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calculus · Applications of multivariable derivatives · Constrained optimization (articles) Lagrange multipliers, examples Examples of the Lagrangian and Lagrange multiplier technique in action. A student’s K-12 math education should also prepare him or her to be free to pursue post-secondary education opportunities. Day Lesson Title Math Learning Goals Expectations 1 What Is the Largest Rectangle? • Use an inquiry process to determine that a square is the largest rectangle that can be constructed for a given perimeter. Email: [email protected] The specific details that make it hard to find and convert from this delimiter type to another are: Math mode can be broken across lines, but need not be. If the sum of a number's digits is a multiple of 3, that number can be divided by. Mathematics IA Worked Examples ALGEBRA: OPTIMISATION AND CONVEX SETS. convex sets, functions, optimization problems 2. The website “PBS for Teachers” offers ready to use lessons based on NTCM standards as an example. Here's a simple example of this type of problem. 2) Chemical properties of consumer products. score at least 70% (14 out of 20) on the special Calculus II placement assessment accesses through Canvas. His current research focus is on convex optimization applications in control, signal processing, and circuit design. Produced in collaboration with a lead for maths. Some of the notes here are for previous versions for the courses: caveat lector. Research in optimisation includes model development, analysis, numerical techniques and applications. We aim to advance the mathematical foundations of both discrete and continuous optimization and to leverage these advances to develop new algorithms with. It can be used to solve large scale, practical problems by quantifying them into a mathematical optimization model. We offer expert private/ home tutors for ib math physics chemistry biology english economics business studies and management. It is not designed to be an evaluation tool of | {
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whatever inputs appear on the other side of the equation. The principal motivation for writing this work has not been the teaching of mathematics per se, but to equip students with the nec. Because of that, we sometimes see the function. May 1, 2013. Properties of arithmetic lessons with lots of worked examples and practice problems. The Department of Mathematics welcomes gifts to a variety of funds, be they general-purpose funds to be used for the department’s greatest needs, donations in memory of our former colleagues, or for specific purposes. Mathematical models are used particularly in the natural sciences and engineering. Optimization. In most instances, requiring students to explain their thinking is at Level 3. Lecture 14 (Oct. Math Jam for BSTEM Prep is for students preparing for math courses in College Algebra, Trigonometry, Business Calculus and review prior to Calculus I. The idea of pairing each member of the domainComplete information about the mapping, definition of an mapping, examples of an mapping, step by step solution of problems involving mapping. Commons Math is a library of lightweight, self-contained mathematics and statistics components addressing the most common problems not available in the Java programming language or Commons Lang. Caution: Sometimes the Math. The oldest mathematics journal in continuous publication in the Western Hemisphere, American Journal of Mathematics ranks as one of the most respected and celebrated journals in its field. At a glance: Mathematics. It has only 2 steps: Step 1. Min on integral types. We help ib students in IB math portfolio HL/SL type I and type II Internal assessment(IA) Task Help Samples Examples. The Caribbean Examinations Council seeks to ensure that the Internal Assessment scores are. Final selected Candidates for PhD program in Department of Mathematics & Statistics at IIT Kanpur Session 2018-19 (Sem II) 11-Dec-2018 Shortlisted Candidates for Written Test and Interview - Ph. Typesetting | {
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2018-19 (Sem II) 11-Dec-2018 Shortlisted Candidates for Written Test and Interview - Ph. Typesetting Simple Formulas. It is simplest to organize these on paper before you start working with the spreadsheet. Multiplication, for example, was achieved by a process of repeated doubling of the number to be multiplied on one side and of one on the other, essentially a kind of multiplication of binary factors similar to that used by modern computers (see the example at right). Institute of Mathematics UP Diliman Optimization Mathematics 21 10 14 Example from MATH 21 at University of the Philippines Diliman. If you are looking for help with your essay then we offer a comprehensive writing service provided by fully qualified academics in your field of study. Berkeley in 1985. 1 Einstein Drive Princeton, New Jersey 08540 USA. Please close and relaunch it. Cross-Entropy Method for Optimization To see how Algorithm 2. mathematics (outside of teaching or academia), your best bet is applied mathematics with computers. Optimization Problems There are many math problems where, based on a given set of constraints, you must minimize something, like the cost of producing a container, or maximize something, like an. Instructional Practices - Resources for understanding Mathematics Instruction. 1a) over x 2 lRn subject to h(x) = 0 (2. This is an example of how an investigation into area optimisation could progress. In the classroom, the tiny bricks are now my favorite possibility-packed math manipulative! Read on for a sampling of math activities that use LEGO pieces to build and reinforce key math concepts. For example, if a paper is rated internally at a 20 and is moderated at an 18, then all papers rated at a 20 (those not included in the sample) will receive a score of 18. ★ Help your child to do math in her head. The example demonstrates the typical work flow: create an objective function, create constraints, solve the problem, and examine the results. Ani is presenting joint work | {
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create constraints, solve the problem, and examine the results. Ani is presenting joint work on DSOS/SDSOS optimization. I am a proud member of the NCSU Numerical Analysis Group and am very active in SIAM. The mathematical concepts applied only accounts for 6 marks at most. Obviously it’s not possible to make a payment to greater precision than one cent, so you’ll be paying either $65. These are pretty cool and exciting topics, but this tutorial is not about this advanced topics. Week 7 of the Course is devoted to identification of global extrema and constrained optimization with inequality constraints. Calculus is the principal "tool" in finding the Best Solutions to these practical problems. Potential barriers to abstract understanding for students who have learning problems and how to manage these barriers. Chen and D. Click on the link with each question to go straight to the relevant page. But neither admission to study nor course design is the direct responsibility of the Faculty of Mathematics rather than DPMMS. Learn more. Let us start with a short list of problems. Min on integral types. Learn Pear Deck Lessons and activities to help you get started. Examples of Acronyms By YourDictionary An acronym is a pronounceable word formed from the first letter (or first few letters) of each word in a phrase or title. Example: optimoptions(@fmincon,'Display','iter','FunctionTolerance',1e-10) sets fmincon options to have iterative display, and to have a FunctionTolerance of 1e-10. If width (x) = 600 feet, then length (y) = 2400- 2x = 2400 – 1200 = 1200 feet. Because of that, we sometimes see the function. For example 4 * 2 = 2 * 4. We aim to be a department which is both intensely research-active and student-centered. In this case we want to optimize the volume and the constraint this time is the amount of material used. ] Optimization over a (finite) Closed Interval: Maximizing Area or Volume, Minimizing Cost [21. Elements of a successf ul IB Internal Assessment. | {
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Maximizing Area or Volume, Minimizing Cost [21. Elements of a successf ul IB Internal Assessment. Introduction to Quantum Chemistry (39 slides) Introduction to GAMESS (32 slides) From Schrodinger to Hartree-Fock (CHEM580 - 43 slides). These are the units that are used most often. A relation R on set A is called Transitive if and implies. Video created by National Research University Higher School of Economics for the course "Mathematics for economists". I am in math hl and I really need a good math IA topic. Hi guys so I thought I'd start working on my IB Maths Exploration (SL). Again, there are many insights from this example into the challenges that must be faced in optimization theory and practice. 25% is a function of the length of time the money is invested. If Tyrolit can achieve this, its inauspicious cost structure due to high level of service would not decrease profit as much, because the company would work generally more efficiently. For example, dynamic search models are used to study labor-market behavior. This free workbook contains nine example models from investment and portfolio management. For an example, though, we might note that y =sin x is a solution of (1. Covered topics include special functions, linear algebra, probability models, random numbers, interpolation, integration, regression, optimization problems and more. Using clear explanations, standard Python. Top class International Baccalaureate(ib) tutors is one of the best ib tutorials available in the world. Show them how they can directly impact the company’s performance and their career path, as well as influence their schedule and other contact center processes. In this post, you will get a gentle introduction to the Adam optimization algorithm for use in deep learning. Department of Mathematics | University of Washington. These options promote competence in abstract thinking, logical rigor, analysis, expository clarity, and critical writing. In other words, 2% of the people who visit | {
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analysis, expository clarity, and critical writing. In other words, 2% of the people who visit your landing page convert to customer. Whether you're a developer, ISV, or technical researcher, if you need to optimize high-performance software on today's leading processors, one book delivers the advanced techniques and code examples you need: Software Optimization for High Performance Computing. 2018 - 2019. The website “PBS for Teachers” offers ready to use lessons based on NTCM standards as an example. For example the function f(x)=x has neither a maximum nor a minimum value for −∞ 0 which shows that −1 2. This is an example of how an investigation into area optimisation could progress. 3); HL syllabus (see syllabus section 6. Published on January 9, 2018 January 9, 2018 • 77 Likes • 8 Comments. Here is a set of practice problems to accompany the Optimization section of the Applications of Derivatives chapter of the notes for Paul Dawkins Calculus I course at Lamar University. Sample IA 1 - this scored a 16/20. Read the latest articles of Applied Mathematics and Computation at ScienceDirect. In this "Maths Exploration" the maths definitely does not have to be complicated, this is a very good example of a maths exploration which scored a perfect 20/20 The assessment criteria are these and I would really appreciate it if you took a look at them. I am a proud member of the NCSU Numerical Analysis Group and am very active in SIAM. While many of Euclid's successors implicitly assumed that all perfect numbers were of the form (Dickson 2005, pp. Thus the rectangular field should be 600 feet wide and 1200 feet long. Escher are the result of his attempts to visually express such mathematical concepts as infinity, duality, dimension, recursion, topological morphing, and self-similarity. You can use the worksheet that most closely models your situation as a starting point. Expressing things differently. SIAM hosts conferences, publishes book and journals, and has a robust | {
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Expressing things differently. SIAM hosts conferences, publishes book and journals, and has a robust membership program. It is no matter what Words you put them into. Algorithmic, data analytic, machine learning and numerical methods which support the modeling and analysis of optimization problems are encouraged. It is a very good tool for improving reasoning and problem-solving capabilities. SIAM hosts conferences, publishes book and journals, and has a robust membership program. Visit BYJU’S to learn different types of equations provided with examples. Note that we can rearrange the error bound to see the minimum number of iterations required to guarantee absolute error less than a prescribed$\epsilon\$:. This construction math self-paced class is intended to develop mathematical skills that can be applied to the construction trade through practice and application. To provide instructional and professional support to local districts and school sites, so that ALL students will achieve proficiency in Algebra 1 and more advanced mathematics courses. Just as mathematics reveals the motions of the stars and the rhythms of nature, it can also shed light on the more mundane decisions of everyday life. 100 majors. Explore the possibilities of math through coursework, research, and public lectures. The structure presented in these internal assessments does not have to be followed strictly. A standard example of motivating constrained optimization are examples where the setup is described in a lot of lines, e. 5 we will consider other possibilities. In these cases, higher-order optimization methods are ill-suited, and discussion in this paper will be restricted to rst-order methods. optimize() is devoted to one dimensional optimization problem. For K-12 kids, teachers and parents. A typical example would be taking the limitations of materials and labor, and then determining the "best" production levels for maximal profits under those conditions. Please add My Skype | {
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the "best" production levels for maximal profits under those conditions. Please add My Skype Address:ykreddy22 20 plus years experienced, highly qualified Indian math teacher offers one to one lesson in maths for IGCSE ,IB all grades up to 12 Grades levels Providing IGCSE IB math lessons in very simple ways including the basics of maths which will helps in your exams and the future studies. Draw a picture of the situation. Alongside 7 examples of exemplary Mathematics SL Internal Assessments, an extensive introduction to IAs provide International Baccalaureate students with tips, resources, and ideas to help students maximize their marks on the portfolio. Examples taken from biology. Math Placement Information. What are good examples of constrained optimization problems (perhaps not simple!) that today's students might actually encounter in their lives? If your goal is to find problems that are more easily accessible, see also the sister question What are easy examples from daily life of constrained optimization?. DEFINITION OF LOCAL MAXIMA AND LOCAL MINIMA 1. They illustrate one of the most important applications of the first derivative. Only one of MATH 151, MATH 160, the sequence MATH 165-MATH 166, or MATH 181 may be counted towards graduation. Basic elements of a good Math Studies Project or Math SL/HL Portfolio piece: • Correct answers throughout. The Journal of Computational Mathematics published bi-monthly. Optimization is a part of Calculus which I certainly do enjoy, but once again I don't have any ideas regarding suitable optimization problems suitable for the investigation. With MATH 75B, equivalent to MATH 75. Powered by Create your own unique website with customizable templates. Optimization Notice 15 DNN Primitives in Intel® MKL Highlights A plain C API to be used in the existing DNN frameworks Brings IA-optimized performance to popular image recognition topologies: – AlexNet, Visual Geometry Group (VGG), GoogleNet, and ResNet. Students majoring in | {
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topologies: – AlexNet, Visual Geometry Group (VGG), GoogleNet, and ResNet. Students majoring in mathematics might wonder whether they will ever use the mathematics they are learning, once they graduate and get a job. The basic idea of the optimization problems that follow is the same. | {
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1. ## Integral convergence proof.
Why $\int^{\infty}_{\pi} \frac{sinx}{lnx}dx$ converges?
2. You can rewrite the integral in the following way:
$\displaystyle \int_\pi^\infty \frac{\sin x}{\log x} \text{ d}x=\int_\pi^{2\pi} \frac{\sin x}{\log x} \text{ d}x+\int_{2\pi}^{3\pi} \frac{\sin x}{\log x} \text{ d}x+\int_{3\pi}^{4\pi} \frac{\sin x}{\log x} \text{ d}x+\ldots=\sum_{k=1}^\infty \int_{k\pi}^{(k+1)\pi} \frac{\sin x}{\log x} \text{ d}x$.
Take note that the terms of the sum are alternating! We can write the sum as
$\displaystyle \sum_{k=1}^\infty (-1)^k \int_{k\pi}^{(k+1)\pi} \left|\frac{\sin x}{\log x}\right| \text{ d}x$
Also observe that as $k\to \infty$ the term $\displaystyle a_k=\int_{k\pi}^{(k+1)\pi} \left|\frac{\sin x}{\log x}\right| \text{ d}x\to 0$ monotonically. Therefore, the sum (and equivalently, the integral) converges via the alternating series test.
3. Wow! I never guess to do this!!!
Maybe there is more simple way to prove that?
The reason I'm asking this is because that is from "special"(true false questions) test and for each question I have something like 7 minutes!
Thank you again!
4. This is probably the cleanest way to do the problem, but you might be able to get away with an integration by parts, first. Let $u=\frac{1}{\log x}$ and $dv=\sin x\text{ d}x$. Then, $du=\frac{1}{x(\log x)^2} \text{ d}x$ and $v=-\cos x$. The integral becomes
$\displaystyle \int_\pi^\infty \frac{\sin x}{\log x}\text{ d}x=\left.\frac{-\cos x}{\log x} \right|_\pi^\infty+\int_\pi^\infty \frac{\cos x}{x (\log x)^2}\text{ d}x$
Can you see how the terms will converge?
5. ## Eureka!
Dirichlet's test for integrals!
Let,
$f(x)=sin(x)$
and
$g(x)=\frac{1}{ln(x)}$
$|\int^b_{\frac{\pi}{2}} sin(x)dx| \leq 2$ for all $b>\frac{\pi}{2}}$
and $g(x)$ is monotonic function and $lim_{x\to\infty}g(x)=0$
So, by the test of Dirichlet the integral converges! | {
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# Moments at which moving points on a circle coincide
Points A $(0,1)$ and B $(1,0)$ start moving along the circumference of a unit circle with center $(0,0)$ in the same, positive (that is, counterclockwise) direction. Every minute, points A and B traverse arcs respectively of $60$° and $42$°. Visually:
Determine moments $t_1, \ldots, t_k,\ldots,$ such that at time $t_k$ points A and B coincide for the $k^\text{th}$ time.
I've been able to determine $t_1$, but cant seem to determine the next moment. I'll describe how I've gotten $t_1$ and hopefully you can suggest how to proceed (or if I'm doing it wrong, how to go about solving for all $t$). We're given the angular velocities of the two points:
• $v_A = 60$° ($\pi\over 3$) per minute;
• $v_B = 42$° ($7\pi\over 30$) per minute.
We also know the starting angles of the two points (shown also on the graph):
• $d_A = {\pi \over 2}$ and $d_B = 0$.
To calculate $t_1$, we just have to solve the following equation for $t$:
$$\left({\pi\over 2} + {\pi \over 3}\cdot t \right)= \left(0 + {7\pi\over 30}\cdot t\right) \tag{T_1}.$$
Calculation yields the value of $5$ for $t$, so $\color{brown}{t_1 = 5}$ minutes.
Now, another basic calculation tells us that at minute $t_1$, points A and B form an angle of $5\pi\over 3$ with respect to $OB$ (sorry, forgot to label 'O' on the graph). So, I figured that to calculate moment $t_2$, it will suffice to solve the following equation for $t$ and add $t_1$ to it:
$$\left({5\pi\over 3} + {\pi \over 3}\cdot t \right)= \left({5\pi\over 3} + {7\pi\over 30}\cdot t\right) \tag{T_2^?}.$$
But, of course, the first summands are canceled out, leaving us with:
$$\left({\pi \over 3}\cdot t \right)= \left({7\pi\over 30}\cdot t\right) \tag{T_2^?}.$$ | {
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$$\left({\pi \over 3}\cdot t \right)= \left({7\pi\over 30}\cdot t\right) \tag{T_2^?}.$$
This solution is true only for $t=0$, so clearly something went wrong with my reasoning. (Of course, $t=0 + t_1 = t_1$, which is a moment of coincidence, but it's not the moment we're looking for). I would appreciate any help with the strategy I've taken or the way I should approach it instead.
• We use degrees, since typing those $\pi$ is a nuisance, and degree intuition is better, and fractions are unpleasant. In $k$ minutes, $A$ travels $18$ degrees more than $B$. Since $B$ starts out $90$ degrees ahead, the first time of coincidence is $5$ minutes. For the next time of meeting, $A$ has to gain $360$ more degrees, which takes $20$ minutes, so $t_2=25$. And $20$ more minutes gets us to the third time of meeting, and so on. The $k$-th time of meeting is $5+20(k-1)$. Jun 2, 2014 at 20:53
• Thank you André! $20k-15$ it is! Jun 2, 2014 at 20:55
• You are welcome. You can see that viewing the problem concretely makes the answer pop out. Jun 2, 2014 at 20:57
• It does! Thank you all for your help. Jun 2, 2014 at 20:58
• Please note that I misread and interchanged $A$ and $B$. It is $15+20(k-1)$. Jun 2, 2014 at 21:29
All your thinking so far is good. What you're missing is that the next time your points meet, A will have gone around the circle an extra time. So their angles won't be equal - $A$'s will be exactly $2\pi$ more than $B$'s, accounting for the extra lap. If you solve $$\frac{\pi}{3} \cdot t = \frac{7\pi}{30} \cdot t + 2\pi,$$ you get $t= 20$. That tells you they'll meet again another 20 minutes later, at a total time of 25 minutes after the start. Can you guess from there what time the third meeting will be? Even better - can you explain why? :)
• I was just thinking about the 2pi. I had the stupid thought to divide by 2pi, but this makes sense. Thank you. Jun 2, 2014 at 20:51
• I'll think about you question and get back to you soon. Jun 2, 2014 at 20:53 | {
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An easy way to do this is to solve this equation $$42t\equiv 90+60t \bmod 360$$
• Ah! That's the same as cool papa's answer, but with degrees. I shouldn't have complicated by converting to radians. Thank you. Jun 2, 2014 at 20:52
• The use of modular arithmetic helps a lot here too, that way, if you see a more convoluted problem in the future where you can't intuit the number of additional laps it will take, you don't have to worry. Jun 2, 2014 at 20:53
• This doesn't lend itself to solving the equation very well. How would you go about solving for t? Just guess and check? Dec 11, 2014 at 23:18
Keep in mind that when you're calculating distance around a circle in this way you have to mod it by $2\pi$. If, for example, A and B start both straight up and A makes a revolution every minute while B makes a revolution every 2 minutes, then they'll coincide after 2 minutes, though the distance traveled is very different.
