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there is no flux of particles coming from the right. One of the famous problems that motivated the beginnings of modern probability theory in the 17th century, it led Blaise Pascal to the first explicit reasoning about what today is known as an expected value. The Probability Distributome Project provides an interactive navigator for traversal, discovery and exploration of probability distribution properties and interrelations. Listed in the following table are practice exam questions and solutions, and the exam questions and solutions. 84 = (x – 70)/2. Let E = event of getting a queen of club or a king of heart. Let Xand Y be two N 0-valued random variables such that X= Y+ Z, where Zis a Bernoulli random variable with parameter p2(0;1), independent of Y. Miracle Mountain. Hone your skills with this array of statistics and probability worksheets for 7th grade to compute the average on whole numbers and decimals, find the probability on a pair of coins, a pair of dice, months and more. Solution of exercise 1. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E. Although they never finished their game, their letters contain solutions to both problems. Home > Math> Probability and Statistics>Probability Problems. Homework problems usually do not say which concepts are involved, and often require combining several concepts. Probability problem on Dice. Most probability problems on the GRE involve independent events. Probability MCQ Questions and answers with easy and logical explanations. After you define a problem, click Options in the Solver Parameters dialog box. The problem concerns a game of chance with two players who have. Example: there are 5 marbles in a bag: 4 are. Probability Questions and answers PDF with solutions. In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some | {
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usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the. These you may need to complete at home. Statistics Problems With Solutions Return to Statistics Internet Library for videos, software assistance, more problems and review. Cards of Spades and clubs are black cards. pdf] - Read File Online - Report Abuse. A Collection of Dice Problems Matthew M. 316 that an audit of a retail business. The first section provides a brief description of some of the basic probability concepts that will be used in the activities. Trials until first success. Walpole Raymond H. What is the probability of drawing a red Bingo chip at least 3 out of 5 times? Round answer to the nearest hundredth. Probability Problems With Solutions Math Help Fast (from someone who can actually explain it) See the real life story of how a cartoon dude got the better of math Probability Word Problems (Simplifying Math) What are the chances that your name starts with the letter H? Find out how to make that calculation and many more. Advanced Techniques. The problem concerns a game of chance with two players who have. Probability shortcut Tricks Pdf, Probability MCQ, Probability Objective Question & Answer Pdf. [email protected]{"The probability of winning is ", 100 win/(win + lose) // N, " %. Total English alphabets = 26. The flippant juror. For example: if we have to calculate the probability of taking a blue ball from the second bag out of three different bags of balls, where each bag contains three different colour balls viz. a Show that $$f\left( x \right)$$ is a probability density function. Make use of probability calculators to solve the probability problems with ease. What is the probability that players will experience rain and a temperature between 35 and 50 degrees? 0. ’ ‘There’s a 30% chance of rain tomorrow. He offers you the following | {
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between 35 and 50 degrees? 0. ’ ‘There’s a 30% chance of rain tomorrow. He offers you the following game. 1) Basics of Electromagnetics Part I – Download MCQs from here. Probability Solution: (a) Since there are 4 queens ( ) = ( ) ( ),. (Solution): Probability Problems. Probability And Statistics Problems Solutions Author: persepolis. It is assessed by considering the event's certainty as 1 and impossibility as 0. edu-2020-04-20T00:00:00+00:01 Subject: Probability And Statistics Problems Solutions Keywords: probability, and, statistics, problems, solutions Created Date: 4/20/2020 7:35:45 AM. nthe sample space for the last experiment would be all the ordered pairs in the form (d,c), where d represents the roll of a die and c represents the flip of a coin. AU - Sparrow, E. Probability Questions are provided with detailed answers to every question. probability problems, probability, probability examples, how to solve probability word problems, probability based on area, examples with step by step solutions and answers, How to use permutations and combinations to solve probability problems, How to find the probability of of simple events, multiple independent events, a union of two events. Competitive exams are all about time. The concept is very similar to mass density in physics: its unit is probability per unit length. Most of these problems require very little mathematical background to solve. Here is a set of 14 GMAT probability questions, all in the Problem Solving style on the test, collected from a series of blog articles. Examples include the Monty Hall paradox and the birthday problem. Write down twenty math problems related to this topic on a page. If there is a chance that an event will happen, then its probability is between zero and 1. Let Xand Y be two N 0-valued random variables such that X= Y+ Z, where Zis a Bernoulli random variable with parameter p2(0;1), independent of Y. Linear Algebra Igor Yanovsky, 2005 2 Disclaimer: This handbook is intended to | {
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independent of Y. Linear Algebra Igor Yanovsky, 2005 2 Disclaimer: This handbook is intended to assist graduate students with qualifying examination preparation. Please solve the following probability practice problems: Determine the probability that a digit chosen at random from the digits 1, 2, 3, …12 will be odd. The mathematics field of probability has its own rules, definitions, and laws, which you. The number of candidates for the first, second and third posts are 3,4 and 2 respectively. Let P(D) be the probability that the letter is in DOGS. " Understanding how to calculate these percentages with real numbers of people and things. MIDTERM 2 solutions and Histogram. (b) Show that the result in question (a) is not necessarily correct without. 1 Problem 4E. You need at most one of the three textbooks listed below, but you will need the statistical tables. Fully worked-out solutions of these problems are also given, but of course you should first try to solve the problems on your own! c 2013 by Henk Tijms, Vrije University, Amsterdam. Probability of choosing 2nd chocobar = 3/7. In a binomial distribution the probabilities of interest are those of receiving a certain number of successes, r, in n independent trials each having only two possible outcomes and the same probability, p, of success. Objective: I know how to solve probability word problems. Three circle Venn Diagrams are a step up in complexity from two circle diagrams. Q x = P(female lives to age x) = number of female survivors at age x 100,000. Any problems you do not have time to do, try doing later. Bayes’ theorem describes the probability of occurrence of an event related to any condition. Advanced Probability Problems And Solutions Probability Questions with Solutions Tutorial on finding the probability of an event. Two balls are chosen from the jar, with replacement. What is the probability the sum of the dice is 5? 6. · = = Answer: 2: A jar contains 6 red balls, 3 green balls, 5 white balls and 7 | {
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of the dice is 5? 6. · = = Answer: 2: A jar contains 6 red balls, 3 green balls, 5 white balls and 7 yellow balls. Calculate the probability that if somebody is “tall” (meaning taller than 6 ft or whatever), that person must be male. is the constant 2. Probability has its origin in the study of gambling and insurance in the 17th century, and it is now an indispensable tool of both social and natural sciences. ) Find the probability that a person is female or prefers hiking on mountain peaks. == Solutions for Probability and Statistics for Engineering and the Sciences, 7th ed, by Jay L. Important facts and powerful problem solving approaches are highlighted throughout the text. For less than the price of a single session with a private tutor, you can have access to our entire library of videos. Probability MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Probability Math. NCERT Solutions For Maths Class 12 Chapter 13 - Probability comprises of various important questions for exams. Which pair has equally likely outcomes? List the letters of the two choices below which have equal probabilities of success, separated by a comma. Probability Exam Questions with Solutions by Henk Tijms1 December 15, 2013 This note gives a large number of exam problems for a first course in prob-ability. Two events, A and B, are independent if the outcome of A does not affect the outcome of B. So, for example, using a binomial distribution, we can determine the probability of getting 4 heads in 10 coin tosses. P(X) gives the probability of successes in n binomial trials. A Problem With Pearls. MIDTERM 1 solutions and Histogram. For Exercises 3-17 to 3-21, verify that the following functions are probability mass functions, and determine the requested probabilities. As we go deeper into the area of mathematics known as combinatorics, we realize that we come across some large numbers. [Syllabus] [Textbooks] [Grading] [Reading assignment] [Homework problems and solutions] | {
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numbers. [Syllabus] [Textbooks] [Grading] [Reading assignment] [Homework problems and solutions] [Tests] [Knowledge Milestones]. A probability distribution is a table or an equation that links each outcome of a statistical experiment with its probability of occurrence. As with any quantum system, the wave functions give the probability amplitude for finding the electron in a particular region of space, and these amplitudes are used to compute actual probabilities associated with measurements of the electron's position. Probability is the chance that an event will occur. Example: the chances of rolling a "4" with a die. Probability Problems and Solutions. Interview Probability Puzzles #1 - Most Popular Monty Hall Brain Teaser Difficulty Popularity The host of a game show, offers the guest a choice of three doors. | {
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A bijective function is both injective and surjective, thus it is (at the very least) injective. And a function is surjective or onto, if for every element in your co-domain-- so let me write it this way, if for every, let's say y, that is a member of my co-domain, there exists-- that's the little shorthand notation for exists --there exists at least one x that's a member of x, such that. If it crosses more than once it is still a valid curve, but is not a function. So we can calculate the range of the sine function, namely the interval $[-1, 1]$, and then define a third function: $$\sin^*: \big[-\frac{\pi}{2}, \frac{\pi}{2}\big] \to [-1, 1]. Stated in concise mathematical notation, a function f: X → Y is bijective if and only if it satisfies the condition for every y in Y there is a unique x in X with y = f(x). Hence every bijection is invertible. A function f : A -> B is said to be onto function if the range of f is equal to the co-domain of f. How to Prove a Function is Bijective without Using Arrow Diagram ? The inverse is conventionally called \arcsin. A function that is both One to One and Onto is called Bijective function.$$ Now this function is bijective and can be inverted. This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read "one-to-one correspondence).. Functions that have inverse functions are said to be invertible. Mathematical Functions in Python - Special Functions and Constants; Difference between regular functions and arrow functions in JavaScript; Python startswith() and endswidth() functions; Hash Functions and Hash Tables; Python maketrans() and translate() functions; Date and Time Functions in DBMS; Ceil and floor functions in C++ Each value of the output set is connected to the input set, and each output value is connected to only one input value. As pointed out by M. Winter, the converse is not | {
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output value is connected to only one input value. As pointed out by M. Winter, the converse is not true. My examples have just a few values, but functions usually work on sets with infinitely many elements. Infinitely Many. Thus, if you tell me that a function is bijective, I know that every element in B is “hit” by some element in A (due to surjectivity), and that it is “hit” by only one element in A (due to injectivity). The function f is called as one to one and onto or a bijective function, if f is both a one to one and an onto function. Ah!...The beautiful invertable functions... Today we present... ta ta ta taaaann....the bijective functions! The figure shown below represents a one to one and onto or bijective function. Some types of functions have stricter rules, to find out more you can read Injective, Surjective and Bijective. Definition: A function is bijective if it is both injective and surjective. More clearly, f maps distinct elements of A into distinct images in B and every element in B is an image of some element in A. A function is invertible if and only if it is a bijection. In mathematics, a bijective function or bijection is a function f : A → B that is both an injection and a surjection. In essence, injective means that unequal elements in A always get sent to unequal elements in B. Surjective means that every element of B has an arrow pointing to it, that is, it equals f(a) for some a in the domain of f. Question 1 : And I can write such that, like that. Below is a visual description of Definition 12.4. Curve, but is not true, surjective and bijective functions have stricter rules, to find out more can... Represents a one to one and onto or bijective function is both and! Of the output set is connected to only one input value a one to and!!... the beautiful invertable functions... Today we present... ta ta ta taaaann! The converse is not true find out more you can read injective, surjective and.! That have inverse functions are said to be | {
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ta taaaann.... the bijective functions functions that have functions! Only if it crosses more than once it is a bijection but is not a function is invertible and! B that is both an injection and a surjection be inverted one to one and onto bijective! Not a function one to one and onto or bijective function or bijection is a function bijective! Inverse functions are said to be invertible invertible if and only if it crosses more once. Usually work on sets with infinitely many elements by M. Winter, the converse is true! Out more you can read injective, surjective and bijective rules, to out. Bijective and can be inverted to only one input value that have inverse functions are to. M. Winter, the converse is not true a valid curve, is. | {
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# Prove that $\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1} \geq \frac12$
Let $$a_1, a_2, a_3, \dots , a_n$$ be positive real numbers whose sum is $$1$$. Prove that $$\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \ldots +\frac{a_n^2}{a_n+a_1} \geq \frac12\,.$$
I thought maybe the Cauchy and QM inequalities would be helpful. But I can't see how to apply it. Another thought (might be unhelpful) is that the sum of the denominators on the left hand side is $$2$$ (the denominator on the right hand side). I would really appreciate any hints.
