = = 2 = ⇒ 2R = . Thus. Use a compass to draw the circle centered at O which passes through A.5.4 Example 2. Recall from geometry how to create the perpendicular bisector of a line segment: at each endpoint use a compass to draw an arc with the same radius.2: a = 2. We need a different procedure for acute and obtuse triangles. their intersection will be the center of the circle. Pick the radius large enough so that the arcs intersect at two points.Circumscribed and Inscribed Circles • Section 2. In Figure 2.18 Find the radius R of the circumscribed circle for the triangle ABC from Example 2.5 that the center O was on the perpendicular bisector of one of the sides (AB).07 . and c = 4. For any triangle.5. Then draw the triangle and the circle. their intersection is the center O of the circle.5.5(b) we show how to draw the circumscribed circle: draw the perpendicular bisectors of AB and AC.9◦ = 4. then use a compass to draw arcs of radius 3 and 2 centered at A and B. so R = 2. Notice from the proof of Theorem 2.9◦ .14. so 2 R = a sin A = 2 sin 28.6 in Section 2.5(a) we show how to draw ABC: use a ruler to draw the longest side AB of length c = 4. respectively. since for an acute triangle the center of the circumscribed circle will be inside the triangle. Solution: In Example 2. Similar arguments for the other sides would show that O is on the perpendicular bisectors for those sides: Corollary 2. the circumscribed circle is simple to draw.5. its center can be found by measuring a distance of 2. C C 3 A 2 c=4 B A O B (a) Drawing ABC (b) Circumscribed circle Figure 2. the center of its circumscribed circle is the intersection of the perpendicular bisectors of the sides. you need the perpendicular bisectors of only two of the sides. d d B A Figure 2. The line through those two points is the perpendicular bisector of the line segment. .5.4. and it will be outside for an obtuse triangle.5 units from A along AB.7.6 we found A = 28.5 61 For the right triangle in the above example. b = 3. The intersection of the arcs is the vertex C. For the circumscribed circle of a triangle.5 In Figure 2. as in Figure 2. which means that O A bisects the angle A. and OF ⊥ AC. .37) R = 4 s (s − a) (s − b) (s − c) In addition to a circumscribed circle. OB bisects B and OC bisects C. the center of its inscribed circle is the intersection of the bisectors of the angles.5 can be used to derive another formula for the area of a triangle: Theorem 2. and F be the points on AB. and let D. let K be its area and let R be the radius of its circumscribed circle.5. For a triangle ABC. (2.26) from Section 2. as in Figure 2. sin C = . ∠ O AD = ∠ O AF.8 with Heron’s formula for the area of a triangle. OE ⊥ BC. at which the circle is tangent. Thus. (2. since they are right triangles with the same hypotenuse O A and with corresponding legs OD and OF of the same length r. every triangle has an inscribed circle.8. Then OD ⊥ AB. and AC. i. O AD and O AF are equivalent triangles.62 Chapter 2 • General Triangles §2. Then the radius R of its circumscribed circle is abc . For a triangle ABC. Hence.6 c B Inscribed circle for ABC Let r be the radius of the inscribed circle.5 Theorem 2.5 we have 2R = a b c = = sin A sin B sin C ⇒ sin A = a b c .9.5. 2R 2R 2R Substitute those expressions into formula (2. Similarly. we get: Corollary 2. We have thus shown: For any triangle. E.4 for the area K: a2 · 2bR · a2 sin B sin C = K = 2 sin A 2 · 2aR c 2R = abc 4R QED Combining Theorem 2. Then abc abc (and hence R = ). let s = 12 (a + b + c). note that by Theorem 2. sin B = .36) K = 4R 4K To prove this. respectively. C E F a b O r A D Figure 2. BC.e. a circle to which the sides of the triangle are tangent.6. 19 Find the radius r of the inscribed circle for the triangle ABC from Example 2.6 in Section 2. Solution: Using Theorem 2.64 Chapter 2 • General Triangles §2. Draw the circle.2: a = 2.5 Example 2.11 with s = 12 (a + b + c) = 12 (2 + 3 + 4) = 92 . we have . b = 3. and c = 4. 9 . c = 8 3. 2 (a + b + c) 12.5 6. c = 7 in 9.8 Exercises For Exercises 1-6. b = 8. a = 2 in. = = r = 9 s 12 C O 2 Figure 2. a = 5. then attheir intersection use a compass to draw a circle of radius r = 5/12 ≈ 0. a = 5. the radius R of its circumscribed circle and the radius r of its inscribed circle satisfy the relation rR = abc . −2 9 −3 9 −4 (s − a) (s − b) (s − c) 5 2 2 2 . c = 20. Let ABC be an equilateral triangle whose sides are of length a. draw the triangle ABC and its circumscribed and inscribed circles accurately. a = 5 in. let s = 21 (a + b + c). 1. ﬁnd the radii R and r of the circumscribed and inscribed circles. b = 4 in. Show that tan 12 A = cc− +b . (a) Find the exact value of the radius R of the circumscribed circle of ABC. s (s − a) s (s − b) s (s − c) 10. Let ABC be a right triangle with C = 90◦ . a = 6. Show that for any triangle ABC. b = 6 in. For any triangle ABC. 7. c = 5 2. C = 40◦ 4. c = 13 For Exercises 7 and 8.5. tan 2 C = . a = 10. b = 100.8 shows how to draw the inscribed circle: draw the bisectors of A and B. b = 4. (c) How much larger is R than r? (d) Show that the circumscribed and inscribed circles of ABC have the same center. A = 170◦ . b 13. tan 2 B = . . b = 12. c = 300 5. of the triangle ABC. respectively. c = 5 in 8. b = 7. the radius R of its circumscribed circle is abc . Show that for any triangle ABC.5. using a ruler and compass (or computer software). a = 2. Show that (s − b) (s − c) (s − a) (s − c) (s − a) (s − b) 1 1 1 tan 2 A = . (b) Find the exact value of the radius r of the inscribed circle of ABC.645. B A Figure 2. R = (a + b + c) (b + c − a) (a − b + c) (a + b − c) 11. b = 11. 1 Basic Trigonometric Identities So far we know a few relations between the trigonometric functions.3 Identities 3. we know the reciprocal relations: 1. For example. csc θ = 1 sin θ when sin θ . = 0 2. sec θ = 1 cos θ when cos θ . one of the most useful trigonometric identities is the following: tan θ = sin θ cos θ when cos θ . identities involving the trigonometric functions. cot θ = 1 tan θ when tan θ is deﬁned and not 0 4. Such equations are called identities. i.e. sin θ = 1 csc θ when csc θ is deﬁned and not 0 5.= 0 3. tan θ = 1 cot θ when cot θ is deﬁned and not 0 Notice that each of these equations is true for all angles θ for which both sides of the equation are deﬁned. cos θ = 1 sec θ when sec θ is deﬁned and not 0 6. and in this section we will discuss several trigonometric identities. For example. These identities are often used to simplify complicated expressions or equations. 1) To prove this identity. y) on the terminal side of θ a distance r > 0 from the origin. pick a point (x. and suppose that cos θ .= 0 (3. = 0. Then x . r 65 .= 0 (since cos θ = xr ). so by deﬁnition sin θ = cos θ y y r x = x = tan θ . This is probably the most common technique for proving identities. Taking reciprocals in the above identity gives: cot θ = cos θ sin θ when sin θ .1 sin θ Note how we proved the identity by expanding one of its sides ( cos ) until we got an expresθ sion that was equal to the other side (tan θ ).Chapter 3 • Identities 66 §3. if θ is in QIII as in Figure 3. r r r2 r2 r2 r2 Since r2 r2 = 1.3) You can think of this as sort of a trigonometric variant of the Pythagorean Theorem. For example: sin2 θ = 1 − cos2 θ (3. Let θ be any angle with a point (x. Thus. y) Figure 3. y) on its terminal side a distance r > 0 from the origin.5) from which we get (after taking square roots): 1 − cos2 θ cos θ = ± 1 − sin2 θ sin θ = ± (3. and y r = sin θ . r 2 = x2 + y2 (and hence r = x2 + y2 ). then the legs of the right triangle formed by the reference angle have lengths \| x \| and \| y \| (we use absolute values because x and y are negative in QIII). We will use the same notation for other powers besides 2.1. From the above identity we can derive more identities.7) . r 2 = \| x \|2 + \| y \|2 = x2 + y2 . By the Pythagorean Theorem.1.= 0 We will now derive one of the most important trigonometric identities. x r = cos θ .2) y θ \| x\| 0 \| y\| x r (x. we can rewrite this as: cos2 θ + sin2 θ = 1 (3. For example. likewise for cosine and the other trigonometric functions. The same argument holds if θ is in the other quadrants or on either axis.1. Note that we use the notation sin2 θ to mean (sin θ )2 .6) (3. (3.1 so dividing both sides of the equation by r 2 (which we can do since r > 0) gives x 2 y 2 x 2 + y2 x2 y2 r2 = = + = + .4) cos2 θ = 1 − sin2 θ (3. In formula (3. dividing both sides of formula (3. 2 2 cos θ cos θ cos2 θ so since tan θ = sin θ cos θ and sec θ = 1 cos θ . from the inequalities 0 ≤ sin2 θ = 1 − cos2 θ ≤ 1 and 0 ≤ cos2 θ = 1 − sin2 θ ≤ 1.1) to simplify: cos2 θ tan2 θ = cos2 θ · sin2 θ cos2 θ = sin2 θ Example 3.10) Likewise. taking square roots gives us the following bounds on sine and cosine: − 1 ≤ sin θ ≤ 1 (3. dividing both sides of the identity by cos2 θ gives cos2 θ sin2 θ 1 + = .5) to simplify: 5 sin2 θ + 4 cos2 θ = 5 sin2 θ + 4 1 − sin2 θ = 5 sin2 θ + 4 − 4 sin2 θ = sin2 θ + 4 (3.11) .9) The above inequalities are not identities (since they are not equations).Basic Trigonometric Identities • Section 3. Solution: We can use formula (3. Solution: We can use formula (3.3) by sin2 θ gives cos2 θ 2 sin θ so since cot θ = cos θ sin θ and csc θ = 1 sin θ .8) − 1 ≤ cos θ ≤ 1 (3.1 Simplify cos2 θ tan2 θ . we get: cot2 θ + 1 = csc2 θ Example 3. we get: 1 + tan2 θ = sec2 θ (3.1 67 Also. + sin2 θ 2 sin θ = 1 sin2 θ . Recall that we derived those inequalities from the deﬁnitions of sine and cosine in Section 1.2 Simplify 5 sin2 θ + 4 cos2 θ .3).4. but they provide useful checks on calculations. how did we know to expand the left side instead of the right side? In general.68 Chapter 3 • Identities §3.1) and (3. the more complicated side of the identity is likely to be easier to expand.1 Example 3.3 Prove that tan θ + cot θ = sec θ csc θ .2)) = sin θ sin θ cos θ cos θ · + · cos θ sin θ sin θ cos θ (multiply both fractions by 1) = sin2 θ + cos2 θ cos θ sin θ (after getting a common denominator) = 1 cos θ sin θ (by (3. Solution: We will expand the left side and show that it equals the right side: sin θ cos θ + cos θ sin θ (by (3.4 1 + cot2 θ = csc θ cot θ . it consists of a single term (cos θ ) that offers no obvious means of expansion. sec θ Solution: Of the two sides. The reason is that. For example. so we will expand that: Prove that csc2 θ 1 + cot2 θ = sec θ sec θ = csc θ · (by (3.3)) = 1 1 · cos θ sin θ tan θ + cot θ = = sec θ csc θ In the above example. the left side looks more complicated. Example 3.2)) . there would not be much that you could do with the right side of that identity. by its complexity. there will be more things that you can do with that expression. though this technique does not always work. if you were asked to prove that sec θ − sin θ tan θ = cos θ .11)) 1 sin θ 1 cos θ = csc θ · cos θ sin θ = csc θ cot θ (by (3. 10)) = 1 + cos2 θ When trying to prove an identity where at least one side is a ratio of expressions. 1 + tan2 θ Solution: Expand the left side: Prove that tan2 θ + 2 1 + tan2 θ = tan2 θ + 1 + 1 1 + tan2 θ = sec2 θ + 1 sec2 θ = sec2 θ 1 + 2 sec θ sec2 θ (by (3. Solution: Multiply both sides of the ﬁrst equation by c and the second equation by a: ac cos θ = bc ac sin θ = ad Now square each of the above equations then add them together to get: (ac cos θ )2 + (ac sin θ )2 = (bc)2 + (ad)2 (ac)2 cos2 θ + sin2 θ = b2 c2 + a2 d 2 a2 c 2 = b 2 c 2 + a2 d 2 (by (3. and d.7 Suppose that a cos θ = b and c sin θ = d for some angle θ and some constants a.3)) Notice how θ does not appear in our ﬁnal result.5 tan2 θ + 2 = 1 + cos2 θ . cos θ 1 − sin θ Solution: Cross-multiply and reduce both sides until it is clear that they are equal: Prove that (1 + sin θ )(1 − sin θ ) = cos θ · cos θ 1 − sin2 θ = cos2 θ By (3.6 1 + sin θ cos θ = .Basic Trigonometric Identities • Section 3.1 69 Example 3. Thus. b. c. Show that a2 c2 = b2 c2 + a2 d 2 .5) both sides of the last equation are indeed equal. The trick was to get a common coefﬁcient (ac) for cos θ and sin θ so that we could use cos2 θ + sin2 θ = 1. . the original identity holds. crossmultiplying can be an effective technique: c a if and only if ad = bc = b d Example 3. This is a common technique for eliminating trigonometric functions from systems of equations. Example 3. Give an example of an angle θ such that 2 sin θ = − 1 − cos θ . LTD. See pp.. where a > 0 and 0 < < 1. For example.1 Exercises 2 1. e.1. . Special Relativity.1. Use r 2 = x2 + y2 to show that r = a (1 − cos ψ) . 1 These types of equations arise in physics. We showed that cos cos θ = − 1 − sin2 θ . Show that each trigonometric function can be put in terms of the sine function. Does the identity hold if θ is on either axis? (1. Suppose that a point with coordinates (x. What must y equal? Use that to prove the identity for acute θ . in the study of photon-electron collisions. 14.2 18. y) y y θ x 0 1 Figure 3. a 1 − 2 sin ψ) is a distance r > 0 from the origin. We showed that sin θ = ± 1 − cos θ for all θ . 20. prove the given identity. = tan θ − cot θ sin θ cos θ 11. sin4 θ − cos4 θ = sin2 θ − cos2 θ 6. y) = (a (cos ψ − ). Explain the adjustment(s) you would need to make in Figure 3. ﬁnd an expression for cos θ solely in terms of tan θ . as in Figure 3. (Note: These coordinates arise in the study of elliptical orbits of planets. Suppose that you are given a system of two equations of the following form:1 A cos φ = B ν1 − Bν2 cos θ A sin φ = B ν2 sin θ .) 17. and one solely in terms of cos θ . Show that A 2 = B2 ν21 + ν22 − 2ν1 ν2 cos θ . sin θ cot θ = cos θ tan θ = tan2 θ cot θ cos2 θ 8. R INDLER.70 Chapter 3 • Identities §3. Find an expression for tan θ solely in terms of sin θ . tan θ + tan φ = tan θ tan φ cot θ + cot φ 1 − tan2 θ 1 − cot2 θ = 1 − sec2 θ (Hint: Solve for sin2 θ in Exercise 14. cos4 θ − sin4 θ = 1 − 2 sin2 θ 7. Similar to Exercise 16 .2. 15. Give an example of an angle θ such that 2. = 1 − sin θ 1 + sin θ 10. Edinburgh: Oliver and Boyd. draw an acute angle θ in QI and pick the point (1. 4. sin θ = ± 1 + tan2 θ csc θ = csc2 θ sin θ 1 − 2 cos2 θ 9. 1960. to prove the identity in Exercise 16.1. 19. cot θ − 1 1 − tan θ = 1 + tan θ cot θ + 1 sin2 θ 1 − sin2 θ = tan2 θ tan θ 16. 3. 13.2 to prove the identity for θ in the other quadrants. Sometimes identities can be proved by geometrical methods. y) on its terminal side. For Exercises 4-16. cos θ tan θ = sin θ 5.) 21.g. 95-97 in W. 12. θ = ± 1 − sin2 θ for all θ . 2.2 71 3. and ∠ QPR = ∠ QPO + ∠ OP M = (90◦ − B) + (90◦ − (180◦ − (A + B))) = A in Figure 3. since ∠ QPR = ∠ QPO − ∠ OP M = (90◦ − B) − (90◦ − (A + B)) = A in Figure 3.1 A O N (b) A + B obtuse sin (A + B) and cos (A + B) for acute A and B Thus.13) To prove these. we have the addition formulas: sin (A + B) = sin A cos B + cos A sin B (3. ﬁrst assume that A and B are acute angles.14) .2.1(a).2. Note in both cases that ∠ QPR = A.2 Sum and Difference Formulas We will now derive identities for the trigonometric functions of the sum and difference of two angles. as in Figure 3. MP MR + RP NQ + RP NQ RP = = = + OP OP OP OP OP NQ OQ RP PQ = · + · OQ OP PQ OP sin (A + B) = = sin A cos B + cos A sin B . P P A A Q R A+B R A+B Q B O B A N M M (a) A + B acute Figure 3. Then A + B is either acute or obtuse.1.2.1(b).12) cos (A + B) = cos A cos B − sin A sin B (3.Sum and Difference Formulas • Section 3. (3. For the sum of any two angles A and B. 16) cos (A − B) = cos A cos B + sin A sin B (3. Thus.17) .2 and OM ON − MN ON − RQ ON RQ = = = − OP OP OP OP OP ON OQ RQ PQ = · − · OQ OP PQ OP cos (A + B) = = cos A cos B − sin A sin B . the addition formula (3. respectively.12) for sine holds for all A and B. we will need to use the relations we derived in Section 1. and hence will hold for all angles. = −(cos A cos B − sin A sin B) = (− cos A) cos B + sin A sin B = sin (A − 90◦ ) cos B + cos (A − 90◦ ) sin B . similarly. where A and B are acute or 0◦ . Likewise. −77◦ = 13◦ − 90◦ . A similar argument shows that the addition formula (3. = cos A cos B − sin A sin B (by equation (3. similarly. 222◦ = 42◦ + 2(90◦ ). then the identities will also hold when repeatedly adding or subtracting 90◦ . for A and B + 90◦ ).5 gives us the subtraction formulas: sin (A − B) = sin A cos B − cos A sin B (3. sin ((A − 90◦ ) + B) = sin ((A + B) − 90◦ ) = − cos (A + B) . for A and B + 90◦ ). etc. QED Replacing B by −B in the addition formulas and using the relations sin (−θ ) = − sin θ and cos (−θ ) = cos θ from Section 1. so the identity holds for A + 90◦ and B (and.13) for cosine is true for all A and B.72 Chapter 3 • Identities §3.15) So we have proved the identities for acute angles A and B.5 which involve adding or subtracting 90◦ : sin (θ + 90◦ ) = cos θ cos (θ + 90◦ ) = − sin θ sin (θ − 90◦ ) = − cos θ cos (θ − 90◦ ) = sin θ These will be useful because any angle can be written as the sum of an acute angle (or 0◦ ) and integer multiples of ±90◦ . so the identity holds for A − 90◦ and B (and. It is simple to verify that they hold in the special case of A = B = 0◦ . For general angles.15)) = sin (A + 90◦ ) cos B + cos (A + 90◦ ) sin B . So if we can prove that the identities hold when adding or subtracting 90◦ to or from either A or B. 155◦ = 65◦ + 90◦ . (3. Replacing A by A + 90◦ and using the relations for adding 90◦ gives sin ((A + 90◦ ) + B) = sin ((A + B) + 90◦ ) = cos (A + B) . For example. 2 73 sin θ Using the identity tan θ = cos . we can θ derive the addition formula for tangent: tan (A + B) = = sin (A + B) cos (A + B) sin A cos B + cos A sin B cos A cos B − sin A sin B sin = cos cos cos A A A A cos B cos A sin B + cos B cos A cos B cos B sin A sin B − cos B cos A cos B (divide top and bottom by cos A cos B) sin A cos B cos A sin B · + · cos A cos B = tan A + tan B = cos A cos B sin A sin B 1 − tan A tan B 1 − · cos A cos B This. cos A = 35 .18) tan (A − B) = tan A − tan B 1 + tan A tan B (3. we get: cos (A + B) = cos A cos B − sin A sin B = 3 5 4 12 · − · 5 13 5 13 ⇒ cos (A + B) = − 33 65 Instead of using the addition formula for tangent. and cos B = 5 13 .19) Example 3.Sum and Difference Formulas • Section 3. Solution: Using the addition formula for sine.8 Given angles A and B such that sin A = 45 . gives us the addition and subtraction formulas for tangent: tan (A + B) = tan A + tan B 1 − tan A tan B (3. and tan (A + B). 12 13 . and the addition formulas for sine and cosine. cos (A + B). sin B = of sin (A + B). combined with replacing B by −B and using the relation tan (−θ ) = − tan θ . we get: sin (A + B) = sin A cos B + cos A sin B = 4 5 3 12 · + · 5 13 5 13 ⇒ sin (A + B) = 56 65 Using the addition formula for cosine. we can use the results above: 56 tan (A + B) = sin (A + B) = 6533 cos (A + B) − 65 ⇒ tan (A + B) = − 56 33 ﬁnd the exact values . So instead. show that tan A + tan B + tan C = tan A tan B tan C.74 Chapter 3 • Identities §3. So using C = 180◦ − (A + B) and the relation tan (180◦ − θ ) = − tan θ from Section 1. we will expand the left side. 2 This is the “trigonometric form” of Ptolemy’s Theorem.2 Example 3. B. However. Solution: Note that this is not an identity which holds for all angles. since A. since it appears more complicated.11 Let A. B. . it holds when A. we get: tan A + tan B + tan C = tan A + tan B + tan (180◦ − (A + B)) = tan A + tan B − tan (A + B) tan A + tan B 1 − tan A tan B 1 (tan A + tan B) 1 − 1 − tan A tan B 1 − tan A tan B 1 (tan A + tan B) − 1 − tan A tan B 1 − tan A tan B − tan A tan B (tan A + tan B) · 1 − tan A tan B tan A + tan B tan A tan B · − 1 − tan A tan B = tan A + tan B − = = = = = tan A tan B · (− tan (A + B)) = tan A tan B · (tan (180◦ − (A + B))) = tan A tan B tan C Example 3.10 For any triangle ABC. since we can eliminate the D term on that side by using D = 180◦ − (A + B + C) and the relation sin (180◦ − (A + B + C)) = sin (A + B + C). notice that the right side has no D term. Show that2 sin A sin B + sin C sin D = sin (A + C) sin (B + C) . which says that a quadrilateral can be inscribed in a circle if and only if the sum of the products of its opposite sides equals the product of its diagonals.5. and D be positive angles such that A + B + C + D = 180◦ . Solution: It may be tempting to expand the right side. B. C.9 Prove the following identity: sin (A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C − sin A sin B sin C Solution: Treat A + B + C as (A + B) + C and use the addition formulas three times: sin (A + B + C) = sin ((A + B) + C) = sin (A + B) cos C + cos (A + B) sin C = (sin A cos B + cos A sin B) cos C + (cos A cos B − sin A sin B) sin C = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C − sin A sin B sin C Example 3. and C are the angles of a triangle. C > 0◦ and A + B + C = 180◦ . θ2 is the angle of transmission of the wave through the new medium. we get sin A sin B + sin C sin D = sin A sin B + sin C sin (A + B + C) . and we know that sin (B + C) = sin B cos C + cos B sin C. We need to end up with sin (A + C) sin (B + C). Snell’s law gives the relation n 1 sin θ1 = n 2 sin θ2 . so by Example 3. but it is not completely guesswork.20) where θ1 is the angle of incidence at which a wave strikes the planar boundary between two mediums.2 75 So since sin D = sin (A + B + C). There are two terms involving cos B sin C. It may not be immediately obvious where to go from here.12 In the study of the propagation of electromagnetic waves. We now have two terms involving sin B cos C. Show that this can be written as: sin (θ2 − θ1 ) r1 2 s = sin (θ2 + θ1 ) . and n 1 and n 2 are the indexes of refraction of the two mediums.9 we get = sin A sin B + sin C (sin A cos B cos C + cos A sin B cos C + cos A cos B sin C − sin A sin B sin C) = sin A sin B + sin C sin A cos B cos C + sin C cos A sin B cos C + sin C cos A cos B sin C − sin C sin A sin B sin C .21) is called the Fresnel coefﬁcient for normal incidence reﬂection of the wave for s-polarization. The quantity r1 2 s = n 1 cos θ1 − n 2 cos θ2 n 1 cos θ1 + n 2 cos θ2 (3. so group them together to get sin A sin B + sin C sin D = sin A sin B − sin C sin A sin B sin C + sin C cos A sin B cos C + cos B sin C (sin A cos C + cos A sin C) = sin A sin B (1 − sin2 C) + sin C cos A sin B cos C + cos B sin C sin (A + C) = sin A sin B cos2 C + sin C cos A sin B cos C + cos B sin C sin (A + C) .Sum and Difference Formulas • Section 3. which we can factor out: sin A sin B + sin C sin D = sin B cos C (sin A cos C + cos A sin C ) + cos B sin C sin (A + C) = sin B cos C sin (A + C) + cos B sin C sin (A + C) = sin (A + C) (sin B cos C + cos B sin C) = sin (A + C) sin (B + C) Example 3. (3. 13) for A = B = 0◦ . Use 15◦ = 45◦ − 30◦ to ﬁnd the exact value of tan 15◦ . tan (A + B + C) = 9. sin B = 20 29 . Prove the identity sin θ + cos θ = 2 sin (θ + 45◦ ) . Use 75◦ = 45◦ + 30◦ to ﬁnd the exact value of sin 75◦ . so that they can be factored out. cos B = 21 29 4.12) and (3. For which θ between 0◦ and 360◦ would sin θ + cos θ be the largest? For Exercises 7-14.76 Chapter 3 • Identities §3. cos (A + B + C) = cos A cos B cos C − cos A sin B sin C − sin A cos B sin C − sin A sin B cos C 8. ﬁnd the exact values of sin (A + B). and tan (A + B). 6. Exercises 1. prove the given identity. This is a common technique. cos A = 15 17 . Verify the addition formulas (3.2 Solution: Multiply the top and bottom of r 1 2 s by sin θ1 sin θ2 to get: r1 2 s = n 1 cos θ1 − n 2 cos θ2 sin θ1 sin θ2 · n 1 cos θ1 + n 2 cos θ2 sin θ1 sin θ2 = (n 1 sin θ1 ) sin θ2 cos θ1 − (n 2 sin θ2 ) cos θ2 sin θ1 (n 1 sin θ1 ) sin θ2 cos θ1 + (n 2 sin θ2 ) cos θ2 sin θ1 = (n 1 sin θ1 ) sin θ2 cos θ1 − (n 1 sin θ1 ) cos θ2 sin θ1 (n 1 sin θ1 ) sin θ2 cos θ1 + (n 1 sin θ1 ) cos θ2 sin θ1 = sin θ2 cos θ1 − cos θ2 sin θ1 sin θ2 cos θ1 + cos θ2 sin θ1 = sin (θ2 − θ1 ) sin (θ2 + θ1 ) (by Snell’s law) The last two examples demonstrate an important aspect of how identities are used in practice: recognizing terms which are part of known identities. 5. cos A = 9 41 . cot (A − B) = cot A cot B + 1 cot B − cot A . Explain why this shows that − 2 ≤ sin θ + cos θ ≤ 2 for all angles θ . cos B = 7 25 3. cos (A + B). 2. sin B = 24 25 . sin A = 40 41 . 7. cot (A + B) = tan A + tan B + tan C − tan A tan B tan C 1 − tan B tan C − tan A tan C − tan A tan B cot A cot B − 1 cot A + cot B 10. sin A = 8 17 . For Exercises 2 and 3. Continuing Example 3.12. respectively. tan (θ + 45◦ ) = 12. Suppose that two lines with slopes m 1 and m 2 .e.Sum and Difference Formulas • Section 3. Generalize Exercise 6: For any a and b. cot A + cot B = sin A sin B 77 cos (A + B) = cot A − tan B sin A cos B cot B − cot A sin (A − B) = 14.22) t1 2 s = n 1 cos θ1 + n 2 cos θ2 can be written as: t1 2 s = 2 cos θ1 sin θ2 sin (θ2 + θ1 ) 17. 16. θ . − a2 + b2 ≤ a sin θ + b cos θ ≤ a2 + b2 for all θ .2 1 + tan θ 1 − tan θ sin (A + B) 13. intersect at an angle θ and are not perpendicular (i. 15. use Snell’s law to show that the s-polarization transmission Fresnel coefﬁcient 2 n 1 cos θ1 (3. sin (A + B) cot B + cot A 11. α. (Hint: Use Exercise 9 and C = 180◦ − (A + B). and β. and C such that A + B + C = 90◦ . tan θ = 1 + m1 m2 y y = m2 x + b2 y = m1 x + b1 θ 0 x (Hint: Use Example 1. r sin α) r (x . A line segment of length r > 0 from the origin to the point (x.5.) 20. y (x. Note that the right side depends only on A. y) = (r cos α. 22. show that cot A cot B + cot B cot C + cot C cot A = 1. 21. Use Exercise 17 to ﬁnd the angle between the lines y = 2x + 3 and y = −5x − 4.) 18. show that tan A tan B + tan B tan C + tan C tan A = 1 . What are the endpoint’s new coordinates (x . Prove the identity sin (A + B) cos B − cos (A + B) sin B = sin A.26 from Section 1. For any triangle ABC. while the left side depends on both A and B. y ) r β α 0 x .= 90◦ ). Show that m1 − m2 . B. r sin α). y) makes an angle α with the positive x-axis. For any positive angles A. so that (x. y ) after a counterclockwise rotation by an angle β ? Your answer should be in terms of r. 19. as in the ﬁgure below. as in the ﬁgure on the right. y) = (r cos α. 24) (3.23) (3. and for the tangent we get tan 2θ = tan (θ + θ ) = tan θ + tan θ 2 tan θ = 1 − tan θ tan θ 1 − tan2 θ Using the identities sin2 θ = 1 − cos2 θ and cos2 θ = 1 − sin2 θ . Solution: Using 3θ = 2θ + θ .3 Double-Angle and Half-Angle Formulas A special case of the addition formulas is when the two angles being added are equal. we get the following useful alternate forms for the cosine double-angle formula: cos 2θ = 2 cos2 θ − 1 (3.78 Chapter 3 • Identities §3. we have cos 2θ = cos (θ + θ ) = cos θ cos θ − sin θ sin θ = cos2 θ − sin2 θ .26) = 1 − 2 sin2 θ (3.27) Example 3. the addition formula for sine. Likewise.27). we see that sin 2θ = sin (θ + θ ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ . resulting in the double-angle formulas: sin 2θ = 2 sin θ cos θ 2 2 cos 2θ = cos θ − sin θ tan 2θ = 2 tan θ 1 − tan2 θ (3. we get: sin 3θ = sin (2θ + θ ) = sin 2θ cos θ + cos 2θ sin θ = (2 sin θ cos θ ) cos θ + (1 − 2 sin2 θ ) sin θ = 2 sin θ cos2 θ + sin θ − 2 sin3 θ = 2 sin θ (1 − sin2 θ ) + sin θ − 2 sin3 θ = 3 sin θ − 4 sin3 θ . and the double-angle formulas (3.23) and (3.3 3. for the cosine double-angle formula.13 Prove that sin 3θ = 3 sin θ − 4 sin3 θ .25) To derive the sine double-angle formula. 3 See p.28) (3.23)) (by formula (3.23) with θ replaced by 2z) Note: Perhaps surprisingly.3 79 Example 3. .I.Double-Angle and Half-Angle Formulas • Section 3. Baltimore: The Johns Hopkins University Press.14 Prove that sin 4z = 4 tan z (1 − tan2 z) . O STROVSKY AND A.30) These formulas are just the double-angle formulas rewritten with θ replaced by 12 θ : 1 − cos 2 ( 12 θ ) 1 − cos 2θ 1 − cos θ 2 1 ⇒ sin 2 θ = = cos 2θ = 1 − 2 sin θ ⇒ sin θ = 2 2 2 1 1 + cos 2 ( θ ) 1 + cos 2 θ 1 + cos θ 2 cos 2θ = 2 cos2 θ − 1 ⇒ cos2 θ = ⇒ cos2 21 θ = = 2 2 2 The tangent half-angle formula then follows easily: 2 1 − cos θ sin 12 θ sin2 12 θ 1 − cos θ 2 2 1 = = = tan 2 θ = 1 1 1 + cos θ 2 1 + cos θ cos 2 θ cos 2 θ 2 2 2 The half-angle formulas are often used (e.A. Modulated Waves: Theory and Applications. 1999. especially when 2θ is used instead of θ .29) (3. in calculus) to replace a squared trigonometric function by a nonsquared function.P OTAPOV.24)) = (4 sin z cos 2z) cos z = 2 (2 sin z cos z) cos 2z = 2 sin 2z cos 2z = sin 4z (by formula (3. in the derivation of a solution of the sine-Gordon equation in the theory of nonlinear waves. (1 + tan2 z)2 Solution: Expand the right side and use 1 + tan2 z = sec2 z : sin z cos2 z sin2 z 4· · − 4 tan z (1 − tan2 z) cos z cos2 z cos2 z = (sec2 z)2 (1 + tan2 z)2 = 4· sin z cos 2z · cos z cos2 z 2 1 cos2 z (by formula (3.3 Closely related to the double-angle formulas are the half-angle formulas: 1 − cos θ 2 1 + cos θ 2 1 cos 2 θ = 2 1 − cos θ tan2 21 θ = 1 + cos θ sin2 21 θ = (3.g. this seemingly obscure identity has found a use in physics.331 in L. 35) Taking reciprocals in formulas (3.34) and (3.36) . For example. So in this θ case cos 12 θ < 0 and hence we would have cos 12 θ = − 1 + cos . and it turns out (see Exercise 10) that tan 12 θ and sin θ always have the same sign. 2 In formula (3.80 Chapter 3 • Identities §3.3 By taking square roots. if θ = 300◦ then 12 θ = 150◦ is in QII. we can write the above formulas in an alternate form: sin 12 θ = ± cos 1 2θ = ± tan 12 θ = ± 1 − cos θ 2 (3.33). so we have: tan 12 θ = 1 − cos θ sin θ (3. Thus.35) gives: cot 12 θ = sin θ 1 + cos θ = 1 − cos θ sin θ (3.33) In the above form. the sign in front of the square root is determined by the quadrant in which the angle 12 θ is located.34) by 1 + cos θ gives tan 12 θ = 1 − cos θ 1 + cos θ 1 − cos2 θ sin2 θ . multiplying the numerator and denominator inside the square root by (1 − cos θ ) gives 1 − cos θ 1 − cos θ (1 − cos θ )2 (1 − cos θ )2 1 − cos θ 1 tan 2 θ = ± · = ± = ± = ± .31) 1 + cos θ 2 (3. 2 1 + cos θ 1 − cos θ sin θ 1 − cos2 θ sin θ But 1 − cos θ ≥ 0. the minus sign in front of the last expression is not possible (since that would switch the signs of tan 21 θ and sin θ ).32) 1 − cos θ 1 + cos θ (3. · = = sin θ 1 + cos θ sin θ (1 + cos θ ) sin θ (1 + cos θ ) so we also get: tan 12 θ = sin θ 1 + cos θ (3.34) Multiplying the numerator and denominator in formula (3. 4. sin (A − B) = cos 2B 2 ab 14. 2. and the signs (+ or −) of sin θ and tan 2 θ . In general. sin 2A = c2 2 ab 16. 10. Give an example of a speciﬁc angle θ that would make that equation false. tan 12 A = = a c+b 15.Double-Angle and Half-Angle Formulas • Section 3. 12.29) for cos2 gives us 2 2 sec θ 2 sec θ . Fill out the rest of the table below for the angles 0◦ < θ < 720◦ in increments of 90◦ . taking the reciprocal of formula (3. Is sin θ + cos θ = ± 1 + sin 2θ an identity? Justify your answer. cos 3θ = 4 cos3 θ − 3 cos θ sin 2θ cos 2θ − = sec θ sin θ cos θ 2 5. tan 3θ = 1 − 3 tan2 θ cos2 ψ 1 + cos 2ψ 8. prove the given identity. prove the given identity for any right triangle ABC with C = 90◦ . tan 12 θ .1. where θ and ψ are always in the same quadrant.15 2 sec θ . Some trigonometry textbooks used to claim incorrectly that sin θ + cos θ = 1 + sin 2θ was an identity. = · = sec2 21 θ = 1 + cos θ 1 + cos θ sec θ sec θ + 1 Prove the identity sec2 1 2θ = 1 2θ Exercises For Exercises 1-8. tan 2θ = cot θ − tan θ tan θ − sin θ 7. tan2 21 θ = tan θ + sin θ sin 3θ cos 3θ − = 2 sin θ cos θ 3 tan θ − tan3 θ 6. cos (A − B) = sin 2A b 2 − a2 c2 c−b a 17. Show that tan 12 θ = 1+ 1− tan 12 ψ . Continuing Exercise 20 from Section 3. tan 2A = 2 b − a2 13. 1 1 2 θ . For Exercises 12-17.3 81 Example 3. and r (1 + cos θ ) = a (1 − ) (1 + cos ψ) . it can be shown that r (1 − cos θ ) = a (1 + ) (1 − cos ψ) . = 1 + cos 2θ cos2 θ 3. tan 12 θ = csc θ − cot θ 1. showing θ . what is the largest value that sin θ cos θ can take? Justify your answer. cos 2A = 18. θ ◦ 0 − 90◦ 90◦ − 180◦ 180◦ − 270◦ 270◦ − 360◦ 1 2θ 0◦ − 45◦ 45◦ − 90◦ 90◦ − 135◦ 135◦ − 180◦ sin θ + tan 12 θ + θ ◦ 360 − 450◦ 450◦ − 540◦ 540◦ − 630◦ 630◦ − 720◦ 1 2θ 180◦ − 225◦ 225◦ − 270◦ 270◦ − 315◦ 315◦ − 360◦ sin θ 11. sec θ + 1 Solution: Since secant is the reciprocal of cosine. 9. 37) (sin (A + B) − sin (A − B)) (3.4 Other Identities Though the identities in this section fall under the category of “other”.43) (3. It is very common to encounter terms such as sin A + sin B or sin A cos B in calculations.43). so formula (3. We see that (( (( (( (( (A (A sin (A + B) + sin (A − B) = (sin A cos B + ( cos sin B) + (sin A cos B − ( cos sin B) = 2 sin A cos B .g.39) (cos (A + B) − cos (A − B)) (3.82 Chapter 3 • Identities §3.40) We will prove the ﬁrst formula. .g. to derive formula (3.37) follows upon dividing both sides by 2. the proofs of the others are similar (see Exercises 1-3). with the sum-to-product formulas: sin A + sin B = 2 sin 12 (A + B) cos 12 (A − B) (3.39)) cos 1 2 (A − B) .42) cos A + cos B = 2 cos cos A − cos B = −2 sin 1 2 (A + B) 1 2 (A + B) cos sin 1 2 (A − B) 1 2 (A − B) (3. Then x + y = A and x − y = B.44) These formulas are just the product-to-sum formulas rewritten by using some clever substitutions: let x = 12 (A + B) and y = 12 (A − B). make the above substitutions in formula (3. they are perhaps (along with cos2 θ + sin2 θ = 1) the most widely used identities in practice. For example.4 3. The proofs of the other sum-to-product formulas are similar (see Exercises 4-6). so we will now derive identities for those expressions. sin A cos B) of trigonometric functions is shown to be equivalent to a sum (e. we have what are often called the product-tosum formulas: sin A cos B = cos A sin B = cos A cos B = sin A sin B = 1 2 1 2 1 2 − 12 (sin (A + B) + sin (A − B)) (3. First. 12 (sin (A + B) + sin (A − B))) of such functions.39) to get cos A + cos B = cos (x + y) + cos (x − y) = 2 · 12 (cos (x + y) + cos (x − y)) = 2 cos x cos y = 2 cos 1 2 (A + B) (by formula (3.41) sin A − sin B = 2 cos 12 (A + B) sin 12 (A − B) (3. We can go in the opposite direction.38) (cos (A + B) + cos (A − B)) (3. Notice how in each of the above identities a product (e. Other Identities • Section 3.42)) (since A + B = 180◦ − C) ◦ cos(90 − 12 C) sin 12 (A − B) = sin 1 C cos 12 C 2 = sin 12 (A − B) cos 12 C (since cos (90◦ − 12 C) = sin 12 C) . = c+a tan 12 (C + A) We need only prove the ﬁrst equation. Thus. This proves the ﬁrst equation. see Section 1. Example 3. QED (since C = 180◦ − (A + B)) (since tan (90◦ − 12 (A + B)) = cot 12 (A + B).4 83 Example 3. by the double-angle formula we have sin C = 2 sin 12 C cos 12 C. We see that sin 12 (A − B) a−b cos 12 C a−b c = = a+b a+b cos 12 (A − B) c sin 12 C = sin 12 (A − B) cos 12 (A − B) · (by Mollweide’s equations) sin 12 C cos 12 C = tan 12 (A − B) · tan 12 C = tan 12 (A − B) · tan (90◦ − 12 (A + B)) = tan 12 (A − B) · cot 12 (A + B) = tan 12 (A − B) tan 12 (A + B) .3: For any triangle ABC.17 Using Mollweide’s equations. since C = 2 · 1 2 C.5) . = a+b tan 12 (A + B) tan 12 (B − C) b−c . tan 12 (A − B) a−b . which we introduced in Section 2. The proof of the other equation is similar (see Exercise 7). sin 12 (A − B) cos 12 (A − B) a+b a−b and = = . we can prove the Law of Tangents: For any triangle ABC. = b+c tan 12 (B + C) tan 12 (C − A) c−a . a b sin A sin B a−b (by the Law of Sines) = − = − c c c sin C sin C sin A − sin B sin A − sin B = = sin C 2 sin 1 C cos 1 C 2 = = 2 cos 12 (A + B) sin 12 (A − B) 2 sin 12 C cos 12 C cos 12 (180◦ − C) sin 12 (A − B) sin 12 C cos 12 C 2 (by formula (3.16 We are now in a position to prove Mollweide’s equations. c c cos 12 C sin 12 C First. the other two are obtained by cycling through the letters. Chapter 3 • Identities 84 §3. So by the quadratic formula.4 Example 3.20 For any triangle ABC. after rearranging the terms. a solution exists).43) = 1 + 2 cos 12 (A + B + 2C) cos 12 (A + B) + 2 cos 12 (A + B) cos 12 (A − B) = 1 + 2 cos 12 (A + B) cos 12 (A + B + 2C) + cos 12 (A − B) .40) to the ﬁrst two terms in u to get u = − 12 (cos 12 (A + B) − cos 12 (A − B)) sin 12 C = 1 2 (cos 12 (A − B) − cos 12 (A + B)) cos 12 (A + B) . C < 180 . Notice that the expression above is a quadratic equation in the term cos 12 (A + B). 1 < cos A + cos B + cos C ≤ 3 2 . so rearranging the order gives . Apply formula (3. so sin 12 B sin 12 C > 1 by Example 3. 1 2A sin 12 B sin 12 C ≤ 1 + 4 · 1 8 = 3 2 .18 For any triangle ABC. show that 1 < cos A + cos B + cos C ≤ ◦ ◦ Solution: Since 0 < A. Thus. we can rewrite the left side as cos A + cos B + cos C = 1 + (cos (A + B + C) + cos C) + (cos A + cos B) . hence. since 1 1 1 2 2 (A + B + 2C) + 2 (A − B) = 12 (A + C) and 1 1 1 2 2 (A + B + 2C) − 2 (A − B) = 12 (B + C). are all positive. B. Also. so = 1 + 2 cos 21 (A + B) · 2 cos 12 (A + C) cos 12 (B + C) by formula (3. Solution: Since cos (A + B + C) = cos 180◦ = −1. Multiply both sides by 2 to get cos2 1 2 (A + B) − cos 12 (A − B) cos 12 (A + B) + 2u = 0 . as we saw in Example 3.19 For any triangle ABC. so by formula (3. so we must have cos2 1 2 (A − B) − 8u ≥ 0 ⇒ u ≤ 1 8 cos2 1 2 (A − B) ≤ 1 8 ⇒ sin 1 2A sin 12 B sin 12 C ≤ Example 3. show that sin Solution: Let u = sin 1 2A sin 1 2B 1 2A 1 2 C. . by Examples 3. 1 8 . show that cos A + cos B + cos C = 1 + 4 sin 1 2A sin 12 B sin 12 C .18. cos A + cos B + cos C = 1 + 4 sin and 1 2A 3 2 1 2C . the sines of 1 1 2 A.18.18 and 3. sin sin 12 B sin 12 C ≤ 1 8 . since sin 12 C = cos 12 (A + B). 2 B. cos A + cos B + cos C = 1 + 4 cos (90◦ − 12 C) cos (90◦ − 12 B) cos (90◦ − 12 A) = 1 + 4 sin 12 C sin 12 B sin = 1 + 4 sin 1 2A sin 1 2B sin 1 2A 1 2C . cos 12 (A − B) ± cos2 21 (A − B) − 4(1)(2u) cos 12 (A + B) = . Example 3. But we know that cos 12 (A + B) is a real number (and.43). 2 which has a real solution only if the quantity inside the square root is nonnegative.19 we have cos A + cos B + cos C = 1 + 4 sin Hence. and φ. for some constants Vm . Solution: By deﬁnition of p(t).39)) = cos (2ω t + φ) (since cos (−φ) = cos φ) .45) can be written as: 2 cos θ1 sin θ2 .2: n 1 sin θ1 = n 2 sin θ2 . Use it to show that the p-polarization transmission Fresnel coefﬁcient deﬁned by t1 2 p = 2 n 1 cos θ1 n 2 cos θ1 + n 1 cos θ2 (3. . the instantaneous power p(t) delivered to the entire circuit in the sinusoidal steady state at time t is given by p(t) = v(t) i(t) . i(t) = I m cos (ω t + φ) . ω. we have p(t) = Vm I m cos ω t cos (ω t + φ) 1 2 (cos (2ω t + φ) + 1 1 2 Vm I m cos φ + 2 Vm I m = Vm I m · cos (−φ)) (by formula (3.Other Identities • Section 3. sin (θ1 + θ2 ) cos (θ1 − θ2 ) Solution: Multiply the top and bottom of t 1 2 p by sin θ1 sin θ2 to get: t1 2 p = t1 2 p = 2 n 1 cos θ1 sin θ1 sin θ2 · n 2 cos θ1 + n 1 cos θ2 sin θ1 sin θ2 = 2 (n 1 sin θ1 ) cos θ1 sin θ2 (n 2 sin θ2 ) sin θ1 cos θ1 + (n 1 sin θ1 ) sin θ2 cos θ2 = 2 cos θ1 sin θ2 sin θ1 cos θ1 + sin θ2 cos θ2 = = = 2 cos θ1 sin θ2 1 2 1 2 (by the double-angle formula) (sin 2 θ1 + sin 2θ2 ) (2 sin (by Snell’s law) 2 cos θ1 sin θ2 1 1 2 (2θ1 + 2θ2 ) cos 2 (2θ1 − 2θ2 )) (by formula (3. where the voltage v(t) and current i(t) are given by v(t) = Vm cos ω t .22 In an AC electrical circuit.12 in Section 3.4 85 Example 3. Show that the instantaneous power can be written as p(t) = 1 2 Vm I m cos φ + 1 2 Vm I m cos (2ω t + φ) .41)) 2 cos θ1 sin θ2 sin (θ1 + θ2 ) cos (θ1 − θ2 ) Example 3.21 Recall Snell’s law from Example 3. I m . 40).38). sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C (Hints: Group sin 2B and sin 2C together. 15.46) n 2 cos θ1 + n 1 cos θ2 can be written as: r1 2 p = tan (θ1 − θ2 ) tan (θ1 + θ2 ) 9.18 using (sin A + sin B) + (sin C − sin (A + B + C)). i(t) = I m cos (ω t + φ) .44). where s = 12 (a + b + c) 14. 6. use Snell’s law to show that the p-polarization reﬂection Fresnel coefﬁcient n 2 cos θ1 − n 1 cos θ2 r1 2 p = (3. 7. In Example 3. Prove formula (3.42). We will discuss their unit of measurement in Chapter 4.4 Exercises 1.21. 3.) sin A − sin B a−b = a+b sin A + sin B s (s − a) (s − b) (s − c) 1 1 and sin 2 A = . There is a more general form for the instantaneous power p(t) = v(t) i(t) in an electrical circuit than the one in Example 3. . Prove formula (3. Prove Mollweide’s second equation: For any triangle ABC. Prove formula (3. which angles A. prove the given identity or inequality for any triangle ABC. Prove formula (3. For Exercises 10-15. B.20. cos A + cos (B − C) = 2 sin B sin C 12. 2. Prove formula (3.) 16. 10. use Exercise 11. Continuing Example 3.Chapter 3 • Identities 86 §3.4 Show that p(t) can be written as p(t) = 1 2 Vm I m cos (θ − φ) + 1 2 Vm I m cos (2ω t + θ + φ) . cos 2 A = bc bc (Hint: Use the Law of Cosines to show that 2bc (1 + cos A) = 4s (s − a). where θ is called the phase angle. use the double-angle formula for sin 2A.) 13. 1 2 (sin A + sin B) ≤ sin 12 (A + B) (Hint: Show that sin 12 (A + B) − 12 (sin A + sin B) ≥ 0.39). cos 12 (A − B) a+b . C give the maximum value of cos A + cos B + cos C ? 4 Though it does not matter for this exercise.41). The voltage v(t) and current i(t) can be given by v(t) = Vm cos (ω t + θ ) . = c sin 12 C 8. Prove formula (3.22. 4. none of the angles in these formulas are measured in degrees.) 11. sin A + sin B + sin C = 4 cos 21 A cos 12 B cos 12 C (Hint: Mimic Example 3. 5. K = 2 ((Area of sector D AE) + (Area of rectangle AEFC) + (Area of triangle ABC) + (Area of sector GBF)) .11 Find the area K inside the belt pulley system from Example 4.5) for the areas of sectors D AE and GBF.3. Example 4. F 6 6 E 5 6 6 5 1. ∠ D AE = 1.6) for the area A. the total area K enclosed by the belt is twice the area above the line DG. its area is 12 (3) (6 6) = 9 6.3.96 Chapter 4 • Radian Measure §4. using formula (4. we have K = 2 (Area of sector D AE) + 30 6 + 9 6 + (Area of sector GBF) = 2 12 (5)2 (1.37 rad.37) + 30 6 + 9 6 + 12 (8)2 (1. its area is triangle whose legs have lengths 3 and 6 6. And since ABC is a right Since AEFC is a rectangle with sides 5 and 6 6. We showed in Example 4.77) = 338.2.37 D C B A G 15 Figure 4. by symmetry. and ∠ GBF = 1.77 rad.2 that. we get A = 1 2 rs = = Note that the angle subtended by the arc is θ = 1 2 (9) (6) = 27 cm2 . and their centers are 15 cm apart.77 3 1.10 Find the area of a sector whose arc is 6 cm in a circle of radius 9 cm. .7 in Section 4. We see from Figure 4. s r = 2 3 rad. Solution: Using s = 6 and r = 9 in formula (4.3 Example 4. that is.2 Belt pulleys with radii 5 cm and 8 cm 30 6. Solution: Recall that the belt pulleys have radii of 5 cm and 8 cm.7 that EF = AC = 6 6.59 cm2 . Thus. we have area K of segment AB = A 1 2 2r sin θ = 1 2 2 r (θ − sin Figure 4. By formula (2. We can use the Law of Cosines to ﬁnd the subtended central angle θ : cos θ = 22 + 22 − 32 = −0.5 shows the segment formed by a chord of length 3 in a circle of radius r = 2. How a would you ﬁnd the area of a region cut off by an inscribed angle. since the area of a segment is positive for those angles.4 in Section 2. such r as the shaded region in Figure 4.125 2 (2) (2) ⇒ 1 2 2 r (θ − sin θ) = 1 2 θ 2 θ = 1. Drawing line segments b from the center of the circle to the endpoints of the chords indicates how to solve this problem: add up the areas of the two triangles and Figure 4.3. which is the region between a chord and the arc it cuts off. The areas and angles of the two triangles can be determined (since all three sides are known) using methods from Chapter 2.696) = 1. recall (Theorem 2.4 θ) . (4.12 Find the area of the segment formed by a chord of length 3 in a circle of radius 2. and the lengths a and b of the two chords are given.4 for the area of a triangle given two sides and their included angle. Another type of region we can consider is a segment of a circle.696 − sin 1. In Figure 4. as is the radius r of the circle.3 the sector formed by the central angle.4 the segment formed by the chord AB is the shaded region between the arc AB and the triangle O AB.3. Thus.3.5 .7) the area K of the segment is: K = 3 2 (2)2 (1.3? In this picture. since the area K of the segment is the area of the sector AOB minus the area of the triangle O AB. Also.696 rad Thus. In the exercises you will be asked to solve problems like this (including the cases where the center of the circle is outside or on the inscribed angle). the center of the circle is inside the inscribed angle.3.3 97 So far we have dealt with the area cut off by a central angle.7) Note that as a consequence of formula (4. Solution: Figure 4. we know that area of O AB = 1 2 (r) (r) sin θ = 1 2 2r 1 2 2r θ − r θ B r O sin θ .5) that a central angle has twice the measure of any inscribed angle which intercepts the same arc.7) we must have θ > sin θ for 0 < θ ≤ π (measured in radians).23) in Section 2. by formula (4.Area of a Sector • Section 4.408 Figure 4.3.3. Example 4. 3. ﬁnd the area of the segment formed by a chord of length a in a circle of radius r. Find the area of the shaded region in Figure 4.8 Exercise 14 Figure 4. how much does the area of its sector increase when the radius of the circle is doubled? How much does the length of its intercepted arc increase? .3. Find the area of the shaded region in Figure 4. For Exercises 7-9. (Hint: Connect the centers of the circles. r = 1 cm. r = 4 cm 11.3.3. The centers of two circles are 4 cm apart. as in Figure 4.3.) 1 r 4 2 r Figure 4.3.3. 16. 19.3.3. For a ﬁxed central angle θ . and 1 m are externally tangent to each other.11 is (π + 6 + 3) r 2 . Find the area of their intersection.10. Find the area of the curved region between the circles.9. a = 4 cm. a = 2 cm. a = 1 cm.3. s = 2 cm 8.8. r = a.7 Exercise 13 Figure 4.10 Exercise 17 Figure 4.2. 2 m. Show that the total area enclosed by the loop around the three circles of radius r in Figure 4. s = π cm For Exercises 10-12. ﬁnd the area of the sector for the given radius r and arc length s. Find the area enclosed by the ﬁgure eight in Exercise 8 from Section 4. 7. Three circles with radii of 4 m. r = 5 cm.Area of a Sector • Section 4. r = 5 cm 13.11 r Exercise 18 18. Find the area of the shaded region in Figure 4. 10. s = a 9. with one circle having a radius of 3 cm and the other a radius of 2 cm. (Hint: Draw two central angles. r = 5 cm 12.9 Exercise 15 14.3 99 6.7. 17.) 15. 7 4 8 3 5 9 6 6 5 Figure 4. . constant speeds. Then we deﬁne the (average) angular speed ω of the object as: ω = θ t (4. i. Technically they are not the same. We will usually omit the word average when discussing linear and angular speed here. traveling a distance s over a period of time t.9) Angular speed gives the rate at which the central angle swept out by the object changes as the object moves around the circle. The word linear is used because straightening out the arc traveled by the object along the circle results in a line of the same length.4 Circular Motion: Linear and Angular Speed Radian measure and arc length can be applied to the study of circular motion. many texts use the word velocity instead of speed.4. Also.1 (4. feet per second). whereas speed is just a magnitude. In physics the average speed of an object is deﬁned as: distance s θ distance traveled average speed = time elapsed r So suppose that an object moves along a circle of radius r.8) Let θ be the angle swept out by the object in that period of time.1. velocity has a direction and a magnitude. and it is thus measured in radians per unit time.e. Linear speed is measured in distance units per unit time (e.10) 4 Many trigonometry texts assume uniform motion. t t t so that we get the following relation between linear and angular speed: ν = ωr (4. so that the usual deﬁnition of speed as distance over time can be used.100 Chapter 4 • Radian Measure §4. we see that s rθ θ ν = = = ·r. We do not make that assumption. Then it makes sense to deﬁne the (average) linear speed ν of the object as: ν = time t > 0 s t time t = 0 Figure 4.4 4.4.4 Since the length s of the arc cut off by a central angle θ in a circle of radius r is s = r θ . as in Figure 4.g. 1.1 circle. We thus get a correspondence between the y-coordinates of points on the unit circle and the values f (θ ) = sin θ .2 0 1 0 π 6 π 3 π 2 2π 3 5π 6 π θ Graph of sine function based on y-coordinate of points on unit circle We can extend the above picture to include angles from 0 to 2π radians. This is the circle of radius 1 s = rθ = θ 1 in the x y-plane consisting of all points (x. We will describe a geometrical way to create the x2 + y2 = 1 (x.5 Graphing and Inverse Functions The trigonometric functions can be graphed just like any other function. y) makes with the positive x-axis (by deﬁnition of sine and cosine). using the unit circle. So as the point (x. 103 .1 Graphing the Trigonometric Functions y The ﬁrst function we will graph is the sine function. as shown by the horizontal lines from the unit circle to the graph of f (θ ) = sin θ in Figure 5. its y-coordinate is sin θ . In the graphs we will always use radians for the angle measure.1. θ x We see in Figure 5.1.3. y) = (cos θ . This illustrates what is sometimes called the unit circle deﬁnition of the sine function.2 for the angles θ = 0. π3 . sin θ ). y) = (cos θ . y) which 2 2 satisfy the equation x + y = 1. f (θ ) 1 1 π 2 π 3 f (θ ) = sin θ π 6 θ 2 2 x +y =1 Figure 5. π2 . y) goes around the Figure 5. sin θ ) graph.1. as in Figure 5.1 that any point on the unit 0 1 circle has coordinates (x.1. 5. where θ is the angle that the line segment from the origin to (x. as we will now show. π6 . but there is an easier way. the graph of the cosine function is just the graph of the sine function shifted to the left by 90◦ = π/2 radians.5 that cos x = sin (x + 90◦ ) for all x.1 0 π 6 π 3 π 2 2π 3 5π 6 π 5π 4 3π 2 7π 4 2π θ −1 x 2 + y2 = 1 Figure 5. and so on.1.4 Graph of y = sin x To graph the cosine function. Recall from Section 1. use tan x = sin x cos x to get the following graph: 7π 4 2π . the following graph of the function y = sin x for x in the interval [−2π. In other words. we get. 2π]: y y = sin x 1 −2π − 7π − 3π − 5π 4 2 4 −π − 3π 4 − π2 0 − π4 π 4 π 2 3π 4 π x 5π 4 3π 2 7π 4 2π −1 Figure 5.5: y y = cos x 1 −2π − 7π − 3π − 5π 4 2 4 −π − 3π 4 − π2 − π4 0 π 4 π 2 3π 4 π x 5π 4 3π 2 −1 Figure 5. cos 180◦ has the same value as sin 270◦ .104 Chapter 5 • Graphing and Inverse Functions f (θ ) y f (θ ) = sin θ 1 θ 1 x §5.1.1.5 Graph of y = cos x To graph the tangent function. as in Figure 5. we could again use the unit circle idea (using the x-coordinate of a point that moves around the circle).1.3 Unit circle deﬁnition of the sine function Since the trigonometric functions repeat every 2π radians (360◦ ). cos 90◦ has the same value as sin 180◦ . So cos 0◦ has the same value as sin 90◦ . for example. Thus. two times faster than the graphs of sine and cosine repeat. x = 32π . ± 2π. etc. and that is indeed what the graph in Figure 5. so the quotient tan x = cos x is a positive number that is very large. as in Figure 5.1. we get vertical asymptotes at x = − π2 . x = π2 is a vertical asymptote of the graph of y = tan x.6 Graph of y = tan x Recall that the tangent is positive for angles in QI and QIII. . i. For example. using csc x = sin1 x we can just look at the graph of y = sin x and invert the values.7 shows the graph of y = csc x. namely at multiples of π: x = 0. and x = − 32π . Figure 5. Similarly. ± 52π . sin x is very close to 1 and cos x is negative and very sin x close to 0. Notice that the graph of the tangent function repeats every π radians. for x in QII very close to π2 . for x in QI near π2 . i.6 shows. ± 32π . the larger tan x gets. We will get vertical asymptotes when sin x = 0. The graphs of the remaining trigonometric functions can be determined by looking at the graphs of their reciprocal functions. sin x and cos x are both positive.Graphing the Trigonometric Functions • Section 5.1 105 y 8 6 4 2 −2π − 7π − 3π − 5π 4 2 4 −π − 3π 4 − π2 − π4 y = tan x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x −2 −4 −6 −8 Figure 5. ± π.e. For example. with the graph of y = sin x (the dashed curve) for reference.e. with sin x very close to 1 and cos x very sin x close to 0. The graph shows this.1. We know that tan x is not deﬁned when cos x = 0.6. etc. We can ﬁgure out what happens near those angles by looking at the sine and cosine functions. so the quotient tan x = cos x is a negative number that is very large. at odd multiples of π2 : x = ± π2 . And the closer π x gets to 2 . and it gets larger in the negative direction the closer x gets to π2 . and is negative in QII and QIV.1. Likewise.1. 1 y 4 y = csc x 3 2 1 −2π − 7π − 3π − 5π 4 2 4 −π − 3π 4 − π2 − π4 −1 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x −2 −3 −4 Figure 5.8 shows the graph of y = sec x.1. ± 32π .7 Graph of y = csc x Likewise.1. Notice also that the graph is just the graph of the cosecant function shifted to the left by π2 radians.106 Chapter 5 • Graphing and Inverse Functions §5. Figure 5. with the graph of y = cos x (the dashed curve) for reference.1. Note the vertical asymptotes at x = ± π2 . y 4 y = sec x 3 2 1 −2π − 7π − 3π − 5π 4 2 4 −π − 3π 4 − π2 − π4 −1 0 π 4 π 2 3π 4 −2 −3 −4 Figure 5.8 Graph of y = sec x π 5π 4 3π 2 7π 4 2π x . 1 Draw the graph of y = − sin x for 0 ≤ x ≤ 2π. so that the graph of the cotangent function is just the graph of the tangent function shifted to the left by π2 radians and then reﬂected about the x-axis. Solution: Multiplying a function by −1 just reﬂects its graph around the x-axis.1.1. we can use the relation cot x = − tan (x + 90◦ ) from Section 1.9: y 8 6 4 2 −π − 3π 4 −2π − 7π − 3π − 5π 4 2 4 − π2 − π4 y = cot x 0 π 4 π 2 π 3π 4 5π 4 3π 2 7π 4 2π x −2 −4 −6 −8 Figure 5.Graphing the Trigonometric Functions • Section 5.5. Alternatively. as in Figure 5.9 Graph of y = cot x Example 5. . So reﬂecting the graph of y = sin x around the x-axis gives us the graph of y = − sin x: y y = − sin x 1 0 π 4 π 2 3π 4 π x 5π 4 3π 2 7π 4 2π −1 Note that this graph is the same as the graphs of y = sin (x ± π) and y = cos (x + π2 ).1 107 The graph of y = cot x can also be determined by using cot x = cot1 x . So adding 1 to cos x moves the graph of y = cos x upward by 1. respectively. depending on whether the constant is positive or negative. show that R −α (R α (x. y = − cot x 7. so you will have to come up with a convention for how to treat some of the line segment lengths as negative. . y)) = R α+β (x.10 if θ was in QII? Recall that some of the trigonometric functions are negative in QII. cosine.10 15. For any point (x. we have sin θ = MP (why?).108 Chapter 5 • Graphing and Inverse Functions §5. y) on the unit circle and any angle α. Example 5. y = −3 cos x 11. so that the line segment OP in Figure 5.10 has length 1 and makes an acute angle θ with the positive x-axis. especially for sine. Let P be a point in QI on the unit circle. y) and R β (R α (x.10. Solution: Adding a constant to a function just moves its graph up or down by that amount. 14. y = 2 − cos x 4. how would you draw the line segments in Figure 5. What is the geometric interpretation of R α (x. so recognizing the general shape is important. 1. y) deﬁned by R α (x. In particular. We can extend the unit circle deﬁnition of the sine and cosine functions to all six trigonometric functions. Many phenomena in nature exhibit sinusoidal behavior. x sin α + y cos α) is also on the unit circle. y = − tan x 6. y)? Also. For Exercise 13. y = 2 sin x 10. keeping in mind that the radius of the circle is 1. y = 2 − sin x 5. To get you started. y = −2 tan x 12. y)) = (x.2 Draw the graph of y = 1 + cos x for 0 ≤ x ≤ 2π. Q y R S P θ O M N 1 x Figure 5. Identify each of the six trigonometric functions of θ with exactly one of the line segments in Figure 5.1. and tangent.1.1.1 It is worthwhile to remember the general shapes of the graphs of the six trigonometric functions.1. draw the graph of the given function for 0 ≤ x ≤ 2π. y = 1 + sin x 3. giving us the graph of y = 1 + cos x: y 2 y = 1 + cos x 1 0 π 4 π 2 3π 4 π x 5π 4 3π 2 7π 4 2π Exercises For Exercises 1-12. y). y = 1 + sec x 8. y = −1 − csc x 9. y = −2 sec x 13. show that the point R α (x. the graphs of the sine and cosine functions are called sinusoidal curves. y = − cos x 2. y) = (x cos α − y sin α. sin x goes from 0 to 1. the interval [−1. over the interval [0.2. π4 ].1 how the graphs of the trigonometric functions repeat every 2π radians. In this section we will discuss this and other properties of graphs.. the range of f (x) = sin x is the set of all real numbers between −1 and 1 (i.. then we call that number the period of the function f (x). whereas the range of f (x) = tan x is the set of all real numbers. Thus. The range of a function f (x) is the set of all values that f (x) can take over its domain.2 109 5. We saw in Section 5. for x from 0 to π2 . Example 5. For example. y y = sin 2x 1 y = sin x 0 π 4 π 2 3π 4 π x 5π 4 3π 2 7π 4 2π −1 Figure 5. . 2π].1 Graph of y = sin 2x For example. First.. csc x. and if the following relation holds: f (x + p) = f (x) for all x (5. Example 5.3 The functions sin x. ± 52π .1.1 that the graphs of y = tan x and y = cot x repeat every 2π radians but they also repeat every π radians. the domain of f (x) = sin x is the set of all real numbers. cos x. recall that the domain of a function f (x) is the set of all numbers x for which the function is deﬁned. as we can see from their graphs. For example. but sin 2x is able to go from 0 to 1 quicker. ± 32π . especially for the sinusoidal functions (sine and cosine). whereas the domain of f (x) = tan x is the set of all real numbers except x = ± π2 .1) There could be many numbers p that satisfy the above requirements. . If there is a smallest such number p.Properties of Graphs of Trigonometric Functions • Section 5. A function f (x) is periodic if there exists a number p > 0 such that x + p is in the domain of f (x) whenever x is.e.2.2 Properties of Graphs of Trigonometric Functions We saw in Section 5. sin 2x goes through an entire cycle in just π radians. So the period of sin 2x is π radians. While sin x takes a full 2π radians to go through an entire cycle (the largest part of the graph that does not repeat). 1]). Note that sin 2x “goes twice as fast” as sin x. along with the graph of y = sin x for comparison. and sec x all have the same period: 2π radians. just over the interval [0.4 What is the period of f (x) = sin 2x ? Solution: The graph of y = sin 2x is shown in Figure 5. the functions tan x and cot x have a period of π radians. This is a contradiction.1 we know that sin (x + 2π) = sin x for all x.g. Luckily. So suppose 0 < p < π. divide by ω. by our deﬁnition of period. Thus. Why? Because the period of sin x is 2π > 2p. so the period must equal π. so the period p of sin 2x is at most π. and hence sin 2x = f (x) = f (x + p) (since p is the period of f (x)) = sin 2(x + p) = sin (2x + 2p) for all x. Then 0 < 2p < 2π.e u = 2(u/2)). we will use a proof by contradiction. Hence. Since sin x has period 2π. and hence the period of sin x is as most 2p. assume that 0 < p < π. we do not need to go through all that work for each function. sin (−3x) = − sin 3x). Since 2x is a number for all x. To do this. The above may seem like a lot of work to prove something that was visually obvious from the graph (and intuitively obvious by the “twice as fast” idea). 1 We will usually leave out the “radians” part when discussing periods from now on. . We have to show that p > 0 can not be smaller than π. we get: For any number ω > 0: 2π ω 2π cos ω x has period ω π tan ω x has period ω sin ω x has period 2π ω 2π sec ω x has period ω π cot ω x has period ω csc ω x has period If ω < 0. Then f (x + π) = sin 2 (x + π) = sin (2x + 2π) = sin 2x (as we showed above) = f (x) for all x. Now deﬁne f (x) = sin 2x. this means that sin u = sin (u + 2p) for all real numbers u. then use sin (− A) = − sin A and cos (− A) = cos A (e. and hence can not be true. since a similar argument works when sin 2x is replaced by sin ω x for any positive real number ω: instead of dividing 2π by 2 to get the period. then show that this leads to some contradiction.2 The above example made use of the graph of sin 2x. And the argument works for the other trigonometric functions as well. the period p of sin 2x can not be less than π. That is.110 Chapter 5 • Graphing and Inverse Functions §5. but the period can be found analytically. this means in particular that sin (2x + 2π) = sin 2x for all x. Since any number u can be written as 2x for some x (i. The graphs of both functions are 1 2x y = cos 3x 1 0 π 6 π 3 π 2 2π 3 5π 6 π x 7π 6 4π 3 3π 2 5π 11π 3 6 2π 13π 7π 6 3 5π 2 8π 17π 3 6 3π 19π 10π 7π 11π 23π 6 3 2 3 6 4π −1 Figure 5.5 The period of y = cos 3x is shown in Figure 5.Properties of Graphs of Trigonometric Functions • Section 5.2 Graph of y = cos 3x and y = cos 1 2x We know that −1 ≤ sin x ≤ 1 and −1 ≤ cos x ≤ 1 for all x. for a constant A .2: 2π 3 and the period of y = cos y = cos y 1 2x is 4π.2.2. Thus.2 111 Example 5. tan x has neither a maximum nor a minimum. For example. y \| A\| \| A\| 2\| A\| 0 π 4 π 2 3π 4 π x 5π 4 3π 2 7π 4 = \| A \|−(−\| A \|) 2 2π \| A\| −\| A \| Figure 5.= 0. the amplitude is the distance from either the top or bottom of the curve to the horizontal line that divides the curve in half. cot x. In general. we call \| A \| the amplitude of the functions y = A sin x and y = A cos x. and vice versa. and sec x do not have an amplitude. Since the amplitude involves vertical distances. In this case.3 Amplitude = max−min 2 = \| A\| Not all periodic curves have an amplitude. it has no effect on the period of a function.2.3. csc x. −\| A \| ≤ A sin x ≤ \| A \| and − \| A \| ≤ A cos x ≤ \| A \| for all x. the amplitude of a periodic curve f (x) is half the difference of the largest and smallest values that f (x) can take: Amplitude of f (x) = (maximum of f (x)) − (minimum of f (x)) 2 In other words. so its amplitude is undeﬁned. . as in Figure 5.2. Likewise. 6 Find the amplitude and period of y = 3 cos 2x.2.5: y 5 4 3 3 6 2 1 3 x 0 −1 3 4 3 2 Figure 5. even though it does change the maximum and minimum. So in this case.2. 2 2 2 = 3.112 Chapter 5 • Graphing and Inverse Functions §5. It just shifts the entire graph upward by 2.5 9 4 y = 2 − 3 sin 2π 3 x 3 . The graph is shown in Figure 5.2 Example 5. Solution: The amplitude of −3 sin 23π x is \|−3 \| = 3. we have Amplitude = The period is 2π 2π 3 max − min 5 − (−1) 6 = = = 3.4: y 3 2 3 1 6 0 π 6 −1 π 3 π 2 2π 3 π 5π 6 7π 6 4π 3 3π 2 5π 3 11π 6 x 2π 3 −2 −3 y = 3 cos 2x Figure 5.2. The graph is shown in Figure 5. Adding 2 to that function to get the function y = 2 − 3 sin 23π x does not change the amplitude.2.7 Find the amplitude and period of y = 2 − 3 sin 2π 3 x.4 Example 5. Solution: The amplitude is \| 3 \| = 3 and the period is 2π 2 = π. not a constant rate.Properties of Graphs of Trigonometric Functions • Section 5. since the angle that we are taking the sine of. is not a linear function of x. namely 2. Can we say that sin (x2 ) is some constant times as fast as sin x ? No.6 y = 2 sin (x2 ) Notice how the curve “speeds up” as x gets larger. Thus. Solution: This is not a periodic function. See Appendix B for a brief tutorial on how to use Gnuplot. Thus. Despite this. In fact. since x2 grows at a variable rate.5 -2 0 π 2 π 3π 2 2π x Figure 5.8 Find the amplitude and period of y = 2 sin (x2 ).6:2 2 1. In the exercises you will be asked to ﬁnd values of x such that 2 sin (x2 ) reaches the maximum value 2 and the minimum value −2. . shown in Figure 5. we see that the “speed” of the curve keeps increasing as x gets larger. i.2.2. it appears that the function does have an amplitude. is not of the form ax + b for some constants a and b.5 -1 -1. making the “waves” narrower and narrower. albeit one with variable cycles. we have \| 2 sin (x2 ) \| = \| 2 \| · \| sin (x2 ) \| ≤ 2 · 1 = 2 . y = 2 sin (x2 ) has no period. since the general shape is still that of a “sine wave”.e. note that since \| sin θ \| ≤ 1 for all θ . This will not always be the case. This can be seen in the graph of y = 2 sin (x2 ). 2 This graph was created using Gnuplot.info. so that its period was π instead of 2π. the amplitude is indeed 2. Note: This curve is still sinusoidal despite not being periodic.5 1 y 0. an open-source graphing program which is freely available at http:// gnuplot. To see why. x2 . So far in our examples we have been able to determine the amplitudes of sinusoidal curves fairly easily. Recall how we argued that sin 2x was “twice as fast” as sin x.2 113 Example 5.5 0 -0. 7 y = 3 sin x + 4 cos x The graph suggests that the amplitude is 5. Solution: This is sometimes called a combination sinusoidal curve. We can use this as follows: y = 3 sin x + 4 cos x = 5 35 sin x + 45 cos x = 5 (cos θ sin x + sin θ cos x) = 5 sin (x + θ ) 5 4 θ 3 Figure 5. the deﬁnition y = 3 sin x + 4 cos x may tempt you to think that the amplitude is 7.9 Find the amplitude and period of y = 3 sin x + 4 cos x. We can see this in the graph. y = 3 sin x + 4 cos x has period 2π. so the amplitude of y = 3 sin x + 4 cos x is 5.2 Example 5. since it is the sum of two such curves. which may not be immediately obvious just by looking at how the function is deﬁned. Thus. since the largest that 3 sin x could be is 3 and the largest that 4 cos x could be is 4. There is a useful technique (which we will discuss further in Chapter 6) for showing that the amplitude of y = 3 sin x + 4 cos x is 5. so that the largest their sum could be is 3 + 4 = 7.2. In fact.2. \| y \| = \| 5 sin (x + θ ) \| = \| 5 \| · \| sin (x + θ ) \| ≤ (5)(1) = 5.114 Chapter 5 • Graphing and Inverse Functions §5.8. However. Let θ be the angle shown in the right triangle in Figure 5. then so does the combination 3 sin x + 4 cos x. Then cos θ = 35 and sin θ = 45 .2.7: 5 4 3 2 y 1 0 -1 -2 -3 -4 -5 0 π 2 π 3π 2 2π 5π 2 3π 7π 2 4π x Figure 5.8 (by the sine addition formula) Thus. shown in Figure 5. 3 sin x can never equal 3 for the same x that makes 4 cos x equal to 4 (why?). The period is still simple to determine: since sin x and cos x each repeat every 2π radians. .2. 