So instead, just look at the difference in distance traveled. A travels a certain distance farther than B every minute. Every time that distance is equal to a multiple of $2\pi$ they will coincide.
• Thank you. That is helpful. Jun 2, 2014 at 21:06 | {
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GroupTheory/IsSubnormal - Maple Help
GroupTheory
IsSubnormal
test whether one group is contained as a subnormal subgroup of another
Calling Sequence IsSubnormal( H, G )
Parameters
H - a group G - a group
Description
• A group $H$ is a subnormal subgroup of a group $G$ if $H$ is a subgroup of $G$, and if there is a chain
$G={G}_{0}▹{G}_{1}▹\dots ▹H$
such that ${G}_{k}$ is normal in ${G}_{k-1}$, for each $i$. Every normal subgroup of a group is subnormal, but not conversely.
• The IsSubnormal( H, G ) command tests whether the group H is a subnormal subgroup of the group G. It returns true if H is subnormal in G, and returns false otherwise. For some pairs H and G of groups, the value FAIL may be returned if IsSubnormal cannot determine whether H is a subnormal subgroup of G.
Examples
> $\mathrm{with}\left(\mathrm{GroupTheory}\right):$
> $G≔\mathrm{Group}\left(\mathrm{Perm}\left(\left[\left[1,2,3,6,4,5,7,8\right]\right]\right),\mathrm{Perm}\left(\left[\left[2,5\right],\left[6,8\right]\right]\right)\right)$
${G}{≔}⟨\left({1}{,}{2}{,}{3}{,}{6}{,}{4}{,}{5}{,}{7}{,}{8}\right){,}\left({2}{,}{5}\right)\left({6}{,}{8}\right)⟩$ (1)
> $\mathrm{GroupOrder}\left(G\right)$
${16}$ (2)
> $H≔\mathrm{Subgroup}\left(\left\{\mathrm{Perm}\left(\left[\left[2,5\right],\left[6,8\right]\right]\right)\right\},G\right)$
${H}{≔}⟨\left({2}{,}{5}\right)\left({6}{,}{8}\right)⟩$ (3)
> $\mathrm{IsSubnormal}\left(H,G\right)$
${\mathrm{true}}$ (4)
Every normal subgroup of a group is subnormal.
> $\mathrm{andmap}\left(\mathrm{IsSubnormal},\mathrm{NormalSubgroups}\left(G\right),G\right)$
${\mathrm{true}}$ (5)
Compatibility
• The GroupTheory[IsSubnormal] command was introduced in Maple 2018. | {
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# Are these two events in $S_{30}$ independent?
Let $S = S_{30}$, the set of permutations on $\{1,...,30\}$.
Suppose $A$ is the event: $\{\pi\in S_{30}: \pi(1) = 1\}$ and $B$ is the event $\{\pi\in S_{30}: \pi(2) < \pi(3)\}$.
I am asked to show that these events are independent.
Now $|A| = 29!$ and $|B| = (1+...+29)\times 28!$
Hence $|A|\times|B| = 29!\times 28! \times (1+...+29)$
However $|A\cap B| = (1+...+28)\times 27!$
So $P(A\cap B)< P(A)\times P(B)$.
So these events don't seem to be independent? What am I doing wrong?
• @AdamHughes Thank you, I was thinking of something else. I've written an answer now. – DonAntonio Jun 6 '16 at 15:44
By definition: since $\;A\cong S_{29}\;$ :
$$P(A)=\frac{29!}{30!}=\frac1{30}$$ , whereas
$$B:=\{\pi\in S_n\;/\;\pi(2)<\pi(3)\}$$
and we can then count:
$$\begin{cases}\pi(3)=30\implies\pi(2)=1,2,...,29\\{}{}\\\pi(3)=29\implies\pi(2)=1,2,...,28\\\ldots\\\pi(3)=2\implies \pi(2)=1\end{cases}$$
For each option in each line we have $\;(n-2)!\;$ permutations, so all in all we have
$$28!\left(29+28+\ldots2+1\right)=28!\frac{29\cdot30}2=\frac{30!}2\implies$$
$$P(B)=\frac{\frac{30!}2}{30!}=\frac12$$
Finally: if $\;\pi\in A\cap B\;$ then
$$\begin{cases}\pi(1)=1,\,\pi(2)=2\implies \pi(3)=2,3,....,30\\{}\\\pi(1)=1,\,\pi(2)=3\implies \pi(3)=4,...,30\\\ldots\\{}\\\pi(1)=1,\,\pi(2)=29\implies \pi(3)=30\end{cases}$$
For each option above we have $\;27!\;$ permutations, so in total we have
$$27!\left(1+2+\ldots+28\right)=\frac{29!}2\implies P(A\cap B)=\frac{\frac{29!}2}{30!}=\frac1{60}=P(A)P(B)$$
and the events are independent
I assume that we are picking a permutation $\pi$ at random, with all permutations equally likely.
Let $A$ be the event $\pi(1)=1$ and $B$ the event $\pi(2)\lt \pi(3)$. To show independence, we show that $\Pr(B\mid A)=\Pr(B)$. This is clear, for by symmetry each is $\frac{1}{2}$. | {
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• Thanks for this explanation. May I ask where did I go wrong in my argument? – fosho Jun 6 '16 at 15:39
• The calculations seem to be correct, but do not address the issue of independence. – André Nicolas Jun 6 '16 at 15:43
• @McFry: At the time I wrote the comment, the OP made no reference to probability. – André Nicolas Jun 6 '16 at 16:02 | {
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# Minimal steps to reach a natural number by a signed arithmetic progression
$$F:=\big\{f\,\big|\,\text{function }f: \mathbf N\rightarrow \{1,2\}\big\}$$. \begin{align} &\min_{f\in F,\,k\in\mathbf N} k, \\ &\sum_{i=1}^k (-1)^{f(i)}i = n \in\mathbf N. \end{align}
1. Is there an analytical or an asymptotic solution to this problem for general $$n$$?
2. For a specifically given $$n$$, what would a good algorithms to solve this problem? The dynamic programming?
Essentially, for any given $$n$$, the task is to select $$+$$'s and $$-$$'s to reach $$n$$ as fast as possible (minimize $$k$$): $$\pm 1 \pm 2 \pm 3 \pm \dots \pm k = n$$
If we don't worry about overshooting $$n$$, the obvious strategy is to only add and never subtract. If $$n$$ is a triangular number, this will give us the answer. This would necessarily be the optimal solution since we did not subtract at all.
If $$n$$ is not a triangular number, at some point $$k'$$, we will overshoot $$n$$: $$1 + 2 + 3 + \dots + k' > n$$
Since we need $$k$$ at least $$k'$$ to even reach $$n$$, we know that $$k=k'$$ is optimal if it works.
Consider the value $$1 + 2 + 3 + \dots + k' - n$$. We will denote this amount as $$a$$. We know that $$a < k'$$. If $$a$$ is even, then we can switch the operator of $$a/2$$ to $$-$$, and we have our optimal solution.
Figuring out the solution for when $$a$$ is odd may be trickier.
EDIT:
Figured out the "odd" case!
If $$a$$ is odd, then we cannot have $$k = k'$$ because there is no expression that will give us exactly $$n$$ using $$k'$$ terms. For any signs we switch in the $$k'$$ terms, the total sum will retain the same parity, so we cannot reach exactly $$n$$. Thus, if we can find a solution with $$k'+1$$ terms, it is necessarily optimal.
Consider $$1 + 2 + 3 + \dots + k' + (k'+1) = n + a + k'+1$$.
If $$k'$$ is even, then consider the value $$\frac{a+k'+1}{2}$$. We have $$\frac{a+k'+1}{2} < \frac{2k'+1}{2} = k' + 1/2 \,.$$ | {
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So, $$\frac{a+k'+1}{2} \leq k'$$. Thus, it is in our series, so we can assign it the operator $$-$$, and we have achieved exactly a sum of exactly $$n$$.
Now, to consider what happens when both $$a$$ and $$k'$$ are odd...
EDIT 2:
First off, is it possible to find a solution with $$k=k'+1$$ when $$k'$$ is odd?
Well if we use all $$+$$'s, we have $$1 + 2 + 3 + \dots + k' + (k'+1) = n + a + k' + 1$$. As before, switching the signs of any of our terms will preserve the parity of the total sum. Since $$a$$ and $$k'$$ are odd, $$n+a+k'+1$$ has opposite parity from $$n$$. So, no rearrangement of signs will enable a series of $$k'+1$$ terms to work.
Our next candidate is $$k=k'+2$$. Let us consider the following sum: $$1 + 2 + 3 + \dots + k' + (k'+1) + (k'+2) = n + a + (k'+1) + (k'+2)$$
We are too high by the quantity $$a + 2k' + 3$$. We can switch the sign of $$k'$$ and switch the sign of $$\frac{a+3}{2}$$, and we arrive at exactly $$n$$ with $$k=k'+2$$.
And that concludes the proof! No dynamic programming needed - a purely analytic solution.
TL;DR
Let $$k'$$ be the least integer such that $$T_{k'} \geq n$$. If $$T_{k'}$$ and $$n$$ have the same parity, then $$k=k'$$. Else, $$k$$ is equal to the least odd integer greater than $$k'$$.
• How do you know that these solutions are optimal ? – Yves Daoust Jan 24 at 22:25
• I will edit my post to make it more clear. – inavda Jan 24 at 22:26
• Is it more clear now? – inavda Jan 24 at 22:36
• Yep, it is, thanks. – Yves Daoust Jan 24 at 22:41
• Can't you say that you start with the smallest $k$ such that $T_k\ge n$ and $T_k$ has the same parity as $n$, as the overshoot correction will be even. – Yves Daoust Jan 24 at 22:44 | {
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Continuing on @inavda's answer, for a given $$n$$ we find the minimum number of terms, which is the index of the smallest triangular number not smaller than $$n$$, and with the same parity. As the parities of the triangular numbers follow the pattern $$e,o,o,e,e,o,o,e,e,\cdots$$, we can find the smallest triangular number not smaller than $$n$$, and increment once or twice as needed to reach the desired parity.
Then we repeat this operation recursively on the half of the residue $$\dfrac{k(k+1)}2-n$$ to obtain the desired correction.
Example: $$n=37$$
$$37\le T_9=45, \text{half residue }4\to+++++++++$$
$$4\le T_3=6, \text{half residue }1\to---+++++$$
$$1=T_1\to+--++++++$$
The smallest triangular number is obtained by
$$\frac{k(k+1)}2\ge n$$ or
$$k\ge\frac{\sqrt{8n+1}-1}2,$$
$$k=\left\lceil\frac{\sqrt{8n+1}-1}2\right\rceil.$$
Notice that the procedure fails for $$n=2$$, as the smallest triangular number is $$T_3=6$$ and the half residue is again $$2$$. Also for $$n=5$$, $$T_5=5$$ and the half residue is $$5$$. These are the only fixed points. We solve these with $$2=1-2+3$$ and $$5=1-2-3+4+5$$.
Notice that the residue is on the order $$O(\sqrt n)$$, so that the successive residues decrease very quickly, and the recursion depth is of order $$O(\log\log n)$$.
Update:
Checking the example, it turns out that this procedure is wrong.
• "We find the minimum number of terms, which is the index of the smallest triangular number not smaller than $n$, and with the same parity." I don't think this phrasing is quite correct. For example, $a(5)=5$, but the index of the smallest triangular number greater than $5$ is $3$. – inavda Jan 25 at 0:10
• @inavda: $T_3$ does not have the right parity. – Yves Daoust Jan 25 at 0:11 | {
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# Is there a built-in function to get the centroid of a table?
Suppose I have a table with some data that are 'lumped' around some central location, such as the dummy example
i0 = 22.3;
j0 = 34.1;
table = Table[
Exp[-((i - i0)^2 + (j - j0)^2)/10^2]
, {i, 1, 50}, {j, 1, 50}
]
which kind of looks like
ListPointPlot3D[table, PlotRange -> Full]
I would like to recover the centroid of the "mass" determined by the signal in my table, i.e. in the case above, the tuplet {i0,j0}, in the cleanest way possible. I've managed to do it in the obvious but expensive way with a bunch of explicit sums and products with explicit Ranges, but it feels like there should be a built-in function that will do this ─ and I've as yet been unable to find it.
Can this be done? If so, how?
wd = WeightedData[Tuples @ Range @ Dimensions @ table, Join @@ table]
Mean @ wd
{22.3232, 33.9072}
Also
Total[MapIndexed[#2 # &, table / Total[table, 2], {2}], 2] (* and *)
Dot[Join @@ table , Tuples[Range @ Dimensions @ table]] / Total[table, 2]
{22.3232, 33.9072}
i0 = 22.3; j0 = 34.1;
table = Table[
Exp[-((i - i0)^2 + (j - j0)^2)/10^2], {i, 1, 50}, {j, 1, 50}];
x = Array[#1 &, Dimensions[table]];
y = Array[#2 &, Dimensions[table]];
pt = {Total@Flatten[x table], Total@Flatten[y table]}/
Total[table, 2]
(* {22.3232, 33.9072} *)
ListDensityPlot[table, Epilog -> {Red, Point@Reverse[pt]},
PlotRange -> All]
• Thanks both. I accepted the 'proper' built-in as that's what the OP asks for, but this is faster for my instance and it is clean enough that it makes for readable code. – Emilio Pisanty Jul 11 '18 at 12:19
I can provide some improvement if it is about speed.
Let's generate a larger data set (I use Compile merely to speed it up a bit). | {
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Let's generate a larger data set (I use Compile merely to speed it up a bit).
i0 = 22.3; j0 = 34.1;
m = 2500;
n = 1500;
x = Subdivide[1., 50, m - 1];
y = Subdivide[1., 50, n - 1];
table2 = Partition[#, n] &@
Compile[{{X, _Real, 1}, {i0, _Real}, {j0, _Real}},
Exp[-((X[[1]] - i0)^2 + (X[[2]] - j0)^2)/10^2],
RuntimeAttributes -> {Listable}
][Tuples[{x, y}], i0, j0];
Szabolcs' approach
AbsoluteTiming[
pt = {
Total@Flatten[Transpose[ConstantArray[x, n]] table],
Total@Flatten[ConstantArray[y, m] table]
}/Total[table, 2]
]
{0.733425, {22.3288, 33.8733}}
The problem is that before summation, some large arrays have to be constructed and multiplied. But the summations and matrix-matrix products can also be expressed by cheaper matrix-vector products:
AbsoluteTiming[
pt2 = With[{buffer = ConstantArray[1., m].table},
{x.(table.ConstantArray[1., n]),
buffer.y
}/(ConstantArray[1., n].buffer)
]
]
{0.044209, {22.3288, 33.8733}} | {
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# Notation for the set of all finite subsets of $\mathbb{N}$
Is there a "standard" notation to denote the set of all finite subsets of $\mathbb{N}$? (or any set, not just $\mathbb{N}$)
Thanks
-
Doubt it. I would go with $\mathcal{P}_{\text{fin}}(\mathbb{N})$ or, depending on what exactly you're doing, just $S$. – Qiaochu Yuan Aug 4 '11 at 14:52
(some) people think of $2^{<\omega}$ (finite length binary strings) as the set of all finite subsets of $\omega.$ – jspecter Aug 4 '11 at 14:54
I think $2^{<\omega}$ is pretty ambiguous (what does a terminal $0$ mean?). The notation I see most often is some version of $[\omega]^{<\omega}$, with one or both occurrences of $\omega$ replaced by $\mathbb{N}$ or $\aleph_0$, depending on personal preference. Alternatively, I have also seen FIN, which has a nice blunt simplicity about it. – user83827 Aug 4 '11 at 15:07
@user10: It looks like some more comprehensive answers have popped up in the meantime, so I'll just keep my comment as a comment unless you feel strongly otherwise. – user83827 Aug 4 '11 at 15:40
Several possible notations for $\{A\subseteq\omega\mid |A|<\omega\}$:
1. $[\omega]^{<\omega}$
2. $P_\omega(\omega)$
3. $\operatorname{Fin}(\omega)$
Where, of course, $\omega=\mathbb N$.
And as usual my advice on the matter: When in doubt, open with "We denote by [the chosen notation here] the set ..." | {
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-
Isn't $\omega$ the usual notation for the ordinal, rather than the cardinal? I'd be tempted to use $|A|\lt\aleph_0$ or $|A|\lt|\omega|$, rather than $|A|\lt\omega$... – Arturo Magidin Aug 4 '11 at 16:07
@Arturo: It is indeed confusing, but initial ordinals function as both ordinals and cardinals. In this case, $|\omega|=\omega$, so it is alright. I agree that it is somewhat confusing, and I do my best to use $\aleph_0$ when I only care about cardinality. The usual notations, however, use $\omega$ - and I do agree that it is less... cluttered than $\mathcal P_{\aleph_0}(\omega)$ or such. – Asaf Karagila Aug 4 '11 at 16:09
Yes, technically, cardinals are particular ordinals; but my impression is that one uses $\omega$ when one wants to emphasize/keep in mind the ordinal structure, and uses $|\omega|$ and $\aleph_0$ when one wants to ignore the ordinal structure. Clearly, my impression was mistaken, I'll have to try to track down where I got it from. – Arturo Magidin Aug 4 '11 at 16:14
@Arturo: Yes, in essence you are correct. However since one uses $\aleph$ notation a lot less than people would expect (as most of the time we use $\kappa,\lambda$ and such to denote cardinals) it is customary to just let them assume their "ordinal role" when needed, and to say an ordinal is of smaller cardinality than some initial ordinal is the same as saying it has a smaller order type. So it works out just fine. – Asaf Karagila Aug 4 '11 at 16:46
The second notation is rather confusing to me. If I saw it in a paper I would not be able to guess what it meant. – Qiaochu Yuan Aug 4 '11 at 16:52
You can find various notations, as mentioned in coments. (I doubt there is some generally accepted notation.)
• You can find $[\omega]^{<\omega}$, e.g. here, which can be considered as a special case of $[A]^{<\kappa}$ - which denotes all subsets of $A$ of cardinality less then $\kappa$, see e.g. p.18 of the same book. In your case you could use $[\mathbb N]^{<\omega}$. | {
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• You can find $\mathrm{Fin}$, e.g. here and here
• You can find $\mathbb N^{[<\infty]}$, e.g. here.