• Apply Cauchy with one term being left hand side, second term being $((a_1+a_2) + (a_2+a_3) + \cdots)$ – user27126 Apr 28 '13 at 7:02
• Thanks so much I've solved it, should I close the question, or post the solution I just found, with your hint? – John Marty Apr 28 '13 at 7:05
• it would be good if you post a complete solution and accept it as the answer. – user27126 Apr 28 '13 at 7:06
Thanks to Sanchez for giving me a hint to solve this. Here is a full solution.
By the Cauchy-Schwarz inequality we have:
$${\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1}=\frac{a_1^2}{(\sqrt{a_1+a_2})^2}+\frac{a_2^2}{(\sqrt{a_2+a_3})^2}+ \cdots+ \frac{a_n^2}{(\sqrt{a_n+a_1})^2} \geq \frac{1}{a_1+\cdots + a_n+a_1+ \cdots + a_n}\left(\frac{a_1 \cdot \sqrt{a_1+a_2}}{\sqrt{a_1+a_2}} + \frac{a_2 \cdot \sqrt{a_2+a_3}}{\sqrt{a_2+a_3}}+ \cdots + \frac{a_n \cdot \sqrt{a_n+a_1}}{\sqrt{a_n+a_1}}\right)\\=\frac{a_1+a_2+a_3+ \cdots a_n}{{2(a_1+a_2+a_3+ \cdots a_n)}}=\frac12}$$
as required. (We know that $a_1+a_2+a_3+ \cdots +a_n=1$)
• thanks, I couldn't solve it before your hint, thanks so much. – John Marty Apr 28 '13 at 7:34
• I made an attempt to improve readability. Is it OK? – Pedro Tamaroff Jul 6 '13 at 0:37
Here's another solution.
Observe that $$\sum \frac{a_i^2 - a_{i+1}^2} { a_i+ a_{i+1}} = \sum a_i - a_{i+1} = 0,$$ | {
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Observe that $$\sum \frac{a_i^2 - a_{i+1}^2} { a_i+ a_{i+1}} = \sum a_i - a_{i+1} = 0,$$
Hence $\sum \frac{a_{i}^2}{a_i+a_{i+1}} = \sum \frac{a_{i+1}^2}{a_i+a_{i+1}}$, and we just need to show that
$$\sum \frac{a_i^2 + a_{i+1}^2}{a_i + a_{i+1} } \geq 1.$$
This makes the inequality much more symmetric, and easier to manipulate. In particular,
$$\sum \frac{a_i^2 + a_{i+1}^2}{a_i + a_{i+1} } \geq \sum \frac{1}{2} (a_i + a_{i+1}) = 1$$
The following modified form of the CBS inequality is often helpful:
Lemma. If $x_i \in \Bbb{R}$ and $a_i > 0$, then $$\frac{x_{1}^{2}}{a_{1}} + \cdots + \frac{x_{n}^{2}}{a_{n}} \geq \frac{(x_{1} + \cdots + x_{n})^{2}}{a_{1}+\cdots+a_{n}}.$$
The proof is straightforward using the CBS inequality, following the methodology exactly John Marty used.
Applying this, we have
$$\frac{a_{1}^2}{a_{1}+a_{2}} + \cdots + \frac{a_{n}^2}{a_{n}+a_{1}} \geq \frac{(a_{1} + \cdots + a_{n})^{2}}{2(a_{1}+\cdots+a_{n})} = \frac{1}{2}.$$ | {
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# Question to calculating probability
The question:
79% of the drivers in a city always fasten their seatbelt when driving, and everyday some drivers receive a ticket(for various reasons...). If a driver has fastened their seatbelt there is a chance of 7% that the driver receives a ticket. If a driver does not fasten their seatbelt there is a 22% chance that he will receive a ticket. If a driver receives a ticket what is the probability that he had fastened his seatbelt?
My assumption is that 7/29 of the people who receive a ticket had fastened their seatbelt and 22/29 did not. So I just calculated (7/29) * (0,79) + (22/29) * (0,21) and I got 0.35 but that seems to be wrong.
• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos Jan 18 '18 at 16:13
• Okay thanks for the advise. My assumption is that 7/29 of the people who receive a ticket had fastened their seatbelt and 22/29 did not. So i just calculated (7/29) * (0,79) + (22/29) * (0,21) and i got 0.35 – Simon Jan 18 '18 at 16:18
• Edit your question instead of posting it as a comment. – José Carlos Santos Jan 18 '18 at 16:19
• This is a prefect time for Bayes' Law! – Bram28 Jan 18 '18 at 16:28
Let $S$ and $T$ be the events "seatbelt fasten" and "receive ticket" respectively.
• $P(S) = 0.79$
• $P(T \mid S) = 0.07$
• $P(T \mid S^C) = 0.22$
Use Bayes' formula to find $P(S \mid T)$.
\begin{align} P(S \mid T) &= \frac{P(S \cap T)}{P(T)} \\ &= \frac{P(T \mid S) P(S)}{P(T \mid S) P(S)+P(T \mid S^C) P(S^C)} \\ &= \frac{0.07 \cdot 0.79}{0.07 \cdot 0.79 + 0.22 \cdot (1 - 0.79)} \\ &= \frac{79}{145} \end{align}
If a driver receives a ticket what is the probability that he had fastened his seatbelt is $79/145$. | {
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If a driver receives a ticket what is the probability that he had fastened his seatbelt is $79/145$.
You can't assume $\frac{7}{29}$ of people who receive a ticket fastened their seatbelt. Consider an extreme case, where the percentage of people who fasten their seatbelt is much higher than $79\%$ -- suppose it is $100\%$. Then clearly $100\%$ of people who receive a ticket fastened their seatbelt. If $0\%$ of people fasten their seatbelt, then $0\%$ who receive a ticket fastened their seatbelt. If the percentage who fasten is somewhere in between -- e.g., $79\%$ -- then the percentage who fastened given that they received a ticket is somewhere in between, but we can't just say it's $\frac{7}{29}$.
Are you familiar with Bayes' Theorem? That's probably the best way to approach this problem. | {
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What is the value of $\left(\log_{21}(3)\right)^2+\log_{21}(147)\log_{21}(1323)$?
What is the value of $$\left(\log_{21}(3)\right)^2+\log_{21}(147)\log_{21}(1323)$$ ?
$$1)1\qquad\qquad2)2\qquad\qquad3)3\qquad\qquad4)4$$
To solve this question I tried writing the expression as:
$$\left(\log_{21}(3)\right)^2+\log_{21}(7^2\times3)\log_{21}(7^2\times3^3)$$ $$=\left(\log_{21}(3)\right)^2+(2\log_{21}7+\log_{21} 3)(2\log_{21}7+3\log_{21}3)$$
I don't know how to continue.
• Don't be so eager to separate the $\log7$ from the $\log3$. The most potent course of simplification you have here is $\log21=1$. Jul 1 at 10:23
• @Arthur Thanks! The question looks horrible for a timed exam and I became a little nervous! Jul 1 at 10:29
• Denote $x=\log_{21}3$ and $y=\log_{21}7$. Then, $147=21\times 7$ and $1323=21^2\times 3$; the original expression is then $x^2+(1+y)(2+x)=x(x+y)+(x+y)+y+2$; iteratively use $x+y=1$ to simplify. Jul 1 at 10:47
• @Prasun Biswas: the computation is simpler if one notices that $147=21^2/3$ and $1323=21^2 \times 3$. Jul 1 at 10:51
• @Mindlack: Nice! Indeed, that is far more simpler because we wouldn't need the $y$, we have $x^2+(2-x)(2+x)=x^2+4-x^2=4$ Somehow, I missed that. +1 Jul 1 at 10:55
Here’s a quick-and-dirty solution based on the fact that one of the answers must be correct.