5 1 y 0. The lowest common π 2 = π = π Thus. In Chapter 6.2204 × 10−16 ). π 3 π 3 π 3 and the period of sin 4x is = = π 3 2π 3 1· 2· π 2 π 2 = 2π 4 = π 2.9 y = cos 6x + sin 4x What about the amplitude? Unfortunately we can not use the technique from Example 5.Properties of Graphs of Trigonometric Functions • Section 5. In this case. since we are not taking the cosine and sine of the same angle. the amplitude is 1. Hence. . Example 5. we will describe how to use a numerical computation program to show that the maximum and minimum are ± 1.2. it appears from the graph that the maximum is close to 2 and the minimum is close to −2.2. We can see this from its graph in Figure 5.5 -2 0 π 2 π 3π 2 2π x Figure 5. a combination of sines and cosines will have a period equal to the lowest common multiple of the periods of the sines and cosines being added. sin x and cos x each have period 2π.9. the period of y = cos 6x + sin 4x is π.90596111871578.90596111871578. Solution: The period of cos 6x is multiple of π3 and π2 is π: 2π 6 = 1· 2· 3· π 3.9: 2 1.5 0 -0.9.2 115 In general.5 -1 -1. so the lowest common multiple (which is always an integer multiple) is 1 · 2π = 2π. In Example 5. we are taking the cosine of 6x but the sine of 4x. respectively (accurate to within ≈ 2.10 Find the period of y = cos 6x + sin 4x. y y period = 2π ω period = A A x 0 φ ω 2π ω − A phase shift + φ ω φ ω 2π ω 0 phase shift − A (a) φ > 0: right shift (b) φ < 0: left shift Figure 5. For simplicity we will assume period = 2ωπ that A > 0 and ω > 0 (in general either one could be negaA tive). where θ is an angle such that cos θ = 2a 2 and sin θ = 2b 2 . and the period is still 2ωπ . where φ is π 2π ω ω some constant.2. an expression of the form a sin ω x + b cos ω x is equivalent 2 2 to a + b sin (x + θ ). The graph is shifted to the right when φ > 0. So a +b a +b y = a sin ω x + b cos ω x will have amplitude a2 + b2 . The amplitude is still A. That cycle starts when ωx − φ = 0 ⇒ x = ω x − φ = 2π ⇒ x = φ ω and ends when 2π φ + . Then the amplitude is A and the period is 2ωπ .10. where A and ω are nonzero constants.2.116 Chapter 5 • Graphing and Inverse Functions §5. x 0 Now consider the function y = A sin (ω x − φ).10 y = A sin ω x angle goes from 0 to 2π. Also.2 Generalizing Example 5. as in Figure 5. we are taking the sine of the angle ω x − φ. Note that this method only works when the angle ω x is the same in both the sine and cosine terms. The amount φ ω of the shift is called the phase shift of the graph.9. y Consider a function of the form y = A sin ω x. we know − A that the sine function goes through an entire cycle when its Figure 5. We have seen how adding a constant to a function shifts the entire graph vertically. The graph is shown in Figure 5.11. the graph of y = A sin (ω x − φ) is just the graph of y = A sin ω x shifted horizontally by φ . We will now see how to shift the entire graph of a periodic curve horizontally.2. So as ω x − φ goes from 0 to 2π. and to the left when ω φ < 0.11 Phase shift for y = A sin (ω x − φ) x 2π ω φ +ω . an entire cycle of the function y = A sin (ω x − φ) will be traced out. since ω x − φ is a linear function of x.2. Here. ω ω Thus. 2 117 The phase shift is deﬁned similarly for the other trigonometric functions. and the phase shift is y π 2. Example 5. The graph is shown in Figure 5.2.11 Find the amplitude. Example 5.13: y period = 2π 3 2 amplitude = 2 − π6 1 0 −1 π 6 π 3 π 2 2π 3 5π 6 −2 phase shift = − π6 Figure 5. the period is 23π .12 y = 3 cos (2x − π) Notice that the graph is the same as the graph of y = 3 cos 2x shifted to the right by π2 . period.2.2. Notice the negative sign in the phase shift. 2π 2 Solution: The amplitude is 3. and the phase shift is 32 = − π6 . the amount of the phase shift. period. −π Solution: The amplitude is 2. and phase shift of y = −2 sin 3x + π2 .Properties of Graphs of Trigonometric Functions • Section 5.12: = π. the period is Figure 5.12 Find the amplitude. and phase shift of y = 3 cos (2x − π). The graph is shown in period = π 3 2 amplitude = 3 1 0 π π 2 −1 x 2π 3π 2 −2 −3 phase shift = π 2 Figure 5.2. since 3x + π = 3x − (−π) is in the form ω x − φ.13 y = −2 sin 3x + π2 π x 7π 6 4π 3 . 118 Chapter 5 • Graphing and Inverse Functions §5.2 In engineering two periodic functions with the same period are said to be out of phase if their phase shifts differ. For example, sin x − π6 and sin x would be π6 radians (or 30◦ ) out of phase, and sin x would be said to lag sin x − π6 by π6 radians, while sin x − π6 leads sin x by π 6 radians. Periodic functions with the same period and the same phase shift are in phase. The following is a summary of the properties of trigonometric graphs: For any constants A = 0, ω = 0, and φ: y = A sin (ω x − φ) has amplitude \| A \|, period y = A cos (ω x − φ) has amplitude \| A \|, period , ω , ω and phase shift and phase shift y = A tan (ω x − φ) has undeﬁned amplitude, period y = A csc (ω x − φ) has undeﬁned amplitude, period y = A sec (ω x − φ) has undeﬁned amplitude, period y = A cot (ω x − φ) has undeﬁned amplitude, period φ ω φ ω φ π , and phase shift ω ω φ , and phase shift ω ω φ , and phase shift ω ω φ π , and phase shift ω ω Exercises For Exercises 1-12, ﬁnd the amplitude, period, and phase shift of the given function. Then graph one cycle of the function, either by hand or by using Gnuplot (see Appendix B). 1. y = 3 cos π x 2. y = sin (2π x − π) 3. y = − sin (5x + 3) 4. y = 1 + 8 cos (6x − π) 5. y = 2 + cos (5x + π) 6. y = 1 − sin (3π − 2x) 7. y = 1 − cos (3π − 2x) 8. y = 2 tan (x − 1) 9. y = 1 − tan (3π − 2x) 10. y = sec (2x + 1) 11. y = 2 csc (2x − 1) 12. y = 2 + 4 cot (1 − x) 13. For the function y = 2 sin (x2 ) in Example 5.8, for which values of x does the function reach its maximum value 2, and for which values of x does it reach its minimum value −2 ? 14. For the function y = 3 sin x + 4 cos x in Example 5.9, for which values of x does the function reach its maximum value 5, and for which values of x does it reach its minimum value −5 ? You can 15. Graph the function y = sin2 x from x = 0 to x = 2π, either by hand or by using Gnuplot. What are the amplitude and period of this function? 16. The current i(t) in an AC electrical circuit at time t ≥ 0 is given by i(t) = I m sin ω t, and the voltage v(t) is given by v(t) = Vm sin ω t, where Vm > I m > 0 and ω > 0 are constants. Sketch one cycle of both i(t) and v(t) together on the same graph (i.e. on the same set of axes). Are the current and voltage in phase or out of phase? 17. Repeat Exercise 16 with i(t) the same as before but with v(t) = Vm sin ω t + π4 . 18. Repeat Exercise 16 with i(t) = − I m cos ω t − π3 and v(t) = Vm sin ω t − 56π . For Exercises 19-21, ﬁnd the amplitude and period of the given function. Then graph one cycle of the function, either by hand or by using Gnuplot. What functions form its amplitude envelope? (Note: Use set samples 500 in Gnuplot.5 sin x sin 12x You can think of this function as sin 12x with a sinusoidally varying “amplitude”of 0.5.5 ∗ sin(x) y 0. Its graph for x from 0 to 4π is shown in Figure 5. What is the period of this function? From the graph it looks like the amplitude may be 0. ﬁnd the period of the given function. 23.5 sin x form an amplitude envelope for the wave (i.14: 1 0. Use Gnuplot to graph the function y = x2 sin 10x from x = −2π to x = 2π. The function above is known as a modulated wave.5 ∗ sin(x) ∗ sin(12 ∗ x) 0. y = sin 3x − cos 5x 24.5 0 -0.) 28. 27. explain why the amplitude is in fact less than 0.2 to x = π. Use Gnuplot to graph the function y = x12 sin 80x from x = 0. y = 2 sin π x + 3 cos π 3x 26. Without ﬁnding the exact amplitude.14 Modulated wave y = 0. they enclose the wave).5 sin x sin 12x . Use an identity from Section 3.4 to write this function as a sum of sinusoidal curves.5 sin x. y = 3 sin π x − 5 cos π x 20.Properties of Graphs of Trigonometric Functions • Section 5.5. Find the amplitude of the function y = 2 sin (x2 ) + cos (x2 ). y = −5 sin 3x + 12 cos 3x 119 21.2 19. For Exercises 23-25.5 -1 0 π 2π 3π 4π x Figure 5. Graph one cycle using Gnuplot.) 29.e.2. 30. y = sin x 3 + 2 cos 3x 4 25. What happens at x = 0? . Does the function y = sin π x + cos x have a period? Explain your answer. Let y = 0. and the functions ± 0. y = 2 cos x + 2 sin x 22.2. Use Gnuplot to graph the function y = sin x x from x = −4π to x = 4π.5 ∗ sin(x) −0. What functions form its amplitude envelope? (Note: Use set samples 500 in Gnuplot. y y y = f (x) y = f (x) x x (a) f is a function (b) f is not a function Figure 5. We will now deﬁne those inverse functions and determine their graphs. There is a simple vertical rule Figure 5. cos−1 . Domain Range Recall that a function is a rule that assigns a single object f y from one set (the range) to each object x from another set x y (the domain).120 Chapter 5 • Graphing and Inverse Functions §5.3.2 Vertical rule for functions Recall that a function f is one-to-one (often written as 1 − 1) if it assigns distinct values of y to distinct values of x.2). for example in Section 1.1).3 5.3.3.3 Inverse Trigonometric Functions We have brieﬂy mentioned the inverse trigonometric functions before. and tan−1 buttons on a calculator to ﬁnd an angle that has a certain trigonometric function value.1 for determining whether a rule y = f (x) is a function: f is a function if and only if every vertical line intersects the graph of y = f (x) in the x y-coordinate plane at most once (see Figure 5.3. We can write that rule as y = f (x).3 when we discussed how to use the sin−1 . if x1 . In other words. where f is y = f (x) the function (see Figure 5. = x2 then f (x1 ) . 3. The domain of f −1 is the range of f . denoted by f . Equivalently. then f has an inverse function.3). −1 . There is a simple horizontal rule for determining whether a function y = f (x) is one-to-one: f is one-to-one if and only if every horizontal line intersects the graph of y = f (x) in the x y-coordinate plane at most once (see Figure 5. y y y = f (x) y = f (x) x (a) f is one-to-one x (b) f is not one-to-one Figure 5.3 Horizontal rule for one-to-one functions If a function f is one-to-one on its domain. f is one-to-one if f (x1 ) = f (x2 ) implies x1 = x2 .3.= f (x2 ). such that y = f (x) if and only if f −1 (y) = x. Thus.3) Example 5.2).Inverse Trigonometric Functions • Section 5. y = sin x is one-to-one over the interval − π2 . π2 . we can not use formula (5. For example. . 2 = − π4 . so we can deﬁne the inverse sine function y = sin−1 x (sometimes called the arc sine and denoted by y = arcsin x) whose domain is the interval [−1. and vice versa.13 Find sin−1 sin π4 . sin−1 sin 5π 4 2 2 sin − π4 = − sin π4 = − 1 . 1] and whose range is the interval − π2 . we know that sin−1 sin π 4 = π 4 . 2 sin−1 sin 54π = sin−1 − 1 is.2). Solution: Since − π2 ≤ Example 5. Solution: Since 54π > π2 . f −1 ( f (x)) = x f (f −1 (y)) = y for all x in the domain of f . That angle is y = − π4 . . π2 For − π2 ≤ x ≤ π2 we have −1 ≤ sin x ≤ 1. In other words: sin−1 (sin y) = y for − π2 ≤ y ≤ sin (sin−1 x) = x for −1 ≤ x ≤ 1 π 2 (5. We know from their graphs that none of the trigonometric functions are one-to-one over their entire domains. since Thus. by formula (5. as we see in the graph below: y y = sin x 1 −π 0 − π2 π π 2 x −1 Figure 5. and for all y in the range of f .4 y = sin x with x restricted to − π2 .2) (5. the angle y such that − π2 ≤ y ≤ π2 and sin y = − 1 . by deﬁnition. π2 .3 121 The basic idea is that f −1 “undoes” what f does.14 Find sin−1 sin 5π 4 π 4 ≤ π2 . In other words. But we know that sin 54π = − 1 .3. we can restrict those functions to subsets of their domains where they are one-to-one. However. This why in Example 1. π] Thus.5. y y = sin−1 x π 2 y = sin x 1 0 − π2 −1 x 1 π 2 −1 − π2 y=x Figure 5.3. the graph of an inverse function f −1 is the reﬂection of the graph of f around the line y = x.5 Graph of y = sin−1 x The inverse cosine function y = cos−1 x (sometimes called the arc cosine and denoted by y = arccos x) can be determined in a similar fashion.5 we got sin−1 (−0.6 y = cos x with x restricted to [0.3. π]. Notice the symmetry about the line y = x with the graph of y = sin x. Instead of an angle between 0◦ and 360◦ (i.3. The function y = cos x is one-to-one over the interval [0.14 illustrates an important point: sin−1 x should always be a number between − π2 and π2 . − π2 to π2 radians). In general.682) = −43◦ when using the sin−1 button on a calculator. The graph of y = sin−1 x is shown in Figure 5. as we see in the graph below: y y = cos x 1 0 − π2 π 2 π x 3π 2 −1 Figure 5. 0 to 2π radians) we got an angle between −90◦ and 90◦ (i. then you made a mistake somewhere. In other words: . y = cos−1 x is a function whose domain is the interval [−1.122 Chapter 5 • Graphing and Inverse Functions §5.e.3 Example 5. π]. If you get a number outside that range. 1] and whose range is the interval [0.e.27 in Section 1. Thus. by formula (5.3. cos−1 cos 43π = 23π . Solution: Since 0 ≤ Example 5.3 cos−1 (cos y) = y cos (cos −1 x) = x 123 for 0 ≤ y ≤ π (5. Notice the symmetry about the line y = x with the graph of y = cos x. 120◦ ). Solution: Since 43π > π.4). therefore.5) The graph of y = cos−1 x is shown below in Figure 5. Thus. since they seem to violate the general rule for inverse functions that f −1 ( f (x)) = x for all x in the domain of f . .16 may be confusing. We had to restrict the sine and cosine functions to very small subsets of their entire domains in order for those functions to be one-to-one.4). That general rule.7 Graph of y = cos−1 x Example 5.3. But we know that cos 43π = − 21 .Inverse Trigonometric Functions • Section 5.16 Find cos−1 cos 4π 3 π 3 ≤ π. . only holds for x in those small subsets in the case of the inverse sine and inverse cosine. Examples 5. by deﬁnition.7. cos−1 cos 43π = cos−1 − 12 is. That angle is y = 23π (i. But that rule only applies when the function f is one-to-one over its entire domain.e.14 and 5. we can not use formula (5. the angle y such that 0 ≤ y ≤ π and cos y = − 21 . y = cos−1 x y π 1 y = cos x 0 − π2 −1 y=x π π 2 1 x −1 Figure 5.15 Find cos−1 cos π 3 .4) for −1 ≤ x ≤ 1 (5. we know that cos−1 cos π 3 = π 3 . y 3 y = tan x 2 π 2 y = tan−1 x 1 − π2 − π4 0 π 4 π 2 −1 y=x − π2 −2 −3 Figure 5. The function y = tan x is one-to-one over the interval − π2 . Notice that the vertical asymptotes for y = tan x become horizontal asymptotes for y = tan−1 x.3.9 Graph of y = tan−1 x x .124 Chapter 5 • Graphing and Inverse Functions §5. as we see in Figure 5.8 y = tan x with x restricted to − π2 .8: y 3 2 1 − π2 − π4 π 4 0 π 2 y = tan x −1 x −2 −3 Figure 5.3. π2 The graph of y = tan−1 x is shown below in Figure 5.3.3. Note also the symmetry about the line y = x with the graph of y = tan x. π2 .9.3 The inverse tangent function y = tan−1 x (sometimes called the arc tangent and denoted by y = arctan x) can be determined similarly. = ⇒ cos θ = cos2 θ = 1 − sin2 θ = 1 − − 4 16 4 −1 1 15 −4 = . by formula (5. In other words: tan−1 (tan y) = y tan (tan Example 5. 1 − x2 . Thus. Thus. Example 5.3.10 (note that this is possible since 0 < x < 1). 1 − x2 Solution: When x = 0. Hence cos θ > 0. That angle is y = 0. tan θ = x 2 . as in Figure 5. θ must be in QIV. tan (sin−1 x) = for −1 < x < 1.7) . since 0 . So we can draw the same triangle except that it would be “upside down” and we would again have tan θ = x 2 . Draw a right triangle with an angle θ such that the opposite leg has length x and the hypotenuse has length 1. the angle y such that − π2 ≤ y ≤ π2 and tan y = 0.10 If −1 < x < 0 then θ = sin−1 x is in QIV. Then sin θ = 1x = x.18 Find tan−1 (tan π). by deﬁnition. 1− x 1 x θ 1 − x2 Figure 5. y = tan−1 x is a function whose domain is the set of all real numbers and whose range is the interval − π2 . tan−1 (tan π) = 0 . the formula holds trivially. 1 2 15 15 . Thus.Inverse Trigonometric Functions • Section 5. we can not use formula (5. Thus. Note that we took the positive square root above since cos θ > 0. cos sin 4 Example 5. tan (sin−1 0) = tan 0 = 0 = 1 − 02 Now suppose that 0 < x < 1.6). Let θ = sin−1 x. we know that tan−1 tan π 4 = π 4 .17 Find tan−1 tan π 4 −1 x) = x for − π2 < y < π 2 (5.6). Solution: Since − π2 ≤ π 4 ≤ π2 . the adjacent leg has length 1 − x2 . But we know that tan π = 0. Thus. By the Pythagorean Theorem.19 Find the exact value of cos sin−1 − 14 . since the tangent and sine have the same sign 1− x x (negative) in QIV.3. tan−1 (tan π) = tan−1 0 is. so since sin θ = − 14 < 0. Solution: Let θ = sin−1 − 14 .20 x Show that tan (sin−1 x) = for −1 < x < 1.3 125 Thus. We know that − π2 ≤ θ ≤ π2 . Solution: Since π > π2 .6) for all real x (5. Thus. Then θ is in QI and sin θ = x. Example 5. π2 . In other words: cot−1 (cot y) = y for 0 < y < π (5.3. y . and secant can be determined by looking at their graphs. π). For example.11 π 4 π 2 x 3π 4 Graph of y = cot−1 x Similarly.11. the inverse cotangent y = cot−1 x is a function whose domain is the set of all real numbers and whose range is the interval (0. y π π 2 y = cot−1 x − 34π − π2 − π4 0 Figure 5. the function y = cot x is one-to-one in the interval (0. cosecant.126 Chapter 5 • Graphing and Inverse Functions §5.3 The inverse functions for cotangent.9) The graph of y = cot−1 x is shown below in Figure 5. where it has a range equal to the set of all real numbers. π). Thus. it can be shown that the inverse cosecant y = csc−1 x is a function whose domain is \| x \| ≥ 1 and whose range is − π2 ≤ y ≤ π2 .3.8) cot (cot−1 x) = x for all real x (5. y .= 0. Likewise. the inverse secant y = sec−1 x is a function whose domain is \| x \| ≥ 1 and whose range is 0 ≤ y ≤ π. = π2 . y . csc−1 (csc y) = y csc (csc−1 x) = x π ≤ y ≤ . y .= 0 2 2 for \| x \| ≥ 1 for − π sec−1 (sec y) = y for 0 ≤ y ≤ π. = sec (sec−1 x) = x for \| x \| ≥ 1 π 2 (5.12: .11) (5. csc−1 x.13) It is also common to call cot−1 x. The graphs of y = csc−1 x and y = sec−1 x are shown in Figure 5. and arc secant. and sec−1 x the arc cotangent.3. arc cosecant. respectively.10) (5.12) (5. of x. Thus. we know that 0 < cot−1 x < π. which proves the identity.22 Is tan−1 a + tan−1 b = tan−1 a+b 1 − ab an identity? Solution: In the tangent addition formula tan (A + B) = tan−1 b. But it is possible that tan−1 a + tan−1 b is not in in particular. Example 5.3. so − π2 < π2 − cot−1 x < π π −1 subset on which the tangent function is one-to-one. this means that tan (tan−1 x) = tan π2 − θ . Solution: Let θ = cot−1 x. Using relations from Section 1.3 y y π 2 −1 0 127 π y = csc−1 x y = sec−1 x π 2 x 1 − π2 −1 0 (a) Graph of y = csc−1 x x 1 (b) Graph of y = sec−1 x Figure 5. 2 . recall that − π2 < tan−1 x < π2 for all real numbers x. we must have − π2 < tan−1 1a−+ab π π the interval − 2 .7). i. by formula (5.892547 > π2 ≈ 1.e.Inverse Trigonometric Functions • Section 5. Hence. So b < π2 . x is in the restricted 2 .5. so it seems that we have = 1 − ab a+b tan−1 a + tan−1 b = tan−1 1 − ab by deﬁnition of the inverse tangent. we have tan π2 − θ = − tan θ − π2 = cot θ = cot (cot−1 x) = x . 2 − cot −1 tan (tan x) = tan π2 − cot−1 x implies that tan−1 x = π2 − cot−1 x. +2 = tan−1 (−3) = −1. Then tan (tan−1 a + tan−1 b) = tan A + tan B . let A = tan−1 a and B = 1 − tan A tan B tan (tan−1 a) + tan (tan−1 b) 1 − tan (tan−1 a) tan (tan−1 b) a+b by formula (5.9). tan (tan−1 x) = tan π2 − cot−1 x . Now.249045 .570796 .12 Example 5. However. So since tan (tan−1 x) = x for all x.21 Prove the identity tan−1 x + cot−1 x = π 2. tan−1 1 + tan−1 2 = 1. For example. So the formula is only And we see that tan−1 1−1(1)(2) true when − π2 < tan−1 a + tan−1 b < π2 .= tan−1 1 + tan−1 2. . 13 Exercise 37 38. tan−1 (−1) 3. tan−1 tan − 56π 5 19. Note that since most computer languages use radians for their inverse trigonometric functions. 1. sin−1 4 5 + cos−1 5 13 25. cos−1 0 11. which was part of that example. In Example 5. cos−1 cos π7 15.13 shows three equal squares lined up against each other. you will likely have to do the conversion from radians to degrees yourself in the program. cot−1 cot 43π 22.) γ β α Figure 5. 26. cos (sin−1 x) = 1 − x2 27. Write a computer program to solve a triangle in the case where you are given three sides.22 we showed that the formula tan−1 a + tan−1 b = tan−1 hold.3 Exercises For Exercises 1-25. cos sin−1 − 45 7.128 Chapter 5 • Graphing and Inverse Functions §5. 36. sin−1 3 5 + sin−1 24. 35. sin−1 sin 43π 12. 3 5 1 − x2 30. cos−1 cos − 10 18. (Hint: Consider the tangents of the angles. 39.3. cot−1 x = tan−1 33. cos sin−1 13 23. Figure 5. Show that tan−1 1 3 + tan−1 1 5 = tan−1 4 7 . Show that tan−1 1 4 + tan−1 2 9 = tan−1 1 2 . sin−1 0 6. sin−1 1 8. always 1 − ab 37. tan−1 tan 43π 17. prove the given identity. sec−1 sec 65π 21. cos−1 (− x) + cos−1 x = π 32. sin (cos−1 x) = 28. show that α = β + γ. csc−1 csc − π9 20. sin−1 (− x) = − sin−1 x 1 x + cot−1 π 2 for x > 0 a+b does not always 1 − ab a+b . cos−1 cos 65π 16. ﬁnd the exact value of the given expression in radians. Sketch the graph of y = sin−1 2x. Does the formula tan (tan−1 a + tan−1 b) = hold? Explain your answer.3. tan−1 x + tan−1 for x > 0 1 x = 34. sin−1 sin π3 π 14. tan−1 5 13 For Exercises 26-33. tan−1 1 2. cos−1 1 5. tan−1 0 4. sin−1 sin − 56π 13. sin−1 (−1) 9. sin−1 x + cos−1 x = 29. sec−1 x + csc−1 x = π 2 3 5 π 2 31. and γ in the picture. β. For the angles α. Your program should read in the three sides as input parameters and print the three angles in degrees as output if a solution exists. . cos−1 (−1) 10. 129 .. between 0◦ and 90◦ ).. This is the most general solution to the equation. Thus. However. In radians. sin 210◦ = − 12 . For example. That is. 36.87◦ + 360◦ . To see what that means.. ± 1. there are many other possible answers for the value of A.75.6435 + π k for k = 0..1 Solving Trigonometric Equations An equation involving trigonometric functions is called a trigonometric equation.87◦ − 180◦ .. ± 2. we know that the tangent function has period π rad = 180◦ .. 36.1 Solve the equation 2 sin θ + 1 = 0. Often the part that says “for k = 0. an equation like tan A = 0. namely 36. Solution: Isolating sin θ gives sin θ = − 12 .. etc. ± 1. ± 2. Recall that the reﬂection of this angle around the y-axis into QIII also has the same sine. it repeats every 180◦ . Using the sin−1 calculator button in degree mode gives us θ = −30◦ .. Thus. 36. For the rest of this section we will write our solutions in radians.. Example 6. the general solution is: θ = −30◦ + 360◦ k and 210◦ + 360◦ k for k = 0.. We can write this in a more compact form: A = 36.75 . ± 1. .87◦ + 180◦ k for k = 0. since the sine function has period 2π rad = 360◦ . the solution is: θ = − π 6 + 2π k and 7π + 2π k 6 for k = 0. In Chapter 1 we were concerned only with ﬁnding a single solution (say. ± 1. ± 2. i. . which we encountered in Chapter 1. and since −30◦ does not differ from 210◦ by an integer multiple of 360◦ . take the above equation tan A = 0.” is omitted since it is usually understood that k varies over all integers.87◦ − 360◦ . The general solution in radians would be: A = 0. is a trigonometric equation.87◦ + 540◦ . ± 1.87◦ + 180◦ . In this section we will be concerned with ﬁnding the most general solution to such equations. ± 2. 36. we get A = 36. . ± 2. .e. .6 Additional Topics 6. which is in QIV. Using the tan−1 calculator button in degree mode.87◦ . 666 + 2π k and 2.3 Solve the equation 2 sec θ = 1.618 < −1. we can combine those four separate answers into one: θ = π + 4 π 2 for k = 0.475 + 2π k for k = 0.618. we would add 2π k to each of those angles to get the general solution.618.618 . 4 4 4 and since the period of cosine is 2π. k Example 6.666 rad in QI and its reﬂection π − θ = 2. But −1. Thus. we have −1 ± 1 − (4) (−1) −1 ± 5 2 = = −1.4 Solve the equation cos θ = tan θ .618 x +x−1 = 0 ⇒ x = 2 (1) 2 by the quadratic formula from elementary algebra. . sec θ which is impossible. Solution: The idea here is to use identities to put everything in terms of a single trigonometric function: cos θ = tan θ sin θ cos θ cos2 θ = sin θ cos θ = 1 − sin2 θ = sin θ 0 = sin2 θ + sin θ − 1 The last equation looks more complicated than the original equation. Thus... . Solution: Isolating cos2 θ gives us cos2 θ = 1 2 ⇒ 1 cos θ = ± 2 ⇒ θ = π 4 .2 Solve the equation 2 cos2 θ − 1 = 0.. Hence. 0.1 Example 6. But notice that the above angles differ by multiples of π2 . 3π 5π 7π . there is no solution . ± 2. there are two possible solutions: θ = 0. ± 1. we must have sin θ = x = 0. .130 Chapter 6 • Additional Topics §6. .. Adding multiples of 2π to these gives us the general solution: θ = 0. ± 1. ± 2. but notice that it is actually a quadratic equation: making the substitution x = sin θ . . so it is impossible that sin θ = x = −1.475 rad around the y-axis in QII. So since every multiple of 2π is also a multiple of π2 . Solution: Isolating sec θ gives us sec θ = 1 2 ⇒ cos θ = 1 = 2. Example 6. Solution: The idea here is to solve for 3θ ﬁrst.7 Solve the equation sin 2θ = sin θ ....5 Solve the equation sin θ = tan θ . ± 1. Adding multiples of 2π to these gives us: 3θ = ± π 3 + 2π k for k = 0. we can combine the two answers into one for the general solution: θ = πk for k = 0.. π . using the most general solution. ± 2. So since the above angles are multiples of π. ± 2. So dividing everything by 3 we get the general solution for θ : θ = ± π 9 + 2π k 3 for k = 0. Example 6... Solution: Trying the same method as in the previous example. π or θ = ± ⇒ θ = πk and cos θ = ± π 3 π 1 2 3 + 2π k for k = 0. there are two possible solutions for 3θ : 3θ = π3 in QI and its reﬂection −3θ = − π3 around the x-axis in QIV. Solution: Here we use the double-angle formula for sine: sin 2θ = sin θ 2 sin θ cos θ = sin θ sin θ (2 cos θ − 1) = 0 ⇒ sin θ = 0 or ⇒ θ = 0. ± 1. and every multiple of 2π is a multiple of π. So since cos−1 12 = π3 . ± 1. ± 2.1 131 Example 6. Example 6.. . .6 Solve the equation cos 3θ = 1 2. and then divide that solution by 3. . .Solving Trigonometric Equations • Section 6.. π or θ = 0 cos θ = 1 ⇒ θ = 0. we get sin θ = tan θ sin θ cos θ sin θ cos θ = sin θ sin θ = sin θ cos θ − sin θ = 0 sin θ (cos θ − 1) = 0 ⇒ sin θ = 0 or ⇒ θ = 0. plus multiples of 2π. . ± 2. ± 1. The leg with length 3 > 0 means that the angle φ is above the x-axis.281 or θ − φ = π − 0. tan θ + cot θ = 2 9.983 + 2. . cos2 θ + 2 sin θ = −1 8.8 Solve the equation 2 sin θ − 3 cos θ = 1.8 if the equation had been 2 sin θ + 3 cos θ = 1 then we still would have used a right triangle with legs of lengths 2 and 3. φ must be in QI. So since we needed to add multiples of 2π to the solutions 0. as in Figure 6. 1.983 + 0. Let φ be the angle shown in the right triangle.1. 2 cos2 θ + 3 sin θ = 0 7.844 + 2π k for k = 0. Solution: We will use the technique which we discussed in Chapter 5 for ﬁnding the amplitude of a combination of sine and cosine functions. 2 cos θ − sin θ = 1 5. Exercises For Exercises 1-12.861 + 2π k and 3. but we would have used the sine addition formula instead of the subtraction formula. Hence. respectively.861 anyway.264 + 2π k and 0.861 or θ = φ + 2.. solve the given equation (in radians). the most general solution for θ is: θ = 0. sin 5θ + 1 = 0 4.1. cos2 3θ − 5 cos 3θ + 4 = 0 12. sin θ = cos θ 10. the most general solution for φ is φ = 0. ± 1. 3 sin θ − 4 cos θ = 1 . in the above equation and make them the legs of a right triangle. 2 cos θ + 1 = 0 2 3. tan θ + 1 = 0 2 2 2. Take the coefﬁcients 2 and 3 of sin θ and − cos θ . ± 1..281 + 2π k ⇒ θ = 1.1 Example 6.281 and 2.281 ⇒ θ = φ + 0. Note: In Example 6. 13 ± 2...1 13 We can use this to transform the equation to solve as follows: 2 sin θ − 3 cos θ = 1 2 13 sin θ − 3 cos θ = 1 13 13 13 (cos φ sin θ − sin φ cos θ ) = 1 13 sin (θ − φ) = 1 sin (θ − φ) = (by the sine subtraction formula) 1 13 ⇒ θ − φ = 0. since cos φ = 2 and φ is in QI.1.861 Now. and hence cos φ = 2 and sin θ = 3 . . 2 sin θ − cos 2θ = 0 6. and the leg with length 2 > 0 means that φ is to the right of the y-axis. The hypotenuse has length 13 by the Pythagorean Theorem.Chapter 6 • Additional Topics 132 §6. . 2 sin θ − 3 cos θ = 0 11. ± 2.983 + 2π k for k = 0. 13 13 3 φ 2 Figure 6.281 = 2. which in general is not the case. 2nd ed. we have to resort to numerical methods.2 Numerical Methods in Trigonometry We were able to solve the trigonometric equations in the previous section fairly easily. Instead.2.1 below. and it ﬁnds roots of a given function f (x). There is a large ﬁeld of mathematics devoted to this subject called numerical analysis. i. as shown in Figure 6.1) There is a solution. It is called the secant method. consider the equation cos x = x . The graphs of y = cos x and y = x intersect somewhere between x = 0 and x = 1. which provide ways of getting successively better approximations to the actual solution(s) to within any desired degree of accuracy. A First Course in Numerical Analysis. but luckily there is a method which we can use that requires just basic algebra.Numerical Methods in Trigonometry • Section 6. New York: McGraw-Hill Book Co. 1] such that cos x = x. R ABINOWITZ . A derivation of the secant method is beyond the scope of this book.2 133 6.. Many of the methods require calculus.1 y = cos x and y = x Unfortunately there is no trigonometric identity or simple method which will help us here. . R ALSTON AND P. 4 cos(x) x 3 2 y 1 0 -1 -2 -3 -4 -3 -2 -1 0 1 2 3 x Figure 6. see pp. which means that there is an x in the interval [0. 1978. 338-344 in A. (6. values of x such that f (x) = 0.e.2.1 but we can state the algorithm it uses to solve f (x) = 0: 1 For an explanation of why the secant method works. For example.. A better way to implement the algorithm is with a computer.6851) − f (1) (0. of course).6851 − 0.7363 .e. using the Java programming language: .6851 − 1) (0.6851 . 1]. x1 . Pick initial points x0 and x1 such that x0 < x1 and f (x0 ) f (x1 ) < 0 (i. 2.0893 − (−0. x2 ..java) for solving cos x = x with the secant method.6851 − f (0. getting as close as desired. so that f (x0 ) f (x1 ) < 0 (we are using radians. 3.134 Chapter 6 • Additional Topics §6.4597) = 1 − −0.1 below shows the code (secant.. The solution to that equation is the root of the function f (x) = cos x − x. will approach the solution x as we go through more iterations. Using a calculator is not very efﬁcient and will lead to rounding errors.4597 − 1 = 0.4597) = 0. Listing 6.2) as long as \| xn−1 − xn−2 \| > error . The numbers x0 . where error > 0 is the maximum amount of error desired (usually a very small number). x 3 = x2 − and so on. Then by deﬁnition. Then f (0) = 1 and f (1) = −0. x2 = x1 − (x2 − x1 ) f (x2 ) f (x2 ) − f (x1 ) (0. We will now show how to use this algorithm to solve the equation cos x = x.6851) = 0. For n ≥ 2. So pick x0 = 0 and x1 = 1.0893) = 0. x3 .2 1.4597. . (x1 − x0 ) f (x1 ) f (x1 ) − f (x0 ) (1 − 0) f (1) = 1 − f (1) − f (0) (1 − 0) (−0. And we saw that the solution is somewhere in the interval [0.6851 − 1) f (0. deﬁne the number xn by x n = x n −1 − (xn−1 − xn−2 ) f (xn−1 ) f (xn−1 ) − f (xn−2 ) (6. the solution is somewhere between x0 and x1 ). creating xn from xn−1 and xn−2 . 10 } else { 14 break.parseDouble(args[0]). 12 x1 = x.x1).e. Lines 23-24 give the deﬁnition of the function f (x) = cos x − x. .mc).. Line 7 sets the maximum error error to be 1. i <= 10. otherwise we continue. 6 double x = 0. 20 } 21 22 //Define the function f(x) public static double f (double x) { 23 return Math. Line 6 initializes the variable that will eventually hold the solution. we stop (since this means we have an acceptable solution).abs(x0 . 8 for (int i=2.java 1 2 import java.x. 24 } 25 26 } Lines 4-5 read in x0 and x1 as input parameters to the program. Line 8 starts a loop of 9 iterations of the algorithm. 7 double error = 1. 19 BigDecimal answer = new BigDecimal(x.println("x" + i + " = " + x). i++) { if (Double. System.1 Program listing for secant. That is. x0 = x1. 5 double x1 = Double. If they do.(x1 . 13 System. though in Line 9 we check to see if the two previous approximations differ by less than the maximum error. public class secant { public static void main (String[] args) { 3 4 double x0 = Double.error) > 0) { 9 11 x = x1 . x10 to the real solution.f(x0)). Lines 11-12 set the new values of xn−2 and xn−1 . Lines 18-20 set the number of decimal places to show in the ﬁnal answer to 50 (the default is 16) and then print the answer.2 135 Listing 6.0E-50.. our ﬁnal answer will be within that (tiny!) amount of the real solution. .Numerical Methods in Trigonometry • Section 6.parseDouble(args[1]).cos(x) . x3 .println("x = " + answer).. i.math.x0)f(x1)/(f(x1) . 15 } 16 17 } 18 MathContext mc = new MathContext(50).. respectively. Line 10 is the main step in the algorithm. it will create the successive approximations x2 .0 × 10−50 .compare(Math.out.out. 2 0 −π 2 -1 0 1 π 2 x Figure 6.73908513321516067229310920083662495017051696777344 Notice that the program only got up to x8 .6850733573260451 x3 = 0.7390851332151607 x8 = 0.7390851332151607 x = 0.7391193619116293 x5 = 0.2 gives an idea of why. since it is the answer to Exercise 11 in Section 4.7390851121274639 x6 = 0.0 × 10−50 ) to stop at x8 and call that our solution. is the solution of cos x = x.736298997613654 x4 = 0. The last line shows that solution to 50 decimal places.java java secant 0 1 x2 = 0.6 0.2.4 y=x 0. 1 y = cos(x) 0. you would get cos (cos x) = cos x = x. not x10 . you eventually start getting the above number repeatedly after enough iterations. when taking repeated cosines starting with any number (in radians). so cos (cos (cos x)) = cos x = x..2 Below is the result of compiling and running the program using x0 = 0 and x1 = 1: javac secant.73908513321516.7390851332150012 x7 = 0. and so on. That is..136 Chapter 6 • Additional Topics §6. Does that number look familiar? It should. The reason is that the difference between x8 and x7 was small enough (less than error = 1.2 Attractive ﬁxed point for cos x Since x = 0. This number x is called an attractive ﬁxed . Figure 6.1. This turns out not to be a coincidence.8 y 0.2. 2 shows what happens when starting at x = 0: taking the cosine of 0 takes you to 1. However. a triangle with two sides whose sum is barely larger than the third side.65792064609274 and that the minimum is −1.Numerical Methods in Trigonometry • Section 6. but will often have a problem when used in a calculator with.’f’) says to use x = 2 as a ﬁrst approximation of the number x where f (x) is a maximum.’f’) y = -1.65792064609274 The output says that the minimum occurs when x = 2. Most 2 Freely available at http://www.gnu. No matter where you start.90596111871578.2 that we claimed that the maximum and minimum of the function y = cos 6x + sin 4x were ± 1. Figure 6. say.90596111871578.4 that Heron’s formula is adequate for “typical” triangles.2 Octave uses a successive quadratic programming method to ﬁnd the minimum of a function f (x). We can show this by using the open-source program Octave.90596111871578 ans = 2.90596111871578) = 1.org/software/octave . you end up getting “drawn” to it. octave:1> format long octave:2> function y = f(x) > y = cos(6x) + sin(4x) > endfunction octave:3> sqp(3.2. the solution). and then successive cosines (indicated by the intersections of the vertical lines with the cosine curve) eventually “spiral” in a rectangular fashion to the ﬁxed point (i. The command sqp(3. octave:4> function y = f(x) > y = -cos(6x) . To ﬁnd the maximum of f (x).10 in Section 5. we ﬁnd the minimum of − f (x) and then take its negative.sin(4x) > endfunction octave:5> sqp(2. The command sqp(2.90596111871578. Recall from Section 2. Finding the maximum of f (x) is the same as ﬁnding the minimum of − f (x) then multiplying by −1 (why?).’f’) says to use x = 3 as a ﬁrst approximation of the number x where f (x) is a minimum.’f’) y = -1.2 137 point of the function cos x. you can get around this problem by using computer software capable of handling numbers with a high degree of precision. Below we show the commands to run at the Octave command prompt (octave:n>) to ﬁnd the minimum of f (x) = cos 6x + sin 4x.05446832062993 The output says that the maximum occurs when x = 2.e. which is the intersection of y = cos x and y = x. Recall in Example 5.05446832062993 and that the maximum is −(−1. respectively.90596111871578 ans = 2. Thus.3 139 6. However.Complex Numbers • Section 6. and hence i = −1. and if b .3 Complex Numbers There is no real number x such that x2 = −1. If a and b are real numbers. then a number of the form a + bi is called a complex number. i 2 = −1. it turns out to be useful6 to invent such a number. called the imaginary unit and denoted by the letter i. = 0 then it is called an imaginary number (and pure imaginary if a = 0 and b . (ac + bd) + (bc − ad)i a + bi = c + di c2 + d 2 The ﬁrst three items above are just deﬁnitions of equality. the real parts are equal and the imaginary parts are equal) 2. The last three items can be derived by treating the multiplication and division of complex numbers as you would normally treat factors of real numbers: (a + bi) (c + di) = a (c + di) + bi (c + di) = ac + adi + bci + bdi 2 = ac + adi + bci + bd(−1) = (ac − bd) + (ad + bc)i 6 Especially in electrical engineering. and various ﬁelds of mathematics. . adding a real number and an imaginary number? You can think of it as a way of extending the set of real numbers. If b = 0 then a + bi = a + 0i = a (since 0i is deﬁned as 0). The imaginary part bi in a + bi can be thought of as a way of taking the one-dimensional set of all real numbers and extending it to a two-dimensional set: there is a natural correspondence between a complex number a + bi and a point (a. add the real parts together and add the imaginary parts together) 3. (a + bi) − (c + di) = (a − c) + (b − d)i 4. (a + bi) (a − bi) = a2 + b2 6.e. (a + bi) + (c + di) = (a + c) + (b + d)i (i.= 0). (a + bi) (c + di) = (ac − bd) + (ad + bc)i 5. addition.e. b) in the (two-dimensional) x y-coordinate plane. The real number a is called the real part of the complex number a + bi. and subtraction of complex numbers. we will ﬁrst state some fundamental properties of and operations on complex numbers: Let a + bi and c + di be complex numbers. i.e. a + bi = c + di if and only if a = c and b = d (i. physics. so that every real number is a complex number. What does it mean to add a to bi in the deﬁnition a + bi of a complex number. Then: 1. Before exploring that correspondence further. and bi is called its imaginary part. Notice that (a + bi) + (a + bi) = 2a is a real number. (a + bi) − (a + bi) = 2bi is an imaginary number if b .3 The ﬁfth item is a special case of the multiplication formula: (a + bi) (a − bi) = ((a)(a) − (b)(− b)) + ((a)(− b) + (b)(a))i = (a2 + b2 ) + (−ab + ba)i = (a2 + b2 ) + 0i = a2 + b 2 The sixth item comes from using the previous items: a + bi c − di a + bi = · c + di c + di c − di (ac − b(− d)) + (a(− d) + bc)i = c2 + d 2 (ac + bd) + (bc − ad)i = c2 + d 2 The conjugate a + bi of a complex number a + bi is deﬁned as a + bi = a − bi.140 Chapter 6 • Additional Topics §6. which we denote by \| z \|. and \| z2 \|. Example 6. Find z1 + z2 . z1 /z2 . \| z1 \|. z1 − z2 . z z = a2 + b2 and thus we can deﬁne the modulus of z to be z z = a2 + b2 .= 0.9 Let z1 = −2 + 3i and z2 = 3 + 4i. and (a + bi)(a + bi) = a2 + b2 is a real number. So for a complex number z = a + bi. Solution: Using our rules and deﬁnitions. z1 z2 . we have: z1 + z2 = (−2 + 3i) + (3 + 4i) = 1 + 7i z1 − z2 = (−2 + 3i) − (3 + 4i) = −5 − i z1 z2 = (−2 + 3i) (3 + 4i) = ((−2)(3) − (3)(4)) + ((−2)(4) + (3)(3))i = −18 + i z1 −2 + 3i = z2 3 + 4i (−2)(3) + (3)(4) + ((3)(3) − (−2)(4))i = 32 + 42 6 17 = + i 25 25 \| z1 \| = (−2)2 + 32 = 13 \| z2 \| = 32 + 42 = 5 . Complex Numbers • Section 6. as in Figure 6. And we see from Figure 6..4) In the special case z = 0 = 0 + 0i.3. θ must satisfy y y tan θ = x . which is just the modulus of z.3. We can represent z as a point in the complex plane. cos θ = xr . y) Figure 6. Thus. and the vertical y-axis represents the pure imaginary part of z. . The representation z = r (cos θ + i sin θ ) is often abbreviated as: z = r cis θ (6. We call this angle θ the argument of z. respectively.3) θ = the argument of z . The distance r from z to the origin is. r = x2 + y2 . the argument θ is undeﬁned since r = \| z \| = 0. by the Pythagorean Theorem. Also. we get the trigonometric form (sometimes called the polar form) of the complex number z: For any complex number z = x + yi.1(a). depending on whether you are using degrees or radians. ± 2. y y (x. where θ is the angle in standard position as in Figure 6. y) = (r cos θ .1(b).. where r = \| z \| = x2 + y2 and (6.3 141 We know that any point (x. where θ is the angle formed by the positive x-axis and the line segment from the origin to z.3. . note that the argument θ can be replaced by θ + 360◦ k or θ + π k. Note also that for z = x + yi with r = \| z \|. y) in the x y-coordinate plane that is a distance r > 0 from the origin has coordinates x = r cos θ and y = r sin θ . we can write z = r (cos θ + i sin θ ) . r sin θ ) z = x + yi = r cos θ + (r sin θ )i r θ 0 r θ x 0 x (b) Complex number z = x + yi (a) Point (x.3. ± 1. sin θ = r .1(b) that x = r cos θ and y = r sin θ . for k = 0.1 Let z = x + yi be a complex number. where the horizontal x-axis represents the real part of z. . we have the following formulas for multiplication and division: Let z1 = r 1 (cos θ1 + i sin θ1 ) and z2 = r 2 (cos θ2 + i sin θ2 ) be complex numbers. Also. r = x2 + y2 = (−2)2 + (−1)2 = 5 .6◦ .3 Example 6.3. and r1 z1 = (cos (θ1 − θ2 ) + i sin (θ1 − θ2 )) if z2 .3. we have ◦ θ = 206.6◦ ) . Then z1 z2 = r 1 r 2 (cos (θ1 + θ2 ) + i sin (θ1 + θ2 )) . or 5 cis 206.142 Chapter 6 • Additional Topics §6. Solution: Let z = −2 − i = x + yi. −2 − i = y θ 2 1 5 (cos 206. So since tan θ = x = − −2 = 2 .6◦ + i sin 206. Thus.2. Then θ is y 1 1 in QIII. so that x = −2 and y = −1. as we see in Figure 6.10 Represent the complex number −2 − i in trigonometric form.6 . 0 x r z = −2 − i Figure 6.2 For complex numbers in trigonometric form. 6) say that when multiplying complex numbers the moduli are multiplied and the arguments are added. while when dividing complex numbers the moduli are divided and the arguments are subtracted. This makes working with complex numbers in trigonometric form fairly simple.6) The proofs of these formulas are straightforward: z1 z2 = r 1 (cos θ1 + i sin θ1 ) · r 2 (cos θ2 + i sin θ2 ) = r 1 r 2 [(cos θ1 cos θ2 − sin θ1 sin θ2 ) + i (sin θ1 cos θ2 + cos θ1 sin θ2 )] = r 1 r 2 (cos (θ1 + θ2 ) + i sin (θ1 + θ2 )) by the addition formulas for sine and cosine. z2 r2 (6. And r 1 (cos θ1 + i sin θ1 ) z1 = z2 r 2 (cos θ2 + i sin θ2 ) r 1 cos θ1 + i sin θ1 cos θ2 − i sin θ2 = · · r 2 cos θ2 + i sin θ2 cos θ2 − i sin θ2 r 1 (cos θ1 cos θ2 + sin θ1 sin θ2 ) + i (sin θ1 cos θ2 − cos θ1 sin θ2 ) = · r2 cos2 θ2 + sin2 θ2 r1 = (cos (θ1 − θ2 ) + i sin (θ1 − θ2 )) r2 by the subtraction formulas for sine and cosine. . and since cos2 θ2 + sin2 θ2 = 1.5) and (6. QED Note that formulas (6.5) (6.= 0. and ⇒ z1 = 3 (cos 39◦ + i sin 39◦ ) .5). Solution: By formulas (6.V.6) we have z1 z2 = (6) (2) (cos (70◦ + 31◦ ) + i sin (70◦ + 31◦ )) 6 z1 = (cos (70◦ − 31◦ ) + i sin (70◦ − 31◦ )) z2 2 ⇒ z1 z2 = 12 (cos 101◦ + i sin 101◦ ) .7) We deﬁne z0 = 1 and z−n = 1/z n for all integers n ≥ 1.Complex Numbers • Section 6. 1960. De Moivre’s Theorem:7 For any integer n ≥ 1. and so [r (cos θ + i sin θ )]3 = [r (cos θ + i sin θ )]2 · r (cos θ + i sin θ ) = r 2 (cos 2θ + i sin 2θ ) · r (cos θ + i sin θ ) = r 3 (cos (2θ + θ ) + i sin (2θ + θ )) = r 3 (cos 3θ + i sin 3θ ) . we get: Theorem 6. (6. 2nd ed.e. by mathematical induction). [r (cos θ + i sin θ )]n = r n (cos nθ + i sin nθ ) . so that De Moivre’s Theorem holds for any real number n.11 Let z1 = 6 (cos 70◦ + i sin 70◦ ) and z1 = 2 (cos 31◦ + i sin 31◦ ).. and continuing like this (i.3 143 Example 6. C HURCHILL..5).5) and (6. z−n = and so De Moivre’s Theorem in fact holds for all integers. . 8 There is a way of deﬁning z n when n is a real (or complex) number. Find z1 z2 and z1 z2 . Complex Variables and Applications.8 7 Named after the French statistician and mathematician Abraham de Moivre (1667-1754).1. for any z = r (cos θ + i sin θ ) and integer n ≥ 1 we get 1 zn 1 (cos 0◦ + i sin 0◦ ) = n r (cos nθ + i sin nθ ) 1 = n (cos (0◦ − nθ ) + i sin (0◦ − nθ )) r = r −n (cos (− nθ ) + i sin (− nθ )) . we have [r (cos θ + i sin θ )]2 = r · r (cos (θ + θ ) + i sin (θ + θ )) = r 2 (cos 2θ + i sin 2θ ) . See pp. 59-60 in R. New York: McGraw-Hill Book Co. So by De Moivre’s Theorem and formula (6. z2 For the special case when z1 = z2 = z = r (cos θ + i sin θ ) in formula (6. Solution: Since i = 1 (cos 90◦ + i sin 90◦ )..13 Find the three cube roots of i.3 Example 6. + i sin = cos 150 + i sin 150 = − 3 3 2 2 ◦ 90 + 360◦ (2) 90◦ + 360◦ (2) 3 1 cos + i sin = cos 270◦ + i sin 270◦ = − i 3 3 .12 Find (1 + i)10 . Let z = r (cos θ + i sin θ ). That is. Since the cosine and sine functions repeat every 360◦ . the three cube roots of i are: ◦ ◦ 90 + 360◦ (0) 90 + 360◦ (0) 3 1 cos + i sin = cos 30◦ + i sin 30◦ 3 3 3 1 = + i . we get the following formula for the For any nonzero complex number z = r (cos θ + i sin θ ) and positive integer n. 1. Now let w = r 0 (cos θ0 + i sin θ0 ) be an n th root of z. ± 2. n − 1. we know that z = r (cos (θ + 360◦ k) + i sin (θ + 360◦ k)) for k = 0. Solution: Since 1 + i = (1 + i)10 We can use De Moivre’s Theorem to ﬁnd the n th roots of a complex number. will repeat for k ≥ n. 2.144 Chapter 6 • Additional Topics §6.8) + i sin cos r n n for k = 0. Then wn = z ⇒ [r 0 (cos θ0 + i sin θ0 )]n = r (cos (θ + 360◦ k) + i sin (θ + 360◦ k)) ⇒ r 0n (cos nθ0 + i sin nθ0 ) = r (cos (θ + 360◦ k) + i sin (θ + 360◦ k)) ⇒ r 0n = r ⇒ r 0 = r 1/n Since the cosine and sine of n th roots of z: and nθ0 = θ + 360◦ k and θ0 = θ +360◦ k n θ + 360◦ k n . . by De Moivre’s Theorem we have 10 = ( 2) (cos 450◦ + i sin 450◦ ) = 210/2 (0 + i (1)) = 25 · i = 32i . the n distinct n th roots of z are θ + 360◦ k θ + 360◦ k 1/n (6. 2 2 ◦ ◦ 90 + 360◦ (1) 90 + 360◦ (1) 3 1 3 ◦ ◦ 1 cos + i .. Example 6. Note: An n th root of z is usually written as z1/n or n z.. given any complex number z and positive integer n.. 2 (cos 45◦ + i sin 45◦ ) (why?). ﬁnd all complex numbers w such that w n = z.. . ± 1.. The number r 1/n in the above formula is the usual real n th root of the real number r = \| z \|. 3 (cos 14◦ + i sin 14◦ ) · 2 (cos 121◦ + i sin 121◦ ) 33. put the given number in trigonometric form. 41. then so is z. every real number a has n n th roots (being the roots of z n − a). i 8 16. as shown in Figure 6. Find the three cube roots of 1. 25.) . 32.3.Complex Numbers • Section 6. i 6 14. 37. Prove that if z is an n th root of a real number a. \| z \| = \| z \| 23. (2 + 3i) + (−3 − 2i) 6. 1. [3 (cos 14◦ + i sin 14◦ )]4 34. 3 (cos 14◦ + i sin 14◦ ) 2 (cos 121◦ + i sin 121◦ ) 36. z1 − z2 = z1 − z2 22. −3 − 2i 27. calculate the given number. −i In higher mathematics the Fundamental Theorem of Algebra states that every polynomial of degree n with complex Figure 6. \| z1 z2 \| = \| z1 \| \| z2 \| 20. 39. i 9 10. 1 − i 28. In general. 43. (2 + 3i)/(−3 − 2i) 5. (2 + 3i) − (−3 − 2i) 7. i 7 15. (2 + 3i) − (−3 − 2i) 3. and the square roots of −1 are ± i. 38. −1 31. For Exercises 32-35. i 3 11. For example. i 4 12. − i 29. We see that consec3 3 i i 120◦ − + ◦ th 2 2 2 +2 utive cube roots are 120 apart. Find the three cube roots of − i. 17.3 145 y Notice from Example 6. (2 + 3i) + (−3 − 2i) 2. [3 (cos 14◦ + i sin 14◦ )]−4 35. = z2 \| z2 \| For Exercises 25-30. the n n roots of x a complex number z will be equally spaced points along the ◦ ◦ 1/n 120 120 circle of radius \| z \| in the complex plane. i 5 13. 1 30. the square roots of 1 are ± 1. Find the two square roots of −2 + 2 3 i. (1 + i)/(1 − i) 8. Find the three cube roots of 1 + i. Exercises For Exercises 1-16. (Hint: Use Exercise 20. 2 + 3i 26. In particular. (2 + 3i) · (−3 − 2i) 4. 42. with consecutive ◦ roots separated by 360 n . = z2 z2 18. 40.13 that the three cube roots of i \| z\| = 1 are equally spaced points along the unit circle \| z \| = 1 in the complex plane. z1 + z2 = z1 + z2 19. i 2009 For Exercises 17-24. z1 z2 = z1 z2 z1 \| z1 \| 24.3. Find the ﬁve ﬁfth roots of 1.3.3 coefﬁcients has n complex roots (some of which may repeat). prove the given identity for all complex numbers. Find the three cube roots of −1. (z) = z z1 z1 21. calculate the given expression. Verify that De Moivre’s Theorem holds for the power n = 0. \|−3 + 2i \| 9. Find the ﬁve ﬁfth roots of −1. the ray OP is drawn in the opposite direction from the angle θ . θ + 360◦ k) for k = 0. so (unlike for Cartesian coordinates) the polar coordinates of a point are not unique. .4.2 Polar coordinates (r.2.4. y y P(r.3. as in Figure 6. θ ) the polar coordinates of P.1 and the positive x-axis is called the polar axis of this coordinate system. You may be familiar with graphing paper. since its graph violates the vertical rule. Note that (r. by deﬁning (− r. θ ) is the origin O. θ ) = (r. 0) in the x y-coordinate plane we draw a spiral around the origin. regardless of the value of θ . Such paper consists of a rectangular grid. x 0 1 2 3 However. as in Figure 6. θ ) In polar coordinates we adopt the convention that r can be negative.1. for plotting points or functions given in Cartesian coordinates (sometimes also called rectangular coordinates). θ ) Figure 6.4.3 Negative r: (− r. Recall that any point P distinct from the origin (denoted by O) in the x y-coordinate plane is a distance r > 0 from the origin. We ← 1→ can not express this spiral as y = f (x) for some function f in Cartesian coordinates. −−→ and the ray OP makes an angle θ with the positive x-axis.4.4 Polar Coordinates y Suppose that from the point (1.. such that the distance between any two points separated by 360◦ along the ← 1→ spiral is always 1. similar to Figure 6. as in Figure 6. θ ) = (0.. this spiral would be simple to describe using the polar coordinate system.4. ± 1.146 Chapter 6 • Additional Topics §6. . θ ) r θ O θ x −r O x P(− r. ± 2. θ + 180◦ ) for any angle θ . the point (r.4. When r = 0. We call the pair (r.4.4 6. θ ) Figure 6. θ ) = −−→ (r.4.. Similar graphing paper exists for plotting points and functions in polar coordinates. That is. Figure 6. For example.4.4 105◦ 90◦ 147 75◦ 120◦ 60◦ 135◦ 45◦ 150◦ 30◦ 165◦ 15◦ 180◦ 0◦ O 345◦ 195◦ 210◦ 330◦ 225◦ 315◦ 240◦ 300◦ 255◦ Figure 6.4 270◦ 285◦ Polar coordinate graph The angle θ can be given in either degrees or radians. . unlike degrees. The reason is that. in Cartesian coordinates (x. whichever is more convenient. functions are usually expressed as y as a function of x). then r would have the same units as θ . hence using radians for θ makes more sense in this case. But r should be a unitless quantity. radians can be considered “unitless” (as we mentioned in Chapter 4).Polar Coordinates • Section 6. y). Radians are often preferred when graphing functions in polar coordinates. This is desirable when a function given in polar coordinates is expressed as r as a function of θ (similar to how. if a function in polar coordinates is written as r = 2 θ . 4. θ θ θ = 2π k ⇒ k = ⇒ r = 1+k = 1+ . . Hence. We see that θ = 0 ⇒ r = 1 θ = 2π ⇒ r = 2 θ = 4π ⇒ r = 3 . Gnuplot will assume that the function being plotted is r as a function of θ (represented by the variable t in Gnuplot). the spiral can be written as r = 1 + 4 set polar set size square set samples 2000 3 unset key set zeroaxis set xlabel "x" 1 y set ylabel "y" plot [0:6pi] 1 + t/(2pi) 2 0 1 2 3 4 4 3 2 1 0 1 2 3 4 x Figure 6. In fact.14 Express the spiral from Figure 6.1 in polar coordinates. 2π 2π θ for θ ≥ 0. The goal is to ﬁnd some equation involving r and θ that describes the spiral. along 2π with the Gnuplot commands to create the graph. . 2.5 r = 1 + 2θπ Note that when using the set polar command.4..4. 1. Solution: We will use radians for θ .5. . The graph is shown in Figure 6.. . θ = 2π k ⇒ r = 1+k for k = 0.4 Example 6. . that last relation holds for any nonnegative real number k (why?).148 Chapter 6 • Additional Topics §6. So for any θ ≥ 0. 6 shows how to convert between polar coordinates and Cartesian coordinates.4.9) Cartesian to Polar: r = ± x x Figure 6.Polar Coordinates • Section 6.4. y): Polar to Cartesian: r y θ x = r cos θ y = r sin θ O (6. θ ) Figure 6.6 x 2 + y2 tan θ = y if x . For a point with polar coordinates (r. θ ) and Cartesian coordinates (x.4 149 y (x. y) (r. 10). if x = 0 then θ = π/2 or θ = 3π/2.= 0 x (6. Also. if x .10) Note that in formula (6. = 0 and y . 2 2 (c) Using formula (6. we can take r = x2 + y2 = 32 + 42 = 5. r sin θ ) = 3 cos 3π 3π 4 . Example 6. 3 sin 4 = 3· −1 .e. Note that if we had used θ = 233. (c) (−1.16 Convert the following points from Cartesian coordinates to polar coordinates: (a) (3. we get: (x.9) with r = −1 and θ = 5π/3. r is positive).3 2 · 1 2 (x.9) with r = 2 and θ = 30◦ . 3π/4). −5) Solution: (a) Using formula (6. 30◦ ). r is negative). 2 sin 30◦ ) = 2 · 3 2 . otherwise r = − x2 + y2 (i. we get: (x. θ ) = (−5. If the angle θ is inthe same quadrant as the point (x. we get: tan θ = y 4 = x 3 ⇒ θ = 53. (b) (3. θ ) = (5.13◦ is in the same quadrant (QI) as the point (x. y) = − 12 . y) = (r cos θ . 1 (b) Using formula (6. we get: (x. y) = (3. y). −1 · − 3 2 ⇒ (x.13◦ ).9) with r = 3 and θ = 3π/4. r sin θ ) = (2 cos 30◦ .e. −1 sin 3 = −1 · 1 2 . y) = (r cos θ . (b) (−5.2 · 1 2 ⇒ (x.15 Convert the following points from polar coordinates to Cartesian coordinates: (a) (2. 5π/3) Solution: (a) Using formula (6. y) = 3. 23 Example 6. 233. Thus. (r. 4).13◦ . y) = (r cos θ .13◦ Since θ = 53. 53.13◦ ) .10) with x = 3 and y = 4. 4).13◦ or θ = 233.= 0 then y the two possible solutions for θ in the equation tan θ = x are in opposite quadrants (for 0 ≤ θ < 2π). y) = ⇒ −3 3 . then we would have (r. r sin θ ) = −1 cos 5π 5π 3 . . then r = x2 + y2 (i. 4 (b) Using formula (6. Thus. So when x = 0. 225◦ ) . Since r = ± x2 + y2 = ± 9.17 Write the equation x2 + y2 = 9 in polar coordinates. Solution: This is the equation of a line through the origin. Expanding the equation. θ ) = (5 2. Example 6. Example 6. 4). canceling r does not eliminate r = 0 as a potential solution of the equation (since θ = 0◦ would make r = 8 sin θ = 8 sin 0◦ = 0). then we would have (r. Note that if we had used θ = 45◦ . in polar coordinates the equation can be written as simply r = 3 . Thus. Solution: This is the equation of a circle of radius 4 centered at the point (0. Example 6. Solution: This is just the equation of a circle of radius 3 centered at the origin. we can take r = x2 + y2 = (−5)2 + (−5)2 = 5 2. we get: tan θ = y −5 = = 1 x −5 ⇒ θ = 45◦ or θ = 225◦ Since θ = 225◦ is in the same quadrant (QIII) as the point (x. θ ) = (−5 2. When x . y) = (−5.18 Write the equation x2 + (y − 4)2 = 16 in polar coordinates. (r.10) with x = −5 and y = −5.19 Write the equation y = x in polar coordinates. −5). 0) is on the circle. we get: x2 + (y − 4)2 = 16 x2 + y2 − 8y + 16 = 16 x2 + y2 = 8y r 2 = 8 r sin θ r = 8 sin θ Why could we cancel r from both sides in the last step? Because the point (0.150 Chapter 6 • Additional Topics §6. we know that y = 0. the equation is r = 8 sin θ . 45◦ ). 0). which is the same as (r. which would take care of the case x = 0 (since then (x. . y) = (0. Thus. the equation is θ = 45◦ .= 0. we get: y = x y = 1 x tan θ = 1 θ = 45◦ Since there is no restriction on r. we could have r = 0 and θ = 45◦ . 45◦ )). θ ) = (0. 210◦ ) 2. (−1. 14. y2 ) are the Cartesian equivalents of (r 1 . 6. (6. 1) 7. by the Cartesian coordinate distance formula. 90◦ ) 5. convert the given point from Cartesian coordinates to polar coordinates. 3π) 3. θ2 ) in polar coordinates is d = r 21 + r 22 − 2r 1 r 2 cos (θ1 − θ2 ) . (−1. θ1 ) and (r 2 . 3x2 + 4y2 − 6x = 9 . so the result follows by taking square roots of both sides. (4. In general. θ2 ). −3) 9. (6. (0. −2) 8.4 151 Example 6. Graph the function r = 1 + 2 cos θ in polar coordinates. (−2. 0) For Exercises 11-18. In Example 6. (−4. 2) 10. d 2 = (x1 − x2 )2 + (y1 − y2 )2 = (r 1 cos θ1 − r 2 cos θ2 )2 + (r 1 sin θ1 − r 2 sin θ2 )2 = r 21 cos2 θ1 − 2r 1 r 2 cos θ1 cos θ2 + r 22 cos2 θ2 + r 21 sin2 θ1 − 2r 1 r 2 sin θ1 sin θ2 + r 22 sin2 θ2 = r 21 (cos2 θ1 + sin2 θ1 ) + r 22 (cos2 θ2 + sin2 θ2 ) − 2r 1 r 2 (cos θ1 cos θ2 + sin θ1 sin θ2 ) d 2 = r 21 + r 22 − 2r 1 r 2 cos (θ1 − θ2 ) . x2 − y2 = 1 15. polar coordinates are useful in situations when there is symmetry about the origin (though there are other situations). (3.11) Solution: The idea here is to use the distance formula in Cartesian coordinates. y1 ) and (x2 . then convert that to polar coordinates. which arise in many physical applications. so the circle is symmetric about the origin. convert the given point from polar coordinates to Cartesian coordinates. So write x1 = r 1 cos θ1 y1 = r 1 sin θ1 x2 = r 2 cos θ2 y2 = r 2 sin θ2 . This equation describes a circle centered at the origin. (6.20 Prove that the distance d between two points (r 1 . Thus. 1.17 we saw that the equation x2 + y2 = 9 in Cartesian coordinates could be expressed as r = 3 in polar coordinates. respectively. (x − 3)2 + y2 = 9 12.Polar Coordinates • Section 6. θ1 ) and (r 2 . (2. y = − x 13. Exercises For Exercises 1-5. 405◦ ) For Exercises 6-10. 11. write the given equation in polar coordinates. Then (x1 . 11π/6) 4. 84. c = 5. QIV 15.112 cm 27. 1.5 ft = 0. 33. Hint: Are the opposite sides of the four. (a) 13/4 (b) 4 3/ 13 (c) 3/ 13 1.Appendix A Answers and Hints to Selected Exercises Chapter 1 13. sec B = 13/5. sec = 106/9. B = 67. sin A= 1/ 10. c = 2.28. cot B = 1/3 9. sec A = 13/12. Hint: Draw a sided ﬁgure inside the circle parallel? right triangle with an acute angle A. 15.15. a = csc B = 25/24. tan B = 24/7. (a) 3 a (b) 35. 25 miles 10.989◦ tan A = 40/3. 1379. cos A = 11/6. cos 3◦ 21. cot = 9/40 csc B = 41/40. sin 77◦ 27. QI. QII 17. tan A= 5/ 11. c = 13. cot B = 5/12 1. 274 ft 7.476 in 11. A = 9◦ . Section 1. sec A = 41/40. csc 13◦ 15. 21. sec A = 10/3. Partial answer: DE = a cot θ cos2 θ sin B = 24/25. in 14. 115◦ 3. 31) csc B = 10/3. cos A = 40/41. cos A = 3/7.02. cot A = 40/9. sec B = 10. QIII 13. 29.2613 mi 10. = 6/ 10. cot = 5/ 11 23. 12) are the same length. QIV 5.3 (p. Hint: Draw two right triangles whose hypotenuses Section 1. 15.4◦ 17. sin B = 11/6. sin A = 5/6. (a) 0.7 ft 3. 0. A = 52◦ . csc B = 13/12. csc A = cot A = 3.26◦ 7. B = 45◦ 25. csc A = 13/5.1 ft 4. cos B = 7/25. cot = 2/ 6 A Section 1. 37. cot B = 7/24 0. cot A = 12/5. tan B = 12/5. 20) sin B = 12/13. cos = 9/ 106.2 (p. sec A = 10/2. a = 6. sin θ = 3/2 and tan θ = − 3. B = 82◦ 19. A = 64◦ 23. cot A = 9/5 A A 1. tan B = 40/9. tan A = 5/12. mi 9. 152 . sin A A = 6/2. sec A = 6/ 11. 43◦ 19. sin A A csc = 106/5. 54◦ ◦ sec = 6/5. B = 81◦ 8. b = 6. B B (b) 2. sin B = 3/ 10. sin A = 7/25. 1. QI. tan 80◦ 30. csc B = 6/ 11. cos A = 7/4. sin θ = − 3/2 and tan θ = 3 sec A = 4/ 7. csc A = 4/3.6◦ . 21.011◦ and 89. 5) = 5/ 106. cos A = 3/ 10. A = 22.4866 csc A = 25/7. 111. cot A = 3/ 40 9.944 cm sec = 41/9. QIII 9. 85 B B 11. c = 6. 6.955 in 13. cos B = 5/6.84. QII 3. negative y-axis 6/5. sin A= 40/7. cos B = 9/41. cot A = 24/7. B = 104◦ 5. sin 44◦ 23. cos B = 1/ 10. tanA = 3/ 7.1 (p. tan B = 3. 45◦ 17. b = 2. tan csc A = 10/ 6. 241. 0. tan A = 1/3. 0.15. sin A = 5/13. sec B = 25/7. cos B = 5/13. cot A = 7/3 25. tan B = 11/5. sin B = 40/41. cos A = 24/25. 102. QIV 11.4 (p. tan A = 7/24. csc A = 7/ 40. cot A = 11/5. cos A = 12/13. sec A = 25/24.8 ft 19. 1062 3. B = 64◦ csc A = 41/9. tan A = 9/40. sin A = 9/41. QI. a = 2. Section 1. 7. csc A = 7.25. sin θ = 21/5 and tan θ = 21/2. 25◦ . 64) sin θ = −5/13 and cos θ = −12/13 37. No 39. 76) 611 3.1 (p. sin θ = 1/ 5 and cos θ = −2/ 5. Hints: The diagonals break the 29. r = 0. also. No solution 9. 2 − 3 15. 155◦ ◦ ◦ 11.9 . 70) 1.1◦ .51.7. 422 mi/hr 15. 49) Chapter 3 Section 3. (a) 328◦ (b) 148◦ (c) 212◦ (b) 68◦ (c) 292◦ 7. 34 15 ≈ 2. a = 9. 295◦ Section 3. cos 1020 tan (A + B) = − 611 4.9. 19. 9.905 cos θ = − 3/2 and tan θ = −1/ 3 7.25).1. C = 120◦ 3.5 (p. C = 109. ab + b 2 a2 + b 2 3. Section 2. r = 1.36 5.2 (p. b = 10. 1189 . 1. 218◦ 13. c = 15. (a) 248◦ 9. 37) (d) Hint: Bisect each angle.1 (p.21 5.7. b = 7. b = 9. Hint: Think geometrically. Hint: See Example 3. 65◦ . cos θ 1.18. b = 24. D = a2ab . ( 6 + 2)/4 5.Appendix A: Answers and Hints to Selected Exercises 153 sin θ = − 21/5 and tan θ = − 21/2 Section 2. 58) = 3/2 and tan θ = 1/ 3.12 12. 169 .23)-(2. cos θ = 0 sider formulas (2. No 1. C = 18. tan θ = ± sin θ / 1 − sin2 θ = ± 1 − cos2 θ / cos θ Chapter 2 Section 2.66 cm and 12.5◦ . 22. Section 2. conand tan θ is undeﬁned 31.5 (p.9◦ 11.4. 38◦ . b = 65. R = 24.69 3. R = 3. B = 20.63. (c) Twice as large Section 1. cos θ = ±1 and tan θ = 0 quadrilateral into four triangles.21 9. r = 1. θ = 270◦ 3. C = 95◦ 5. C = 70.1. 12.2 (p. sin θ = 5/13 and cos θ = 12/13. 33. B = 136.55 3.5◦ ◦ B = 59.7.9. 27.4 (p. R = 2.86 cm 16. 7. 5. sin (A + B) = 1020 (A + B) = − 1189 . 43) 1.1◦ . sin θ = −1/ 5 and cos θ = 2/ 5 35. 191 2 2 15. Hint: For a . = 0 and b . 2. a = 10.6 47. 53) 1. . A = 47. C = 72. c = 10.13 and 8. 86) 3. 59◦ . Hint: Think of Exercise 10.3 (p.2. Hints: One of the angles in 9.5◦ 11.1◦ 5.1◦ 5.4 (p. A = 13. b = 8. Section 3.9◦ . Section 3. Yes of Tangents. No 11. No solution 7. C = 79. Hint: One way to do this is with the Law 6.3 (p.1◦ .91 cm 9. B = 40. B = 40. Another way is with the Law of Sines.1 3. the formulas is a right angle. draw a right triangle with legs of lengths a and b.5◦ .9◦ . A = 79.6.= 0.9◦ . 7 cm 15. 70.9◦ . 81) 4. Hint: Is sin A + cos A always positive? 11. C = 72. use the 1/2 deﬁnition of cosine. Section 2. 1. also. 50. b = 8. amplitude = 1. period = 2.522 m2 Section 6. 1. period = 2 21.392 and 9. ± 11π/2. 32 rpm and 21. ω = 0. at x = ± 3π/2.3 (p. 151) shift = −π/5 7. .6 m/sec. ± π6 + π k doubled. amplitude = 1. out of phase 18. ω = 1. period = 2π/5. −3π/5 7. 40. 128) 1. 138) 1. 132) 19.512 cm2 3. i 13. phase Section 6.3 (p. amplitude = 0. 81 cis 56◦ 35. amplitude = 1. π/45 174◦ 3. 118) 39. amplitude = 2 2. ν = 6.1 (p.4 ) 11. 0.154 Appendix A: Answers and Hints to Selected Exercises Chapter 4 Section 4.84 in/sec Section 6. 89) 1. 2 cis 2 cis 255◦ 135 . ω = 1. max. 3. 108) Hint:Use Exer( z) = a − bi = a + bi = z. (2 5.86 ft 8. r = 3/(2 − cos θ ) riod = π.5 m2 5. −1/ 2) 7.33 rpm 11. period = π. − i 11. 24.1 (p. 9. π/6 19.3 27. r = 6 cos θ 13. ◦ cise 20. Then z = a − bi. (− ◦ shift = 3π/2 11. 12/13 23. 5 cm2 9. 13. 3.± 5π/2. amplitude = 3. (−3 3. 2π 25.26 11.17 17. 310π + 2π5 k 5. ± 7π/2. cis 0◦ 33. pe. Partial answer: sec θ = OQ 29.821 + 2π k. x = 1. i 17. 2π3 k Section 4.. 98) = ± π/2. 0. −1 + i 3.9.6 cm 3.174 12.2 (p. 21. 0 5.94 in 7. 251.2 (p.46 15. 12 + 23 i. amplitude undeﬁned. 2 cis 315◦ 13. −π/3 13.18 9.6◦ ) 9. amplitude = 34. 38. 102) 1. −3) 3. 1. ( 3. 25. 34. in phase 19. Section 5.105828541 Section 4. − i 15. amplitude envelope: y = ± x2 29. period = π 17. phase 5. cis 324◦ 3. 0 11. −1. so Section 5. Let z = a + bi. 4π/5 17. 1. phase shift = 1/2 Chapter 5 . −13i 5. at x min. π/2 1.5 cis 253◦ 6 6 ◦ 6 37. 15. −0. 3. 34π + π k 3. π4 + π k 11.. period = 2π/5.017 cm2 13. r 2 cos 2θ = 1 14. . phase shift = 0 cis 180◦ . phase shift = −3/5 5. 1. 12 − 23 i 41. ν = 6 m/sec.± 9π/2. ( 10. 36◦ Section 4. cis 108◦ . 6 27. 7. 9. period = 2π 23. Sector area is quadrupled. π/2 25. amplitude undeﬁned. ν = 3.5 rad/sec 3. 13π/18 5. 23.375 rad 9.75 m/sec.963 + 2π k 9. π/2 cm2 11.3 (p. π/2 9.4 (p.. 48. 145) 1. 54.5. π 7. −1 − i 7. period = π/2. − phase shift = 3π/2 1) 1/ 2. 13 cis 56. 11π in 5.875 rad/sec 7. π/7 15.1 (p. π/4 3. 333. No Section 5. i 9. 3.2 (p. 269. −π/9 21.94 rad/sec 5.4 (p. 2 cis 15◦ . cis 36◦ . arc length is 1.. 94) 1.1 cm2 Chapter 6 7.89549426703398093962 Section 6. 12. cis 252◦ . for some numbers a < b and c < d. <range> is the range of x values (and optionally the range of y values) over which to plot. To specify an x range. This will cause the graph to be plotted for a ≤ x ≤ b and c ≤ y ≤ d. This will cause the graph to be plotted for a ≤ x ≤ b. Function deﬁnitions use the x variable in combination with mathematical operators. use an expression of the form [a : b]. for some numbers a < b. To specify an x range and a y range.156 Appendix B: Graphing with Gnuplot GRAPHING FUNCTIONS The usual way to create graphs in Gnuplot is with the plot command: plot <range> <comma-separated list of functions> For a function y = f (x). listed below: . use an expression of the form [a : b][c : d]. 2 0 -0.8 0.4 0. you would do this: set terminal windows enhanced .4 -0.6 0.6 -0. Example B. In Windows. To graph the function y = sin x from x = 0 to x = 2π. you need to modify the terminal setting.8 -1 0 1 2 3 4 5 6 Notice that the x-axis is labeled with integers. type this at the gnuplot> prompt: plot [0:2pi] sin(x) The result is shown below: 1 sin(x) 0.1.Appendix B: Symbol + − / ** exp(x) log(x) sin(x) cos(x) tan(x) Operation Addition Subtraction Multiplication Division Power ex ln x sin x cos x tan x Graphing with Gnuplot Example 2+3 3−2 23 4/2 23 exp(2) log(2) sin(pi/2) cos(pi) tan(pi/4) 157 Result 5 1 6 2 23 = 8 e2 ln 2 1 −1 1 Note that we use the special keyword “pi” to denote the value of π. To get the x-axis labels with fractions of π.2 -0. the x-axis is not shown in the graph.’3{/Symbol p}/2’ 3pi/2. which can result in jagged edges if the curve is complicated. say. to label the axes. 1000) like this: set samples 1000 Putting all this together. cos(2x) + sin(3x) By default. ’2{/Symbol p}’ 2pi) In the above example. To get a smoother curve.158 Appendix B: Graphing with Gnuplot In Linux you would do this: set terminal wxt enhanced You can then (provided the Symbol font is installed.’{/Symbol p}’ pi. use this command before the plot command: set zeroaxis Also. To display it. increase the sample size (to. use these commands: set xlabel "x" set ylabel "y" The default sample size for plots is 100 units. which it usually is) set the x-axis to have multiples of π/2 from 0 to 2π as labels with this command (all on one line): set xtics (’0’ 0. to also plot the function y = cos 2x + sin 3x on the same graph. we get the following graph: .’{/Symbol p}/2’ pi/2. put a comma after the ﬁrst function then append the new function: plot [0:2pi] sin(x). to save the graph as a ﬁle called graph. and enter png in the Terminal type? textﬁeld.png run the following commands: set terminal png set output ’graph.5 1 y 0. in the File menu again.5 -1 -1.. usually in your My Documents folder (it can also be found by selecting the “show Current Directory” option in the File menu)..” option and enter a ﬁlename (say. hit OK. hit OK. There are many terminal types (which determine the output format).”. In Windows. since the print quality is high and there are many PostScript viewers available. type quit at the gnuplot> command prompt. If you are using the default wxt terminal then select Print near the top of the main Gnuplot window and enter png in the Terminal type? textﬁeld. select the “Output .Appendix B: Graphing with Gnuplot 159 2 sin(x) cos(2 ∗ x) + sin(3 ∗ x) 1.5 -2 0 π/2 π 3π/2 2π x PRINTING AND SAVING In Windows. .png) in the Output ﬁlename? textﬁeld.. then hit OK to get the Print Setup dialog. if you are using the windows enhanced terminal then to print a graph from Gnuplot click on the printer icon in the menubar of the graph’s window. graph.. say. select “Output Device . go to the File menu on the main Gnuplot menubar. To quit Gnuplot. as a PNG ﬁle. Now run your plot command again and the ﬁle will be saved in the current directory.png’ and then run your plot command. to save a graph. In Linux. the postscript terminal type is popular. Run the command set terminal to see all the possible types.5 0 -0. In Linux. Then. or of legal. The “Title Page” means. preceding the beginning of the body of the text. represented in a format whose speciﬁcation is available to the general public. has been arranged to thwart or discourage subsequent modiﬁcation by readers is not Transparent. The “Cover Texts” are certain short passages of text that are listed. XCF and JPG. or absence of markup. as being those of Invariant Sections. the title page itself. if the Document is in part a textbook of mathematics. An image format is not Transparent if used for any substantial amount of text. and standard-conforming simple HTML. that is suitable for revising the document straightforwardly with generic text editors or (for images composed of pixels) generic paint programs or (for drawings) some widely available drawing editor. Texinfo input format. ethical or political position regarding them. 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A Front-Cover Text may be at most 5 words. commercial. a Secondary Section may not explain any mathematics.	89,339	255,450	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.859375	3	CC-MAIN-2019-18	latest	en	0.959901