• Hindman and Strauss use $\mathcal P_f(\mathbb N)$ in this book, which is similar to Qiaochu's suggestion $\mathcal P_{\mathrm{fin}}(\mathbb N)$.
- | {
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# Relation between matrix power and Jordan normal form
(a) Assume $$A\in\mathbb{C}^{n\times n}$$ has $$n$$ distinct eigenvalues. Prove that there are exactly $$2^n$$ distinct matrices $$B$$ such that $$B^2 = A$$ (i.e., in particular, there are no more than $$2^n$$ matrices with this property).
(b) How many such matrices $$B\in\mathbb{C}^{3\times3}$$ exist if $$A=\begin{pmatrix}2&0&0\\0&2&0\\0&0&1\end{pmatrix}$$? Why?
(a) It is clear that if $$\lambda_i$$ is an eigenvalue of $$A$$, then $$\pm\sqrt\lambda_i$$ is an eigenvalue of $$B$$, therefore $$B = \begin{pmatrix}\pm\sqrt\lambda_1&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&\pm\sqrt\lambda_n\end{pmatrix}$$ satisfies $$B^2=A$$ and there are $$2^n$$ such matrices $$B$$ because of $$2^n$$ possibilities of rearranging different numbers of plus and minus signs on the diagonal.
(b) We can construct $$2^3=8$$ such matrices $$B$$ using approach of (a), but $$B$$ is not necessarily diagonal. Consider $$B=\begin{pmatrix}\sqrt2&-1&0\\0&-\sqrt2&0\\0&0&1\end{pmatrix}$$ which satisfies $$B^2=A$$ as well. So, there exist more than $$2^3$$ such matrices.
I can't give an explanation why there are no more than $$2^n$$ possibilities in the first case and more than $$2^3$$ in the second. It definitely follows from the fact that Jordan normal form is unique in (a) and isn't unique in (b) because of an eigenvalue with algebraic multiplicity $$2$$, however, I can't formulate it into a self-contained statement.
• Did you mean $$B=\begin{pmatrix}\sqrt2&-1&0\\0&-\sqrt2&0\\0&0&1\end{pmatrix}?$$ – Angina Seng Jun 29 '19 at 16:50
• @LordSharktheUnknown, yes, it was a typo and I edited the question. Thank you. – Hasek Jun 29 '19 at 17:06
Can we reduce finding matrix roots to finding roots of Jordan blocks?
In (b), your matrix $$A$$ has $$2$$ Jordan blocks associated to the same eigenvalue $$2$$. Then $$A$$ admits an infinity of square roots. | {
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Here the infinity of matrices $$B\in M_2(\mathbb{C})$$ s.t. $$tr(B)=0,\det(B)=-2$$ are among the square roots of $$2I_2$$.
EDIT. I forgot to write that, in case (a), there are no more than $$2^n$$ square roots because, necessarily $$AB=BA$$.
You showed that $$A$$ has at least $$2^n$$ square roots. Assume $$B^2 = A$$. Since $$\sigma(B)^2 =\sigma(B^2) = \sigma(A)$$, it follows that $$\sigma(B) = \{\varepsilon_1\sqrt{\lambda_1}, \ldots, \varepsilon_n\sqrt{\lambda_n}\}$$ for some $$\varepsilon_1, \ldots, \varepsilon_n \in \{-1,1\}$$ where $$\sigma(A) = \{\lambda_1, \ldots \lambda_n\}$$. Set $$D = \operatorname{diag}(\varepsilon_1\sqrt{\lambda_1}, \ldots, \varepsilon_n\sqrt{\lambda_n})$$. $$A$$ has $$n$$ distinct eigenvalues so it is diagonalizable, i.e. there exists an invertible matrix $$P$$ such that $$P^{-1}AP = \operatorname{diag}(\lambda_1, \ldots, \lambda_n) = D^2$$.
Assume that $$C^2 = A$$ and that $$\sigma(C) = \sigma(B)$$. $$B$$ and $$C$$ both have $$n$$ distinct eingenvalues so there exist invertible matrices $$Q,R$$ such that $$Q^{-1}BQ = D = R^{-1}CR$$ It follows $$Q^{-1}AQ = Q^{-1}B^2Q = (Q^{-1}BQ)^2 = D^2 = (R^{-1}CR)^2 = R^{-1}C^2R = R^{-1}AR$$
If $$e_1, \ldots, e_n$$ is the standard basis, it follows that $$A(Pe_i) = \lambda_iPe_i, A(Qe_i) = \lambda_iQe_i, A(Re_i) = \lambda_iRe_i$$.
Since $$\lambda_1, \ldots, \lambda_n$$ are all distinct, the eigenvectors are unique up to scalar multiplication so it follows that $$Qe_i = \sigma_i Pe_i, Re_i = \pi_i Pe_i$$ for some nonzero scalars $$\sigma_i, \pi_i$$, or $$Q = P\Sigma, R = P\Pi$$ for some invertible diagonal matrices $$\Sigma, \Pi$$.
Finally, $$B = QDQ^{-1} = (P\Sigma)D(P\Sigma)^{-1} = P(\Sigma D\Sigma^{-1})P^{-1} =\\ PDP^{-1} = P(\Pi D\Pi^{-1})P^{-1} = (P\Pi)D(P\Pi)^{-1} = R^{-1}DR = C$$ | {
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Hence $$B$$ is uniquely determined by its spectrum $$\sigma(B)$$, which is uniquely defined by choosing $$\varepsilon_i \in \{-1,1\}$$ which can be done it $$2^n$$ ways. We conclude that there are $$2^n$$ such matrices $$B$$. | {
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The L2 norm calculates the distance of the vector coordinate from the origin of the vector space. As such, it is also known as the Euclidean norm as it is calculated as the Euclidean distance from the origin. The result is a positive distance value. It can help in calculating the Euclidean Distance between two coordinates, as shown below. There can be performance gain due to optimization. Norm(x) is the Euclidean length of a vecor x; same as Norm(x, 2). Chapter 8 Euclidean Space and Metric Spaces 8.1 Structures on Euclidean Space 8.1.1 Vector and Metric Spaces The set K n of n -tuples x = ( x 1;x 2:::;xn) can be made into a vector space by introducing the standard operations of addition and scalar multiplication 2. To take this point home, let’s construct a vector that is almost evenly distant in our euclidean space, but where the cosine similarity is much lower (because the angle is larger): The "-norm" (denoted with an uppercase … On R2 let jjjjbe the usual Euclidean norm and set jj(x;y)jj0= max(jxj;jyj). The squared Euclidean norm is widely used in machine learning partly because it can be calculated with the vector operation xᵀx. Example 1 (Euclidean norm on IR2). In simple terms, Euclidean distance is the shortest between the 2 points irrespective of the dimensions. 2-Norm. Python Code The norm is a scalar value. Input array. The L2 norm or Euclidean norm of an array is calculated using the following formula: Note that we will use the default value for the ord parameter for most of our code examples. I haven't found the equivalent to norm(v) from MATLAB. linalg import norm #define two vectors a = np.array([2, 6, 7, 7, 5, 13, 14, 17, 11, 8]) b = … There is nothing special about the Euclidean norm. Vector norms are any function that fulfil the following criteria: 1. JonnyJohnson May 29, 2013, 6:21am #1. euclidean_distances (X, Y = None, *, Y_norm_squared = None, squared = False, X_norm_squared = None) [source] ¶ Compute the distance matrix between each pair from | {
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= False, X_norm_squared = None) [source] ¶ Compute the distance matrix between each pair from a vector array X and Y. Its documentation and behavior may be incorrect, and it is no longer actively maintained. The 2-norm of a vector is also known as Euclidean distance or length and is usually denoted by L 2. Moreover, this equals zero only when both x = 0 and y = 0. The L2 norm is calculated as the square root of the sum of the squared vector values. There can be performance gain due to the optimization See here and here for more details. In order for a matrix norm to be consistent with the linear operator norm, you need to be able to say the following: norm() is a vector-valued function which computes the length of the vector. Let’s discuss a few ways to find Euclidean distance by NumPy library. As a measure One of the most useful features of orthonormal bases is that they a↵ord a very simple method for computing the coordinates of a vector over any basis vector. Transformational function Syntax: Definition 8. The -norm is also known as the Euclidean norm.However, this terminology is not recommended since it may cause confusion with the Frobenius norm (a matrix norm) is also sometimes called the Euclidean norm.The -norm of a vector is implemented in the Wolfram Language as Norm[m, 2], or more simply as Norm[m].. Euclidean distance is the L2 norm of a vector (sometimes known as the Euclidean norm ) and by default, the norm() function uses L2 - the ord parameter is set to 2. Transformational function Syntax: It is, also, known as Euclidean norm, Euclidean metric, L2 norm, L2 metric and Pythagorean metric. Matrix Norms • If A is a Matrix and p is included in the calling sequence, p must be one of 1, 2, infinity, Frobenius, or Euclidean. Search all packages and functions. PROBLEM 1{5. Building on @GonzaloMedina's answer, I suggest you create a macro called \norm in the document's preamble, using either of the following two approaches: auto-size the double-bar "fence" | {
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document's preamble, using either of the following two approaches: auto-size the double-bar "fence" symbols: \newcommand{\norm}[1]{\left\lVert #1 \right\rVert} This will place double vertical bars around the command's argument. Vote. Euclidean Norm. The euclidean norm of a matrix considered as a vector in m2-space is a matrix norm that is consistent with the euclidean vector norm. It is defined as the root of the sum of the squares of the components of the vector. So, for our given vector x, the L² norm would be: Answer (1 of 2): The Euclidean Norm is our usual notion of distance applied to an n-dimensional space. The numpy module can be used to find the required distance when the coordinates are in the form of an array. Vector Norms: a. Vector Norms Given vectors x and y of length one, which are simply scalars xand y, the most natural notion of distance between xand yis obtained from the absolute value; we de ne the distance to be jx yj. ‖ is called a normed vector space. For efficiency reasons, the euclidean distance between a pair of row vector x and y is computed as: Is there a block that finds the norm of a vector in simulink? It can be calculated by numpy.linalg.euc(). Then the first-order norm normalization method is employed to achieve vector orthogonality. Answer for Euclidean length of a vector (k-norm) with scaling to avoid destructive underflow and overflow is. We calculated the Euclidean norm of this vector with the norm() command by simply type the variable ‘a’ inside the norm(). These vectors are usually denoted ˆ→s Let’s say we have a vector, . In this norm, all the components of the vector are weighted equally. In 2-D complex plane, the norm of a complex number is its modulus , its Euclidean distance to the origin. Euclidean space 5 PROBLEM 1{4. Another familiar norm would be the Euclidean norm for vectors x ∈ IR2. Roughly right. The L2 norm, represented as ||v||2 is calculated as the square root of the sum of the squared vector values.Clearly, the | {
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as ||v||2 is calculated as the square root of the sum of the squared vector values.Clearly, the norm is a calculation of … Matrix 2 … Let us instantiate the definition of the vector $$p$$ norm for the case where $$p=2 \text{,}$$ giving us a matrix norm induced by the vector 2-norm or Euclidean norm: Definition 1.3.5.1 . In this article to find the Euclidean distance, we will use the NumPy library. The Euclidean norm of a vector \vecu of coordinates (x, y) in the 2-dimensional Euclidean space, can be defined as its length (or magnitude) and is calculated as follows : norm(vecu) = sqrt(x^2+y^2) The norm (or length) of a vector \vecu of coordinates (x, y, z) in the 3-dimensional Euclidean space is defined by: The relationhip between the norm of a vector and the Euclidean distance between two vectors appears in several machine learning scenarios. Calculates the Euclidean vector norm (L_2 norm) of ARRAY along dimension DIM.Standard:. So every inner product space inherits the Euclidean norm and becomes a metric space. The two-norm of a vector in ℝ 3 vector = {1, 2, 3}; magnitude = Norm [vector, 2] √14 Norm [vector] == Norm [vector, 2] True Another important example of matrix norms is given by the norm induced by a vector norm. Gives the largest magnitude among each element of a vector. 8.207 NORM2 — Euclidean vector norms Description:. Working in a Euclidean plane, he made equipollent any pair of line segments of the same length and orientati… In this article to find the Euclidean distance, we will use the NumPy library. “The L2 norm of a vector can be calculated in NumPy using the norm() function with a parameter to specify the norm order, in this case 1.” Also, even though, not something I would do while programming in the real world, the ‘l” in l1, l2, might be better represented with capital letters L1, L2 for the python programming examples. Vote. Norms are Use the NumPy Module to Find the Euclidean Distance Between Two Points. In Euclidean space the length of a | {
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NumPy Module to Find the Euclidean Distance Between Two Points. In Euclidean space the length of a vector, or equivalently the distance between a point and the origin, is its norm, and just as in R, the distance between two points is the norm of their di erence: De nitions: The Euclidean norm of an element x2Rn is the number kxk:= q x2 1 + x2 2 + + x2 n: The Euclidean distance between two points x;x0 2Rn is Many equivalent symbols Now also note that the symbol for the L2 norm is not always the same. We recognize them as. De nition 2 (Norm) Let V, ( ; ) be a inner product space. Norm An inner product space induces a norm, that is, a notion of length of a vector. The most common norm, calculated by summing the squares of all coordinates and taking the square root. Chapter 8 Euclidean Space and Metric Spaces 8.1 Structures on Euclidean Space 8.1.1 Vector and Metric Spaces The set K n of n -tuples x = ( x 1;x 2:::;xn) can be made into a vector space by introducing the standard operations of addition and scalar multiplication The matrix 2-norm is the maximum 2-norm of m.v for all unit vectors v: This is also equal to the largest singular value of : The Frobenius norm is the same as … This library used for manipulating multidimensional array in a very efficient way. The Euclidean norm (two norm) for matrices is a natural norm to use, but it has the disadvantage of requiring more computation time than the other norms. For each and in , ∑ {∑ } {∑ } | | | | Remark: | | | | Definition. Consider the vector hx,yi ∈ IR2. I need to calculate the two image distance value. Given a Euclidean space E, any two vectors u,v 2 E are orthogonal i↵ ku+vk2 = kuk2 +kvk2. Note that the answer of Dznrm2 is a real value. If it overflows, then you find a large power of two M, divide all numbers by M, calculate the norm, and multiply by M. If overflow happens at 2 1023, then you have numbers greater than 2 500. In 2-D complex plane, the norm of a complex number is its modulus , its Euclidean | {
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than 2 500. In 2-D complex plane, the norm of a complex number is its modulus , its Euclidean distance to the origin. In this tutorial, we looked at different ways to calculate vector lengths or magnitudes, called the vector norms. ⋮ . The infinity norm of a matrix is the maximum row sum, and the 1-norm is the maximum column sum, all after taking absolute values. State-of-the-art dual-frame phase recovery techniques are evaluated, and it shows that the first-order norm normalization method outperforms the second-order norm normalization method. Definition 8. Euclidean length of a vector with no scaling: The squared Euclidean norm is widely used in machine learning partly because it can be calculated with the vector operation $\bs{x}^\text{T}\bs{x}$. In L-infinity norm, only the largest element has any effect. 0. In 1835, Giusto Bellavitis abstracted the basic idea when he established the concept of equipollence. Euclidean distance = √ Σ(A i-B i) 2. You can first calculate the sum of squares. The length of a vector is most commonly measured by the "square root of the sum of the squares of the elements," also known as the Euclidean norm. 1. In N-D space (), the norm of a vector can be defined as its Euclidean distance to the origin of the space. Example 1.2. When np.linalg.norm() is called on an array-like input without any additional arguments, the default behavior is to compute the L2 norm on a flattened view of the array.This is the square root of the sum of squared elements and can be interpreted as the length of the vector in Euclidean space.. Some, but not all, norms are based on inner products. CUDA Programming and Performance. The L² norm measures the shortest distance from the origin. The length of a vector is a nonnegative number that describes the extent of the vector in space, and is sometimes referred to as the vector’s magnitude or the norm. The Euclidean norm is the square root of the sum of the squares of the magnitudes in each dimension. 2. Another | {
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norm is the square root of the sum of the squares of the magnitudes in each dimension. 2. Another important example of matrix norms is given by the norm induced by a vector norm. The numpy norm of a vector or matrix is the maximum absolute value of all its components. In Euclidean spaces, a vector is a geometrical object that possesses both a magnitude and a direction defined in terms of the dot product. >> a = [4 1 5] b = norm(a) c = 3*b a = 4 1 5 b = 6.4807 c = 19.4422 >> For example, we created a vector that has three elements called ‘a’ as shown above in Matlab®. It might turn out to be quite helpful to recall the basic knowledge about In mathematics, a norm is a function from a real or complex vector space to the nonnegative real numbers that behaves in certain ways like the distance from the origin: it commutes with scaling, obeys a form of the triangle inequality, and is zero only at the origin.In particular, the Euclidean distance of a vector from the origin is a norm, called the Euclidean norm, or 2-norm, which may … A matrix norm defined in this way is said to be vector-bound'' to the given vector norm. Returns the Euclidean norm of the vector as a double. n o r m o f V e c t o r L 1 = n ∑ i = 1 | x i | L 2 = √ n ∑ i = 1 x 2 i L ∞ = m a x ( | x i | ) n o r m o f V e c … calculate the L2 norm that is calculated as the square root of the sum of the squared vector values. As a measure This is also called the spectral norm This is called the Frobenius norm, and it is a matrix norm compatible with the Euclidean vector norm. However, the two norm is compatible with the Frobenius norm, so when computation time is an issue, the Frobenius norm should be used instead of the two norm. Since the ravel() method flattens an array without making any copies and … (This proves the theorem which states that the medians of a triangle are concurrent.) This is perhaps the matrix norm that occurs most frequently in the literature. See here and here for more details. We will not | {
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norm that occurs most frequently in the literature. See here and here for more details. We will not define it … On R2 let jjjjbe the usual Euclidean norm and set jj(x;y)jj0= max(jxj;jyj). The norm (and therefore the inner product) measures the length, size, magnitude, or strength of the vector depending on what interpretation you are giving the vector. In the infinite-dimensional case, the sum is infinite or is replaced with an integral when the number of dimensions is uncountable. That is, the number of non-zero elements in a vector. The Euclidean norm of a vector measures the “length” or “size” of the vector. 1. If you were to set the ord parameter to some other value p, you'd calculate other p-norms. 1 Norms and Vector Spaces 2008.10.07.01 The induced 2-norm. — Page 112, No Bullshit Guide To Linear Algebra, 2017. This is the Euclidean norm which is used throughout this section to denote the length of a vector. (It would be more precise to use rather than here but the surface of a sphere in finite-dimensional space is a compact set, so the supremum is attained, and the maximum is correct.) Calculate euclidean norm of a vector. In the triangle depicted above let L1 be the line determined by x and the midpoint 1 2 (y + z), and L2 the line determined by y and the midpoint 12 (x + z).Show that the intersection L1 \L2 of these lines is the centroid. Properties of Euclidean distance are: There is an unique path between two points whose length is equal to Euclidean distance Norm type, specified as 2 (default), a different positive integer scalar, Inf, or -Inf.The valid values of p and what they return depend on whether the first input to norm is a matrix or vector, as shown in the table. Euclidean Norm of a vector. The L² norm of a single vector is equivalent to the Euclidean distance from that point to the origin, and the L² norm of the difference between two vectors is equivalent to the Euclidean distance between the two points. L² Norm / Euclidean Norm. I have the two | {
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to the Euclidean distance between the two points. L² Norm / Euclidean Norm. I have the two image values G=[1x72] and G1 = [1x72]. The L2 norm of a vector can be calculated in NumPy using the norm() function with default parameters.First, a 1×3 vector is defined, then the L2 norm of the vector is calculated.. What is L2 norm squared? computes the euclidean norm of vector containing double-complex elements NRM2 = sqrt ( X**H * X ) Parameters: N ( int [in]) – Number of elements in vector X. NRM2 ( pyopencl.Buffer [out]) – Buffer object that will contain the NRM2 value. In particular, the Euclidean distance of a vector from the origin is a norm, called the Euclidean norm, or 2-norm, which may also be defined as the square root of the inner product of a vector with itself. | {
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euclidean norm of vector | {
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# In how many ways can the students answer a 10-question true false examination?