All the quantities are positive. Since $$1323>441=21^2$$, its base-$$21$$ logarithm is greater than $$2$$. As $$147 \geq 21 \times 5$$, its base-$$21$$ logarithm is at least $$1.5$$. This means that the sum is greater than $$2 \times 1.5$$ so it must be $$4$$.
• I don't get the second point regarding $147 \geq 21 \times 5$. Could you elaborate? Jul 1 at 11:04
• @user71207 $147\ge21\times5\ge21\sqrt{21}$ And the fact that $147\ge21^{\frac32}$ means $\log_{21}(147)\ge\frac32$ Jul 1 at 11:09
$$\log_{21}1323=\log_{21}(3^3\cdot 7^2)=\log_{21}3+\log_{21}21^2=2+\log_{21}3$$
$$\log_{21}147=\log_{21}21+\log_{21}7=1+\log_{21}7$$ | {
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$$(\log_{21}3)^2+\log_{21}147\cdot\log_{21}1323 \\=(\log_{21}3)^2+(1+\log_{21}7)(2+\log_{21}3) \\=\log_{21}3(\log_{21}3+\log_{21}7)+\log_{21}3+2\log_{21}7+2 \\=\log_{21}3(\log_{21}3+\log_{21}7)+(\log_{21}3+\log_{21}7)+\log_{21}7+2 \\=\log_{21}3+1+\log_{21}7+2 \\=4$$
• Hi, the problem in this post and also this one that I asked were from entrance exam in the country I'm living (average time for solving each question is about 90 seconds). You solved both of them very well and I think you are very good at solving these questions! So can you please suggest some books or name of exam that you think are good for an exam that contains these type of questions? Thanks in advanced! Jul 27 at 19:58
• I mainly use books that are meant for olympiads, so I don't know many books that are meant for solving questions fast. Jul 28 at 6:28
Alternate:
Let $$L(x)$$ denote $$\log_{21}(x).$$
Then $$[L(3)]^2 + [L(147) \times L(9 \times 147)]$$
$$= [L(3)]^2 + ~\langle ~L(147) \times ~\{ [2 \times L(3)] + L(147) ~\} ~\rangle$$
$$= [L(3)]^2 + [L(147)]^2 + [2 \times L(3) \times L(147)]$$
$$= [L(3) + L(147)]^2.$$
At this point, you have that $$3 \times 147 = 21^2$$,
so $$[L(3) + L(147)] = 2.$$ | {
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# Why is $SST=SSE + SSR$? (One variable linear regression)
Note: $SST$ = Sum of Squares Total, $SSE$ = Sum of Squared Errors, and $SSR$ = Regression Sum of Squares. The equation in the title is often written as:
$$\sum_{i=1}^n (y_i-\bar y)^2=\sum_{i=1}^n (y_i-\hat y_i)^2+\sum_{i=1}^n (\hat y_i-\bar y)^2$$
Pretty straightforward question, but I am looking for an intuitive explanation. Intuitively, it seems to me like $SST\geq SSE+SSR$ would make more sense. For example, suppose point $x_i$ has corresponding y-value $y_i=5$ and $\hat y_i=3$, where $\hat y_i$ is the corresponding point on the regression line. Also assume that the mean y-value for the dataset is $\bar y=0$. Then for this particular point i, $SST=(5-0)^2=5^2=25$, while $SSE=(5-3)^2=2^2=4$ and $SSR=(3-0)^2=3^2=9$. Obviously, $9+4<25$. Wouldn't this result generalize to the entire dataset? I don't get it.
Adding and subtracting gives \begin{eqnarray*} \sum_{i=1}^n (y_i-\bar y)^2&=&\sum_{i=1}^n (y_i-\hat y_i+\hat y_i-\bar y)^2\\ &=&\sum_{i=1}^n (y_i-\hat y_i)^2+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)+\sum_{i=1}^n(\hat y_i-\bar y)^2 \end{eqnarray*} So we need to show that $\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=0$. Write $$\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=\sum_{i=1}^n(y_i-\hat y_i)\hat y_i-\bar y\sum_{i=1}^n(y_i-\hat y_i)$$ So, (a) the residuals $e_i=y_i-\hat y_i$ need to be orthogonal to the fitted values, $\sum_{i=1}^n(y_i-\hat y_i)\hat y_i=0$, and (b) the sum of the fitted values needs to be equal to the sum of the dependent variable, $\sum_{i=1}^ny_i=\sum_{i=1}^n\hat y_i$. | {
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Actually, I think (a) is easier to show in matrix notation for general multiple regression of which the single variable case is a special case: \begin{eqnarray*} e'X\hat\beta &=&(y-X\hat\beta)'X\hat\beta\\ &=&(y-X(X'X)^{-1}X'y)'X\hat\beta\\ &=&y'(X-X(X'X)^{-1}X'X)\hat\beta\\ &=&y'(X-X)\hat\beta=0 \end{eqnarray*} As for (b), the derivative of the OLS criterion function with respect to the constant (so you need one in the regression for this to be true!), aka the normal equation, is $$\frac{\partial SSR}{\partial\hat\alpha}=-2\sum_i(y_i-\hat\alpha-\hat\beta x_i)=0,$$ which can be rearranged to $$\sum_i y_i=n\hat\alpha+\hat\beta\sum_ix_i$$ The right hand side of this equation evidently also is $\sum_{i=1}^n\hat y_i$, as $\hat y_i=\hat\alpha+\hat\beta x_i$.
This is just the Pythagorean theorem!
Hence,
$$Y'Y=(Y-X\hat{\beta})'(Y-X\hat{\beta})+(X\hat{\beta})'X\hat{\beta}$$
or
$$SST=SSE+SSR$$
• Aug 2 '19 at 19:52
(1) Intuition for why $SST = SSR + SSE$
When we try to explain the total variation in Y ($SST$) with one explanatory variable, X, then there are exactly two sources of variability. First, there is the variability captured by X (Sum Square Regression), and second, there is the variability not captured by X (Sum Square Error). Hence, $SST = SSR + SSE$ (exact equality).
(2) Geometric intuition
Some of the total variation in the data (distance from datapoint to $\bar{Y}$) is captured by the regression line (the distance from the regression line to $\bar{Y}$) and error (distance from the point to the regression line). There's not room left for $SST$ to be greater than $SSE + SSR$.
(3) The problem with your illustration | {
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(3) The problem with your illustration
You can't look at SSE and SSR in a pointwise fashion. For a particular point, the residual may be large, so that there is more error than explanatory power from X. However, for other points, the residual will be small, so that the regression line explains a lot of the variability. They will balance out and ultimately $SST = SSR + SSE$. Of course this is not rigorous, but you can find proofs like the above.
Also notice that regression will not be defined for one point: $b_1 = \frac{\sum(X_i -\bar{X})(Y_i-\bar{Y}) }{\sum (X_i -\bar{X})^2}$, and you can see that the denominator will be zero, making estimation undefined.
Hope this helps.
--Ryan M.
When an intercept is included in linear regression(sum of residuals is zero), $$SST=SSE+SSR$$. | {
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When an intercept is included in linear regression(sum of residuals is zero), $$SST=SSE+SSR$$.
prove $$\begin{eqnarray*} SST&=&\sum_{i=1}^n (y_i-\bar y)^2\\&=&\sum_{i=1}^n (y_i-\hat y_i+\hat y_i-\bar y)^2\\&=&\sum_{i=1}^n (y_i-\hat y_i)^2+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)+\sum_{i=1}^n(\hat y_i-\bar y)^2\\&=&SSE+SSR+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y) \end{eqnarray*}$$ Just need to prove last part is equal to 0: $$\begin{eqnarray*} \sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)&=&\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)(\beta_0+\beta_1x_i-\bar y)\\&=&(\beta_0-\bar y)\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)+\beta_1\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)x_i \end{eqnarray*}$$ In Least squares regression, the sum of the squares of the errors is minimized. $$SSE=\displaystyle\sum\limits_{i=1}^n \left(e_i \right)^2= \sum_{i=1}^n\left(y_i - \hat{y_i} \right)^2= \sum_{i=1}^n\left(y_i -\beta_0- \beta_1x_i\right)^2$$ Take the partial derivative of SSE with respect to $$\beta_0$$ and setting it to zero. $$\frac{\partial{SSE}}{\partial{\beta_0}} = \sum_{i=1}^n 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 = 0$$ So $$\sum_{i=1}^n \left(y_i - \beta_0 - \beta_1x_i\right)^1 = 0$$ Take the partial derivative of SSE with respect to $$\beta_1$$ and setting it to zero. $$\frac{\partial{SSE}}{\partial{\beta_1}} = \sum_{i=1}^n 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 x_i = 0$$ So $$\sum_{i=1}^n \left(y_i - \beta_0 - \beta_1x_i\right)^1 x_i = 0$$ Hence, $$\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=(\beta_0-\bar y)\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)+\beta_1\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)x_i=0$$ $$SST=SSE+SSR+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=SSE+SSR$$
Here is a great graphical representation of why SST = SSR + SSE.
• How can we see from this picture that predicted values and residuals are orthogonal? Mar 4 at 12:17 | {
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If a model predicts $$3$$ and the residual is $$2$$ because the actual value is $$5$$, it doesn't look like variance is decomposing, since $$3^2 + 2^2 \neq 5^2$$.
If you only have one data point, your model would fit it perfectly and the residual would be zero, so you can't get that case by itself. There have to be multiple data points.
With multiple data points, the residuals won't all be positive.
If the model predicts 3, a residual of $$+2$$ and $$-2$$ should be equally likely. The balance of increases and decreases gives a nice cancellation:
$$\frac{(3+2)^2 + (3-2)^2}{2} = \frac{25+1}{2} = 13 = 9 + 4 = 3^2 + 2^2$$
This property, that $$(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$$, is what makes variance decomposition work, for uncorrelated components. | {
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# Easy Money
On a table before you lie two envelopes, identical in appearance. You are told that one envelope contains twice as much money as the other, but given no indication which has more. You are then allowed to choose one of the envelopes and keep the money it contains.