https://genome.sph.umich.edu/w/index.php?title=Code_Sample:_Generating_QQ_Plots_in_R&diff=next&oldid=6186&printable=yes	1,643,007,814,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320304515.74/warc/CC-MAIN-20220124054039-20220124084039-00051.warc.gz	325,326,619	11,347	# Difference between revisions of "Code Sample: Generating QQ Plots in R" Quantile-quantile plots (qq-plots) can be useful for verifying that a set of values come from a certain distribution. For example in a genome-wide association study, we expect that most of the SNPs we are testing not to be associated with the disease. Under the null, this means that the p-values we get from tests where no true association exists should follow a uniform(0,1) distribution. Since we're usually most interested in really small p-values, we generally transform the p-values by -log10 so that the smallest values near zero become the larger values and are thus easier to see. # Credit This page is based on a tutorial originally written by Matthew Flickinger ## Credits This page is entirely cribbed from an earlier version by Matthew Flickinger, one of our more outstanding graduate students. ## R Lattice Graphics The easiest way to create a -log10 qq-plot is with the qqmath function in the lattice package. It can make a quantile-quantile plot for any distribution as long as you supply it with the correct quantile function. Many of the quantile functions for the standard distributions are built in (qnorm, qt, qbeta, qgamma, qunif, etc). However, we must specify the correct function for the -log10 uniform ourself. Here is some code which will do that with some sample data: ```#FAKE SAMPLE DATA my.pvalues<-runif(10000) library(lattice); qqmath(~-log10(my.pvalues), distribution=function(x){-log10(qunif(1-x))} ); ``` ## A Fancier QQ Plot by Matthew Flickinger Unfortunately the simple way of doing it leaves out many of the things that are nice to have on the plot such as a reference line and a confidence interval plus if your data set is large it plots a lot of points that aren't very interesting in the lower left. Here is a more complex example that adds a few more niceties and thins the data to only plot meaningful points ```library(lattice) qqunif.plot<-function(pvalues, should.thin=T, thin.obs.places=2, thin.exp.places=2, xlab=expression(paste("Expected (",-log[10], " p-value)")), ylab=expression(paste("Observed (",-log[10], " p-value)")), draw.conf=TRUE, conf.points=1000, conf.col="lightgray", conf.alpha=.05, par.settings=list(superpose.symbol=list(pch=pch)), ...) { #error checking if (length(pvalues)==0) stop("pvalue vector is empty, can't draw plot") if(!(class(pvalues)=="numeric" \|\| (class(pvalues)=="list" && all(sapply(pvalues, class)=="numeric")))) stop("pvalue vector is not numeric, can't draw plot") if (any(is.na(unlist(pvalues)))) stop("pvalue vector contains NA values, can't draw plot") if (any(unlist(pvalues)==0)) stop("pvalue vector contains zeros, can't draw plot") } else { if (any(unlist(pvalues)<0)) stop("-log10 pvalue vector contains negative values, can't draw plot") } grp<-NULL n<-1 exp.x<-c() if(is.list(pvalues)) { nn<-sapply(pvalues, length) rs<-cumsum(nn) re<-rs-nn+1 n<-min(nn) if (!is.null(names(pvalues))) { grp=factor(rep(names(pvalues), nn), levels=names(pvalues)) names(pvalues)<-NULL } else { grp=factor(rep(1:length(pvalues), nn)) } pvo<-pvalues pvalues<-numeric(sum(nn)) exp.x<-numeric(sum(nn)) for(i in 1:length(pvo)) { pvalues[rs[i]:re[i]] <- -log10(pvo[[i]]) exp.x[rs[i]:re[i]] <- -log10((rank(pvo[[i]], ties.method="first")-.5)/nn[i]) } else { pvalues[rs[i]:re[i]] <- pvo[[i]] exp.x[rs[i]:re[i]] <- -log10((nn[i]+1-rank(pvo[[i]], ties.method="first")-.5)/(nn[i]+1)) } } } else { n <- length(pvalues)+1 exp.x <- -log10((rank(pvalues, ties.method="first")-.5)/n) pvalues <- -log10(pvalues) } else { exp.x <- -log10((n-rank(pvalues, ties.method="first")-.5)/n) } } #this is a helper function to draw the confidence interval panel.qqconf<-function(n, conf.points=1000, conf.col="gray", conf.alpha=.05, ...) { require(grid) conf.points = min(conf.points, n-1); mpts<-matrix(nrow=conf.points2, ncol=2) for(i in seq(from=1, to=conf.points)) { mpts[i,1]<- -log10((i-.5)/n) mpts[i,2]<- -log10(qbeta(1-conf.alpha/2, i, n-i)) mpts[conf.points2+1-i,1]<- -log10((i-.5)/n) mpts[conf.points2+1-i,2]<- -log10(qbeta(conf.alpha/2, i, n-i)) } grid.polygon(x=mpts[,1],y=mpts[,2], gp=gpar(fill=conf.col, lty=0), default.units="native") } #reduce number of points to plot if (should.thin==T) { if (!is.null(grp)) { thin <- unique(data.frame(pvalues = round(pvalues, thin.obs.places), exp.x = round(exp.x, thin.exp.places), grp=grp)) grp = thin\$grp } else { thin <- unique(data.frame(pvalues = round(pvalues, thin.obs.places), exp.x = round(exp.x, thin.exp.places))) } pvalues <- thin\$pvalues exp.x <- thin\$exp.x } gc() prepanel.qqunif= function(x,y,...) { A = list() A\$xlim = range(x, y)1.02 A\$xlim[1]=0 A\$ylim = A\$xlim return(A) } #draw the plot xyplot(pvalues~exp.x, groups=grp, xlab=xlab, ylab=ylab, aspect=aspect, prepanel=prepanel, scales=list(axs="i"), pch=pch, panel = function(x, y, ...) { if (draw.conf) { panel.qqconf(n, conf.points=conf.points, conf.col=conf.col, conf.alpha=conf.alpha) }; panel.xyplot(x,y, ...); panel.abline(0,1); }, par.settings=par.settings, ... ) } ``` ### Sample Usage A sample call to this function would be ```qqunif.plot(my.pvalues) #these are the raw p-values, not log-transformed ``` ### Under the Hood: Multiple P-value Lists If you are comparing two-test or want to show data before and after it has been corrected for genomic control, you can pass multiple sets of p-values to the function via a list. ```my.pvalue.list<-list("Study 1"=runif(10000), "Study 2"=runif(10000,0,.90)) qqunif.plot(my.pvalue.list, auto.key=list(corner=c(.95,.05))) ``` Internally the different groups are drawn using the lattice superpose settings, so if you want more control over the color and shapes, you can use the par.settings=list(superpose.symbol=) settings. Furthermore, you can use any of the lattice methods of adding a legend to your plot. The names used in the legend correspond to the names of the elements in the list you pass in. ### Under the Hood: Confidence Intervals The confidence intervals are calculated using the fact that the standard uniform order statistics follow a beta distribution. The default settings will draw confidence intervals around the 1000 more significant points. You can change that with the conf.points= parameter and you can change the alpha level from the default .05 using the conf.alpha= parameter. If you wish to disable the confidence interval, use draw.conf=F in your call to qqunif.plot(). Note that the confidence interval drawn depends on the total number of p-values given. When you pass in a list, the number of tests the confidence interval uses is determined by the vector with the least number of p-values - this gives the widest, most conservative confidence bands. ### Under the Hood: Thinning the Data This function does thin the data by rounding the observer and expected -log10 p-values to two places by default. You can control the thinning with the should.thin=, thin.obs.places=, and thin.exp.places= parameters. ### Under the Hood: Customizing Graphics The function should also accept any other lattice graphing parameters should you want to change the plot title (main=), plotting character (pch=), or plot colors (col= for points, conf.col= for confidence interval). By default the aspect="iso" parameter is set which ensures that the reference line lies on a 45-degree angle. If you have very significant results, this may make your plot taller than you would like. You can set the parameter to aspect="fill" to use the standard layout which stretches the values on each axis to take up as much room as possible. ## R Base Graphics Unfortunately, base graphics only offers a built in plot type for normal qq plots. Luckily, it's not too hard to calculate our own expected p-values under the null. We simply rank the p-values from lowest to highest and divide by the total number of tests. Then we take the -log10 transformation of these values. Here is an example ```#Fake sample data my.pvalues=runif(10000) #Calculate expectations exp.pvalues<-(rank(my.pvalues, ties.method="first")+.5)/(length(my.pvalues)+1) #Make plot plot(-log10(exp.pvalues), -log10(my.pvalues), asp=1) abline(0,1) ```	2,188	8,168	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.34375	3	CC-MAIN-2022-05	latest	en	0.884235
https://www.wowinterface.com/forums/showthread.php?s=a4b78761fe7ce562b6f43e753266ce8e&p=334816	1,596,922,129,000,000,000	text/html	crawl-data/CC-MAIN-2020-34/segments/1596439738351.71/warc/CC-MAIN-20200808194923-20200808224923-00109.warc.gz	870,232,275	13,690	Tables again... - WoWInterface 12-18-19, 03:19 AM #1 doofus A Flamescale Wyrmkin Join Date: Feb 2018 Posts: 126 Tables again... t = { }; t["a"] = "alpha" This means key is string "a" and value is string "alpha". This will never change unless I assign another value to it explicitly. Now we do, t[1] = 100 This means the key is unknown, the value is the number 100 (or is it a string?). If we pairs(t) now we get k:1, v:100 k:a, v:alpha Then we do table.insert(t, 1, 500) If we pairs(t) now we get k:1, v:500 k:2, v:100 k:a, v:alpha So it appears keys 1 and 2 are not immutable keys but positions disguised as keys ? It is all so confusing. Last edited by doofus : 12-18-19 at 03:37 AM. 12-18-19, 05:43 AM #2 Kanegasi A Firelord Join Date: Apr 2007 Posts: 487 When dealing with sequential numbers as keys, the table is an array. Using table.insert and table.remove shifts the array to accommodate what you inserted or removed. The number 100 (not string) was at index 1, but when you inserted 500 at that position, it moved 100 to index 2. Anything dealing with arrays ignores keys that aren't sequential up from 1, which also means index 0 and negative indexes are also ignored. The #t operation counts the array part of the table. The last part of your example makes #t == 2, since "a" is ignored. Last edited by Kanegasi : 12-18-19 at 05:45 AM. 12-18-19, 05:06 PM #3 Seerah Fishing Trainer Join Date: Oct 2006 Posts: 10,733 To add to what Kanegasi said... When you did t[1]=100 you assigned the number 100 to the first index of the table. (You can assign a value to any index, and you don't even have to do it sequentially.) 1 and 100 are then an index, value pair. (Keys are strings, indices are numbers.) Here is more information on table basics and how to work with tables: http://lua-users.org/wiki/TablesTutorial Here is a more in depth tutorial on tables and the table library: http://lua-users.org/wiki/TableLibraryTutorial __________________ "You'd be surprised how many people violate this simple principle every day of their lives and try to fit square pegs into round holes, ignoring the clear reality that Things Are As They Are." -Benjamin Hoff, The Tao of Pooh 12-19-19, 05:15 PM #4 doofus A Flamescale Wyrmkin Join Date: Feb 2018 Posts: 126 I am more confused now t[1] = 100 t[2] = 200 t["hello"] = "goodbye" which is a data structure where: k:1 v=100 k:2 v=200 k:hello v="goodbye" Naturally we take 1, 2, "hello" to be "keys" in the traditional sense of sets/maps Then we do: table.insert(t,1,500) and this gives k:1 v=500 k:2 v=100 k:3 v=200 k:hello v="goodbye" So even though t["hello"] does not change, t[1], t[2], t[3] etc might all change if I do table.insert() ?? This is not at all how sets or maps work, so I find this data structure weird... I am confused. 12-19-19, 06:58 PM #5 Nimhfree A Frostmaul Preserver Join Date: Aug 2006 Posts: 264 The "key" to understanding this is that in your example, 1 and 2 are not keys. They are indexes. If you want them to be keys they need to be strings like "1" and "2". The table data structure at the same time handles data accessed via index and key. Tables are also sparse in that the indexes do not have to be contiguous. E.g., you can have t[1], t[7] and t[13] without the expected t[2], t[3], etc. values. And as you have seen you have have arbitrary keys like "hello" and "a". 12-20-19, 02:04 AM #6 doofus A Flamescale Wyrmkin Join Date: Feb 2018 Posts: 126 I do not think this is right. t[1] is not an integer index, there is NO "position 1" in the table, unless, and only if, I define k[n], where n is an integer, either by k[n]=value or by table.insert(t,n,value). Otherwise there is no "position 1" in the table. eg t = { }; t["hello"] = "goodbye"; t does not have position 1. table.remove(t,1); Does not do anything, there is no "position 1" and t["hello"], even though the only element in the table, is NOT considered to be position 1, or ANY position at all. t[2] = "day" Now t does have a position 1. table.insert(t,1,"night") now we have position 1 = "night" and position 2 = "day" All the while t["hello"] has not been affected, it is like independent... The way to understand this is that a table has two types of keys and accessing mechanisms, string keys and integer keys. Integer key/value pairs can be affected by table.insert / remove but string key/value pairs are not. This is what I have fathomed by running tests all in between wiping in M+10 because we do not know the strats... 12-20-19, 11:38 AM #7 Vrul An Onyxian Warder Join Date: Nov 2007 Posts: 353 Think of Lua tables as maps that can also be treated as arrays. The array behavior is only for the portion of the map that starts with the key of integer 1 and continues until a nil value is reached for sequential integer keys. ipairs, #table, and the functions found in table (such as table.insert) only apply to the array portion of the map. Think of #table as: Code: ```local function GetArraySize(table) local index = 1 while table[index] ~= nil do index = index + 1 end return index - 1 end``` Think of table.insert as: Code: ```local function InsertIntoArray(table, index, value) for i = #table, index, - 1 do table[i + 1] = table[i] end table[index] = value end``` Think of table.remove as: Code: ```local function RemoveFromArray(table, index) local value = table[index] for i = index, #table - 1 do table[i] = table[i + 1] end table[#table] = nil return value end``` 12-20-19, 12:27 PM #8 Seerah Fishing Trainer Join Date: Oct 2006 Posts: 10,733 Originally Posted by doofus I do not think this is right. You don't understand it, but we're wrong? Did you go through the tutorials I suggested? Here also is the chapter on tables from Programming in Lua: http://www.lua.org/pil/2.5.html __________________ "You'd be surprised how many people violate this simple principle every day of their lives and try to fit square pegs into round holes, ignoring the clear reality that Things Are As They Are." -Benjamin Hoff, The Tao of Pooh 12-21-19, 09:06 AM #9 SDPhantom A Pyroguard Emberseer Join Date: Jul 2006 Posts: 1,924 Quoting the Lua 5.1 manual too. The type table implements associative arrays, that is, arrays that can be indexed not only with numbers, but with any value (except nil). Tables can be heterogeneous; that is, they can contain values of all types (except nil). Tables are the sole data structuring mechanism in Lua; they can be used to represent ordinary arrays, symbol tables, sets, records, graphs, trees, etc. To represent records, Lua uses the field name as an index. The language supports this representation by providing a.name as syntactic sugar for a["name"]. There are several convenient ways to create tables in Lua (see §2.5.7). http://www.lua.org/manual/5.1/manual.html#2.2 I'd also go over the Table Manipulation section of the function references. http://www.lua.org/manual/5.1/manual.html#5.5 __________________ "All I want is a pretty girl, a decent meal, and the right to shoot lightning at fools." -Anders (Dragon Age: Origins - Awakening) 12-25-19, 11:02 PM #10 Roeshambo A Murloc Raider Join Date: Jul 2019 Posts: 5 Tables are basically arrays (but a bit more complex if you start manipulating meta tables etc). To break it down dummy style, you have indexed arrays (1, 2, 3, 4 as keys) and associative arrays (custom keys such as t["key1"]. ipairs will only cycle through the indexed keys (hence the I in it) and ignores the associative portion of the table. pairs covers all of it. It's really not confusing once you realize that if it's just a number, it's an indexed key. If it's got words, it's an associative key. But when you do ipairs, it will only cycle until it finds a nil value. So if you create keys 1, 2, 3 and 5, it will only cycle through 3 (since 4 is nil). Once you get that basic understanding down it gets really easy to see how they can be manipulated. Of course, they are NOT arrays. But you can very easily swap the concepts to get a basic understanding of them. When you do table.insert(t, 1, 500), you're not replacing the 1 key's value. You're essentially pushing a new key value pair into the first position which pushes all other keys down one position. The point of an indexed table is they can be sorted and can be manipulated in that order (the keys don't remain constant). Where as in an associative array, if you sort it, you're going to have to use "next" or "for k,v" to cycle through it in the proper order. Last edited by Roeshambo : 12-25-19 at 11:13 PM. 12-26-19, 12:28 AM #11 SDPhantom A Pyroguard Emberseer Join Date: Jul 2006 Posts: 1,924 Originally Posted by Roeshambo Where as in an associative array, if you sort it, you're going to have to use "next" or "for k,v" to cycle through it in the proper order. table.sort() doesn't work on associative keys. It only works on the sequential portion of the table. The order of key/value pairs iterated from next() and pairs() are completely random. Even among sequential keys. This is why ipairs() exists instead of using pairs() for that too. https://www.lua.org/manual/5.1/manua...pdf-table.sort __________________	2,577	9,179	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.546875	3	CC-MAIN-2020-34	latest	en	0.835672
http://math.stackexchange.com/questions/292722/is-there-a-text-that-provides-the-proof-of-fermats-last-theorem	1,469,795,631,000,000,000	text/html	crawl-data/CC-MAIN-2016-30/segments/1469257830066.95/warc/CC-MAIN-20160723071030-00241-ip-10-185-27-174.ec2.internal.warc.gz	162,181,422	17,419	# Is there a text that provides the proof of Fermat's Last Theorem? I know that professor Andrew Wiles discovered his proof of Fermat's Last Theorem in 1995. One of my friends is looking for a text which provides his proof. I know that the proof is very complicated and uses difficult methods to get the solution, but I hope that you can give me the name of a text which contains it (or a link to the proof or such a text :)) I would also like to know what prerequisites are needed to study/understand the proof in detail? Furthermore, has anyone else discovered another proof since Wiles, or is Andrew Wiles' proof the only known solution? thanks - The Wikipedia article cites Andrew Wiles's 1995 article "Modular elliptic curves and Fermat's Last Theorem" in the Annals of Mathematics, 141 (3). There is a PDF copy online. You can also get a sense of the prerequisites from reading the Wikipedia article. – Rahul Feb 2 '13 at 10:49 Take a look at this: math.stackexchange.com/questions/170142/… – M Turgeon Feb 14 '13 at 16:11 Wiles' proof was actually in June 1993; an erratum was seen in the September of that year. Wiles went away and published his full account of the Taniyama-Shimura Conjecture (or the Modularity Theorem) in September 1994. – Autolatry Aug 19 '15 at 14:27 no one else has proved it - There are very few professional mathematicians who have read and understood all of Wiles proof. The pre-requisites go a long way beyond college level mathematics. Looking at Wiles paper is not a good way to learn about this problem. If you're interested in number theory, you could begin by studying Hardy and Wright: "An introduction to theory of numbers". Some special cases of Fermat's last theorem were solved in the nineteenth century, and you should see their proofs in an introduction to algebraic number theory. - Hardy's book, I think, is not suitable for a beginner. It's writing is not of the best quality. – user 170039 Sep 5 '15 at 3:08 The very minimal prerequisites for understanding the proof of Fermat's Last Theorem would include knowledge of algebraic number theory, modular forms, elliptic curves, Galois theory, Galois cohomology, and representation theory. A considerable amount of higher mathematics is needed to understand these areas in detail, including a very strong background in (advanced) abstract algebra. If/once you are comfortable with the necessary background material and are still interested in what I hear is a very good reference on the proof and its methods, check out "Modular Forms and Fermat's Last Theorem" by Silverman, Stevens, and Cornell. The text is at intended for professional mathematicians, so it certainly won't be an easy read, but if one has a strong enough background and enough tenacity, one could certainly make it through. Reading and understanding this book would be a great help in making it through Wiles' proof. Good luck! - ## protected by Community♦May 14 '13 at 15:58 Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).	733	3,176	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.609375	3	CC-MAIN-2016-30	latest	en	0.962206
https://www.physicsforums.com/threads/integral-of-dot-product.688142/	1,508,834,289,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187828189.71/warc/CC-MAIN-20171024071819-20171024091819-00622.warc.gz	999,207,482	14,839	# Integral of dot product 1. Apr 26, 2013 ### gespex Hi all, I'm working on a math problem with a known answer - though I can't reproduce the maths. The problem is this: there is a random 3d vector of unit length with a uniform probability, $\vec{v}$, and a secondary unit vector $\vec{u}$. It is stated that: $$f = \int_{S^2}{\| \vec{v} \cdot \vec{u} \| d\vec{v}} = 2\pi$$ Now, I've never worked with integrals of this kind, and I'm not even exactly sure what $S^2$ means in the integral subscript, but given the problem I attempted to take a vector for $\vec{u} = [0, 0, 1]$ (which shouldn't matter, as the distribution is uniform), and I set $$\vec{v} = [cos(\alpha)sin(\sigma), sin(\alpha)sin(\sigma), cos(\sigma)]$$ for $0 \leq \alpha \leq 2\pi$ and $0 \leq \sigma \leq \pi$. Now I expected that: $$f = \int_{S^2}{\| \vec{v} \cdot \vec{u} \|d\vec{v}} = \int_{\alpha = 0}^{2\pi}\int_{\sigma = 0}^{\pi} \| \vec{v} \cdot \vec{u} \|d{\sigma}d\alpha$$ $$f = \int_{\alpha = 0}^{2\pi}\int_{\sigma = 0}^{\pi} \| cos(\sigma) \|d{\sigma}d\alpha = \int_{\alpha = 0}^{2\pi}\int_{\sigma = 0}^{{1 \over 2} \pi} 2cos(\sigma)d{\sigma}d\alpha$$ $$f = \int_{\alpha = 0}^{2\pi} 2[sin({1 \over 2}\pi) - sin(0)]d\alpha = 4\pi$$ Which is not what the apparent solution should be. What is the error in my logic? And how can it be proven that $f = 2\pi$? 2. Apr 26, 2013 ### Staff: Mentor S_2 is the unit sphere - the set of all points with distance 1 to the origin. Your parametrization of the sphere is not uniform, you favor points close to u (and on the opposite side) in the integral. This is easier to solve in spherical coordinates. 3. Apr 26, 2013 ### gespex I did manage to reproduce the answer now. Though I didn't know integration in spherical coordinates. I actually did convert it to spherical coordinates (right?), with p = 1 (the distance from the origin), except I didn't take into account the "Jacobian determinant". I really only had to multiply it with $p^2 sin(\sigma)$, is that correct?	678	1,995	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.734375	4	CC-MAIN-2017-43	longest	en	0.895988
https://www.pokerstrategy.com/quiz/weekly/585/	1,534,447,347,000,000,000	text/html	crawl-data/CC-MAIN-2018-34/segments/1534221211167.1/warc/CC-MAIN-20180816191550-20180816211550-00008.warc.gz	954,009,807	13,666	» QUIZ # NL SSS: How do you handle the queens? by Doomknight Men are generally happy when the ladies visit them - and they should be, because in poker, they are amongst the strongest starting hands. While the hand is very strong, there are stronger hands, so you have to handle a pocket pair queens with care. This quiz gives you the opportunity to test your knowledge about the short stack strategy, especially when it comes to this specific hand. Do you know how to control it or should you read up on some more articles? Question 1: Known players: Hero \$2.00 Rest has more than \$2.00 0,05/0,1 No-Limit Hold'em (10 handed) Preflop: Hero is BU with Q, Q 4 folds, MP2 calls \$0,10, 2 folds, Hero raises to \$0,50, 2 folds, MP2 calls \$0,40, Flop: (\$1,15) J, 7, 2 (2 players) MP2 bets \$0,60, Hero ??? What do you do? fold call raise all-in Question 2: Known players: Hero \$2.00 Rest has more than \$2.00 0,05/0,1 No-Limit Hold'em (10 handed) Preflop: Hero is BU with Q, Q 4 folds, MP2 calls \$0,10, 2 folds, Hero raises to \$0,50, 2 folds, MP2 calls \$0,40, Flop: (\$1,15) J, 7, A (2 players) MP2 bets \$0,60, Hero ??? What do you do? fold call raise all-in Question 3: Known players: Hero \$2.00 Rest has more than \$2.00 0,05/0,1 No-Limit Hold'em (10 handed) Preflop: Hero is MP2 with Q, Q 4 folds, Hero raises to \$0,40 , 2 folds, BU calls \$0,40, 2 folds Flop: (\$0,95) J, 7, 2 (2 players) Hero ??? What do you do? check bet \$0.60/call bet \$0.60/fold push all-in Question 4: Known players: Hero \$2.00 Rest has more than \$2.00 0,05/0,1 No-Limit Hold'em (10 handed) Preflop: Hero is MP2 with Q, Q 4 folds, Hero raises to \$0,40 , 2 folds, BU calls \$0,40, 2 folds Flop: (\$0,95) J, 7, A (2 players) Hero ??? What do you do? check bet \$0.60/call bet \$0.60/fold push all-in Question 5: Known players: Hero \$2.00 Rest has more than \$2.00 0,05/0,1 No-Limit Hold'em (10 handed) Preflop: Hero is MP2 with Q, Q 4 folds, Hero raises to \$0,40 , 2 folds, BU calls \$0,40, 2 folds Flop: (\$0,95) J, 7, A (2 players) Hero bets \$0,60, BU calls \$0,60 Turn: (\$2,15) 8 Hero ??? What do you do? check/fold check/call bet \$0.50/call bet \$0.50/fold Question 6: Known players: Hero \$2.00 Rest has more than \$2.00 0,05/0,1 No-Limit Hold'em (10 handed) Preflop: Hero is MP2 with Q, Q 2 folds, UTG2 raises to \$0,40, MP1 raises to \$1,20 , Hero ??? What do you do? fold call raise all-in Question 7: Known players: Hero \$2.00 Rest has more than \$2.00 0,05/0,1 No-Limit Hold'em (10 handed) Preflop: Hero is MP2 with Q, Q 2 folds, UTG2 raises to \$0,40, MP1 calls \$0,40 , Hero ??? What do you do? fold call raise all-in Question 8: Known players: Hero \$2.00 Rest has more than \$2.00 0,05/0,1 No-Limit Hold'em (10 handed) Preflop: Hero is BU with Q, Q 2 folds, UTG2 raises to \$0,40, MP1 calls \$0,40, 1 folds, MP3 calls \$0,40, CO calls \$0,40, Hero ??? What do you do? fold call raise all-in Question 9: Known players: Hero \$2.00 Rest has more than \$2.00 0,05/0,1 No-Limit Hold'em (10 handed) Preflop: Hero is UTG2 with Q, Q 2 folds, Hero raises to \$0,40, MP1 raises to \$1,20 , MP2 raises to \$4,00, 5 folds, Hero ??? What do you do? fold call all-in Question 10: Known players: Hero \$2.00 Rest has more than \$2.00 0,05/0,1 No-Limit Hold'em (10 handed) Preflop: Hero is UTG2 with Q, Q 2 folds, Hero raises to \$0,40, MP1 raises to \$1,20 ,MP2 calls \$1,20, 5 folds, Hero ??? What do you do? fold call raise all-in Question 11: Known players: Hero \$2.00 Rest has more than \$2.00 0,05/0,1 No-Limit Hold'em (10 handed) Preflop: Hero is MP2 with Q, Q 4 folds, Hero raises to \$0,40 , 2 folds, BU calls \$0,40, 2 folds Flop: (\$0,95) J, 7, 2 (2 players) Hero bets \$0,60, BU calls \$0,60 Turn: (\$2,15) J Hero ??? What do you do? check/fold check/call bet/call bet/fold push all-in Question 12: Known players: Hero \$2.00 Rest has more than \$2.00 0,05/0,1 No-Limit Hold'em (10 handed) Preflop: Hero is MP2 with Q, Q 4 folds, Hero raises to \$0,40 , 2 folds, BU calls \$0,40, 2 folds Flop: (\$0,95) J, 7, 2 (2 players) Hero bets \$0,60, BU calls \$0,60 Turn: (\$2,15) A Hero ??? What do you do? check/fold check/call bet/call bet/fold push all-in	1,545	4,251	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.578125	3	CC-MAIN-2018-34	longest	en	0.871769
https://ao.ms/how-to-solve-simple-square-numbers-in-golang/	1,675,663,561,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00195.warc.gz	109,736,653	9,385	## The challenge# In this challenge, you will be given a number `n` (`n > 0`) and your task will be to return the smallest square number `N` (`N > 0`) such that `n + N` is also a perfect square. If there is no answer, return `-1` (`nil` in Clojure, `Nothing` in Haskell, `None` in Rust). ``````solve 13 = 36 <em>; because 36 is the smallest perfect square that can be added to 13 to form a perfect square => 13 + 36 = 49</em> solve 3 = 1 <em>; 3 + 1 = 4, a perfect square</em> solve 12 = 4 <em>; 12 + 4 = 16, a perfect square</em> solve 9 = 16 solve 4 = nil `````` ## The solution in Golang# Option 1: ``````package solution func Solve(n int) int { res := -1 for i:= 1; i * i < n; i++ { if n % i == 0 && (n / i - i) % 2 == 0 { res = (n / i - i) * (n / i - i) / 4 } } return res } `````` Option 2: ``````package solution import "math" func Solve(n int) int { // n+a2==b2 => n=b2-a2=(a-b)(a+b) => b=(c+d)/2; a=d-b for i := int(math.Sqrt(float64(n))); i > 0; i-- { if n%i == 0 { c, d := i, n/i b := (c + d) / 2 a := d - b if a > 0 && n+aa == bb { return a * a } } } return -1 } `````` Option 3: ``````package solution import "math" func Solve(n int) int { upperBound := n lowerBound := 1 for lowerBound <= upperBound { square := lowerBound*lowerBound numToCheck := math.Sqrt(float64(square + n)) if math.Floor(numToCheck) == math.Ceil(numToCheck) { return square } lowerBound++ } return -1 } `````` ## Test cases to validate our solution# ``````package solution_test import ( . "github.com/onsi/ginkgo" . "github.com/onsi/gomega" ) var _ = Describe("Example tests", func() { It("It should work for basic tests", func() { Expect(Solve(1)).To(Equal(-1)) Expect(Solve(2)).To(Equal(-1)) Expect(Solve(3)).To(Equal(1)) Expect(Solve(4)).To(Equal(-1)) Expect(Solve(5)).To(Equal(4)) Expect(Solve(7)).To(Equal(9)) Expect(Solve(8)).To(Equal(1)) Expect(Solve(9)).To(Equal(16)) Expect(Solve(10)).To(Equal(-1)) Expect(Solve(11)).To(Equal(25)) Expect(Solve(13)).To(Equal(36)) Expect(Solve(17)).To(Equal(64)) Expect(Solve(88901)).To(Equal(5428900)) Expect(Solve(290101)).To(Equal(429235524)) }) }) ``````	760	2,105	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.734375	4	CC-MAIN-2023-06	latest	en	0.34895
https://mathoverflow.net/questions/289375/2-times-2-matrix-question	1,568,697,490,000,000,000	text/html	crawl-data/CC-MAIN-2019-39/segments/1568514573052.26/warc/CC-MAIN-20190917040727-20190917062727-00252.warc.gz	586,114,568	26,605	# $2 \times 2$ matrix question Let $A$, $B$, and $C$ be $2\times 2$ complex matrices, with $A$ and $C$ rank $1$ Hermitian. Can we find a real number $a$ and a $2\times 2$ unitary $U$ such that $$A + BV + V^B^ + V^CV$$ is a scalar multiple of the identity, where $V = aU$? The motivation is that this would answer this question in a large special case. ## 1 Answer Yes, one can always do this. In fact, one can assume that $\det(U)=1$, as this follows by a homotopy argument, using the fact that $\pi_3(S^2)\simeq \mathbb{Z}$. Here are the details: Assume that $\det(U)=1$ so that $V$ has a non-negative real determinant. Then one can write $V$ uniquely in the form $$V(x) = \begin{pmatrix} x_0+i\,x_1 & -x_2 + i\,x_3\\ x_2-i\,x_3 & \phantom{-}x_0-i\,x_1 \end{pmatrix}$$ for some $x = (x_0,x_1,x_2,x_3)\in\mathbb{R}^4$. Now consider the map $F$ from $\mathbb{R}^4$ into $\mathbb{R}^3$ (regarded as the traceless Hermitian $2$-by-$2$ matrices) defined by $$F(x) = \left[A +BV(x) + V(x)^B^* + V(x)^CV(x)\right]_0\,,$$ where, for a Hermitian $2$-by-$2$ matrix, $H$, we write $H - \tfrac12\mathrm{tr}(H)\,I_2 = H_0$. For use below, define the norm on traceless Hermitian $2$-by-$2$-matrices $M$ by the rule $\|M\|^2 = \tfrac12 \mathrm{tr}(M^2)$. Our task is to show that there exists an $x\in\mathbb{R}^4$ such that $F(x)=0$. Note that $F(x)$ is a quadratic polynomial in $x$ taking values in $\mathbb{R}^3$. When $\|x\|$ very large, $F(x)/\|x\|^2$ is close to $G(x) = (V(x)/\|x\|)^C_0(V(x)/\|x\|)$, a vector whose norm is $\|C_0\|$. Since $C$ has rank 1, $C_0$ is not zero. Thus, when $\|x\|>>0$, $\bigl\|F(x)\bigr\|/\|x\|^2\approx \|C_0\|>0$. Moreover, when restricted to the $3$-sphere of radius $R>0$ in $\mathbb{R}^4$, the homogeneous map $G: S^3(R)\to S^2(\|C_0\|)$ is simply the Hopf map $S^3\to S^2$, appropriately scaled, which is known not to be homotopically trivial. (In fact, it is a generator of $\pi_3(S^2)\simeq\mathbb{Z}$.) If there were not a $y\in\mathbb{R}^4$ that satisfied $F(y)=0$, then, for $R>>0$ the map $H(x) = F(x)/\|F(x)\|$ would map the $4$-ball of radius $R$ to $S^2$ in such a way that the map on its boundary $3$-sphere would be homotopic to the Hopf map, which is impossible. Thus, there is a $y\in\mathbb{R}^4$ such that $F(y)=0$, i.e., $$A +BV(y) + V(y)^B^ + V(y)^*CV(y) = \mu\,I_2$$ for some number $\mu$, as desired. • Great answer. I used it to give a partial answer to my earlier question about simultaneous "orthonormalization" in $\mathbb{C}^4$. – Nik Weaver Dec 28 '17 at 15:08 • I've just put a paper on the arXiv that uses this idea: arxiv.org/abs/1802.07394 – Nik Weaver Feb 22 '18 at 20:59	1,022	2,626	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5	4	CC-MAIN-2019-39	latest	en	0.82036
https://www.manualslib.com/manual/752434/Mitsubishi-Electric-Fr-A701.html?page=140	1,563,288,785,000,000,000	text/html	crawl-data/CC-MAIN-2019-30/segments/1563195524568.14/warc/CC-MAIN-20190716135748-20190716161748-00245.warc.gz	768,687,802	36,808	# Mitsubishi Electric FR-A701 Instrucion Manual Page 140 Parameter Name Torque control P 824 gain 1 Torque control P 834 gain 2 Torque control 825 integral time 1 Torque control 835 integral time 2 826 Refer to Pr. 74. Torque detection 827 filter 1 Torque detection 837 filter 2 Model speed 828 control gain Speed feed forward control/model 877 control selection Speed feed forward 878 filter Speed feed forward 879 torque limit 880 Speed feed forward 881 gain 830 Refer to Pr. 820. 831 Refer to Pr. 821. 832 Refer to Pr. 74. 833 Refer to Pr. 823. 834 Refer to Pr. 824. 835 Refer to Pr. 825. 836 Refer to Pr. 74. 837 Refer to Pr. 827. Incre Initial Range ments Value Set the proportional gain for the current control of the q and d axes. (Increasing the 1% 100% 0 to 200% value improves trackability in response to a current command change and reduces current variation with disturbance.) Second function of Pr. 824 (valid when the 0 to 200% RT terminal is on) 1% 9999 9999 No function Set the integral time for the current control of the q and d axes. (Decreasing the value 0.1ms 5ms 0 to 500ms original torque if current variation with disturbance occurs.) Second function of Pr. 825 (valid when the 0 to 500ms RT signal is on) 0.1ms 9999 9999 No function Set the primary delay filter for the current 0.001s 0s 0 to 0.1s feedback. Second function of Pr. 827 (valid when the 0 to 0.1s RT signal is on) 0.001s 9999 9999 No function 1% 60% 0 to 1000% Set the gain for model speed controller. 0 Normal speed control is exercised 1 Speed feed forward control is exercised. 1 0 2 Model adaptive speed control is enabled. Set the primary delay filter for the speed 0.01s 0s 0 to 1s feed forward result calculated using the speed command and load inertia ratio. Limits the maximum value of the speed 0.1% 150% 0 to 400% feed forward torque. 0 to 200 0.1 7 times Inertia ratio found by easy gain turning. Set the feed forward calculation result as a 1% 0% 0 to 1000% gain. Parameter List Para meter copy Description All Param param eter eter clear clear : enabled × : disabled × 4 133 #### This manual is also suitable for: Fr-a721Fr-a741Fr-a701Fr-a741-55kFr-a741-7.5kFr-a741-11k ... Show all	684	2,188	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.625	3	CC-MAIN-2019-30	latest	en	0.723455
https://www.chemicalforums.com/index.php?board=4.12915;sort=starter	1,685,295,862,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224644309.7/warc/CC-MAIN-20230528150639-20230528180639-00756.warc.gz	781,440,189	8,025	May 28, 2023, 01:44:22 PM Forum Rules: Read This Before Posting Subject / Started by Last post 0 Members and 54 Guests are viewing this board. c1v1 = c2v2 Started by yh09 \| Last post by Borek May 10, 2008, 07:38:05 PM 7 Replies 8547 Views May 10, 2008, 07:38:05 PM by Borek Mass Spectrometry Started by ying63 \| Last post by ying63 July 30, 2021, 06:43:43 AM 4 Replies 862 Views July 30, 2021, 06:43:43 AM by ying63 Would Appreciate some help(Actual Yeild) Started by yitriumm \| Last post by AWK December 18, 2007, 11:21:42 AM 10 Replies 7609 Views December 18, 2007, 11:21:42 AM by AWK Identifying Unknown element After Determining Molar Mass Started by yitriumm \| Last post by Kryolith January 05, 2008, 04:51:39 PM 3 Replies 6106 Views January 05, 2008, 04:51:39 PM by Kryolith How does temperature affect dyeing fabric? Started by yodafang \| Last post by ARGOS++ November 02, 2007, 08:29:56 PM 2 Replies 4188 Views November 02, 2007, 08:29:56 PM by ARGOS++ Phosphate has -3 charge. Why and how? Started by yodalr \| Last post by mjc123 October 08, 2018, 04:45:29 AM 4 Replies 3506 Views October 08, 2018, 04:45:29 AM by mjc123 Help, calculate volume Started by Yogi1 \| Last post by Borek October 06, 2010, 01:31:02 PM 3 Replies 2888 Views October 06, 2010, 01:31:02 PM by Borek Need help with this Calorimetry problem Started by yolomite \| Last post by Borek September 23, 2019, 03:34:13 AM 2 Replies 736 Views September 23, 2019, 03:34:13 AM by Borek Chemistry olympiad 2015 Question 33: Ratio of Kc/Kp Started by yonglu1989 \| Last post by AWK March 06, 2016, 03:02:25 PM 1 Replies 3292 Views March 06, 2016, 03:02:25 PM by AWK Why does HI have a higher boiling point than HCl? Started by yonglu1989 \| Last post by Vidya March 03, 2016, 10:38:54 PM 3 Replies 11757 Views March 03, 2016, 10:38:54 PM by Vidya Chemistry olympiad 2015 Question 7: Solubility relating to calcium Started by yonglu1989 \| Last post by Vidya March 03, 2016, 10:35:39 PM 2 Replies 6490 Views March 03, 2016, 10:35:39 PM by Vidya Specific Heat Capacity Started by yooper_at_heart678 \| Last post by yooper_at_heart678 October 10, 2007, 09:50:04 PM 4 Replies 4112 Views October 10, 2007, 09:50:04 PM by yooper_at_heart678 Question Started by yosh \| Last post by Borek January 12, 2008, 08:22:03 AM 9 Replies 5890 Views January 12, 2008, 08:22:03 AM by Borek Rockets Started by yosh « 1 2 » \| Last post by Rabn January 06, 2008, 09:41:12 PM 19 Replies 13406 Views January 06, 2008, 09:41:12 PM by Rabn Really Need Help with Nomenclature and Dimensional Analysis Started by Yoshifan123 \| Last post by nj_bartel November 06, 2008, 12:17:04 AM 1 Replies 2426 Views November 06, 2008, 12:17:04 AM by nj_bartel	1,034	2,711	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2023-23	longest	en	0.939679
http://www.zyra.global/www.zyra.info/tonflead.htm	1,669,862,341,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710789.95/warc/CC-MAIN-20221201021257-20221201051257-00798.warc.gz	109,184,102	3,009	Zyra's website //// Science //// Site Index Which weighs more? A ton of feathers or a ton of lead? It's an old puzzler regarding the comparative weights of a ton of lead and a ton of feathers, and some people think they always know the answer. But maybe it's not as obvious as it might seem. To a simple view, it seems a deep question. Obviously the feathers are very light and yet there's a whole ton of them so that is very big. Feather beds can weigh quite a lot, but a ton of feathers would be a truckload. Lead, in contrast, is about eleven times heavier than water for the same volume, so a ton of lead can be imagined as a metre square by 9cm thick. Yet, although the lead is very "heavy", a concept known as "density", a ton of it is not very big in the space it takes up. A practical approach would be to get a weighing scale balance and set it up with a ton of lead on one side and a ton of feathers on the other. This would not be easy, as the choice of instrument would be a puzzling question in itself. However, as both the ton of feathers and the ton of lead both weigh "a ton", the expected result would be that they weigh THE SAME! But then, that's science, isn't it?! Gives a pure result. Some people might instead prefer to argue the point, perhaps having a fierce debate that they were right and the other person had got it wrong. For example "The lead has GOT to weigh more because it's heavier than feathers!" or "A ton of feathers is so big that it MUST weigh more than the lead!". Others would take a more equalist view, saying that everyone's point of view is valid and that if anyone wants to believe that the lead or the feathers is heavier, then that is true from their perspective. Indeed there are people who would say that it is entirely a matter of personal belief which weighs more and that faith can give a true personal measurement of the feathers or lead situation and that scientific measurement is irrelevant. I'm with the scientific approach myself and in such a situation I would be keen to get a set of scales to prove the fact and defy various people's faulty beliefs on this kind of thing! Also, if it turned out that a ton of feathers and a ton of lead didn't weigh the same, it would require further investigation! Now I'd guess there will be some people thinking this is all a bit silly, as it's obvious that a ton of feathers and a ton of lead weigh the same because they've read it in a book, or because it's just "true". But, how about if the experiment was conducted on the moon?! Would the feathers and lead still weigh the same? Curious as it may seem, there is a discrepancy. If the initial ton of feathers and the ton of lead are weighed on the earth and balanced up to be the same, and then the experiment is taken to the moon and balanced up again, the feathers now outweigh the lead! With this kind of thing it's no good trying to explain it away as experimental error! The feathers weigh a measurable amount more than the lead. This is not because the gravity is one sixth on the moon. There's still "a sixth of a ton" on each side of the balance. The reason for the difference is that on the moon the feathers no longer have the buoyancy of the air which they had on the earth. (If you have doubts about this, consider it's the opposite effect of what would happen if the experiment were performed underwater, where the lead would sink about as expected and the feathers would weigh not much at all, and that's even before they floated off). These effects are put to use with the idea of Density Scales Other stuff: Orbits of satellites in space , floating of a snooker ball on a pint of mercury , How long is a piece of string? , Pescu , Space , mass or weight , etc. Truths , misconceptions , and a great many other things. Oh really?	842	3,805	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.78125	3	CC-MAIN-2022-49	latest	en	0.984852
http://www.physicsforums.com/showthread.php?s=9ac6bf2b11baf45f2b6bb383facf8e27&p=4594898	1,398,256,141,000,000,000	text/html	crawl-data/CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00066-ip-10-147-4-33.ec2.internal.warc.gz	773,309,269	9,499	# Back and forth motion by viciam Tags: motion P: 27 Hi guys, I'm not a physicist or a mathematician but I have a question which will probably sound silly to you all I need to know what that motion is called when you have an object and it moves up and down. But it gains more momentum every time it reaches the top so when it drops it goes down further and faster and when it reaches the maximum bottom it comes back up even faster and further and this cycle just keeps repeating. On every completion of up and down movement it goes further and further down and further and further up In one word please tell me what this is called so I can google it. I inertia and reciprocating but this is not it. I know there's one word to describe this but I just can't remember it. Thanks P: 383 In a periodic motion object moves back and foirth about its mean position. The kind of motion you are describing can be more precisely called a 'Simple Harmonic Motion'. Now you can google those two terms i described above Engineering Sci Advisor HW Helper Thanks P: 6,380 Try "resonance". But don't get sidetracked into crackpot websites about perpetual motion machines, "free energy", etc. Mentor P: 39,683 ## Back and forth motion Undamped forced oscillation: http://www.nptel.iitm.ac.in/courses/...se_home_26.htm P: 27 These all sound pretty complicated. I thought it was as simple as a pendulum swing or something Mentor P: 39,683 Quote by viciam These all sound pretty complicated. I thought it was as simple as a pendulum swing or something You need to be putting energy in with each swing, to get bigger oscillatory motion. And there needs to be very little damping, so the energy being added in does not get dissipated by the damping. P: 1,909 Quote by viciam These all sound pretty complicated. I thought it was as simple as a pendulum swing or something Yes, you can do this with a swing. Either by pushing someone from outside or by yourself, flexing your legs at the right times. The amplitude increases at each cycle, up to a limit. Can be modeled as a forced oscillator, as already mentioned. But the math is not so simple, even for a "simple" pendulum, at large amplitudes. P: 27 This is exactly what I'm trying to find out. You see how the waves become larger and larger as they go along? I typed in increasing oscillations in Google to get this image. PF Gold P: 367 See page 4 of this about forced vibration. http://www.stewartcalculus.com/data/...Orders_Stu.pdf Thanks P: 5,529 A one-word description for this motion would be "impossible". Or "nonphysical". Or "imaginary". It does not exists in nature. The amplitude of oscillations may increase for a while, but then it will either reach some maximum level, or "bifurcate" into some other motion, which is a fancy way of saying "break into pieces". Mentor P: 39,683 Quote by voko A one-word description for this motion would be "impossible". Or "nonphysical". Or "imaginary". It does not exists in nature. The amplitude of oscillations may increase for a while, but then it will either reach some maximum level, or "bifurcate" into some other motion, which is a fancy way of saying "break into pieces". One example of this oscillatory buildup which is very physical and non-destructive is the startup of a crystal oscillator. It starts with noise deviating the output of the oscillator amp a small amount, and the combination of positive & negative feedback builds up the oscillation slowly over many cycles until the oscillation limits itself at the nominal oscillator output amplitude. Thanks P: 5,529 Quote by berkeman One example of this oscillatory buildup which is very physical and non-destructive is the startup of a crystal oscillator. It starts with noise deviating the output of the oscillator amp a small amount, and the combination of positive & negative feedback builds up the oscillation slowly over many cycles until the oscillation limits itself at the nominal oscillator output amplitude. So this reaches a maximum level, just like I said. It does not grow infinitely. Mentor P: 39,683 Quote by voko So this reaches a maximum level, just like I said. It does not grow infinitely. Yes, absolutely. It just doesn't break into pieces like the Tacoma Narrows Bridge! Related Discussions Mechanical Engineering 4 Special & General Relativity 14 Introductory Physics Homework 1 Mechanical Engineering 1 High Energy, Nuclear, Particle Physics 11	1,015	4,431	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.03125	3	CC-MAIN-2014-15	latest	en	0.960851
https://prezi.com/wzffi0yfgguf/logistic-growth-and-decay/	1,542,282,634,000,000,000	text/html	crawl-data/CC-MAIN-2018-47/segments/1542039742666.36/warc/CC-MAIN-20181115095814-20181115121814-00273.warc.gz	732,658,224	24,347	### Present Remotely Send the link below via email or IM • Invited audience members will follow you as you navigate and present • People invited to a presentation do not need a Prezi account • This link expires 10 minutes after you close the presentation Do you really want to delete this prezi? Neither you, nor the coeditors you shared it with will be able to recover it again. # Logistic Growth and Decay No description by ## Alexandra Shlosman on 16 March 2015 Report abuse #### Transcript of Logistic Growth and Decay Logistic Decay Examples Wood Products History P.F. Verhulst was a Belgian mathematician who studied growth and decay in the 19th century. How this relates to class In class we learned about exponential growth and decay which has the same purpose as logistic growth and decay, except logistic growth and decay has more exact calculations. This topic uses properties of logarithms, exponential growth, exponential decay, and the natural base e. Logistic Growth Examples Rabbit Population Logistic Growth and Decay The EFISCEN wood product model classifies wood products according to their life span. There are four classifications: short (1 year), medium short (4 years), medium long (16 years), long (50 years). Based on data obtained from the European Forest Institute the percentage of remaining wood products after "t" years for wood products with long life-spans (such as those used in the building industry) is given by: P(t) = 100.3952/(1+0.0316e^0.0581t ) 2 years ago, a population of rabbits in Japan on Rabbit Island was 500. Rabbit Island is a predator- free zone, making it inevitable for growth to occur. The population has grown to 5,000. What will the population of rabbits be 1 year from now? How long will it take for the population to grow from 500 to 10,000? Solution How to Solve Logistic Growth and Decay Growth Decay Properties P(t)=500e^kt Given information is P_0(initial population) is 500 rabbits. First step is to solve for "k" the growth constant. Second step is answering the first question, which asks what the population will be in 1 year, meaning when t=1. Third step is answering the second question, which asks for the time it takes for the population to grow from 500 to 10,000 which is telling to solve for t. 1) Solve for k 5000=500e^2k 10=e^2k 2k=ln⁡〖(10) k=0.5 ln⁡〖(10) 1. Domain: {all real numbers} Range: {0,c} where c is the carrying capacity 2. No x-intercepts; y-intercept is P(0) 3. 2 Horizontal Asymptotes at y=0 and y=c 4. P(t) is an increasing function if b>0 and a decreasing function if b<0 5. Inflection point where P(t)= 0.5 of c growth: curve up to curve down decay: curve down to curve up 6. Smooth continuous graphs *Typically used for calculation of population growth Properties Graphs Achievements Differential Equation If dy/dt=ky, then y=Ae^kt for some constant A. The rate of change is y=ky "k" is the constant variable Differential Eq. of Growth Differential Eq. of Decay Differential Equation of Growth: dP/dt=kP "k" is the growth constant Exponential Growth Equation: P(t)= P_0e^kt P(t) is the population at time "t" P_0 is the initial population The Differential Equation of Growth: dP/dt=-kP "-k" is the decay constant Exponential Decay Equation: P(t)= P_0e^-kt Limitations In logistic growth, there are limitations that are taken into account. They are represented by either the variable of "c" or "k" Growth Model Logistic functions provide a more realistic model of population growth. P(t) = c/1 + ae^bt t= time (in years, months, days, etc.) P(t)= population after "t" years a= the number that shifts the graph b= positive growth rate c= carrying capacity Education and teaching Secondary education at Athenaeum of Brussels Studied at the University of Ghent and received a degree in exact sciences Taught calculus at the Belgium Military Academy Became professor of mathematics at Université Libre of Brussels Elected president of the Belgium Academy of Science Published a historical essay on an eighteenth century patriot Contributions to logistic growth and decay This graph compares exponential growth and logistic growth and shows how logistic growth is more realistic because of the limitations or "carrying capacity" "Logistic Growth - Boundless Open Textbook." Boundless. N.p., n.d. Web. 08 Mar. 2015. Alexandra Shlosman Mrs. Slavoski Algebra 2B Per.4 15 March 2015 When the "carrying capacity" "c" is between 0 and 1, the graph represents logistic decay. When the "carrying capacity" " is greater than 1, the graph represents logistic growth Decay Growth "Logistic Functions." N.p., n.d. Web. 08 Mar. 2015 <http://wmueller.com/precalculus/families/1_81.html>. "Logistic Population Growth." Encyclopedia Britannica Online. Encyclopedia Britannica, n.d. Web. 09 Mar. 2015. <http://www.britannica.com/EBchecked/topic/470416/population-ecology/70579/Logistic-population-growth>. This is a graph of logistic growth which follows the basic equation of f(x)=1/(1+e^-x) Levenbach, Hans. "Forecasting Trending Time Series with Relative Growth Rate Models." Technometrics 18.3 (1976): 261-72. Web. 9 Mar. 2015. <http://www.pps.k12.or.us/schools/benson/files/htrinh/5144_Demana_Ch03pp275-348.pdf>. First published Logistic Growth Equation in 1838 Used data of several countries in comparison to Belgium to estimate the unknown parameters Continued population studies in 1845 When "carrying capacity" "c" is between 0 and 1, decay is represented. 324. "Ch.4 Exponential and Logarithmic Functions." 4.8 (n.d.): n. pag. Web. 10 Mar. 2015. <http://faculty.uncfsu.edu/fnani/FicamsFrontpage/ch4.8.pdf>. Growth Decay b>0 for growth and b<0 for decay where c is greater than 0 for both functions. 1) What is the decay rate? 2)What is the percentage of remaining wood products after 10 years? 3) How long does it take for the percentage of remaining wood products to reach 50%? In this problem, we will be using the logistic decay model where P(t)=c/1+ae^-bt The decay rate is equal to the absolute value of "b". The value of "b" in the given equation is 0.0581. \|b\|=\|-0.0581\|=5.81% The question asks to solve when "t" is 10 years. P(10)=100.3952/(1+0.0316e^(0.0581(10)) ) P(10) is approximately 95.0, so 95% of the wood products remain after 10 years To solve, make P(t)=50 and solve for "t". P(10)=100.3952/(1+0.0316e^0.0581(10) )=~ 50=100.3952/(1+0.0316e^0.0581t ) 100.3952=50(1+0.0316e^0.0581t ) 2.0079=1+0.0316e^0.0581t 1.0079=0.0316e^0.0581t 31.8956=e^0.0581t ln⁡〖(31.8956)=0.0581t〗 t=59.6 years It will take approximately 59 years and 7 months for the percentage of wood products remaining to reach 50%. 324. "Ch.4 Exponential and Logarithmic Functions." 4.8 (n.d.): n. pag. Web. 10 Mar. 2015. <http://faculty.uncfsu.edu/fnani/FicamsFrontpage/ch4.8.pdf>. 2) Population in 1 year Since the data was taken 2 years ago and is asking for the population 1 year later, the time will actually be 3. P(3)=500e^(0.5 ln⁡(10)×3) =500e^(1.5ln⁡(10)) =500e^(ln⁡(10^1.5)) =500e^(ln⁡(31.62)) =500(31.62) =15,811 The population of rabbits is approximately 15,811. 3) Time for population to grow from 500 to 10,000 1000=500e^(0.5 ln⁡(10)t) 20=e^(0.5 ln⁡(10)t) 0.5 ln⁡(10)t=ln⁡(20) t=(ln⁡(20))/(0.5 ln⁡(10) ) t=2 years,7 months,6 days The time for the population to grow from 500 to 10,000 is approximately 2 years, 7 months, and 6 days. 1. Banner, Adrian. The Calculus Lifesaver. Princeton: Princeton UP, 2007. Print. 2. "Calculus 131, Supplemental Sections 11.1-11.2 Logistic Growth" . (n.d.): n. pag. University of Maryland. Web. 8 Mar. 2015. 3. Exponential Functions: Population Growth, Radioactive Decay, and More (n.d.): n. pag. University of North Texas. Web. 8 Mar. 2015. 4. La Rosa, Myrna. 6.8, Section, Exponential Growth And, Decay Models, and Newton’s Law. Models OBJECTIVE 1 (n.d.): n. pag. Web. 22 Feb. 2015. 5. Levenbach, Hans. "Forecasting Trending Time Series with Relative Growth Rate Models." Technometrics 18.3 (1976): 261-72. Web. 8 Mar. 2015. 6. "Logistic Growth - Boundless Open Textbook." Boundless. N.p., n.d. Web. 08 Mar. 2015. 7. "Logistic Growth, Part 1." Logistic Growth, Part 1. Duke University, n.d. Web. 22 Feb. 2015. 8. "Logistic Population Growth." Encyclopedia Britannica Online. Encyclopedia Britannica, n.d. Web. 8 Mar. 2015. 9. Mueller, William. "Logistic Functions." Logistic Functions. N.p., n.d. Web. 8 Mar. 2015. 10. "Pierre François Verhulst." Verhulst Biography. University of St. Andrews, Scotland, Jan. 2014. Web. 8 Mar. 2015. Works Cited Full transcript	2,545	8,499	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.03125	4	CC-MAIN-2018-47	latest	en	0.923928
http://luxseattle.com/canaccord-jersey-wpzwla/cardinality-of-a-set-e48fde	1,713,171,253,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296816954.20/warc/CC-MAIN-20240415080257-20240415110257-00582.warc.gz	19,750,473	39,560	# cardinality of a set Set A contains number of elements = 5. In other words, it was not defined as a specific object itself. The set of natural numbers is an infinite set, and its cardinality is called (aleph null or aleph naught). Subsets. To see that the function $$f$$ is injective, we take $${x_1} \ne {x_2}$$ and suppose that $$f\left( {{x_1}} \right) = f\left( {{x_2}} \right).$$ This yields: ${f\left( {{x_1}} \right) = f\left( {{x_2}} \right),}\;\; \Rightarrow {\frac{1}{{{x_1}}} = \frac{1}{{{x_2}}},}\;\; \Rightarrow {{x_1} = {x_2}.}$. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. The cardinality of a set is a measure of a set's size, meaning the number of elements in the set. We first discuss cardinality for finite sets and then talk about infinite sets. Two infinite sets $$A$$ and $$B$$ have the same cardinality (that is, $$\left\| A \right\| = \left\| B \right\|$$) if there exists a bijection $$A \to B.$$ This bijection-based definition is also applicable to finite sets. Therefore the function $$f$$ is injective. This lesson covers the following objectives: In mathematics, the cardinality of a set is a measure of the "number of elements" of the set.For example, the set = {,,} contains 3 elements, and therefore has a cardinality of 3. Hence, there is a bijection between the two sets. cardinality definition: 1. the number of elements (= separate items) in a mathematical set: 2. the number of elements…. Cardinality of a set S, denoted by \|S\|, is the number of elements of the set. We show that any intervals $$\left( {a,b} \right)$$ and $$\left( {c,d} \right)$$ have the equal cardinality. To prove this, we need to find a bijective function from $$\mathbb{N}$$ to $$\mathbb{Z}$$ (or from $$\mathbb{Z}$$ to $$\mathbb{N}$$). Definition. This lesson covers the following objectives: This is a contradiction. Let $$\left( {a,b} \right)$$ and $$\left( {c,d} \right)$$ be two open finite intervals on the real axis. What is the cardinality of a set? Simply said: the cardinality of a set S is the number of the element(s) in S. Since the Empty set contains no element, his cardinality (number of element(s)) is 0. The sets N, Z, Q of natural numbers, integers, and ratio-nal numbers are all known to be countable. \end{array}} \right..}\]. There is an ordering on the cardinal numbers which declares ∣A∣≤∣B∣\|A\| \le \|B\|∣A∣≤∣B∣ when there exists an injection A→BA \to BA→B. Therefore, the sets $$\mathbb{R}$$ and $$\left( {0,1} \right)$$ have equal cardinality: $\left\| \mathbb{R} \right\| = \left\| {\left( {0,1} \right)} \right\|.$. The cardinality (size) of a nite set X is the number jXjde ned by j;j= 0, and jXj= n if X can be put into 1-1 correspondence with f1;2;:::;ng. Thus, the mapping function is given by, $f\left( x \right) = \left\{ {\begin{array}{{20}{l}} {\frac{1}{{n + 1}}} &{\text{if }\; x = \frac{1}{n}}\\ {x} &{\text{if }\; x \ne \frac{1}{n}} \end{array}} \right.,$, $\left\| {\left( {0,1} \right]} \right\| = \left\| {\left( {0,1} \right)} \right\|.$, Consider two disks with radii $$R_1$$ and $$R_2$$ centered at the origin. Cardinality of a Set in mathematics, a generalization of the concept of number of elements of a set. However, the cardinality of these indexes is greater than that of the single column indexes, which could reduce their chances of being used by the query optimiser. These cookies will be stored in your browser only with your consent. The contrapositive statement is $$f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$$ for $${x_1} \ne {x_2}.$$ If so, then we have, ${f\left( {{x_1}} \right) = f\left( {{x_2}} \right),}\;\; \Rightarrow {c + \frac{{d – c}}{{b – a}}\left( {{x_1} – a} \right) }={ c + \frac{{d – c}}{{b – a}}\left( {{x_2} – a} \right),}\;\; \Rightarrow {\frac{{d – c}}{{b – a}}\left( {{x_1} – a} \right) = \frac{{d – c}}{{b – a}}\left( {{x_2} – a} \right),}\;\; \Rightarrow {{x_1} – a = {x_2} – a,}\;\; \Rightarrow {{x_1} = {x_2}.}$. So conceptually: 1. cardinality(Bool) = 2 2. cardinality(Color) = 3 3. cardinality(Int) = ∞ 4. cardinality(Float) = ∞ 5. cardinality(String) = ∞ This gets more interesting when we start thinking about types like (Bool, Bool)that combine sets together. Some interesting things happen when you start figuring out how many values are in these sets. His argument is a clever proof by contradiction. We'll assume you're ok with this, but you can opt-out if you wish. This is common in surveying. All finite sets are countable and have a finite value for a cardinality. \end{array}} \right..}\]. This website uses cookies to improve your experience. Nevertheless, as the following construction shows, Q is a countable set. Solution: The cardinality of a set is a measure of the “number of elements” of the set. Formula 1 : n(A u B) = n(A) + n(B) - n(A n B) If A and B are disjoint sets, n(A n B) = 0 Then, n(A u B) = n(A) + n(B) Formula 2 : n(A u B u C) = n(A) + n(B) + n(C) - n(A … For a set SSS, let ∣S∣\|S\|∣S∣ denote its cardinal number. (Georg Cantor) A useful application of cardinality is the following result. Cardinality of a set Intersection. Cardinality of a Set. Make sure that $$f$$ is surjective. CARDINALITY OF INFINITE SETS 3 As an aside, the vertical bars, jj, are used throughout mathematics to denote some measure of size. But this means xxx is not in the list {a1,a2,a3,…}\{a_1, a_2, a_3, \ldots\}{a1,a2,a3,…}, even though x∈[0,1]x\in [0,1]x∈[0,1]. The mapping between the two sets is defined by the function $$f:\left( {0,1} \right] \to \left( {0,1} \right)$$ that maps each term of the sequence to the next one: ${f\left( {{x_n}} \right) = {x_{n + 1}},\;\text{ or }\;}\kern0pt{\frac{1}{n} \to \frac{1}{{n + 1}}. But opting out of some of these cookies may affect your browsing experience. Let Z={…,−2,−1,0,1,2,…}\mathbb{Z} = \{\ldots, -2, -1, 0, 1, 2, \ldots\}Z={…,−2,−1,0,1,2,…} denote the set of integers. Solving the system for $$n$$ and $$m$$ by elimination gives: \[\left( {n,m} \right) = \left( {\frac{{a + b}}{2},\frac{{b – a}}{2}} \right).$, Check the mapping with these values of $$n,m:$$, ${f\left( {n,m} \right) = f\left( {\frac{{a + b}}{2},\frac{{b – a}}{2}} \right) }={ \left( {\frac{{a + b}}{2} – \frac{{b – a}}{2},\frac{{a + b}}{2} + \frac{{b – a}}{2}} \right) }={ \left( {\frac{{a + \cancel{b} – \cancel{b} + a}}{2},\frac{{\cancel{a} + b + b – \cancel{a}}}{2}} \right) }={ \left( {a,b} \right).}$. Below are some examples of countable and uncountable sets. For example, if the set A is {0, 1, 2}, then its cardinality is 3, and the set B = {a, b, c, d} has a cardinality of 4. Forums. LEARNING APP; ANSWR; CODR; XPLOR; SCHOOL OS; answr. A map from N→Q\mathbb{N} \to \mathbb{Q}N→Q can be described simply by a list of rational numbers. Thus, we get a contradiction: $$\left( {{n_1},{m_1}} \right) = \left( {{n_2},{m_2}} \right),$$ which means that the function $$f$$ is injective. The cardinality of this set is 12, since there are 12 months in the year. It is mandatory to procure user consent prior to running these cookies on your website. For each aia_iai, write (one of) its binary representation(s): ai=0.di1di2di3…2,a_i = {0.d_{i1} d_{i2} d_{i3} \ldots}_{2}, ai=0.di1di2di3…2, where each di∈{0,1}d_i \in \{0,1\}di∈{0,1}. The examples are clear, except for perhaps the last row, which highlights the fact that only unique elements within a set contribute to the cardinality. Power object. We also use third-party cookies that help us analyze and understand how you use this website. {n – m = a}\\ Take a number $$y$$ from the codomain $$\left( {c,d} \right)$$ and find the preimage $$x:$$, ${y = c + \frac{{d – c}}{{b – a}}\left( {x – a} \right),}\;\; \Rightarrow {\frac{{d – c}}{{b – a}}\left( {x – a} \right) = y – c,}\;\; \Rightarrow {x – a = \frac{{b – a}}{{d – c}}\left( {y – c} \right),}\;\; \Rightarrow {x = a + \frac{{b – a}}{{d – c}}\left( {y – c} \right). Log in here. The equivalence class of a set $$A$$ under this relation contains all sets with the same cardinality $$\left\| A \right\|.$$, The mapping $$f : \mathbb{N} \to \mathbb{O}$$ between the set of natural numbers $$\mathbb{N}$$ and the set of odd natural numbers $$\mathbb{O} = \left\{ {1,3,5,7,9,\ldots } \right\}$$ is defined by the function $$f\left( n \right) = 2n – 1,$$ where $$n \in \mathbb{N}.$$ This function is bijective. Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. }$, ${f\left( {{x_1}} \right) = f\left( 1 \right) = {x_2} = \frac{1}{2},\;\;}\kern0pt{f\left( {{x_2}} \right) = f\left( {\frac{1}{2}} \right) = {x_3} = \frac{1}{3}, \ldots }$, All other values of $$x$$ different from $$x_n$$ do not change. Example 2.3.6. Cardinality of a set is the number of elements in that set. 11th. For example, the cardinality of the set of people in the United States is approximately 270,000,000; the cardinality of the set of integers is denumerably infinite. If a set S' have the empty set as a subset, this subset is counted as an element of S', therefore S' have a cardinality of 1. Determine the power set of S, denoted as P: The power set P is the set of all subsets of S including S and the empty set ∅. The mapping from $$\left( {a,b} \right)$$ and $$\left( {c,d} \right)$$ is given by the function, ${f(x) = c + \frac{{d – c}}{{b – a}}\left( {x – a} \right) }={ y,}$, where $$x \in \left( {a,b} \right)$$ and $$y \in \left( {c,d} \right).$$, ${f\left( a \right) = c + \frac{{d – c}}{{b – a}}\left( {a – a} \right) }={ c + 0 }={ c,}$, $\require{cancel}{f\left( b \right) = c + \frac{{d – c}}{\cancel{b – a}}\cancel{\left( {b – a} \right)} }={ \cancel{c} + d – \cancel{c} }={ d.}$, Prove that the function $$f$$ is injective. Partition of a set, say S, is a collection of n disjoint subsets, say P 1, P 1, ...P n that satisfies the following three conditions −. Cardinality is a measure of the size of a set.For finite sets, its cardinality is simply the number of elements in it.. For example, there are 7 days in the week (Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday), so the cardinality of the set of days of the week is 7. This is actually the Cantor-Bernstein-Schroeder theorem stated as follows: If ∣A∣≤∣B∣\|A\| \le \|B\|∣A∣≤∣B∣ and ∣B∣≤∣A∣\|B\| \le \|A\|∣B∣≤∣A∣, then ∣A∣=∣B∣\|A\| = \|B\|∣A∣=∣B∣. Consider a set $$A.$$ If $$A$$ contains exactly $$n$$ elements, where $$n \ge 0,$$ then we say that the set $$A$$ is finite and its cardinality is equal to the number of elements $$n.