(a) In how many ways can the students answer a 10-question true false examination?
(b) In how many ways can the student answer the test in part (a) if it is possible to leave a question unanswered in order to avoid an extra penalty for a wrong answer
For part (a) I've got the answer, it is $2^{10}$.
For part (b) I think the answer is $10 \times 2^9$ because the number of ways to choose the question to answer is 10 and in each selection the number of ways to answer the question is $2^9$ but the answer provided in the book is $3^{10}$.
Can someone explain to me?
-
Probably the book suggest that the student can leave as many questions unanswered as he wants. – barto Feb 10 '13 at 14:43
@barto. I agree. On (b) she/he has 3 possibilities to each question. So $3^{10}$. – Sigur Feb 10 '13 at 14:44
For your response to part b you should have $10 \times 2^9 + 2^{10}$ since "it is possible to leave a question unanswered" does not require an unanswer. But it is more likely to mean there can be any number of unanswers from 0 through to 10. – Henry Feb 10 '13 at 15:08
If the answer has to be $3^{10}$, then this means that in case (b) it is intended that for each question the student can choose 1`out of 3 possibilites: true, false, not telling.
-
In part b), for simplicity, let's reduce the problem to just two questions.
There are $2^2$ ways in which both questions may be answered true/false.
If a student does not answer question 1, there are still $2^1$ ways in which they can answer question 2, and vice versa. Thus there are $2\times 2^1$ ways in which only one question is answered.
Finally, there is just one way in which neither question is answered.
Putting this together, there are $2^2 + 2\times 2^1 + 1 = (2+1)^2 = 3^2$ ways of answering the questions. | {
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Extending this to ten questions, there are $2^{10}$ ways to answer all ten questions, $10\times 2^9$ for answering all but one question, $45 \times 2^8$ ways of answering all but two questions, etc, giving a total of $(2+1)^{10} = 3^{10}$.
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# Lemma used to prove $\left|HK\right|=\frac{\left|H\right|\left|K\right|}{\left|H \cap K\right|}$
Given a group $$G$$ and $$H,K \le G$$,then :
$$\left|HK\right|=\frac{\left|H\right|\left|K\right|}{\left|H \cap K\right|}$$
Where $$HK:=\left\{hk:h \in H ,k \in K\right\}$$
Lemma:
For $$h_1,h_2 \in H$$
$$hK=h'K \iff h(H \cap K)=h'(H \cap K)$$
We have:
$$HK=\bigcup_{h \in H}hK$$
Not every such left cosets of $$K$$ in $$H$$ are distinct, on the other hand the function $$\phi:hK \to K$$ with $$hk \mapsto k$$ is a bijection, so the number of elements in $$hK$$ is the same as that $$K$$'s , here I showed that the set of left cosets (equivalently right cosets) partitions the group.
By this we see that :$$\left|HK\right|=\left|\color{blue}{\text{the set consiting of all distinct left cosets }}hK\right|\left|K\right|$$
One concludes from the lemma that the number of such distinct left cosets is the same as $$\left|H: H \cap K\right|$$ but I don't know how such a conclusion is possible, how the lemma helps us?
It looks that $$hK \ne h^{'} K$$ iff $$h(H \cap K) \ne h^{'}(H \cap K)$$ and the order of the set of all such distinct $$h(H \cap K)$$ for $$h \in H$$ is $$\left|H: H \cap K\right|$$...
Also, it would be appreciated if someone gives me an example where such left cosets $$hk$$ are identical.
• For finding example identical left cosets look at monogen groups and its sub-groups.
– EDX
Oct 26 '20 at 18:34
• The formula only makes sense when the quantities are finite. But the equality $|HK||H\cap K| = |H||K|$ holds in the sense of cardinalities in all cases, so it should be used instead. Oct 26 '20 at 18:42
Consider the map $$\varphi: H/H\cap K\longrightarrow HK/K$$ by $$h(H\cap K)\mapsto hK$$.
1. This is a well defined map by your lemma $$\impliedby$$.
2. This map is injective by your lemma $$\implies$$.
3. This map is surjective by definition of $$HK$$. | {
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3. This map is surjective by definition of $$HK$$.
Therefore this is a natural one-to-one correspondence between these cosets, and the Product Formula follows immediately.
I happen to have written about this yesterday, so here's a link for you https://ml868.user.srcf.net/ExpositoryWritings/Groups3.pdf. There are a few typos I haven't fixed but I hope it is readable and somewhat inspiring.
You noted that in the union $$\bigcup_{h \in H} hK$$, some cosets appear more than once. If you are able to show that each distinct coset appears $$|H \cap K|$$ times in the union, then you can arrive at the desired conclusion.
The lemma implies that the only way $$hK=h'K$$ can happen (for $$h,h' \in H$$) is if $$h' = gh$$ for some $$g \in H \cap K$$. In particular, for a given coset $$hK$$, it appears in the union $$|H \cap K|$$ times as $$(gh)K$$ for each $$g \in H \cap K$$.
For simplicity: $$I=\{hK|h\in H\}$$ $$J=\{h(H\cap K)|h\in H\}$$ Notice that: $$|J|=|H:(H\cap K)|$$ And we have simply to prove that $$|I|=|J|$$ Thanks to the lemma the application: $$\omega: I \to J$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ hK \mapsto h(H\cap K)$$ Is a bijection, in fact firstly the application is well defined since: $$hK=h'K \Rightarrow^{\text{Lemma}} h(H\cap K)=h'(H\cap K)\Rightarrow \omega(hK)=\omega(h'K)$$ The application is also injective: $$\omega(hK)=\omega(h'K)\Rightarrow h(H\cap K)=h'(H\cap K)\Rightarrow^{\text{Lemma}} hK=h'K$$ And it's clearly surjective because for every $$h(H\cap K)\in J, \omega (hK)=h(H\cap K)$$ It follows $$|I|=|J|$$.
• Should not the function be defined as $\phi :I \to J$? and how do we know such a function is bijective? Oct 26 '20 at 18:41
• It's a matter of notation. Briefly this application associates to every $hK$, $h(H\cap K)$, Now I'll write the proof that the application is a bijection. Oct 26 '20 at 18:43 | {
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The equivalence relation $$(h,k)\sim (h',k')\stackrel{(def.)}{\iff} hk=h'k'$$ induces a partition of $$H\times K$$ into equivalence classes each of cardinality $$|H\cap K|$$, and the quotient set $$(H\times K)/\sim$$ has cardinality $$|HK|$$. Therefore, $$|H\times K|=|H||K|=|H\cap K| |HK|$$, whence (if $$H$$ and $$K$$ are finite, in particular if they are subgroups of a finite group) the formula in the OP. Hereafter the details.
(Note that the formula holds irrespective of $$HK$$ being a subgroup.)
Let's define in $$H\times K$$ the equivalence relation: $$(h,k)\sim (h',k')\stackrel{(def.)}{\iff} hk=h'k'$$. The equivalence class of $$(h,k)$$ is given by:
$$[(h,k)]_\sim=\{(h',k')\in H\times K\mid h'k'=hk\} \tag 1$$
Now define the following map from any equivalence class:
\begin{alignat*}{1} f_{(h,k)}:[(h,k)]_\sim &\longrightarrow& H\cap K \\ (h',k')&\longmapsto& f_{(h,k)}((h',k')):=k'k^{-1} \\ \tag 2 \end{alignat*}
Note that $$k'k^{-1}\in K$$ by closure of $$K$$, and $$k'k^{-1}\in H$$ because $$k'k^{-1}=h'^{-1}h$$ (being $$(h',k')\in [(h,k)]_\sim$$) and by closure of $$H$$. Therefore, indeed $$k'k^{-1}\in H\cap K$$.
Lemma 1. $$f_{(h,k)}$$ is bijective.
Proof.
\begin{alignat}{2} f_{(h,k)}((h',k'))=f_{(h,k)}((h'',k'')) &\space\space\space\Longrightarrow &&k'k^{-1}=k''k^{-1} \\ &\space\space\space\Longrightarrow &&k'=k'' \\ &\stackrel{h'k'=h''k''}{\Longrightarrow} &&h'=h'' \\ &\space\space\space\Longrightarrow &&(h',k')=(h'',k'') \\ \end{alignat}
and the map is injective. Then, for every $$a\in H\cap K$$, we get $$ak\in K$$ and $$a=f_{(h,k)}((h',ak))$$, and the map is surjective. $$\space\space\Box$$
Now define the following map from the quotient set:
\begin{alignat}{1} f:(H\times K)/\sim &\longrightarrow& HK \\ [(h,k)]_\sim &\longmapsto& f([(h,k)]_\sim):=hk \\ \tag 3 \end{alignat}
Lemma 2. $$f$$ is well-defined and bijective.
Proof. | {
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Lemma 2. $$f$$ is well-defined and bijective.
Proof.
• Good definition: $$(h',k')\in [(h,k)]_\sim \Rightarrow f([(h',k')]_\sim)=h'k'=hk=f([(h,k)]_\sim)$$;
• Injectivity: $$f([(h',k')]_\sim)=f([(h,k)]_\sim) \Rightarrow h'k'=hk \Rightarrow (h',k')\in [(h,k)]_\sim \Rightarrow [(h',k')]_\sim=[(h,k)]_\sim$$;
• Surjectivity: for every $$ab\in HK$$ , we get $$ab=f([(a,b)]_\sim)$$. $$\space\space\Box$$
Finally, the formula holds irrespective of $$HK$$ being a subgroup, which was never used in the proof. | {
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# Homework Help: Linear algebra question
1. Jan 20, 2013
### Mdhiggenz
1. The problem statement, all variables and given/known data
Explain why each of the following algebraic rules will not work in general when the real numbers a and b are placed by nxn matrix A and B
(a+b)2=a2+2ab+b2
This question is a bit confusing to me and I have no idea how to even start this problem
2. Relevant equations
3. The attempt at a solution
2. Jan 20, 2013
### Ray Vickson
Try a couple of small examples: let A and B be some specific 2x2 matrices. Compute both sides and see if you get equality or not.
3. Jan 20, 2013
### Mdhiggenz
So lets say if I make A the identity matrix, and B some random matrix. I would computer one side (a+b)2 where a=A and b=B?
and do the same for the other side?
Thanks
4. Jan 20, 2013
### Ray Vickson
Don't pick one of them to be the identity; pick two different matrices, neither one and identity matrix. (Try it both ways to see why; that is, try it first when one is the identity, and then do it again with two non-identity matrices. Look at what happens, and that will give you a clue for solving the problem.)
5. Jan 20, 2013
### LCKurtz
Two matrices I wouldn't ever use for trying to find a counterexample would be the Identity and Zero matrices. They have too many special properties of their own.
6. Jan 20, 2013
### Mdhiggenz
Lckurtz: thanks for the tip. You think you can give me a matrix that you would always use to prove, something simple.
And Ray here is my work, would you agree that the reason they could not be equal is due to the 2ab term?
7. Jan 20, 2013
### SammyS
Staff Emeritus
In particular compare AB with BA .
8. Jan 20, 2013
### Mdhiggenz
So what I did was incorrect?
9. Jan 20, 2013
### SammyS
Staff Emeritus
What you did was fine, and yes, it is the 2AB that messes you up.
Did you compare AB with BA ? --- to see why it's the 2AB that messes you up ?
10. Jan 20, 2013
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10. Jan 20, 2013
### LCKurtz
No, there isn't a single special matrix. In the given example, as I think you are now aware, the point is that matrices don't generally satisfy $AB=BA$ so you don't get $2AB$ in the expansion when you multiply them out. So if you are looking for a counterexample, you don't want to choose, for example, A = the zero matrix, because it commutes with everything just because it always gives all zeroes. That is a general idea: If you are trying to prove some identity doesn't always work, look for an example by choosing variables that are unlikely to "accidentally" work because they have special properties. So if I were looking for examples in the current case, I wouldn't use the zero matrix, the identity matrix, a lower triangular matrix or a symmetric matrix. I would try a more "random" matrix that doesn't have any special properties. Does that make sense to you?
11. Jan 20, 2013
### Mdhiggenz
I did AB and compared it to BA they are different. This I understand, what I don't understand is what that has to do with 2AB?
Or is what your saying is that in order to show that something is 100% true we must look at both cases. For instance A2 no matter what will be A2. Same thing goes with B, and A+B will be the same as B+A. However 2AB does not equal 2BA therefore it cant possibly be equal.
12. Jan 20, 2013
### Mdhiggenz
Absolutely thanks for the clear explanation.
13. Jan 20, 2013
### LCKurtz
Re. your comment in red above. Expand $(A+B)(A+B)$ symbolically the long way (distributive law) being careful about the order. What do you get?
14. Jan 20, 2013
### Mdhiggenz
Not quite sure what you mean, it says math processing error
15. Jan 20, 2013
### LCKurtz
Do it just by manipulating the A and B's with a pencil and paper. Expand (A+B)(A+B).
16. Jan 20, 2013
### Mdhiggenz
Oh I see now A2+AB+BA+B2. | {
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16. Jan 20, 2013
### Mdhiggenz
Oh I see now A2+AB+BA+B2.
So making AB+BA=2AB that is implying that AB=BA which means you would be able to add them and get 2AB which is incorrect because AB does not necessarily equal BA.
?
17. Jan 20, 2013
### vela
Staff Emeritus
Yup, that's right.
Multiplication of real numbers has a property called commutativity, which means the order you multiply two numbers in doesn't matter, e.g. $2\times 4 = 4\times 2$. Hence, if a and b are real numbers, you can always say that (a+b)2 equals a2+2ab+b2.
Matrix multiplication doesn't have this property, as you found. When you're multiplying matrices, you have to remember that the order in which they appear is important. You can't just move them around like you are used to from working with real numbers.
18. Jan 23, 2013
### EM_Guy
It is worth considering all of this in light of vector spaces. Scalars (aka - 1-vectors, aka - 1x1 matrices) don't exist in the same vector spaces in which N x N matrices exist (where N does not equal 1). Addition and scalar multiplication are defined distinctly in each distinct vector space. Matrix multiplication is something else altogether. So, when studying linear algebra, you can't just assume that "plus" means "plus" or that "times" means "times." You can't even assume that "zero" means "zero." The addition operation is defined differently for each vector space. Thus, the idea of adding a 3x1 vector to a 2x2 matrix together doesn't make sense, because those two elements exist in different vector spaces; therefore, "addition" between two such elements is not even defined. Likewise, if v exists in a vector space V, then 0v = 0 is true, but the first 0 and the second 0 are not the same. The first 0 is a scalar. The second 0 is the null vector that exists in the vector space V. | {
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Again, matrix multiplication is very different than the multiplication between two scalars, and it is also different than scalar multiplication in any given vector space. Matrix multiplication actually maps matrices from one vector space to another vector space (or to the same vector space when you have an NxN matrix multiplied by another NxN matrix). It should therefore not surprise us that matrix multiplication is not commutative.
Here's a challenging question: Why do you suppose that matrix multiplication is defined as it is? That is, what was the motivating reason to define matrix multiplication the way it has been defined? But perhaps that is a question for a different thread. | {
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# Geometry notation: what does $m\angle ABC$ mean?
I see in some math formulation that a certain angle is called, let's say
$$\angle ABC$$
but there is a letter put in front of the angle notation.
$$m\angle ABC$$
What does the $m$ represent here? A factor?
• Try to use LaTeX to write accurately mathematics here. You also didn't write what the angle is called, "let say..." – DonAntonio Oct 30 '13 at 17:23
• <ABC and m<ABC.. I dont know why it doesnt show up in the question.. – Hilbert Oct 30 '13 at 17:24
• Do you mean $\theta(x)$? – Don Larynx Oct 30 '13 at 17:24
• You're going to give more context, for example to write down a complete exercise. The letter $\;m\;$ is many times reserved for slope in analytic geometry. – DonAntonio Oct 30 '13 at 17:25
• It's because the < hides the subsequent text when the post is rendered. I have fixed it. – Cameron Buie Oct 30 '13 at 17:26
$\angle ABC$ : The angle ABC
$m\angle ABC$: The measure of $\angle ABC$
So, when $\angle ABC \cong \angle DFG$ , that means, $m\angle ABC = m\angle DFG$
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The notation $m\angle ABC$ typically denotes "the measure of angle $ABC$."