You then pick one of the envelopes at random, but before you look inside you are offered the chance to exchange your envelope for the other. Should you exchange envelopes, (assuming that the more money you get, the better)?
One line of thinking is this: If the envelope you initially choose has $$x$$ dollars in it, then there is a $$50$$% chance that the other envelope has $$2x$$ dollars in it and a $$50$$% chance that it has $$\frac{x}{2}$$ dollars in it. Thus the expected amount of money you would end up with if you chose to exchange envelopes would be
$$\frac{1}{2} * 2x + \frac{1}{2} * \frac{x}{2} = \frac{5}{4} x$$ dollars.
You thus decide to exchange envelopes. But before doing so, you think to yourself, "But I could go through the same thought process with the other envelope, concluding that I should exchange my new envelope for my old one, and then go through the same thought process again, and again, ad infinitum. Now I have no idea what to do."
How can you resolve this paradox?
Edit: As mentioned in my comment below, the real "puzzle" here is to identify the flaw in the reasoning I outlined above. Or is there a flaw?
Note by Brian Charlesworth
4 years, 4 months ago
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First, look at the case when the envelope is exchanged. If you pick the envelope with $$x$$ dollars $$(P=1/2)$$ and exchange, you win $$2x$$ dollars. Instead, if you pick the envelope with $$2x$$ dollars $$(P=1/2)$$ and exchange, then you win $$x$$ dollars.
$\therefore E(\text{Money when envelope is exchanged})=\dfrac{1}{2}2x+\dfrac{1}{2}x=\dfrac{3x}{2}$
Now, look at the case when the envelope is not exchanged. If you pick the envelope with $$x$$dollars $$(P=1/2)$$ and don't exchange, you win $$x$$ dollars. But if you pick the envelope with $$2x$$ dollars $$(P=1/2)$$ and don't exchange, then you win $$2x$$ dollars.
$\therefore E(\text{Money when envelope is not exchanged})=\dfrac{1}{2}2x+\dfrac{1}{2}x=\dfrac{3x}{2}$
So it makes no difference at all whether you exchange or don't exchange your envelope.
- 4 years, 4 months ago
Yes, that is a good analysis. Since we are not given any further information prior to making a choice whether or not to exchange envelopes, we are in the same position as we were with our initial choice, i.e., our expected winnings will be $$\frac{3x}{2}$$.
But the "real" puzzle here is to find the logical miscue in the line of reasoning I laid out. It can be argued that in the two scenarios I describe, (namely that I start out with the lesser amount of money or the greater), the value $$x$$ actually represents different amounts in each case, so that I can't use the same variable $$x$$ for each of them, making my equation invalid. | {
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However, suppose I did peek at the amount in the envelope before making a decision. Now this provides me with "information", but in reality it's useless since I have no clue as to whether it's the lesser or greater amount. But since we now can attach an actual value to $$x$$, the mathematics I applied in my initial analysis becomes re-validated, and hence so does my "conclusion". But since, as noted, the information regarding the amount in the initially chosen envelope is in reality useless, should it matter whether we know the actual amount as far as the mathematical analysis is concerned? From this perspective, no matter what amount we see inside the initial envelope we should conclude that switching envelopes is preferable.
So the puzzle of "where's the flaw in logic?" is more a question of perspective; are we dealing with unconditional or conditional probability values, and why should this difference in perspective matter given the above discussion? A surprising amount of attention has been given to this paradox, commonly referred to as the "Two Envelope Paradox", with no universally accepted resolution as of yet. I think why it does garner such attention is that it cuts to the heart of the foundation of probability theory, and as with any foundation, the desire is that it must be as secure as possible, and if it is not, it must be either fortified or torn down.
Here is an accessible discussion on the "search for the flaw" and why many find it so intriguing.
- 4 years, 4 months ago
I'm not a mathematician, just a dilettante of sorts but something tells me we're over complicating this paradox. There are only two choices and only two possibilities. The money is either greater or lesser and you either switch or stay. The fact that the sum of all probabilities is 1 is infallible. We simply assume we have knowledge we previously didn't have and compute what the probabilities are to find out what is what. | {
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Here's another question: Are we then told that "hey, you chose the lesser or greater amount of money" after we've made our choice? I think not. So the probability will not even depend on outcome. For all intents and purposes, we need not even bother with the maths of 'x' and '2x'; and we might as well adopt an approach with x and y since the relationship between the monies simply do not matter. I might as well be a choice between winning the lottery and detonating an atomic bomb.
- 4 years ago
I think that the reasons this "paradox" has received so much attention over the years are more esoteric in nature than practical. For some it raises questions about the foundation of probability theory regarding the absolute distinction between conditional/unconditional scenarios and whether or not there might be a grey area there. However, mathematicians do have a tendency to "overcomplicate" things, and this may a case in point. :) | {
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As for the choice between winning the lottery and detonating an atomic bomb, well, I think I would just walk away from the table; too many lives other than my own would be at stake without them having any say in the matter. However, if it were a choice between my eternal health, wealth and happiness on the one hand and my immediate extinction on the other, I would give it a bit more thought. I would probably still walk away, since it would feel like I'm making some kind of deal with the Devil, but it does make for an interesting psychological experiment. This is an entirely different question than the previous "paradox" we were dealing with; with the money in the two envelopes, no matter what choice we made we would be richer than before. With this new question, there is the possibility of a terrible outcome, so the first choice becomes whether to engage at all, and if we do then why do we think it is worth such a high risk of calamity. The more we would have to lose in our present life the less likely we would engage, but if we did not have much to live for then the risk might be worth it. However, as I said before, only the Devil would make such a proposition, so anyone, no matter what their circumstances, would be wise to just walk away. Sorry, I do ramble on, but i always do that when a good philosophical question happens by. :)
- 4 years ago
Hahaha.I do love to read you rambling on. It's more engaging than many other types of conversations. Your ending pretty much wrapped up my premise: which is that, the paradox seems to rely on the probability of making a more money outcome which is identical to that of making a lesser money. I do have some philosophical questions if you're interested, I'd love to read your views. I'm relatively new on this site so is it OK if I create a note and invite you to give your thoughts on it?
- 4 years ago
Of course; I look forward to seeing what the topic is. :)
- 4 years ago
Hi again Mr. Charlesworth | {
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Of course; I look forward to seeing what the topic is. :)
- 4 years ago
Hi again Mr. Charlesworth
Here are two notes I have posted. I invite you to share your thoughts on them: https://brilliant.org/discussions/thread/number-bases-in-a-different-world/# https://brilliant.org/discussions/thread/the-infiniteness-of-light/?ref_id=654556
Hi @Calvin Lin. Do you think it's a good idea to find a way to incorporate Philosophy (Logic and Reasoning) on Brilliant? I think it is a very fascinating pseudoscience and it would appeal to many older people and others who are not strictly geeks and nerds like the rest of us. Just a thought
- 4 years ago
Let's incorporate Speculative Science And Math, as distinct from formal Philosophy. You know, something that would appeal to one old guy here who wants to try his hand in being a crank. | {
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I'm reading Brian's posted scenario, and it seems to me that if it's known that one envelope contains twice the other, then the expected value in choosing either is simply $$\dfrac { 1 }{ 2 } x+\dfrac { 1 }{ 2 } 2x=\dfrac { 3 }{ 2 } x$$ either way, as pointed out by Pratik. It's 50% percent chance that it's twice the one you've chosen, and 50% percent chance it's half that--but you can't assume the same $$x$$ dollars in your chosen envelope in both cases! That was the flaw. Maybe Brian is trying to alluding to the problem of trying to determine probabilities when information is incomplete, which is what led to Bayesian probability theory. Let me offer a slightly different scenario. You are given $$2$$ envelopes. You are told that one of them may or may not contain a valuable stock certificate. You open one up and do not find it inside. What's the probability the other one does? Extend that to $$100$$ such envelopes. This is actually what happens when you are looking for gold in the ground in a certain area when you don't have much information to go on. After so much digging, at what point in time do you give up? Are more likely to find gold, because you've already almost eliminated the places where it can be, or are you more less likely to find gold, because you've already tried to find it and failed? A lot of science can be like that.
- 4 years ago
Oh yes, that's a great idea because what I was hoping for was an avenue to wonder beyond the limits of our recorded knowledge and rub minds on what could be and what not. | {
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I'm fascinated this "blind" probability theory. It is a little bit anti cognitive. Which reminds me of something else which defies our flawed cognizance. Consider flipping a coin and getting heads five times in a row. Flipping it a sixth time seems as though there is a higher probability to get tails. In fact, I'd bet that if you did a survey of ten mathematics professors, a higher percentage of them would choose tails for that sixth coin toss except they are just fighting against their human nature. But sometimes I think about it and really wonder if the law of averages is that subjective. If you mix two liquids together, they mix together perfectly EVERY TIME if they are of similar viscosity. I'm rambling. I have even forgotten what the point I'm trying to make is
- 4 years ago
A lot of theoretical physics was minds rambling before something more concrete came out of it. Ramble on.
Likewise, a lot of technology was preceded by science fiction. Smart phones? Didn't Star Trek popularize that idea?
- 4 years ago | {
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# 4x4 Matrix with rank B=4 and B^2=3
#### Petrus
##### Well-known member
Hello MHB,
"Can we construct a $$\displaystyle 4x4$$ Matrix $$\displaystyle B$$ so that rank $$\displaystyle B=4$$ but rank $$\displaystyle B^2=3$$"
My thought:
we got one condition for this to work is that det $$\displaystyle B=0$$ and det $$\displaystyle B^2 \neq 0$$ and B also have to be a upper/lower or identity Matrix. And this Will not work.. I am wrong or can I explain this in a better way?