$$ The cardinality of a set $$A$$ is denoted by $$\left\| A \right\|.$$ For example, $A = \left\{ {1,2,3,4,5} \right\}, \Rightarrow \left\| A \right\| = 5.$, Recall that we count only distinct elements, so $$\left\| {\left\{ {1,2,1,4,2} \right\}} \right\| = 3.$$. Describe memberships of sets, including the empty set, using proper notation, and decide whether given items are members and determine the cardinality of a given set. The following corollary of Theorem 7.1.1 seems more than just a bit obvious. Set A contains number of elements = 5. In this video we go over just that, defining cardinality with examples both easy and hard. Of course, finite sets are "smaller" than any infinite sets, but the distinction between countable and uncountable gives a way of comparing sizes of infinite sets as well. What is more surprising is that N (and hence Z) has the same cardinality as the set Q of all rational numbers. Applied Mathematics. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. Read more. Click hereto get an answer to your question ️ What is the Cardinality of the Power set of the set {0, 1, 2 } ? It is clear that $$f\left( n \right) \ne b$$ for any $$n \in \mathbb{N}.$$ This means that the function $$f$$ is not surjective. www.Stats-Lab.com \| Discrete Mathematics \| Set Theory \| Cardinality How to compute the cardinality of a set. Types as Sets. For example, the cardinality of the set of people in the United States is approximately 270,000,000; the cardinality of the set of integers is denumerably infinite. By Cantor's famous diagonal argument, it turns out [0,1][0,1][0,1] is uncountable. We need to find a bijective function between the two sets. For a rational number ab\frac abba (in lowest terms), call ∣a∣+∣b∣\|a\| + \|b\|∣a∣+∣b∣ its height. The smallest infinite cardinal is ℵ0\aleph_0ℵ0, which represents the equivalence class of N\mathbb{N}N. This means that for any infinite set SSS, one has ℵ0≤∣S∣\aleph_0 \le \|S\|ℵ0≤∣S∣; that is, for any infinite set, there is an injection N→S\mathbb{N} \to SN→S. Cardinality used to define the size of a set. The given set A contains "5" elements. Is Z\mathbb{Z}Z countable or uncountable? 8. Declaration. Hence, there is no bijection from $$\mathbb{N}$$ to $$\mathbb{R}.$$ Therefore, $\left\| \mathbb{N} \right\| \ne \left\| \mathbb{R} \right\|.$. It is interesting to compare the cardinalities of two infinite sets: $$\mathbb{N}$$ and $$\mathbb{R}.$$ It turns out that $$\left\| \mathbb{N} \right\| \ne \left\| \mathbb{R} \right\|.$$ This was proved by Georg Cantor in $$1891$$ who showed that there are infinite sets which do not have a bijective mapping to the set of natural numbers $$\mathbb{N}.$$ This proof is known as Cantor’s diagonal argument. This canonical example shows that the sets $$\mathbb{N}$$ and $$\mathbb{Z}$$ are equinumerous. public int cardinality() Parameters. Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between the different types of infinity, and to perform arithmetic on them. The cardinality of a set is the property that the set shares with all sets (quantitatively) equivalent to the set (two sets are said to be equivalent if there is a one-to-one correspondence between them). These definitions suggest that even among the class of infinite sets, there are different "sizes of infinity." This is a contradiction. This means that both sets have the same cardinality. Aleph null is a cardinal number, and the first cardinal infinity — it can be thought of informally as the "number of natural numbers." A minimum cardinality of 0 indicates that the relationship is optional. For finite sets, these two definitions are equivalent. As it can be seen, the function $$f\left( x \right) = \large{\frac{1}{x}}\normalsize$$ is injective and surjective, and therefore it is bijective. For example, let A = { -2, 0, 3, 7, 9, 11, 13 } Here, n(A) stands for cardinality of the set A And n (A) = 7 That is, there are 7 elements in the given set A. Thus, the list does not include every element of the set [0,1][0,1][0,1], contradicting our assumption of countability! Read more. Cardinality of sets : Cardinality of a set is a measure of the number of elements in the set. If this list contains each rational number at least once, we can remove repeats to obtain a bijection N→Q\mathbb{N} \to \mathbb{Q}N→Q. Cardinality can be finite (a non-negative integer) or infinite. These cookies do not store any personal information. We have seen primitive types like Bool and String.We have made our own custom types like this: type Color = Red \| Yellow \| Green. n [P (A)] = 2ⁿ Here "n" stands for the number of elements contained by the given set A. Cardinality is the ability to understand that the last number which was counted when counting a set of objects is a direct representation of the total in that group.. Children will first learn to count by matching number words with objects (1-to-1 correspondence) before they understand that the last number stated in a count indicates the amount of the set. We conclude Z\mathbb{Z}Z is countable. The union of the subsets must equal the entire original set. Describe the relations between sets regarding membership, equality, subset, and proper subset, using proper notation. [ P i ≠ { ∅ } for all 0 < i ≤ n ]. We can choose, for example, the following mapping function: $f\left( {n,m} \right) = \left( {n – m,n + m} \right),$, To see that $$f$$ is injective, we suppose (by contradiction) that $$\left( {{n_1},{m_1}} \right) \ne \left( {{n_2},{m_2}} \right),$$ but $$f\left( {{n_1},{m_1}} \right) = f\left( {{n_2},{m_2}} \right).$$ Then we have, ${\left( {{n_1} – {m_1},{n_1} + {m_1}} \right) }={ \left( {{n_2} – {m_2},{n_2} + {m_2}} \right),}\;\; \Rightarrow {\left\{ {\begin{array}{{20}{l}} Let SSS denote the set of continuous functions f:[0,1]→Rf: [0,1] \to \mathbb{R}f:[0,1]→R. So math people would say that Bool has a cardinalityof two. Since $$f$$ is both injective and surjective, it is bijective. Set theory. A relationship with cardinality specified as 1:1 to 1:n is commonly referred to as 1 to n when focusing on the maximum cardinalities. Learn more. Assuming the axiom of choice, the formulas for infinite cardinal arithmetic are even simpler, since the axiom of choice implies ∣A∪B∣=∣A×B∣=max⁡(∣A∣,∣B∣)\|A \cup B\| = \|A \times B\| = \max\big(\|A\|, \|B\|\big)∣A∪B∣=∣A×B∣=max(∣A∣,∣B∣). Example 14. The cardinality of this set is 12, since there are 12 months in the year. P i does not contain the empty set. This means that, in terms of cardinality, the size of the set of all integers is exactly the same as the size of the set of even integers. Cardinality. Let Q\mathbb{Q} Q denote the set of rational numbers. For instance, the set of real numbers has greater cardinality than the set of natural numbers. When AAA is infinite, ∣A∣\|A\|∣A∣ is represented by a cardinal number. {2\left\| z \right\|,} & {\text{if }\; z \lt 0} More formally, this is the bijection f:{integers}→{even integers}f:\{\text{integers}\}\rightarrow \{\text{even integers}\}f:{integers}→{even integers} where f(n)=2n.f(n) = 2n.f(n)=2n. In the sense of cardinality, countably infinite sets are "smaller" than uncountably infinite sets. If A has only a finite number of elements, its cardinality is simply the number of elements in A. 7.3. A set of cardinality n or @ 0 is called countable; otherwise uncountable or non-denumerable. To formulate this notion of size without reference to the natural numbers, one might declare two finite sets AAA and BBB to have the same cardinality if and only if there exists a bijection A→BA \to B A→B. As seen, the symbol for the cardinality of a set resembles the absolute value symbol — a variable sandwiched between two vertical lines. Cardinality used to define the size of a set. But, it is important because it will lead to the way we talk about the cardinality of in nite sets (sets that are not nite). The cardinality of a set is the same as the cardinality of any set for which there is a bijection between the sets and is, informally, the "number of elements" in the set. So, \[\left\| R \right\| = \left\| {\left( {1,\infty } \right)} \right\|.$, To build a bijection from the half-open interval $$\left( {0,1} \right]$$ to the open interval $$\left( {0,1} \right),$$ we choose an infinite sequence $$\left\{ {{x_n}} \right\}$$ such that all its elements belong to $$\left( {0,1} \right].$$ We can choose, for example, the sequence $$\left\{ {{x_n}} \right\} = \large{\frac{1}{n}}\normalsize,$$ where $$n \ge 1.$$. It can be shown that there are as many points left behind in this process as there were to begin with, and that therefore, the Cantor set is uncountable. Here we need to talk about cardinality of a set, which is basically the size of the set. The continuum hypothesis actually started out as the continuum conjecture , until it was shown to be consistent with the usual axioms of the real number system (by Kurt Gödel in 1940), and independent of those axioms (by Paul Cohen in 1963). ${f\left( {r,\theta } \right) = \left( {\frac{{{R_2}r}}{{{R_1}}},\theta } \right) = \left( {a,b} \right),}\;\; \Rightarrow {\left\{ {\begin{array}{{20}{l}} {\frac{{{R_2}r}}{{{R_1}}} = a}\\ {\theta = b} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{{20}{l}} {r = \frac{{{R_1}a}}{{{R_2}}}}\\ {\theta = b} \end{array}} \right..}$, Check that with these values of $$r$$ and $$\theta,$$ we have $$f\left( {r,\theta } \right) = \left( {a,b} \right):$$, ${f\left( {r,\theta } \right) = \left( {\frac{{{R_2}r}}{{{R_1}}},\theta } \right) }={ \left( {\frac{{\cancel{R_2}}}{{\cancel{R_1}}}\frac{{\cancel{R_1}}}{{\cancel{R_2}}}a,b} \right) }={ \left( {a,b} \right).}$. It can be written like this: How to write cardinality; An empty set is one that doesn't have any elements. However, it can be shown that the cardinality of a straight line (i.e., the number of points on a line) is the same as the cardinality of any segment of that line, of the entire plane, and … [ P 1 ∪ P 2 ∪ ... ∪ P n = S ]. Cardinality. \|S7\| = \| \| T. TKHunny. Cardinality of a set is the number of elements in that set. The cardinality of a set is denoted by $\|A\|$. Noun (cardinalities) (set theory) Of a set, the number of elements it contains. Thus, this is a bijection. Remember subsets from the preceding article? To see this, we show that there is a function f from the Cantor set C to the closed interval that is surjective (i.e. Hence, the function $$f$$ is injective. Let’s arrange all integers $$z \in \mathbb{Z}$$ in the following order: $0, – 1,1, – 2,2, – 3,3, – 4,4, \ldots$, Now we numerate this sequence with natural numbers $$1,2,3,4,5,\ldots$$. For instance, the set of real numbers has greater cardinality than the set of natural numbers. f maps from C onto ) so that the cardinality of C is no less than that of . Otherwise it is inﬁnite. The cardinality of the empty set is equal to zero: $\require{AMSsymbols}{\left\| \varnothing \right\| = 0.}$. There is nothing preventing one from making a similar definition for infinite sets: Two sets AAA and BBB are said to have the same cardinality if there exists a bijection A→BA \to BA→B. Consider an arbitrary function $$f: \mathbb{N} \to \mathbb{R}.$$ Suppose the function has the following values $$f\left( n \right)$$ for the first few entries $$n:$$, We now construct a diagonal that covers the $$n\text{th}$$ decimal place of $$f\left( n \right)$$ for each $$n \in \mathbb{N}.$$ This diagonal helps us find a number $$b$$ in the codomain $$\mathbb{R}$$ that does not match any value of $$f\left( n \right).$$, Take, the first number $$\color{#006699}{f\left( 1 \right)} = 0.\color{#f40b37}{5}8109205$$ and change the $$1\text{st}$$ decimal place value to something different, say $$\color{#f40b37}{5} \to \color{blue}{9}.$$ Similarly, take the second number $$\color{#006699}{f\left( 2 \right)} = 5.3\color{#f40b37}{0}159257$$ and change the $$2\text{nd}$$ decimal place: $$\color{#f40b37}{0} \to \color{blue}{6}.$$ Continue this process for all $$n \in \mathbb{N}.$$ The number $$b = 0.\color{blue}{96\ldots}$$ will consist of the modified values in each cell of the diagonal. Only with your consent Georg Cantor ) a useful application of cardinality can be written like this: to... Defining cardinality with examples both easy and hard 9, 10 } eliminate the variables \ ( )... A list of rational numbers < =Infinity } would the cardinality of:! ; an empty set is a measure of the same number of elements in the set require some care \mathbb! Finitely many rational numbers uses the cardinality of a proper class would ORD... 1, 4, 5 }, ⇒ \| a \| = 5 of elements in AAA your! \| a \| = 5 minimum cardinality, they are said to be of the set \. Specific object itself of its elements ) method set a and set B both have a = { <... On sets, we need to find at least one bijective function the. Have a = left { { 1,2,3,4,5 } right }, Rightarrow left\| right\|... In math, science, and ratio-nal numbers are sparse and evenly spaced, whereas rational. Sets regarding membership, equality, subset, using proper notation aleph naught ) set has an set. Or aleph naught ) this video we go over just that, defining with... Are some examples of countable and uncountable sets actually the Cantor-Bernstein-Schroeder Theorem as! Following is true of S? S? S? S? S? S??... Simply the number of related rows for each of the “ number of cardinal ( basic ) members in set! Security features of the set ≠ { ∅ } for all 0 < i ≤ ]! Infinite sets, there are 12 months in the set: to avoid fact! The interval [ 0,1 ] countable or uncountable definition creates some initially counterintuitive results specific object itself an number!, 9, 10 } Theory ) of a set is one that does n't have elements. ) and \ ( f\ ) is surjective distinct sets is empty uncountable sets $\|A\|$ this. Numbers of each height { n } \to \mathbb { R } S⊂R denote the set {,... Finite ( a non-negative integer ) or infinite avoid double-counting fact data m_1, \ ) we both... Option to opt-out of these cookies on your website packed into the number of elements ” of the of! See that the relationship i ≠ { ∅ } for all 0 < i ≤ ]!, Q is countable case, two or more sets are smaller '' than uncountably infinite or. A ) with finite sets, but infinite sets require some care since \ ( f\ ) surjective. Use third-party cookies that ensures basic functionalities and security features of the to... =Infinity } would the cardinality of a set seemingly straightforward definition creates some initially counterintuitive.! The Power set of real and complex numbers are densely packed into the number of elements in year. When you start figuring out How many values are in these sets fact data is [ 0,1 ] 0,1., meaning the number of elements in the set { true, False } contains two.. The following result cardinality definition, ( of a set resembles the absolute value —. A contains 5 '' elements S⊂R denote the set of countable and uncountable sets the cardinality of! Discrete Mathematics \| set Theory \| cardinality How to compute the cardinality of a set real and complex are... Your browsing experience and ∣B∣≤∣A∣\|B\| \le \|A\|∣B∣≤∣A∣, then ∣A∣=∣B∣\|A\| = \|B\|∣A∣=∣B∣ contains two values: What is number... Relationship in the sense of cardinality can be generalized to infinite sets some! N is the number of elements in $a$ A→BA \to BA→B straightforward definition creates some initially counterintuitive.... Cookies are absolutely essential for the website suggest that even among the of! A bijection between the two sets have the same cardinality any elements, 4 8. Contained in the above section, cardinality '' of a set is bijection! The function \ ( f\ ) is surjective '' elements engineering topics following construction shows, Q of numbers! Suggest that even among the class of infinite sets, there is measure! Understand How you use this website uses cookies to improve your experience while you navigate through website... And C of real numbers has the same cardinality as the set uses cookies to your! Can opt-out if you wish with positive integers wikis and quizzes in math, science, and is. \|B\|∣A∣≤∣B∣ and ∣B∣≤∣A∣\|B\| \le \|A\|∣B∣≤∣A∣, then ∣A∣=∣B∣\|A\| = cardinality of a set map from N→Q\mathbb n. Let Q\mathbb { Q } Q denote the set such an object can be generalized to infinite sets can that! In Mathematics, the function \ ( f\ ) is injective the year countably infinite sets are using. A finite number of elements in the set and is denoted by n ( a ) no integer is to., which is basically the size of the concept of number of elements of the “ number of of. Hence Z ) has the same cardinality ) we add both equations together A→BA... Sets \ ( f\ ) is injective Cognos® software uses the cardinality of set... Number of cardinal ( basic ) members in a set \mathbb { }. Theorem 7.1.1 seems more than just a bit obvious in 0:1, 0 is called countable ; otherwise uncountable non-denumerable... Since there are 12 months in the set of natural numbers is an infinite set AAA finite... Definitions are equivalent, defining cardinality with examples both easy and hard, 3, 4, 8,,! Opt-Out if you wish minimum cardinality, and proper subset, using proper notation the declaration for (. This lesson covers the following construction shows, Q is countable cardinality How to write cardinality ; an empty is! Procure user consent prior to running these cookies all known to be countable below are some examples of and! Numbers, integers, and proper subset, using proper notation lesson covers the following ways: to double-counting! And B are two subsets of a set ) the cardinal number indicating the number of in! Spaced, whereas the rational numbers a rational number ab\frac abba ( in terms... That even among the class of infinite sets [ 0,1 ] [ ]! When you start figuring out How many values are in these sets do cardinality of a set resemble each other in! \ ) we add both equations together R } S⊂R denote the set of rational numbers are all to. On the cardinal number than that of cookies may affect your browsing experience entire set. A rational number ab\frac abba ( in lowest terms ), call +... Thanks the cardinality be Inifinity - 9, Rightarrow left\| a right\| = 5 ] countable or?... Elements by trying to pair the elements up can tell that two sets =x < =Infinity } would the of! Discrete Mathematics \| set Theory \| cardinality How to compute the cardinality of set a contains ''... 'S size, meaning the number of elements in the above section, cardinality '' a! Q countable or uncountable for example, if we have a cardinality of a set was defined functionally finite. 1 is the cardinality of a set $a$ ; Home ; an empty set cardinality of a set number... It turns out [ 0,1 ] [ 0,1 ] then ∣A∣=∣B∣\|A\| = \|B\|∣A∣=∣B∣ written like this: How to cardinality! Includes cookies that help us analyze and understand How you use this website these sets do resemble! Java.Util.Bitset.Cardinality ( ) method returns the number of elements in a geometric sense finite. Indicates that the cardinality cardinality of a set a set is roughly the number of it. Left { { 1,2,3,4,5 } right }, ⇒ \| a \| = 5 cardinal. Returns the number is also referred as the number of cardinal ( basic ) members in a.... Which declares ∣A∣≤∣B∣\|A\| \le \|B\|∣A∣≤∣B∣ when there exists no bijection A→NA \to \mathbb n. Than the set ) \ ( m_2, \ ) we add equations!, we can say that Bool has a cardinalityof two following objectives: Types as sets ( cardinalities ) set... Than the set ∣B∣≤∣A∣\|B\| \le \|A\|∣B∣≤∣A∣, then ∣A∣=∣B∣\|A\| = \|B\|∣A∣=∣B∣ to the number is also referred the. P 2 ∪... ∪ P 2 ∪... ∪ P n = S ] ), call ∣a∣+∣b∣\|a\| \|b\|∣a∣+∣b∣... Of cardinality is the cardinality of sets: cardinality of 0 indicates that the relationship is the declaration java.util.BitSet.cardinality... Using proper notation a non-negative integer ) or infinite ensures basic functionalities and security of! Each of the number of elements in that set a is defined a! You can opt-out if you wish consent prior to running these cookies may affect your browsing.... Both have a = { 1, 4, 5 }, ⇒ a! Through the website to function properly least one bijective function between the two sets $. All 0 < i ≤ n ] some of these cookies on your website following objectives: Types sets... Is defined as the set and is denoted by$ \|A\| \$ ≤ n.! Both equations together C onto ) so that the function \ ( f\ ) is both and... But you can opt-out if you wish the year by some natural number and! When there exists no bijection A→NA \to \mathbb { Q } Q or. Sets, cardinal numbers may be identified with positive integers \| set Theory cardinality... Venn-Diagram as: What is the following is true of S? S? S? S? S S! Shows, Q of natural numbers to be countable measure of the set, 2019 Mishal... Out of some of these cookies 1,2,3,4,5 } right }, Rightarrow left\| a right\| =.!	9,553	31,561	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.53125	5	CC-MAIN-2024-18	latest	en	0.883296
https://johnloomis.org/ece564/notes/flow/flow2.html	1,696,040,553,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233510575.93/warc/CC-MAIN-20230930014147-20230930044147-00676.warc.gz	353,178,185	2,771	# flow, random points on a plane ```% calculate the scene - a field of points clear close all N = 2000; x = rand(N,1)40-20; y = rand(N,1)80-20; z = zeros(size(x)); ``` ```% draw the scene figure('Position',[320 240 320 240]); plot3(x,y,z,'ko'); axis equal axis off camproj('perspective'); campos([0 -20 5]); camtarget([0 -10 0]); camva(30); pos = campos(); tgt = camtarget(); ``` ```% move the camera delta = [0 0.2 0]; aviobj = avifile('rand_flow1.avi','fps',20,'compression','Cinepak'); for k=1:81 campos(pos+(k-1)delta); camtarget(tgt+(k-1)delta); drawnow; F = getframe(gcf); end aviobj = close(aviobj); ``` ```% do flow without MATLAB 3D graphpics figure(2) up = [0 0 1]; vn = pos - tgt; vz = vn/norm(vn); vx = cross(up,vz); vx = vx/norm(vx); vy = cross(vz,vx); R = [vx; vy; -vz;]; T2 = [R -Rpos']; % test disp('transform target point:'); disp(T2[tgt 1]') xp = x(:); yp = y(:); zp = z(:); xform = T2[xp'; yp'; zp'; ones(size(zp'))]; va = 30; scl = 1/tan(0.5vapi/180); idx = find(xform(3,:)>0); % select points in front of camera xpts = xform(1,idx)./xform(3,idx)scl; ypts = xform(2,idx)./xform(3,idx)scl; plot(xpts,ypts,'ko'); axis([-1 1 -1 1]); ``` ```transform target point: 0 0 11.1803 ``` ```% traditional optical flow diagram yp = y(:) - 0.2; xform = T2[xp'; yp'; zp'; ones(size(zp'))]; xnew = xform(1,idx)./xform(3,idx)scl; ynew = xform(2,idx)./xform(3,idx)scl; quiver(xpts,ypts,xnew-xpts,ynew-ypts,10,'k'); axis([-1 1 -1 1]); ```	574	1,462	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.78125	3	CC-MAIN-2023-40	latest	en	0.417832
https://stats.stackexchange.com/questions/454043/random-likelihood-function-during-maximizing-process	1,716,465,502,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058625.16/warc/CC-MAIN-20240523111540-20240523141540-00770.warc.gz	472,158,793	39,327	# Random likelihood function during maximizing process Mostly common definition of Maximum Likelihood Estimator in certain parametric problem $$(\mathcal{X},\mathcal{B}(\mathbb{R}^n),\mathcal{P}=\{\mathbb{P}_\theta\mid\theta\in\Theta\})$$is what follows: $$\hat\theta_{MLE}(\mathbf{x})\equiv\underset{\theta\in\Theta}{\operatorname{argmax}}L_n(\theta\mid\mathbf{x}),\quad\forall\mathbf{x}\in\mathcal{X}$$ It means that for all n-dimensional data points $$\mathbf{x}$$ we have unique solution, which maximize Likelihood function. Then, during the process of maximizing Likelihood is considered as non-random function of $$\theta$$, with data given. I am interested under what assumption we can treat the Likelihood as random in process of deriving estimator- i look for more formal, measure-theoretic approach to this problem. I think that the likelihood should be measurable function, differentiable in $$\theta\quad a.s\quad\mathcal{P}$$. Am I right? What else? • There is no need for the likelihood to be either measurable nor differentiable in $\theta.$ Insisting on that would introduce unnecessary restrictions and rule out the applicability of this theory in some significant situations. – whuber Mar 13, 2020 at 16:37 • The random function $L_n(\theta\mid\mathbf{x})$ should be measurable in $\theta$ (at the very least, to talk about existence of argmax you need more structure) $\mathcal{P}$-a.s. The argmax $\hat\theta_{MLE}$ should be a measurable function of data $\mathbf{x}$, as any estimator should be. There is no requirement that $L_n(\theta\mid\mathbf{x})$ be jointly measurable in $\theta$ and $\mathbf{x}$. Mar 13, 2020 at 20:59 Let $$\left(\mathcal X, \mathcal A, \left(\mathbb P_\vartheta\right)_{\vartheta \in \Theta}\right)$$ be a statistical model consisting of • a set $$\mathcal X$$ (the sample space), • a $$\sigma$$-algebra $$\mathcal A$$ on $$\mathcal X$$, • a family of probability measures $$\left(\mathbb P_\vartheta\right)_{\vartheta \in \Theta}$$ on $$\mathcal A$$ with • index set (the parameter space) $$\Theta$$ of cardinality bigger than one. If, for all $$\vartheta \in \Theta$$, $$\mathbb P_\vartheta$$ is absolutely continuous w.r.t. a $$\sigma$$-finite measure $$\mu$$ on $$\mathcal A$$, then the likelihood function $$\mathcal L$$ is defined as the Radon–Nikodym derivative of $$\mathbb P_\vartheta$$ w.r.t. $$\mu$$: $$\mathcal L(\vartheta, x) \mathrel{:=} \frac{\mathrm d \mathbb P_\vartheta}{\mathrm d \mu}(x), \; \forall \, \vartheta \in \Theta, x \in \mathcal X,$$ and often viewed as a function in $$\vartheta$$.	749	2,564	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.125	3	CC-MAIN-2024-22	latest	en	0.767538
http://guidanceportal.com/economics/use-of-mathematical-tools-in-economics.html	1,529,831,737,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267866926.96/warc/CC-MAIN-20180624083011-20180624103011-00196.warc.gz	138,198,955	5,860	# Use of Mathematical Tools in Economics: “Various fundamental mathematical tools are really helpful in modern economics to calculate different quantities and to measure some relationships. To view such relationships we study the following mathematical tools in economics:" ## Variable: Generally the quantities which change their values are called variable. In mathematics the alphabets (x, y, z) are used to represent variables. But in economics usually the first letter of the quantity is used as its variable. For example: P for price, S for savings and u for utility etc. some symbols are also being used to represent some quantities in economics such as π for profit etc. ### Types of Variables: Variables are divided into some important parts which are described as follows: #### Continuous Variable: Continuous variable never leaves even a slightest gap within its range in the values. For example: time is a continuous variable as we had noticed that the needle of the clock or watch pass through all the points on it. #### Discontinuous Variable: A variable which leaves gaps while changing its values. Price of a good is a common example of discontinuous variable. A change in the price of good changes from Rs. 50 to Rs. 60 per Kg. #### Independent Variable: A variable whose value does not depend on any certain condition and it can assume its value on its own is known as independent variable. These variables cannot get affected from any influence. #### Dependent variables: Those variables whose occurrence and non-occurrence are dependent on independent variables are known as dependent variables. ## Constants: It is also a mathematical tool which is really helpful in advanced economics. Constant is defined as a quantity which does not change its value in a given period. For example: 2, 5, 8 or 56, 98 etc. ## Parameters: The third mathematical tool which we use in economics is called parameter. The certain quantities which do change in real life but in economics we assume them as constant. For example: price of a raw material, condition of weather etc.	407	2,095	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2018-26	latest	en	0.949949
https://communities.sas.com/t5/Base-SAS-Programming/Decision-tree-Array-or-Hash-solution/td-p/306551?nobounce	1,534,729,994,000,000,000	text/html	crawl-data/CC-MAIN-2018-34/segments/1534221215487.79/warc/CC-MAIN-20180820003554-20180820023554-00329.warc.gz	654,909,726	32,134	Valued Guide Posts: 864 Decision tree - Array or Hash solution? I have a list of 447 students, each of these students have classes that they have taken over the last 4 years. I am interested in somehow joining all of them to find groups of classes that they have all taken. I'm not that familiar with hash and I'm new to array. What I'm thinking is a do loop, from 1 to 447. I'd like to join 1 to 1 - 447, then 2 to 1 - 447 and so on. I have a two variable dataset, student_name and course_name, I've added student_num (1-447). The output would be sorted in descending order by how many similar classes there are alike between students. If I get through the first iteration I'll most likely do it again and again. I'm doing this in WEKA software for an Applied Stats class, but I thought I'd take the opportunity to learn something in SAS as well. Cheers, Mark Valued Guide Posts: 540 Re: Decision tree - Array or Hash solution? In essence you are looking for a Cartesian product of a table with itself. This can be done using SQL. An query joining two tables without a WHERE or ON clause will combine every row of the left table with every row of the right one. Usually this happens by mistake (and SAS warns you) but here it's what you want. ``````proc sql; select * from A, A; quit;`````` You can add an ORDER BY to do the sorting. Hope this helps, - Jan. Super User Posts: 6,934 Re: Decision tree - Array or Hash solution? How many are there? With only 10 classes, you can construct as many as 1,000 possible groups. Are you going to try to roll up the groups? For example, if one student took classes A, B, C, and D, but anothe student took classes A, B, C, and E, will you try to count both of them under the group ABC? Valued Guide Posts: 864 Re: Decision tree - Array or Hash solution? There are 130 college courses and 448 students. The end game is to cluster groups of 3 or 4 classes that have comminality between students. It's a project where I will recommend minors or concentrations, classes that have a history of students taking them together. Super User Posts: 24,012 Re: Decision tree - Array or Hash solution? MBA looks at things people purchase together, For example the beer and diaper example from WalMart, which is actually false. Super User Posts: 6,934 Re: Decision tree - Array or Hash solution? That helps. Here are some steps I would take. (Feel free to ask about any of them in more detail.) Come up with a master list of classes, numbering them in an order of your choice 1 - 130. In the students data set, create a character variable 130 characters long. A set of 0/1 values indicate whether the student took that course or not. For example, "11010" means the student took classes 1, 2, and 4, but not classes 3 and 5. Go through the students data set, and output each student multiple times. (I'll use groups of 3 here.) There will be one observation for each group of 3 classes the student took. For example: data halfway_there; set have; do class1=1 to 128; do class2 = class1 + 1 to 129; do class3 = class2 + 1 to 130; if substr(_130_classes, class1, 1) = '1' and substr(_130_classes, class2, 1)='1' and substr(_130_classes, class3, 1) = 1 then output; end; end; end; run; Now a final PROC FREQ will count the occurrences of each group of 3: proc freq data=halfway_there; tables class1 * class2 * class3 / list; run; Posts: 1,848 Re: Decision tree - Array or Hash solution? [ Edited ] According to your statement "... each of these students have classes that they have taken ..." and assuming that most of them have not taken all classes, then you probably have a table of classes with the students in each class, that is: a table with 2 variables: class, student name (or student number) then you can run PROC FREQ like: proc freq data=have; table class*student / out=freqcount; run; proc sort data=freqcount; by descending count; run; proc print data=freqcount; run; if you need the list of students having those max freq. classes, it is possible to subset the table by: proc sql; select h.class, h.student from have as h left join freqcount (obs=1) as f where h.class = f.class and h.student = f.student; quit; Super User Posts: 10,850	1,119	4,256	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.71875	3	CC-MAIN-2018-34	latest	en	0.945212
https://www.physicsforums.com/threads/another-family-question.530720/	1,527,164,682,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794866276.61/warc/CC-MAIN-20180524112244-20180524132243-00019.warc.gz	792,742,059	14,195	# Another family question 1. Sep 16, 2011 ### tramp Hi I have some difficulties with following question. A family has five children. Assuming that the probability of a girl on each birth was 1/2 and that the five births were independent, what is the probability the family has at least one girl, given they have at least one boy? My solution is 1-(1/2)^4 = 15/16 However according to the book correct answer it 30/31. Any ideas? 2. Sep 18, 2011 ### Stephen Tashi Look at material on the binomial distribution. The conditions in the problem are satisified if the family has any of the following combinations of genders: (exactly 1 boy and 4 girls) (exactly 2 boys and 3 girls) (exactly 3 boys and 2 girls) (exactly 4 boys and 1 girl) The conditions are not satisfied by the combinations of genders: (exactly 5 boys and 0 girls ) (exactly 0 boys and 5 girls ) The problem says we are "given" that the family has at least one boy, so you should look at the formula for conditional probability. 3. Sep 20, 2011 ### tramp Thanks a lot Stephen. I figured it out. Your post was a great help.	287	1,099	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.359375	3	CC-MAIN-2018-22	latest	en	0.969223
https://www.studymode.com/essays/Student-1656014.html	1,586,403,585,000,000,000	text/html	crawl-data/CC-MAIN-2020-16/segments/1585371829677.89/warc/CC-MAIN-20200409024535-20200409055035-00416.warc.gz	1,140,645,548	26,427	# Student Pages: 5 (1186 words) Published: May 2, 2013 For use only in [the name of your school] S1 Sample Definitions for S1 Statistical Experiment A text/investigation/process adopted for collecting data to provide evidence for or against a hypothesis. “Explain briefly why mathematical models can help to improve our understanding of real world problems” Simplifies a real world problem; enables us to gain a quicker / cheaper understanding of a real world problem Advantage and disadvantage of statistical model Advantage : cheaper and quicker Disadvantage : not fully accurate “Statistical models can be used to describe real world problems. Explain the process involved in the formulation of a statistical model.” • Observe real-world problem • Devise a statistical model and collect data • Compare and observe against expected outcomes and test model; • Refine model if necessary. A sample space A list of all possible outcomes of an experiment Event Sub-set of possible outcomes of an experiment. Normal Distribution Bell shaped curve symmetrical about mean; mean = mode = median 95% of data lies within 2 standard deviations of mean 2 conditions for skewness Positive skew if ( Q3 − Q2 ) − ( Q2 − Q1 ) > 0 and if Mean − Median > 0 . Negative skew if ( Q3 − Q2 ) − ( Q2 − Q1 ) < 0 and if Mean − Median < 0 . Independent Events P ( A ∩ B) = P( A) × P( B) Mutually Exclusive Events P( A ∩ B) = 0 Explanatory and response variables The response variable is the dependent variable. It depends on the explanatory variable (also called the independent variable). So in a graph of length of life versus number of cigarettes smoked per week, the dependent variable would be length of life. It depends (or may do) on the number of cigarettes smoked per week. Copyright www.pgmaths.co.uk - for GCSE, IGCSE, AS and A2 notes www.XtremePapers.net For use only in [the name of your school] S1 Sample Data Discrete Discrete data can only take certain values in any given range. Number of cars in a household is an example of discrete data. The values do not have to be whole numbers (e.g. shoe size is discrete). Continuous Continuous data can take any value in a given range. So a person’s height is continuous since it could be any value within set limits. Categorical Categorical data is data which is not numerical, such as choice of breakfast cereal etc. Data may be displayed as grouped data or ungrouped data. We say that data is “grouped” when we present it in the following way: Weight (w) 6570Or Score (s) 5-9 10-14 Frequency 2 5 Frequency 3 7 NB: We can group discrete data or continuous data. We must know how to interpret these groups, So that Weight (w) 6570Or Score (s) 5-9 10-14 4.5 ≤ s < 9.5 9.5 ≤ s < 14.5 65 ≤ w < 70 70 ≤ w < 75 Copyright www.pgmaths.co.uk - for GCSE, IGCSE, AS and A2 notes www.XtremePapers.net For use only in [the name of your school] S1 Sample ⎛ 18 − 20 X − 20 23 − 20 ⎞ P (18 < X < 23) = P ⎜ < < ⎟ = P ( −1 < Z < 1.5 ) . 2 2 ⎠ ⎝ 2 If we now had a set of tables showing us all possible values for P ( Z < z ) then we could calculate this since P ( −1 < Z < 1.5 ) = P ( Z < 1.5 ) − P ( Z < −1) . = So we have two curves (1) N (µ , σ 2 ) is the general normal distribution σ 2 (the variance ). We use the variable X. For example, with parameters µ (the mean) and (2) N (0 ,1) is the standard normal distribution with parameter 0 (the mean) and. 1 (the variance). We use the variable Z to distinguish it form the general normal case. Copyright www.pgmaths.co.uk - for GCSE, IGCSE, AS and A2 notes www.XtremePapers.net For use only in [the name of your school] S1 Sample Standard Normal Distribution The cumulative distribution function for the random variable Z is written as Φ(z) . In other words Φ ( z ) = P( Z < z ) = 1 2π −∞ ∫e z − 1t2 2 dt . From the tables we have Φ ( 0 ) = 0.5, Φ (1) = 0.8413, Φ ( 2 ) = 0.9772, Φ ( 3) = 0.9987 etc. We see from the above that to calculate Φ(z) when z < 0 we use symmetry, i.e. we use...	1,121	3,988	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.984375	4	CC-MAIN-2020-16	latest	en	0.812839
http://oeis.org/A117148/internal	1,618,122,013,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618038061562.11/warc/CC-MAIN-20210411055903-20210411085903-00117.warc.gz	73,014,841	3,386	The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A117148 Number of parts in all partitions of n in which no part occurs more than 3 times. 4 %I %S 1,3,6,8,15,24,36,50,75,102,143,197,264,349,467,606,789,1016,1299, %T 1656,2100,2634,3302,4117,5106,6306,7772,9523,11639,14185,17216,20839, %U 25166,30280,36361,43551,52022,62004,73753,87510,103638,122507,144496,170133 %N Number of parts in all partitions of n in which no part occurs more than 3 times. %C a(n) = sum(A117147(n,k), k>=1). %H Vaclav Kotesovec, <a href="/A117148/b117148.txt">Table of n, a(n) for n = 1..15000</a> (terms 1..1000 from Alois P. Heinz) %F G.f.: product(1+x^j+x^(2j)+x^(3j), j=1..infinity) * sum((x^j+2x^(2j)+3x^(3j)) / (1+x^j+x^(2j)+x^(3j)), j=1..infinity). %F a(n) ~ log(2) * exp(Pisqrt(n/2)) / (Pi 2^(1/4) * n^(1/4)). - _Vaclav Kotesovec_, May 27 2018 %e a(4) = 8 because the partitions of 4 in which no part occurs more than 3 times are [4], [3,1], [2,2] and [2,1,1] ([1,1,1,1] does not qualify) with a total of 1+2+2+3=8 parts. %p g:=product(1+x^j+x^(2j)+x^(3j),j=1..55) sum((x^j+2x^(2j)+3x^(3j))/ (1+x^j+x^(2j)+x^(3j)), j=1..55): gser:=series(g,x=0,53): seq(coeff(gser,x^n),n=1..50); %p # second Maple program: %p b:= proc(n, i) option remember; `if`(n=0, [1, 0], `if`(i<1, [0, 0], %p add((l->[l[1], l[2]+l[1]j])(b(n-ij, i-1)), j=0..min(n/i, 3)))) %p end: %p a:= n-> b(n, n)[2]: %p seq(a(n), n=1..50); # _Alois P. Heinz_, Jan 08 2013 %t b[n_, i_] := b[n, i] = If[n == 0, {1, 0}, If[i<1, {0, 0}, Sum[Function[{l}, {l[[1]], l[[2]] + l[[1]]j}][b[n-ij, i-1]], {j, 0, Min[n/i, 3]}]]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 50}] ( _Jean-François Alcover_, May 26 2015, after _Alois P. Heinz_ *) %Y Cf. A001935, A117147. %Y Column k=3 of A210485. - _Alois P. Heinz_, Jan 23 2013 %K nonn %O 1,2 %A _Emeric Deutsch_, Mar 07 2006 Lookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam Contribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent The OEIS Community \| Maintained by The OEIS Foundation Inc. Last modified April 11 02:02 EDT 2021. Contains 342886 sequences. (Running on oeis4.)	985	2,310	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.234375	3	CC-MAIN-2021-17	latest	en	0.56803
http://www.numbersaplenty.com/2208151	1,591,016,550,000,000,000	text/html	crawl-data/CC-MAIN-2020-24/segments/1590347417746.33/warc/CC-MAIN-20200601113849-20200601143849-00258.warc.gz	195,184,313	3,565	Search a number 2208151 = 111911051 BaseRepresentation bin1000011011000110010111 311011012000101 420123012113 51031130101 6115154531 724524521 oct10330627 94135011 102208151 111279020 128a5a47 135c40ca 14416a11 152d9401 hex21b197 2208151 has 8 divisors (see below), whose sum is σ = 2423808. Its totient is φ = 1995000. The previous prime is 2208131. The next prime is 2208203. The reversal of 2208151 is 1518022. 2208151 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times. 2208151 is a nontrivial binomial coefficient, being equal to C(2102, 2). It is a sphenic number, since it is the product of 3 distinct primes. It is a cyclic number. It is not a de Polignac number, because 2208151 - 211 = 2206103 is a prime. It is a Duffinian number. It is a congruent number. It is not an unprimeable number, because it can be changed into a prime (2208131) by changing a digit. It is a pernicious number, because its binary representation contains a prime number (11) of ones. It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 1576 + ... + 2626. It is an arithmetic number, because the mean of its divisors is an integer number (302976). 22208151 is an apocalyptic number. 2208151 is the 2101-st triangular number and also the 1051-st hexagonal number. 2208151 is the 701-st centered nonagonal number. 2208151 is a deficient number, since it is larger than the sum of its proper divisors (215657). 2208151 is a wasteful number, since it uses less digits than its factorization. 2208151 is an odious number, because the sum of its binary digits is odd. The sum of its prime factors is 1253. The product of its (nonzero) digits is 160, while the sum is 19. The square root of 2208151 is about 1485.9848586039. The cubic root of 2208151 is about 130.2195697667. The spelling of 2208151 in words is "two million, two hundred eight thousand, one hundred fifty-one". Divisors: 1 11 191 1051 2101 11561 200741 2208151	606	2,046	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.125	3	CC-MAIN-2020-24	latest	en	0.874568
https://pastebin.com/13AUSyrL	1,638,261,396,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964358966.62/warc/CC-MAIN-20211130080511-20211130110511-00270.warc.gz	528,794,646	5,680	# Leetcode_one_without dp Oct 21st, 2021 826 Never Not a member of Pastebin yet? Sign Up, it unlocks many cool features! 1. class Solution { 2. int dp[21][1001]; 3. int fun(int i,int n,int target,vector<int>&a) 4. { 5. //if(csum == target) return 1; 6. if(target == 0) 7. { 8. return 1; 9. } 10. if(target < 0) return 0; 11. if(i == n-1) 12. { 13. if(target - a[i] == 0) return 1; return 0; 14. } 15. 16. //if(dp[i][target]!=-1) return dp[i][target]; 17. int op1 = a[i] > 0? fun(i+1,n,target-a[i],a):0; 18. int op2 = fun(i+1,n,target,a); 19. return op1 + op2; 20. } 21. public: 22. int findTargetSumWays(vector<int>& a, int target) { 23. int n = a.size(); 24. memset(dp,-1,sizeof dp); 25. int sum = 0 , c = 0; 26. for(int x:a) {sum += x; if(x == 0) c++;} 27. if( (target + sum) & 1) return 0; 28. if(target > sum) return 0; 29. return fun(0,n,(target+sum)>>1,a) * (1<<c); 30. } 31. }; RAW Paste Data	387	1,102	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.71875	3	CC-MAIN-2021-49	latest	en	0.065584
https://elsmar.com/elsmarqualityforum/threads/assignable-causes-rules-clarification.81745/	1,623,909,628,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623487629209.28/warc/CC-MAIN-20210617041347-20210617071347-00342.warc.gz	215,567,904	28,646	# Assignable causes rules clarification #### rnsvasan ##### Involved In Discussions In control chart there is a rule that if 7 points continuously falls on either upper or below the average value, then it is assignable causes. Can any one clarify why 7 is taken as a guideline and why not 6, 5 or 4 etc. #### Johnny Quality ##### Quite Involved in Discussions The guideline depends on which rules you are using. The chart below is taken from here. It seems you are using the AIAG rules, someone more experienced with SPC will hopefully post and show you where the numbers come from. I would say the numbers come from the % chance of the phenomenon coming from common causes, as a point beyond UCL or LCL is something like 99.8% from the top of my head. #### Bev D ##### Heretical Statistician Staff member Super Moderator “What are the odds of certain patterns occurring simply by chance?” Many of the rules were deceptively derived through the use of PROBABILITY (Not inferential statistics of any theoretical distributional model, ie the NOT the Normal distribution or the central limit theorem). Some of the probability calculations are exact, some are based loosely on the empirical rule and Tchebycheff’s Theorem. Of course while the Probability calculation can be made exact or precise there is no precise or exact ‘cutoff’ of when to say that a ‘change’ is suspicious enough to warrant investigation. These rules have stood the empirical test of time and are based on the users 'risk' tolerance of a wrong answer. Remember the alarm doesn't mean that an assignable cause has created a change; it means you should investigate to see if a change has occurred. This is subtle but critical. the calculations can be found at: Western Electric, "Statistical Quality Control Handbook", 1956, 2nd Edition, 10th Printing, May 1984, Part B, pp. 23-30 and Part F, pp. 149-183 http://www.contesolutions.com/Western_Electric_SQC_Handbook.pdf Lloyd S. Nelson, "Technical Aids," Journal of Quality Technology vol. 16, no. 4 (October 1984), pp.238-239 The Western Electric Handbook is available for FREE. I strongly recommend it. #### Mike S. ##### Happy to be Alive Trusted Information Resource In "Understanding Variation" Dr. Wheeler states that 8 points above or below the avg is "roughly the same as getting 8 or more heads or tails in succession when tossing a coin" with the odds "commonly cited at less than 1 out of 128". I suppose 7 points are 1 in 64 odds and for some that is enough to suspect a signal. Assignable cause/corrective action list for SPC Software Statistical Analysis Tools, Techniques and SPC 3 Corrective Action for a scenario where Assignable Cause is not confirmed Nonconformance and Corrective Action 4 O Is it an Assignable Cause only if the Control Chart says so? Statistical Analysis Tools, Techniques and SPC 16 W How to justify Widened Control Limits - No Assignable Cause scenario Statistical Analysis Tools, Techniques and SPC 26 The Perfect audit? External Audit causes a significant negative impact in a company General Auditing Discussions 9 Root causes - operator's actions Problem Solving, Root Cause Fault and Failure Analysis 17 The known set of all root causes Nonconformance and Corrective Action 11 A special Supermoon comes—but what causes it? 14 November 2016 Coffee Break and Water Cooler Discussions 0 Preparing a Root Cause Report - Possible Root Causes Document Control Systems, Procedures, Forms and Templates 11 B Process Capability - Changing Limits to Improve Cpk or find Root Causes? Capability, Accuracy and Stability - Processes, Machines, etc. 6 M What are the major causes of Forklift Accidents? Occupational Health & Safety Management Standards 13 Major Nonconformance on Stage 2 Audit causes to start over Stage 1? IATF 16949 - Automotive Quality Systems Standard 2 FDA 483 Listing Observations - Mostly Corrective Action and Root Causes? 21 CFR Part 820 - US FDA Quality System Regulations (QSR) 5 A Standard Root Causes in a process orientated engineering company Problem Solving, Root Cause Fault and Failure Analysis 17 E Corrective Actions Root Causes from Surveillance Audit Findings Nonconformance and Corrective Action 3 P What are the causes of painting defects? Manufacturing and Related Processes 2 Linking the FMEA Causes & Action Plan with DOE Experiments or Experimentation Design FMEA and Control Plans 1 PMMA IRK304 Resin and EVA (Ethylene Vinyl Acetate) Tape causes a Stain Mark Manufacturing and Related Processes 3 D Root Cause Analysis and "Comprehensive List of Causes" Nonconformance and Corrective Action 4 B Process FMEA (PFMEA) Same Multiple Causes for Different Effects FMEA and Control Plans 6 J Causes of Porosity of Die-Castings and How to Improve Die-Casting Manufacturing and Related Processes 10 A Resistance Welding Customer Complaints - Cracking and Pitting Root Causes Customer Complaints 4 Causes of Disbond Defects in Composites Manufacturing Process Manufacturing and Related Processes 7 S On Time Delivery - Non conformance? Root Causes, Corrective Actions ISO 9000, ISO 9001, and ISO 9004 Quality Management Systems Standards 10 R Lyophilization: Causes of process failure? Manufacturing and Related Processes 3 Possible Causes of Invert Sugar Recrystallising Manufacturing and Related Processes 1 Failure Modes & Causes in dFMEA & pFMEA FMEA and Control Plans 3 PFMEA (Process FMEA - Potential Causes and Failure) - Operator's Role FMEA and Control Plans 11 B Seeking advice on Recurring Findings with Different Causes Internal Auditing 13 A Common causes of Corrosion on Alm 7075 Material Annodized, Tumbled, and Plated Manufacturing and Related Processes 6 Wavelength Shift in Emitter Assemblies - Causes Manufacturing and Related Processes 1 M Cl 8.5.2 Corrective Action - "Cause" vs. "Causes" ISO 9000, ISO 9001, and ISO 9004 Quality Management Systems Standards 11 February 2010 Windows Update Causes BSOD for Alureon Rooted XP Users After Work and Weekend Discussion Topics 8 Going "green" causes problem for public works crews in West Bend Coffee Break and Water Cooler Discussions 14 S External Audit - Nonconformance Root causes ISO 9000, ISO 9001, and ISO 9004 Quality Management Systems Standards 23 Documenting the Root Causes for Internal Audit Non-Conformance Findings Problem Solving, Root Cause Fault and Failure Analysis 5 C Identifying Causes (book chapter 4 on problem solving) The Reading Room 26 K Should we rate according to the prob. a risk will occur or that risk causes hazards? ISO 14971 - Medical Device Risk Management 5 L How to distinguish two types of causes (common cause & special cause) Statistical Analysis Tools, Techniques and SPC 4 M Process FMEA (PFMEA) Assumptions Regarding Causes Identification FMEA and Control Plans 6 B Rational Subgroup and Estimation of Variability Due to Common Causes Statistical Analysis Tools, Techniques and SPC 7 C Quality causes business demise World News 10 J The Red Plague - Improper copper wire coating process causes oxidation ISO 9000, ISO 9001, and ISO 9004 Quality Management Systems Standards 3 M System Level Root Causes for each individual CAR generated in the system? Problem Solving, Root Cause Fault and Failure Analysis 10 A How to organize DFMEA for Failures that have many Causes and Effects FMEA and Control Plans 2 M Problem resolution for Common Causes.. Misc. Quality Assurance and Business Systems Related Topics 4 TS 16949 Certificate - What are causes for which we can lose our certificate? IATF 16949 - Automotive Quality Systems Standard 22 A Removing the effect of special causes on Xbar-R chart - Centerless grinding process Statistical Analysis Tools, Techniques and SPC 6 Definition Statistical Control - A stable process free of special causes Definitions, Acronyms, Abbreviations and Interpretations Listed Alphabetically 0 A FDA will not make a company comply with a regulation if it causes undue hardship ISO 13485:2016 - Medical Device Quality Management Systems 2	1,810	7,980	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.921875	3	CC-MAIN-2021-25	latest	en	0.932443
http://mathoverflow.net/revisions/72998/list	1,371,703,285,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368710299158/warc/CC-MAIN-20130516131819-00052-ip-10-60-113-184.ec2.internal.warc.gz	163,486,456	5,292	MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4). ## Return to Question 2 added 358 characters in body; added 28 characters in body Question. Is there an example of a compact $3$-dimensional Calabi-Yau manifold with finite fundamental group $G$ that does not admit a free action on $S^3$? This question is motivated by the following: it is known that many simply-connected Clabi-Yau 3-folds admit a singular Lagrangian torus fibration over $S^3$. I don't know if there are exceptions. On the other hand, if $\pi_1$ is finite and we still have a lagrangean torus fibration, one can expect that the base is a lens space. But in this case probably $\pi_1$ of the CY-manifold will be equal to $\pi_1$ of the base. PS. As Tony Pantev explains, the answer to this question is YES -- there are such examples. On the other hand, if we assume that a finite group $G$ is acting freely on a CY 3-manifold preserving the volume form and preserving a Lagrangian torus fibration, this should impose some very strong restrictions on $G$. I wonder if anyone bothered to work out what is the restriction :). 1 # Finite fundamental groups of 3-dimensional Calabi-Yau manifolds Question. Is there an example of a compact $3$-dimensional Calabi-Yau manifold with finite fundamental group $G$ that does not admit a free action on $S^3$? This question is motivated by the following: it is known that many simply-connected Clabi-Yau 3-folds admit a singular Lagrangian torus fibration over $S^3$. I don't know if there are exceptions. On the other hand, if $\pi_1$ is finite and we still have a lagrangean torus fibration, one can expect that the base is a lens space. But in this case probably $\pi_1$ of the CY-manifold will be equal to $\pi_1$ of the base.	472	1,848	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.796875	3	CC-MAIN-2013-20	latest	en	0.915018
https://physics.stackexchange.com/questions/484222/mathematical-rigorous-definition-for-an-electrical-dipole/484233	1,571,751,454,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570987822098.86/warc/CC-MAIN-20191022132135-20191022155635-00386.warc.gz	643,274,543	33,932	# Mathematical rigorous definition for an electrical dipole I've been reading Laurent Schwartz's Mathematics for the physical sciences, and in the chapter on distributions he makes many cool examples of ways to define in a mathematical rigorous way certain entities given for granted in physics (e.g. a distribution of mass that may be discrete using the Dirac's delta). What is unclear to me is the definition of electrical dipole of moment +1 in one dimension as a distribution. The way he goes is the following: a dipole is the "limit" of thd system $$T_\epsilon$$ of two charges $$\frac{1}{\epsilon},-\frac{1}{\epsilon}$$ respectively at the positions $$0,\epsilon$$ for $$\epsilon \to 0$$. He then says that this corresponds to the distribution $$$$\langle T_\epsilon \| \phi \rangle = \frac{1}{\epsilon} \phi(\epsilon) - \frac{1}{\epsilon} \phi(0) = \phi'(0).$$$$ Would there be someone patient enough as to make it a bit clearer for me? I understand the reasoning of the dipole as a limit, but I can't understand what he means by the undefined function $$\phi$$ • So you're reading a chapter about distribution theory and you got this far while still being unclear about the role of test functions? To be honest, it sounds like you should do a whole lot of in-depth revision of the earlier parts of that chapter. – Emilio Pisanty Jun 4 at 19:28 • Yes, I knkw what a test function is! Anyhow, I'll make myself clearer: when he makes the example of mass distribution, he says thag evaluating the total mass of a distribution $\rho$ means evaluating $\langle \rho \| 1 \rangle$. He also makes the example of the moment of inertia of a body with said mass distribution: $I=\langle \rho \| r^2 \rangle$. As for the dipole, he doesn't specify what that $\phi$ is. And yes, his conclusion is that a dipole should be identified with $- \delta'(0)$, but I can't really figure out why he comes to this conclusion (heuristically) – Operatore_Nabla Jun 5 at 11:04 ## 2 Answers I prefer to think of dipoles as irreducible representations of the SO(3) group. Now you could have such representations over different fields, e.g. it could be electromagnetic field, it could be current density, it could be gravitational potential etc. For me, the most familiar one is the dipoles that occur in electromagnetism. So let us think about the dipole that is the component of some charge density $$\rho\left(\mathbf{r}\right)$$. Assuming you cannot actually go an poke that charge density, the only way you can observe it, is through its potential: $$\phi\left(\mathbf{r}\right)=\int d^3 r' G\left(\mathbf{r}-\mathbf{r'}\right)\rho\left(\mathbf{r}\right)$$ Where $$G\left(\mathbf{r}-\mathbf{r'}\right)$$ is the relevant Greens function. That greens function is well-behaved as long as the observer does not go into the region actually occupied by charge. As a result all the necessary information about the charge density can be expressed as: $$\rho\left(\mathbf{r}\right)=\rho_0\delta\left(\mathbf{r}\right)+\rho_1^\mu\partial_\mu \delta \left(\mathbf{r}\right)+\rho_2^{\mu\nu}\partial_{\mu\nu}\delta\left(\mathbf{r}\right)...$$ So that: $$\phi\left(\mathbf{r}\right)=\rho_0G\left(\mathbf{r}\right)+\rho_1^\mu\partial_\mu G\left(\mathbf{r}\right)+\rho_2^{\mu\nu}\partial_{\mu\nu}G\left(\mathbf{r}\right)+\dots$$ Basically you do Taylor expansion on the Greens function in the region where charge density is not zero (and then integrate the charge density, e.g. $$\rho_0=\int d^3 r' \rho\left(\mathbf{r'}\right)$$), but you package it as 'Taylor expansion' of the delta function. So then the question comes about what to do with the tensors $$\rho_{0,1,2,\dots}$$. If you choose to decompose them into the irreducible representations of SO(3) group (representation follows from the derivatives), then $$\rho_0$$ will be a monopole (trivial), $$\rho_1$$ will be the dipole (trivial), $$\rho_2$$ will have a quadrupole component and some other stuff which may or may not be discarded (depends on treatment). Another way to approach this is to think about charges. The charge density of a point charge is $$\rho=q\delta\left(\mathbf{r}\right)$$. Now consider the charge density due to two opposite charges at $$\mathbf{r}$$, separated by vector $$\mathbf{\epsilon\mathbf{\hat{p}}}$$: $$\rho=+q\delta\left(\mathbf{r}-\frac{\epsilon}{2}\mathbf{\hat{p}}\right)+(-q)\delta\left(\mathbf{r}-\left(-\frac{\epsilon}{2}\mathbf{\hat{p}}\right)\right)=-q\epsilon\left(\frac{\delta\left(\mathbf{r}+\frac{\epsilon}{2}\mathbf{\hat{p}}\right)-\delta\left(\mathbf{r}-\frac{\epsilon}{2}\mathbf{\hat{p}}\right)}{\epsilon}\right)=-\mathbf{p}.\boldsymbol{\nabla}\delta\left(\mathbf{r}\right)$$ In the process we took the limit $$\epsilon\to0$$ and $$\epsilon q\to p$$ and defined $$p\mathbf{\hat{p}}=\mathbf{p}$$, which we call the dipole moment. Clearly, limit of a delta function works only in the generalized sense. The quantity $$\phi$$ is his smooth test function. The dipole distribution itself is $$T=-\delta'(x)$$ with $$\langle T\|\phi\rangle\equiv -\int \delta'(x) \phi(x) \,dx= \int \delta(x) \phi'(x)\,dx= \phi'(x).$$ The integration by parts is really the definition of the derivative of $$\delta(x)$$ rather than a real integration by parts. Does not Schwartz describe test functions in his book? After all he invented them and whole idea of a rigorous approach to Dirac's delta. • Yes I know what test functions are, read my comment above, I think I made a bit clearer what I'm having trouble with. Note that the "intuitive" side of the reasoning is unclear to me, rather than the rigorous one. – Operatore_Nabla Jun 5 at 11:09	1,574	5,617	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 27, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0}	3.71875	4	CC-MAIN-2019-43	latest	en	0.906829
http://www.jiskha.com/9th_grade/?page=28	1,498,542,085,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128320995.93/warc/CC-MAIN-20170627050500-20170627070500-00307.warc.gz	557,680,606	6,152	what is the modern atomic theory how to use a hundred chart to subtract How do you turn eleven hundredths in standard form? How do i do the multi-step equation, 2(7+x)=22? how do you figure the perimeter of a rectangle? -6d/3 when d=-10 name 2 scientest who inherited traits The skydiver had fries and a milkshake. Why was he nervous?. how does aother properties affect physical property 1. (y^2+6y+1)(3y^2-3) 2. (4x^2-2x-1)(3x-1) 3. (n^2-n+2)(n+2) Help!! I'm trying to help my daughter with her homework. grammar what is a predicate in 3 grade? what is a stressed syllable HELP MATH Thank you for taking the time to respond. I am in the 5th grade. However word problems are a problem. On my homework sheet it says, Strategies Used.... Visual Thinking... My soloution First I... Then I... Next I... After that I... Finally I... My answer is... I think my answer... what is an algebraic expression for the value of 12 how do u solve this in analysis and funtion 7(b-6)=-12+b the equation y= 6 sin(3x-1)pi+4 has a fundamental period at what is the squrae root of 729??????? whwat does 12g mean? 2b/4 + b/3 = 10 solve for b. solve for x. \|x+9\|=17 solve for n. \|3n+5\|=7 2b/4+b/3=10 solve for b. solve the equation 2a=3a+4c for a solve the equation 2x+y=9 for y solve the equation for x: 3x+y=6 name the 50 states in alphabetical order? What is Mary Ludwig Education as an child What is chemical formula of copper(ll) sulfide? how can a map help ypu find places 3 1/4 - 1 1/2 = ? I got the answer 1 3/4. is 6.12 greater than 5.99 rectangular arrays for 8. write prime or composit formula to find the circumference end radious 15divided by 10 = science need to write 5 liquid and 5 solids for grade 2 solve a perfect square trinomial equation. (4x+1)(x+4)=49 name the weakness for article of conferation (-4k)2+k2 solve what is the sum 0f 198 and 864? 44 TIMES N =368....FIND N 368/44 211 DIVIDED BY 36? What is the least common multiple for the following? 1.) 7,8,10 2.) 2,5,6 3.) 3,4,7 Thank you how do you do cross word puzzles? unscramble math words:goatnoc how to you say FeSO4(aq) in words need help with edhelperku puzzle Forms of communications of the andes? what is a compass rose and legend? What is a mood/tone map? 1 1 1 - . - . - 5 5 2 help me answer this fraction problem what are 2 sets of patterns that may not be part of visual arts ? how to model each rule with a table of values and a graph how do i find antony for words in my weglish class. what are complex animals i need help with unit rates how is ignous rock formed how do you estimate a product when multiplying fractions? what are two numbers that have the quotient of 2 and the sum of 14 what solids and liquids used i house what is the shrinking patterns? what is the algebraic expression for 200 less than 3? arrary 4 x 36 What is the main purpose of a flower's Aroma? It is greater than 30 and less than 40.If you add the digits,the sum is 10 How do you graph f(x)= ((x+5)(x-4)^(2))/((x-2)(x^(4)) WHAT IS A CHEMICAL SOLUTION AND GIVE AN EXAMPLE? I need help with 6 math unscrambler problems. Show me a plot diagram example please. How did Anne Frank change the world? Change each measurement to the given unit what is the antonym for legacy what is the answer -14.88 - (-20.88? do you have a anwser to this question ,68, ,94, ,120 and the rule 3y-1=13-4y what is y? What did the spanish farmer say to his chicken how do I do a cover page for my book report 5y=2x+3 can you put it in AX+By=C form how do you work this out? 5x-6x and 6x+2x-3 -5+(5) Problem: c=45.00 (m) M=8 Do I times 45 by 8? math Between 50 and 100 grade 7 students at school are having a bbq. Each student will receive one hot dog. Hot dogs come in packages of 12 and buns in packages of 8. How many students are in grade 7, if all the packages of hot dogs and buns purchased are used up? plot elements: cause of conflict I need help on this equation: -5(-3n+4)-12=13 What is a PRIME NUMBER? arithmetic I have a Q that im trying to solve: it says: the 1st , 2nd, and 3rd term of a geo. sequence is the same as the 1st, 7th and 9th terms of an arith. sequence.: I have to find the common ratio. So i know: a+8d = a(r^2) (1) a+6d = ar (2) but i don't know how to solve this ... i need a picture for X+2Y=13 to pile up or collect how can I make the number 40 in two parts. 1/7+3/5 what is a sentence using the word sacred? the perimeter of a rectangle is 12 units.what is the possible area? Were could I find LA. How do you measure volume and mass? Why are federal judges appointed and not elected?	1,420	4,632	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.0625	3	CC-MAIN-2017-26	latest	en	0.877702
metropolice.jp	1,516,157,385,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084886794.24/warc/CC-MAIN-20180117023532-20180117043532-00415.warc.gz	225,250,587	3,902	# QuaterMaster ### Overview For the approach to the machine learning, we adopted a way to use physical theory like quantum annealing and neural network to enforce existing device and /or tool solutions such as conventional computing way, GPU and FPGA. ### Problems to solve Machine Learning Deep Learning Crytography Data Analysis Finite Element Method ### Problems to research Automatic layout and wiring system for integrated circuit Route search method based on quantum annealing Finding optimal formulation for drug and chemical substance development ## About Quantum Annealing Simulation ### Overview We are simulating quantum annealing on a classical computor based on physical theory to discretization the Schrodinger equation. ### Path Integral We discretize the Schrodinger equation using path integral to several Trotter Slice and calcurating the interaction between adjoining slices. ### Annealing We control two way of annealing (Temperature and Quantum effect) by controlling two parameter of beta and gamma. ## About Machine Learning adopted to logical circuit ### Overview In a basic physical theory, many phenomenon is written in Ising model . Basically these Ising Model has {-1,1} value of each spins. To interpret these Ising model to Quantum annealing logical circuit, we are using QUBO to transform {-1,1} to {0.1}. ### Bit notation To write a decimal number, we have to use some binary qubit to write these numbers. ### Logical Operations For expample, to solve prime factorization ,we need to solove (N-pq)^2.Each of "p" and "q" is decimal number by binary qubit. N^2 + 2pqN + (pq)^2 has calcuration of interaction over 2 factors. To solve 3 or 4 interaction of qubits, we need to decomposite these to 2 interaction. ## QuaterMaster{logical simulator based on Quantum Annealing} Machine Spec Tuning Mac Pro for Quantum annealing -3.7 GHz quad Core Intel Xeon E5 -12Gb 1,866Mz DDR3 ECC memory -Dual AMD FirePro D300 (2GB GDDR5 VRAM) -256GB flash storage (PCIe)	463	2,004	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2018-05	longest	en	0.805086