• To clarify, the $m$ is sometimes used to distinguish between the measure of the angle ($m\angle ABC$ = a number, in degrees/radians) and the actual angle itself (the geometric object $\angle ABC$). – angryavian Oct 30 '13 at 17:29
• Thanks for all your replies and Cameron has understood the badly formulated question correct. But if the angle is, lets say <ABC = 50 degrees. What is the need of m<ABC then? Doesnt <ABC already denote the measure of the angle? – Hilbert Oct 30 '13 at 17:31
• @Hilbert No, $\angle ABC$ is used to denote the angle itself, as a geometric object. Of course, this is a minor distinction, and some authors may choose to be lax in the notation. – angryavian Oct 30 '13 at 17:36
• @Hilbert: As blf points out, $\angle ABC$ denotes the angle, itself, while $m\angle ABC$ is its measure. For example, suppose we have an equilateral triangle with vertices $A,B,C.$ Then $\angle ABC$ occurs at the intersection of the segments $AB$ and $BC$, while $m\angle ABC$ is the measure of that angle. The distinction isn't always important, but sometimes it is. – Cameron Buie Oct 30 '13 at 17:36
• Thanks Cameron & blf.. Perfectly explained. I have been searching for geometry textbooks that explains things like this and deals with elementary things like Congruence, Arc, Geometric shapes like triangles and so on. Any recommendations? – Hilbert Oct 30 '13 at 19:33
∠ABC refers to the physical angle itself while m∠ABC refers to its measure. There isn't any difference between these two and it is absolutely correct to use any of them...Some people just want to formalize things; these people think-"I have a car that is priced at 50000dollars, so i am driving a car and not 50000dollars" and so want to differentiate between little everything. | {
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Question
# From a two digit number N, we subtract the number with the digits reversed and find that the result is a positive perfect cube. Then:
A
N cannot end in 5
B
N can end in any digit other than 5
C
N does not exist
D
there are exactly 7 values for N
E
there are exactly 10 values for N
Solution
## The correct option is D there are exactly 7 values for NLet the two digit number be 10a+bN = 10a+bN' = reversed number = 10b+aN-N' = 9(a-b)N-N' is positive perfect cube.$$\therefore$$ a>bFor 9(a-b) be perfect cube, a-b = 3$$\therefore$$ b $$\epsilon$$ [0, 6] $$\longrightarrow$$ 7 values$$\therefore$$ a $$\epsilon$$ [3, 9] $$\longrightarrow$$ 7 values$$\therefore$$ Total 7 values are possible for N.Mathematics
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# set operations proofs | {
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\] The empty set, $$\emptyset$$, is the set with no elements: ... Logicians sometimes describe ordinary mathematical proofs as informal, in contrast to the formal proofs in natural deduction. \mathbf{R} = \mathbf{Q} \cup \overline{\mathbf{Q}}\,.\], Written $$A\cap B$$ and defined But from the definition of set difference, we see that though they can be proven also using some of these properties (after those properties are proven, needless to say). Theorem: For any sets, $$\overline{A\cup B}= \overline{A} \cap \overline{B}$$. 13. . Then the set of students taking classes this semester is When writing informal proofs, the focus is on readability. Hence . Proof for 11: Let x be an arbitrary element in the universe. The set $$\overline{B}$$ is the set of all values not in $$B$$. Theorem: $$A-(B\cup C)= (A-B)\cap(A-C)$$. but also for others. &= \{x\mid x\in A \wedge x\in \overline{B}\} \\ when we're working with real numbers, probably $$U=\mathbf{R}$$. Back to Schedule We can use the set identities to prove other facts about sets. Hence . For example, (b) can be proven as follows: A &= \{x\mid x \in (\overline{A}\cap\overline{B}) )\} \\ If , then and . 1 - 6 directly correspond (See section 2.2 example 10 for that. x A A The standard notation for irrational numbers should now make a lot of sense: with universal set $$\mathbf{R}$$, the irrationals ($$\overline{\mathbf{Q}}$$) are the complement of the rationals ($$\mathbf{Q}$$). x\in S \wedge x\notin{S}\,. x Back to Table of Contents. A $A\cup B\cup C \cup D\,,\\A\cap B\cap C \cap D\,.$. A to identities &= \{x\mid x\in (A \cap \overline{B})\} \\ If we need to do union/intersection of a lot of things, there is a notation like summation that is used occasionally. Theorem: For any sets, $$|A\cap B|\le|A|$$ and $$|A\cap B|\le|B|$$. B B, by 7 6. and that of ------- Identity Laws 5. is also in This name is used since the basic method is to choose an arbitrary element from one set and “chase it” until you prove it must be in | {
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method is to choose an arbitrary element from one set and “chase it” until you prove it must be in another set. \overline{A\cup B} \] Alternative proof For example: Those identities should convince you that order of unions and intersections don't matter (in the same way as addition, multiplication, conjunction, and disjunction: they're all commutative operations). First by 15 by the commutativity of 4. ( cf. ) by the definition of set union. Proof for 4: and Proof for 12: (a) ? Proof for 8: (a) If then . The primary purpose of this section is to have in one place many of the properties of set operations that we may use in later proofs. &= \overline{A}\cap\overline{B}\,.\quad{}∎ ) by the definition of ( B - A ) . Since , since . A Since (use "addition" rule), and between follows. The properties 1 6 , and 11 Thus, A−B = A∩Bc. By definition of set difference, x ∈ A− B. by the definition of . ( B - A ) 2. 12. if and only if and Proof for 13: Since , . &= \{x\mid \neg(x \in A)\wedge \neg(x\in B )\} \\ Properties of Set Operation Subjects to be Learned . 9. Theorem For any sets A and B, A∩B ⊆ A. B Often not explicitly defined, but implicit based on the problem we're looking at. ( B Since we're doing the same manipulations, we ended up with the same tables. A-B ------- Domination Laws $\sum_{i=1}^{n} i^2\,.$. and vice versa. if and only if We are going to prove this by showing that every element that is in by 1. Let the sets $$S_1,S_2,\ldots ,S_n$$ be the students in each course. For any one of the set operations, we can expand to set builder notation, and then use the logical equivalences to manipulate the conditions. = A by 3, For example, suppose there are $$n$$ courses being offered at ZJU this semester. Like the domain for quantifiers, it's the set of all possible values we're working with. Hence . Next -- Recursive Definition Notice the similarity between the corresponding set and logical operators: $$\vee,\cup$$ and $$\wedge,\cap$$ and | {
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between the corresponding set and logical operators: $$\vee,\cup$$ and $$\wedge,\cap$$ and $$\overline{\mbox{S}},\neg$$. B ------- Distributive Laws We have used the choose-an-element method to prove Propositions 5.7, 5.11, and 5.14. \overline{\mathbf{Q}} = \mathbf{R}-\mathbf{Q} \,.\]. Look familiar? Then there must be an element $$x$$ with $$x\in(A-B)$$, but $$x\notin A$$. by the definition \begin{align*} 1. Then by the definition of the operators, Hence A satisfies the conditions for the complement of . The “more formal” version has more steps and leaves out the intuitive reason (that might help you actually remember why). Alternative proof: This proof might give a hint why the equivalences and set identities tables are so similiar. Less Formal Proof: The set $$A-B$$ is the values from $$A$$ with any values from $$B$$ removed. Note here the correspondence of Here is an example. = ( A (See example 10 for an example of that too.). B ) Here the only if part is going to be proven. &= A\cap \overline{B}\cap A\cap \overline{C} \\ Be careful with the other operations. of propositional logic, and 7 - 11 also follow immediately from them as illustrated below. &= \{x\mid x\notin (A\cup B)\} \\ With similar proofs, we could prove these things: When doing set operations we often need to define a. ( cf. ) \[\bigcup_{i=1}^{n} S_i\,. There is no logical version of set difference, or set version of exclusive or (at least as far as we have defined). x \end{align*}. Since A Additional properties: Then there is an element x that is in , i.e. For any one of the set operations, we can expand to set builder notation, and then use the logical equivalences to manipulate the conditions. The students taking, This is exactly analogous to the summation notation you have seen before, except with union/intersection instead of addition: A. The if part can be proven similarly. Hence . If and , then , Also since , . the commutativity of Proof: Suppose for contradiction that there is an | {
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and , then , Also since , . the commutativity of Proof: Suppose for contradiction that there is an element $$x\in S\cap\overline{S}$$. x Could have also given a less formal proof. Proof: By definition of the set operations, A These can also be proven using 8, 14, and 15. equalities involving set operations intersection of sets subset relations proofs of equalities proofs of subset relations Contents . Proof for 9: Let x be an arbitrary element in the universe. by the definition of . Set. A. x\in S \wedge x\in\overline{S} \\ \[A\cap B = \{x \mid x\in A\wedge x\in B\}\,\\ Theorem For any sets A and B, B ⊆ A∪ B. Proof… by the distribution Also . A Hence . B . Proof for 6: By the definition of the equality of sets, we need to prove that Let x be an arbitrary element in the universe. Here are some basic subset proofs about set operations. Hence does not hold. Since , . • Applying this to S we get: • x (x S x S) which is trivially True • End of proof Note on equivalence: • Two sets are equal if each is a subset of the other set. Copyright © 2013, Greg Baker. if and only if | {
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Strategy Vs Tactics Marketing, 1966 F100 Engine Swap, Inheriting A House From Your Parents Uk, Bank Of America Mastercard Login, Lola Car Price, Georgia Highlands College Basketball Coach, | {
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# Mahler measure of a totally positive, expanding algebraic integer
Consider a degree-$d$ algebraic integer $\alpha$ all of whose conjugates (including itself) are real numbers greater than 1. Its Mahler measure $M(\alpha)$ is simply equal to the norm $N(\alpha)$. Since $N(\alpha-1) \geq 1$, an application of $H\ddot{o}lder's$ inequality gives the lower bound $N(\alpha) \geq 2^d$, with equality achieved only by $\alpha = 2$.
Question. Does there exist $C > 2$ such that $N(\alpha) \geq C^d$ for all $\alpha \neq 2$?
• Interesting question. I assume you're familiar with work of Smyth, Flammang, etc. on lower bounds for the Mahler measure of totally positive algebraic integers? (These give a lower bound of $C=1.722...$ for $M(\alpha)^{1/d}$ with finitely many exceptions, but valid for all totally positive integers $\alpha$, not just those with all conjugates $>1$). Apr 20, 2016 at 21:13
• On the hypothesis that $\alpha$ is algebraic of degree $d$, the value $\alpha=2$ can only arise in the case $d=1$, so the equality $N(\alpha)=2^d$ is achieved only for $d=1$, $\alpha=2$. Apr 20, 2016 at 22:58
• @BobbyGrizzard Is the "with finitely many exceptions" part effective? Apr 21, 2016 at 5:50
• @FanZheng yes. Here is one effective result: ams.org/journals/mcom/1996-65-213/S0025-5718-96-00664-3/… edit: and here is the one I was probably looking at yesterday, due to Wu and Mu (Quanwu Mu, not the OP!): sciencedirect.com/science/article/pii/S0022314X12001989 Apr 21, 2016 at 11:58
• @BobbyGrizzard NOT the OP LOL Apr 22, 2016 at 4:19
Suppose that $\alpha$ is totally real algebraic integer, and that all its conjugates are greater than $1$. Suppose also that $\alpha \ne 2$.
Note the elementary inequality for $x \in (1,\infty) \setminus \{2\}$:
$$\ln|x| \ge \frac{\ln|x-1| + \ln|x-2|}{3} + \ln C,$$
where $C = 2^{1/3} 3^{1/2} = 2.18 \ldots > 2$. (Equality holds precisely at $3 \pm \sqrt{3}$.) | {
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where $C = 2^{1/3} 3^{1/2} = 2.18 \ldots > 2$. (Equality holds precisely at $3 \pm \sqrt{3}$.)
Denote the conjugates of $\alpha$ by $\alpha_i$. Assuming that $\alpha \ne 1,2$, we see that, because $\alpha - 1$ and $\alpha - 2$ are non-zero algebraic integers, we have inequalities
$$\sum_{i=1}^{d} \ln | \alpha_i - 1| = \ln N(\alpha -1) \ge 0,$$ $$\sum_{i=1}^{d} \ln |\alpha_i - 2| = \ln N(\alpha -2) \ge 0.$$
Hence we deduce that $\displaystyle{\ln N(\alpha) = \sum_{i=1}^{d} \ln |\alpha_i| \ge d \cdot \ln C}$, and thus $N(\alpha) \ge C^{d}$ where $C > 2$.
Gypsum's argument is really nice. In the same spirit, we have the following inequality: $$\ln|x| \geq \frac{2\ln|x-1| + \ln|x-2|}{5} + \ln\sqrt{5}$$ which is achieved at $x = \frac{5\pm \sqrt{5}}{2}$. This way we obtain the optimal constant $C = \sqrt{5}$. Perhaps some further tweaks can yield even larger $C$ with finitely many exceptions, as in the work of Smyth and Flammang.
I wonder how far this approach can be extended. Smyth showed that for totally positive algebraic integers (whose conjugates are not necessarily greater than 1), their $M(\alpha)^{\frac{1}{d}}$ are dense beyond 1.73, which is very close to the lower bound cited by @BobbyGrizzard. Do we have a similar situation here? As a first step, it would be good to construct infinitely many $\alpha$ that give upper bound to $C$. For example, consider the $n$-th $Chebyshev$ polynomial $T_n(x)$. Then the monic polynomial $(-x)^nT_n(\frac{2}{x}-1)$ has all its roots greater than 1. In this case $N(\alpha) = 2^{2n-1}$, suggesting $C \leq 4$. | {
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• Welcome to mathoverflow! This is really a great question, but as you are new here, please be informed that the answers are just "answers"; new questions included in answers may not get enough attention. If you want to probe further, please post a separate question; you may include a link back to this question. Apr 22, 2016 at 6:25
• One can usually extend this approach a certain amount, but not as far as the limit. The fact that equality is achieved at $x = (5+\sqrt{5})/2$ with the optimal bound is related to the fact that $x - 1$ and $x - 2$ are both units. But there is no reason this had to happen. BTW, to improve the bound beyond this point (with the exceptions $x = 2$ and $(5 \pm \sqrt{5})/2$, you simply need to add a very small multiple of the term $\log(x^2 - 5 x + 5)$. Apr 22, 2016 at 13:11
• @FanZheng Thanks for your advice :) Apr 22, 2016 at 17:33 | {
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+0
# What is a number 12, 24, and 30 have in common?
0
193
4
+13
A store would like to set up fish tanks that contain equal numbers of angle fish, sword fish, and guppies. What is the greatest number of tanks that can be set up if the store has 12 angle fish, 24 sword fish, and 30 guppies?
Aug 29, 2018
#1
+374
+4
The GCF for $$30, 24,$$ and $$12$$ is $$6$$. This means that there will be $$6$$ tanks. (Correct me if I'm wrong, I'm not so sure about this one. I may have read it wrong).
- Daisy
Aug 29, 2018
#3
+13
+2
Yep ur right it was a question on my math hw I thought I’d post. Lol
TruppaGirl Aug 29, 2018
#2
+3994
+3
GCF common factor is the way to solve it! The GCF is 6, because it's the greatest number that evenly divided 12,24, and 30.
Aug 29, 2018
#4
+21848
+4
A store would like to set up fish tanks that contain equal numbers of angle fish, sword fish, and guppies.
What is the greatest number of tanks that can be set up if the store has 12 angle fish, 24 sword fish, and 30 guppies?
Formula:
$$\begin{array}{|rcll|} \hline \gcd(a,b,c) &=& \gcd( \gcd(a,b), c ) \\ \gcd(a,b ) &=& \gcd (a-b,b ) & a \ge b \\ \gcd(a,a) &=& a & a \ge 0 \\ \gcd(a,b) &=& \gcd(b,a) \\ \hline \end{array}$$
$$\begin{array}{|rcll|} \hline \gcd(30,24,12) &=& \gcd( \gcd(30,24), 12 ) \\\\ && \gcd(30,24) \\ && = \gcd(30-24,24) \\ && = \gcd(6,24) \\ && = \gcd(24,6) \\ && = \gcd(24-6,6) \\ && = \gcd(18,6) \\ && = \gcd(18-6,6) \\ && = \gcd(12,6) \\ && = \gcd(12-6,6) \\ && = \gcd(6,6) \\ && = 6 \\\\ \gcd(30,24,12) &=& \gcd( \gcd(30,24), 12 ) \\\\ &=& \gcd( 6, 12 ) \\ &=& \gcd( 12, 6 ) \\ &=& \gcd( 12-6, 6 ) \\ &=& \gcd( 6, 6 ) \\ &=& 6 \\ \hline \end{array}$$
Aug 30, 2018
edited by heureka Aug 30, 2018 | {
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# triangular_lattice_graph#
triangular_lattice_graph(m, n, periodic=False, with_positions=True, create_using=None)[source]#
Returns the $$m$$ by $$n$$ triangular lattice graph.
The triangular lattice graph is a two-dimensional grid graph in which each square unit has a diagonal edge (each grid unit has a chord).
The returned graph has $$m$$ rows and $$n$$ columns of triangles. Rows and columns include both triangles pointing up and down. Rows form a strip of constant height. Columns form a series of diamond shapes, staggered with the columns on either side. Another way to state the size is that the nodes form a grid of m+1 rows and (n + 1) // 2 columns. The odd row nodes are shifted horizontally relative to the even rows.
Directed graph types have edges pointed up or right.
Positions of nodes are computed by default or with_positions is True. The position of each node (embedded in a euclidean plane) is stored in the graph using equilateral triangles with sidelength 1. The height between rows of nodes is thus $$\sqrt(3)/2$$. Nodes lie in the first quadrant with the node $$(0, 0)$$ at the origin.
Parameters:
mint
The number of rows in the lattice.
nint
The number of columns in the lattice.
periodicbool (default: False)
If True, join the boundary vertices of the grid using periodic boundary conditions. The join between boundaries is the final row and column of triangles. This means there is one row and one column fewer nodes for the periodic lattice. Periodic lattices require m >= 3, n >= 5 and are allowed but misaligned if m or n are odd
with_positionsbool (default: True)
Store the coordinates of each node in the graph node attribute ‘pos’. The coordinates provide a lattice with equilateral triangles. Periodic positions shift the nodes vertically in a nonlinear way so the edges don’t overlap so much.
create_usingNetworkX graph constructor, optional (default=nx.Graph)
Graph type to create. If graph instance, then cleared before populated. | {
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Graph type to create. If graph instance, then cleared before populated.