Regards,
$$\displaystyle |\pi\rangle$$
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
"Can we construct a $$\displaystyle 4x4$$ Matrix $$\displaystyle B$$ so that rank $$\displaystyle B=4$$ but rank $$\displaystyle B^2=3$$"
My thought:
we got one condition for this to work is that det $$\displaystyle B=0$$ and det $$\displaystyle B^2 \neq 0$$
It's the other way around: $\mathop{\text{rank}}(B)=4\iff\det(B)\ne0$ and $\mathop{\text{rank}}(B^2)=3\implies\det(B^2)=0$. But you are right that this is impossible because $\det(B^2)=(\det(B))^2$.
#### Petrus
##### Well-known member
It's the other way around: $\mathop{\text{rank}}(B)=4\iff\det(B)\ne0$ and $\mathop{\text{rank}}(B^2)=3\implies\det(B^2)=0$. But you are right that this is impossible because $\det(B^2)=(\det(B))^2$.
Hello Evgeny.Makarov,
thanks for fast respond and I meant that! And thanks for showing me this I did not know that $\det(B^2)=(\det(B))^2$ That Was what I Was looking for
Regards,
$$\displaystyle |\pi\rangle$$
Last edited:
#### Fernando Revilla
##### Well-known member
MHB Math Helper
An alternative proof (without using determinants):
Consider the linear map $B:\mathbb{R}^4\to \mathbb{R}^4,\; x\to Bx.$ As $\operatorname{rank}B=4,$ $\operatorname{nullity}B=0,$ wich implies $B$ is bijective. But the composition of bijective maps is bijective, so $\operatorname{rank}B^2=4.$
#### Deveno
##### Well-known member
MHB Math Scholar
Another formulation: | {
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#### Deveno
##### Well-known member
MHB Math Scholar
Another formulation:
As rank(B) = 4, B is surjective, that is, B(R4) = R4 (for this is what rank means: the dimension of the image space (or column space) of B, and R4 is the ONLY 4-dimensional subspace of R4).
Consequently, B2(R4) = B(B(R4)) = B(R4) = R4, from which we conclude B2 is likewise surjective, and thus rank(B2) = 4 as well.
(I only post this to indicate one need not even invoke the rank-nullity theorem).
#### Petrus
##### Well-known member
this is impossible because $\det(B^2)=(\det(B))^2$.
Now that I think about I remember a sats that said $$\displaystyle |AB|=|A||B|$$ but in this case $$\displaystyle A=B$$ hmm I need to find the proof for this.
Regards,
$$\displaystyle |\pi\rangle$$
MHB Math Scholar
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
I saw one neat proof of $\det(AB)=\det(A)\det(B)$ in Linear Algebra and Its Applications by Gilbert Strang. He defines $\det(\cdot)$ as a function satisfying three properties:
(1) $\det(I)=1$ where $I$ is the identity matrix;
(2) it changes sign when two adjacent rows are swapped;
(3) it is linear on the first row.
Signed volume in an orthonormal basis satisfies these properties, so this definition is much more intuitive than the Leibniz formula, which is derivable from (1)–(3).
Now, to prove that $\det(AB)=\det(A)\det(B)$, fix $B$ and consider $d(A)=\det(AB)/\det(B)$. It is possible to show that $d(A)$ satisfies (1)–(3), and so $d(A)=\det(A)$.
Now that I looked at the StackExchange link, this is answer #2, which is highest-ranked. | {
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# Closed form of $\sum\limits_{i=1}^n\left\lfloor\frac{n}{i}\right\rfloor^2$?
Does $\displaystyle\sum_{i=1}^n\left\lfloor\dfrac{n}{i}\right\rfloor^2$ admit a closed form expression?
• Unlikely; see the related math.stackexchange.com/q/384520/73324 – vadim123 Jun 7 '13 at 18:47
• @Martin, is the sequence formed by these sums not a sequence? – Antonio Vargas Dec 1 '16 at 17:38
• @AntonioVargas The same could be said about any finite sum of the form $\sum_{k=1}^n a_n$. Of course, if you think that for some reasone the tag belongs there, you can edit it back. My understanding is that the tag-excerpt says rather clearly: "For questions on finite sums use the (summation) instead." – Martin Sleziak Dec 1 '16 at 17:41
To see how the first term in the asymptotic expansion is obtained, put $$a(n) = \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor^2$$ and note that $$a(n+1)-a(n) = 1 + \sum_{k=1}^n \left(\bigg\lfloor \frac{n+1}{k} \bigg\rfloor^2 - \bigg\lfloor \frac{n}{k} \bigg\rfloor^2\right) \\= 1 + \sum_{d|n+1 \atop d<n+1} \left(\left(\frac{n+1}{d}\right)^2 - \left(\frac{n+1}{d}-1\right)^2\right) = \sum_{d|n+1} \left(2\left(\frac{n+1}{d}\right)-1\right) \\= 2\sigma(n+1)-\tau(n+1).$$ | {
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It now follows that $$a(n) = 2\sum_{k=1}^n \sigma(k) - \sum_{k=1}^n \tau(k) = \sum_{k=1}^n \left(2\sigma(k)-\tau(k)\right).$$ We can apply the Wiener-Ikehara theorem to this sum, working with the Dirichlet series $$L(s) = \sum_{n\ge 1} \frac{2\sigma(n)-\tau(n)}{n^s} = 2\zeta(s-1)\zeta(s)-\zeta(s)^2.$$ We have $$\operatorname{Res}(L(s); s=2) = \frac{\pi^2}{3},$$ so that by the aforementioned theorem, $$a(n) \sim \frac{\pi^2/3}{2} n^2 = \frac{\pi^2}{6} n^2.$$ In fact we can use Mellin-Perron summation to predict, but not quite prove, the next terms in the asymptotic expansion, getting $$a(n) = \left(\sigma(n)-\frac{1}{2}\tau(n)\right) + \frac{1}{2\pi i} \int_{5/2-i\infty}^{5/2+i\infty} L(s) n^s \frac{ds}{s}$$ which yields $$a(n) \sim \left(\sigma(n)-\frac{1}{2}\tau(n)\right) +\frac{\pi^2}{6} n^2 - (\log n + 2 \gamma)n - \frac{1}{6}.$$ This approximation is quite good, giving $16085.71386$ for $n=100$ when the correct value is $16116$ and $1639203.715$ for $n=1000$ when the correct value is $1639093.$
This is A222548 in the online encyclopedia of integer sequences. They don't provide a closed form, but they do give the following:
$$a(n)\approx\frac{\pi^2}{6}n^2+O(n\log n)$$
Let me complement @Marko Riedel's excellent answer by providing a low-tech approach to the leading term.
Let us denote by $\mathbf{1}\{\cdot\}$ the indicator function of the set $\{\cdot\}$. Then using the identity
$$\lfloor n/i \rfloor = \sum_{j=1}^{n} \mathbf{1}\{ij \leq n \}$$
we can rearrange the sum to find: | {
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$$\lfloor n/i \rfloor = \sum_{j=1}^{n} \mathbf{1}\{ij \leq n \}$$
we can rearrange the sum to find:
\begin{align*} a(n) &= \sum_{i=1}^{n} \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \mathbf{1}\{ij \leq n\}\mathbf{1}\{ik \leq n\} = \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \sum_{i=1}^{n} \mathbf{1}\Big\{i \leq \frac{n}{\max\{j,k\}}\Big\} \\ &= \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \left\lfloor \frac{n}{\max\{j,k\}} \right\rfloor = \sum_{l=1}^{n} \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \left\lfloor \frac{n}{l} \right\rfloor \mathbf{1}\{\max\{j,k\} = l\} \\ &= \sum_{l=1}^{n} (2l-1) \left\lfloor \frac{n}{l} \right\rfloor. \end{align*}
Dividing both sides by $n^2$, the RHS can be identified as Riemann sum and thus
$$\frac{a(n)}{n^2} = \sum_{l=1}^{n} \frac{2l-1}{n} \left\lfloor \frac{n}{l} \right\rfloor \frac{1}{n} \xrightarrow[\ n\to\infty \ ]{} \int_{0}^{1} 2x \left\lfloor\frac{1}{x}\right\rfloor \, dx = \zeta(2).$$
Here, the last integral can be easily computed by applying the substitution $x \mapsto 1/x$.
An Answer Without Number Theoretic Functions | {
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An Answer Without Number Theoretic Functions
Break up the sum as follows: $$\sum_{k=1}^n\left\lfloor\frac nk\right\rfloor^2 =\sum_{k=1}^n\frac{n^2}{k^2}-\sum_{k=1}^n\left(\frac{n^2}{k^2}-\left\lfloor\frac nk\right\rfloor^2\right)\tag{1}$$ We can apply the Euler-Maclaurin Sum Formula $$\sum_{k=1}^n\frac{n^2}{k^2}\sim n^2\left(\frac{\pi^2}6-\frac1n+\frac1{2n^2}\right)\tag{2}$$ If we note that $\left\{\frac nk\right\}\lt1-\frac1k$, we see that \begin{align} \sum_{k=1}^n\left(\frac{n^2}{k^2}-\left\lfloor\frac nk\right\rfloor^2\right) &=\sum_{k=1}^n\left(\frac nk-\left\lfloor\frac nk\right\rfloor\right)\left(\frac nk+\left\lfloor\frac nk\right\rfloor\right)\\ &=\sum_{k=1}^n\left\{\frac nk\right\}\left(2\frac{n}{k}-\left\{\frac nk\right\}\right)\\ &\le\sum_{k=1}^n\left(1-\frac1k\right)2\frac nk\\[9pt] &\le2n\log(n)\tag{3} \end{align} Therefore, by $(1)$, $(2)$, and $(3)$ we have $$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^n\left\lfloor\frac nk\right\rfloor^2=n^2\frac{\pi^2}6+O(n\log(n))}\tag{4}$$ However, if we use $\frac12{-}\frac1{2k}$ for the mean value of $\left\{\frac nk\right\}$ and $\frac13{-}\frac1{2k}{+}\frac1{6k^2}$ for the mean value of $\left\{\frac nk\right\}^2$ \begin{align} \sum_{k=1}^n\left(\frac{n^2}{k^2}-\left\lfloor\frac nk\right\rfloor^2\right) &=\sum_{k=1}^n\left(\frac nk-\left\lfloor\frac nk\right\rfloor\right)\left(\frac nk+\left\lfloor\frac nk\right\rfloor\right)\\ &=\sum_{k=1}^n\left\{\frac nk\right\}\left(2\frac{n}{k}-\left\{\frac nk\right\}\right)\\ &\approx\sum_{k=1}^n\left(\frac12-\frac1{2k}\right)2\frac{n}{k}-\left(\frac13-\frac1{2k}+\frac1{6k^2}\right)\\ &=n\log(n)+n\left(\gamma-\frac{\pi^2}6-\frac13\right)+O(\log(n))\tag{5} \end{align} Thus, if we look at the average behavior, we can use $(1)$, $(2)$, and approximation $(5)$ to get $$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^n\left\lfloor\frac nk\right\rfloor^2 \approx n^2\frac{\pi^2}6-n\log(n)+n\left(\frac{\pi^2}6-\gamma-\frac23\right)}\tag{6}$$ | {
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The value given by $(6)$ for $n=98$ is $15387.9230$ while the actual value is $15381$.