https://ch.mathworks.com/matlabcentral/answers/42337-i-have-a-20x1-vertical-matrix-i-want-to-take-the-averages-of-the-pairs-in-order-and-display-them-i	1,569,315,023,000,000,000	text/html	crawl-data/CC-MAIN-2019-39/segments/1568514572896.15/warc/CC-MAIN-20190924083200-20190924105200-00070.warc.gz	429,699,773	18,393	## I have a 20x1 vertical matrix. I want to take the averages of the pairs in order and display them in another matrix. How? on 28 Jun 2012 ### Image Analyst (view profile) To elaborate. I want to average the first and second numbers, third and fourth numbers, etc.. and display them again in a 10x1 matrix. ### Image Analyst (view profile) on 28 Jun 2012 Edited by Image Analyst ### Image Analyst (view profile) on 28 Jun 2012 Try this: m = randi(5, [20 1]) % Generate sample data % Reshape to a 10 by 2 array. reshapedMatrix = reshape(m, [10,2]) % Get the means going across columns (within a row) meansOfPairs = mean(reshapedMatrix, 2) ### Honglei Chen (view profile) on 28 Jun 2012 Edited by Honglei Chen ### Honglei Chen (view profile) on 28 Jun 2012 x = rand(20,1); y = transpose(mean(reshape(x,2,10)))	241	821	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.703125	3	CC-MAIN-2019-39	latest	en	0.832401
https://testbook.com/blog/rrb-ntpc-exam-analysis-shift-2-4th-jan-2021/	1,695,346,724,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233506320.28/warc/CC-MAIN-20230922002008-20230922032008-00641.warc.gz	626,264,120	47,932	# RRB NTPC Exam Analysis Shift 2 4th Jan 2021 – Section-wise Analysis Here! 0 Save RRB NTPC Exam Analysis Shift 2 4th Jan 2021: The most renowned Railway Recruitment Board (RRB) is conducting the NTPC Exam all across the country at various examination centers. The RRB NTPC is set to take place from 28th December 2020 till 13th January 2021 to select deserving candidates for its various posts. Recently, RRB has completed the second shift of the day i.e of 4th Jan 2021, check important questions asked, difficulty level, and good attempts here. Know about RRB NTPC Exam Analysis Shift 2 4th Jan 2021 related important information here. • The RRB NTPC exam took place following COVID 19 safety precautions issued by the govt like wearing masks, etc. • According to experts the difficulty level for the exam easy to moderate, Questions asked in General awareness have more questions on important personalities, static Gk, and modern history. Whereas questions asked in mathematics were easy and were direct formula based. • The candidates who have their exam on the other days can go through the exam analysis for today and can have an idea regarding the questions being asked. Check RRB JE Recruitment here. ## RRB NTPC Exam Analysis Shift 2 4th Jan 2021 The RRB NTPC Exam comprises 3 subjects are, General Intelligence and Reasoning, Mathematics, and General Awareness. The table below shows the distribution of questions topic wise and the difficulty level of difficulty under each section. Section Marks Shift 2 Difficulty Level Shift 2 Good Attempts General Awareness 40 Easy to Moderate 27-35 Mathematics 30 Easy 22-27 General Intelligenceand Reasoning 30 Easy to Moderate 23-27 Overall 100 Easy to Moderate 67-79 Check RRB JE Exam Language here. ### RRB NTPC Exam Analysis Shift 2 4th Jan 2021 Mathematics The Numerical Ability section was easy to moderate in level. To know the number of questions asked in shift 2, have a look at the table below. Topic Number of Que- Shift 2 S.I, CI 2 Ratio & Proportion 3-4 Profit/Loss 4-5 Mensuration 3-4 Trigonometry 4-5 Number System – Time and Work/ Pipe & Cistern 1-2 Geometry 3-4 Speed and Distance 2-3 Average & Percentage 2-3 MSC 13-15 Total 30 Important Questions asked in RRB NTPC Exam Analysis Shift 2 4th Jan 2021 • The principal value is Rs 5000, Simple Interest is Rs 1500 and the rate is 10 % than time will be Check RRB JE Exam Date here. ### RRB NTPC Exam Analysis Shift 2 4th Jan 2021: Reasoning The Reasoning section included questions on Blood-Relation, Direction, verbal reasoning, coding-decoding, etc. The overall level of reasoning section is easy-moderate. Topic Number of Que- Shift 2 Syllogism 3-4 Matrix – Mathematical Operations 2-3 Blood Relations 2-3 Non-Verbal (Mirror Image, Counting figure, Embedded figure) – Odd one out 2-3 Coding-Decoding 3-4 Direction Sense Test – Venn Diagram – Series 3-4 Miscellaneous 12-15 Total 30 Check RRB JE Syllabus here. ### RRB NTPC Exam Analysis Shift 2 4th Jan 2021: General Awareness General Awareness sections constitute Static GK, Current Affairs, and General Awareness of India & its neighboring countries. The level of this section today was moderate. Topic Number of Que- Shift 2 History 3-4 Geography 2-3 Polity 3-4 Economics 2 Current Affairs 8-15 Science 5 Static GK 5-8 Total 40 Important Questions asked in RRB NTPC Exam Analysis Shift 2 1st Jan 2021 • Which pass connects Leh with Jammu kashmir • The bloodsucking animal is known as • Question on RBC Count was asked • Question on Article 21 • Question on White revolution • Nuclear reactors till Nov 2020? • First Bank of India is? • How many nuclear weapons have been tested in India? • Question on PH was asked • MSP is decided by? • SMPTP Full form is? Know RRB JE Apply Online process here. ## RRB NTPC Exam Pattern Those candidates who clear the written exam will be eligible for the document verification process. The time duration of Paper I is 90 minutes divided into three sections viz. General Awareness, Mathematics and, General Intelligence & Reasoning. Subject No. of Questions Max. Marks General Awareness 40 40 General Intelligence and Reasoning 30 30 Mathematics 30 30 Total 100 questions 100 marks Check RRB NTPC Exam Pattern here ## RRB NTPC Cut Off Marks Previous Year Let’s know the RRB NTPC Cut Off Marks for previous years’. ## RRB NTPC Previous Year Exam Analysis Railway RRB Exam Analysis 18th April 2016 Shift 2 Railway RRB Exam Analysis 12th April 2016 Shift 1 Railway RRB Exam Analysis 18th April 2016 Shift 1 Railway RRB Exam Analysis 12th April 2016 Shift 2 Railway RRB Exam Analysis 16th April 2016 Shift 2 Railway RRB Exam Analysis 11th April 2016 Shift 3 Railway RRB Exam Analysis 16th April 2016 Shift 3 Railway RRB Exam Analysis 11th April 2016 Shift 2 Railway RRB Exam Analysis 16th April 2016 Shift 1 Railway RRB Exam Analysis 11th April 2016 Shift 1 To get updated about the Latest Current Affairs Read the table below Current Affairs 2021 Current Affairs Today’s Current Affairs Current Affairs in Hindi Current Affairs Quiz We hope you found this article informative and helpful. If you have any queries regarding the RRB NTPC exam analysis 2020-21, please write in the comment section below. You can also download our Testbook App which is free and start preparing for the competitive examination by taking the mock tests before the examination to upgrade your preparations. Practice is the key to success therefore, boost your preparation by starting your practice now.	1,389	5,538	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.84375	3	CC-MAIN-2023-40	latest	en	0.932386
http://ilcasarosf.com/long-division-worksheets-no-remainders/	1,534,602,332,000,000,000	text/html	crawl-data/CC-MAIN-2018-34/segments/1534221213689.68/warc/CC-MAIN-20180818134554-20180818154554-00658.warc.gz	192,868,831	9,500	Categories Recent Files Archives # Long Division Worksheets No Remainders By Susan J. Ward on August 12 2018 07:20:41 In the simplest cases, the two fractions will already have a common denominator. In this case, add the numerators and then reduce the resulting fraction. In KS2 children are taught that decimals are another way of writing fractions. The hundred number square is a really good way of showing children the equivalence between fractions and decimals. Use flashcards. Make multiplication cards for each number set. Although this may seem tedious, the process of making the cards will actually help you to learn them. Once youâ€™ve made them, spend some time each day studying until you know them all. Focus on one number set at a time. When you go through the cards, put the ones you get wrong back into the pile so you see them multiple times. The last set of worksheets deals with commonly encountered fractions, including percentage values. These are values that students should be able to site reduce when they encounter them. Mastering these reducing problems will make many other fraction problems go much faster.	231	1,140	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.9375	3	CC-MAIN-2018-34	latest	en	0.937982
http://physicshelpline.net/entries/electrostatics/electrostatics-coulomb-s-law-4	1,568,918,967,000,000,000	text/html	crawl-data/CC-MAIN-2019-39/segments/1568514573570.6/warc/CC-MAIN-20190919183843-20190919205843-00289.warc.gz	142,336,076	7,260	UA-78309841-1 ## Solution Search • Abhigyan Martin Ninama September 8, 2017 • Rohit Dewal July 8, 2017 • Ajay kumar January 14, 2017 • September 27, 2016 ### Electrostatics: Coulomb's Law Q- Two equally charged point mass of mass m each are suspended from a point by massless threads of length L. In equilibrium position each thread makes and angle ɵ with vertical. (a) Find the ratio of magnitudes of horizontal electric force and vertical gravitational force on a the masses. (b) What is the ratio if the angle ɵ = 150 ? (C) If each mass is 0.15 gm and the length L =14 inch, what is the magnitude of excess charge on each mass? Solution:	184	648	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.859375	3	CC-MAIN-2019-39	longest	en	0.865684
http://www.pagalguy.com/@ojas	1,462,144,316,000,000,000	text/html	crawl-data/CC-MAIN-2016-18/segments/1461860117244.29/warc/CC-MAIN-20160428161517-00103-ip-10-239-7-51.ec2.internal.warc.gz	729,519,268	13,135	# @ojas ##### Ojas Doshi 2 karma Sana Sharma @sana9sharma2 red light flashes 5 times per minute and green 4 times per minute. if light starts flashing off at the same time, how many times do they flash together in a day? 1260 1350 1440 1530 Ojas Doshi @ojas since its "in a day" ....answer would have been 1440 had it been "everyday" Sana Sharma @sana9sharma2 red light flashes 5 times per minute and green 4 times per minute. if light starts flashing off at the same time, how many times do they flash together in a day? 1260 1350 1440 1530 Ojas Doshi @ojas ansher shud b 1441 Sana Sharma @sana9sharma2 ABC is an equilateral triangles. AGFC and BCDE are two squares drawn on side AC and BC respectively. H is any point where BF and AD intersects within the triangle. Q1 what is the measure of angle DHF? (options 105, 90, 75, 45) Q2 what is the ratio of area of triangle( ABH+DHF) to triangle( BHD+... Ojas Doshi @ojas 90 and 2:root(3) Priya Singh @priyasingh0113 How many three-digit numbers in base 10 are three-digit numbers in base 8 but not in base 7? a)147 b)169 c)153 d)177 Sana Sharma @sana9sharma2 what is the product of the roots of y= \|x\|^2 - 3\|x\|+2 ?? 4 -4 2 -2 Ojas Doshi @ojas 1,-1,2,-2 so 4 i think Sagar Gupta @sagarcat In a 5 by 12 rectangle, one of the diagonals is drawn and circles are inscribed in both right triangles thus formed. Find the distance between the centers of the two circles. Ojas Doshi @ojas root (65) Sagar Gupta @sagarcat A triangle with sides of 5, 12 and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles. Ojas Doshi @ojas root(16.25) ?? ###### CATOfficial Quant Thread for CAT 2014!! K @KPuy What is the remainder when 10^1283 is divided by 514? Ojas Doshi @ojas @ashishmakkar : calculations dekhlo ek baar ###### CATOfficial Quant Thread for CAT 2014!! K @KPuy What is the remainder when 10^1283 is divided by 514? Ojas Doshi @ojas Use CRT by splitting into 2 and 257. Rem with 2 is 0 and with 257 its 229. Thus 486 ###### CATOfficial Quant Thread for CAT 2014!! K @KPuy What is the remainder when 10^1283 is divided by 514?	688	2,153	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.65625	4	CC-MAIN-2016-18	latest	en	0.864216
https://es.mathworks.com/matlabcentral/profile/authors/3298841?detail=cody&page=2	1,674,901,449,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764499541.63/warc/CC-MAIN-20230128090359-20230128120359-00251.warc.gz	255,417,163	4,828	Resuelto Project Euler: Problem 2, Sum of even Fibonacci Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 te... casi 2 años hace Resuelto Project Euler: Problem 1, Multiples of 3 and 5 If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23... más de 2 años hace Resuelto Can we make a triangle? Given three positive number, check whether a triangle can be made with these sides length or not. remember that in a triangle su... más de 2 años hace Resuelto Find the sides of an isosceles triangle when given its area and height from its base to apex Find the sides of an isosceles triangle when given its area and the height from its base to apex. For example, with A=12 and ... más de 2 años hace Resuelto Height of a right-angled triangle Given numbers a, b and c, find the height of the right angled triangle with sides a and b and hypotenuse c, for the base c. If a... más de 2 años hace Resuelto Is the Point in a Triangle? Check whether a point or multiple points is/are in a triangle with three corners Points = [x, y]; Triangle = [x1, y1; x... más de 2 años hace Resuelto Calculate the centroid of a triangle Info: https://en.wikipedia.org/wiki/Centroid Example Input: x = [0 0 1]; % x-coordinate y = [0 1 0]; % y-coordinat... más de 2 años hace Resuelto Right Triangle Side Lengths (Inspired by Project Euler Problem 39) If _p_ is the perimeter of a right angle triangle with integral length sides, { _a_, _b_, _c_ }, there are exactly three solutio... más de 2 años hace Resuelto Find my daddy long leg (No 's') Given the ratio of the two legs (longer / shorter), and the hypotenuse length, find the value of the bigger leg. más de 2 años hace Resuelto Calculate the area of a triangle between three points Calculate the area of a triangle between three points: P1(X1,Y1) P2(X2,Y2) P3(X3,Y3) these three points are the vert... más de 2 años hace Resuelto Natural numbers in string form Create a cell array of strings containing the first n natural numbers. _Slightly_ harder than it seems like it should be. Ex... más de 2 años hace Resuelto Fix the last element of a cell array Note: this is lifted directly from <http://www.mathworks.com/matlabcentral/answers/82825-puzzler-for-a-monday Puzzler for a Mond... más de 2 años hace Resuelto String Array Basics, Part 1: Convert Cell Array to String Array; No Missing Values <http://www.mathworks.com/help/matlab/characters-and-strings.html String array> and cell array are two types of containers for s... más de 2 años hace Resuelto Cell Counting: How Many Draws? You are given a cell array containing information about a number of soccer games. Each cell contains one of the following: * ... más de 2 años hace Resuelto Remove element(s) from cell array You can easily remove an element (or a column in any dimension) from a normal matrix, but assigning that value (or range) empty.... más de 2 años hace Resuelto Convert a Cell Array into an Array Given a square cell array: x = {'01', '56'; '234', '789'}; return a single character array: y = '0123456789' más de 2 años hace Resuelto Split a string into chunks of specified length Given a string and a vector of integers, break the string into chunks whose lengths are given by the elements of the vector. Ex... más de 2 años hace Resuelto Convert a numerical matrix into a cell array of strings Given a numerical matrix, output a cell array of string. For example: if input = 1:3 output is {'1','2','3'} whic... más de 2 años hace Resuelto Create a cell array out of a struct Create a cell array out of a (single) struct with the fieldname in the first column and the value in the second column: in: ... más de 2 años hace Resuelto Cell joiner You are given a cell array of strings and a string delimiter. You need to produce one string which is composed of each string fr... más de 2 años hace Resuelto There are 10 types of people in the world Those who know binary, and those who don't. The number 2015 is a palindrome in binary (11111011111 to be exact) Given a year... más de 2 años hace Resuelto Converting binary to decimals Convert binary to decimals. Example: 010111 = 23. 110000 = 48. más de 2 años hace Resuelto Binary numbers Given a positive, scalar integer n, create a (2^n)-by-n double-precision matrix containing the binary numbers from 0 through 2^n... más de 2 años hace Resuelto Binary code (array) Write a function which calculates the binary code of a number 'n' and gives the result as an array(vector). Example: Inpu... más de 2 años hace Resuelto Relative ratio of "1" in binary number Input(n) is positive integer number Output(r) is (number of "1" in binary input) / (number of bits). Example: * n=0; r=... más de 2 años hace Resuelto Bit Reversal Given an unsigned integer _x_, convert it to binary with _n_ bits, reverse the order of the bits, and convert it back to an inte... más de 2 años hace Resuelto Given an unsigned integer x, find the largest y by rearranging the bits in x Given an unsigned integer x, find the largest y by rearranging the bits in x. Example: Input x = 10 Output y is 12 ... más de 2 años hace Resuelto Find out sum and carry of Binary adder Find out sum and carry of a binary adder if previous carry is given with two bits (x and y) for addition. Examples Previo... más de 2 años hace Resuelto Convert given decimal number to binary number. Convert given decimal number to binary number. Example x=10, then answer must be 1010. más de 2 años hace Resuelto Find the longest sequence of 1's in a binary sequence. Given a string such as s = '011110010000000100010111' find the length of the longest string of consecutive 1's. In this examp... más de 2 años hace	1,599	5,867	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.625	4	CC-MAIN-2023-06	latest	en	0.484307
https://www.clutchprep.com/physics/practice-problems/141282/an-object-of-mass-m-attached-to-a-spring-of-force-constant-k-oscillates-with-sim-1	1,637,970,151,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964358074.14/warc/CC-MAIN-20211126224056-20211127014056-00526.warc.gz	747,403,807	29,050	Energy in Simple Harmonic Motion Video Lessons Concept # Problem: An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E.What is the object's velocity when its potential energy is 2/3E?Answer in terms of m, k, A ###### FREE Expert Solution Maximum velocity: $\overline{){{\mathbf{v}}}_{\mathbf{m}\mathbf{a}\mathbf{x}}{\mathbf{=}}\sqrt{\frac{\mathbf{k}}{\mathbf{m}}}{\mathbf{A}}}$ Potential energy is given as, P.E = 2/3E But, E = P.E + K.E 88% (443 ratings) ###### Problem Details An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E. What is the object's velocity when its potential energy is 2/3E? Answer in terms of m, k, A	244	927	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.71875	4	CC-MAIN-2021-49	latest	en	0.788465
https://chemistry.stackexchange.com/questions/82670/how-to-determine-the-highest-freezing-point-by-vant-hoff-factor-when-concentrat/82671	1,611,378,379,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703533863.67/warc/CC-MAIN-20210123032629-20210123062629-00142.warc.gz	276,990,578	28,703	# How to determine the highest freezing point by van't hoff factor when concentration is same? [closed] Whch of the following will have the highest freezing point? a) 0.1M KCl b) 0.1M glucose c) 0.1M BaCl2 d) 0.1M AlCl3 Raoult's law originally was meant to describe the properties of non-electrolytes soltions. When it comes to the solutions of electrolytes, such as salts, you need to introduce the van't Hoff factor, sometimes called isotonic coefficient. To deal with it, you need to know the folmula: i=1+α(n-1), where i - van't Hoff factor α - dissociation degree n - number of ions formed from 1 ionic formula of electrolyte In the case of strong electrolytes, we just assume α=1, so i=n. Thus, you just need to calculate the number of ions which are formed from 1 ionic formula (for instance, it's 5 for Al2(SO4)3, as it goes into 5 ions - 2 aluminums and 3 sulfates), and multiply the cryoscopic constant and molality by it. So in your case the biggest number of ions comes from d) • What if concentration and van't hoff factor is same? Can i consider molar mass then? – Jui Sep 15 '17 at 8:36 • No, colligative properties are about the number of particles, not their nature by definition. Like, 0.01 M solutions of KCl and NaBr would have same freezing points, as both dissociate giving 2 ions (if we assume their dissociation is 100% complete) – MEL Science Sep 15 '17 at 8:38 • But, depression in freezing point is a colligative property, not freezing point. Am I right? – Jui Sep 15 '17 at 8:41 • You are right. If depressions for both compounds are the same, aqueous solutions of these compounds would have the same FP. – MEL Science Sep 15 '17 at 9:02 • Let's say the depression for both KCl and NaBr of some concentration is 0.5 degrees. Then the freezing point of both solutions is (FP of water)-(depression)=-0.5 degrees Celcius – MEL Science Sep 15 '17 at 9:03	532	1,881	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.15625	3	CC-MAIN-2021-04	latest	en	0.928828
https://unm.ge/tags/7c8b39-c%2B%2B-program-to-find-inverse-of-a-4x4-matrix	1,653,634,127,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652662636717.74/warc/CC-MAIN-20220527050925-20220527080925-00076.warc.gz	641,860,819	5,212	I'm implementing a 4x4 matrix class and all is going well until the inverse function turned up. We are working with a 4x4 matrix, so it has 4 rows and 4 columns. So I was wondering, how would you calculate the inverse of a 4x4 â¦ filter_none. Program : Finding Inverse of a 3 X 3 Matrix [crayon-5f8135ba158a8503949924/] Output : [crayon-5f8135ba158b5911112260/] Explanation : Suppose we have to find Inverse of â [crayon-5f8135ba158b8153207791/] Step 1 : Create One Matrix of Size 3 x 6 i.e Create 3 x 3 Matrix and Append 3 x 3 Unit Matrix Step 2 : Factor = â¦ Write a c program for scalar multiplication of matrix. Read more about C Programming Language . Just leaving some code here to invert either column or row major 4x4 matrices. 6. To find the Matrix Inverse, matrix should be a square matrix and Matrix Determinant is should not Equal to Zero. 7. C Program #include #include float [â¦] C Program to find the Inverse of the Matrix C code for 4x4 matrix inversion - 02/2013 . Upper triangular matrix in c 10. MultiMedia. play_arrow. In this tutorial we are going to implement this method using C programming language. C program to find inverse of a matrix 8. 4x4 Matrix Inverse Calculator . if A is a Square matrix and \|A\|!=0, then AAâ=I (I Means Identity Matrix). Below is the C++ program to find the inverse of a matrix using the Gauss-Jordan method: CPP. 4x4 Matrix Inverse calculator to find the inverse of a 4x4 matrix input values. Lower triangular matrix in c 9. Write a c program to find out transport of a matrix. Big list of c program â¦ Strassen's matrix multiplication program in c 11. The inverse matrix C/C++ software. Earlier in Matrix Inverse Using Gauss Jordan Method Algorithm and Matrix Inverse Using Gauss Jordan Method Pseudocode, we discussed about an algorithm and pseudocode for finding inverse of matrix using Gauss Jordan Method. C++ :: 4x4 Matrix Inverse Implementation Sep 19, 2013. ... C/C++ :: Program To Calculate Inverse Of Matrix ; Contribute to md-akhi/Inverse-matrix development by creating an account on GitHub. The matrix has four rows and columns. #include #include using namespace std; // Function to Print matrix. Hello, I am trying to calculate the inverse of a 4x4, I have been thinking about it endlessly yet I can't seem to be able to do it. General C++ Programming; Calculate the inverse of a 4x4 matrix - Calculate the inverse of a 4x4 matrix - Why is this code not working? The inverse of a square n x n matrix A, is another n x n matrix, denoted as A-1. edit close. It is a matrix when multiplied by the original matrix yields the identity matrix. 5. C Program to find the Inverse of a Matrix. I'm trying to implement the inverse function, but I can't seem to get my head around it. Contribute to md-akhi/Inverse-matrix.c-cpp development by creating an account on GitHub. The inverse matrix C/C++ software. link brightness_4 code // C++ program to find the inverse of Matrix. DEFINITION The matrix A is invertible if there exists a matrix â¦ ... Intel's optimized SSE matrix inverse routine described here. If you need to invert larger â¦ C program to find determinant of a matrix 12. Yields the Identity matrix ) matrix inverse Calculator denoted as A-1 // function to Print matrix ( I Means matrix. Matrix should be a square n x n matrix a, is another n n. Inverse, matrix should be a square matrix and matrix Determinant is should not Equal to Zero denoted! Matrix Determinant is should not Equal to Zero matrix 8 function, I! It is a matrix 8 is another n x n matrix, denoted as A-1 ca... < vector > using namespace std ; // function to Print matrix 4x4 matrix inverse, matrix should a! Implement the inverse of a square matrix and matrix Determinant is should not Equal to Zero yields the Identity )! Is going well until the inverse of a matrix square matrix and \|A\|! =0 then... Creating an account on GitHub major 4x4 matrices Sep 19, 2013 inverse a... ( I Means Identity matrix inverse, matrix should be a square matrix and!! Md-Akhi/Inverse-Matrix development by creating an account on GitHub â¦ 4x4 matrix class and all is going well until inverse. Column or row major 4x4 matrices this method using c programming language going well until the inverse of a.... Is a square n x n matrix, so it has 4 rows 4... Working with a 4x4 â¦ 4x4 matrix inverse, matrix should be square... Implementing a 4x4 matrix input values wondering, how would you calculate the function... 4X4 matrices matrix 8 denoted as A-1 is going well until the inverse function turned up an account GitHub... So I was wondering, how would you calculate the inverse function turned up denoted as A-1 the original yields... Â¦ c program â¦ C++:: 4x4 matrix inverse Implementation c++ program to find inverse of a 4x4 matrix 19, 2013 namespace std ; function. To find the matrix inverse Calculator of matrix has 4 rows and 4 columns md-akhi/Inverse-matrix.c-cpp development by creating account. Of c program to find the inverse function turned up, 2013 larger â¦ program! Iostream > # include < vector > using namespace std ; // function to Print matrix to. I ca n't seem to get my head around it should not Equal to Zero namespace ;... With a 4x4 matrix, so it has 4 rows and 4.! A 4x4 â¦ 4x4 matrix inverse Calculator to find the matrix inverse Calculator to the... With a 4x4 matrix class and all is going well until the inverse of a matrix when multiplied by original... Account on GitHub ; // function to Print matrix well until the inverse function but! When multiplied by the original matrix yields the Identity matrix n x n matrix,... I Means Identity matrix ) this tutorial we are working with a 4x4 matrix inverse Calculator to this. As A-1 it has 4 rows and 4 columns find the matrix inverse Calculator we are going to the! Implementation Sep 19, 2013 big list of c program to find out transport of a when... A is a matrix 12 // function to Print matrix this method using c language... Matrix yields the Identity matrix ) to md-akhi/Inverse-matrix development by creating an account on GitHub a matrix.. The original matrix yields the Identity matrix ) Intel 's optimized SSE matrix Implementation! To Zero a 4x4 matrix inverse Calculator to find the inverse function, I! You need to invert larger â¦ c program â¦ C++:: 4x4 matrix inverse.. Until the inverse function turned up function to Print matrix rows and 4 columns it has rows... Print matrix and all is going well until the inverse of a matrix 12 Sep 19, 2013 's! Matrix yields the Identity matrix ) n't seem to get my head around it AAâ=I ( Means. It has 4 rows and 4 columns by the original matrix yields the matrix! \|A\|! =0, then AAâ=I ( I Means Identity matrix ) â¦ matrix! Either column or row major 4x4 matrices programming language scalar multiplication of matrix a, is another n n... Md-Akhi/Inverse-Matrix development by creating an account on GitHub is a matrix 8 AAâ=I ( I Identity. Development by creating an account on GitHub big list of c program for scalar multiplication of matrix code... Print matrix a, is another n x n matrix, denoted as A-1 # include < >! Development by creating an account on GitHub c programming language matrix should be a square matrix \|A\|... Of c++ program to find inverse of a 4x4 matrix program to find Determinant of a matrix described here working with 4x4!	1,821	7,325	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.203125	3	CC-MAIN-2022-21	latest	en	0.787252
https://www.physicsforums.com/threads/double-check-the-derivation-integral-representation-of-bessel-function.700648/	1,508,745,962,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187825812.89/warc/CC-MAIN-20171023073607-20171023093607-00863.warc.gz	958,154,342	17,760	# Double check the derivation integral representation of Bessel Function 1. Jul 9, 2013 ### yungman I am reading the article Mirela Vinerean: http://www.math.kau.se/mirevine/mf2bess.pdf On page 6, I have a question about $$e^{\frac{x}{2}t} e^{-\frac{x}{2}\frac{1}{t}}=\sum^{\infty}_{n=-\infty}J_n(x)e^{jn\theta}=\sum_{n=0}^{\infty}J_n(x)[e^{jn\theta}+(-1)^ne^{-jn\theta}]$$ I think there is a mistake at the last term. If you look at n=0, the equation will be: $$J_0(x)[e^0+(-1)^0e^{-0}]\;=\;j_0(x)[1+1]\;=\;2J_0(x)$$ Which is not correct. The problem is n=0 is being repeated in both the n=+ve and n=-ve. The equation should be: $$e^{\frac{x}{2}t} e^{-\frac{x}{2}\frac{1}{t}}=\sum_{n=0}^{\infty}[J_n(x)e^{jn\theta}]\;+\;\sum_{n=1}^{\infty}[(-1)^ne^{-jn\theta}]$$ With this, the first term covers the original n=+ve from 0 to ∞. The second term covers the original from -∞ to -1. Now you only have one term contain n=0. Am I correct? Thanks 2. Jul 9, 2013 ### yungman Also in the same page right below the equations of the first post: $$\int_0^{\pi} \sin n\theta \sin m\theta d\theta\;=\;\frac{\pi}{2}\delta_{mn}$$ 1) What is $\delta_{mn}$? 2) If m≠n, the result should be zero. But $\int_0^{\pi} \sin n\theta \sin m\theta d\theta\;≠0$ because the integration is from 0 to $\pi$, not from -$\pi$ to +$\pi$. Thanks 3. Jul 9, 2013 ### Infrared I can't help you with your first post. $\delta_{mn}$ is the Kronecker delta. It is 1 is m=n and 0 if m≠n. https://en.wikipedia.org/wiki/Kronecker_delta That integral is zero if m≠n. One to show that is using complex exponentials $$\int_0^\pi sin(n\theta)sin(m\theta)d\theta = \int_0^\pi \frac{e^{in\theta}-e^{-in\theta}}{2i}\frac{e^{im\theta}-e^{-im\theta}}{2i} d\theta= -\frac{1}{4} \int_0^\pi e^{i(m+n)\theta}-e^{i(m-n)\theta}-e^{i(n-m)\theta}+e^{-i(m+n)\theta}d\theta$$. Since we know that this integral will be real, we can just consider the real parts of each of the expressions. $\mathrm{Re}(e^{ix})=cos(x)$ so we can write the integral as $$-\frac{1}{4} \int_0^\pi cos((m+n)\theta)-cos((n-m)\theta)-cos((m-n)\theta)+cos(-(m+n)\theta) d\theta$$. This is zero since cosine is odd about $\pi/2$ in sense that $cos(\pi/2+x)=-cos(\pi/2-x)$. Note that you need m≠n. Else, the $cos((n-m)\theta)$ is constantly one and does not integrate to zero. Last edited: Jul 9, 2013 4. Jul 9, 2013 ### yungman I know the cosine part is fine. But if you look at page 6 of the provided link in the first post, the author claimed orthogonality properties with the sine function. That's the part I am challenging. Thanks 5. Jul 9, 2013 ### Infrared I am confused. Didn't I just show that $\int_0^\pi sin(mx)sin(nx) dx =0$ where $m \neq n$, which is the orthogonality condition? Last edited: Jul 9, 2013 6. Jul 9, 2013 ### yungman An integral does not have to be real, I don't think you can assume it's real and get rid of the imaginary part. A very simple test is not use exponential and just go with integration with n=0 and m=1 $$\int_0^{\pi} \sin \theta d\theta =-\cos\theta\|_0^{\pi}=-[-1-1]=2$$ Last edited: Jul 9, 2013 7. Jul 9, 2013 ### Infrared It is real. If you wanted to verify this, you could expand all of the $e^{ik\theta}$ (where k is one of the linear combinations of m and n) terms as $cos(k\theta)+isin(k\theta)$ and find that all of the imaginary terms drop out. I skipped this step because the original integral with just sines must be real, and the integral with complex exponentials is only different by a factor of -1/4, so it too must be real. Edit: I just saw your edit. If n=0, then the integrand is 0 and the integral is trivially zero. Last edited: Jul 9, 2013 8. Jul 9, 2013 ### yungman The forum is so slow that we are cross posting. I just edited the last post, I don't even use exponential, just use n=0 and m=1. that will be just integrate a sine function and show it's not zero. 9. Jul 9, 2013 ### Infrared Yes, it is a bit annoying. See my last edit. Look more carefully at what happens to the integrand if one of m or n is zero. 10. Jul 9, 2013 ### micromass Staff Emeritus You're taking the integral of a real function. That will always be real. 11. Jul 9, 2013 ### yungman Yes, I was wrong on the integration. I kept thinking if n=0, then it's an integration of a sine function and it's not zero. But if n=0, the whole thing is zero to start with!!! Thanks Can you help with my first post? 12. Jul 9, 2013 ### Infrared Sorry, I can't as I don't know anything about Bessel functions. I hope that you get the answer you need though. 13. Jul 9, 2013 ### yungman Actually the question has not much to do with Bessel Function. All you need to know is if n is an integer, $J_{-n}(x)=(-1)^n J_n(x)$, the rest is a series problem. It should be $$\sum^{\infty}_{n=-\infty}J_n(x)e^{jn\theta}\;=\;\sum_{n=0}^{\infty}[J_n(x)e^{jn\theta}]\;+\;\sum_{n=1}^{\infty}[(-1)^nJ_n(x)e^{-jn\theta}]$$ Not as the article that: $$\sum^{\infty}_{n=-\infty}J_n(x)e^{jn\theta}=\sum_{n=0}^{\infty}J_n(x)[e^{jn\theta}+(-1)^ne^{-jn\theta}]\;=\;\sum_{n=0}^{\infty}[J_n(x)e^{jn\theta}]\;+\;\sum_{n=0}^{\infty}[(-1)^nJ_n(x)e^{-jn\theta}]$$ According to the article, n=0 are being repeated in both and result in twice the value for n=0. 14. Jul 12, 2013	1,824	5,240	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2017-43	longest	en	0.657701
https://www.r-bloggers.com/2013/06/quartiles-deciles-and-percentiles/	1,721,371,473,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00313.warc.gz	809,220,104	21,789	Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. The measures of position such as quartiles, deciles, and percentiles are available in quantile function. This function has a usage, where: • x – the data points • prob – the location to measure • na.rm – if FALSE, NA (Not Available) data points are not ignored • names – for attributes, FALSE means no attributes, hence speeds-up the computation • type – type of the quantile algorithms • – further arguments Example 1. The junior BS Stat students of MSU-IIT have the following SASE scores: 88, 84, 83, 80, 94, 90, 81, 79, 79, 81, 85, 87, 86, 89, and 92. Determine and interpret the quartiles of these scores. Interpretation: Therefore, \$Q_1\$=25% implies that, 25% of the SASE scores fall below or equal to 81.0, while the other 75% of it is above 81.0. \$Q_2\$=50% is the median, and thus half of the scores are below or equal to 85.0, while the other half, scores more than 85.0. \$Q_3\$=75%, implies that three-fourth of the data are below or equal to 88.5, while the remaining one-fourth is above 88.5. And the minimum and maximum values are 79.0 and 94.0, respectively. Example 2. The surveyed weights (in kilograms) of the students in Stat 131 were the following: 69, 70, 75, 66, 83, 88, 66, 63, 61, 68, 73, 57, 52, 58, and 77. Compute and interpret the deciles of these weights. Notice the difference between the codes of quartiles and deciles computations. This time the function quantile has an argument type which is set to 5. With this, the quantile algorithm between the quartiles and deciles differ. Hence, the appropriate algorithm for decile is type 5, while the quartile is type 7, which is the default one. For further reading about the quantile algorithm run ?quantile. In addition, the prob argument above is the position to be measured, and since deciles divide the data points into ten parts, then a sequence function, seq, is used for prob‘s value that is from 0 to 1 of length 11 (length = 11, 11 because zero is included, which is the minimum of the data points). Interpretation: The first decile is \$D_1\$=10%, implies that one-tenth of the weights fall below or equal to 57.0, and the remaining nine-tenth fall above 57.0. The \$D_5\$=50% is the median, thus half of the students’ weights weigh below or equal to 68.0, while the other half fall above this. And so on. Example 3. Compute the \$15^{th}\$, \$25^{th}\$, and \$35^{th}\$ percentiles of weights in Example 2. Interpretation: The fifteenth percentile \$P_{15}\$=15% is interpreted as 15% of the samples fall below or equal to 58.3 while 85% fall above 58.3. The thirty-fifth percentile \$P_{35}\$=35%, implies that 35% of the weights fall below or equal to 65.7, and that 65% of it fall above 65.7. Reference: Yau, Chi. R Tutorial: Percentile.	815	2,837	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2024-30	latest	en	0.888018
https://goprep.co/a-tortoise-moves-a-distance-of-1000-metres-in-15-mintues-i-1njhph	1,624,578,970,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623488560777.97/warc/CC-MAIN-20210624233218-20210625023218-00082.warc.gz	259,089,953	37,116	A tortoise moves Given: Total distance=1000 m Total time taken=15 minutes= 15/60=0.25 hour formula used: We know 1 km=1000 m Total distance traveled=1 km 1 hour=60 minutes 15 minutes=0.25 hours Total time taken=0.25 hours putting the values of distance and time in the equation: hence, the speed of tortoise is 4 km/h Rate this question : How useful is this solution? We strive to provide quality solutions. Please rate us to serve you better. Related Videos Mastering the Numericals43 mins Acceleration44 mins Equations of Motion41 mins Speed and Velocity42 mins Satellite Motion45 mins Circular Motion37 mins Newton's Second Law46 mins Significance of Newton's Laws in daily life42 mins Newton's First Law45 mins Miscellaneous questions42 mins Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts Dedicated counsellor for each student 24X7 Doubt Resolution Daily Report Card Detailed Performance Evaluation view all courses RELATED QUESTIONS : Which of the follLakhmir Singh & Manjit Kaur - Physics	271	1,069	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.578125	3	CC-MAIN-2021-25	latest	en	0.772849
https://mano.xyz/406/	1,726,586,842,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00544.warc.gz	352,494,873	5,375	233 Lattice points on a circle ポイントは (N-2x)^2 + （N-2y)^2 = 2N^2 という式と， C++で実装． ```#include <iostream> #include <vector> #include <cmath> #define FOR(x, xs) for (__typeof( (xs).begin() ) x = (xs).begin(); x != (xs).end(); x++) #define WHILE(x, xs, p) for (__typeof( (xs).begin() ) x = (xs).begin(); (p) && x != (xs).end(); x++) using namespace std; static const long long u = pow(10, 11); static const int v = u / ( pow(5, 3) * pow(13, 2) ); static const int w = u / ( pow(5, 3) * pow(13, 2) * 17); int main() { vector<bool> isPrime( v + 1); //primes up to v FOR (q, isPrime) q = true; isPrime[] = false; isPrime[1] = false; for (int q = 2; q <= sqrt(v) ; q++) if ( isPrime[q] ) for (int i = q q; i <= v; i += q) isPrime[i] = false; vector<int> primes1; // primes form of 4k+1 vector<long long> multi( w + 1 ); for (int c = ; c <= w; c++) multi = c; for (int i = 1; i <= v ; i++) if ( isPrime[i] && i % 4 == 1 ) { primes1.push_back(i); for (int j = i; j <= w; j += i) multi[j] = ; } for (int c = 1; c <= w; c++) multi += multi[c-1]; long long sum = ; // n = p * q^2 * r^3 WHILE (r, primes1, r <= pow( u / (13 5 * 5), 1.0 / 3 ) ) WHILE (q, primes1, q <= sqrt ( u / ( 5 pow(r, 3) ) ) ) { if ( r == q ) continue; WHILE (p, primes1, p <= u / ( pow(q, 2) pow(r, 3) ) ) { if ( r == p \|\| q == p ) continue; long long n = p * pow(q, 2) pow(r, 3); sum += n multi[ u / n ]; } } // n = q^3 * r^7 WHILE (r, primes1, r <= pow ( u / pow(5, 3), 1.0 / 7 ) ) WHILE (q, primes1, q <= pow( u / pow(r, 7), 1.0 / 3) ) { if ( r == q ) continue; long long n = pow(r, 7) * pow(q, 3); sum += n multi[ u / n ]; } // n = q^2 * r^10 WHILE (r, primes1, r <= pow ( u / pow(5, 2), 0.1 ) ) WHILE (q, primes1, q <= sqrt( u / pow(r, 10) ) ){ if ( r == q ) continue; long long n = pow(r, 10) * pow(q, 2); sum += n multi[ u / n ]; } cout << sum << endl; } ``` はじめは，n = pq^2r^3 だけしか，考えておらず， n ＝ q^3r^7, q^2r10 などのことを完全に忘れていた．どうも，答えがあわない．で，いろいろと悩んだあげくに辿り着いた．まぁ，詰めが甘いですな．	905	1,996	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.328125	3	CC-MAIN-2024-38	latest	en	0.195914
https://www.riddlesandanswers.com/v/236035/robert-and-cindy-were-stranded-on-a-desert-island-he-has-two-blue-pills-and-two-red-pills-that-are/	1,638,275,535,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964358973.70/warc/CC-MAIN-20211130110936-20211130140936-00565.warc.gz	1,078,779,657	22,393	ROBERT AND CINDY WERE STRANDED ON AN ISLAND RIDDLE Trending Tags Terms · Privacy · Contact Robert And Cindy Were Stranded On An Island Riddle Robert and Cindy were stranded on A desert island. He has two blue pills and two red pills that are identical in shape. He must take exactly one red pill and one blue pill or he will die. How does he do it? Hint: Dissolve all 4 pills in water and drink the half water. The Robert And Cindy will dissolve all the pills in water and will drink the half water so that he each person can get 1 red and 1 blue pills and they both get save. Did you answer this riddle correctly? YES NO Solved: 61%	162	640	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.609375	3	CC-MAIN-2021-49	latest	en	0.932044
https://stacks.math.columbia.edu/tag/0G6M	1,721,484,317,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763515164.46/warc/CC-MAIN-20240720113754-20240720143754-00705.warc.gz	459,504,824	6,596	Lemma 15.59.16. Let $R$ be a ring. Let $a : K^\bullet \to L^\bullet$ be a map of complexes of $R$-modules. If $K^\bullet$ is K-flat, then there exist a complex $N^\bullet$ and maps of complexes $b : K^\bullet \to N^\bullet$ and $c : N^\bullet \to L^\bullet$ such that 1. $N^\bullet$ is K-flat, 2. $c$ is a quasi-isomorphism, 3. $a$ is homotopic to $c \circ b$. If the terms of $K^\bullet$ are flat, then we may choose $N^\bullet$, $b$, and $c$ such that the same is true for $N^\bullet$. Proof. We will use that the homotopy category $K(R)$ is a triangulated category, see Derived Categories, Proposition 13.10.3. Choose a distinguished triangle $K^\bullet \to L^\bullet \to C^\bullet \to K^\bullet [1]$. Choose a quasi-isomorphism $M^\bullet \to C^\bullet$ with $M^\bullet$ K-flat with flat terms, see Lemma 15.59.10. By the axioms of triangulated categories, we may fit the composition $M^\bullet \to C^\bullet \to K^\bullet [1]$ into a distinguished triangle $K^\bullet \to N^\bullet \to M^\bullet \to K^\bullet [1]$. By Lemma 15.59.5 we see that $N^\bullet$ is K-flat. Again using the axioms of triangulated categories, we can choose a map $N^\bullet \to L^\bullet$ fitting into the following morphism of distinghuised triangles $\xymatrix{ K^\bullet \ar[r] \ar[d] & N^\bullet \ar[r] \ar[d] & M^\bullet \ar[r] \ar[d] & K^\bullet [1] \ar[d] \\ K^\bullet \ar[r] & L^\bullet \ar[r] & C^\bullet \ar[r] & K^\bullet [1] }$ Since two out of three of the arrows are quasi-isomorphisms, so is the third arrow $N^\bullet \to L^\bullet$ by the long exact sequences of cohomology associated to these distinguished triangles (or you can look at the image of this diagram in $D(R)$ and use Derived Categories, Lemma 13.4.3 if you like). This finishes the proof of (1), (2), and (3). To prove the final assertion, we may choose $N^\bullet$ such that $N^ n \cong M^ n \oplus K^ n$, see Derived Categories, Lemma 13.10.7. Hence we get the desired flatness if the terms of $K^\bullet$ are flat. $\square$ There are also: • 4 comment(s) on Section 15.59: Derived tensor product In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).	695	2,268	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0}	2.96875	3	CC-MAIN-2024-30	latest	en	0.739319
http://physics.stackexchange.com/questions/87539/some-hints-for-special-case-of-metric-tensor-in-gr	1,469,730,359,000,000,000	text/html	crawl-data/CC-MAIN-2016-30/segments/1469257828313.74/warc/CC-MAIN-20160723071028-00308-ip-10-185-27-174.ec2.internal.warc.gz	197,649,302	18,109	Some hints for special case of metric tensor in GR Let's have metric $$ds^2 = dt^2 - dx^2 - dy^2 - dz^2 - 2f(t - z, x, y)(dt - dz)^2.$$ I need to prove that it is an exact solution for Einstein equations in vacuum for $\partial_{x}^{2}f + \partial_{y}^{2}f = 0$. The straightforward method is obvious, but does some method especially for this metric exist? Edit. As I see, first is replacing the variables as $$u = (t - z), \quad v = (t + z).$$ - Hi Andrew McAdams. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. – Qmechanic Nov 27 '13 at 12:37 The metric is well-known, and can be shown to describe a gravitational wave. I will help prove the claim using a slightly different metric which is more convenient for the tetrad formalism. The metric: $$ds^2=dt^2-dr^2+H(t-r,x^1,x^2)(dt-dr) - d(x^i)^2$$ where $i=1,2$ and $x^1,x^2$ are generic coordinates. We define an orthonormal basis, $$\omega^t=dt+\frac{1}{2}H(...)(dt-dr) \quad \omega^r = dr +\frac{1}{2}H(...)(dt-dr)$$ and $\omega^i=dx^i$ such that $\underline{g}=\eta^{ab}\omega_a \omega_b$. Taking exterior derivatives of the basis yields, $$d\omega^t=\frac{1}{2}H_{,i} dx^i \wedge (dt-dr)=-\frac{1}{2}H_{,i}(\omega^t -\omega^r)\wedge \omega^i$$ and similarly for the other basis; of course $d\omega^i = 0$. By Cartan's first equation, $$d\omega^a=-\theta^a_b \wedge \omega^b$$ for the spin connection $\theta^a_b$. Applying the equation, we may deduce the non-zero components, $$\theta^t_i=\frac{1}{2}H_{,i}(\omega^t-\omega^r)$$ and identically for the $\omega^r$ case; the rest vanish. Cartan's second equation dictates, $$R^a_b = d\theta^a_b +\omega^a_c \wedge \omega^c_b$$ You should be able to take over the computation from here. As we are only interested in the intrinsically flat case, the Ricci scalar should vanish in any basis, hence there is no need to convert back to the coordinate basis. Eventually, you will find the condition, $$\Delta H = 0$$ where $\Delta$ is the Laplace operator, the desired result. To ensure it is a gravitational wave, we demand $T_{\mu\nu}=0$, otherwise we can't, at least by inspection, immediately conclude the wave is gravitational, rather than for example electromagnetic. By the Einstein field equations, that implies $R=0$. -	729	2,359	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.484375	3	CC-MAIN-2016-30	latest	en	0.804132
https://takeawildguess.net/blog/fcnn/fcnn02/	1,603,252,795,000,000,000	text/html	crawl-data/CC-MAIN-2020-45/segments/1603107875980.5/warc/CC-MAIN-20201021035155-20201021065155-00478.warc.gz	565,541,202	8,727	## 1. Introduction This post belongs to a new series of posts related to a huge and popular topic in machine learning: fully connected neural networks. The series scope is three-fold: 1. visualize the model features and characteristics with schematic pictures and charts 2. learn to implement the model with different levels of abstraction, given by the framework used 3. have some fun with one of the hottest topics right now! In this post, we give some geometric insight into what occurs in a single neuron. Note that, if the activation is a sigmoid function, it performs a logistic regression to the inputs. In the following post, we extend this geometric intuition to a neuron network. ## 2. What is a neural network We start analyzing a toy problem and understanding why logistic regression (LR) is no suitable and we need more powerful and advanced functions to try to solve the problem. Below there is a chart showing nine points. Different colours mean different classes, so is a binary classification because we have only two colours, blue and red. It’s easy to see it would be impossible to separate the two regions, blue and red, with only one line and this is why we need a more complex and advanced function. We, in fact, need to find a narrow area between the two blue regions where we are going to specify the red class. For that, we need to combine multiple logistic regressions. Figure 1 - NN neuron intuition The main idea is to try to overcome the limit and see how far we can go by just combining more of such functions. If you want to understand more about logistic regression, please see my previous series. Now we are going to focus on how we can interpret LR geometrically and what we need to understand if you want to combine multiple regressions altogether. ## 3. Decision boundary Geometrically we can interpret this function as a simple classifier that is going to draw a line, which is called decision boundary. That line is going to split the entire domain into two sub-areas, one for each class. This decision line is perpendicular to the direction given by the weights if you interpret the weights as a vector. We can have a look at the below chart and see that we have two weights, one for each input, $x_1$ and $x_2$, in the 2D domain. That weight vector is giving the direction (line r in the chart). Figure 2 - Decision boundary How a combination of weights and biases is going to define the decision line? Mathematically we know that the line defining the decision boundary is: $$z = w\cdot X + b =$$ $$= w_1\cdot x_1 + w_2\cdot x_2 + b = 0$$ This set of points represents where we don’t know exactly how we can classify the input. But if $z$ becomes greater than 0, we get more confident that it belongs to the first class and if $z$ becomes less than 0, to the other class. Now we draw a line $s$, which is perpendicular to the previous line and is crossing the origin $(0, 0)$, whose equation is: $$s: x_2 = \frac{w_2}{w_1}\cdot x_1$$ You can prove it that the vector along $s$ is $\big(1, \frac{w_2}{w_1}\big)$ and that it is perpendicular to $r$ since the perpendicular vector of $r$, $(w_1, w_2)$ is proportional to $\big(1, \frac{w_2}{w_1}\big)$. We got a system of two equations in $x_1$ and $x_2$. We can solve it using very simple mathematical steps and find out that the intersection of these two lines is giving us the point B which is exactly at the point on our decision line and whose distance from the origin is the minimum possible. We end up with: $$r\cap s: x_1 = -\frac{w_1\cdot b}{w^2}$$ $$x_2 = -\frac{w_2\cdot b}{w^2}$$ We combine these two coordinates into the point P and realize that direction is given by the weight unit vector $\vec{u_w} = \vec{w}/w$ and distance from the origin is the ratio of bias and weight module $b/w$: $$P = (x_1, x_2) = -\frac{b}{w^2}\cdot \vec{w} =$$ $$= -\frac{b}{w}\cdot \vec{u_w}$$ We can also determine the distance of P from origin by using the two coordinates of P and the Euclidean formulation: $$d = \sqrt{\frac{(w_1\cdot b)^2+(w_2\cdot b)^2}{w^4}}$$ $$= \sqrt{\frac{b^2\cdot w^2}{w^4}} = \sqrt{\frac{b^2}{w^2}}$$ $$d = \frac{\|b\|}{w}$$ It means that, given a certain level of bias, if we increase the weight magnitude $w$ we are going to shift your decision line toward the origin; if we instead fix the weights’ magnitude and increase bias, the decision line moves away from the origin The direction will always be defined by the weight vector. ## 4. Affine transformation Here we see how to draw the decision line. The weights’ vector direction is perpendicular to the decision line. There are two points on such a line whose distance is $\frac{\|b\|}{w}$, one is $P(b<0)$ on the right-hand side when bias is negative, the other is $P(b>0)$ on the left-hand side when bias is positive. Figure 3 - Affine transformation Another property that we can see in the next figure is that if you collect as many points as you want on the same line $s_1$ perpendicular to $r$, say $x^{(1)}$, $x^{(2)}$, $x^{(3)}$, and feed those points to the affine transformation (geometric projection), we end up with the same output. If we instead take the yellow points $x^{(4)}$, $x^{(5)}$, $x^{(6)}$ that lie on $s_2$ and feed those points to the same transformation, we got a common projected point again. It means that basically what our logistic regression is doing is projecting every point on a given line perpendicular to $r$ to the same point. It just squishes the 2-dimensional space into 1-dimensional space. That’s the idea to get the decision line. We know we still need to do one step. If we apply the activation function, say a sigmoid, to the output of the affine transformation, we really got the complete process output of a network layer. ## 5. Uncertainty area By analyzing the sigmoid function with respect to the received input, it squishes to $(0, 1)$. The output is either zero or one for most of the input domain and is continuously moving from 0 to 1 only in a really narrow area. Figure 4 - 2D visualization of the sigmoid function In practice, we can just assume the function returns 0 for input less than -5, continuously increases from 0 to 1 in the area ranging from -5 to 5 and returns 1 for input greater than 5. In the above chart, there are three lines: $$r: w\cdot \vec{x} + b = -5 \quad \text{red}$$ $$y: w\cdot \vec{x} + b = 0 \quad \text{yellow}$$ $$g: w\cdot \vec{x} + b = +5 \quad \text{green}$$ Line $r$ defines the area where the model is very confident to assign the 0-class. The decision boundary, $y$, is the yellow line where we are completely uncertain about the two classes (50-50% chances). The third line, $g$, delimits the area where we are quite confident the points there belong to the other class. The point of this chart is to show this yellow stripe, which could be narrower or wider according to the model parameters, where we are less sure about anything. we can still say that, if a point is on the left-hand side of the yellow line, it belongs to class 0 and to class 1 if it is on the other side, but the probability of this statement is lower. The yellow area is confined between the two lines r and g, whose distance is double the distance from the central line and one of the two lines, $2\cdot \frac{5}{w}$. This means that, if the weight magnitude increases, the area in which we are somehow uncertainty is going to decrease to become really narrow. If weights go down closer to 0, the yellow area is going to be wider, i.e., the model becomes more confident. The tradeoff can be determined with a proper regularization technique, which usually tends to keep weights as close as possible to 0 to prevent weights to become extremely large and, therefore, unreliable to generalize to unseen data. From this geometrical interpretation we derive that, if we want to keep weights small, we need to prevent the neural networks to build some super narrow and confident decision boundaries, which then translates into overfitting over new data points. As soon as there is a new point from one class which is slightly shifting to the other class decision area, the model immediately switches to the other class, since there is no uncertain area in between (yellow). This behaviour reflects the inability of the model to generalize to new scenarios.	2,061	8,333	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.328125	3	CC-MAIN-2020-45	latest	en	0.935306
https://discuss.codecademy.com/t/exponentiation-11-explanation/18995	1,534,484,392,000,000,000	text/html	crawl-data/CC-MAIN-2018-34/segments/1534221211719.12/warc/CC-MAIN-20180817045508-20180817065508-00396.warc.gz	670,690,141	4,194	# Exponentiation - 11 explanation #1 Hi guys new here hope you are well. I have finished this but need some clarification. eight = 2 3 In the above example, we create a new variable called eight and set it to 8, or the result of 2 to the power to 3 (2^3). Am I right in thinking that to get to 8 you would follow this process, 2 3 = 2x2= 4 x2 = 8 ?. Create a new variable called eggs and use exponents to set eggs equal 100. Try raising 10 to the power of 2. 10 2 =10x10 = 100 ? #2 Yes @er0rrb1ankpag3! Your terminology is defintely correct! #3 this is how mine looked and it said i got it correct( Minus the " ) "#Set eggs equal to 100 using exponentiation on line 3! eggs = 10 2 print eggs #4 Oh my word thank you so much.	233	753	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.796875	4	CC-MAIN-2018-34	latest	en	0.922794
https://www.unitconverters.net/length/cubit-uk-to-rope.htm	1,718,839,945,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00077.warc.gz	926,020,997	4,067	Home / Length Conversion / Convert Cubit (UK) to Rope # Convert Cubit (UK) to Rope Please provide values below to convert cubit (UK) to rope, or vice versa. From: cubit (UK) To: rope ### Cubit (UK) to Rope Conversion Table Cubit (UK)Rope 0.01 cubit (UK)0.00075 rope 0.1 cubit (UK)0.0075 rope 1 cubit (UK)0.075 rope 2 cubit (UK)0.15 rope 3 cubit (UK)0.225 rope 5 cubit (UK)0.375 rope 10 cubit (UK)0.75 rope 20 cubit (UK)1.5 rope 50 cubit (UK)3.75 rope 100 cubit (UK)7.5 rope 1000 cubit (UK)75 rope ### How to Convert Cubit (UK) to Rope 1 cubit (UK) = 0.075 rope 1 rope = 13.3333333333 cubit (UK) Example: convert 15 cubit (UK) to rope: 15 cubit (UK) = 15 × 0.075 rope = 1.125 rope	273	688	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.78125	3	CC-MAIN-2024-26	latest	en	0.489844
https://sudocue.net/windoku.php?id=qql6_$tbEk_whx4ZIsLOKg	1,695,512,444,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233506539.13/warc/CC-MAIN-20230923231031-20230924021031-00538.warc.gz	614,744,712	3,694	# Daily Windoku Here is your Daily Windoku puzzle. This the entry for Sunday, September 17, 2023. Windoku, also known as the NRC Sudoku variant, Hyper(su)Doku or Sudoku Hyper, has all the rules of a “vanilla” Sudoku, with one addition: The four window panes with the blue background color must also contain digits 1 through 9. Many solving techniques for normal Sudoku can be used, but you will also need to use the 4 windows panes to check for singles, pairs, triples, and their interaction with rows, columns and boxes. The way the 4 window panes are positioned allow you to find 5 additional groups of cells where only digits 1 through 9 can go. These implied rules can prove very useful when you try to solve this puzzle. To facilitate you using a Sudoku helper program, you can copy the number string below the puzzle and paste it into your program. Before you paste it, switch the program to the Windoku variant. If you paste it as a regular Sudoku, you will receive a warning that it has multiple solutions. 500071000301500200060000000004000000090000048000000601000000000030800000000000070 level: 22, requires: N2,H2	272	1,128	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2023-40	latest	en	0.882749
http://esvc006636.swp0002ssl.server-secure.com/tbb/solobid/gametry/gtr_facts.htm	1,718,899,325,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00266.warc.gz	9,948,724	1,554	Game-try DBL : Facts sheet Law of Total Tricks When playing in a trump contract, the total number of tricks that can be made in the same Deal by the two sides (each in its best suit) is equal to the total number of trumps held in these two suits. Jean-Rene Vernes 1. The Law of Total tricks infers that when the HCP are not unequally divided between the two sides : the tricks a side will make is equal to their total number of trumps. 2. Variations of the above guide may occur depending on the relative position of Kings with respect to their Aces in the four hands. 3. After either side has bid and raised a suit : 1. another raise of that suit is competitive only, and not invitational For example : 1♥ - (1♠) - 2♥ - (2♠) - 3♥ 2. a double is a Game-try DBL and invites Partner to Game For example : 1♥ - (1♠) - 2♥ - (2♠) - DBL (The Game-try Double replaces the normal Invitation bid of 3♥ or 3♠.) 3. Partner responds to a Game-try DBL 1. with 3♥ or 3♠ : when minimum 2. with 4♥ or 4♠ : when maximum	289	1,013	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.203125	3	CC-MAIN-2024-26	latest	en	0.90451
https://orinanobworld.blogspot.fr/2014_02_01_archive.html	1,505,847,758,000,000,000	text/html	crawl-data/CC-MAIN-2017-39/segments/1505818685993.12/warc/CC-MAIN-20170919183419-20170919203419-00489.warc.gz	709,264,058	24,423	## Friday, February 28, 2014 ### Ridge Regression Revisited I've been neglecting the blog a bit lately, partly because I haven't had much to say and partly because I've been a bit busy with other things. One of the things keeping me occupied is the excellent Statistical Learning course offered by Stanford. (In one of those coming-full-cycle things, now that I'm retired from teaching, I'm going back to my student roots and taking MOOCs in subjects of interest.) During one of the lectures I learned something new that was both interesting and had the side effect of making me feel old (a job my knees are entirely up to on their own, thank you very much). Ridge regression has apparently been picked up from the scrap heap of statistical history and repurposed. #### Stiffness: it's not just for us old people To explain ridge regression as I originally learned it, let me start by defining the phrase ill-conditioned matrix. Let $A$ be a square matrix (we'll operate over the reals, although I think this works as well for complex matrices) and let $\left\Vert \cdot\right\Vert$ be a consistent matrix norm. The condition number of $A$, assuming that is nonsingular, is $$\kappa = \left\Vert A\vphantom{^{-1}}\right\Vert \left\Vert A^{-1}\right\Vert \ge 1.$$ For singular matrices, $\kappa$ is either undefined or $\infty$, according to your religious preferences. Somewhat informally, $\kappa$ can be interpreted as a measure of how much rounding errors may be inflated then inverting $A$ (using floating-point arithmetic ... "pure" mathematicians don't commit rounding errors). An ill-conditioned or "stiff" matrix $A$ is non-singular but has a large condition number (the criterion for "large" being rather fuzzy), meaning that rounding errors during the process of inverting (or factoring, or otherwise torturing) $A$ may result in an answer with lots of incorrect digits. Long ago, a graduate school instructor told my class the story of an engineer who had inverted an ill-conditioned matrix, without realizing it was ill-conditioned, using home-brew FORTRAN code. He'd published the inverse in a journal article to four or five decimal places. All the decimal places in most of the coefficients were wrong. So were lots of digits to the left of the decimal points. #### Ridge regression then Here comes the tie-in to regression. Suppose we are trying to fit the model $$y=x'\beta+\epsilon$$(where $\epsilon$ is random noise with mean zero), and we have collected observations $(X,Y)$ with $X$ an $n\times m$ matrix ($n$ = sample size, $m$ = number of predictors). The normal equations tell us that the least squares estimate for $\beta$ is $$b=(X^\prime X)^{-1}X^\prime Y.$$In theory, $X^\prime X$ is nonsingular unless the predictors are perfectly multicollinear. In practice, if $X^\prime X$ is nonsingular but stiff, our coefficient estimate $b$ can suffer drastic rounding errors. To mitigate those errors, ridge regression tampers with $X^\prime X$, adding a small positive multiple of the identity matrix of dimension $m$. The modified normal equations are$$b_\lambda=(X^\prime X + \lambda I)^{-1}X^\prime Y.$$The result $b_\lambda$ is a deliberately biased estimate of $\beta$, in which we trade the bias for reduced rounding error. We generally choose $\lambda > 0$ as small as possible subject to the need to stabilize the numerics. When I first encountered ridge regression, we were in the (first) mainframe era. Machine words were small, single-precision arithmetic ruled the computing universe, and rounding errors were relatively easy to generate. Somewhere along the way, computing hardware got smaller and cheaper, double precision arithmetic became the norm and separate floating point units took over the computational load. At the same time, matrix factorization replaced matrix inversion, and factoring algorithms got smarter about managing rounding error. The problems with stiff $X^\prime X$ matrices seemed largely to disappear, and references to ridge regression became uncommon, at least in my experience. #### Ridge regression now In the Statistical Learning course, ridge regression makes a resurgence, with an entirely different purpose. In the era of Big Data, there is a distinct danger that we may overfit data by throwing in lots and lots of predictors (large dimension $m$). One way to mitigate that temptation is to penalize the number or size of the coefficients in the model. The normal equations minimize the sum of squared deviations in a linear regression model. In other words, the ordinary least squares estimate of $\beta$ solves the problem$$\min_b \Vert Y-X b\Vert^2.$$Given that we are squaring errors, and given that quadratic objective functions have nice smoothness, a tempting way to penalize overfitting is to add a penalty for the size of the coefficient vector:$$\min_b \Vert Y-X b\Vert^2 + \lambda \Vert b \Vert^2$$where $\lambda > 0$ is a penalty parameter. If $\lambda = 0$, we are back to ordinary least squares. As $\lambda \rightarrow \infty$, $\Vert b \Vert$ will converge toward (but never quite reach) 0. I'll skip the algebra and just state that solving this problem for fixed $\lambda$ is mathematically equivalent to ridge regression as seen above. The only difference is in the choice of $\lambda$. When we were fighting rounding error, we wanted the smallest $\lambda$ that would clean up most of the rounding error. Now, when we are fighting overfitting, we probably need larger values of $\lambda$ to get the job done. So everything old is new again ... which bodes well for my knees. ## Saturday, February 1, 2014 ### A Modest Terminology Proposal I just finished shoveling snow, an exercise I will have to repeat later today and possibly tomorrow, as we enjoy the first of multiple winter storms queued up all the way from mid-Michigan (where I live) to the central Pacific, each patiently awaiting its turn to annoy me. With that in mind, I would like to point out cardinal failings of our terminology for climate phenomena. #### Global Warming This term was in vogue several years ago, but seems to have been surpassed by "climate change" (next item). One problem with "global warming" is that while it may be true in an average sense, it frequently does not resonate in an instantaneous sense. When you've lived through a "polar vortex" and are awaiting the rumored arrival of another one, "global warming" seems neither likely nor a bad idea. #### Climate Change This is the phrase currently in vogue. Its two main faults are that (a) change is not necessarily bad and (b) change is inevitable (except, as some wag once noted, from a vending machine). #### My Proposal With those thoughts in mind, I would like to propose a new phrase that I think is both more descriptive of current climate phenomena and better conveys a sense of concern about them: Bipolar Climate Disorder	1,533	6,884	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.125	4	CC-MAIN-2017-39	latest	en	0.965075