Returns:
NetworkX graph
The m by n triangular lattice graph. | {
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# How To Find Total Distance Traveled By Particle Calculus | {
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For example, you might find a distance of 10. The distance traveled in each interval is thus 4 times 20, or 80 feet, for a total of 80 + 80 = 160 feet. At time t=2, the position of the particle is x(2)=0. Exercise 1: Calculate the total distance traveled given the velocity equation. 45m if you calculate in 1-minute chunks, 10. So the particle has gone over 10 seconds 12. A particle moves along the x-axis. But no, to find x you have to use physics equations, not curve equations. (a) Find the instantaneous velocity at time t and at t = 3 seconds. The total DISPLACEMENT would be the ∫v (t) from 1 to 6. (b) Find the acceleration of the particle at time t = 1. For each problem, find the maximum speed and times t when this speed occurs, the displacement of the particle, and the distance traveled by the particle over the given interval. You can also find Total distance traveled by a particle - Mathematics ppt and other Engineering Mathematics slides as well. How to Find Total Distance Calculus: Steps. behind the fifth method of approximation called Simpson's Rule. Remember: Velocity is the rate of change in position with respect to time. For example, D 2 and D 3 are =. Include units. However, it was a long time ago with crappy looking graphs. 5? Is the velocity of the particle increasing at time t = 1. A particle moves in a straight line with velocity t^-2 - 1/9 ft/s. Advanced Placement Calculus AB APCD. (d) Find the total distance traveled by the particle from t = 0 to t = 2. 4) v(t) = 3t2 — 18t; ì)ó4anQ. of a particle moving on a horizontal axis is shown below. How do you find the total displacement for the particle whose position at time #t# is given by How many values of t does the particle change direction if a particle moves with acceleration What is the position of a particle at time #t=2# if a particle moves along the x axis so that at. A Dodge Neon and a Mack truck leave an intersection at the same time. A person is standing on top of the Tower of Pisa and throws | {
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leave an intersection at the same time. A person is standing on top of the Tower of Pisa and throws a ball directly upward with an initial velocity of 96 feet per second. AP Calculus Particle Motion Worksheet For #6 - 10: A particle moves along a line such that its position is s ( t ) = t 4 - 4 t 3. c) Set up an integral to find the total distance traveled by the particle in the interval [0, 4]. EDIT - I made a slight mistake the first time I posted this. Thus, its average speed = distance/time = 2π/3 and its average velocity = displacement/time = 0. To summarize, we see that if velocity is sometimes negative, a moving object's change in position different from its distance traveled. If a body moves along a straight line with velocity v = t 3 + 3t 2, find the distance traveled between t. The Neon heads east at an average speed of 30 mph, while the truck heads south at an average speed of 40 mph. These deriv-atives can be viewed in four ways: physically, numerically, symbolically, and graphically. The velocity function (in meters per second) is given for a particle moving along a line. The Attempt at a Solution I cannot think of a way to do it keeping it in terms of t. 45m if you calculate in 1-minute chunks, 10. Homework Equations Can't think of any 3. 25 t and positive on the interval 1. (a) Find the acceleration of the particle at time t 3. A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. Finding the total distance {eq}(\Delta x){/eq} traveled by the particle. What is the velocity after 3 seconds? C. With this information, it's possible to find the distance the object has traveled using the formula d = s avg × t. Find the speed of the particle at time t 3 seconds. 1 Integral as Net Change Calculus. B) Find the total distance travelled by the particle. The Riemann sum approximating total distance traveled is v t k Δt, and we are led to the. Att =1, theparticleisattheorigin. Distance is the measure of “how | {
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v t k Δt, and we are led to the. Att =1, theparticleisattheorigin. Distance is the measure of “how much ground an object has covered” during its motion while displacement refers to the measure of how far out of place is an object. Find the velocity when t = 3 D. Distance and displacement are different quantities, but they are related. Justify your answer. Given the position function, find the total distance. (e) The displacement of the particle. (b) Find the average velocity of the particle for the time period 06. 5 meters to the right and then 12. When you try to find the distance a moving object has traveled, two pieces of information are vital for making this calculation: its speed (or velocity magnitude) and the time that it has been moving. Please show detailed step-by-step explanations for both parts show more A particle is moving along the x-axis at velocity v(t) = 3t^2 - 12t + 9, 0 ≤ t ≤ 3. Let f(x) = e2x. We have to evaluate this to find the velocity at any particular time. (b) When is the particle at rest? Moving to the right? Moving to the left? Justify your answers. Approximately where does the particle achieve its greatest positive acceleration on the interval 0, b ? t a 16. The particle may be a “particle,” a person, a car, or some other moving object. B) Find all for which the velocity is increasing. Video transcript. (f) Find the displacement of the particle during the first five seconds. When calculating the total distance traveled by the particle, consider the intervals where v(t) ≤ 0 and the intervals where v(t) ≥ 0. Using the result from part (b) and the function V Q from part (c), approximate the distance between particles P and Q at time t = 2. The velocity function is v(t) = - t^2 + 6t - 8 for a particle moving along a line. c) Find the particle's total distance traveled by setting up ONE integral and using your calculator. Chapter 10 Velocity, Acceleration, and Calculus The first derivative of position is velocity, and the second derivative is | {
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Acceleration, and Calculus The first derivative of position is velocity, and the second derivative is acceleration. (e) Use geometry to nd the distance traveled to the left. I said 8 seconds instead of 8 feet. To maximize the distance traveled, take the derivative of the coefficient of i with respect to θ and set it equal to zero: d d θ ( v 0 2 sin 2 θ g ) = 0 2 v 0 2 cos 2 θ g = 0 θ = 45 °. ) but you can also compute traveled distance having time and average speed (given in different units of speed mph, kmh, mps yds per second etc. travels to point 2, and reverses direction and travels to point 3, then its distance travelled is 2 + 1 + 1 = 4. In a physics equation, given a constant acceleration and the change in velocity of an object, you can figure out both the time involved and the distance traveled. b) Use your an swer to part (a) to find the position of the particle at time t = 4. In this case, we can use the two triangles in the figure to. Is the speed of the particle. If the graph dips below the x-axis, you’ll need to integrate two or more parts of the graph and add the absolute values. You can read about it in your book if you find yourself just dying of curiousity, but it's not in the AP curriculum. 01s chunks, and 10. Here is a set of practice problems to accompany the Arc Length with Parametric Equations section of the Parametric Equations and Polar Coordinates chapter of the notes for Paul Dawkins Calculus II course at Lamar University. How many bushels were consumed from the beginning of 1972 to the end of 1973?. Displacement is a vector quantity as it has both magnitude and direction. Solution: The displacement is the net area bounded by v(t), and the total distance traveled is the total area. The Travel Distance Calculator will calculate instantly the total distance you traveled during your trip based on your average speed and the amount of time you traveled. Your acceleration is 26. Since a = DIV = 2t— I is equal to 3 t = 2, the position s of the | {
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traveled. Your acceleration is 26. Since a = DIV = 2t— I is equal to 3 t = 2, the position s of the particle is a relative minimum when t = 2. To find the position of a particle given its initial position and the velocity function, add the initial position to the displacement (integral of velocity). Find the position of the particle at time t = 3. The displacement or net change in the particle's position from t = a to t = b is equal, by the Fundamental Theorem of Calculus (FTC), to. 9: Velocity & Acceleration SOLUTION KEY. Sample Test Questions: A particle moves along a horizontal line and its position at time is. Sometimes it's a particle, sometimes a car, or a rocket. f) Draw a diagram to illustrate the motion of the particle. (c) Find the average velocity of the particle over the interval. 5, you get the total distance. behind the fifth method of approximation called Simpson's Rule. Be sure to label the time, [position, and velocity at each change and at the beginning. are unit vectors in the x and y directions, it is possible to fi nd the position or coordinates of the particle at a given value of t. (a) Use a de nite intergal and the Fundamental Theorem of Calculus to compute the net signed area between the graph of f(x) and the x-axis on the interval [1;4]. How do you find the total displacement for the particle whose position at time #t# is given by How many values of t does the particle change direction if a particle moves with acceleration What is the position of a particle at time #t=2# if a particle moves along the x axis so that at. Hence, when calculating the distance, we split the interval of integration into two intervals where the velocity has a constant sign. Find the velocity vector at the time when the particle’s horizontal position is x = 25. To calculate the total distance traveled, integrate the absolute. 0 ms What is the amplitude if the maximum displacement is 26. Let’s say we are given the position of a particle P in three-dimensional | {
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maximum displacement is 26. Let’s say we are given the position of a particle P in three-dimensional Cartesian ( x , y , z ) coordinates, with respect to time, where. Displacement vs Total Distance Traveled Given a Derivative Function; Page 11. zz go HEY Athina. 5 The Substution Rule: Students have trouble with this topic. The velocity of the particle at time t is 6t t2. (a) Find the speed of the particle at time t = 2, and find the acceleration vector of the particle at time t = 2. B) Find the total distance travelled by the particle. Both x and y are measured in meters, and t is measured in seconds. 5 seconds to t = 7 seconds. Video Examples: Acceleration and. The total distance that the car is from its starting location is -32 feet, which means that the car ends up 32 feet behind where it started. Multiply velocity by time to get distance covered in meters (m). ii Find the distance of the particle from the origin at any time t. The distance traveled between times t and t + h is f(t + h) − f(t). The graph below shows the velocity, v, of an object (in meters/sec). What is the total distance traveled by the particle?. b) Find the particle's displaçement for the given time interval. Note that displacement is not the same as distance traveled; while a particle might travel back and forth or in circles, the displacement only represents the difference between the starting and ending position. (d) Find the total distance traveled by the particle during the first 8 seconds. " So begins a number of AP Calculus questions. -24 m (b) Find the distance traveled by the particle during the given time interval. The position of a particle at any time tt0 is given by 233 and. (a) When the particle is at rest. Where is the particle located at the end of the trip (t = 10)? b. We can also calculate force,. AP Calculus Worksheet: Rectilinear Motion 1. The position of the particle at time t is x(t) and its position at time t = 0 is (a) Find the acceleration of the particle at time t = | {
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t is x(t) and its position at time t = 0 is (a) Find the acceleration of the particle at time t = 3. At what is the particle changing direction? Find the total distance traveled by the particle from time t = O to r = 4. miles, yards, meters, kilometers, inches etc. ? The motion of a particle is described by the postion function s = t^3 - 12t^2 + 45t + 3 , when t is greater than or equal to zero. Note that if the car changes direction midway and heads south after five seconds, the distance covered, too, changes. A particle’s velocity is represented by the graph below. It is known that. (a) Graph the function v(t). A particle moves along a horizontal line. Find the velocity at time t. Find an expression that may be used to determine how far the carrier will travel and how long it will take to stop. The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. Find the speed when. Since the idea of substitution is so important in Calculus II, the instructor. If it did, we wouldn't need calculus at all, we could just read the value for x right off the graph, for any and all curves. D the distance traveled by the truck from t = 3 to t = 15 E The average position of the truck in the interval t = 3 and t = 15. Here's an example: If the position of a particle is given by: x(t)! 1 3 t3 "t2" 3t # 4,. also check your result geometrically. These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum. In physics the average speed of an object is defined as: $$\text{average speed} = \frac{\text{distance traveled}}{\text{time elapsed}}$$. The corresponding average velocity is then 146 3 = 48. What is the total distance traveled by the particle from t - 0 to t = 3? Show Step-by-step Solutions. c) Ifs(0) = 3, what is the particle's final position? d) Find the total distance traveled by the particle. the | {
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= 3, what is the particle's final position? d) Find the total distance traveled by the particle. the distance positive. (a) What is the velocity of the particle at t = 0? (b) During what time intervals is the particle moving to the left? (c) What is the total distance traveled by the particle from t = 0 to t= 2?. We can integrate the given velocity function to arrive at the position function. 9: Velocity & Acceleration SOLUTION KEY. In particular, when velocity is positive on an interval, we can find the total distance traveled by finding the area under the velocity curve and above the t-axis on the given time interval. (b) Find the total distance traveled by the particle from time t 0. Find the intervals on which the velocity is increasing. Find the intervals on which the particle is slowing down. 6) Find all t for which the distance s is increasing. In Exercises 1-5, the function v(t) is the velocity in m/sec of a particle moving long the x-axis. ≤t ≤ (c) Find the total distance traveled by the particle from time t =0 to t =6. This Displacement Calculator finds the distance traveled or displacement (s) of an object using its initial velocity (u), acceleration (a), and time (t) traveled. The distance traveled is a reasonable 14 km, but the resultant displacement is a mere 2. Calculates the free fall distance and velocity without air resistance from the free fall time. We have to evaluate this to find the velocity at any particular time. A particle moves along the x-axis with position at time t given by x(t) = e-t sin t for 0 :5: t :5: 2JC. Velocity Equation in these calculations: Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v. Textbook solution for Calculus: Early Transcendentals 8th Edition James Stewart Chapter 10. (c) Find the acceleration of the particle at time t. (d) Find the total distance traveled by the particle over the time interval 0 ≤ t ≤ 2. (e) Draw | {
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t. (d) Find the total distance traveled by the particle over the time interval 0 ≤ t ≤ 2. (e) Draw a diagram to represent the motion of the particle. A particle moves in a straight line with velocity t^-2 - 1/9 ft/s. (c) Find the average velocity of the particle over the interval 0 § t § 5. Get an answer for 'The velocity function is v(t)= -(t^2)+6t-8for a particle moving along a line. b) Find the particle's displaçement for the given time interval. Find the body's acceleration each time the velocity is zero B. Is the direction of motion of the particle toward the left or toward the right at that time? Give a reason for your answer. If you look carefully, we've used a boldface 0 because velocity is a vector. Find the total distance traveled from t = 0 to t = 4. A particle moves along the x-axis so that its velocity at time t, , is given by v(t) = 3(t - 1)(t - 3). Find the distance traveled by a particle with position (x,y) as varies in the given time interval. To find that number, we'd need to add the absolute value of each interval. AP Calculus Particle Motion Worksheet For #6 – 10: A particle moves along a line such that its position is s ( t ) = t 4 – 4 t 3. When velocity = 0 Divide into intervals; 0 2 and 2 4 At any time t, the position of a particle moving along an axis is: A. Estimate the total dis-tance the object traveled between t= 0 and t= 6. v (t) ≤ 0, the particle moves to the. Here is a set of practice problems to accompany the Arc Length with Parametric Equations section of the Parametric Equations and Polar Coordinates chapter of the notes for Paul Dawkins Calculus II course at Lamar University. The distance traveled between times t and t + h is f(t + h) − f(t). (a) When the particle is at rest. displacement = -66. Answer: 450 feet. The total distance traveled by the particle from time to time is For the time interval, the situation is a little bit more complicated since the particle changes direction. f) Draw a diagram to illustrate the motion of the | {
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complicated since the particle changes direction. f) Draw a diagram to illustrate the motion of the particle. SOLUTION Solve Analytically We partition the time interval as in Example 2 but record every position shift as positive by taking absolute values. Find the total distance traveled by the particle. e) Find the total distance traveled during the first 8 sec. 0 \leq t \leq 7. If a body moves along a straight line with velocity v = t 3 + 3t 2, find the distance traveled between t. Enter the required values know the unknown value of work or force or distance. Include units. Your acceleration is 26. We can also calculate force,. To calculate the speed and angular velocity of objects. B) Find the total distance travelled by the particle. will have a horizontal tangent? (7 Points) 15) Find the slope of the tangent line to the curve. CALCULUS I Worksheet #74 1. Distance and displacement are two quantities that seem to mean the same but are distinctly different with different meanings and definition. (c) Find the displacement of the particle after the first 8 seconds. We have step-by-step solutions for your textbooks written by Bartleby experts! Find the distance traveled by a particle with position | bartleby. vt is negative on the interval 01. (a) Find the time t at which the particle is farthest to the left. 2 3 x t t y t t (a) Find the magnitude of the velocity vector at time t = 5. 1 t millions of bushels per year, with t being years since the beginning of 1970. b) Find the average value ofg(x)intermsofA over the interval [1 ,3]. 5 miles (or 13,200 feet or 158,400 inches ,etc. I found out that the total displacement is. If ( )= 3−4 2+5 −6 gives the position of a point P as it travels along the x-axis, describe the. Motion in Two and Three Dimensions Conceptual Problems 1 • [SSM] Can the magnitude of the displacement of a particle be less than the distance traveled by the particle along its path? Can its magnitude be more than the distance traveled? Explain. Let's | {
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by the particle along its path? Can its magnitude be more than the distance traveled? Explain. Let's assume that we were given. Now you might start, you might start to be appreciating what the difference between displacement and distance traveled is. A particle moves with a position function s(t) = t3 - 12t2 + 36t for t ≥ 0, where t is measured in seconds and s in feet. Write a polynomial expression for the position of the particle at any time r > O if the position of the particle at t = 0 is 5 At what time(s) is the particle changing direction? Find the total distance traveled by the particle fmm time r = O to r 4. Its position function is s(t) for t ≥≥ ≥ 0 ≥ 000. Calculus Total Distance Particle Traveled? I'm in college, and I stumbled into a problem that deals with a particle. To find the position of a particle given its initial position and the velocity function, add the initial position to the displacement (integral of velocity). The graph of v on the left shows that the velocity of the particle is 16 at time and 0 at time The total distance traveled by the particle is given by the definite integral. Find the total distance traveled from t = 0 seconds to t = 4 seconds. Motion Along a Line; Page 8. This picture is helpful: The positions of the words in the triangle show where they need to go in the equations. Find the speed when. (b) Set up an integral expression to find the total distance traveled by the particle from t 0to t 4. To solve for c1, we know that at t = 0, the initial velocity was 4. Sample Test Questions: A particle moves along a horizontal line and its position at time is. Apply the fundamental theorem of calculus to evaluate integrals and to di erentiate integrals with respect to a limit of integration. Claudette responded. 2 The key to finding the total distance traveled in the last example in a method similar to the first example is to break the time. Justify your answer. Therefore, the average velocity is 8. seconds? (2 points) When is the | {
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the time. Justify your answer. Therefore, the average velocity is 8. seconds? (2 points) When is the particle speeding up? (2 points) When is the particle slowing down? (2 points) (7 Points) 14) Find all values of so that the graph of. To find the total distance traveled by an object, regardless of direction, we need to integrate the absolute value of the velocity function. When you try to find the distance a moving object has traveled, two pieces of information are vital for making this calculation: its speed (or velocity magnitude) and the time that it has been moving. Here is a set of practice problems to accompany the Arc Length with Parametric Equations section of the Parametric Equations and Polar Coordinates chapter of the notes for Paul Dawkins Calculus II course at Lamar University. The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. Find the displacement and the total distance traveled by the particle from t = 1. Distance-traveled-by-a-particle Page history last edited by [email protected] Find the initial velocity and displacement. Calculus is im-portant because most of the laws of science do not provide direct information about the values of variables but only about their rate of change. SOLUTION Solve Analytically We partition the time interval as in Example 2 but record every position shift as positive by taking absolute values. (a) Find the intervals where the function is increasing or decreasing. (b) Find the total distance traveled by the particle from time t = 0 to t = 3. Thus, to calculate the total distance, you need to find the area of the entire region under the v vs. Calculus- find total distance of a particle given its velocity equation? Here's the problem: Find the total distance traveled by a particle moving along a straight line with a velocity v = sin (pi*t) for ( 0 | {
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# A combinatorial interpretation of a counting problen
This question is about a surprising formula that is the answer to a counting problem. To me it suggests that there might just be a nice combinatorial interpretation of the problem that I have not yet found. Even if there is not, I'm sure there are much more elegant solutions than mine.
## The problem
Suppose we have an $n\times m$ grid of squares with $n$ rows and $m$ columns. In every square we write the number of rectangles that can be made inside the grid and in which it is contained. Consider for example the green square in the image below.
We see that it is contained in $8$ rectangles that can be made inside this $2 \times 3$ grid. If we count this number for every square we get.
Now let $f(n,m)$ denote the sum of these numbers, so $f(2,3)=4 \cdot 6 + 2 \cdot 8=40$. The task is to find a general formula for $f(n,m)$.
## My solution
Consider the square with coordinates $(i,j)$, as in the image below. Any rectangle that it is contained in must have its top left corner in the area shown in red. Its bottom right corner must be in the area shown in blue. So the square $(i,j)$ is contained in exactly $\color{red}{ i \cdot j} \cdot \color{blue}{ \left( n - i + 1 \right) \cdot \left( m - j + 1 \right)}$ rectangles. | {
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Now we only need to sum over all squares to get $$f(n,m)= \sum_{i=1}^n \sum_{j=1}^m \left [ i \cdot j \cdot \left( n - i + 1 \right) \cdot \left( m - j + 1 \right) \right ].$$ To make this calculation a little easier we can introduce $S_n = \displaystyle \sum_{i=1}^n i$ and $T_n = \displaystyle \sum_{i=1}^n i^2$ so that \begin{align*} f(n,m) &= \sum_{i=1}^n \sum_{j=1}^m \left [ i \cdot j \cdot \left( n - i + 1 \right) \cdot \left( m - j + 1 \right) \right ] \\ &= \sum_{i=1}^n \sum_{j=1}^m \left [ i^2 j^2 - m\cdot i^2 j -i^2 j - n \cdot i j^2 -i j^2+ n m \cdot i j+ m \cdot i j + n \cdot i j +i j \right ] \\ &= T_n T_m - m\cdot T_n S_m - T_n S_m - n \cdot S_n T_m - S_n T_m + n m \cdot S_n S_m + m \cdot S_n S_m + n \cdot S_n S_m + S_n S_m \\ &= \left(S_n + n\cdot S_n - T_n\right)\cdot \left(S_m + m\cdot S_m - T_m\right) \end{align*}
Now we can use the well known formulas for $S_n$ and $T_n$ to find that $$S_n + n\cdot S_n - T_n = \frac{1}{6} n (n+1) (n+2) = \binom{n+2}{3}.$$ Putting this together we get that $$f(n,m) = \binom{n+2}{3} \cdot \binom{m+2}{3}.$$
I was pretty surprised by the simple nature of this answer. Especially by the fact that it seems to suggest that somewhere along the line $3$ things get chosen out of $n+2$ things and also from $m+2$ things. It gives me a small spark of hope that there exists a nice counting argument that solves this problem. Can anyone give a combinatorial interpretation of the formula for $f(n,m)$? More elegant solutions to the problem are also welcome.