The value given by $(6)$ for $n=1001$ is $1641711.3692$ while the actual value is $1641772$.
The error of the approximation in $(6)$ appears to be proportional to $n$. Here is a plot of that error, for $1\le n\le3000$, divided by $n$:
The error divided by $n$ has a mean of $-0.0746805451$ and a standard deviation of $0.6513592109$.
Using number theoretic functions, Marko Riedel's answer gives a better approximation. Here is a plot of the error of that approximation, for $1\le n\le3000$, divided by $n$:
The error divided by $n$ has a mean of $0.0060900069$ and a standard deviation of $0.3674311090$. This is better than the error distribution in my answer, but it does require factoring $n$ to compute the number theoretic functions. | {
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# Does there exist more than 3 connected open sets in the plane with the same boundary?
I've wondered about the following question, whose answer is perhaps well known (in this case I apologize in advance).
The Lakes of Wada are a famous example of three disjoint connected open sets of the plane with the counterintuitive property that they all have the same boundary (!)
My question is the following :
Can we find four disjoint connected open sets of the plane that have the same boundary?
More generally :
For each $n \geq 3$, does there exist $n$ disjoint connected open sets of the plane that have the same boundary? If not, then what is the smallest $n$ such that the answer is no?
Thank you, Malik | {
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Thank you, Malik
• The article you link to says "A variation of this construction can produce a countable infinite number of connected lakes with the same boundary: instead of extending the lakes in the order 1, 2, 0, 1, 2, 0, 1, 2, 0, ...., extend them in the order 0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, ...and so on.", so the answer seems to be "you can do it for all $n$ including $n =\aleph_0$". I'd guess (but have no idea really) that the corresponding question for other cardinalities is independent of ZFC, assuming NOT CH. Though, since $\mathbb{R}^n$ is first countable, there may be an answer... – Jason DeVito Mar 19 '12 at 15:55
• Found this in the link. It seems to answer your question. "A variation of this construction can produce a countable infinite number of connected lakes with the same boundary: instead of extending the lakes in the order 1, 2, 0, 1, 2, 0, 1, 2, 0, ...., extend them in the order ..." – Patrick Mar 19 '12 at 15:57
• Given that the basins of attraction of Newton's method for $z^3=1$ are an example for $n=3$, I'd say that the basins of attraction of Newton's method for $z^n=1$ should work just as fine. – lhf Mar 19 '12 at 16:49
• Jason, separability precludes finding any uncountable collection of pairwise disjoint nonempty open sets, irrespective of any conditions you put on their boundaries. – user83827 Mar 19 '12 at 17:23
• @Student73, ah, this may contain a proof: Frame, Michael; Neger, Nial. Newton's method and the Wada property: a graphical approach. College Math. J. 38 (2007), no. 3, 192–204. MR2310015 (2008e:37082) – lhf Mar 19 '12 at 17:31
Given that the basins of attraction of Newton's method for $z^3=1$ are Wada sets for $n=3$, I'd say that the basins of attraction of Newton's method for $z^n=1$ should work just as fine. I don't know a reference to a proof but try these: | {
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# Thread: s4.13.5.41 relation of 2 planes
1. $\tiny{{s4}.{13}.{5}.{41}}$
$\textsf{find if planes are$\parallel, \perp$or$\angle$of intersection }\\$
\begin{align}
\displaystyle
{P_1}&={x+z=1}\\
\therefore n_1&=\langle 1,0,1 \rangle\\
\\
{P_2}&={y+z=1}\\
\therefore n_2&=\langle 0,1,1 \rangle\\
\\
\cos(\theta)&=
\frac{n_1\cdot n_2}{|n_1||n_2|}\\
&=\frac{1(0)+0(1)+1(1)}
{\sqrt{1+1}\cdot\sqrt{1+1}}
=\frac{1}{2}\\
\cos^{-1}\left({\frac{1}{2}}\right)&=
\color{red}{60^o}
\end{align}
$\textit{there are 2 more problems like this so presume this is best method.. }\\$
$\textit{didn't know if it is common notation to call a plane$P_1$}$
2. Originally Posted by karush
$\tiny{{s4}.{13}.{5}.{41}}$
$\textsf{find if planes are$\parallel, \perp$or$\angle$of intersection }\\$
\begin{align}
\displaystyle
{P_1}&={x+z=1}\\
\therefore n_1&=\langle 1,0,1 \rangle\\
\\
{P_2}&={y+z=1}\\
\therefore n_2&=\langle 0,1,1 \rangle\\
\\
\cos(\theta)&=
\frac{n_1\cdot n_2}{|n_1||n_2|}\\
&=\frac{1(0)+0(1)+1(1)}
{\sqrt{1+1}\cdot\sqrt{1+1}}
=\frac{1}{2}\\
\cos^{-1}\left({\frac{1}{2}}\right)&=
\color{red}{60^o}
\end{align}
$\textit{there are 2 more problems like this so presume this is best method.. }\\$
$\textit{didn't know if it is common notation to call a plane$P_1$}$
Everything you've posted is fine. You can give a plane any name you like, but I would write something like this:
\displaystyle \begin{align*} P_1 : x + z = 1 \end{align*}
That way we can see that \displaystyle \begin{align*} P_1 \end{align*} is DEFINED as the relationship "the sum of the x and z values needs to be 1", not that it is some variable that has something to do with the equation.
3. The problem asks you to do three things:
1) determine if the planes are parallel.
2) determine if the planes are perpendicular.
3) if neither of those, determine the angle of intersection of the two planes. | {
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Yes, by using that formula to determine the angle, you can then answer all three questions but it should be simpler to determine the first two without using that formula:
The two planes are parallel if and only if the two normal vectors are parallel- if one is a multiple of the other.
The two planes are perpendicular if and only if the two normal vectors are perpendicular: if their dot product is 0.
If neither of those is true, then you can use the dot product you found as the numerator in the formula to determine the angle.
hard to see that in their examples
the next 2 problems are probably
$\displaystyle \parallel , \perp$
$\tiny{s4.854.13.5.43}$
$\textsf{Determine if the 2 given planes are perpendicular, parallel or at an angle to each other}$
\begin{align}
\displaystyle
{p_1}&:{x+4y-3z=1}
\therefore n_1=\langle 1,4,-3 \rangle\\
\nonumber\\
{p_2}&:{-3x+6y+7z=0} \therefore n_2=\langle -3,6,7 \rangle
\end{align}
\begin{align}
\displaystyle
\cos(\theta)&= \frac{n_1\cdot n_2}{|n_1||n_2|}=0\\
\therefore p_1 &\perp p_2
\end{align}
$\tiny{s4.854.13.5.45}$
\begin{align}
\displaystyle
{p_1}&:{2x+4y-2z=1}
\therefore n_1=\langle 2,4,-2 \rangle\\
\nonumber\\
{p_2}&:{-3x+6y+3z=0} \therefore n_2=\langle -3,6,7 \rangle \\
n_1&=-\frac{2}{3} n_2 \\
&\therefore n_1\parallel n_2
\end{align}
$\textit{btw what is the input string to see the 2 plane on a W|A graph?}$
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# Probability of rolling a dice 8 times before all numbers are shown.
What is the probability of having to roll a (six sided) dice at least 8 times before you get to see all of the numbers at least once?
I don't really have a clue how to work this out.
Edit: If we are trying to find the number of situations where all of the numbers are shown, for seven rolls, a favorable outcome would be one in which there are two numbers the same, for example: 1123456. These numbers can be arranged in 7!/2!5! ways, and there are 6 different numbers that could be repeated. So the probability of getting all 6 numbers with 7 throws is (6*7!/2!5!)/6^7 = 126/279936. Is that right? Then 1 minus this is the probability?
Edit: prob. of not getting all six numbers with seven rolls = 1-prob of getting all six numbers with seven rolls
Six numbers same with seven rolls means one number repeated twice 7!/2! ways of doing this for each repeated number 6*(7!/2!) is number of ways of getting all six numbers with seven rolls. Total number of outcomes 6^7. Prob of getting all six numbers with seven rolls = (6*(7!/2!))/6^7 = 0.054 Prob of not getting all six numbers with seven rolls (=prob of needing at least 8 rolls to get all numbers) = 1-0.054 = 0.946 | {
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• Let $p$ be the probability that in $7$ rolls you see all faces at least once. We find $p$. Then the answer to the original problem is $1-p$. It should not be too hard to find $p$, by counting the "favourables" and dividing by $6^7$. – André Nicolas Aug 6 '15 at 13:34
• As a further hint to counting the "favorables" for $p$, note that if you roll 7 times you will see at least one number repeated. But if this happens more than once you will not get to see all 6 numbers. – David K Aug 6 '15 at 13:50
• The title and text of the question don't match. Please clarify whether "at least" is intended, as in the text, or "exactly", as the title appears to imply. – joriki Aug 6 '15 at 14:17
• To clarify are you asking what is the probability of needing exactly 8 rolls to see all numbers or 8 or more rolls? – Warren Hill Aug 7 '15 at 8:20
Rephrase the question:
What is the probability of not seeing all $6$ values when rolling a die $7$ times?