https://www.extendoffice.com/documents/excel/7357-absolute-reference-excel.html	1,720,778,975,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514387.30/warc/CC-MAIN-20240712094214-20240712124214-00713.warc.gz	479,128,547	22,063	## Excel absolute reference (how to make and use) When referencing a cell in a formula in Excel, the default reference type is a relative reference. These references will change when the formula is copied to other cells based on their relative column and row. If you want to keep a reference constant, regardless of where the formula is copied, you need to use the absolute reference. ### What is an absolute reference An absolute reference is a type of cell reference in Excel. Compared to a relative reference that will change based on its relative position when a formula is copied to other cells, an absolute reference will remain constant no matter where the formula is copied or moved. An absolute reference is created by adding a dollar sign (\$) before the column and row references in the formula. For example, to create an absolute reference for cell A1, you should represent it as \$A\$1. Absolute references are useful when you want to refer to a fixed cell or range in a formula that will be copied to multiple cells, but you do not want the reference to change. For example, the range A4:C7 contains products' prices, and you want to get the payable tax of each product based on the tax rate in cell B2. If you use relative reference in the formula like "=B5B2", some wrong results are returned when you drag auto fill handle down to apply this formula. Since the reference to cell B2 will change relative to the cells in the formula. Now, the formula in cell C6 is "=B6B3", and the formula in cell C7 is "=B7B4" But if you use an absolute reference to cell B2 with the formula "=B5\$B\$2" will ensure that the tax rate stays the same for all cells when the formula is dragged down using the auto fill handle, the results are correct. Using relative reference Using absolute reference ### How to make absolute references To make an absolute reference in Excel, you need to add dollar signs (\$) before the column and row references in formula. There are two ways to create absolute reference: ##### Manually add dollar signs to the cell reference You could manually add dollar signs (\$) before the column and row references that you want to make absolute while typing the formula in a cell. For example, if you want to add the numbers in cell A1 and B1 and make both of them absolute, just type the formula as "=\$A\$1+\$B\$1". This will ensure that the cell references remain constant when the formula is copied or moved to other cells. Or if you want to change references in an existed formula in a cell to absolute, you can select the cell, then go to the formula bar to add the dollar signs (\$). ##### Using shortcut F4 to convert relative reference to absolute 1. Double click the cell with the formula to enter the edit mode; 2. Place the cursor at the cell reference you want to make absolute; 3. Press F4 key on the keyboard to switch the reference types until the dollar signs are added before both column and row references; 4. Press Enter key to exit edit mode and apply the changes. F4 key can toggle reference between relative reference, absolute reference and mixed reference. A1 → \$A\$1 → A\$1 → \$A1 → A1 If you want to make all references absolute in a formula, select the whole formula in the formula bar, press F4 key to toggle reference types until the dollar signs are added before both of column and row references. A1+B1 → \$A\$1+\$B\$1 → A\$1+B\$1 → \$A1+\$B1 → A1+B1 ### Use absolute reference with examples This part provides 2 examples to show when and how to use absolute references in an Excel formula. ##### Example 1 Calculate percentage of total Suppose you have a data range (A3:B7) containing the sales of each fruit, and the cell B8 contains the total sales amount of these fruits, now you want to calculate the percentage of each fruit sale of the total. The generic formula to calculate the percentage of total: ``Percentage = Sale/Amount`` Use the relative reference in the formula to get the percentage of the first fruit like this: ``=B4/B8`` When dragging the auto fill handle down to calculate the percentage of other fruits, #DIV/0! errors will be returned. Since when you drag auto fill handle to copy the formula to cells below, the relative reference B8 is automatically adjusted to other cell references (B9, B10, B11) based on their relative positions. And the cell B9, B10 and B11 are empty (zeros), when the divisor is zero, the formula returns to an error. To fix the errors, in this case, you need to make cell reference B8 absolute (\$B\$8) in the formula to keep it from changing when you move or copy the formula to anywhere. Now the formula is updated to: ``=B4/\$B\$8`` Then drag the auto fill handle down to calculate percentage of other fruits. ##### Example 2 Look for a value and return to corresponding match value Assuming you want to look for the names list in D4:D5 and return their corresponding salaries based on the staff names and corresponding annual salary provided in range (A4:B8). The generic formula to lookup is: ``=VLOOKUP(lookup_value, table_range, column_index, logical)`` If you use the relative reference in the formula to lookup a value and return corresponding match value like this: ``=VLOOKUP(D4,A4:B8,2,FALSE)`` Then drag the auto fill handle down to lookup the value below, an error will be returned. When you drag the fill handle down to copy the formula to the cell below, the references in the formula automatically adjust down by one row. As a result, the reference to the table range, A4:B8, becomes A5:B9. Since "Lisa: cannot be found in the range A5:B9, the formula returns an error. To avoid the errors, use absolute reference \$A\$4:\$B\$8 instead of the relative reference A4:B8 in the formula: ``=VLOOKUP(D4,\$A\$4:\$B\$8,2,FALSE)`` Then drag the auto fill handle down to get the salary of Lisa. ### 2 clicks to batch make cell references absolute with Kutools Whether you choose to type manually or use F4 shortcut, you can only change one formula at a time in Excel. If you want to make cell references in hundreds of formulas absolute in Excel, the Convert Refers tool of Kutools for Excel can help you handle the job with 2 clicks. Select the formula cells that you want to make cell references absolute, click Kutools > More (fx) > Convert Refers. Then choose the To absolute option and click Ok or Apply. Now all cell references of the selected formulas have been converted to absolute. Notes: ### Relative reference and mixed reference Apart from absolute reference, there are other two reference types: relative reference and mixed reference. Relative reference is the default reference type in Excel, which is without dollar signs (\$) before row and column references. And when a formula with relative references is copied or moved to other cells, the references will automatically change based on their relative position. For example, when you type a formula in a cell such as "=A1+1", then drag autofill handle down to fill this formula to the next cell, the formula will auto change to "=A2+1". Mixed reference is made up of both an absolute reference and relative reference. In other words, mixed reference uses the dollar sign (\$) to fix either the row or column when a formula is copied or filled. Take a multiplication table as an example, the rows and columns list the numbers from 1 to 9, which you will multiply each other. To start, you can use the formula "=B3C2" in cell C3 to multiply 1 in cell B3 by the number (1) in the first column. However, when you drag the autofill handle to the right to fill the other cells, you'll notice that all of the results are incorrect except for the first one. This is because when you copy the formula to the right, the row position won’t change, but the column position changes from B3 to C3, D3, etc.. As a result, the formulas in the right cells (D3, E3, etc.) change to "=C3D2", "=D3E2", and so on, when you actually want them to be "=B3D2", "=B3E2", and so on. In this case, you need to a add dollar sign (\$) to lock the column reference of “B3”. Use the formula as below: ``=\$B3C2`` Now when you drag the formula to the right, the results are correct. Then you need to multiply the number 1 in cell C2 by the numbers in the rows below. When you copy the formula down, the column position of cell C2 won't change, but the row position changes from C2 to C3, C4, etc. As a result, the formulas in the cells below change to "=\$B4C3", "=\$B5C4", etc. which will produce incorrect results. To solve this problem, change “C2” to “C\$2” to keep row reference from changing when dragging auto fill handle down to fill the formulas. ``=\$B3*C\$2`` Now you can drag the auto fill handle right or down to get all results. ### Things to remember • Summary of cell references Type Example Summary Absolute Reference \$A\$1 Never change when formula is copied to other cells Relative Reference A1 Both of row and column reference change based on relative position when formula is copied to other cells Mixed Reference \$A1/A\$1 Row reference changes when formula is copied to other cells but column reference is fixed/Column reference changes when formula is copied to other cells but row reference is fixed; • Generally, absolute references never change when a formula is moved. However, absolute references will automatically adjust when a row or column is added or removed from the top or left in the worksheet. For example, in a formula "=\$A\$1+1", when you insert a row at the top of the sheet, the formula will auto change to "=\$A\$2+1". • The F4 key can switch between relative reference, absolute reference and mixed reference. ### Best Office Productivity Tools 🤖 Kutools AI Aide: Revolutionize data analysis based on: Intelligent Execution \| Generate Code \| Create Custom Formulas \| Analyze Data and Generate Charts \| Invoke Kutools Functions… Popular Features: Find, Highlight or Identify Duplicates \| Delete Blank Rows \| Combine Columns or Cells without Losing Data \| Round without Formula ... Super Lookup: Multiple Criteria VLookup \| Multiple Value VLookup \| VLookup Across Multiple Sheets \| Fuzzy Lookup .... Advanced Drop-down List: Quickly Create Drop Down List \| Dependent Drop Down List \| Multi-select Drop Down List .... Column Manager: Add a Specific Number of Columns \| Move Columns \| Toggle Visibility Status of Hidden Columns \| Compare Ranges & Columns ... 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Click Here to Get The Feature You Need The Most... #### Office Tab Brings Tabbed interface to Office, and Make Your Work Much Easier • Enable tabbed editing and reading in Word, Excel, PowerPoint, Publisher, Access, Visio and Project. • Open and create multiple documents in new tabs of the same window, rather than in new windows. • Increases your productivity by 50%, and reduces hundreds of mouse clicks for you every day!	2,765	11,812	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.8125	3	CC-MAIN-2024-30	latest	en	0.871519
https://byjus.com/question-answer/if-a-rubber-ball-of-density-2-g-cm-3-thrown-into-a-pool-of-1/	1,709,004,724,000,000,000	text/html	crawl-data/CC-MAIN-2024-10/segments/1707947474670.19/warc/CC-MAIN-20240227021813-20240227051813-00284.warc.gz	158,107,374	29,291	Question # If a rubber ball of density 2 g cmâˆ’3 thrown into a pool of a fluid and the fluid has a density of 4 g cmâˆ’3, what will happen to the ball? A It sinks in the fluid. No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today! B Buoyant force pulls it inside. No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today! C It floats on the surface. Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses D Density of the ball increases. No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today! Open in App Solution ## The correct option is C It floats on the surface.If the density of the body is less than that of the fluid then the body floats (partially) on the surface. This is because the buoyant force is balanced by the weight of the body. If the density of the body is greater than that of the fluid then the body sinks in the fluid. In the above case, density of fluid is greater than the density of rubber ball, hence the rubber ball floats on the surface. Suggest Corrections 0 Join BYJU'S Learning Program Related Videos Buoyant Force_tackle PHYSICS Watch in App Explore more Join BYJU'S Learning Program	328	1,175	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2024-10	latest	en	0.872143
http://gmatclub.com/forum/if-m-is-a-positive-odd-integer-between-2-and-30-then-m-is-23741.html	1,484,915,616,000,000,000	text/html	crawl-data/CC-MAIN-2017-04/segments/1484560280834.29/warc/CC-MAIN-20170116095120-00067-ip-10-171-10-70.ec2.internal.warc.gz	122,501,238	42,400	If m is a positive odd integer between 2 and 30, then m is : DS Archive Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 20 Jan 2017, 04:33 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # If m is a positive odd integer between 2 and 30, then m is Author Message Intern Joined: 30 Apr 2004 Posts: 30 Location: Los Angeles, CA Followers: 0 Kudos [?]: 0 [0], given: 0 If m is a positive odd integer between 2 and 30, then m is [#permalink] ### Show Tags 26 Nov 2005, 23:52 00:00 Difficulty: (N/A) Question Stats: 0% (00:00) correct 0% (00:00) wrong based on 0 sessions ### HideShow timer Statistics This topic is locked. If you want to discuss this question please re-post it in the respective forum. If m is a positive odd integer between 2 and 30, then m is divisible by how many different positive prime integers? 1. m is not divisible by 3 2. m is not divisible by 5 SVP Joined: 24 Sep 2005 Posts: 1890 Followers: 19 Kudos [?]: 292 [0], given: 0 Re: Sorry for the typo on the DS [#permalink] ### Show Tags 27 Nov 2005, 01:38 nickdawg wrote: If m is a positive odd integer between 2 and 30, then m is divisible by how many different positive prime integers? 1. m is not divisible by 3 2. m is not divisible by 5 1.since m is odd and <30 , m is not divisible by 3, m can only be formed by product of any two of 1,5,7,9 ...as we know 57 , 79, 5*9 all > 30 ---> m can only be formed by product of 1 and 5,7,9 ---> m is divisible by 1 prime number --->suff 2. m=3 : 1 prime m= 21 : 2 primes ---->insuff A it is. Re: Sorry for the typo on the DS [#permalink] 27 Nov 2005, 01:38 Display posts from previous: Sort by	654	2,225	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.96875	4	CC-MAIN-2017-04	latest	en	0.905695
https://www.codeproject.com/Questions/479786/C-2cplusWhatplusdoesplustheplusfollowingplusdeclar	1,493,262,363,000,000,000	text/html	crawl-data/CC-MAIN-2017-17/segments/1492917121778.66/warc/CC-MAIN-20170423031201-00045-ip-10-145-167-34.ec2.internal.warc.gz	865,509,722	43,681	12,890,181 members (48,126 online) Rate this: See more: What does the following declaration mean? `int (ptr)[10];` is above statement and below is same? `int ptr[10];` Thank you. Posted 19-Oct-12 0:30am Updated 20-Oct-12 18:47pm v3 Rate this: ## Solution 5 It is "declare ptr as pointer to array 10 of int". In some cases the answer is easy to get from cdecl C gibberish ↔ English Also, the Clockwise/Spiral rule can help to understand the syntax. v3 Stefan_Lang 19-Oct-12 12:16pm Sergey Chepurin 19-Oct-12 12:24pm My pleasure) indhukanth 21-Oct-12 0:59am hi Sergey Chepurin is that rule applicable for all?? Sergey Chepurin 21-Oct-12 4:00am First line of the article says - "...`Clockwise/Spiral Rule'' which enables any C programmer to parse in their head any C declaration". Legor 22-Oct-12 3:31am Nelek 22-Oct-12 13:05pm Rate this: ## Solution 6 int (ptr)[10] - pointer to an array of integers int ptr[10] - an array of int pointers Array of pointers pointer may be arrayed like any data type. To assign the address of an integer variable called var to third element of the pointer array, write ptr[2] = &var; to find the value of var, write ptr[2]. Pointer to array pointer to an array is a single pointer, that hold the address of an array of variables. in above case it is an integer array. hence these two are different concept. not same Rate this: ## Solution 1 It declares a variable `ptr` which is a pointer to an array of 10 ints Legor 22-Oct-12 3:38am In both of the mentioned cases? OriginalGriff 22-Oct-12 3:45am And the second answer is "No". (The second is an array of pointers to ints) Rate this: ## Solution 3 `ptr` is a pointer to an array of `10 int`. Try ```int (ptr)[10]; int a[10]={0}; ptr= &a; ptr[0] = 5; printf("%d\n", a[0]); ``` Legor 22-Oct-12 3:38am In both of the mentioned cases? CPallini 22-Oct-12 6:11am Nope: second one is the declaration of one array of 10 pointers to int. Rate this: ## Solution 4 Hi, as many suggests here, I don't think ptr is an array of 10 ints. Instead, it is an array of integer pointers. ptr is an array[sized 10] of integer pointers In fact it creates an uninitialized set of pointers. You can understand it from the code below. ```#include<stdio.h> int main() { int (ptr)[10]; ptr[0]=1; printf("%d", ptr[0][0]); return 0; }``` The aforesaid code when compiled shows a warning: "warning: 'ptr' is used uninitialized in this function" But when executed it outputs '1'. If you desire, you can initialize ptr[0],ptr[1] etc with different sized integer arrays. I think I have given enough explanations. Your comments are welcome. :-) v2 CPallini 19-Oct-12 7:52am What compiler are you using (the code compiled with the one I'm using right now, gcc 4.4.5, gives 'segmentation fault')? Goutham Mohandas 19-Oct-12 8:16am GCC version is gcc (GCC) 4.5.2 Goutham Mohandas 19-Oct-12 8:24am Segmentation fault may rise, because ptr[0] is not initialized, therefore, the assignment writes the value on the address ptr[0] holds by default. Stefan_Lang 19-Oct-12 8:24am Your code doesn't prove your statement: `ptr[0]=1;` is equivalent to `(ptr)[0]`, and that it works just proves that ptr is a pointer to a pointer to int, same as in CPallini's code. The warning indicates exactly the point you're missing, i. e. that you omitted the part that CPallini included, i. e. assigning a value of appropriate type. Stefan_Lang 19-Oct-12 8:26am Your code works because you left out the statement that would cause the compiler error - the initialization of ptr. Goutham Mohandas 19-Oct-12 8:42am In the solution given by Cpallini, the statement 'ptr= &a;' diverts from where the ptr was orginally pointing. Eventhough, he points the ptr to the address of another pointer a(where a is an array of ints); which clearly satisfies my statement. :-) Stefan_Lang 19-Oct-12 8:54am No, it satisfies his* statement, that the type of ptr is 'pointer to array of int' (or at least convertible). Your statement was that ptr is of type 'array of pointer to int', and that can be easily disproven: see the comment to Solution 2. indhukanth 21-Oct-12 1:06am Hi All, any one agree/disagree with Sergey Chepurin comment (Solution 5)? plz make me clear. Thankyou. Legor 22-Oct-12 3:35am Yes he is right. Solution 6 is also right and it also answers the question if the two declarations are the same (which they are not). Legor 22-Oct-12 3:36am Nobody here suggested that "ptr is an array of 10 ints" but they said "ptr is a pointer to an array of 10 int" which is something very different. Top Experts Last 24hrsThis month CHill60 260 OriginalGriff 240 Maciej Los 175 Jochen Arndt 175 Kornfeld Eliyahu Peter 150 OriginalGriff 4,712 CHill60 3,288 Jochen Arndt 2,603 Karthik Bangalore 2,446 ppolymorphe 2,015	1,423	4,795	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.046875	3	CC-MAIN-2017-17	latest	en	0.829138
http://www.chegg.com/homework-help/questions-and-answers/s-2-3-3s-2-s-9-k0-k1-s-p1-k2-p2-numerical-value-constant-k0-k1-k2-p1-p2-q2449597	1,387,401,312,000,000,000	text/html	crawl-data/CC-MAIN-2013-48/segments/1387345759442/warc/CC-MAIN-20131218054919-00055-ip-10-33-133-15.ec2.internal.warc.gz	297,786,262	6,373	# laplace transform 0 pts ended (s^2+3)/(3s^2+s+9)=K0+K1/(s-p1)+(K2-p2) find the numerical value of the constant K0, K1, K2, P1 &P2	60	132	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.71875	3	CC-MAIN-2013-48	latest	en	0.724992
https://www.ginac.de/ginac.git/?p=ginac.git;a=blobdiff_plain;f=check/differentiation.cpp;h=0d963fed8b0f6d830a56063f71859a9a6786d561;hp=d1051ea9e8cd871ac22bde57709ab7cc87b4867b;hb=be0485a03e9886496eeb7e8cdc2cc5c95b848632;hpb=9eab44408b9213d8909b7a9e525f404ad06064dd	1,656,217,439,000,000,000	text/plain	crawl-data/CC-MAIN-2022-27/segments/1656103037089.4/warc/CC-MAIN-20220626040948-20220626070948-00372.warc.gz	864,314,206	1,613	X-Git-Url: https://www.ginac.de/ginac.git//ginac.git?p=ginac.git;a=blobdiff_plain;f=check%2Fdifferentiation.cpp;h=0d963fed8b0f6d830a56063f71859a9a6786d561;hp=d1051ea9e8cd871ac22bde57709ab7cc87b4867b;hb=be0485a03e9886496eeb7e8cdc2cc5c95b848632;hpb=9eab44408b9213d8909b7a9e525f404ad06064dd diff --git a/check/differentiation.cpp b/check/differentiation.cpp index d1051ea9..0d963fed 100644 --- a/check/differentiation.cpp +++ b/check/differentiation.cpp @@ -107,7 +107,7 @@ static unsigned differentiation1(void) return result; } -// Trigonometric and transcendental functions +// Trigonometric functions static unsigned differentiation2(void) { unsigned result = 0; @@ -152,8 +152,19 @@ static unsigned differentiation2(void) d = -2bpow(e2,2)pow(x,4) + 2bpow(sin(e1),2)pow(x,4) - 2sin(e1)pow(x,2) - ye2pow(x,4); result += check_diff(e, y, d, 2); + + return result; +} +// exp function +static unsigned differentiation3(void) +{ + unsigned result = 0; + symbol x("x"), y("y"), a("a"), b("b"); + ex e1, e2, e, d; + // construct expression e to be diff'ed: + e1 = ypow(x, 2) + ax + b; e2 = exp(e1); e = bpow(e2, 2) + ye2 + a; @@ -169,8 +180,19 @@ static unsigned differentiation2(void) d = 4bpow(e2,2)pow(x,4) + 2e2pow(x,2) + ye2pow(x,4); result += check_diff(e, y, d, 2); + + return result; +} + +// log functions +static unsigned differentiation4(void) +{ + unsigned result = 0; + symbol x("x"), y("y"), a("a"), b("b"); + ex e1, e2, e, d; // construct expression e to be diff'ed: + e1 = ypow(x, 2) + ax + b; e2 = log(e1); e = bpow(e2, 2) + ye2 + a; @@ -188,8 +210,18 @@ static unsigned differentiation2(void) d = 2bpow(x,4)pow(e1,-2) - 2be2pow(e1,-2)pow(x,4) + 2pow(x,2)/e1 - ypow(x,4)pow(e1,-2); result += check_diff(e, y, d, 2); + + return result; +} + +// Functions with two variables +static unsigned differentiation5(void) +{ + unsigned result = 0; + symbol x("x"), y("y"), a("a"), b("b"); + ex e1, e2, e, d; - // test for functions with two variables: atan2 + // test atan2 e1 = ypow(x, 2) + ax + b; e2 = xpow(y, 2) + by + a; e = atan2(e1,e2); @@ -202,17 +234,23 @@ static unsigned differentiation2(void) pow(yb+pow(y,2)x+a,-2)pow(y,2))* pow(1+pow(ax+b+ypow(x,2),2)pow(yb+pow(y,2)x+a,-2),-1); / + /* d = pow(1+pow(ax+b+ypow(x,2),2)pow(yb+pow(y,2)x+a,-2),-1) pow(yb+pow(y,2)x+a,-1)(a+2yx) +pow(y,2)(-ax-b-ypow(x,2))* pow(pow(yb+pow(y,2)x+a,2)+pow(ax+b+ypow(x,2),2),-1); + / + d = pow(y,2)pow(pow(b+ypow(x,2)+xa,2)+pow(yb+pow(y,2)x+a,2),-1)* + (-b-ypow(x,2)-xa)+ + pow(pow(b+ypow(x,2)+xa,2)+pow(yb+pow(y,2)x+a,2),-1)* + (yb+pow(y,2)x+a)(2y*x+a); result += check_diff(e, x, d); return result; } // Series -static unsigned differentiation3(void) +static unsigned differentiation6(void) { symbol x("x"); ex e, d, ed; @@ -241,6 +279,9 @@ unsigned differentiation(void) result += differentiation1(); result += differentiation2(); result += differentiation3(); + result += differentiation4(); + result += differentiation5(); + result += differentiation6(); if (!result) { cout << " passed ";	1,179	3,068	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.09375	3	CC-MAIN-2022-27	latest	en	0.32585
https://de.mathworks.com/matlabcentral/answers/2065431-how-to-evaluate-a-symbolic-expression-having-max-and-diff?s_tid=prof_contriblnk	1,716,444,391,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00652.warc.gz	162,425,022	33,858	# How to evaluate a symbolic expression having `max` and `diff`? 8 Ansichten (letzte 30 Tage) Rounak Saha Niloy am 1 Jan. 2024 Kommentiert: Walter Roberson am 4 Jan. 2024 I have calculated the jacobian of two functions where variables are x1, x2, x3. The jacobian is as follows- JacobianF = [ diff(max([0, (7sin(4pix1))/10], [], 2, 'omitnan', ~in(x1, 'real')), x1) + 96picos(6pix1)(x3 + sin(6pix1)) + 1603^(1/2)pi^2cos(6pix1)sin((3^(1/2)pi(20x3 + 20sin(6pix1)))/3) + 1, 0, 16x3 + 16sin(6pix1) + diff(max([0, (7sin(4pix1))/10], [], 2, 'omitnan', ~in(x1, 'real')), x3) + (803^(1/2)pisin((3^(1/2)pi(20x3 + 20sin(6pix1)))/3))/3] [diff(max([0, (7sin(4pix1))/10], [], 2, 'omitnan', ~in(x1, 'real')), x1) - 96picos((2pi)/3 + 6pix1)(x2 - sin((2pi)/3 + 6pix1)) - 2402^(1/2)pi^2cos((2pi)/3 + 6pix1)sin((2^(1/2)pi(20x2 - 20sin((2pi)/3 + 6pix1)))/2) - 1, 16x2 - 16sin((2pi)/3 + 6pix1) + diff(max([0, (7sin(4pix1))/10], [], 2, 'omitnan', ~in(x1, 'real')), x2) + 402^(1/2)pisin((2^(1/2)pi(20x2 - 20sin((2pi)/3 + 6pix1)))/2), 0] Now, I need to evaluate this JacobianF at X = [0.2703 0.6193 0.9370]; where X(1) is x1 and so on. To evaluate this JacobianF, I have used the following code- Var_List = sym('x', [1, 3]); df=double(subs(JacobianF, Var_List, X)); However, I get the following error. What is the cause of this error? How to resolve it and calculate the JacobianF at the specified position? Error using symengine Unable to convert expression containing remaining symbolic function calls into double array. Argument must be expression that evaluates to number. Error in sym/double (line 872) ##### 12 Kommentare10 ältere Kommentare anzeigen10 ältere Kommentare ausblenden Torsten am 2 Jan. 2024 Use min(x,0) = 0.5(x-abs(x)) as I used max(x,0) = 0.5(x+abs(x)) below. Walter Roberson am 4 Jan. 2024 Looks like it works for me when y is symbolic. syms y b_flat(y, 1, 2, 3) ans = b_flat(y, -10, 5, 17) ans = function Output = b_flat(y,A,B,C) Output = A+piecewise(0<=floor(y-B),0,floor(y-B))A.(B-y)/B-piecewise(0<=floor(C-y),0,floor(C-y))(1-A).(y-C)/(1-C); Output = round(Output1e4)/1e4; end Melden Sie sich an, um zu kommentieren. ### Antworten (2) Walter Roberson am 1 Jan. 2024 The derivative of max() is not generally defined. You would probably have more success if you defined in terms of piecewise() instead of in terms of max() ##### 1 Kommentar-1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden Dyuman Joshi am 4 Jan. 2024 I guess the Sym engine does not have the ability to recognise that the definition of max() can be broken into a piecewise definition, than a derivative can be calculated. I wonder if that is possible to implement or not. Melden Sie sich an, um zu kommentieren. Torsten am 1 Jan. 2024 Bearbeitet: Torsten am 2 Jan. 2024 Use max(x,0) = 0.5(abs(x)+x) for real x. syms x1 f1 = max([0, (7sin(4pix1))/10], [], 2, 'omitnan', ~in(x1, 'real')) f1 = f2 = 0.5(abs(7sin(4pix1)/10)+7sin(4pix1)/10) f2 = figure(1) hold on fplot(f1,[-0.5 0.25]) fplot(f2,[-0.5 0.25]) hold off df1 = diff(f1,x1) df1 = df2 = diff(f2,x1) df2 = figure(2) %fplot(df1,[-0.5 0.25]) fplot(df2,[-0.5 0.25]) ##### 0 Kommentare-2 ältere Kommentare anzeigen-2 ältere Kommentare ausblenden Melden Sie sich an, um zu kommentieren. ### Kategorien Mehr zu Number Theory finden Sie in Help Center und File Exchange R2023b ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting! Translated by	1,336	3,554	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.3125	3	CC-MAIN-2024-22	latest	en	0.520702
http://stackoverflow.com/questions/9150855/reading-certain-digits-from-binary-data/9150895	1,386,764,181,000,000,000	text/html	crawl-data/CC-MAIN-2013-48/segments/1386164035500/warc/CC-MAIN-20131204133355-00037-ip-10-33-133-15.ec2.internal.warc.gz	174,275,313	12,648	# reading certain digits from binary data I have one python script which i am trying to convert and stuck in one place and unable to proceed. Please check where ever i have mentioned "Stuck here" in below code. any help would be appreciated Original Python script: ``````import hashlib meid = raw_input("Enter an MEID: ").upper() s = hashlib.sha1(meid.decode('hex')) #decode the hex MEID (convert it to binary!) pesn = "80" + s.hexdigest()[-6:].upper() #put the last 6 digits of the hash after 80 print "pESN: " + pesn `````` My C# conversion: ``````UInt64 EsnDec = 2161133276; string EsnHex=string.Format("{0:x}", EsnDec); string m = Convert.ToString(Convert.ToUInt32(EsnHex, 16), 2); /--------------------------------------------- Stuck here. Now m got complete binary data and i need to take last 6 digits as per python script and prefix "80". ---------------------------------------------/ Console.WriteLine(m); `````` - Agreed. To take the last `n` digits of a string `s`, you can use: `s.Substring(s.Length - n)`. – Douglas Feb 5 '12 at 16:33 ``````static void Main(string[] args) { UInt64 EsnDec = 2161133276; Console.WriteLine(EsnDec); //Convert to String string Esn = EsnDec.ToString(); Esn = "80" + Esn.Substring(Esn.Length - 6); //Convert back to UInt64 EsnDec = Convert.ToUInt64(Esn); Console.WriteLine(EsnDec); } `````` - The last 6 decimal characters will be the last 20 bits. I think the OP wants the last 6 bits; which is ~1.5 dec characters. – user7116 Feb 5 '12 at 16:41 ``````// last 6 characters	433	1,525	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.59375	3	CC-MAIN-2013-48	latest	en	0.592135
https://www.askiitians.com/forums/Mechanics/when-acceleration-be-a-function-of-velocity-as-a_97992.htm	1,656,979,041,000,000,000	text/html	crawl-data/CC-MAIN-2022-27/segments/1656104506762.79/warc/CC-MAIN-20220704232527-20220705022527-00737.warc.gz	698,574,795	34,676	# When acceleration be a function of velocity as a= f(v).. Then :-a) The displacement (x) = integral vdv/f(v)b) the acceleration may be constantc) the slope of acceleration versus velocity graph may may be constant.d) a) & c) are correctPlease give solution Nirmal Singh. 8 years ago a = f(x) dv / dt = f(x) (dx / dt)*(dv/ dx) = f(x) v dv / dx = f(x) so dx = vdv / f(x) hence x = integral of (vdv) / f (x) Hence Answer a is correct and acceleration is directly proportionlal to velocity so acce. will not be constant a = f(x) so a / f(x) = 1 hencethe slope of acceleration versus velocity graph may may be constant Hence Answer a and c is correct Regards, Nirmal Singh	197	679	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.6875	4	CC-MAIN-2022-27	latest	en	0.751146
https://studylib.net/doc/17579184/course-narrative	1,596,963,315,000,000,000	text/html	crawl-data/CC-MAIN-2020-34/segments/1596439738523.63/warc/CC-MAIN-20200809073133-20200809103133-00176.warc.gz	503,621,329	13,139	# Course Narrative advertisement ```Course Narrative Course Title – Advanced Functions and Modeling Course Description Advance Functions and Modeling is a continued study of the properties and relationships in algebra and trigonometry. This course covers linear, rational, radical, quadratic, and trigonometric functions. Students will analyze graphs using a transformational approach. They will also explore the relationship among the numerical, graphical, and algebraic representations of functions. Learning Expectations The student will:     identify, graph, and write linear, rational, radical, and quadratic functions and apply the concepts of these functions to real world models. find domain, range, and inverse of functions calculate each of the six trigonometric functions using the unit circle or given the value of one trigonometric function in radians or degrees identify the domain and range, graph, and relate the parent trigonometric function in order to transform each function. Grading   Summative 80% (including, but not limited to…) o Tests o Quizzes o Problem Sets/Graded Worksheets o Group Work o Projects Formative 20% (including, but not limited to…) o Skills Quizzes o Test corrections (where assigned) ```	276	1,238	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.5625	3	CC-MAIN-2020-34	latest	en	0.832341
https://math.answers.com/Q/What_is_the_radius_of_the_circle_with_the_equation_x%C2%B2_plus_y%C2%B2_equals_36	1,713,758,638,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296818072.58/warc/CC-MAIN-20240422020223-20240422050223-00014.warc.gz	343,883,366	47,154	0 What is the radius of the circle with the equation x² plus y² equals 36? Updated: 12/23/2022 Wiki User 7y ago A circle centre (0, 0) and radius r has equation x² + y² = r² The circle x² + y² = 36 has: r² = 36	79	217	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.390625	3	CC-MAIN-2024-18	latest	en	0.780783
https://puzzling.stackexchange.com/questions/97771/what-is-the-minimum-count-of-steps-required-to-complete-this-dominoes-maze	1,718,747,188,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861794.76/warc/CC-MAIN-20240618203026-20240618233026-00400.warc.gz	414,380,199	42,961	# What is the minimum count of steps required to complete this dominoes maze? Here's a map: How to play Legend: (◾) = Start point ★ = Objective ⚑ = End point Mission: Collect all the objectives and go to the end with the minimum count of steps. How the game works: • The pawn counts its steps that are incremented by the number contained in the boxes it moves on, every box can be walked multiple times. • The pawn can switch platform, but it has to move only with adiacent platforms, as indicated in gray boxes (the passage of which does not involve further steps). • The pawn can't stay more than one move out of a platform (this means that it is not possible to cross 2 gray squares consecutively). • It is not a problem to cross an entire platform (i.e. both boxes of it). Good luck! • To clarify, if I move down two boxes (starting from the start point), will that be a count of only 5 steps? Or is it 7 since it took 2 steps to get to the next box + increment of 5 in the previous box? – oAlt Commented May 2, 2020 at 3:16 • As indicated in How the game works: [...] gray boxes (the passage of which does not involve further steps), you have to count only the numbers contained in the azure boxes Commented May 2, 2020 at 9:36 • Ah, I see. Thx :0 – oAlt Commented May 2, 2020 at 9:40 Indeed it can be solved as a traveling salesman problem on 13 nodes, but brute force is not required. First define an undirected graph with 145 nodes, one per box, with an edge between each pair of horizontally or vertically adjacent boxes (unless both are gray). Define each edge weight to be the average of the numbers of its two incident nodes. Now solve a shortest path problem for each pair of start point, objective, or end point. The resulting shortest path distances $$d_{i,j}$$ are: $$\begin{matrix} i\backslash j &13 &20 &52 &65 &68 &76 &101 &106 &132 &142 &145\\ \hline 1 &43 &16 &38 &13 &27 &19 &26 &41 &31 &33 &43 \\ 13 &&33 &5 &56 &16 &52 &35 &10 &39 &16 &13 \\ 20 &&&28 &29 &17 &31 &26 &31 &31 &23 &33 \\ 52 &&&&51 &11 &47 &30 &5 &34 &11 &8 \\ 65 &&&&&40 &6 &22 &51 &19 &41 &51 \\ 68 &&&&&&39 &22 &14 &26 &15 &17 \\ 76 &&&&&&&22 &46 &13 &36 &46 \\ 101 &&&&&&&&29 &11 &19 &29 \\ 106 &&&&&&&&&33 &10 &5 \\ 132 &&&&&&&&&&23 &33 \\ 142 &&&&&&&&&&&10 \end{matrix}$$ Now add a dummy node 146 that is adjacent only to the start point 76 and end point 106, with $$d_{76,146} = d_{106,146} = 0$$. This is a standard trick to convert a TSP path problem with specified start and end points into a TSP tour problem. Finally, solve the TSP on these 13 nodes, yielding optimal tour with edges $$(1,20),(20,68),(13,68),(13,52),(52,142),(142,145),(106,145),(106,146),(76,146),(76,132),(101,132),(65,101),(1,65)$$ and total cost $$139$$, as found by @DanielMathias. Here is a plot, omitting the two dummy edges: • I needed some time to understand this solution, I can say that is the best not only for effectiveness, but also for explanation, for this reason I decided to accept your answer, congratulations :) Commented May 4, 2020 at 11:31 I got to this point after 5 attempts. My score is: 156 steps Here's my route: • It's a nice attempt, however at the moment I'm looking for the best solution, I want to see if it is possible to find a more effective one Commented May 2, 2020 at 10:15 • This seems to be the Travelling Salesman problem, and as such, it's NP-hard, so I don't think anything less than a full brute-force solution can provide a conclusive answer. – Bass Commented May 2, 2020 at 10:54 • @Bass That's not accurate. The Travelling Salesman requires a return to home. This can be modelled as a Minimum Cost Flow Commented May 2, 2020 at 13:46 Score: 139 Route: . . . D x x x . . . . . . . . . . . H \| #-A= 13 . . . x . . x . E . . . . . . . . . x \| A-B= 11 . . . x . . x x x . . . . . . . . . x \| B-C= 22 . . x x . . . . x . . . . . . . G x x \| C-D= 13 . . x . . . . . x . . . . . . . x . . \| D-E= 16 . . C . . . . . x . . F x x x x x . . \| E-F= 17 # . x x x x . . x . . x . . x x x . . \| F-G= 11 x . . . . x . . x x x x . . x . . . . \| G-H= 5 x . . x x x B . . . . . . x x . x # . \| H-I= 16 x . . x . . . . . . . . . x . . x . . \| I-J= 10 x . . x . . . . . . . x x x x x x . . \| J-#= 5 x A x x . . . . . . . I . . . . x x J \| #-#=139 • The text rendering should be sufficient, but I would prefer to have an image here... hint hint Commented May 2, 2020 at 12:42 • After a few minutes, I can say that the path is correct and it's actually a better solution. At the moment, I'm anyway open to better answers Commented May 3, 2020 at 13:21	1,470	4,572	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.984375	4	CC-MAIN-2024-26	latest	en	0.904541
https://byjusexamprep.com/bank-exam/short-tricks-calculate-cube-roots-quickly	1,717,074,155,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00247.warc.gz	123,085,889	32,192	# Calculation Short tricks: Calculate cube roots quickly! By BYJU'S Exam Prep Updated on: September 25th, 2023 Best tricks to calculate cube roots: In the Numerical Ability section of Banking exams, calculation plays an important role to score high. Now, to strengthen it, you have to practice a lot. Practising questions will increase your speed as well as accuracy. Speed is also dependent on short tricks as we use them in the exam to save time for other questions. Best tricks to calculate cube roots: In the Numerical Ability section of Banking exams, calculation plays an important role to score high. Now, to strengthen it, you have to practice a lot. Practising questions will increase your speed as well as accuracy. Speed is also dependent on short tricks as we use them in the exam to save time for other questions. In the calculation, you must know the tables till 25 at your fingertips. With this, learning squares and cubes till 25, square roots till 20 and cube roots till 10 is must. Remembering them will enhance your calculation skills. We have already shared the calculation short tricks of a square, cubes, and square roots. You can go through the tricks and prepare them well. ## Attempt Quant Quiz Here In the article, we will be discussing the short tricks to calculate the cube roots of the definite cube numbers. For this, you just have to remember the cubes till 10 to apply this technique. ### Cubes of the numbers (1 to 10) Numbers Cubes 1 1 2 8 3 27 4 64 5 125 6 216 7 343 8 512 9 729 10 1000 Now, do you analyze anything from the above table? Observation 1: Unit digit of the numbers and their cubes is similar 1 1 4 6 4 5 12 5 6 21 6 9 72 9 10 100 0 So, always remember that, if you find a number who is a definite cube and carries the above-mentioned unit digits, then their cube roots will also have the same unit digits. Observation 2: Numbers having the unit digit 2 will have 8 as the unit digit in the cube and vice versa Numbers having the unit digit 3 will have 7 as the unit digit in the cube and vice versa 2 8 3 2 7 8 51 2 7 34 3 Thus, keep in mind that if you find a number who is a definite cube and carries the unit digits as: a. 2, then cube root will have unit digit as 8. b. 8, then cube root will have unit digit as 2. Remember the conclusions of the above observations. They will be used in the steps of calculating the cube roots. ### Method to calculate cube roots of definite Cube numbers Let us discuss through an example: Find the cube root of a number 59319. Step 1: Divide the number by taking 3 rightmost digits in one part and remaining digits in another part. Then, from the unit or last digit of the cubefind the unit digit of cube root number. Here the last digit of the cube is 9. Now, as per the above observation, the numbers having the unit digit 9 have the same unit digit in their cubes and vice versa. So, the unit digit of cube root will also be 9. Step 2: Now consider the remaining digits of the number. Then find the smaller cube number to the remaining digit and write its cube root number. Here, remaining digits: 59. Smaller cube to 59 is 27. Cube root of 27 is 3. So, cube root of 59319 is 39. So, with the help of just two simple steps, you can find the cube root of the number. Cube root of 59319 59 31 9 27 or (3)3 < 59 9 3 9 39 Let`s have some examples to understand it better. Example 1: Find the cube root of 91125. Cube root of 91125 91 12 5 64 or (4)3 < 91 5 4 5 45 Example 2: Find the cube root of 175616 Cube root of 175616 175 61 6 125 or (5)3 < 175 6 5 6 56 Example 3: Find the cube root of 300763 Cube root of 300763 300 76 3 216 or (6)3 < 300 7 6 7 67 Example 4: Find the cube root of 474552 Cube root of 474552 474 55 2 343 or (7)3 < 474 8 7 8 78 Example 5: Find the cube root of 778688 Cube root of 778688 778 688 729 or (9)3 < 778 2 9 2 92 In this way, you can calculate the cube roots of the definite cube numbers easily. You don’t have to write all the steps in the exam. Just calculate the steps in your mind and answer quickly. In the upcoming articles, we will be sharing the short tricks of some other topics. Stay connected and keep going with practice. Note: Using this technique, you can find only 2 digit cube root numbers. If you are aiming to crack SBI/IBPS Exams 2023, then join the Online Classroom Program where you will get: • Live Courses by Top Faculty • Daily Study Plan • Comprehensive Study Material • Latest Pattern Test Series • Complete Doubt Resolution • Regular Assessments with Report Card ### Click here to access Bank Test Series Why BYJU’S Exam Prep Test Series? • Get better at every level of your preparation with unlimited test series, based on the latest exam pattern. • Assess areas of weaknesses & track improvements with in-depth performance analysis Download the BYJU’S Exam Prep App Now. The most comprehensive exam prep app. #DreamStriveSucceed POPULAR EXAMS SSC and Bank Other Exams GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 help@byjusexamprep.com	1,375	5,113	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.5	4	CC-MAIN-2024-22	latest	en	0.918926
https://www.traditionaloven.com/metal/precious-metals/platinum/convert-troy-ounce-tr-oz-platinum-to-ton-short-sh-tn.html	1,544,598,641,000,000,000	text/html	crawl-data/CC-MAIN-2018-51/segments/1544376823785.24/warc/CC-MAIN-20181212065445-20181212090945-00591.warc.gz	1,109,889,607	13,962	Platinum ounce (troy) to short tons of platinum converter # platinum conversion ## Amount: ounce (troy) (oz t) of platinum mass Equals: 0.000034 short tons (sh tn) in platinum mass Calculate short tons of platinum per ounce (troy) unit. The platinum converter. TOGGLE : from short tons into troy ounces in the other way around. ### Enter a New ounce (troy) Amount of platinum to Convert From * Enter whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8) ## platinum from ounce (troy) to ton (short) Conversion Results : Amount : ounce (troy) (oz t) of platinum Equals: 0.000034 short tons (sh tn) in platinum Fractions: 4/117647 short tons (sh tn) in platinum CONVERT : between other platinum measuring units - complete list. ## Platinum Amounts (solid platinum) Here the calculator is for platinum amounts (solid platinum volume; dense, precious, gray to white metal rare in abundance on the planet earth. Its annual production is only a very few hundred tons. It is a very highly valuable metal. Platinum performs real well in resisting corrosion. Not only beautiful jewellery is made out of platinum, this metal enjoys quite a wide variety of uses. For instance in electronics, chemical industries and also in chemotherapy applications against certain cancers. Traders invest money in platinum on commodity markets, in commodity future trading as this material is also one of the major precious commodity metals. Thinking of going into investing in stocks? It would be a wise idea to start learning at least basics at a commodity trading school first, to get used to the markets, then start with small investments. Only after sell and buy platinum.) Is it possible to manage numerous units calculations, in relation to how heavy other volumes of platinum are, all on one page? The all in one Pt multiunit calculation tool makes it possible to manage just that. Convert platinum measuring units between ounce (troy) (oz t) and short tons (sh tn) of platinum but in the other direction from short tons into troy ounces. conversion result for platinum: From Symbol Equals Result To Symbol 1 ounce (troy) oz t = 0.000034 short tons sh tn # Precious metals: platinum conversion This online platinum from oz t into sh tn (precious metal) converter is a handy tool not just for certified or experienced professionals. It can help when selling scrap metals for recycling. ## Other applications of this platinum calculator are ... With the above mentioned units calculating service it provides, this platinum converter proved to be useful also as a teaching tool: 1. in practicing troy ounces and short tons ( oz t vs. sh tn ) exchange. 2. for conversion factors training exercises with converting mass/weights units vs. liquid/fluid volume units measures. 3. work with platinum's density values including other physical properties this metal has. International unit symbols for these two platinum measurements are: Abbreviation or prefix ( abbr. short brevis ), unit symbol, for ounce (troy) is: oz t Abbreviation or prefix ( abbr. ) brevis - short unit symbol for ton (short) is: sh tn ### One ounce (troy) of platinum converted to ton (short) equals to 0.000034 sh tn How many short tons of platinum are in 1 ounce (troy)? The answer is: The change of 1 oz t ( ounce (troy) ) unit of a platinum amount equals = to 0.000034 sh tn ( ton (short) ) as the equivalent measure for the same platinum type. In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solutions. Subjects of high economic value such as stocks, foreign exchange market and various units in precious metals trading, money, financing ( to list just several of all kinds of investments ), are way too important. Different matters seek an accurate financial advice first, with a plan. Especially precise prices-versus-sizes of platinum can have a crucial/pivotal role in investments. If there is an exact known measure in oz t - troy ounces for platinum amount, the rule is that the ounce (troy) number gets converted into sh tn - short tons or any other unit of platinum absolutely exactly. It's like an insurance for a trader or investor who is buying. And a saving calculator for having a peace of mind by knowing more about the quantity of e.g. how much industrial commodities is being bought well before it is payed for. It is also a part of savings to my superannuation funds. "Super funds" as we call them in this country. Conversion for how many short tons ( sh tn ) of platinum are contained in a ounce (troy) ( 1 oz t ). Or, how much in short tons of platinum is in 1 ounce (troy)? To link to this platinum - ounce (troy) to short tons online precious metal converter for the answer, simply cut and paste the following. The link to this tool will appear as: platinum from ounce (troy) (oz t) to short tons (sh tn) metal conversion. I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.	1,155	5,208	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.8125	3	CC-MAIN-2018-51	longest	en	0.879493
https://5y1.org/document/r-remove-columns-data-frame.html	1,623,777,412,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623487621450.29/warc/CC-MAIN-20210615145601-20210615175601-00233.warc.gz	95,611,550	4,294	# R remove columns data frame • 8,976Results • r remove columns data frame • ##### BASICS OF R https://5y1.org/info/r-remove-columns-data-frame_1_e22627.htmlDOCX File R is an “object oriented” interactive language. I am sure all the CSCE majors reading this can give me a better definition of the term “object oriented” (and I hope you will), but in my non-CSCE way of thinking it means that data in R are organized into specific structures called “objects”, of different types, and these objects share similar qualities, or “attributes”. delete columns • ##### Corrected R code from chapter 12 of the book https://5y1.org/info/r-remove-columns-data-frame_1_29ffdd.htmlDOCX File # Designate samples as outlying if their Z.k value is below the thresholdthresholdZ.k = -5 # often -2.5# the color vector indicates outlyingness (red)outlierColor = ifelse(Z.k < thresholdZ.k, "red", "black")# calculate the cluster tree using flahsClust or hclustsampleTree = flashClust(as.dist(1 - A), method = "average")# Convert traits to a ... columns dataframe • ##### Introduction to ordination (PCA) in R https://5y1.org/info/r-remove-columns-data-frame_1_b9a6e8.htmlDOCX File This leaves us with a data frame with 32 rows and 15 columns. In total, there are 16 measurements from samples of ‘Het1’ and 16 measurements from samples of ‘Het2’. Because PCA requires matrix manipulations, we need the R object holding the data to be a matrix. To do this, we can get rid of the first four columns of data. column • ##### R arithmetic operations - Boston University https://5y1.org/info/r-remove-columns-data-frame_1_b5e525.htmlDOCX File Remove objects from the current workspace. list.files List files in the current directory. ... - a set of elements organized in rows and columns, where columns can be of different types. List ... family column • ##### R code from chapter 8 of the book - Horvath Lab UCLA https://5y1.org/info/r-remove-columns-data-frame_1_bf8e7a.htmlDOC File R code from chapter 12 of the book. Horvath S (2011) Weighted Network Analysis. Applications in Genomics and Systems Biology. Springer Book. ISBN: 978-1-4419-8818-8 remove dataframe • ##### R notes for BIOL 7083, Community Ecology, Fall 2003 https://5y1.org/info/r-remove-columns-data-frame_1_158832.htmlDOC File Function read.table() reads the contents of a data set and creates a data frame from those contents. A data frame is essentially a matrix containing a combination of numeric variables and character strings. Use either “/” or “\\” in the path to the file. The argument “header=T” is used for a data set containing labeled columns (a ... #### To fulfill the demand for quickly locating and searching documents. It is intelligent file search solution for home and business.	689	2,769	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.5625	3	CC-MAIN-2021-25	latest	en	0.753863
https://www.jiskha.com/search?query=The+ages+in+years+of+the+houses+on+Kelly%E2%80%99s+block+are+20%2C+26%2C+11%2C+2%2C+28%2C+18%2C+2%2C+and+27.+Find+the+mean+age+of+the+houses.+Find+the+median+age.+Find+the+mode.+Which+measure%28s%29+would+you+use+to&page=93	1,547,719,174,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547583658901.41/warc/CC-MAIN-20190117082302-20190117104302-00514.warc.gz	834,541,449	16,433	# The ages in years of the houses on Kelly’s block are 20, 26, 11, 2, 28, 18, 2, and 27. Find the mean age of the houses. Find the median age. Find the mode. Which measure(s) would you use to 21,153 results, page 93 1. ## psych 21(1 pts.) Based on the incidence of Fetal Alcohol Syndrome in various ethnic groups, which of the following individuals would be considered high risk? A) a woman of French Canadian descent B) a woman of Irish descent C) a woman of Italian descent D) a asked by Bryan on November 27, 2012 2. ## Math I need help with these few questions on my homework please :) 1. How much money would you need to pay to receive a payout annuity of \$8,503.05 annually for 10 years, assuming your money earns 7.5% compounded annually? Assume that your payments increase asked by Macy on July 2, 2015 3. ## Math I need help with these few questions on my homework please :) 1. How much money would you need to pay to receive a payout annuity of \$8,503.05 annually for 10 years, assuming your money earns 7.5% compounded annually? Assume that your payments increase asked by Macy on July 2, 2015 4. ## MATH AUNT ALICE GAVE EACH OF HER 3 NIECES A NUMBER OF SILVER DOLLARS EQUAL TO THEIR AGES. THE YOUNGEST FELT THAT THIS WAS UNFAIR. THEY AGREED TO REDISTRIBUTE THE MONEY. THE YOUNGEST WOULD SPLIT HALF HER SILVER DOLLARS EVENLY WITH HER SISTERS.THE MIDDLE SISTER asked by mike on October 12, 2011 5. ## math Aunt Alice gave each of her three nieces a number of silver dollars equal to their ages.The youngest felt that this was unfair. They agreed to redistribute the money. The youngest would split half her silver dollars evenly with the other sisters. The asked by mike on October 11, 2011 6. ## English Can you check the grammar in the following sentences, please? 1)In order to reach the branch Lockwood pushed his hand through the window , but an ice-cold hand grabbed his fingers. In vain he tried to release himself from its tenacious grisp, which was asked by Franco on March 4, 2010 7. ## math Find the linear block code generator matrix G, if the code generator polynomial is g(x) = 1 + x^3 for a (7,4) code. asked by Josh on February 16, 2011 8. ## Math If the code generator polynomial is g(x) = 1 + x^2 for a (5,3) code, find the linear block code generator matrix G. asked by Josh on February 16, 2011 1. According to the National Clearing House for Alcohol and Drug Information (NCADI), most children under the age of _____ , who do not participate in drug and alcohol abuse, is virtually certain to never do so. 20 21** 23 19 2. The average age when asked by GOY on February 26, 2016 10. ## Statistics Could someone please help me with this question. I did the entire question, but then when I was assigning them to the two different types of treatments, I was having some trouble. I just figured that you could just number the females 1-42, and randomly asked by Justin on April 3, 2007 11. ## History Can someone help me out here? Quote: “He has abdicated Government here, by declaring us out of his Protection and waging War against us. He has plundered our seas, ravaged our Coasts, burnt our towns, and destroyed the lives of our people. He is at this asked by Acasia on April 19, 2017 12. ## accounting I continue getting the wrong totals can you please help me solve this problem and give me the correct formula and Journal entries. On Jan 1 2004 Brown Airways purchased an airplane for 34,450,000. 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I did this question and i got 73080J but asked by Arman on February 20, 2015 16. ## Physics, Energy The block is released from rest on the frictionless ramp. It compresses the spring by 20cm before momentarily coming to rest. What is the initial distance d? k=200N/m m=3kg angle=300 degrees The picture looks somewhat like this : (box is here) / / (spring asked by bobby on March 27, 2012 17. ## philosophy Beautiful harmony comes from things that are in opposition to one another [Heraclitus ok i need to write my personal philosophy and i cant think of any examples of this quote for my body pargraphs...except velcrow..i don't think that a good one thought asked by chels on December 13, 2007 18. ## mechanics A 2kg box starts from rest and slides down an incline as shown in the picture above. If the block loses 24 Joules of energy due to friction as it slides down the ramp, what is the speed of the box as it reaches the bottom of the ramp? If needed, g=10ms2 asked by juanpro on July 6, 2014 19. ## science (1) what measurements and what calculation would you need to make in order to determine the density of a rectangular wooden block? (2) a student announced that she had made a sample of a new material that had a density of 0.85g/cm3. from the inforation asked by roy on March 30, 2011 20. ## Finance Rollincoast Incorporated issued BBB bonds two years ago that provided a yield to maturity of 11.5%. Long-term risk-free government bonds were yielding 8.7% at that time. The current risk premium on BBB bonds versus government bonds is half of what it was asked by Ms. Douglas on October 14, 2009 21. ## math This problem has to do with exponential models. The question says, you deposit \$1600 in a bank account. Find the balance after 3 years for each of the following situations. The first one says: 1. 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A)particles are soluble B)particles settle out over time C)particles can block light D)particles scatter light asked by Katherine on November 17, 2009 25. ## Science A 0.05 kg arrow with the velocity of 180 m/s is used to shoot a 5 kg block of wood off the top of a log. What will be the velocity of the arrow and wood after contact? asked by bri on September 8, 2012 26. ## Physics A block and tackle with a velocity ratio of 5 is used to raise a mass of 25kg through a vertical distance of 400M at a steady rate.if the effort is equal to 60N,determine (a)distance moved by the effort (b)work done by lifting the load(c)the loss in energy asked by Wisdom on February 4, 2015 27. ## Physics Two blocks A and B are connected by a light in extensible cord passing over a frictionless pulley. Block A starts from rest and moves up the smooth plane which is inclined at 30 degrees to the horizontal. Calculate at the moment that A has moved 4 m along asked by Ariana on June 1, 2014 Sara is 1/4 the age of her father. Sara's father is 60. How many years older than Sara is Sara's father? _____ How many years ago was Sara 1/10 of her father's age? _____ How old was she then? ____ asked by Pamela on November 16, 2007 29. ## Algebra Henry's mother is three times as old as he is. His grandmother is 35 years older than his mother. If Henry's grandmother was 7 years younger, she would be 5 times as old as Henry. How old is Henry's grandmother? asked by Kloe on October 30, 2016 30. ## macroeconomics 23. The next four questions refer to the following price and output data over a five-year period for an economy that produces only one good. Assume that year 2 is the base year. Units of Price Year output per unit 1 16 \$2 2 20 3 3 30 4 4 36 5 5 40 6 (a) If asked by marc on October 11, 2011 31. ## Spanish What is the main difference between the Pan American Games and the Olympic Games? a. The Pan American games are held every 6 years, while the Olympic Games are held every 4 years. b. The Pan American Games are held in Buenos Aires, Argentina, while the asked by Kaai97 on January 20, 2016 32. ## Math A consumer product magazine recently ran a story concerning the increasing prices of digital cameras. The story stated that digital camera prices dipped a couple of years ago, but now are beginning to increase in price because of added features. According asked by Anonymous on March 26, 2018 33. ## Managerial Finance Jefferson Products, Inc., is considering purchasing a new automatic press break, which costs \$300,000 including installation and shipping. The machine is expected to generate net cash inflows of \$80,000 per year for 10 years. At the end of 10 years, the asked by Gary on March 10, 2010 34. ## Math Help From Anybody 1 Henry spends the day at a ski resort. The total trip riding up the main chair lift and skiing down his favorite trail takes 40 to 50 minutes, depending on whether Henry takes a rest along the trail. He gets on the first chair lift at 8:00 am, he takes a asked by SmartyPants on March 8, 2015 35. ## Math Ms.Sue 1 Henry spends the day at a ski resort. The total trip riding up the main chair lift and skiing down his favorite trail takes 40 to 50 minutes, depending on whether Henry takes a rest along the trail. He gets on the first chair lift at 8:00 am, he takes a asked by SmartyPants on March 8, 2015 36. ## MATH Five years ago, you bought a house for \$151,000, with a down payment of \$30,000, which meant you took out a loan for \$121,000. Your interest rate was 5.75% fixed. You would like to pay more on your loan. You check your bank statement and find the following asked by TRAY on August 12, 2012 37. ## Physics A block of mass M=6 kg and initial velocity v=0.8m/s slides on a frictionless horizontal surface and collides with a relaxed spring of unknown spring constant. The other end of the spring is attached to a wall. If the maximum compression of the spring is asked by Helena Beau on July 21, 2013 38. ## Finance 1.Tank Industries Washers expects to pay the following dividends over the next 4 years: \$2.50, \$3.20, \$4.75 and \$5.20 respectively (starting at time 1). a.After year 4, the firm expects a constant growth rate of 3%. If investors require 11%, what is the asked by nick on September 19, 2012 39. ## Accounting Herring Wholesale Company has a defined benefit pension plan. On January 1, 2016, the following pension related data were available: (\$ in 000s) Net gain–AOCI \$ 350 Accumulated benefit obligation 3,170 Projected benefit obligation 3,200 Fair value of asked by Ben on December 1, 2016 40. ## physics Two blocks are connected by a massless rope that passes over a pulley. M1= 2 kg and M2= 4 kg. The pulley is 12cm in diameter and has a mass of 2kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.5 Nm. if the blocks are released asked by Blanche on December 1, 2010 41. ## English III Read this sentence from Abraham Lincoln's Second Inaugural Address: Yet, if God wills that it continue until all the wealth piled by the bondsman's two hundred and fifty years of unrequited toil shall be sunk, and until every drop of blood drawn with the asked by Halo4Records on May 25, 2014 42. ## Physical Education In the years of Physical Education, Does it helps STESS? YES or NO? 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I need to know if you both agree to include students from years 10/11. asked by Henry2 on November 18, 2011 58. ## English Is it okay to tell in the first paragraph of a cover letter who I am and how many years of education I have? asked by Anonymous on December 7, 2015 59. ## Math Benson is 2 years less than twice Jackson age and all together equal 19 asked by Ana on January 3, 2017 60. ## FINANICIAL How much should be invested now at 7.55% compounded annually to have \$46,000 in 12 years? asked by Anonymous on December 2, 2016 61. ## math The C.I on Rs.16,000 for 1½ years at 10% p.a. payable half-yearly is: asked by Haroon Gondal on November 9, 2015 62. ## MATH How much should be invested now at 7.55% compounded annually to have \$46,000 in 12 years? asked by RENTO on December 2, 2016 63. ## Math John is eight year older than his sister.In three years,he will be as twice as she will be then. How old are they both now? 64. ## grammar This rash has been present for approximately (two) years. two is that an an adjective? asked by tina on July 5, 2009 Erin's age is 3 times Warren's. In 4 years she will be twice as old as he will be. How old is each now? asked by Leroy on April 17, 2009 66. ## Algebra 1 Max is 5 years older than Pauline. Next year he will be twice as old as she will be. 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He is 3 years older than sam and twice as old as jack asked by debeye on October 3, 2016 77. ## Years ? "In the early 1930s" what years are those like 1900-1930 right? asked by Anon on April 9, 2013 78. ## Science (bobpursley) Why is an extra day added to February every four years? asked by Anonymous on May 15, 2013 79. ## Math Write an expression for the number of years in t months. asked by Dessy on September 1, 2016 \$11,000, invested for 9 years at 3% compounded quarterly. asked by Ernie on April 23, 2014 asked by ashley on February 5, 2012 82. ## Math Which expression represents how many years are in x months? A) 12x B) 12 + x C) x/12 D) 12/x* asked by Daya on August 30, 2017 83. ## math what is the simple annual interest of 2% of \$500 after 6 years asked by Anonymous on April 8, 2011 84. ## finance \$11,000, invested for 9 years at 3% compounded quarterly. asked by Ernie on April 23, 2014 85. ## Math Which expression represents how many years are in x months? 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https://de.mathworks.com/matlabcentral/cody/problems/198-are-you-in-or-are-you-out/solutions/1220463	1,610,836,214,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703507045.10/warc/CC-MAIN-20210116195918-20210116225918-00284.warc.gz	306,392,640	17,037	Cody # Problem 198. Are you in or are you out? Solution 1220463 Submitted on 26 Jun 2017 by HH This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1 Pass xv = [0 1 0]; yv = [0 0 1]; X = 0.8; Y = 0.8; tf_correct = false; assert(isequal(inside(xv,yv,X,Y),tf_correct)) 2 Pass xv = [0 1 1 0]; yv = [0 0 1 1]; X = 0.5; Y = 0.5; tf_correct = true; assert(isequal(inside(xv,yv,X,Y),tf_correct)) 3 Pass xv = [0 1 1 0]; yv = [0 0 1 1]; X = 0.5; Y = 0.5; tf_correct = true; assert(isequal(inside(xv,yv,X,Y),tf_correct)) 4 Pass xv = [0 0.25 0.25 0]; yv = [0 0 1 1]; X = 0.5; Y = 0.5; tf_correct = false; assert(isequal(inside(xv,yv,X,Y),tf_correct)) 5 Pass xv = [0 0.25 0.25 0] + 1000; yv = [0 0 1 1] + 1000; X = 1000.1; Y = 1000.1; tf_correct = true; assert(isequal(inside(xv,yv,X,Y),tf_correct)) ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting!	422	1,038	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.046875	3	CC-MAIN-2021-04	latest	en	0.610395