## 2 Answers
Each combination of rectangle and contained cell determines and is completely determined by a pair of ordered triples: the first lists the rows of the top edge of the rectangle, the cell, and the bottom edge of the rectangle, and the second does the same for the columns. The row triples correspond to multisets of $3$ elements chosen from the set $[n]=\{1,\ldots,n\}$, and there are
$$\left(\!\!\binom{n}3\!\!\right)=\binom{n+3-1}3=\binom{n+2}3$$ | {
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$$\left(\!\!\binom{n}3\!\!\right)=\binom{n+3-1}3=\binom{n+2}3$$
of these. Similarly, there $\binom{m+2}3$ column triples, so there are
$$\binom{n+2}3\binom{m+2}3$$
combinations of cell and containing rectangle.
Your sum is [the number of ways to choose a square and a rectangle containing it],
which equals [the number of ways to choose a rectangle and a square inside it].
The row parts of the ways to choose [a rectangle and point inside it]
correspond to length-(n+2) strings consisting of
a beginning-of-rectangle symbol
and
a green-row symbol
and
an end-of-rectangle symbol
and
n-1 blank-row symbols
where the 3 special symbols must be in that order,
giving exactly 3+n-1 choose 3 possibilities for the rows.
Similarly, there are exactly m+2 choose 3 possibilities for the columns.
The product of those gives your expression. | {
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# Challenging "Find the Hashed Area" Question
• July 12th 2012, 09:16 AM
danielmc
Challenging "Find the Hashed Area" Question
See attached question. It asks: "A circle is inside a square, which is also found inside another square with with 6 meter sides. Find the hashed area".
Is the correct answer: 9pi/2 meters^2 or 9pi meters^2?
Can you please systematically outline the steps that you took to arrive at the answer?
Thank you for your help and explanation.
• July 12th 2012, 09:20 AM
richard1234
Re: Challenging "Find the Hashed Area" Question
The side length of the smaller square is $3 \sqrt{2}$ (do you see why?).
Therefore the diameter of the circle is $3 \sqrt{2}$, so the radius is $\frac{3 \sqrt{2}}{2}$. The area of the circle is $(\frac{3 \sqrt{2}}{2})^2 \pi = \frac{9 \pi}{2}$.
• July 12th 2012, 09:33 AM
danielmc
Re: Challenging "Find the Hashed Area" Question
Hi Richard,
Thanks for the quick response. I do not see how you got the side length of the smaller square to be http://latex.codecogs.com/png.latex?3%20\sqrt{2}.
Can you please explain how you got to that step? | {
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Can you please explain how you got to that step?
Thanks a million!
• July 12th 2012, 09:34 AM
richard1234
Re: Challenging "Find the Hashed Area" Question
Hint: The side length of the larger square is the diagonal of the smaller square.
• July 12th 2012, 09:43 AM
danielmc
Re: Challenging "Find the Hashed Area" Question
I understand the rest of it but just can't the length of the inner square. I'm sorry my brain isn't working today. I know it has something to do with the Phytagoreans Theorem. Can you please tell me how you got the side length of the inner square.
• July 12th 2012, 09:55 AM
richard1234
Re: Challenging "Find the Hashed Area" Question
A square is composed of two 45-45-90 triangles. The side length of the large square is the diagonal of the small square, so the diagonal of the small square is 6. The diagonal is like the "hypotenuse" of the 45-45-90 triangles, you should be able to find the side length now.
• July 12th 2012, 10:01 AM
danielmc
Re: Challenging "Find the Hashed Area" Question
Thank you Richard for the step-wise explanation. I now understand. Since the diagonal of the small square is 6; it would be b^2 + b^2 = 6^2; so solving for b, you would take square root of 36 so 6, then another square root of 6 so "b" (the side of the square) would be 3 sqrt 2. From here, we know that the diameter of the circle is 3 sqrt 2 and we can find radius and then area as you had outlined. Thanks for your help!
• July 12th 2012, 11:17 AM
Soroban
Re: Challenging "Find the Hashed Area" Question
Hello, danielmc!
Quote:
A circle is inside a square, which is also found inside another square with with 6 m sides.
Find the shaded area.
Code: | {
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E o 3 * * 3 * * * * A * * B o-------* * *-------o * | *:::::::::::* | * 3 * | *:::::::::::::::* | * 3 * |*:::::::::::::::::*| * * |:::::::::::::::::::| * * *:::::::::::::::::::* * H o *:::::::::o:::::::::* o F * *:::::::::::::::::::* * * |:::::::::::::::::::| * 3 * |*:::::::::::::::::*| * 3 * | *:::::::::::::::* | * * | *:::::::::::* | * o-------* * *-------o D * * C * * 3 * * 3 * * o G
Is the correct answer: $\tfrac{9\pi}{2}\:m^2\,\text{ or }\,9\pi\:m^2\,?$
I'd like to know how you got those two answers . . .
The circle is inscribed in square $ABCD$
. . which is inscribed in square $EFGH$ as shown.
The top triangle $ABE$ is an isosceles right triangle with legs of length 3.
. . Hence: $AB = 3\sqrt{2}$
But $AB$ is the diameter of the circle.
. . Hence, the radius is: . $r \:=\:\frac{3\sqrt{2}}{2}$
Now you can find the area of the circle . . . right? | {
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# Inductive step for graph proofs: reduce from n+1 to n or from n to n-1?
I was reading a practice problem set for a discrete maths course. It says:
By far, the most common mistake in this homework was using induction incorrectly for graph problems. When proving something about graphs by induction, you want to reduce from $n$ to $n − 1$. This is because you are proving a statement for all graphs of size $n$, and not just a specific graph. By proving something by constructing an $n + 1$ size graph using an $n$ size graph, you have only shown it for “one particular” graph of size $n + 1$, whereas you wanted to show it for all graphs of size $n + 1$. (By size, I mean either the number of vertices or the number of edges, whatever you are inducting on.)
Then they give an example:
We prove by induction on $n$ that if $|V| = n$ and $G$ is acyclic and $|E| = n − 1$, then $G$ is connected.
Base case ($n = 1$): This is trivial; $G$ consists of a single vertex and no edges.
Induction step ($n \geq 2$): First we will prove that $G$ has a leaf. Since $E \neq \emptyset$, we can pick a vertex $v \in V$ with $deg(v) > 0$. Start walking from $v$ until you reach a leaf, say $u$. This will happen because the graph is acyclic. Now, $G − u$ has $n − 1$ vertices and $n − 2$ edges, so we can use the induction hypothesis to conclude that it is connected. Adding $u$ to $G$ with the original edge keeps it connected.
Common Mistake: Constructing $n + 1$ vertex graph using an $n$ vertex graph in induction step. If you did this, you missed showing existence of a leaf, which is crucial.
I would have done it a bit different (rough outline):
Induction step $(n + 1)$: $G$ has $n+1$ nodes. We can find a leaf node $v$ with $deg(v) = 1$. $v$ has to exists because $G$ is acyclic. We obtain a subgraph $G'$ by removing $v$ from $G$. By induction we can assume $G'$ is connected. Adding $v$ again to $G'$ keeps the resulting graph $G$ connected, which proves the theorem for $n + 1$. | {
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Is mine the wrong approach and I should have started from $n$? I thought in the inductive step for graph proofs, you usually start from a $n+1$ graph and for example reduce it to $n$ by removing a certain node/edge/...
• – Misha Lavrov Nov 28 '17 at 3:00
The "common mistake" described in the problem set instructions has nothing to do with whether your inductive step is based on $n$ and $n-1$ or $n+1$ and $n.$ It has everything to do with which number (the smaller one or the larger one) is the number of nodes in the arbitrary graph that you invoke in the inductive step. If you go from $n$ to $n+1$ (or if you go from $n-1$ to $n,$ for that matter!) by adding a node to an arbitrary graph, the proof is wrong. You must remove a node instead, as indeed you did.
Another vital thing to make sure of is that the inductive step can use the base case as its "input." In the problem set, this is done by making the inductive step use the assumption $P(n-1)$ to prove $P(n)$ for $n\geq 2.$ This means that when $n = 2$ in the inductive step, we use $P(1)$ (the base case) to prove $P(2).$ In your proof you need to specify that the inductive step applies for all $n \geq 1,$ not only for $n\geq 2,$ so that when $n=1$ you use $P(1)$ to prove $P(2).$ There are some classic fake proofs that rely on inductive steps that don't "connect" with the base case. | {
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# Best method for proving that $7\times11^{2n+1}-3^{4n-1}$ is divisible by $10$
I am asked to prove by induction that $$7\times11^{2n+1}-3^{4n-1}$$ is divisible by $$10$$.
I wonder whether there is a more direct method, for example factorizing by $$10$$. If an expression is divisible by $$10$$, does this mean that I can factorize it by $$10$$?
• Just think about the last digit. Sep 19, 2020 at 17:04
• With any statement about the set $\Bbb N$ (especially, when the claim doesn't hold for a larger set), a proof by induction is arguably the most direct one. This is somewhat by definition of $\Bbb N$ Sep 19, 2020 at 17:05
• Are you familiar with congruences such as $\,11^2\equiv 3^4\pmod{10}\,?\ \$ Sep 19, 2020 at 18:37
• I take it you mean for $n\ge 1$. The persistence of divisibility can be shown in various ways. The numbers $k_n=Aa^n+Bb^n$ satisfy the recurrence $k_{n+1}=(a+b)k_n-abk_{n-1}$ (you can easily verify this by substitution) so (in appropriate cases) once you show that two successive elements are divisible by $d$ then they all are. These problems are quite often used as examples for induction. Sep 19, 2020 at 19:39
• Thank you all so much for those answers that will help me more deeply understand this kind of problems. I just still hesitate on this : if an expression in N is divisible by 10, does this mean that I can factorize it by 10 ? Sep 19, 2020 at 22:06
A direct and intuitive way. Every integer power of $$11$$ has $$1$$ as last digit, so every number of the form $$7 \cdot 11^{2n+1}$$ ends with the digit $$7$$.
The last digits of the integer powers of $$3$$ follows the four-step cycle $$(3,9,7,1)$$, so that every number of the form $$3^{4n-1}$$ has $$7$$ as last digit.
Therefore, subtracting this second number from the first one we get a number ending with the digit $$0$$. This implies that this last number is divisible by $$10$$.
Below are four proofs using various methods. | {
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Below are four proofs using various methods.
You seem to seek a direct proof showing a factor of $$10$$ so let's do that first.
\begin{align} x &\,=\ \ \ \ \, 7\cdot 11^{\large 2n+1}\ -\ 3^{\large 4n-1}\\[.2em] \Rightarrow\ \ 3x &\,=\ \ \, 21\cdot 11\cdot 121^n - 81^n\\[.2em] &\,=\, (21\cdot11\!-\!1)121^n+ \color{#0a0}{121^n-81^n}\\[.2em] &\,=\, \color{#c00}{10}\,(23\cdot 121^n + \color{#c00}4(121^{n-1} + \cdots + 81^{n-1}))\ \end{align}\qquad
where we used the Factor Theorem to deduce $$\,\color{#c00}{10\cdot 4} = 121\!-\!81\,$$ divides $$\,\color{#0a0}{121^n-81^n}.\,$$ Thus $$\,10\mid 3x\Rightarrow 10\mid x\,$$ by Euclid (or directly $$\,10\mid 7(3x)\!-\!20x = x,\,$$ or cancel $$3$$ from $$121$$'s and $$81$$'s).
It's much easier by modular arithmetic (congruences)
\begin{align}\bmod 10\!:\ \ 3x &\equiv 21\cdot 11^{\large 2n+1} - 81^{\large n}\\ \iff\ 3x &\equiv \ \ 1\ \cdot\ 1^{\large 2n+1}\ -\ 1^{\large n} \equiv\color{#0a0} 0\\ \iff\ \ \ x &\equiv\,3^{-1}\cdot\color{#0a0} 0\equiv 0 \end{align}\qquad
by basic congruence laws. We used the fact that scaling by an invertible (here $$3$$) yields an equivalent congrence (recall by Bezout that $$3$$ is invertible being coprime to the modulus $$10)$$
By induction: base case $$\,n=1\,$$ is $$\!\bmod 10\!:\ 7\cdot 11^3\equiv 3^3\,$$ (or $$\,7\cdot 11\equiv 1/3\,$$ for $$\,n=0)\,$$ which are both true, and the induction step follows conceptually by simply by multiplying the first two congruences below using $$\rm\color{#0a0}{CPR} =$$ Congruence Product Rule,
\begin{align}\bmod 10\!:\qquad\ \ \ \color{#c00}{11^{\large 2}}\ &\equiv\ \color{#c00}{3^{\large 4}}\\[.2em] {\rm times}\ \ \ \ \ \ \ 7\cdot 11^{\large 2n+1}&\equiv3^{\large 4n-1}\quad \ P(n)_{\phantom{|}}\\[.2em] \hline \Longrightarrow\ \ \ \ \ 7\cdot 11^{\large 2n+\color{#c00}3}&\equiv 3^{\large 4n+\color{#c00}3}\quad\ P(n\!+\!\color{#c00}1), \ \ \rm by \ \,\color{#0a0}{CPR}^{\phantom{|^|}}\end{align}\qquad | {
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If congruences are unfamiliar we can preserve the arithmetical essence of this simple proof by using an analogous product rule for divisibility (DPR), as explained here.
Or as here use Binomial Theorem on $$\,(1\!+\!10)^{2n+1}$$ and $$(-1\!+\!10)^{4n-1}$$ (or $$\,(1\!+\!80)^n\,$$ in $$3x)$$
Remark All these methods do in fact use induction (on $$n),\,$$ but it may be hidden (encapsulated) in the proof of a theorem that is invoked, e.g. the Factor Theorem or Binomial theorem, or the Congruence Power Rule $$\,a\equiv b\Rightarrow\, a^n\equiv b^n$$.
• Thank you all so much for those answers that will help me more deeply understand this kind of problems. I just still hesitate on this : if an expression in N is divisible by 10, does this mean that I can factorize it by 10 ? Sep 19, 2020 at 22:11
• @Romain $\,10\mid n,\,$ i.e. $10$ divides $n$ means by definition that $\, n = 10k\,$ for some integer $k,\,$ It is often quicker to prove that in ways that don't require explicitly calculating the integer cofactor $k = n/10$ that is witness to the divisibility, e.g. using modular arithmetic (congruences) as above. Sep 19, 2020 at 22:14
• " if an expression in N is divisible by 10, does this mean that I can factorize it by 10 ?" Yes, but you might not always know how. Sep 20, 2020 at 1:02
• Sadly there seems to be at least one user who consistently downvotes ansers that have multiples proofs. In the unlikely chance that this has anything to do with matehmatics then please leave a constructive comment so that any deficiency you perceive may have a chance to be addressed. Sep 20, 2020 at 19:30
we have $$7{(10+1)}^{2n+1}-3{(10-1)}^{2n-1}$$
indeed by binomial theorem it is equivalent to:
$$7(10k+1)-3(10m-1)=10+70k-30m$$
which is div by $$10$$
Using modular arithmetic: \begin{align}3(7\cdot 11^{2n+1}-3^{4n-1})&=21\cdot 11^{2n+1}-9^{2n}\\&\equiv 1\cdot (1)^{2n+1}-(-1)^{2n}\\&=0 \pmod {10}.\end{align}
We have $$3(3^3)=81 \equiv 1 \pmod{10}$$ | {
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We have $$3(3^3)=81 \equiv 1 \pmod{10}$$
$$3^{-1}\equiv 3^3 \equiv 7 \pmod{10}$$
\begin{align} 7(11^{2n+1})-3^{4n-1} &\equiv 7 (1^{2n+1}) - (3^4)^n3^{-1}\\ &\equiv 7(1)-(1)(7) \\ &\equiv 0 \pmod{10} \end{align}
Best method? Depends on what you know.
If you know modular arithmetic and Eulers Theorem then
$$7\cdot 11^{2n+1} - 3^{4n-1}\equiv 7\cdot 1^{2n+1} - (3^{-1})\cdot 3^{4n}\equiv 7-(3^{-1})\pmod{10}$$ and as $$3\cdot 7 \equiv 1 \pmod {10}$$ then $$3^{-1}\equiv 7 \pmod {10}$$ and $$7-7\equiv 0\pmod {10}$$ so $$10|7\cdot 11^{2n+1} - 3^{4n-1}$$.
But that assumes you are comfortable with many concepts
If you don't know any modular arithmetic:
$$11^k = (10+1)^k = 10^k + k10^{k-1} + ...... + k\cdot 10 + 1 = 10 M + 1$$.
And $$3^{4n-1} = 3\times 3^{4n-2}= 3\times 9^{2n-1}$$. And if $$k=2n-1$$ is odd the $$9^k = (10-1)^k = 10^k - k10^{k-1} + ....... + k\cdot 10 - 1= 10N-1$$.
So $$7\cdot 11^{something} - 3^{1+2\times something\ odd} = 7(10M + 1)- 3(10N-1)= 70M -30N +10$$.
• You didn't invole Euler (maybe you meant to say Euclid for computing the inverse?) Sep 20, 2020 at 1:29
• I meant Euler and I meant that $3^{4n} \equiv 1\pmod {10}$ so $3^{4n-1} \equiv 3^{-1}$. Sep 20, 2020 at 2:15
• Ah, I didn't expect use of Euler in such a simple case since it is much easier by repeated squaring $\!\bmod 10\!:\ (3^2)^2\equiv (-1)^2\equiv 1,\,$ or even directly $\,(3^2)^2\equiv 9^2\equiv 81\equiv 1.\ \$ Sep 20, 2020 at 2:24
Since
$$\quad 11^k \equiv 1 \pmod{10} \quad \forall k \in \Bbb Z$$
and
$$\quad 3^{4n-1} \equiv {(3^2)}^{2n} \times 3^{-1} \equiv (-1)^{2n} \times 7 \equiv 7 \pmod{10}$$
we can write
$$\quad 7\times11^{2n+1}-3^{4n-1} \equiv 7 - 7 \equiv 0 \pmod{10}$$
Elementary Number Theory:
The invertible elements in $$\mathbb{Z}/{10}\mathbb{Z}$$ are
$$\quad [1], [3], [7] \text{ and } [9]$$
allowing one to quickly determine that $$3^{-1} \equiv 7 \pmod{10}$$. | {
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allowing one to quickly determine that $$3^{-1} \equiv 7 \pmod{10}$$.