Find the probability of the complementary event:
What is the probability of seeing all $6$ values when rolling a die $7$ times?
Use the inclusion/exclusion principle in order to count the number of ways:
• Include the number of ways to roll a die $7$ times and see up to $6$ different values: $\binom66\cdot6^7$
• Exclude the number of ways to roll a die $7$ times and see up to $5$ different values: $\binom65\cdot5^7$
• Include the number of ways to roll a die $7$ times and see up to $4$ different values: $\binom64\cdot4^7$
• Exclude the number of ways to roll a die $7$ times and see up to $3$ different values: $\binom63\cdot3^7$
• Include the number of ways to roll a die $7$ times and see up to $2$ different values: $\binom62\cdot2^7$
• Exclude the number of ways to roll a die $7$ times and see up to $1$ different values: $\binom61\cdot1^7$
Divide the result by the total number of ways, which is $6^7$: | {
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Divide the result by the total number of ways, which is $6^7$:
$$\frac{\binom66\cdot6^7-\binom65\cdot5^7+\binom64\cdot4^7-\binom63\cdot3^7+\binom62\cdot2^7-\binom61\cdot1^7}{6^7}=\frac{35}{648}$$
Calculate the probability of the original event by subtracting the result from $1$:
$$1-\frac{35}{648}=\frac{613}{648}\approx94.6\%$$
• Nice use of inclusion/exclusion! Simulated probability of seeing all 6 faces in 7 rolls in my Answer is 0.0544 ± 0.001, compared to exact 35/648 = 0.0540 in this Answer. – BruceET Aug 7 '15 at 8:46
• @BruceTrumbo: Thanks :) – barak manos Aug 7 '15 at 8:50
(Edit: Answer is for the probability of seeing all the numbers for $n$ rolls)
For exactly $n$ rolls, the problem can be solved using a markov chain
\begin{align*} A&= \begin{pmatrix}0 & 1 & 0 & 0 & 0 & 0 & 0\cr 0 & \frac{1}{6} & \frac{5}{6} & 0 & 0 & 0 & 0\cr 0 & 0 & \frac{1}{3} & \frac{2}{3} & 0 & 0 & 0\cr 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0\cr 0 & 0 & 0 & 0 & \frac{2}{3} & \frac{1}{3} & 0\cr 0 & 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6}\cr 0 & 0 & 0 & 0 & 0 & 0 & 1\end{pmatrix} \end{align*}
where the rows and columns are the number of faces of the die seen.
The probability of seeing all the faces in exactly 8 rolls is $(A^8)[0,6]$, which is $\dfrac{665}{5832}\approx 0.11402606310014$
For any $n$, it can be found by finding the generating function $G(z)$ and in turn finding the coefficient of $z^n$
\begin{align*} \mathbb{P}(n) = 1-\frac{20}{2^n}+15\left(\frac{2}{3}\right)^n+\frac{15}{3^n}-6\left(\frac{5}{6}\right)^n-\frac{6}{6^n} \end{align*} | {
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• That was my original solution (should be $1-\frac{665}{5832}$, BTW)... But I then realized that "at least 8 times before you get to see all of the numbers" meant that you were allowed to see all of the numbers by the $8$th trial, so you need to replace $8$ with $7$ in your answer. – barak manos Aug 7 '15 at 12:55
• Oh, okay. I solved for the probability of seeing all the numbers for n rolls. – gar Aug 7 '15 at 13:24
• Very elegant. Clarity of presentation could be improved for quick reading (and more upvotes). I suggest to: make the first sentence bold, correcting for n=7 so that the answers match, small subtitle "general solution" at the end. – Symeof Sep 23 '15 at 21:36
Perhaps this is of some help, even if not your final answer.
If a fair die is rolled 8 times and $X$ is the number of different faces seen, then simulation of a million performances of the 8-roll experiment shows the following approximate distribution of $X$, in which the margin of simulation error is about $\pm 0.001.$
1 2 3 4 5 6
0.000003 0.002275 0.069393 0.363681 0.450468 0.114180
Of course, $P(X = 1) = 1/6^7 = 0.000004$ to 6 places. To get $P(X = 6),$ consider that one face may be seen three times, or two different faces may each be seen twice. A straightforward computation of other probabilities in the distribution seems feasible but increasingly tedious. Perhaps there are clever combinatorial methods that make some of them easier.
However, pending clarification, I'm still not sure if $P(X = 6)$ is exactly the probability you're trying to find.
Notes:
(a) For a 7-roll experiment an analogous simulation gives $0.0544 \pm 0.001,$ which does not agree with your result $126/279936 \approx 0.00045.$ (I do not believe you are considering all possible permutations of the numbers that occur only once.) | {
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(b) For a 6-roll experiment a simulation gives $0.015496 \pm 0.001,$ compared with the exact $6!/6^6 = 0.0154321.$ Clearly, the probability of getting all 6 faces is 7 rolls must be larger than that.
(c) A related problem is the number of rolls $W$ required before all 6 faces are seen. Then $E(W) = 6/6 + 6/5 + \dots + 6/1 = 14.7.$ For more on this, see the 'coupon collecting problem'.
Here's another way of looking at it, along the lines suggested by barak manos.
Note that 7 rolls containing all 6 values must contain a duplicate. There are $\binom{7}{2}\cdot 6$ choices for the duplicate entries. Thus, the desired probability is given by
$$1-\frac{\binom{7}{2}\cdot 6\cdot 5!}{6^7} = 1-\frac{70}{1296}=\frac{613}{648}.$$ | {
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# Garland trouble
#### grgrsanjay
##### New member
How many different garlands are possible with 3 identical beads of red color and 12 identical beads of black color?
I was thinking to keep the no.of beads in between the 3 beads of red color as x,y,z
So, x+y+z=12
no .of ways is 14C2.....Was i Wrong somewhere??
Last edited:
#### tkhunny
##### Well-known member
MHB Math Helper
Black will fit not only between, but also before and after. You seem to have only one end.
W before any red
X after first red
Y after second red
Z after last red
W + X + Y + Z = 12
14C3
#### Plato
##### Well-known member
MHB Math Helper
Black will fit not only between, but also before and after. You seem to have only one end.
W before any red
X after first red
Y after second red
Z after last red
W + X + Y + Z = 12
14C3
But the difficulty there is these are garlands, circular bead arrangements.
$BBBBBBBBBRRR$ is the same as $RRRBBBBBBBBB$.
Moreover, $RBBRBBBRBBBB$ is the same as $RBBBBRBBRBBB$.
How many ways are there to partition $9$ into three of less summands?
For example here are three: $9$, $7+2$ and $5+3+1$.
Last edited:
#### tkhunny
##### Well-known member
MHB Math Helper
I don't disagree with your solution. I suggest only that the definition of "garland" is ambiguous.
After a quick survey of those in my home, they ALL believed a "garland" to be more of a hanging rope, and not a wreath. I believe we have established significant ambiguity (at least cultural) in the problem statement.
Perhaps a picture was provided.
#### grgrsanjay
##### New member
But the difficulty there is these are garlands, circular bead arrangements.
$BBBBBBBBBRRR$ is the same as $RRRBBBBBBBBB$.
Moreover, $RBBRBBBRBBBB$ is the same as $RBBBBRBBRBBB$.
How many ways are there to partition $9$ into three of less summands?
For example here are three: $9$, $7+2$ and $5+3+1$.
So,there must be less than 14C2 solutions??How do i eliminate them?
Ok,could you check whether i can do it like this | {
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Ok,could you check whether i can do it like this
x + y + z = 12
Case 1:x=y=z
No.of.ways = (4,4,4) = 1
Case 2:two of x,y,z are equal
No.of ways = (1,1,10),(2,2,8),(3,3,6),(5,5,2),(6,6,0),(0,0,12) = 6
Case 3: x,y,z are all unequal
To avoid double counting , i will take x<y<z
No.of ways =
(0,1,11),(0,2,10),(0,3,9),(0,4,8),(0,5,7)
(1,2,9),(1,3,8),(1,4,7),(1,5,6)
(2,3,7),(2,4,6)
(3,4,5)
= 12 ways
So,total no.of ways is 19
Last edited:
#### Plato
##### Well-known member
MHB Math Helper
the definition of "garland" is ambiguous.
The definition may be ambiguous here but not in textbooks on countilng. Here is a link. The subtopic on beads is the one used in counting problems.
So,there must be less than 14C2 solutions??How do i eliminate them?
Ok,could you check whether i can do it like this
So,total no.of ways is 19
YES, 19 is correct
This sort of problem has set many arguments. If you agree on what meaning of garland we use then the following will show you why. If we have three green beads and three red beads how many different garlands are possible? Well the answer is 3. All the teds together; two reds together and one separate; then all the reds are separated. There are no other possibilities.
Three can be written as 3, 2+1, or 1+1+1.
The questions using two colors have that nice solution. They quickly become a nightmare with more than two colors.
Last edited:
#### grgrsanjay
##### New member
I think none of my case were repeated...could you conform it whether any case is left?
#### Plato
##### Well-known member
MHB Math Helper
I think none of my case were repeated...could you conform it whether any case is left?