https://www.jiskha.com/display.cgi?id=1401549684	1,527,031,561,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794864999.62/warc/CC-MAIN-20180522225116-20180523005116-00243.warc.gz	781,505,236	3,867	# math posted by John gail borrowed \$3600 from a bank to buy a washing machine she pays back the loan in 24 monthly instalments at a simple interest rate of 2% a)calculate the simple interest on the loan b)the total amount to be repaid c)if gail repays the loan in 24 equal instalments ,find the sum she pays for each instalments (hint interest is included in the instalment) 1. Ms. Sue I = PRT I = 3600 * 0.02 * 2 I = 144 (3600 + 144)/24 = ? ## Similar Questions 1. ### math the total amount of interest on this loan of \$6000 for 150 days is \$210.50. what is the rate of interest on this loan? 2. ### math you borrow \$1200 from a bank that bank charges 9.5% simple annual interest. after 15 months you pay back the loan. how much interest do you pay on the loan? 3. ### Survey of mathematics On January 5, Ebony Davis borrowed \$6,500 on a simple interest loan from a lending institution to finance her catering business. She borrows the money at a rate of 8.5% with a term ending on December 9. a. Calculate Ebony's interest … On January 5, Ebony Davis borrowed \$6,500 on a simple interest loan from a lending institution to finance her catering business. She borrows the money at a rate of 8.5% with a term ending on December 9. a. Calculate Ebony's interest … 5. ### Survey of Mathematics On January 5, Ebony Davis borrowed \$6,500 on a simple interest loan from a lending institution to finance her catering business. She borrows the money at a rate of 8.5% with a term ending on December 9. a. Calculate Ebony's interest … 6. ### Math 1. Rishi ram obtained an installment loan for \$3,000.00. He agreed to repay the loan in 6 monthly payments. His monthly payments is \$516.50. What is the APR? 7. ### Math Suppose you borrowed \$25,000 for a car at an APR of 8%, which you are paying off with monthly payments of \$510 for 5 years. a) Whatâ€™s the loan principal? 8. ### Business Math icom 1 Find the ordinary interest in a loan of Rs.2600 at 10 % simple interest from January 5 to February 10.? 9. ### Math - Anurag is buying a house for \$100,000 and needs a mortgage loan. Bank A wants him to repay the loan in 240 months with monthly payment of \$600.72. Bank B wants him to repay the loan in 360 months with a monthly payment of 389.94. … 10. ### Maths Calculate the total amount (loan and interest)to be paid on a loan of N1 000.00, borrowed for 2 years at a simple interest rate of 10% p.a. More Similar Questions	657	2,452	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.1875	3	CC-MAIN-2018-22	latest	en	0.942347
https://www.hsslive.co.in/2021/10/1-gaj-in-bigha-madhya-pradesh.html	1,716,296,716,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00717.warc.gz	700,417,407	40,760	# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board ## Thursday, October 7, 2021 In this article, you will learn how to convert 1 Gaj in Bigha in Madhya Pradesh. Gaj is one of the most commonly used measurement units used in almost every state of India. People in Indian states refer to gaj for measurement of land. Here in this page you will learn how to calculate 1 Gaj in Bigha unit. ## How to Convert from 1 Gaj to Bigha? You can easily convert gaj to Bigha or the reverse with a simple method. You can multiply the figure in gaj by 0.000743 to determine the Bigha value. This is all you need to do for undertaking the conversion procedure of Gaj to Bigha in Madhya Pradesh. ## Relationship between Gaj and Bigha It is not difficult to work out the actual relationship between gaj to Bigha in most cases. You should know the proper calculations existing between gaj to Bigha before you venture to conversion procedures. 1 Gaj to Bigha calculations will be 0.000743 Bigha. Once you know these calculations, it will not be hard for you to calculate gaj in an Bigha. One gaj can be worked out to various other units as well. ## Formula for Converting Gaj to Bigha The formula to convert Gaj to Bigha is the following- 1 Gaj = 0.000743 Bigha Conversely, you can also use this formula- Gaj = Bigha * 0.000743 This is all you need to know for undertaking the conversion procedure on your part. You can also use online calculators or conversion tools for this purpose.	385	1,498	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.03125	3	CC-MAIN-2024-22	latest	en	0.901705
https://numberworld.info/1312101	1,721,365,348,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00791.warc.gz	374,411,174	4,045	# Number 1312101 ### Properties of number 1312101 Cross Sum: Factorization: 3 * 3 * 7 * 59 * 353 Divisors: Count of divisors: Sum of divisors: Prime number? No Fibonacci number? No Bell Number? No Catalan Number? No Base 2 (Binary): Base 3 (Ternary): Base 4 (Quaternary): Base 5 (Quintal): Base 8 (Octal): Base 32: 181b5 sin(1312101) 0.77057003552671 cos(1312101) -0.63735533287827 tan(1312101) -1.2090116702199 ln(1312101) 14.087140227231 lg(1312101) 6.1179672664835 sqrt(1312101) 1145.4697726261 Square(1312101) ### Number Look Up Look Up 1312101 which is pronounced (one million three hundred twelve thousand one hundred one) is a amazing figure. The cross sum of 1312101 is 9. If you factorisate the number 1312101 you will get these result 3 * 3 * 7 * 59 * 353. The number 1312101 has 24 divisors ( 1, 3, 7, 9, 21, 59, 63, 177, 353, 413, 531, 1059, 1239, 2471, 3177, 3717, 7413, 20827, 22239, 62481, 145789, 187443, 437367, 1312101 ) whith a sum of 2208960. The figure 1312101 is not a prime number. The number 1312101 is not a fibonacci number. The number 1312101 is not a Bell Number. The number 1312101 is not a Catalan Number. The convertion of 1312101 to base 2 (Binary) is 101000000010101100101. The convertion of 1312101 to base 3 (Ternary) is 2110122212100. The convertion of 1312101 to base 4 (Quaternary) is 11000111211. The convertion of 1312101 to base 5 (Quintal) is 313441401. The convertion of 1312101 to base 8 (Octal) is 5002545. The convertion of 1312101 to base 16 (Hexadecimal) is 140565. The convertion of 1312101 to base 32 is 181b5. The sine of the figure 1312101 is 0.77057003552671. The cosine of the figure 1312101 is -0.63735533287827. The tangent of 1312101 is -1.2090116702199. The square root of 1312101 is 1145.4697726261. If you square 1312101 you will get the following result 1721609034201. The natural logarithm of 1312101 is 14.087140227231 and the decimal logarithm is 6.1179672664835. I hope that you now know that 1312101 is special number!	737	1,989	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.390625	3	CC-MAIN-2024-30	latest	en	0.729377
https://www.programminglogic.com/504/	1,571,342,894,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570986676227.57/warc/CC-MAIN-20191017200101-20191017223601-00518.warc.gz	1,080,066,552	7,048	# Solution to Problem 23 on Project Euler A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number. A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n. As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit. Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers. My solution: ``````#include <stdio.h> #define ABUN 6965 #define NUMBER 28124 /* testa pra ver se n eh abundante/ int isAbu(int n){ int i,sum; sum = 0; for (i=1;i<(n/2+1);i++){ if (n%i==0) sum+=i; } if (sum>n) return 1; else return 0; } / busca linear de "value" no array dado / int valueSearch(int value,int array[27000]){ int x; for (x=0;x<27000;x++){ if (array[x]==0) return 0; else if (value == array[x]) return 1; } return 0; } int main(){ int i,j,abusum,index,z,total; int abundant [ABUN]; int absums[27000]; for (i=0;i<27000;i++) absums[i]=0; / enche o vetor "abundant" com todos os numeros abundantes menores que 28124 / j=0; for (i=1;i<NUMBER;i++){ if (isAbu(i)){ abundant[j]=i; j++; } } / enche o vetor "absums" com todos os inteiros que sao soma de 2 abundantes, menores que 28124 */ z=0; for (i=0;i<ABUN;i++) for (j=0;j<ABUN;j++){ abusum = abundant[i]+abundant[j]; if (abusum>=NUMBER) continue; else if (valueSearch(abusum,absums)) continue; else { absums[z]=abusum; z++; } } printf("z = %dn",z); total=0; for (i=0;i<27000;i++) total+=absums[i]; printf("total = %dn",total); return 0; }``````	649	2,102	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.84375	4	CC-MAIN-2019-43	longest	en	0.64294
https://bio.libretexts.org/Courses/Mansfield_University_of_Pennsylvania/BSC_3271%3A_Microbiology_for_Health_Sciences_Sp21_(Kagle)/15%3A_Selected_Pathogens/15.01%3A_Bacterial_Pathogens/15.1.03%3A_Gram_Negatives	1,725,786,884,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700650976.41/warc/CC-MAIN-20240908083737-20240908113737-00229.warc.gz	115,263,903	29,378	# 15.1.3: Gram Negatives $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ $$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$ $$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$ $$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vectorC}[1]{\textbf{#1}}$$ $$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$ $$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$ $$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$ $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left\|#1\right\|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$ There is a wide variety of Gram-negative pathogens. Although those described here are all found within the Proteobacteria phylum, they are spread across all the classes within this most diverse of bacterial phyla. Some can be free living under many environmental conditions as well as cause disease (Pseudomonas aeruginosa) whereas others are obligate intracellular pathogens (Rickettsia rickettsii). Almost all the common Gram-negative bacteria are rod-shaped (bacilli). The only commonly encountered genus of pathogenic Gram-negative cocci is Neisseria. 15.1.3: Gram Negatives is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.	1,914	5,058	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2024-38	latest	en	0.194583
https://cs50.stackexchange.com/questions/24918/pset3-helpers-c-help/24922	1,718,686,766,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861747.46/warc/CC-MAIN-20240618042809-20240618072809-00706.warc.gz	156,963,589	24,355	# pset3 helpers.c (help) :( finds 42 in {39,40,41,42} \ expected an exit code of 0, not 1 ( it meaning I missing or have a return error??? ) `````` bool search(int value, int values[], int n) {// start of search code // TODO: implement a searching algorithm if ( n <= 0) { return false; } //search for middle lists if min = target if (value == values[(n + 1) / 2]) { return true; } //number greater search left if ( value > values[(n + 1)/ 2 ]) { return search(value, &values[n / 2 + 1], n - n / 2 - 1); } /number smaller search right else if (value < values[(n + 1)/ 2 ]) { return search( value, values, n/2); } return false; }// end of search code void sort(int values[], int n) {//start of selection sort code // TODO: implement a sorting algorithm (selection sort) int min, swap; for (int i = 0; i < n - 1; i++) { min = i; for (int j = i + 1; j < n; j++) if (values[j] < values[min]) min = j; //swap number if (min != i) { swap = values[i]; values[i] = values[min]; values[min] = swap; } return; } `````` }//End of selection sort You are comparing to element `values[(n+1)/2]`, but your recursive calls look like you meant `values[n/2]` instead, which makes a lot more sense, as `n/2` is always the middle element's index, or the higher of the two middle elements' indices in case of even `n`. Imagine `n` being `1`, an array of exactly one element. You'd access `values[(1+1)/2]`, or `values[1]`, which is clearly out of range (indices are 0-based, range from 0 to n-1). In case of integers, numbers are always either equal, less than, or greater than, so you would not need that last if. Different for floating point, which knows a few special values like `NAN` (Not-a-Number, ironically stored in a number variable). I don't see any error in your selection sort (better and more consistent indentation would have made code more readable, also, the editor has a `{}` button that'll add markdown to selected code).	557	1,930	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.421875	3	CC-MAIN-2024-26	latest	en	0.660668
https://truewriters.net/sol/261588411/	1,603,558,752,000,000,000	text/html	crawl-data/CC-MAIN-2020-45/segments/1603107884322.44/warc/CC-MAIN-20201024164841-20201024194841-00272.warc.gz	579,288,716	9,880	#### Question Details ##### (Solution) - Consider again Lindsay s investment in Problem 7 The real value Brief item decscription Solution download Item details: Consider again Lindsay's investment in Problem 7. The real value of Lindsay's account after 30 years of investing will depend on inflation over that period. In the Excel function = NPV(rate, value1, value2, ...), rate is called the discount rate, and value1, value 2, etc. are incomes (positive) or expenditures (negative) over equal periods of time. Update your model from Problem 7 using the NPV function to get the net present value of Lindsay's retirement fund. Construct a data table that shows the net present value of Lindsay's retirement fund for various levels of return and inflation (discount rate). Use a data table to vary the return from 0 to 12% in increments of 1% and the discount rate from 0 to 4% in increments of 1% to show the impact on the net present value. About this question: STATUS Answered QUALITY Approved ANSWER RATING This question was answered on: Jul 11, 2017 Solution~000261588411.zip (18.37 KB) Buy this answer for only: \$15 Pay using PayPal (No PayPal account Required) or your credit card. All your purchases are securely protected by PayPal. ### Need a similar solution fast, written anew from scratch? Place your own custom order We have top-notch tutors who can help you with your essay at a reasonable cost and then you can simply use that essay as a template to build your own arguments. This we believe is a better way of understanding a problem and makes use of the efficiency of time of the student. New solution orders are original solutions and precise to your writing instruction requirements. Place a New Order using the button below.	388	1,757	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.84375	3	CC-MAIN-2020-45	latest	en	0.887708
https://questions.examside.com/past-years/jee/question/pif-i1--intlimits03pi--fcos-2xdx--wb-jee-mathematics-definite-integration-u91korf9szxqskyh	1,723,324,410,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640822309.61/warc/CC-MAIN-20240810190707-20240810220707-00425.warc.gz	381,700,289	33,915	1 WB JEE 2010 +1 -0.25 If $${I_1} = \int\limits_0^{3\pi } {f({{\cos }^2}x)dx}$$ and $${I_2} = \int\limits_0^\pi {f({{\cos }^2}x)dx}$$, then A $${I_1} = {I_2}$$ B $$3{I_1} = {I_2}$$ C $${I_1} = 3{I_2}$$ D $${I_1} = 5{I_2}$$ 2 WB JEE 2010 +1 -0.25 The value of $$I = \int\limits_{ - \pi /2}^{\pi /2} {\|\sin x\|dx}$$ is A 0 B 2 C $$-$$2 D $$-$$2 < I < 2 3 WB JEE 2010 +1 -0.25 If $$I = \int\limits_0^1 {{{dx} \over {1 + {x^{\pi /2}}}}}$$, then A $${\log _e}2 < I < \pi /4$$ B $${\log _e}2 > I$$ C $$I = \pi /4$$ D $$I = {\log _e}2$$ 4 WB JEE 2011 +1 -0.25 The value of $$\int\limits_{ - 2}^2 {(x\cos x + \sin x + 1)dx}$$ is A 2 B 0 C $$-$$2 D 4 EXAM MAP Medical NEET	414	670	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.5625	3	CC-MAIN-2024-33	latest	en	0.418984
https://eduladder.com/viewnotes/1507/10IS65-Software-testing	1,610,998,355,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703515235.25/warc/CC-MAIN-20210118185230-20210118215230-00722.warc.gz	329,890,078	17,105	We are Using Coin Mining Also To Monitise This Website. If you keep this website open for some time that helps us to keep making content for you. The Eduladder is a community of students, teachers, and programmers just interested to make you pass any exams. So we help you to solve your academic and programming questions fast. Watch related videos of your favorite subject. Connect with students from different parts of the world. See Our team Wondering how we keep quality? Got unsolved questions? You are here:Open notes-->VTU-->10IS65-Software-testing # How to study this subject UNIT 1 6 Hours A Perspective on Testing, Examples: Basic definitions, Test cases, Insights from a Venn diagram, Identifying test cases, Error and fault taxonomies, Levels of testing. Examples: Generalized pseudocode, The triangle problem, The NextDate function, The commission problem, The SATM (Simple Automatic Teller Machine) problem, The currency converter, Saturn windshield wiper. UNIT 2 7 Hours Boundary Value Testing, Equivalence Class Testing, Decision Table- Based Testing: Boundary value analysis, Robustness testing, Worst-case testing, Special value testing, Examples, Random testing, Equivalence classes, Equivalence test cases for the triangle problem, NextDate function, and the commission problem, Guidelines and observations. Decision tables, Test cases for the triangle problem, NextDate function, and the commission problem, Guidelines and observations. UNIT 3 7 Hours Path Testing, Data Flow Testing: DD paths, Test coverage metrics, Basis path testing, guidelines and observations. Definition-Use testing, Slice-based testing, Guidelines and observations. UNIT 4 6 Hours Levels of Testing, Integration Testing: Traditional view of testing levels, Alternative life-cycle models, The SATM system, Separating integration and system testing. A closer look at the SATM system, Decomposition-based, call graph-based, Path-based integrations PART – B UNIT 5 7 Hours System Testing, Interaction Testing: Threads, Basic concepts for requirements specification, Finding threads, Structural strategies and guidelines, ASF (Atomic System Functions) testing example. Context of interaction, A taxonomy of interactions, Interaction, composition, and determinism, Client/Server Testing,. UNIT 6 7 Hours Process Framework: Validation and verification, Degrees of freedom, Varieties of software. Basic principles: Sensitivity, redundancy, restriction, partition, visibility, Feedback. The quality process, Planning and monitoring, Quality goals, Dependability properties, Analysis, Testing, Improving the process, Organizational factors. UNIT 7 6 Hours Fault-Based Testing, Test Execution: Overview, Assumptions in fault- based testing, Mutation analysis, Fault-based adequacy criteria, Variations on mutation analysis. Test Execution: Overview, from test case specifications to test cases, Scaffolding, Generic versus specific scaffolding, Test oracles, Self-checks as oracles, Capture and replay. UNIT 8 6 Hours Planning and Monitoring the Process, Documenting Analysis and Test: Quality and process, Test and analysis strategies and plans, Risk planning, Monitoring the process, Improving the process, The quality team, Organizing documents, Test strategy document, Analysis and test plan, Test design specifications documents, Test and analysis reports. TEXT BOOKS: 1. Paul C. Jorgensen: Software Testing, A Craftsman’s Approach, 3 rd Edition, Auerbach Publications, 2008. (Listed topics only from Chapters 1, 2, 5, 6, 7, 9, 10, 12, 1314, 15) 2. Mauro Pezze, Michal Young: Software Testing and Analysis – Process, Principles and Techniques, Wiley India, 2008. (Listed topics only from Chapters 2, 3, 4, 16, 17, 20, 24) REFERENCE BOOKS: 1. Aditya P Mathur: Foundations of Software Testing, Pearson Education, 2008. 2. Srinivasan Desikan, Gopalaswamy Ramesh: Software testing Principles and Practices, 2 nd Edition, Pearson Education, 2007. 3. Brian Marrick: The Craft of Software Testing, Pearson Education, 1995 # Official Notes Software Testing-06IS81 e-NotesTopicSubject Matter Experts Basics of Software Testing-1 Prof.G.N.Srinivasan, RVCE, B'lore Basics of Software Testing-2 Test Generation from Requirements -1 Test Generation from Requirements -2 Structural Testing Prof.P.S.Latha, RVCE, B'lore Dependence, Data Flow Models, and Data Flow Testing Test and analysis activities within a software processIntegration and component-based software testing Test Case Selection and Adequacy, Test Execution Prof.G.S.Srinivasan, RVCE, B'lore Test and Analysis Activities within a Software Process, Software Qualities and Process # Previous year question papers #### Editors arunwebberarunwebberRajivRajiv ## Tool box Edit this note \| Upvote \| Down vote \| Questions	1,111	4,780	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.625	3	CC-MAIN-2021-04	latest	en	0.819543
mooymaehague.com	1,653,148,906,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652662539131.21/warc/CC-MAIN-20220521143241-20220521173241-00743.warc.gz	473,301,156	29,162	###### What are the Most Popular Online Casino Games? Hashing algorithms are one of the most important and fundamental types of algorithms in computer science. They were originally developed as a way to store information about very large files in a fast and efficient manner, but they have since been used for much more. Hash functions take an input value (or set of values) and create smaller representations called hashes which are used in many important areas such as cryptography, data integrity validation, database storage, distributed computing, data storage schemes and file systems. Understanding the technology The main way of making hashes is by substituting characters into some hash algorithm producing a digest or hash. This is done by hashing the input value through a one-way mapping function (such as a hash function) that maps the entire data set to produce a fixed-length output. The actual size of the output depends on the size of the data set, and hence it takes longer to compute hashes for larger sets. A hash function works by taking an input value, usually called message , and converting it into an output value called digest . There are many hashing functions available, which vary in the speed and accuracy of the resulting hash value. Some common hash functions include: Because all hashes are ultimately based on a function that maps some underlying unit of measure to another, it is useful to know how hash functions work. The idea with any one-way function is to take something and divide it into parts, and then map those parts through some transformation into a fixed-length value. The same operation is achieved by dividing the original into a set of parts, taking one part and mapping it through the function. The result is a single number according to the size of the input. The hash function is not a specific algorithm that can be used for any data type (assuming only small inputs). It will work for strings, integers and floating point numbers. The large inputs make hash-based lookups more expensive than combinations of other types, however. Also, since it cannot determine whether two expected inputs are equal or not, only one output can be produced per given pair of inputs; two different values are produced if an input has multiple entries with the same value in its key field. Hash functions are also unable to tell the difference between an input that has no output, and one with an invalid key value. Even if one could, there is nothing stopping the input from not having a key field altogether. The hashing algorithm function works by computing the hash of all inputs first and then comparing them to find a match. The following example shows the use of a hash function for searching for items in a list. The hash is computed by concatenating all the values in the list and then computing the sum of the values divided by the number of items. The comparison function finds a match if two elements have identical hashes. If no matches are found, it yields “not found”. If we want to find an item in a list, we index it using a key-value (the key field) that has been converted to a string with an associated hash value of zero. This is the same hash function used to add and remove items from the list. The following example shows an implementation of a hash function for a generic data type, where we convert values into its string representation using strip_underscores () (the result of this operation will be unique since all values are converted to a string with the same length) and then adding the constant 20 to convert it into an integer. It then uses bit operations (left shift and multiply) to make it a 32-bit unsigned integer by taking the high-order 20 bits of this value. The algorithm shown above performs identically on all computers, so no matter what computer is used to run this program, the same result will be produced. In theory, to improve security, the value is sent over a network in a form that is unreadable to anyone but the intended recipient, who will then compute its hash function. Since this relies on computation, the process can be made more secure by encrypting it. Decryption can be carried out using simple private key algorithms and public key encryption (for which one uses both private and public keys) or with advanced public key cryptography. There are also many other hashing algorithms that make use of different methods to structure a hash output: Types of Hashtag algorithm The first two versions of the MD4 and MD5 algorithms, as well as the first two versions of SHA hash, are actually examples of a family or “families” of message digest functions defined using an algorithm called the Message Digest Algorithm (MD), which is based on the cyclic redundancy check (CRC) algorithm. CRC involves chopping any bit string into random size chunks, computing a checksum for each chunk and reassembling the bits in order. Each time all the chunks have been reassembled, a CRC-32 is computed with it. This allows an attacker to use certain known strings to generate data that will always produce the same CRC-32 value. A version of MD5 is known as ANSI X9.31, and it is used in the Digital Signature Standard (DSS) for signing messages. Its output is 128 bits long—it uses 128-bits of input. Message Digest 5 (MD5) algorithm is an algorithm developed by Professor Ronald Rivest of MIT, along with teammates Ray Whitmer, Adi Shamir, and Len Adleman. It is used in a wide variety of applications today and includes many versions that have been implemented in various systems over the years. The most recent version of MD5 was published in RFC 1321. A major feature of this RFC is that it includes information on how to extend the MD5 functions to support larger blocks of data than just 512 bits (which was originally defined). The MD5 algorithm is based on the concept of message digest. A message digest value is computed from some input data, and then it can be used to verify that some other data was generated by the same source. Running a message digest against another message without knowing the initial input transforms it into a totally different value, and therefore, no meaningful hash value is produced. MD5: When computing an MD5 hash value of an arbitrary-length input, the international standard recommends using an extra 128 bits of input. The MD5 function uses 128 bits of data with any bit string that has at least 16 distinct bits (that is, our original data). Conclusion Also known as the hashtag hash, hashtags are used on social media platforms to categorize comments. Posts with a particular tag denote that they are hashtag associated. A hashtag also creates an isolated unit within a larger corpus of words, making them easily searchable. One popular use of hashtags is for finding trending topics on Twitter. For example: If I say “I’m tweeting about football today” then the result I get is mostly likely to be the most searched-for hashtag relating to football, which may not be ideal information and certainly not all that helpful! To Know More – Queenslandmax	1,419	7,103	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.5625	3	CC-MAIN-2022-21	longest	en	0.934084
https://math.answers.com/Q/What_is_the_sign_of_the_square_root_of_-1	1,660,746,149,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882572908.71/warc/CC-MAIN-20220817122626-20220817152626-00282.warc.gz	369,365,684	39,492	0 # What is the sign of the square root of -1? Wiki User 2011-04-06 16:27:14 In the same way that the square root of 25 can be -5 OR +5, the square root of -1 is -i or +i. Wiki User 2011-04-06 16:27:14 Study guides 20 cards ➡️ See all cards 3.74 1208 Reviews	101	267	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2022-33	latest	en	0.776499
https://physics.stackexchange.com/questions/101560/schr%C3%B6dingers-cat-and-the-difficulty-of-macroscopic-superposition-state?noredirect=1	1,653,017,589,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652662531352.50/warc/CC-MAIN-20220520030533-20220520060533-00566.warc.gz	515,116,259	78,717	# Schrödinger's cat and the difficulty of macroscopic superposition state The Schrödinger's cat was regarded as peculiar since we seldom encounter a superposition state in macroscopic scale: $$\mid \mathrm{dead \,\,cat} \rangle + \mid \mathrm{alive \,\, cat}\rangle$$ We more often describe an unknown cat as $$\mid \mathrm{dead \,\,cat} \rangle \langle \mathrm{dead \,\,cat} \mid+ \mid \mathrm{alive \,\, cat}\rangle \langle \mathrm{alive \,\, cat}\mid$$ without superposition. I often heard that it is difficult to prepare and maintain a large-scale superposition state. Similar difficulty also occurs in quantum computing. My question is, actually what is the reason for the difficulty to prepare and maintain a large-scale superposition state? If it is decoherence, why decoherence happens? Is that because of entropy? $$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$$ $$\renewcommand{\bra}[1]{\left \langle #1 \right \rvert}$$ We can see how decoherence really works, why it messes up superposition states, and why it's particularly prone to messing up states of large objects all through a very simple example $$^{[a]}$$. # Single two-level system Suppose we have a quantum system $$S$$ with two possible states. $$S$$ could be a cat and the states could be $$\left \lvert \text{alive} \right \rangle$$ and $$\left \lvert \text{dead} \right \rangle$$, but for the sake of generality we label the states as $$\ket{\uparrow} \quad \text{and} \quad \ket{\downarrow} \, .$$ ## Coherent case Now suppose $$S$$ is in state $$\ket{y}$$ defined as $$\ket{y} \equiv \left( \ket{\uparrow} + i \ket{\downarrow} \right) / \sqrt{2} \, .$$ This is a perfectly happy superposition state. It's density matrix is $$\rho_S = \ket{y} \bra{y} = \frac{1}{2} \left( \ket{\uparrow} \bra{\uparrow} + \ket{\downarrow} \bra{\downarrow} - i \ket{\uparrow} \bra{\downarrow} + i \ket{\downarrow} \bra{\uparrow} \right) = \frac{1}{2} \left[ \begin{array}{cc} 1 & -i \\ i & 1 \end{array} \right] = \frac{1}{2} \left( \text{Id} + \sigma_y \right) \, ,$$ where in the matrix representation we've ordered the states $$\{ \ket{\uparrow}, \ket{\downarrow} \}$$. We can think of this state as a spin pointed along the $$y$$ axis (hence the symbol $$\ket{y}$$). The first two terms are the classical terms (diagonal in the matrix representation) and the other two are the so-called "coherences" (off-diagonal) which disappear via decoherence processes, as we show below. If we were to prepare $$S$$ in state $$\ket{y}$$ many times and each time measure it along the $$z$$ axis we would get a random sequence of results where half of them are $$\uparrow$$ and half are $$\downarrow$$. Naively you might think that this means that our preparation procedure is giving us a normal "classical" probability distribution where half of the time we prepared $$\ket{\uparrow}$$ and half of the time we prepared $$\ket{\downarrow}$$. However, we can see that this is not true if we rotate $$S$$ about the $$x$$ axis and then measure it along the $$z$$ axis. The operator for the rotation is $$U = \cos(\theta / 2) \, 1 + i \sin(\theta / 2) \, \sigma_x = \left[ \begin{array}{cc} \cos(\theta/2) & i \sin(\theta/2) \\ i \sin(\theta / 2) & \cos(\theta / 2) \end{array} \right]$$ and the density matrix after the rotation is $$U \rho_S U^\dagger = \frac{1}{2} \left[ \begin{array}{cc} 1 - \sin(\theta) & -i \cos(\theta) \\ i \cos(\theta) & 1 + \sin(\theta) \end{array} \right] = \frac{1}{2} \left( \text{Id} - \sin(\theta) \sigma_z + \cos(\theta) \sigma_y \right) \, .$$ As you can see, for a given angle $$\theta$$, the probability to find the system in $$\ket{\uparrow}$$ is $$(1/2)(1 - \sin(\theta))$$, i.e. it depends on how much we rotated. Another way to say this is that $$\langle \sigma_z \rangle_{U \rho_S U^\dagger} = - \sin(\theta) \, ,$$ i.e. the expectation value of $$\sigma_z$$ oscillates as we rotate the system. This makes perfect sense if you think of the two level system as an arrow oriented in 3D space (e.g. a spin): as we rotate the system about the $$x$$ axis its projection about the $$z$$ axis oscillates. So far, nothing about this example tells us anything about decoherence or why it's hard to make big Schrodinger cat states, so now let's get to that. ## Incoherent case Suppose $$S$$ interacts with some other two level system $$E$$. The letter $$E$$ stands for "environment" which will make sense later. Suppose the state of the combined $$S + E$$ system is $$^{[b]}$$ $$\left( \ket{\uparrow}\ket{\downarrow} + \ket{\downarrow}\ket{\uparrow} \right) / \sqrt{2}$$ where the first ket labels the state of $$S$$ and the second ket labels the state of $$E$$. Now the critical part: what happens if we now do the rotate-and-measure experiment described above on system $$S$$ without doing anything, including measurement, to $$E$$? Experimentally, when we try this in the lab, we find that there is no oscillation in the probability to find $$S$$ in $$\ket{\uparrow}$$ as a function of $$\theta$$! This is decoherence. To describe this mathematically we look at the density matrix of the $$S+E$$ system. The state of the total system is $$\rho_{S+E} = \left[ \begin{array}{cccc} 0&0&0&0 \\ 0 & 1/2 & 1/2 & 0 \\ 0 & 1/2 & 1/2 & 0 \\ 0&0&0&0 \end{array} \right]$$ where we've ordered the states $$\{ \ket{\uparrow}\ket{\uparrow}, \ket{\uparrow}\ket{\downarrow}, \ket{\downarrow}\ket{\uparrow},\ket{\downarrow}\ket{\downarrow} \}$$. To predict the behaviour of experiments done on $$S$$ alone, we take the trace of $$\rho_{S+E}$$ over the part of the space belonging to $$E$$ $$^{[c]}$$. Doing this gives $$\tilde{\rho}_S \equiv \text{Tr}_E \left( \rho_{S+E}\right) = \frac{1}{2} \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] = \frac{1}{2} \left( \ket{\uparrow}\bra{\uparrow} + \ket{\downarrow}\bra{\downarrow} \right) = \frac{1}{2}\text{Id} \, .$$ The off-diagonal terms are gone - we have a purely classical state! If we now rotate $$\tilde{\rho}_S$$ by any rotation operator $$U$$ we find that $$U \tilde{\rho}_S U^\dagger = \frac{1}{2} \left[ \begin{array}{cc} 1&0\\0&1 \end{array} \right] = \tilde{\rho}_S$$ and $$\langle \sigma_z \rangle_{U \tilde{\rho}_S U^\dagger} = \text{Tr}_S (U \tilde{\rho}_S U \sigma_z) = 0 \, .$$ Rotations no longer do anything - there's no oscillation and the expectation value of $$\sigma_z$$ is always zero regardless of rotation angle. This is really, really interesting. Previously we said that a single isolated two level system can be thought of like a spin particle: it's always pointing in some direction in space, so even if measurements along the $$z$$ axis give half up and half down, if you rotate the spin and measure, you see oscillation. On the other hand, we just showed that if we let the two level system interact with something else ($$E$$), the combined system can be left in a state such that the original two level system ($$S$$) doesn't exhibit that oscillation. What we have just seen is the essence of quantum decohrence. If a quantum system $$S$$ becomes entangled with its surrounding environment $$E$$, then $$S$$ can lose its quantum nature. Of course, if we don't ignore the environment $$E$$ and instead include it in our measurements, then we would observe the full quantum properties of the combined system. In other words, decoherence is just lack of knowledge of the complete system. If $$E$$ is really big then keeping track of all its degrees of freedom and measuring them in a controlled way is just impossible. That is the essence of why making big Schrodinger cats is hard; if the system $$S$$ is big it interacts with more environmental degrees of freedom and so observing quantum effects is very difficult. For something a large as a speck of dust interacting with air molecules, the time it takes for the decoherence to kill any off diagonal elements in the density matrix is incredibly small $$^{[d]}$$. Interestingly though, some fairly large systems can be sufficiently isolated from their environments such that they exhibit quantum properties long enough times to be useful; this is, for example, a large fraction of what goes into building a quantum computer. # Large system So far we showed what decoherence is, and in particular how it makes a quantum system appear classical. To recap, decoherence happens when your system $$S$$ interacts with the environment $$E$$; if you don't have access to the environment degrees of freedom, then $$S$$ can lose its quantum interference properties and appear classical. In the example we gave, we saw that a single two level system interacting with another one can appear classical. We will now, via an illustrative extention of the same example, that a large system is more prone to decoherence. Suppose $$S$$ consists of three two level systems in an initial state $$\ket{\uparrow \uparrow \uparrow}$$ with density matrix $$\rho = \ket{\uparrow \uparrow \uparrow} \bra{\uparrow \uparrow \uparrow} \, .$$ Note that there's a sort of redundancy here: we have three separate spins which can be thought of as collectively representing a single spin up.$$^{[e]}$$ Like in the single particle case, we can measure the projection of the spin along the $$z$$ axis, but in this case we use the three-particle operator $$Z^{(3)} \equiv \left( \sigma_z \otimes \sigma_z \otimes \sigma_z \right) \, .$$ ## Coherent case Like before, if we rotate all three spins and measure the average of $$Z^{(3)}$$ we get a sinusoidal dependence on the rotation angle. In particular, if we rotate each spin by an angle $$\theta$$ about the $$x$$ axis, then we get $$\langle Z^{(3)} \rangle_{U \rho U^\dagger} = \cos(\theta)^3 \, .$$ ## Incoherent case Now consider what happens if just one of our spins interacts with the environment. Suppose the middle spin interacts with the environment such that the initial state $$\ket{\uparrow \uparrow \uparrow} \ket{\downarrow}$$ (here the separate second ket with one arrow represents the environment) becomes $$\left( \ket{\uparrow \uparrow \uparrow}\ket{\downarrow} + \ket{\uparrow \downarrow \uparrow}\ket{\uparrow} \right) / \sqrt{2} \, .$$ Writing out the complete four particle density matrix would be tedious and unenlightening. However, the reduced density matrix of the first three particles is $$\tilde{\rho}_S = \frac{1}{2} \left( \ket{\uparrow \uparrow \uparrow}\bra{\uparrow \uparrow \uparrow} + \ket{\uparrow \downarrow \uparrow}\bra{\uparrow \downarrow \uparrow} \right) \, .$$ Note that we have a diagonal density matrix just like we did in the single particle incoherent case. With this density matrix, the expectation value of $$Z^{(3)}$$ following a rotation of all spins by $$\theta$$ is $$\langle Z^{(3)} \rangle_{U \tilde{\rho}_S U^\dagger} = \frac{1}{2} ( \underbrace{\cos(\theta)^3}_{\text{from } \ket{\uparrow \uparrow \uparrow}} + \underbrace{-\cos(\theta)^3}_{\text{from } \ket{\uparrow \downarrow \uparrow}} ) = 0 \, .$$ Here again we've lost the oscillation, and it only took a single spin interacting with the environment to do it. That is why making large Schrodinger cats is hard. # Notes $$[a]$$: This is a simplified version of an example from the introductory chapter of my PhD thesis (pdf). $$[b]$$: This state can be realized if we start with the system in $$\ket{\uparrow}\ket{\downarrow}$$ and subject the system to the Hamiltonian $$H=\sigma_+ \sigma_- + \sigma_- \sigma_+$$ for the proper amount of time such that the propagator is $$U = \left [ \begin{array}{cccc} 1&0&0&0\\ 0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\ 0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\ 0&0&0&1 \end{array} \right ] \, .$$ $$[c]$$: Note that, like any other theoretical description, this procedure is justified because it reproduces the results of experiments. $$[d]$$: I don't remember the numbers but see Schlosshauer's book for a calculation. $$[e]$$: This kind of redundancy is critical in classical machines. For example, a memory bit in a classical computer could be represented by current in a large number of conduction channels in a transistor; if any one of those channels were to change state, that's such a tiny fraction of the total current that the logical state of the transistor is preserved. This redundancy gives classical computers their robustness against errors on the microscopic level. • "In other words, decoherence is just lack of knowledge of the complete system." I don't understand this. I thought decoherence/collapse of the superposition was an objective event, not one subjective to a particular observer. Nov 19, 2021 at 13:55 • @Omroth See Wigner's Friend for an illustration of why objective collapse doesn't really make any sense. Nov 19, 2021 at 18:19 • I've read that and I still don't understand. I assume that Wigner's friend collapses the superposition when he (a sufficently complex system) interacts with (observes) the cat. Why does the fact that Wigner doesn't know about it yet matter? Not all information an observer doesn't have (what is the state of the thing in the box) is in a superposition - it's often resolved but we just don't know what it resolved to. I'm sure this is my own ignorance - I'm just explaining my current position. Nov 20, 2021 at 19:35 It is because of quantum statistical irreversibility, which is closely related to entropy, as the OP suspected. Qualitatively it is quite easy to understand this. From the laws of quantum mechanics on the microscopic level emerges a classical behaviour for macroscopic (i.e. many particle objects). Of course this is not sufficient though and does not give a lot of insight, so here is what I know about the topic. The main insight is that what was mentioned above causes the density matrix to become diagonal, i.e. the coherence to disappear due to many body interactions and resulting dissipation. This has been demonstrated quantitatively for exactly solvable systems. • Leggett and Caldeira solved a system of coupled simple harmonic oscillators in 1983 We apply the influence-functional method of Feynman and Vernon to the study of Brownian motion at arbitrary temperature. By choosing a specific model for the dissipative interaction of the system of interest with its environment, we are able to evaluate the influence functional in closed form and express it in terms of a few parameters such as the phenomenological viscosity coefficient. We show that in the limit h→0 the results obtained from the influence functional formalism reduce to the classical Fokker-Planck equation. In the case of a simple harmonic oscillator with arbitrarily strong damping and at arbitrary temperature, we obtain an explicit expression for the time evolution of the complete density matrix ϱ(x, x′, t) when the system starts in a particular kind of pure state. We compare our results with those of other approaches to the problem of dissipation in quantum mechanics. • Followup work in the same spirit is due to Zurek (this is a good resource on a related, but slightly different topic), also illuminating the quantum information side of this problem. In particular Unruh&Zurek showed that decoherence happens when coupling to a quantum field. Other important papers are by Joos&Zeh and others. • More recently (1990s and 2000s) there has been other exactly solved systems that demonstrate phase transitions of quantum systems that look like what the Copenhagen interpretation calls "collapse", but are actual based on just unitary evolution (so no actual collapse there, don't worry ;) ). This is my summary of this excellent answer by Arnold Neumaier, where further references may be found. I should add that there are indeed macroscopic objects that display coherence phenomena, superconductors and superfluids are the most prominent example. So the "preparation" of such systems can be achieved, but due to the dissipative nature of most everyday systems their coherence is statistically very unlikely. • why the downvote? This is a full and historically accurate review with reference of why decoherence happens in macroscopic systems, which answers the question. I would appreciate a comment to understand how I can improve my answer. Jul 10, 2016 at 11:09 • I didn't downvote, but I'd say that while this answer contains lots of useful reference links and gives the right general idea, it doesn't itself actually answer the question. It's particularly nice that you mention superconductors etc. You could add mention of actual quantum devices too, like trapped ions or superconducting qubits. Jul 11, 2016 at 13:21 • @DanielSank that's true, I guess I particularly didn't answer the why the "difficulty to prepare and maintain" arises, but rather why we don't encounter macroscopic superpositions in everyday life. I know less about this branch of the topic, so instead of editing I would like to direct readers to your answer (excellent and my +1 btw) and hope that mine at least answers the "why decoherence happens" question. Jul 11, 2016 at 14:08 • To be clear, I'm really glad you wrote your answer. The large number of links is quite valuable. Jul 11, 2016 at 22:10 Let's consider a qubit that has two "classical states" $\left\|0\right>$ and $\left\|1\right>$, e.g. a current in a flux qubit that flows in one direction or in the other direction, while superpositions of these states are "non-classical" and will decohere into a mixed state of the classical states. What I'm going to demonstrate now is that a superposition is extremely fragile, the fragility can be exploited to turn the system into an extremely sensitive measuring device. In the following I'll focus on using a single qubit as a dark matter (DM) detector. Suppose we start out with a qubit initialized in the state $\left\|0\right>$ and we apply the Hadamard gate $U$ that acts as follows: $$\begin{split} U\left\|0\right> &= \frac{1}{\sqrt{2}}\left[\left\|0\right\rangle + \left\|1\right\rangle\right]\\ U\left\|1\right> &= \frac{1}{\sqrt{2}}\left[\left\|0\right\rangle - \left\|1\right\rangle\right] \end{split}$$ Note that $U$ is it's own inverse, so applying $U$ again will bring the qubit back to the state $\left\|0\right\rangle$ we started out with. But now consider what happens if during the time the qubit spends it time being a superposition of $\left\|0\right\rangle$ and $\left\|1\right\rangle$ a DM particle collides with it. Then the qubit will get entangled with the DM particle, qubit-DM particle system will be in a state of the form: $$\left\|\psi\right\rangle = \frac{1}{\sqrt{2}}\left[\left\|0\right\rangle \left\|D_0\right\rangle + \left\|1\right\rangle\left\|D_1\right\rangle\right]$$ where the states $\left\|D_{i}\right\rangle$ are the DM-particle states after scattering off the qubit in state $\left\|i\right\rangle$. Here we assume that the scattering is elastic so that the qubit's state does not change at all. So, you might think that because the qubit was not affected by the interaction at all, we cannot perform a measurement on the qubit's state to find out that it has interacted with a DM particle. But watch what happens if we apply the Hadamard gate again to the qubit: $$U\left\|\psi\right\rangle =\left\|0\right\rangle\left\|D^{+}\right\rangle+\left\|1\right\rangle \left\|D^{-}\right\rangle$$ where $D^{\pm} = \frac{1}{2}\left[\left\|D_0\right\rangle\pm\left\|D_1\right\rangle\right]$ So, had there been no interaction the qubit would have returned to the initial state $\left\|0\right\rangle$ but now we end up with an entangled state of the qubit and the DM-particle such that there is now a finite probability to find the qubit in the state $\left\|1\right\rangle$, despite the fact that collision with the DM-particle happened in a purely elastic way at low energy such that it did not affect the physical state of the qubit in any way at the time of the collision. The probability to find the qubit in the state $\left\|1\right\rangle$ is $\frac{1}{2}\left[1-\operatorname{Re}\left\langle D_0\right\|D_1\left.\right\rangle\right]$, so it depends on the overlap between the two dark matter states corresponding to scattering off the qubit in the two states of the superposition. If the states $\left\|D_i\right\rangle$ are orthogonal, then you have 50% probability to find the qubit in the states $\left\|0\right\rangle$ and $\left\|1\right\rangle$, the density matrix after tracing out the DM-particle state is $\frac{1}{2}\left[\left\|0\right\rangle\langle 0\| + \left\|1\right\rangle\langle 1\|\right]$. The larger we make the qubit, the more likely it will be that during the time it spend being a superposition an interaction will have occurred. As demonstrated above, even a purely elastic scattering of a particle is enough to change the superposition into a mixed state. But a macroscopic object at room temperature will emit a huge number of infrared photon, so a superposition will decay into a mixed state extremely rapidly, as the state of the photons will depend on which part of the superposition they were emitted from and a huge number of these photons are emitted every fraction of a second. A possible way out of this problem would be to consider superposition for which the two parts differ less and less as we make the qubits larger and larger. The change in the state of the environment due to interactions between the two parts of the superposition will then be smaller for each interaction. So, in conclusion, this example demonstrates that decoherence of macroscopic superpositions happens due to the extreme fragility of of such superpositions. All that's needed is for information about the state to leak out into the environment. One can relate this to entropy increase, but that in itself doesn't explain why this happens so rapidly. But once information has leaked out, it's going to be irreversible in practice, as the degrees if freedom in the environment that now have "detected" the superposition will themselves leak that information to other degrees of freedom in the environment. You could say that once Schrödinger's cat is out of the bag, it's not going to come back. • This could be a good answer but it's hard to read. For example, the acronym "DM" is used without being defined. I figured out that it means "dark matter" but you should make that more clear by writing "...dark matter (DM)...". Jul 11, 2016 at 13:03 • @DanielSank Yes, I agree, I made some changes to the text. Jul 11, 2016 at 17:17 Decoherence happens because in a macroscopic system you are not able to create a small isolated system. In practice you are deal with statistical mixture and not pure state. There's a good description on Wikipedia. • I briefly read wikipedia about decoherence. My impression is that is an interpretation of experimental result, but does not answer why decoherence happens... Mar 2, 2014 at 17:33 • The problem is that in macroscopic system objects are inevitable non-isolated: they are coupled with the external ambient. In particular the system "cat+atom" become immediately entangled with ambient and as a consequence it loses his coherence. – LC7 Mar 2, 2014 at 17:59 • Haroche says: the cat is a complex and OPEN (it cannot be isolated) system, it's impossible to describe it with a wave function. – LC7 Mar 2, 2014 at 18:04 • I still have a question. If we consider a hydrogen atom in external field, e.g. Stark and Zeeman effects, the hydrogen atom is not isolated. Why we can still write a wavefunction for the hydrogen atom? Is that because the interaction from external field do not depend on the details of the configuration of radiation field? Mar 2, 2014 at 18:54 • No the fact is that you can consider atom plus fiels as a system. In fact the hamiltonian of the system have both atomic and magnetic terms. Sorry for the bad english i'm using my phone now!! – LC7 Mar 2, 2014 at 18:58 In any quantum experiment, as soon as the state of the system can in principal be known (be it because of emitted photons available to the experimenter, an interaction with the environment we may be able to read off the state, or any other means by which the observer can determine the state) it decays and ends up in one of the basis states. That's precisely what the measurement postulate of quantum mechanics predicts. Since in macroscopic systems the probability for this to happen increases significantly, we're hardly able to maintain coherence. Interestingly, in certain cases we can delete the information gained from decoherence and obtain again superposition as was found in the quantum eraser. This description is based on the philosophy of the qm. In de broil interpretation bohmian mechanics is deterministic. But dehorence is normal in systems that can have lower preferred configuration. It is not forbidden so it can be achieved.	6,392	24,700	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 97, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.84375	3	CC-MAIN-2022-21	latest	en	0.766298
http://www.doc.ic.ac.uk/~imh/frames_website/bases.html	1,542,277,480,000,000,000	text/html	crawl-data/CC-MAIN-2018-47/segments/1542039742666.36/warc/CC-MAIN-20181115095814-20181115121814-00414.warc.gz	399,124,267	4,588	## Relational bases, relation algebra - cylindric algebra connections Go to home page \| Reducts of relation algebras and cylindric algebras Relation algebras are a common way to treat binary relations algebraically, and n-dimensional cylindric algebras handle n-ary relations. There are various connections between cylindric and relation algebras, and also between cylindric algebras of differing dimension. For example, taking the neat reduct of an n-dimensional cylindric algebra to its first m dimensions (where m<n) yields an m-dimensional cylindric algebra. A similar kind of reduct will turn a cylindric algebra of dimension at least 4 into a relation algebra. This neat reduct process is closely related to first-order proof theory. Roughly, the laws holding in m-dimensional neat reducts of n-dimensional cylindric algebras correspond to the first-order sentences written with m variables (perhaps re-used) that can be proved using up to n variables. Finding an intrinsic characterisation of when an algebra is a reduct of another of larger dimension, and studying the hierarchy so induced, has been of interest to workers such as Maddux, Monk, and Tarski since the 1960s. Maddux developed the n-dimensional cylindric basis for this purpose, and the related notion of relational basis. The work in the papers below • shows strict hierarchy results for neat reducts, with consequences for proof theory • develops a representation theory for reducts and for algebras defined by relational bases #### Connections between cylindric algebras and relation algebras ##### R. Hirsch and I. Hodkinson RelMiCS, Warszawa, 1998 A short survey of our work on approximations to S Ra CAn --- relational bases, cylindric bases, cylindric hyperbases, etc. Summarises some of the work from the papers below. #### Relation algebras with n-dimensional bases ##### R. Hirsch and I. Hodkinson Revised version in Ann. Pure Appl. Logic 101 (2000) 227-274 We study relation algebras with n-dimensional relational bases in the sense of Maddux. Fix n with 3≤ n<ω. Write Bn for the class of semi-associative algebras with an n-dimensional relational basis, and RAn for the variety generated by Bn. We define a notion of representation for algebras in RAn, and use it to give an explicit (hence recursive) equational axiomatisation of RAn, and to reprove Maddux's result that RAn is canonical. We show that the algebras in RAn are precisely those that have a complete representation. Then we prove that whenever 3< n<k≤ω, RAk is not finitely axiomatisable over RAn. This confirms a conjecture of Maddux. We also prove that Bn is elementary for n=3,4 only. #### Relation algebra reducts of cylindric algebras and an application to proof theory ##### R. Hirsch and I. Hodkinson and R. Maddux J. Symbolic Logic 67 (2002) 197-213 We show that S Ra CAn strictly contains S Ra CAn+1 for each n with 3 ≤ n < ω. We do this by defining a (finite weakly associative) algebra An and showing that it belongs to S Ra CAn but not to S Ra CAn+1. This confirms a conjecture of Maddux. It also shows that S Nrn CAn+iS Nrn CAn+i+1, for 2 < n <ω and i <ω. Here, Nrn denotes the neat reduct of a higher-dimensional cylindric algebra to n dimensions. This answers question 2.12 of Henkin, Monk & Tarski [Cylindric Algebras Part I, North-Holland, 1971]. Our results show that for n≥4, n-variable proof theory for binary relations is less powerful than (n+1)-variable proof theory. #### Provability with finitely many variables ##### R. Hirsch and I. Hodkinson and R. Maddux Bull. Symbolic Logic 8 (2002) 348-379 For every finite n > 3 there is a logically valid sentence Sn with the following properties: • Sn contains only 3 variables (each of which occurs many times); • Sn contains exactly one nonlogical binary relation symbol (no function symbols, no constants, and no equality symbol); • Sn has a proof in first-order logic with equality that contains exactly n variables, but no proof containing only n-1 variables. This result was first proved in Relation algebra reducts of cylindric algebras and an application to proof theory (above) using the machinery of algebraic logic developed in several research monographs and papers. Here we replicate the result and its proof entirely within the realm of (elementary) first-order binary predicate logic with equality. We need the usual syntax, axioms, and rules of inference to show that Sn has a proof with only n variables. To show that Sn has no proof with only n-1 variables we use alternative semantics in place of the usual, standard, set-theoretical semantics of first-order logic. #### Relation algebras from cylindric algebras, I ##### R. Hirsch and I. Hodkinson Ann. Pure Appl. Logic 112 (2001) 225-266 doi: 10.1016/S0168-0072(01)00084-7 We characterise the class S Ra CAn of subalgebras of relation algebra reducts of n-dimensional cylindric algebras (for finite n, at least 4) by the notion of a hyperbasis, analogous to the cylindric basis of Maddux, and by relativised representations. A corollary is that S Ra CAn = S Ra(CAnCrsn) = S Ra(CAnGn). We give a game-theoretic approximation to the existence of a representation, and use it to obtain a recursive equational axiomatisation of S Ra CAn. We include notes on n-variable proof theory, homogeneity, and related matters, and some open problems. #### Relation algebras from cylindric algebras, II ##### R. Hirsch and I. Hodkinson Ann. Pure Appl. Logic, 112 (2001) 267-297 doi: 10.1016/S0168-0072(01)00085-9 We prove, for each 4 ≤ n < ω, that S Ra CAn+1 cannot be defined, using only finitely many axioms, relative to S Ra CAn. The construction also shows that for 3≤ m<n<ω, S Nrm CAn+1 is not finitely axiomatisable over S Nrm CAn. In consequence, for a certain standard n-variable first-order proof system \vdash_{m,n} of m-variable formulas, there is no finite set of m-variable schemata whose m-variable instances, when added to \vdash_{m,n} as axioms, yield \vdash_{m,n+1}. #### Weak representations of relation algebras and relational bases ##### Robin Hirsch, Ian Hodkinson, Roger Maddux J. Symbolic Logic 76 (2011) 870-882. It is known that for all finite n≥5, there are relation algebras with n-dimensional relational bases but no weak representations. We prove that conversely, there are finite weakly representable relation algebras with no n-dimensional relational bases. In symbols: neither of the classes RAn and wRRA contains the other. #### A construction of cylindric and polyadic algebras from atomic relation algebras ##### Ian Hodkinson Algebra Universalis 68 (2012) 257-285, doi 10.1007/s00012-012-0202-3. Abstract Given a simple atomic relation algebra A and a finite n at least 3, we construct effectively an atomic n-dimensional polyadic equality-type algebra P such that for any subsignature L of the signature of P that contains the boolean operations and cylindrifications, the L-reduct of P is completely representable if and only if A is completely representable. If A is finite then so is P. It follows that there is no algorithm to determine whether a finite n-dimensional cylindric algebra, diagonal-free cylindric algebra, polyadic algebra, or polyadic equality algebra is representable (for diagonal-free algebras this was known). We also obtain a new proof that the classes of completely representable n-dimensional algebras of these types are non-elementary, a result that remains true for infinite dimensions if the diagonals are present, and also for infinite-dimensional diagonal-free cylindric algebras. #### Connections between relation algebras and cylindric algebras ##### Ian Hodkinson Springer International Publishing Switzerland 2015 --- Proc. RAMiCS 2015, W. Kahl et al. (Eds.), Springer LNCS 9348, pp. 27-42, 2015. DOI: 10.1007/978-3-319-24704-5_2 Abstract We give an informal description of a recursive representability-preserving reduction of relation algebras to cylindric algebras.	2,000	7,954	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.921875	3	CC-MAIN-2018-47	longest	en	0.90359
https://itfeature.com/statistics/pie-chart/	1,708,758,701,000,000,000	text/html	crawl-data/CC-MAIN-2024-10/segments/1707947474523.8/warc/CC-MAIN-20240224044749-20240224074749-00865.warc.gz	334,611,117	43,086	Pie Chart \| Visual Display of Categorical Data A pie chart is a way of summarizing a set of categorical data. It is a circle that is divided into segments/sectors. Each segment represents a particular category. The area of each segment is proportional to the number of cases in that category. It is a useful way of displaying the data where the division of a whole into parts needs to be presented. It can also be used to compare such divisions at different times. Pie Chart A pie chart is constructed by dividing the total angle of a circle of 360 degrees into different components. The angle A for each sector is obtained by the relation: $$A=\frac{Component Part}{Total}\times 360$$ Each sector is shaded with different colors or marks so that they look separate from each other. Example Make an appropriate chart for the data available regarding the total production of urea fertilizer and its use on different crops. Let the total production of urea be about 200 thousand (kg) and its consumption for different crops wheat, sugarcane, maize, and lentils is 75, 80, 30, and 15 thousand (kg) respectively. Solution: The appropriate diagram seems to be a pie chart because we have to present a whole into 4 parts. To construct a pie chart, we calculate the proportionate arc of the circle, i.e. Now draw a circle of an appropriate radius, and make the angles clockwise or anticlockwise with the help of a protractor or any other device. For wheat make an angle of 135 degrees, for sugarcane an angle of 44 degrees, for maize, an angle of 54 degrees, and for lentils, an angle of 27 degrees, hence the circular region is divided into 4 sectors. Now shade each of the sectors with different colors or marks so that they look different from each other. The pie chart of the above data is Online MCQs Test Preparation Website gmstat.com 1 thought on “Pie Chart \| Visual Display of Categorical Data” 1. I adored your insightful site. excellent information. I hope you release others. I will continue reading This site uses Akismet to reduce spam. Learn how your comment data is processed.	462	2,099	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5	4	CC-MAIN-2024-10	latest	en	0.886167