The OP is looking for a factorization argument and so we give a problem/hint that can be directly solved without elementary number theory using simple algebra; this is the step case of the induction proof.
Problem: Show that if $$k \ge 1$$ and
$$\tag 1 \displaystyle 7\times11^{2k+1}-3^{4k-1} = 10q$$
then
$$\tag 2 \displaystyle 7\times11^{2(k+1)+1}-3^{4(k+1)-1} = 10 \times (11^2 q + 4 \times 3^{4k-1})$$
Multiply $$n=7\times11^{2n+1}-3^{4n-1}$$ by $$3$$ to get $$3n=21\times11^{2n+1}-81^{n}$$, which is divisible by $$10$$
(ones digit is $$0$$, because it is the difference of two numbers whose ones digit is $$1$$).
Now if $$3n$$ is divisible by $$10$$, for an integer $$n$$, then $$n$$ is divisible by $$10$$. QED. | {
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# Solids of Revolution Question (Method of Cylinders vs Disc/Washers)
Find the volume of the solid formed by revolving the region bounded by y=x^2+1, y=0, x=0, and x=1 about the y-axis.
I was practicing this concept and I came across this problem. I did it using the shell method and got 2pi*Integral of x(x^2+1) from 0 to 1, which yielded the answer 3pi/2.
I tried checking this with the disc/washer method, but this gave me a different answer. Pi*Integral of [Sqrt(y-1)]^2 from 1 to 2 = pi/2.
Likewise, I tried a similar problem with the functions: y = Sqrt(x), x=axis, and x=4 roated about the x=8
Cylinders: 2piIntegral of Sqrt(x)(8-x) from 0 to 4 = 896pi/15 Disk/Washers: pi*Integral of (8-y^2)^2 from 0 to 2 = 1376pi/15
Which ones are correct? Where did I make mistakes?
• What can I do to improve the quality of this question? I wasn't aware of anything wrong about it besides the improper formatting. Thank you. – bandicoot12 Jan 19 '15 at 21:03
• The problem is that $\int_1^2\pi\sqrt{y-1}^2dy$ represents the volume formed by spinning the region bounded by $y=x^2+1, y=0,$ and $y=2$ about the y-axis. – John Joy Jan 19 '15 at 21:08 | {
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For the particular example you have, it clearly seems that cylindrical shells will be easier. The interval of integration will be $x \in [0,1]$. For a particular representative radius $x$ from the $y$-axis, the height will be $f(x) = y = x^2 + 1$. The circumference is $2\pi x$, and the differential thickness of the shell is $dx$. So the differential volume is $$dV = 2 \pi x f(x) \, dx = 2 \pi x (x^2 + 1) \, dx,$$ and the total volume is $$V = \int_{x=0}^1 2 \pi x (x^2 + 1) \, dx = \frac{3\pi}{2}.$$
Now using the method of washers, the problem you have is that on the interval $y \in [0,1]$, the volume is just a cylinder of radius $1$, whereas on the interval $y \in (1,2]$, we have washers with outer radius $1$ and inner radius $g(y) = \sqrt{y-1}$, since the inverse function of $y = x^2 + 1$ is $x = \sqrt{y-1}$.
So combined, we would have: $$V = \int_{y=0}^1 \pi(1^2) \, dy + \int_{y=1}^2 \pi (1^2 - (\sqrt{y-1})^2) \, dy = \pi + \pi \int_{y=1}^2 2-y \, dy = \frac{3\pi}{2}.$$ | {
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MAT334-2018F > End of Semester Bonus--sample problem for FE
FE Sample--Problem 5
(1/2) > >>
Victor Ivrii:
Show that the equation
$$e^{z}=e^2z$$
has a real root in the unit disk $\{z\colon |z|<1\}$.
Are there non-real roots?
Min Gyu Woo:
Let $f(z) = e^2z$ and $g(z) = e^z$.
We have
$|f(z)| = e^2 > e = |g(z)| \text{ when} |z|<1$
So, $f(z) = 0$ has the same number of zeros as $f(z)=g(z)$. (Q16 in textbook)
$$f(z) = e^2z=0$$
$$z = 0$$
Thus, $f(z)$ has one zero $\implies$ $f(z)=g(z)$ has one zero.
Let's look at the case where z is real. I.e. $z = x$ where $x\in\mathbb{R}$
Then,
Call $h(z) = f(z) - g(z) = e^x - e^2 x$
Note that:
$$h(0) = 1$$
$$h(1) = e - e^2 <0$$
By Mean Value Theorem, there exists $x$ where $0 < x < 1$ such that $h(x) = 0$.
I.e. there is a REAL ROOT x where $0<x<1$.
We know that there is only one real root in $\{z:|z|<1\}$ because $|h(\overline{z_0})| = |h(z_0)| =0$ is only true when $z_0$ is real.
Thus, within $|z| <1$ there is only one root, and that root is real.
Victor Ivrii:
That the only root in $\{z\colon |z|<1\}$ is real follows from the fact that if $z$ is a root, then $\bar{z}$ it also a root. However the last statement do you mean $\{z\colon |z|<1\}$ ? Otherwise it is wrong due to Picard great theorem
Min Gyu Woo:
Fixed
Nikita Dua:
I don't quite understand why the only root is real?. Can you explain more | {
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+0
# Need help with grpahing circles
0
111
1
The equation of the circle shown in the following diagram can be written as $x^2 + Ay^2 + Bx + Cy + D = 0$. Find $A+B+C+D$.
Guest Feb 22, 2018
### Best Answer
#1
+7155
+2
the center of the circle = (-1, 1)
the radius of the circle = $$\sqrt{(-1-1)^2+(1-2)^2}\,=\,\sqrt{(-2)^2+(-1)^2}\,=\,\sqrt{4+1}\,=\,\sqrt5$$
So the equation of the circle is...
$$(x--1)^2+(y-1)^2\,=\,\sqrt5^2 \\~\\ (x+1)^2+(y-1)^2\,=\,5 \\~\\ (x+1)(x+1)+(y-1)(y-1)\,=\,5 \\~\\ x^2+2x+1+y^2-2y+1\,=\,5 \\~\\ x^2+2x+y^2-2y+2\,=\,5 \\~\\ x^2+2x+y^2-2y-3\,=\,0 \\~\\ x^2+y^2+2x-2y-3\,=\,0$$
A + B + C + D = 1 + 2 - 2 - 3 = -2
Here's a graph to check the equation of the circle:
https://www.desmos.com/calculator/rnauqvgqrg
hectictar Feb 22, 2018
#1
+7155
+2
Best Answer
the center of the circle = (-1, 1)
the radius of the circle = $$\sqrt{(-1-1)^2+(1-2)^2}\,=\,\sqrt{(-2)^2+(-1)^2}\,=\,\sqrt{4+1}\,=\,\sqrt5$$
So the equation of the circle is...
$$(x--1)^2+(y-1)^2\,=\,\sqrt5^2 \\~\\ (x+1)^2+(y-1)^2\,=\,5 \\~\\ (x+1)(x+1)+(y-1)(y-1)\,=\,5 \\~\\ x^2+2x+1+y^2-2y+1\,=\,5 \\~\\ x^2+2x+y^2-2y+2\,=\,5 \\~\\ x^2+2x+y^2-2y-3\,=\,0 \\~\\ x^2+y^2+2x-2y-3\,=\,0$$
A + B + C + D = 1 + 2 - 2 - 3 = -2
Here's a graph to check the equation of the circle:
https://www.desmos.com/calculator/rnauqvgqrg
hectictar Feb 22, 2018
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# Verify integration of $\int\frac{\sqrt{2-x-x^2}}{x^2}dx$
This is exercise 6.25.40 from Tom Apostol's Calculus I. I would like to ask someone to verify my solution, the result I got differs from the one provided in the book.
Evaluate the following integral: $\int\frac{\sqrt{2-x-x^2}}{x^2}dx$
$x \in [{-2,0}) \cup ({0,1}]$
As suggested in the book, we multiply both numerator and denumerator by $\sqrt{2-x-x^2}$. This removes the endpoints from the integrand's domain, but the definite integral we calculate with the antiderivative of this new function will be still the proper integral between any two points of the original domain: we only remove a finite number of points from the original domain and the domain of the resulting antiderivative will be the same as the original domain.
$$I=I_1+I_2=\int\frac{2-x}{x^2\sqrt{2-x-x^2}}dx-\int\frac1{\sqrt{2-x-x^2}}dx \tag{1}$$
Evaluating first $I_1$ by substituting $t=\frac1{x} \; \text, \; dx=-\frac1{t^2}dt \text:$
$$I_1=-\frac1{\sqrt2}\int\frac{2t-1}{\frac{t}{|t|}\sqrt{\left(t-\frac14\right)^2-\left(\frac34\right)^2}}dt \tag{2}$$
Substituting again $\frac34\sec{u}=t-\frac14 \; \text, \; dt=\frac34\sec{u}\tan{u}du \; \text, \; t=\frac{3\sec{u}+1}{4} \; \text, \; u=\operatorname{arcsec}{\frac{4t-1}{3}}$, by considering the sign of $t$ and $\tan{u}$ in the integrand's two sub-domains:
a) $x \in (-2, 0): t<-\frac12 \; \text, \; \frac{4t-1}{3}<-1 \; \text, \; u \in(\frac\pi2,\pi) \; \text, \; \tan{u}<0$
b) $x \in (0, 1): t>1 \; \text, \; \frac{4t-1}{3}>1 \; \text, \; u \in(0,\frac\pi2) \; \text, \; \tan{u}>0$
$$I_1=-\frac{1}{2\sqrt2}\int3\sec^2{u}-\sec{u}=-\frac1{2\sqrt2}\left(3\tan{u}-\log\left|\tan{u}+\sec{u}\right|\right)+C_1 \tag{3}$$
$$\sec{u}=\sec\operatorname{arcsec}\frac{4t-1}{3}=\frac{4t-1}{3}=\frac{4-x}{3x} \tag{4}$$
$$\tan^2{u}=\tan^2\operatorname{arcsec}\frac{4t-1}{3}=\left(\frac{4t-1}{3}\right)^2-1=\frac89\frac{2-x-x^2}{x^2} \tag{5}$$ | {
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Considering again cases a) and b): $$\tan{u}=\frac{2\sqrt2}{3}\frac{\sqrt{2-x-x^2}}{x} \tag{6}$$
$$I_1=-\frac{\sqrt{2-x-x^2}}{x}+\frac1{2\sqrt2}\log\left|\frac{2\sqrt2}{3}\left(\frac{\sqrt{2-x-x^2}+\sqrt2}{x}-\frac1{2\sqrt2}\right)\right|+C_1= \tag{7}$$
$$-\frac{\sqrt{2-x-x^2}}{x}+\frac1{2\sqrt2}\log\left|\frac{\sqrt{2-x-x^2}+\sqrt2}{x}-\frac1{2\sqrt2}\right|+C'_1 \tag{8}$$
Evaluating now $I_2$: $$I_2=-\int\frac1{\sqrt{\left(\frac32\right)^2-\left(x+\frac12\right)^2}}dx \tag{9}$$
Substituting $\frac32\sin{z}=x+\frac12 \; \text, \; dx=\frac32\cos{z}dz \; \text, \; z=\operatorname{arcsin}{\frac{2x+1}{3}}$:
$$I_2 = -\int dz = -\operatorname{arcsin}{\frac{2x+1}{3}}+C_2 \tag{10}$$
The final result: $$I = I_1 + I_2 = -\frac{\sqrt{2-x-x^2}}{x}+\frac1{2\sqrt2}\log\left|\frac{\sqrt{2-x-x^2}+\sqrt2}{x}-\frac1{2\sqrt2}\right|-\operatorname{arcsin}{\frac{2x+1}{3}}+C \tag{11}$$
The solution provided in the book:
$$I = -\frac{\sqrt{2-x-x^2}}{x}+\frac1{2\sqrt2}\log\left(\frac{\sqrt{2-x-x^2}}{x}-\frac1{2\sqrt2}\right)-\operatorname{arcsin}\frac{2x+1}{3}+C$$
• The difference between the two solutions is not constant, so one of them is wrong. – egreg Dec 11 '15 at 0:28
• Right, and there is also the difference in the absolute sign for the second term. – Imre Deák Dec 11 '15 at 0:35
The solution in the book is most certainly a typo, your proof seems fine to me. As a confirmation, Mathematica evaluates the integral to be: $$I=-\dfrac {\sqrt {2 - x - x^2}} x + \dfrac1 {2\sqrt {2}}\left[\log\left |4 - x + 2\sqrt {2}\sqrt {2 - x - x^2} \right| - \log |x| \right] \qquad - \arcsin\left (\dfrac {2 x + 1} {3} \right) + \rm C_1,$$ which is the same as your proposed solution since \begin{align} \log\left |4 - x + 2\sqrt {2}\sqrt {2 - x - x^2} \right| - \log |x|&=\log\left|\dfrac{{2\sqrt{2}\sqrt{2-x-x^2}}+{4-x}}{2\sqrt{2}x}\right|+{\rm C_2} \\ &=\log\left|\frac{\sqrt{2-x-x^2}+\sqrt2}{x}-\frac1{2\sqrt2}\right|+{\rm C_2},\end{align} so they only differ by a constant. | {
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• (2) seems ok to me: $I1=\int\frac{2-x}{x^2\sqrt{2-x-x^2}}dx=\int\frac{2-1/t}{1/t^2\sqrt{2-1/t-1/t^2}}(-1)\frac1{t^2}dt=-\int\frac{2t-1}{t\sqrt{2-1/t-1/t^2}}dt=-\int\frac{2t-1}{t/|t| \sqrt{2t^2-t-1}}dt$ – Imre Deák Jan 28 '16 at 17:59
• @ImreDeák See my edit. – Workaholic Jan 28 '16 at 19:50 | {
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abragas@abragas.com.br
+55 (21) 2221-6695
if two triangles are equal in area, they are congruent
26 de janeiro de 2021, às 3:11 | {
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They can have the same angles but have sides of different lengths. Now, the area of triangle ABC is AB*CM/2 and the area of triangle ABD is AB*DN/ 2 But CM=DN so these two areas are equal. In the simple case below, the two triangles PQR and LMN are congruent because every corresponding side has the same length, and every corresponding angle has the same measure. (Why? In this case, two triangles are congruent if two sides and one included angle in a given triangle are equal to the corresponding two sides and one included angle in another triangle. find the price at which it was sold. The two triangles have two angles congruent (equal) and the included side between those angles congruent. You can replicate the SSS Postulate using two straight objects -- uncooked spaghetti or plastic stirrers work great. (ii) If two squares have equal areas, they are congruent. #2 That is true, they must both have the same side length to have the same perimeter, therefore they will also have the same area. You now have two triangles, △SAN and △SWA. Definition: Triangles are congruent when all corresponding sides and interior angles are congruent.The triangles will have the same shape and size, but one may be a mirror image of the other. 2 triangles are congruent if they have: exactly the same three sides and SSSstands for "side, side, side" and means that we have two triangles with all three sides equal. If we can show, then, that two triangles are congruent, we will know the following: 1) Their corresponding sides are equal. An included side is the side between two angles. Two triangles are congruent if their corresponding sides are equal in length and their corresponding interior angles are equal in measure. Join now. We could also think this angle right over here. So also, sides d and e must be equal because DEF is isosceles. It can be shown that two triangles having congruent angles (equiangular triangles) are similar, that is, the corresponding sides can be proved to be | {
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angles (equiangular triangles) are similar, that is, the corresponding sides can be proved to be proportional. I could have one triangle with sides measuring 7 in + 5 in + 8 in (perimeter = 20 in) and another triangle with sides of 6 in + 4 in + 10 in (perimeter = 20 in). Triangles are congruent when all corresponding sides & interior angles are congruent. Only if the two triangles are congruent will they have equal areas. The Angle Side Angle Postulate (ASA) says triangles are congruent if any two angles and their included side are equal in the triangles. Theorem 1 : Hypotenuse-Leg (HL) Theorem. So we have: a=d. Hence all squares are not congruent. ... maths. = True (iii) If two figures have equal areas, they are congruent. Two triangles, ABC and A′B′C′, are similar if and only if corresponding angles have the same measure: this implies that they are similar if and only if the lengths of corresponding sides are proportional. We use the symbol ≅ to show congruence. they have to be congruent to have same perimeters AND area con. …, how much loss and profit computed on his original capital. Technically speaking, that COULD almost be the end of the proof. When triangles are similar, they could be congruent. prove that they are congruent. Move to the next side (in whichever direction you want to move), which will sweep up an included angle. AAS (Angle-Angle-Side): If two pairs of angles of two triangles are equal in measurement, and a pair of corresponding non-included sides are equal in length, then the triangles are congruent. The congruence of two objects is often represented using the symbol "≅". ALL of this is based on a single concept: That the quality that we call "area" is an aspect of dimensional lengths and angles. Light1729 Light1729 If two triangles are congruent then they must overlap each other completely, so, they have exactly equal area. Yes, they are similar. And for sensible cases. Only if the two triangles are congruent will they have equal areas. | {
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similar. And for sensible cases. Only if the two triangles are congruent will they have equal areas. Furthermore, your question is about congruent triangles and not similar triangles.l. If two triangles have the same area, they must be congruent. (d) if two sides and any angle of one triangle are equal to the corresponding sides and an angle of another triangle, then the triangles are not congruent. You can check polygons like parallelograms, squares and rectangles using these postulates. Log in. e.g. …, nswer then don't give) give me with full steps.don't spam ❌❌❌❌❌. Asked on January 17, 2020 by Premlata Pandagre In a squared sheet, draw two triangles of equal areas such that (i) the triangles are congruent. However, different squares can have sides of different lengths. Two or more triangles that have the same size and shape are called congruent triangles.. Thus, a=d. Two triangles are congruent if their corresponding sides are equal in length and their corresponding interior angles are equal in measure. Asked on December 26, 2019 by Yugandhara Jhunjhunwala. #2 That is true, they must both have the same side length to have the same perimeter, therefore they will also have the same area. All the sides of a square are of equal length. Click hereto get an answer to your question ️ If the area of two similar triangles are equal, prove that they are congruent. Triangles can be considered congruent if following conditions are satisfied. All three triangle congruence statements are generally regarded in the mathematics world as postulates, but some authorities identify them as theorems (able to be proved). (iv) If two triangles are equal in area, they are congruent. The two are not the same. one could look upside-down compared to the other but if they have the same dimensions (i.e. monojmahanta4 monojmahanta4 09.04.2020 Math Secondary School If the area of two triangles are equal. (iii) If two rectangles have equal area, they are congruent. But that does not mean that | {
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"url": "https://www.abragas.com.br/4j94m/997ad1-if-two-triangles-are-equal-in-area%2C-they-are-congruent"
} |
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