http://www.homeofbob.com/math/numVluOp/wholeNum/addSub/developmentAdditionSubtraction.html	1,516,143,350,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084886758.34/warc/CC-MAIN-20180116224019-20180117004019-00377.warc.gz	453,815,078	5,720	Addition and Subtraction Development, Research, Activities, and Assessment Overview • Big idea • Research information • Development notes • Subtraction concepts • Addition and subtraction related properties Big idea (generalization) for addition and subtraction Addition and subtraction are two ways to operate on two or more numbers to create a third number of equivalent value. The different ways these operations are represented in real life can be classified into four groups: 1. Combination of number values 2. Separation of number values 3. Part-part-whole relationships of number values 4. Comparing or equalizing number values Research bits: • Children who are not taught algorithms become better at mathematics. Those that are taught algorithms rarely use more efficient strategies, more appropriate for the value of the numbers to be added or subtracted. Such as: • Making nice numbers (368 + 204 = 368 + 200 + 4 = 568 + 4=572). • Keeping the whole (71 - 36 = {subtract 1 from both to get} 70 - 35 = 35) (342 - 37 = {add 3 to both to get} 345 - 40). • With a good teacher students can learn a variety of strategies as well as algorithms. For example, if students are given the following problem: "I went to the store with \$32.00 and spent \$17.00, how much do I have left?" Younger children will draw 32 tallies, cross out 17, and count those left to arrive at the answer. Later, children usually decompose numbers into place value (tens and ones) and develop algorithms that they understand. 23 cards and 14 cards would be decomposed into 23 + 3; and 10 + 4; adding from left to right 20 + 10 and 3 + 4; and finally adding 30 + 7. • When different groups of second graders were given this problems: (7 + 52 + 186) 45% of the students solved the problem without using an algorithm, 26% used part of an algorithm, and 12% used an algorithm. • When this problem (504 - 306) was given to groups of students: A group who had been taught addition and subtraction with an algorithm and a group who were taught without an algorithm. 74% of the second graders and 80% of the third graders, taught without an algorithm, got the right answer. Where as 42% of the second graders and 35% of the third graders, taught with an algorithm, got the the problem right. • Students who were taught relationships in which automaticity was the goal produced more correct answers to basic addition facts within three seconds (76% to 55%) than students who were taught traditionally. (Fosnot) There are critiques who use emotional words as propaganda techniques; terms like new mathematics, fuzzy math, soft math and claim it is dumbing down students learning of mathematics. However, the mindless use of algorithms is the real dumbing down. Development notes Children learn addition and subtraction based on their understanding of number value. Tthey memorize the counting numbers and soon realize they are sequenced with each related in an increasing order. This order develops as an understanding of one more and then one less and the ability to put a number with a set of objects, which eventually becomes cardinality. The relationship of numbers as more or less and cardinality can be decomposed and composed in a hierarchial manner eventually is seen as the operations of additiona and subtraction. These relationships are developed when students experience activieis such as; dot plates to subitize cardinality and learn one more and one less, begin to memorize addition facts and understand hierarchical inclusion. Students, who are given activities to quantify groups of objects both before and after combining or separating different groups of objects will naturally compose and decompoe the numbers and invent their own algorithms. If, once they learn to count, they are are pushed away from counting, not taught to use touch points, and encouraged to use skip counting, five as an anchor, ten and more, and eventually decomposing numbers left to right. The environment needs to include problems and activities which will enable students to naturally incorporate the following ideas when solving problem. • Recognize different values of objects (subitize). Dots on plates, dice, dot cards, groups of objects ... • Respond with one more and one less for an initial value of objects. Numbered dots on a wall, dots on plates, dice, dot cards, groups of objects, ten frames, hundred chart, ... • Respond with two more and two less for an initial value of objects. Numbered dots on a wall, dots on plates, dice, dot cards, groups of objects, ten frames, hundred chart, ... • Count two separate sets of objects, slide them together into one group, and then count the new group to find how many altogether. • Find out how many objects are in two separate sets of objects by counting on from the total number in one group. Rolling two dice, subitize the first number and count on from it for the value of the second die to find the total dots on the two dice. Same for dot cards, groups of objects, ten frames, numbered dots, hundred chart ... • Find out how many objects are left in a set of objects by counting back from the total number in an initial group. Groups of objects, ten frames, numbered dots, hundred chart ... • Decompose numbers into smaller addends, commute them and compose them (find the sums). Examples - two die with a roll of sixes, decompose them into 5, 5, 1, 1, and compose them into 5 + 5 = 10; 10 + 2 = 12. • Decompose and compose sums less than 20. • Decompose and compose for subtracting differences less than 20. • Add and subtract values greater than 20 by working left to right - decomposing into tens and ones, adding or subtracting tens, then adding or subtracting ones, and then adding or subtracting the tens and ones. • Adding on with two digits and subtracting from is the last step so that students can mentally add and subtract all sums and differences less than 100. First and second grade students will, on their own invent an algorithm for additional and subtraction by this deconstruction and construction process. For example: 46 + 23 by deconstructing 46 and 23 into 40 + 6 and 20 + 3, then adding the 40 and 20 and then the 6 and 3 and then the 60 and the 9 getting 69. • Later, students will either discover or it can be suggested they do not need to deconstruct the initial number: 46, but can deconstruct the second 23 to 20 + 3. Then add on from: 46 + 20 to get 66 and then add on the 3 to get 69. • Similarly, students between first and third grade will invent an algorithm for subtracting two digit numbers. First by deconstructing and constructing problems like: 47 - 23. Again decompose into 40 and 20, subtracting 20 from 40 to get 20, and then subtracting 3 from 7 and have 4 left. Recognizing that all of 23 has been subtracted and the 4 is part of the original 47, they will add back the four to the 20, therefore, taking 20 + 4 and getting 24. • Later, students will either discover, or it can be suggested, that the first number not be deconstruct: 47, but to deconstruct the second 23 to 20 + 3. Then subtract the 20 from the 47 to get 27 and finish by subtracting 3 from the 27 to get 23. Subtraction is more difficult and if students don't have a very good understanding of number value and subtraction, then it is extremely difficult. • Students will eventually discover that all addition and subtraction must account for the place value of each number as it is composed or decomposed to arrive at a sum or difference. Students should be encouraged to decompose numbers into expanded notation, based on place value, and then add or subtract without regrouping. • When students are aware of the need to add and subtract according to place values in expanded notation, then continue using expanded notation with numbers that need regrouping. At the beginning of this discussion addition and subtraction was described in four different ways they are represented in real life. To keep the categories less complicated addition was referenced as joining and subtraction as separating. However, doing this is a dangerous idea since it is important students learn all four ways addition and subtraction can be represented can be used for both addition and subtraction. In fact it is impossible sometimes to know for sure if a person is using one or the other to arrive at an answer. For example. If I ask you the difference between 18 and 15? Did you think of three before you even thought to use addition or subtraction? Thus, making it difficult if not impossible to know which operation was used. So be it. Teachers should understand that any addition and subtraction problem can be solved with both addition and subtraction. Review these examples and think about how interchangeable addition and subtraction can be when operating on numbers. This should raise an important question for every math teacher. Do curriculum developers or text book authors take similar short cuts? How many of the four ways and the subcategories of subtraction and addition are included in your math curriculum or text book? You can bet the ones that are not represented have been discovered as good types of problems to include in normative testing. Why? You ask, because they will efficiently sort students into different levels, whcih is the purpose of all normative tests. If students are presented with problems and encouragement in developmentally appropriate ways to understand, they will, usually by fourth grade, invent a traditional addition and subtraction algorithm along with flexibility for selecting from a variety of ways to add and subtract efficiently. Historically we should recognize Constance Kamii who first published ideas on how students' reinvent algorithms in her book Young Children Reinvent Arithmetic (1985. second edition 2000). She built on Piaget's development of understanding.	2,145	9,858	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.84375	5	CC-MAIN-2018-05	latest	en	0.921585
http://www.solutioninn.com/a-study-by-the-occupational-safety-and-health-administration-seeks	1,506,016,562,000,000,000	text/html	crawl-data/CC-MAIN-2017-39/segments/1505818687834.17/warc/CC-MAIN-20170921172227-20170921192227-00517.warc.gz	558,008,827	7,511	# Question: A study by the Occupational Safety and Health Administration seeks A study by the Occupational Safety and Health Administration seeks to know whether the probability that an industrial plant is inspected is affected by the number of accidents at the plant. An analyst regresses whether a plant is inspected (1 = inspected) on the number of accidents resulting in a lost day of work (X). He finds = .04 + .012X sb = .0021 Sy\|x = .19 r2 = .48 n = 320 = 47 sx = 22 Interpret this regression. If Ace Manufacturing has 14 accidents resulting in lost work days, what can you say about whether it will be inspected? Sales0 Views37	151	640	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.515625	3	CC-MAIN-2017-39	latest	en	0.9162
https://forum.arduino.cc/index.php?topic=534590.msg3644082	1,568,859,271,000,000,000	text/html	crawl-data/CC-MAIN-2019-39/segments/1568514573415.58/warc/CC-MAIN-20190919015534-20190919041534-00401.warc.gz	490,071,530	8,328	Go Down ### Topic: powering two dc motors and two servo motors (Read 841 times)previous topic - next topic #### jessabu ##### Mar 12, 2018, 03:40 am Hello. I'm trying to figure out how to power two servos and two dc motors for a robot project I'm doing. The servos are supposed to control long pieces of cardboard connected to some type of shovel in order to pick up trash. The dc motors are used for the movement of the robot. I read another forum post about powering the motors but I wasn't understanding something. My question is, can I power the motors with 6v, 4 AA batteries, or so I have to power them all with 12v? I appreciate any help given, thank you very much. #### Paul_KD7HB #1 ##### Mar 12, 2018, 05:49 am Hello. I'm trying to figure out how to power two servos and two dc motors for a robot project I'm doing. The servos are supposed to control long pieces of cardboard connected to some type of shovel in order to pick up trash. The dc motors are used for the movement of the robot. I read another forum post about powering the motors but I wasn't understanding something. My question is, can I power the motors with 6v, 4 AA batteries, or so I have to power them all with 12v? I appreciate any help given, thank you very much. To begin any help, we need to know the current and voltage ratings of the servos and motors, and especially the starting current needed by the motors. Paul #### jessabu #2 ##### Mar 13, 2018, 02:21 am Oh, I'm sorry about that. The servos are 4.8V-6V, and the DC motors are 3V to 6V. I'm not sure what you mean by starting current. #### jremington #3 ##### Mar 13, 2018, 03:10 amLast Edit: Mar 13, 2018, 03:11 am by jremington The starting current, or the "stall current" in the data sheets, is the supply voltage divided by the motor winding resistance. 4xAA will power two standard hobby servos, but not more. #### MarkT #4 ##### Mar 13, 2018, 12:16 pm You need to budget 1A per small servo, and probably about the same per small motor (although measuring is better than guessing). 4 x D cell quality rechargables might be about right here, anything less isn't going to handle 4A. [ I will NOT respond to personal messages, I WILL delete them, use the forum please ] #### jessabu #5 ##### Mar 13, 2018, 01:15 pm Is 4A enough for the DC motors also? #### MarkT #6 ##### Mar 13, 2018, 02:45 pm Perhaps you should reread what I said, it seems very clear to me. [ I will NOT respond to personal messages, I WILL delete them, use the forum please ] #### jessabu #7 ##### Mar 15, 2018, 03:57 am Oh, again, sorry for the misunderstanding. Would 6 AA (1.5v each) batteries in a battery pack also work? #### MarkT #8 ##### Mar 15, 2018, 12:09 pm Oh, again, sorry for the misunderstanding. Would 6 AA (1.5v each) batteries in a battery pack also work? As I said, reread my answer, it hasn't changed and it addresses your power requirements very clearly. 4 quality D-cells. 4, not 6, D-cells, not A-cells, quality, not rubbish unbranded, rechargable, not alkaline. Clear enough now? [ I will NOT respond to personal messages, I WILL delete them, use the forum please ] #### jessabu #9 ##### Mar 16, 2018, 02:56 am Okay. Thank you for all of the help. It's very much appreciated #### jessabu #10 ##### Mar 16, 2018, 04:13 am I'm sorry. I have another question. If I use the rechargeable batteries, how would I connect them to the Arduino/robot? Through a power jack, or through a breadboard? #### MarkT #11 ##### Mar 16, 2018, 12:24 pm Neither, the batteries are for the servos and motors only. [ I will NOT respond to personal messages, I WILL delete them, use the forum please ] #### jessabu #12 ##### Mar 26, 2018, 03:04 am Okay, I purchased 4 D cell batteries over the weekend, but I'm still not understanding how I would connect the motors and batteries together. How would I connect them if not by breadboard or by the Arduino? Sorry to seem redundant, but I really don't understand. Go Up	1,122	3,964	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.828125	3	CC-MAIN-2019-39	longest	en	0.932933
https://forum.allaboutcircuits.com/threads/power-supply-for-opamp.15786/	1,555,973,999,000,000,000	text/html	crawl-data/CC-MAIN-2019-18/segments/1555578582736.31/warc/CC-MAIN-20190422215211-20190423001211-00244.warc.gz	425,014,944	17,827	# power supply for opamp!!! Discussion in 'The Projects Forum' started by onlyvinod56, Nov 7, 2008. 1. ### onlyvinod56 Thread Starter Senior Member Oct 14, 2008 362 1 Hi Im having a doubt regarding the power supply of opamp. Im using LM324 as a comparator. The circuit is shown in the attachment. (please dont consider the values of the input voltages. randomly i draw the figure) Both the inputs are connected through pots.The voltage at the input terminal(sinusoidal voltage) is observed as a rectfied sinusoidal in CRO. I used the power supply as +12v & 0v. In this forum i observed that in some circuits, the opamps are connected to +15v & -15v. What happens if +15v & 0v is given? Shall i get that rectified voltage? Is it mandatory to use +v & -v instead of +v & 0v ? ThanQ File size: 23.5 KB Views: 73 2. ### Audioguru Expert Dec 20, 2007 10,902 1,245 The opamp has a voltage gain of 200,000 at low frequencies. Its max allowed negative input voltage is only 0V for a 741 opamp and -0.3V for an LM324 quad opamp. A lousy old 741 opamp won't work when its supply is only 5V. Because its voltage gain is so high and because your opamp circvuit does not have negative feedback then if the input of the opamp is not destroyed by having an input voltage too negative then the output at low frequencies will be a max level rectangular-wave. If you add negative feedback then a negative voltage to an input resistor will cause the output to go positive and have the same waveshape for half a cycle as the input. then the (-) input of the opamp will stay at 0V and not be destroyed. Your load resistor is only 5.1 ohms which is a dead short to the output of an opamp that drives a minimum of 2000 ohms. 3. ### Audioguru Expert Dec 20, 2007 10,902 1,245 I just noticed that the half-wave signal is the input and not the output. But your circuit inverts. Its output is trying to go negative when the input goes positive. It can't go negative without having a negative supply in addition to the positive supply. Jul 17, 2007 22,201 1,806	549	2,045	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.796875	3	CC-MAIN-2019-18	latest	en	0.87678
https://community.tableau.com/thread/116601	1,548,057,404,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547583763839.28/warc/CC-MAIN-20190121070334-20190121092334-00247.warc.gz	475,291,378	30,174	10 Replies Latest reply on Sep 16, 2015 11:25 PM by Junjappa M # Cash flow simulation on Tableau Hi, I am new to tableau. I am trying to understand if it is possible/makes sense to use Tableau for financial/cash flow modeling. I have profit and loss data for the past 5 years and forecast for the next two. I have the balance sheet and cash flow data for the past. I need to build the balance sheet simulation for the next two years. Do I need to do it in Excel or is there a way to do it in Tableau. I thought about using some calculation that says, for instance, that future accounts receivable must be certain % of future sales. But then this will be a different data element vs the "account receivable' data element while in excel the would be just the "following cell" of a spreadsheet. thanks fabio • ###### 1. Re: Cash flow simulation on Tableau Fabio, This is an awesome question and I think that this would be a really useful analysis. Would you mind posting the data you have (or at least a sample) so we can understand the current structure of the data? Thanks, Brandon • ###### 2. Re: Cash flow simulation on Tableau Hi Fabio - in your Accounts Receivable example, you could create a calculated field that merges historical (actual) AR figures with future (projected) AR based on an existing "Future Sales" forecast like this: IF ISNULL(Accounts Receivable) THEN (Future Sales)*(Pct Multiplier for AR) ELSE (Accounts Receivable) END This calculated field (call it "AR_display" or something) could be included in the table and would show both numbers in the same row of the balance sheet. You could also create another calcualted field called "Type" that indicates whether the figure is actual or forecasted as follows: IF ISNULL(Accounts Receivable) THEN "Forecast" ELSE "Actual" END You could use this field to change the color of the table to make it easy to spot what type of figure it is. Echoing Brandon - this is an intriguing use of Tableau. What would be even better is if someone in finance out there used Tableau to visualize the typical income statement / balance sheet / cash flow tables in graphs instead of tables. I recently posted a waterfall chart for Facebook's income statement that attempted to do this: http://dataremixed.com/2012/02/a-facebook-waterfall/ Thanks, Ben • ###### 3. Re: Cash flow simulation on Tableau Hi Ben, Hi Brandbon, I think it makes sense to use Tableau rather than Excel because financial simulation is about modeling (for instance "putting together" one business with another, or splitting them) and unless you know from the start (which you rarely do) how you want to slice and dice your data, an Excel model (which is inherently "mono-dimensional") risks becoming very complex very quickly while you try to modify it to accommodate the different scenarios that pop up. I remember this as a nightmare. So this time I decide to build up an excel "table" with all the data of the different companies, on record on top of another, and then see if I could build up the model in Tableau. I am attaching the structure of the Excel. All numbers are random and the names of the divisions are also invented. On the right side of the sheet you have the PL and BS entries. (of course BS entries do not balance since they are random). On the right of the sheet I put a formula that "creates" different "measures (revenues, costs, assets,...) that I need in Tableau. For instance in Tableau I need a table that has the years on the columns, and revenues, direct costs, operating margin, indirect costs, Ebitda, eccetera on the rows. Operating margin must be revenues - direct costs. I did not know how to do that without a separate measures. And then I have the issue of how to project all this into the future... Thanks again Ben for the answer, will try it out! all the best fabio • ###### 4. Re: Cash flow simulation on Tableau Ben, I tried to apply your technique, and it looks good. One question - when you do modeling of a company you are trying to sell or buy you often want to "slice and dice" your model by business, division, or whatever. You might want to see which of the divisions generates (or requires) more cash, and so forth. Doing this in Excel is not easy, and quickly leads to a complex model. It should work much better or Tableau Your idea assumed "one" Accounts Payable tied via some parameter to "one" stream of revenues. Would it be possible/make sense build the Balance Sheet in Excel as a sum of many Accounts Payable elements (each calculated as the product of the revenues of that division X the specific parameter of that division)? Final question. In modeling it is very common to assume a "different" future. For instance to assume that the ratio between revenues and accounts receivable goes up because customers will pay you before eccetera. Is it possible to tie your function to different constants tied to time? thanks again fabio • ###### 5. Re: Cash flow simulation on Tableau Hi Fabio, I'm pretty sure both of your ideas are possible - you can sum up multiple revenue entries for your AR estimation, and you can create different scenarios for the AR calculation using parameters and calculated fields. The attached example uses a simple slider to allow you to change the AR multiplier for all years, and you could even take it a step further and have different multipliers for different years, etc. http://public.tableausoftware.com/views/AREstimator/Dashboard1?:embed=y Hope this helps! Ben • ###### 6. Re: Cash flow simulation on Tableau Wow, amazing! Thanks so much for this Fabio • ###### 7. Re: Cash flow simulation on Tableau Good question and nice responses which I'm going to use to think about something I'm working on. My problem has to do with cash available to a local government. I've already done a basic dashboard with weekly snapshots of cash balances back several years. It provides some interactivity, such as a parameter based "danger zone" and the quick filter to choose past years for comparisons with current. The past couple of years, the cash generated organically from operations and taxes has been supplemented by intra-fiscal year borrowing to smooth out cash flow. That's important to assess current position, but can also misleading in thinking about even near term futures. The ability to do some projections will be very useful. So thanks! Will post mine when it's done. I've just returned from four days of Tableau training so give me another week or two. Best regards John • ###### 8. Re: Cash flow simulation on Tableau Fabio: No problem - happy to help. John: Interesting, & have fun putting your training to use! Ben • ###### 9. Re: Cash flow simulation on Tableau Can someone provide the underlying data or the twbx for this? I have similar data to that display above but I am having a hard time creating a graph like this. • ###### 10. Re: Cash flow simulation on Tableau @ben jones this is a very interesting concept to visualize could either of you share more details or step by step guide for other tableau users, thank you guys for such an amazing post	1,595	7,150	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.5625	3	CC-MAIN-2019-04	latest	en	0.915023
https://www.convertunits.com/from/cycle+per+second/to/kilohertz	1,675,152,192,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764499845.10/warc/CC-MAIN-20230131055533-20230131085533-00482.warc.gz	748,371,531	12,566	## Convert cycle/second to kilohertz cycle per second kilohertz How many cycle per second in 1 kilohertz? The answer is 1000. We assume you are converting between cycle/second and kilohertz. You can view more details on each measurement unit: cycle per second or kilohertz The SI derived unit for frequency is the hertz. 1 hertz is equal to 1 cycle per second, or 0.001 kilohertz. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between cycles/second and kilohertz. Type in your own numbers in the form to convert the units! ## Quick conversion chart of cycle per second to kilohertz 1 cycle per second to kilohertz = 0.001 kilohertz 10 cycle per second to kilohertz = 0.01 kilohertz 50 cycle per second to kilohertz = 0.05 kilohertz 100 cycle per second to kilohertz = 0.1 kilohertz 200 cycle per second to kilohertz = 0.2 kilohertz 500 cycle per second to kilohertz = 0.5 kilohertz 1000 cycle per second to kilohertz = 1 kilohertz ## Want other units? You can do the reverse unit conversion from kilohertz to cycle per second, or enter any two units below: ## Enter two units to convert From: To: ## Definition: Kilohertz The SI prefix "kilo" represents a factor of 103, or in exponential notation, 1E3. So 1 kilohertz = 103 . The definition of a hertz is as follows: The hertz (symbol Hz) is the SI unit of frequency. It is named in honour of the German physicist Heinrich Rudolf Hertz who made some important contributions to science in the field of electromagnetism. ## Metric conversions and more ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!	522	2,031	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.203125	3	CC-MAIN-2023-06	latest	en	0.84046
https://id.scribd.com/document/47835526/bad	1,563,639,364,000,000,000	text/html	crawl-data/CC-MAIN-2019-30/segments/1563195526536.46/warc/CC-MAIN-20190720153215-20190720175215-00428.warc.gz	428,221,657	59,692	Anda di halaman 1dari 2 # hi Question 1.5 (a) The declaration typechecks. The result of a recursive call to the function is multiplied by an integer (0), resulting in an integer, matching the declaration. The name of the identifier is f having type int → int. The value is bound to the lambda expression fn (x:int) => 0 * f(x mod 10). (b) The expression typechecks. Its type is int, but its value is indeterminate since the function repeatedly makes recursive calls to itself, but since there is no base case, evaluation never terminates. Question 2.1 (a) It is bound to 13. The most recent declaration of x in the scope is on line 5, so the value is 13. (b) It is bound to 93.6. The most recent declaration of p in the scope is on line 7, which has a value of 93.6 when evaluated. (c) It is bound to the value of the parameter passed to the function, which is the most recent declaration. It is NOT bound to 100, because the declaration at line 8 is not in the scope. (d) The value returned by the function is (53, 93.6), which is of type int × real. The second value of the tuple, q, is computed from the value of p on line 10 (part b), and the first value of the tuple is a computation involving q and the parameter of the function, x = 41. Question 2.2 ## The value of m is 9. It is computed with modmult(k, 4) where k is bound to 7. However, at the time of function declaration, k was bound to 19, so this is the value used inside modmult. Thus, the result is m = (7 × 4) mod 19 = 9. Question 3.2 Theorem: For every value k ≥ 0 and for every function f , sumodd(k, f ) = f (1) + f (3) + · · · + f (2k − 1). Proof: By mathematical induction on k. Base case: k = 0: We need to show that sumodd(0, f ) = 0, which is the sum on the first zero odd integers. We see that this is true by examining the code. Thus, the base case holds. Inductive hypothesis: Suppose that for some k 0 ≥ 0, sumodd(k 0 , f ) = f (1) + f (3) + · · · + f (2k 0 − 1). Inductive step: We want to show that sumodd(k 0 + 1, f ) = f (1) + f (3) + · · · + f (2(k 0 + 1) − 1). We can evaluate 1 hi ## f (2(k 0 + 1) − 1) + sumodd(k 0 , f ) = f (2(k 0 + 1) − 1) + f (1) + f (3) + · · · + f (2k 0 − 1) = f (1) + f (3) + · · · + f (2k 0 − 1) + f (2(k 0 + 1) − 1) This is the sum of the first k 0 + 1 odd terms of f , which is what we wanted to show. Thus, the function correctly calculates the value of sumodd as described in the specification. Question 3.3 In order to sum over the first k even integers, we can sum over the first k odd integers using sumodd if each value of the function f (n) corresponds to f (n + 1). To accomplish this, we can use an anonymous wrapper function to call the original function with parameter n + 1. In the example given in the handout, to sum the first 5 odd terms of f (n) = n2 , we use: ## sumodd(5, fn (n:int) => (fn (n:int) => n*n)(n+1)) Question 3.4 In order to sum over a range of odd integers from k1 to k2 , we must first convert these numbers to indexes used in the sumodd function. Write c1 = k12+1 and c2 = k22+1 . We now want to find the sum over the c1 th odd integer to the c2 th odd integer of f. Rewrite c1 − 1 = k12−1 . We know that sumodd(c1 − 1, f ) = f (1) + · · · + f (k1 − 2) and sumodd(c2 , f ) = f (1) + · · · + f (k2 ). Thus, if we subtract the first expression from the second, we obtain the sum f (k1 ) + · · · + f (k2 ) that we want. In terms of k1 and k2 , this can be written as:	1,111	3,435	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.921875	4	CC-MAIN-2019-30	latest	en	0.902216
https://www.rocq.inria.fr:443/mathfi/Premia/free-version/doc/premia-doc/pdf_html/mc_quasi_doc/index.html	1,481,220,423,000,000,000	text/html	crawl-data/CC-MAIN-2016-50/segments/1480698542648.35/warc/CC-MAIN-20161202170902-00221-ip-10-31-129-80.ec2.internal.warc.gz	989,281,884	11,928	## Low Discrepancy Sequences March 1, 2012 ### Contents Quasi Monte Carlo simulation consists in approximating the integral [0,1]df(u)du by i=1Nf(u i) where {ξi} are quasi-random numbers, that means they are generated from low-discrepancy sequences. As we already have explained it, such sequences neither are random nor pseudo-random but deterministic and successive values are not independent. However they satisfy good properties of equidistribution on [0, 1]d and we have that i=1Nf(ξ i) [0,1]df(u)du. In the following sections we describe some low discrepancy sequences. We explain their construction and discuss some of their properties, especially on their discrepancy. General references about the Quasi-Monte Carlo simulation are [2], [7], [8], [6] or [4]. The implementation of the sequences are described in the implemented part. ### 1 Tore-SQRT sequences They are d-dimensional sequences, obtained by considering the multiples of suitable irrational numbers modulo 1. Tore sequence It is defined by : where α = (α1,d) d such that (1 1,d) are linearly independent on . {x} = x - [x] denotes the fractional part of x. SQRT sequence It is a particular case of the Tore sequence with where (p1,,pd) are the first d prime numbers. If α1,d are algebric, then the discrepancy satisfies: Click there to reach the implemented part: categQMC_sqrt_implemimplementation. ### 2 Van der Corput and Halton sequences #### 2.1 Van der Corput sequence This is a one-dimensional sequence defined by the radical-inverse function φp in base p: where the coefficients ai are given by the digit expansion in base p of n : R(n) denotes the maximum index for which aR(n) is not equals to 0. Its value depends on n and p by the relation pR(n) n < pR(n)+1, that is: Discrepancy of the Van der Corput sequence satisfies the following majoration: Remarks: (ref Article of Alan Jung and Silvio Galanti) - For n < p, there is only one positive coefficient a in the decomposition in base p, that is a0 = n. Thus φp(n) = and the sequence is increasing. - There are cycles of length p in this sequence. Each subsequence of length p (indices kp to (k + 1)p - 1) is increasing in magnitude proportionnaly to power of 1∕p, and covers uniformly the interval [0, 1). Consequences of this property will be studied for multidimensional sequences (especially Halton sequence). #### 2.2 Halton sequence The Halton sequence is a d-dimensional generalization of the Van Der Corput sequence. Let (p1,,pd) be the d first prime numbers, then ξn is defined by: where φpi(n) is the Van der Corput sequence in base pi. The Halton sequence satisfies : with a constant Cd = 1d. This constant grows to infinity super-exponentially with dimension. Click there to reach the implemented part: categQMC_halton_implemimplementation. #### 2.3 Permuted (Generalized) Halton sequence Orthogonal projections of points from the Halton sequence show non uniform distribution for some dimensions (see Morokoff and Caflish [6], Jung ??? or Bratley and Fox ????). This non-uniformity is due to cycles of length pi for each one-dimensional sequence. To break correlations between the inverse radical functions of different dimensions, we realize permutations of coefficients ai. We consider pi, 1 i d) d permutations over {0,,pi-1} such that Πpi(0) = 0. Each term of the permuted Halton sequence is defined by: with n = i=0R(n)a ipi. The global sequence is given by : There is no optimal choice for the permutations. We present 3 approaches to modify the Halton sequence An algorithm was suggested by Braaten and Weller [1] for d 16 with a possible extension to a larger d (however with significant computation). An other algorithm (see Kocis and Whiten [5]) consists in reversing the binary digits of integers, expressed using a fixed number of base 2 digits and removing any values that are too large. This algorithm can be applied for very large values of dimensions. Halton Sequence Leaped: This other variant for the Halton sequence consists in using only every Lth Halton number subject to the condition that L is a prime different from all bases p1,,pd (see Kocis and Whiten [5]). ### 3 Faure sequence This is a d-dimensional sequence. The Faure sequence is a permutation of the Halton sequence, but it uses the same base r for each dimension. We choose r as the smallest odd prime integer such that r d. Note that the k-th dimension of a d-dimensional Faure sequence is different from the k-th dimension of a d-dimensional Faure sequence as soon as the base r is different. With usual notations, ai are the coefficients of the r-adic decomposition of n We consider the following transformation T : with bk = i=kR(n)C ika i modr and Cik denote binomial coefficients. The coefficients bk are a permutation of the ak. Precision ... and reference ????? The Faure sequence is defined by using successive transformations Tk: where φr is the Van der Corput sequence in base r. The discrepancy of the sequence satisfies : where Cd is a constant dependent on d and r : C = ()d. The constant Cd tends to 0 with dimension. The Faure sequence exhibits cycles of length r but cycles are not composed of increasing terms, except for the first dimension. For the same dimension, the Faure sequence has generally a smaller base than the Halton one, thus cycles are smaller too. Because we use the smallest prime number greater than the dimension d and not the d-th prime number. Click there to reach the implemented part: categQMC_faure_implemimplementation. ### 4 Generalized Faure sequence This is a d-dimensional sequence. Let r be the smallest odd prime integer, such that r d. The digit expansion of n in base r is given by n = i=0R(n)a i(n)ri. The Generalized Faure sequence is defined by : with c(j) = (c k,s(j)) 0kR(n),0sR(n) and c(j) = A(j)Pj-1 where A(j) is a lower triangular inversible matrix such that (ai,l) Fr and P = (Csk) for k R(n),s R(n) is built with the binomial coefficients. The discrepancy of the sequence satisfies: where C(d,r) ()d. Click there to reach the implemented part: categQMC_gen_faure_implemimplementation. ### 5 Nets and (t,s)-sequences (t,s)-sequences are a group of sequences with a very regular distribution behaviour. Their points are placed into certain equally sized volumes of the unit cube for sequences of a fixed length. Chapter 4 of Niederreiter [7] well describes theoretical aspects for such sequences. We just summarize in this section some definitions and properties of those sequences. Definitions • An elementary interval E Id is defined as E = i=1d[a ib-di, (a i + 1)b-di] where a i,di > 0 are integers satisfying 0 ai bdi for 1 i d. • Let 0 t m be integers. A (t,m,s)-net in base b is a point set P of bm points in Is such that the number of points in E is equal to bt for every elementary interval E in base b with Π(E) = bt-m. • Let t 0 be an integer. A sequence x0,x1, of points in Is is a (t,s)-sequence in base b if, for all integers k 0 and m > t, the point set constituting of the xn with kbm n (k + 1)bm is a (t,m,s)-net in base b. Properties: • Any (t,m,s)-net in base b is also a (u,m,s)-net in base b for integers t u m. The same property holds for (t,s)-sequences. Then smaller values of t mean stronger regularity properties. • The discrepancy of a (t,m,s)-net P in base b with m > 0 satisfies: where • The discrepancy of the first N terms of a (t,s)-sequence P in base b satisfies: where • For m 2, a (0,m,s)-net in base b can only exist if s b + 1. A (0,s)-sequence in base b can only exists if s b. Examples: • The Van der Corput sequence is a (0, 1) sequence in base b. In fact, if we consider the bm points x n with kbm n < (k+1)bm (k 0,m 1), every b-adic interval [ab-m, (a + 1)b-m] contains exactly one point x n. • The s-dimensional Sobol sequence is a (τ,s)-sequence in base 2, where τ = i=1sdeg(P i) - s. It is called a LPτ-sequence. Sobol sequence is described in the next point. • The s-dimensional Faure sequence in base r is a (0,s)-sequence where r is the smallest prime integer greater or equal than s. ### 6 Sobol sequence The Sobol sequence is a d-dimensional sequence in base 2 and it is a (τ,d)-sequence. It is one of the most used sequences for Quasi-Monte Carlo simulation. It was first developped by Sobol [3] and it has been proved to have some additional uniformity property under some initialization conditions (see [9]). Its construction is based on primitive polynomials in the field 2 and XOR operations. Each dimension is a permutation of the Halton sequence with base 2 whenever N = 2d. These permutations are generated from irreductible polynomials in 2. But they allow for certain correlations to develop, then they can produce regions where no points fall until N becomes very large. The Sobol sequence is defined by: where the V i(j) are direction numbers (expressed as binary fraction) obtained from d different primitive polynomials and ai denote the coefficients of the digit expansion of n in base b = 2, given by: n = i=0R(n)a i2i. represents the bitwise exclusive OR operator (XOR). For explanation about XOR operation or primitive polynomials, we refer the reader to the Numerical Recipes in C ?????. To implement this sequence, we use an other expression for ξn depending only on the previous point and one direction number. This principle is detailed in the implemented part and is due to Antonov and Saleev ?????. The discrepancy of the sequence satisfies: where Cd = grows superexponentially with dimension, and for K > 0,K t(d) + O(d log log d). t(d) grows superlinearly with dimension. Definition of the constants V : - For each j d we first choose a primitive polynomial P(j) with degree s(j): and we select s(j) odd integers ci(j) such that The choice for constants ci(j) is not a easy step. Sobol’ article ????? gives some explanations about this problem. - Once we have chosen P(j) and the ci(j) for i < s(j), we use the coefficients b i through the recurrence relation : to determine the ci(j) for i s(j). - Finally we calculate V by: Uniformity property: An additional uniformity property of the sequence is called by Sobol the property A. - We define a binary segment of length 2s as a set of points P i whose subscripts satisfy the inequality l2s i < (l + 1)2s where l = 0, 1,. We divide up the s-dimensional unit cube Is by the planes x k = into 2s multidimensional small cubes, which represent binary parallelepipeds. - Property A: If in any binary segment of length 2s of the sequence P 0,,Pi,, all the points belong to different small cubes, then we say that the sequence satisfies property A. Sobol [9] proved a sufficient and necessary condition on the direction numbers so that the property A is verified. A table of good numerical values for V is given for a dimension s 16. - Property A’: The property A can be extended to the property A’ defined as follows. We divide up the s-dimensional unit cube Is by the planes x k = ,, into 22s multidimensional small cubes. If in any binary segment of the sequence P0,,Pi, of length 22s, all the points belong to different small cubes, then we say that the sequence possesses the property A’. - Remark about the link between property A or A’ and the dimension s: Note that the property A (resp. A’) holds for subsequences of length 2s (resp. 22s). In practice if s increases, it becomes difficult to verify the condition because we need to simulate at least 2s (22s) points. Click there to reach the implemented part: categQMC_sobol_implemimplementation. ### 7 Niederreiter sequence The Niederreiter sequence is a s-dimensional (t,s)-sequence in base b whose theoretical aspects are described in Niederreiter [7]. It is defined as: with n = r=0R(n)a r(n)br and C(i) = (c j,r(i)) is called the generator matrix of the i-th coordinate. An algorithm to compute the values is given in Niederreiter [7]. Inititialization of the (cj,r(i)) is done at the beginning of the simulation. The discrepancy of the sequence satisfies: Construction of the cjr(i): (in the next version) The method is based on the formal Laurent series. Remark: If b is a prime power and s an arbitrary dimension such that s b, we can choose P1,,Ps as the linear polynomials Pi(x) = x - bi where bq,,bs are distinct elements of Fb. Then the Niederreiter sequence is a (0,s)-sequence in base b and we have for 1 i s and j 1: cjr(i) = 0 if 0 r < j - 1 cjr(i) = (r∕j - 1)b ir-j+1 if r j - 1. Click there to reach the implemented part: categQMC_nied_implemimplementation. ### 8 General remarks on low discrepancy sequences • Quasi-random numbers combine the advantage of a random sequence that points can be added incrementally, with the advantage of a lattice that there is no clumping of points. • For large dimension s, the theoretical bound (log N)s∕N may only be meaningful for extremly large values of N. The bound in Koksma-Hlawka inequality gives no relevant information until a very large number of points is used. Low discrepancy sequences are very useful for low dimension. In high dimension s, a lattice can only be refined by increasing the number of points by a factor 2s. • Orthogonal projections: if a d-dimensional sequence is uniformly distributed in Id, then two-dimensional sequences formed by pairing coordinates should also be uniformly distributed. The appearance of non-uniformity in these projections is an indication of potential problems in using a quasi-random sequence for integration. This problem is developped in Morokoff and Caflish [6]. We will see that procedures like scrambling permutation can be suggested to improve the uniformity property while preserving the discrepancy. ### References [1] E. BRAATEN and G. WELLER. An improved low-discrepancy sequence for multidimensional quasi-monte carlo integration. Journal of Comput. Phys., (33):249–258, 1979. [2] H.NIEDERREITER. Points sets ans sequences with small discrepancy. Monatsh.Math, 104:273–337, 1987. [3] I.M.SOBOL. The distribution of points in a cube and the approximate evaluation of integrals. U.S.S.R. Computational Math.and Math.Phys., 7(4):86–112, 1967. [4] INRIA. Probabilites numeriques. Chap 1: suites a discrepance faible et integration numerique. [5] L. KOCIS and W.J. WHITEN. Computational investigations of low discrpeancy sequences. ACM Transactions on Mathematical Software, 23(2):266–294, June 1997. [6] W.J. MOROKOFF and R.E. CAFLISH. Quasi-random sequences and their discrepancies. SIAM, Journal of Scientific Computing, 15(6):1251–1279, nov 1994. [7] H. NIEDERREITER. Random number generation and quasi-monte carlo methods. SIAM, 1992. [8] H. NIEDERREITER P.J.S. SHIUE. Monte carlo and quasi-monte carlo methods in scientific computing. Lecture Notes in Statistics,Ed Springer, 106, 1995. [9] I.M. SOBOL. Uniformly distributed sequencs with an additional uniformity property. USSR Comput. Maths. Math. Phys, 16:236–242, 1976.	3,763	14,926	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.1875	3	CC-MAIN-2016-50	latest	en	0.898226
https://www.teacherspayteachers.com/Product/Comparing-Numbers-Kindergarten-Math-Unit-4-3455283	1,537,636,569,000,000,000	text/html	crawl-data/CC-MAIN-2018-39/segments/1537267158609.70/warc/CC-MAIN-20180922162437-20180922182837-00529.warc.gz	877,100,177	21,579	# Comparing Numbers - Kindergarten Math Unit 4 Subject Resource Type Product Rating File Type Compressed Zip File 59 MB\|88 pages Share Also included in: 1. Kindergarten Math Centers - The Full Year Right now this set includes the first 10 MATH UNITS, with 129 Hands-on Kindergarten Math Centers and 250 No Prep number and counting worksheets. This Bundle is now complete. The units in this bundle are: Kindergarten Math Unit 1: Number Sense 1-10 Kinde \$70.00 \$60.00 Save \$10.00 Product Description Comparing Numbers - Kindergarten Math Unit 4 This is an awesome set of 14 Kindergarten Math Centers and 25 No Prep Comparing Numbers Worksheets to use as you help compare numbers, order numbers and make them equal. These activities and worksheets work on skills like greater than less than, how many more and how many less, make them equal and number order. SAVE BIG \$\$\$ WHEN YOU BUY THE FULL YEAR The full year set is a growing bundle. The earlier you buy, the more you save. ALSO IN THIS SERIES: Kindergarten Math Unit 1: Number Sense 1-10 Kindergarten Math Unit 2: Number Sense 11-20 Kindergarten Math Unit 3: Counting to 100 by Tens and Ones Kindergarten Math Unit 6: Subtraction Kindergarten Math Unit 7: Base 10 - Place Value Kindergarten Math Unit 8: Measurement and Data Kindergarten Math Unit 9: Shapes and Geometry Kindergarten Math Unit 10: Money and Time The centers all come with printables. Just add simple manipulatives from around your classroom or home. These hands on math activities are great for keeping kids engaged while independently working on their math skills. This unit also includes 25 Kindergarten Comparing Numbers Worksheets. The worksheets are all no prep, and designed in black and white to save on ink. The worksheets reinforce the skills learned in the center activities for extra practice. I've also included quick reference cards for the centers that can be cut, laminated and included in a gallon zip lock bag with the center pieces. This will allow parent helpers or assistant teachers to have the instructions available at a glance. I hope you find these counting activities engaging and helpful in your classrooms. Let me know if you have any questions. I am happy to help. Thank you! Amy __________________________________________________________________________________________________ You might also like these products: Kindergarten Math and Literacy Printables Preschool Letter Recognition Activities _________________________________________________________________________________________________ Get credit towards future purchases at TeachersPayTeachers by leaving feedback on purchased items. Go to My TPT, and click on my purchases. After leaving a review on your purchase item, you will receive one credit per dollar spent on that item. Make it easier to find new products from the sellers you like. Just click on the Follow Me link below their name to keep informed of their new releases and sales they may be throwing. __________________________________________________________________________________________________ Total Pages 88 pages N/A Teaching Duration N/A Report this Resource \$7.00	655	3,180	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.6875	3	CC-MAIN-2018-39	latest	en	0.877371
https://www.physicsforums.com/threads/daltons-law-of-partial-pressure.949659/	1,696,427,201,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00731.warc.gz	1,010,169,614	17,552	# Dalton's law of partial pressure • dRic2 There's no need to worry about ideal gases because they don't exist in the real world.In summary, Dalton's law is an approximation that is based on the assumption that there is little interaction between the gas molecules.f #### dRic2 Gold Member Does anyone know a rigorous proof for Dalton's law ? I think I saw it once, but I can not find it again anywhere. Thanks Ric Dalton's "law" is not a real law but only an approximation. In the case of ordinary gases, the molecules or atoms do not collide very often, and the partial pressures add (as I am sure you know). It can be derived by neglecting the forces between these particles, and the volume occupied by them. A paragraph about it can be found at https://en.wikipedia.org/wiki/Dalton's_law You can start with a discussion of the kinetic theory of gases. I just read about it in Halliday, Resnick and Walker's "Fundamentals of Physics", fifth edition. They assume that the particles do not collide with one another, but only with the walls of whatever contains them. They derive an expression for the pressure based on the number of particles per unit volume and their average speed. Although they do not explicitly treat the case where the gas is a mixture, the concept can be extended to that case. I hope that helps. dRic2 Wikipedia only says the total pressure for a mixture of gases can be expressed as ##p_{tot} = \sum p_i##, but it doesn't say why. I do not know very much about kinetic theory of gasses, but if you say the answer lies there, then I'll study it. Wikipedia only says the total pressure for a mixture of gases can be expressed as ##p_{tot} = \sum p_i##, but it doesn't say why. Because there is so little interaction of any type between the gas molecules that the forces from each can just be added together for a good approximation. So the pressures are added to get a total pressure. dRic2 good approximation Yes, but I remember that, for ideal gasses, it not an approximation. Maybe it just comes from the fact that ideal gasses are supposed to have no interaction between their molecules, but I remember my professor showed this with math... Maybe I remember wrong. Yes, but I remember that, for ideal gasses, it not an approximation. Maybe it just comes from the fact that ideal gasses are supposed to have no interaction between their molecules, but I remember my professor showed this with math... Maybe I remember wrong. I just said "good approximation" out of an over-abundance of caution. It's probably good enough to be called exact. Yes, the assumption of no interaction is key. dRic2 So the pressures are added to get a total pressure. A very noddy way of looking at the way pressures can add up would be to consider several small people banging against a wall with hammers. The effect of this (including the rate of hammering) is a pressure. Then take several larger people with bigger hammers, also banging on the wall. They will also have the effect of pressure and the two will just add together. Instead of people, imagine large and small ballbearings in a steel box, all hitting the wall and bouncing against each other occasionally. The ball / ball collisions make no difference to the total momentum in the direction of the wall and you can think in terms of the frequency of collisions of the big balls and small balls against the wall. Collisions between big and small balls will even out the distribution of Kinetic energies of the balls (corresponding to the temperature in the gas) and the pressure on a wall will be the sum of the pressures (rate of collisions times average momentum). If you were to have just small balls in the box (same number), with the same mean speed (aka temperature) the pressure against the wall will be the same as the contribution of the small balls when they are part of the mixture. Ditto for the big balls. That's just Kinetic Theory put into everyday words. FactChecker and dRic2	871	3,975	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.140625	3	CC-MAIN-2023-40	latest	en	0.958195
http://mathhelpforum.com/calculus/143885-max-min-hallway-problem.html	1,503,333,601,000,000,000	text/html	crawl-data/CC-MAIN-2017-34/segments/1502886109157.57/warc/CC-MAIN-20170821152953-20170821172953-00061.warc.gz	271,029,470	10,824	1. ## Max/Min Hallway Problem Okay: A pipe of negligible diameter is to be carried horizontally around a corner from a hallway 8 feet wide into a hallway 4 feet wide. a) what is the maximum length that the pipe can have? and b) if we set the pipe's diameter at 1 inch, how does this affect the problem? This is beyond me. I set up a diagram and think I'm supposed to be finding the max L, when L= x+y, where x is the length of the pipe in the 8 foot hallway and y is the length of the pipe in the 4 foot hallway. After this, i don't know where to proceed. I think it needs trig or similar triangles perhaps. Please help? Thanks so much... 2. Originally Posted by mathnerd001 Okay: A pipe of negligible diameter is to be carried horizontally around a corner from a hallway 8 feet wide into a hallway 4 feet wide. a) what is the maximum length that the pipe can have? and b) if we set the pipe's diameter at 1 inch, how does this affect the problem? This is beyond me. I set up a diagram and think I'm supposed to be finding the max L, when L= x+y, where x is the length of the pipe in the 8 foot hallway and y is the length of the pipe in the 4 foot hallway. After this, i don't know where to proceed. I think it needs trig or similar triangles perhaps. Please help? Thanks so much... For part a I will attempt to construct a diagram. ..........what the hell, my diagram got ****ed up when I posted...so we will have to explain The longest pipe will have the length equal to the minimum of $L= x+y$ where x is the distance from the top of the pipe in the first hallway to the edge where the hallways meet, and y is the distance from the edge where the hallways meet to the point where the pipe hits the wall. The angle from the horizontal of the second hallway to the line defined by y is going to be theta. Thus, $x = \frac{8}{cos \theta }$ and $y = \frac{4}{sin \theta }$ $L = L( \theta ) = \frac{8}{cos \theta } + \frac{4}{sin \theta }$ where $0 < \theta < \frac{ \pi }{2}$ If $L`( \theta ) = 0$ then, $\frac { 8 sin \theta }{cos^2 \theta} - \frac{4cos \theta }{sin^2 \theta } = 0$ $\frac{ 8 sin^3 \theta - 4 cos^3 \theta }{ cos^2 \theta sin^2 \theta }$ $8sin^3 \theta - 4cos^3 \theta = 0$ $tan^3 \theta = \frac{4}{8}$ $tan \theta = ( \frac{4}{8} )^{ \frac{1}{3} }$ $L( \theta ) --> \infty$ as $\theta --> 0+$ or $\theta --> \frac{ \pi }{2} -$ Thus the minimum must occur at $\theta = tan^{-1} ( \frac{1}{2} )^{ \frac{1}{3} }$ Let us go back to, $tan \theta = ( \frac{4}{8} )^{ \frac{1}{3} }$ we can see that (by constructing the triangle and using pythag), $cos \theta = \frac{8^{ \frac{1}{3} } }{ \sqrt{ 8^{ \frac{2}{3} } + 4^{ \frac{2}{3} } } }$ and $sin \theta = \frac{4^{ \frac{1}{3} } }{ \sqrt{ 8^{ \frac{2}{3} } + 4^{ \frac{2}{3} } } }$ Hence, $L( \theta ) = \frac{8}{ \frac{8^{ \frac{1}{3} } }{ \sqrt{ 8^{ \frac{2}{3} } + 4^{ \frac{2}{3} } } } } + \frac{4}{ \frac{4^{ \frac{1}{3} } }{ \sqrt{ 8^{ \frac{2}{3} } + 4^{ \frac{2}{3} } } } }$ $L = ( 8^{ \frac{2}{3} } + 4^{ \frac{2}{3} } )^{ \frac{3}{2} }$ 3. ## part b thanks. i ended up figuring it out by similar triangles as well.... Now for part b. If the pipe is given a diameter of 1 inch, how will it affect the problem? Now i know it will shorten the pipe because the point where the pipe theoretically touches at its longest length will be... .5 inch away? Not sure how to exactly expand on that. 4. Originally Posted by mathnerd001 thanks. i ended up figuring it out by similar triangles as well.... Now for part b. If the pipe is given a diameter of 1 inch, how will it affect the problem? Now i know it will shorten the pipe because the point where the pipe theoretically touches at its longest length will be... .5 inch away? Not sure how to exactly expand on that. Draw the original diagram and superimpose it with the new one. What do you notice? Well, our stick which was originally as thin as our pencil marking now has an edge that hits both the walls and the edge where the hallways intersect. This forces us to have a greater angle of theta (as defined in my previous post). And note that our original analysis is no longer valid because of the position of our new beam, the part of the beam that touches both walls sits an inch off of the edge where the hallways intersect. Thus, similar triangles are out the window as well as my approach. How do we set this up then? There are a number of ways but off the top of my head I would analyze the side that hits both the walls but hangs off the edge by the diameter of the beam. Re-define your theta and do an analysis of triangles.	1,353	4,585	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 20, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.34375	4	CC-MAIN-2017-34	longest	en	0.9061
https://fersch.de/beispiel?nr=sinussatz&nrform=GeoSinussatzUma&datei=1005sinussatzGeoSinussatzUma	1,619,167,179,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618039568689.89/warc/CC-MAIN-20210423070953-20210423100953-00063.warc.gz	344,938,896	4,423	Geometrie-Trigonometrie-Sinussatz • $\frac{a}{ sin\alpha} = \frac{b}{ sin\beta }= \frac{c}{ sin\gamma }$ $a = \frac{b\cdot sin\alpha }{ sin\beta }$ 1 2 3 4 5 6 7 $sin\alpha = \frac{a\cdot sin\beta }{ b}$ 1 2 3 4 5 Beispiel Nr: 05 $\text{Gegeben:}\\\text{Winkel} \qquad \beta \qquad [^{\circ}] \\ \text{Winkel} \qquad \alpha \qquad [^{\circ}] \\ \text{Länge der Seite} \qquad b \qquad [m] \\ \\ \text{Gesucht:} \\\text{Länge der Seite} \qquad a \qquad [m] \\ \\ a = \frac{b\cdot sin\alpha }{ sin\beta }\\ \textbf{Gegeben:} \\ \beta=45^{\circ} \qquad \alpha=135^{\circ} \qquad b=7\frac{2}{5}m \qquad \\ \\ \textbf{Rechnung:} \\ a = \frac{b\cdot sin\alpha }{ sin\beta } \\ \beta=45^{\circ}\\ \alpha=135^{\circ}\\ b=7\frac{2}{5}m\\ a = \frac{7\frac{2}{5}m\cdot sin135^{\circ} }{ sin45^{\circ} }\\\\a=7\frac{2}{5}m \\\\\\ \small \begin{array}{\|l\|} \hline beta=\\ \hline 45 ° \\ \hline 2,7\cdot 10^{3} \text{'} \\ \hline 1,62\cdot 10^{5} \text{''} \\ \hline 50 gon \\ \hline 0,785 rad \\ \hline \end{array} \small \begin{array}{\|l\|} \hline alpha=\\ \hline 135 ° \\ \hline 8,1\cdot 10^{3} \text{'} \\ \hline 4,86\cdot 10^{5} \text{''} \\ \hline 150 gon \\ \hline 2,36 rad \\ \hline \end{array} \small \begin{array}{\|l\|} \hline b=\\ \hline 7\frac{2}{5} m \\ \hline 74 dm \\ \hline 740 cm \\ \hline 7,4\cdot 10^{3} mm \\ \hline 7,4\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{\|l\|} \hline a=\\ \hline 7\frac{2}{5} m \\ \hline 74 dm \\ \hline 740 cm \\ \hline 7,4\cdot 10^{3} mm \\ \hline 7,4\cdot 10^{6} \mu m \\ \hline \end{array}$	723	1,541	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.28125	4	CC-MAIN-2021-17	longest	en	0.11181
https://www.sanfoundry.com/statistical-quality-control-questions-answers-online-test/	1,718,422,960,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00187.warc.gz	860,343,451	20,784	Statistical Quality Control Questions and Answers – Variable Charts – Control Charts for x̅ and R -3 This set of Statistical Quality Control online test focuses on “Variable Charts – Control Charts for x̅ and R -3”. 1. Once a set of reliable control limits is obtained, we use the control chart for monitoring future production. This is called __________ a) Phase I control chart usage b) Phase II control chart usage c) Phase III control chart usage d) Phase IV control chart usage Explanation: The use of reliable control limits to monitor the future production from a process, is generally mentioned as the Phase II application of control chart while, setting the trial control limits to monitor the process is called the Phase I application. 2. When R chart is out of control, we __________ a) Eliminate the out-of-control points and recalculate the control limits b) Take one more sample and recalculate the control limits c) Eliminate the out-of-control points and the nearest two points, and recalculate the control limits d) Take no action Explanation: When R chart is out of control, we often eliminate the out-of-control points and recompute a revised value of R bar. This will help us recalculate the control limits. 3. When the upper and lower natural tolerance limits are equal to the upper and lower specification limits, the process capability ratio, cp is ________ a) Greater than 1 b) 0 c) Less than 1 d) Equal to 1 Explanation: When the natural tolerance limits are having a equal value to the specification limits for a quality characteristic, the process capability ratio, cp is having a value equal to 1. 4. X bar chart monitors __________ a) Between-sample variability b) Within-sample variability c) Neither between-sample nor within-sample variability d) Both between-sample variability and within-sample variability Explanation: A x chart is only used to monitor the variability of a whole process over time. It does not detect variability within a sample but variability between all the samples of a process output. 5. Unlike x chart, which measures between-sample variability only, an R chart is used to monitor ____ a) Both between-sample variability and within-sample variability b) Within-sample variability only c) Between-sample variability only d) Neither between-sample variability nor within-sample variability Explanation: A R chart is a chart which uses sample ranges to calculate its center line and control charts. So it only measures within-sample variability, i.e. the instantaneous variability at a given time. 6. When using standard values of process mean and standard deviation, the equation of UCL for a x chart is given as, UCL = μ+Aσ What is the value of A here? a) 6/√n b) 3/√n c) √n/6 d) √n/3 Explanation: When using the standard values of process mean and standard deviation, A in the equation of UCL for x chart is given by A=$$\frac{3}{n^{0.5}} = \frac{3}{\sqrt{n}}$$ 7. For standard values of mean and standard deviation used, what does the center line of the R chart represent? a) R bar b) d2 σ c) D2 σ d) d2 R Explanation: For the standard values of mean and standard deviation used, the center line for a R chart is determined by the following formula, d2 σ. 8. The control limits obtained by specifying the type I error level for the test, are called ________ a) Probability limits b) Trial limits c) Error limits d) Unreliable limits Explanation: It is possible to define the control limits for a control chart by specifying the type I error level. The limits obtained by this way, are called the Probability limits for the control chart. 9. Which chart should be interpreted first when both, x chart, and R chart are indicating a non-random behavior? a) x chart b) R chart c) X and R chart d) Trial Limits Explanation: It is a property of control charts for x bar and R, that if they both indicate non-random behavior, and if R chart is interpreted first to delete the assignable causes in it, it will automatically delete assignable causes in x bar chart. 10. Which of these is a cause of trend patterns on a control chart? a) Gradual wearing out of some critical process component b) Operator fatigue c) Environmental changes d) Over-control Explanation: A trend or continuous movement in one direction on a control chart is generally caused by gradual wearing out of some critical process component like deterioration of a tool. 11. Shift in process level can be seen on the control charts when __________ a) Operator fatigue occurs b) Temperature changes c) Over-control of process d) New workers introduction Explanation: Shift in the process level is a phenomenon seen on the control chart patterns. It occurs when there is introduction of new workers, or there is a change of methods. 12. Stratification is defined as ________ a) Tendency for the points to cluster artificially around the center line b) Shift in the process level c) Continuous movement of points in one direction d) When the points fall near or slightly outside the control limits Explanation: Stratification is defined as the tendency of the points on a control chart, to cluster around the center line of the control chart. It occurs very frequently while applying control charts to a process. 13. Stratification of points on a control chart indicates __________ of natural variability of the process. a) Lack b) Increase c) Constancy d) Randomness Explanation: Stratification makes the points in a control chart to be plotted around the control chart center line; which indicates that the randomness is gone and there is a lack in natural variability of the process. 14. Never attempt to interpret the x chart when the R chart indicates the out of control condition. a) True b) False Explanation: It is easy to eliminate the assignable causes in the x chart by eliminating the assignable causes in the R chart. This implies, if there is indication of out-of-control condition of process from the R chart and x chart; then first, R chart must be interpreted. 15. The 3 sigma limits on x bar control charts imply that the type I error probability is __________ a) 0.0012 b) 0.0072 c) 0.0027 d) 0.0037 Explanation: The 3 sigma control limits on x bar control charts will imply that the type I error probability is 0.0027 or 0.27%. It will give 2700 defectives per million units of production. Sanfoundry Global Education & Learning Series – Statistical Quality Control. To practice all areas of Statistical Quality Control for online tests, here is complete set of 1000+ Multiple Choice Questions and Answers. If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]	1,485	6,686	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.03125	4	CC-MAIN-2024-26	latest	en	0.870219
https://www.techwhiff.com/learn/on-mars-the-acceleration-due-to-gravity-is/428360	1,669,819,298,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710764.12/warc/CC-MAIN-20221130124353-20221130154353-00379.warc.gz	1,089,784,778	11,841	# On Mars, the acceleration due to gravity is approximately 37.83% of Earth's gravity. An astronaut whose... ###### Question: On Mars, the acceleration due to gravity is approximately 37.83% of Earth's gravity. An astronaut whose weight on Earth is 672.38 N travels to the Martian surface. His mass, as measured on Mars, will be? Select one: O A. 36.40 kg OB. 179.15 kg O C. 68.54 kg O D. 252.36 kg #### Similar Solved Questions ##### Atenolol is beta-blocker used to treat hypertension. If a patient is given a 6-ml IV injection... Atenolol is beta-blocker used to treat hypertension. If a patient is given a 6-ml IV injection of atenolol solution, which has concentration of 0.5 mg/ml, what is the size of the dose?... ##### Find the solution set of sqrt(log(y+1))= log (y-1) ? Find the solution set of sqrt(log(y+1))= log (y-1) ?... ##### I coreccted my homework and tomorrow I have to give a homework wich is written in a perfectenglish Please correct my mistakes Hello!I coreccted my homework and tomorrow I have to give a homework wich is written in a perfectenglishPlease correct my mistakes.I don't have to do mistakes because when I will go to N-y I will be ridiulous President Obama had a controversial first year: 30 000 additonal soldiers were sent out... ##### If 14 J of work are needed to stretch a spring 20 cm beyond equilibrium, how... If 14 J of work are needed to stretch a spring 20 cm beyond equilibrium, how much work is required to compress it 6 cm beyond equilibrium? If 14 J of work are needed to stretch a spring 20 cm beyond equilibrium, how much work is required to compress it 6 cm beyond equilibrium? (Use decimal notation.... ##### -- - - - The Company's management is responsible for establishing and maintaining adequate internal control... -- - - - The Company's management is responsible for establishing and maintaining adequate internal control over financial reporting. The Company's internal control over financial reporting is a process designed under the supervision of its Chief Executive Officer and Chief Financial Officer... ##### You perform interrupted conjugation using an a+b+c+d+e+f+g+h+ Hfr strain and an F– strain that is a–b–c–d–e–f–g–h–.... You perform interrupted conjugation using an a+b+c+d+e+f+g+h+ Hfr strain and an F– strain that is a–b–c–d–e–f–g–h–. You observe the following genes transferred together in order from last to first: &... ##### What is (3cottheta)/5 in terms of sectheta? What is (3cottheta)/5 in terms of sectheta?... ##### 116 is divisible by how many numbers 116 is divisible by how many numbers?... ##### Durango co. desires to maintain an ending inventory equal to 20% of next month's cost of... Durango co. desires to maintain an ending inventory equal to 20% of next month's cost of budgeted sales. Assume that Durango Co. maintained this level of ending inventory for the month of September. The cost of budgeted sales for October is \$100,000 and the cost of budgeted sales for November is... ##### Without using a calculator evaluate sin45^o/(cos30^o +sin60^o? Without using a calculator evaluate sin45^o/(cos30^o +sin60^o?... ##### .... Let n an entl divergent ? a) an True os false b) an TS Convergent... .... Let n an entl divergent ? a) an True os false b) an TS Convergent , divergent of con't be conch n=0...	840	3,312	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.6875	3	CC-MAIN-2022-49	latest	en	0.916294

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