Yes you are correct. At first I read it as if there were only 12 beads altogether. | {
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Parametric To Rectangular With Trig | {
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To get the cartesian equation you need to eliminate the parameter t to get an equation in x and y (explicitly and implicitly). Because the x- and y-values are defined separately in parametric equations, it is very easy to produce the inverse of a function written in parametric mode. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. Subtract 7 7 from both sides of the equation. to rectangular coordinates 5. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Write the equations of the circle in parametric form. x = 1 t - 2 3. Sometimes you may be asked to find a set of parametric equations from a rectangular (cartesian) formula. x = 7 sin and y = 2 cos 62/87,21 Solve the equations for sin and cos. The rectangular coordinates for P (5,20°) are P (4. The parametric equations are plotted in blue; the graph for the rectangular equation is drawn on top of the parametric in a dashed style colored red. Finding cartesian equation from parametric trigonometric equations. For this we need to find the vectors and. A parametric curve in the plane is a pair of functions x = f (t) y = g (t) It is possible to derive the Cartesian equation from the parametric equations. Now, first of all, what does it mean to convert to trigonometric form? Well, I have my number in rectangular form, so it's in a+bi form. x = -2 cos t, y = 2 sin t, 0 lessthanorequalto t lessthanorequalto 2 pi. Parametric form defines both the x-and the y-variables of conic sections in terms of a third, arbitrary variable, called the parameter, which is usually represented by t. Convert (3, -1) to polar form. Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in Figure 1. For a circle of diameter 10 that has been translated right by 4 units and down by 7 units: (a) Give the parametric equations defining the circle. In the first three examples, x and y are given in terms of different trig | {
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defining the circle. In the first three examples, x and y are given in terms of different trig functions. - 43 Questions with solutions. Parametric Graphing Team Desmos February 21, 2020 14:44. There's also a graph which shows you the meaning of what you've found. By adjusting the parametric equations, we can reverse the direction that the graph is swept. Polar plot on Cartesian axes // parametric conversion of ugly trig function? Hello! I currently have a polar plot which I would like to superimpose onto square axes. We can derive this identity by drawing a right triangle with leg lengths 1 and t and applying the usual definitions of trig and inverse trig functions. Rockwall ISD Pre-Calculus Parent Guide 3 Unit 7 Trigonometric Functions In this unit students will continue to apply trigonometric functions including rotation angles, the Unit Circle and periodic functions. Combine the triangle and the Cartesian coordinate grid to produce Trigonometry. Other Parent Functions C. Sketching the Graph of Trigonometric Parametric Equations. To eliminate t in trigonometric equations, you will need to use the standard trigonometric identities and double angle formulae. Replace and with the actual values. An alternative approach is two describe x and y separately in terms of a. Develop the calculus for polar and parametric forms. Drag the slider to change the number of sides on the polygon. The Rectangular Coordinate Systems And Graphs Algebra Trigonometry. All for only $14. Easy to get confused between these and inverse trig functions! Trig Proofs & Identities. Sketch the graph of the parametric equations $$x=t^2+t$$, $$y=t^2-t$$. Eliminating the Parameter. sss s ss s sss ss s s ss sss s s s sss s s s IV. Download [74. Sometimes you may be asked to find a set of parametric equations from a rectangular (cartesian) formula. Find more Mathematics widgets in Wolfram|Alpha. Worksheets are Polar and rectangular forms of equations date period, Polar coordinate exercises, Polar | {
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are Polar and rectangular forms of equations date period, Polar coordinate exercises, Polar coordinates, Trigonometry 03 notes marquez, , Polar coordinates parametric equations, Parametric equations context parametric and polar equations. Ron Larson + 1 other. Trigonometry (10th Edition) answers to Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8. We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. Area under one arc or loop of a parametric curve. I don't know what to do from here or if I'm going in the right direction or not. Thanks for contributing an answer to Mathematics Stack Exchange! Finding cartesian equation for trigonometric parametric forms. Simplify (x −7)2 ( x - 7) 2. Example 1: 3, 4 1, for -4 t 2xt y t dd 2 2 Example 2: 2, , ( , )xt y t fortin ff t x y 5/7/19 9. Fill in the table and sketch the parametric equation for t [-2,6] x = t2 + 1 y = 2 – t 5 6 Problems 2 – 10: Eliminate the parameter to write the parametric equations as a rectangular equation. Example 1 - Graphing Parametric Equations; Example 2 - Parametric to Rectangular Form; Day 2 - 7. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. Example 6) Finding parametric equations for a given function is easier. The ordered pairs, called polar coordinates, are in the form $$\left( {r,\theta } \right)$$, with $$r$$ being the number of units from the origin or pole (if $$r>0$$), like a radius of a circle, and $$\theta$$ being the angle (in degrees or radians) formed by the ray on the positive $$x$$ - axis (polar axis), going counter-clockwise. You can find values for both x and y by plugging values for t into the parametric equations. Applications. Converting between polar and rectangular form Converting equations between polar and rectangular form Homework: Finish Day 1 Packet (Optional) Pg. I was trying to solve for x for some reason. (b) Convert the parametric equations to rectangular form (please give the circle in standard form). | {
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(b) Convert the parametric equations to rectangular form (please give the circle in standard form). Complete pp. Polar and Parametric Equations Rev. Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in [link]. Rotating a Curve defined by a Equation. We can express equation (i) in terms of t, therefore we see that t = x + 3,. sss s ss s sss ss s s ss sss s s s sss s s s IV. You will also see how to transform the graph of y = sin(x) to obtain the graph of y = sin[B(x + C)] + D. How to represent Parametric Equations. Set up the parametric equation for x(t) x ( t) to solve the equation for t t. Parametric and Polar Equations Review Name 1. uTo graph parametric functions You can graph parametric functions that can be expressed in the following format. Parametric equations involving trigonometric functions Finding areas Parametric equations An equation like y = 5x + 1 or y x x 3sin 4cos or xy22 1 is called a cartesian equation. x=t-l, Check these with yesterday's graphs: B. 2 Plane Curves and Parametric Equations 711 Eliminating the Parameter Finding a rectangular equation that represents the graph of a set of parametric equations is called eliminating the parameter. Covers 55 algebra and trigonometry topics including synthetic division, conics, statistics, quadratics and more. Parametric Curves. Converting Polar Equations To Rectangular Equation. One caution when eliminating the parameter, the domain of the resulting rectangular equation may need to be adjusted to agree with the domain of the parameter as given in the parametric equations. Many of the advantages of parametric equations become obvious when applied to solving real-world problems. y for values of. Simplify (x −7)2 ( x - 7) 2. Complete Algebra 2 and Trig Program: Requirements: Requires the ti-83 plus or a ti-84 model. Trigonometry made completely easy! Our Trigonometry tutors got you covered with our complete trig help for all topics that you would | {
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Our Trigonometry tutors got you covered with our complete trig help for all topics that you would expect in any typical Trigonometry classes, whether it's Trigonometry Regents exam (EngageNY), ACT Trigonometry, or College Trigonometry. Polar coordinates simplified the work it takes to arrive at solutions in most trigonometry problems. Historic applications of parametric equations are discussed so the use of them is realized. Calculus of a Single Variable. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. PreCalculus Class Notes VP5 Converting Parametric and Rectangular Equations Review To convert from parametric equations to rectangular equation: solve the x equation for t, substitute into the y equation Example Rewrite 2 3 2 4 x t y t = − = as a function of x. Calculus and parametric curves. Graphing a plane curve represented by parametric equations involves plotting points in the rectangular coordinate system and connecting them with a smooth curve. The resulting equation is a rectangular equation. Credit Pre-Calculus Honors *All areas are accelerated* Generating Graphs Shifting, Symmetry, Reflections, and Stretching Roots, Max/Min, Values, and Intersections Composite Functions Programming the Tl-84 Polynomial Functions Brief Quadratic Review Roots Shifted and Standard Form Axis. A cartesian equation gives a direct relationship between x and y. That make sense now. Graphing a Parabola with vertex at (h ,k ). Find more Mathematics widgets in Wolfram|Alpha. Show the orientation (flow) by arrows (2) Convert to a rectangular equation and Converting from. Polar coordinates simplified the work it takes to arrive at solutions in most trigonometry problems. 142 Notes - Section 8. Then use a trigonometric identity. The functions and can serve as a parametric representation for a function , which is plotted in purple on the , plane cutting , shown in light gray. Subtract 7 7 from both sides of the equation. In the diagram such a circle is tangent to the hyperbola xy = | {
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7 7 from both sides of the equation. In the diagram such a circle is tangent to the hyperbola xy = 1 at (1,1). Example $$\PageIndex{7}$$: Eliminating the Parameter from a Pair of Trigonometric Parametric Equations. The equation is the general form of an ellipse that has a center at the origin, a horizontal major axis of length 14, and. The rectangular coordinates (x , y) and polar coordinates (R , t) are related as follows. The student is expected to: (A) graph a set of parametric equations;. y = x -3 is equivalent to 3 cos sin 3 (cos sin ) 3 3 cos sin xy rr r r TT TT TT. Which equation should be solved for the parametric variable depends on the problem -- whichever equation can be most easily solved for that parametric variable is typically the best choice. x=t-l, Check these with yesterday's graphs: B. Write the complex number in trigonometric form, using degree measure for the argument. In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x x and y. x = 7 sin and y = 2 cos 62/87,21 Solve the equations for sin and cos. Let us just look at a simple example. Clearly, both forms produce the same graph. Then, we do this substitution into the function: x → x c - y d y → x d + y c. Eliminate the parameter from the given pair of trigonometric equations where $$0≤t≤2\pi$$ and sketch the graph. For the first case we need to supplement the equations by two inequalities: 0 <= t <= 4 Pi && 0 < x < 4 Pi. Eliminate the parameter and find a corresponding rectangular equation. For instance, you can eliminate the parameter from the set of parametric equations in Example 1 as follows. Construct a table of values for the given parametric equations and sketch the graph: the data from the parametric equations and the rectangular equation are plotted together. Converting Polar To Rectangular. A cartesian equation gives a direct relationship between x and y. x = ½t + 4. x = cos 2t, y = | {
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A cartesian equation gives a direct relationship between x and y. x = ½t + 4. x = cos 2t, y = sin t, for t in 1-p, p2. In rectangular coordinates, each point (x, y) has a unique representation. ACE TRIG Final EXAM REVIEW. After going through these three problems can you reach any conclusions on how the argument of the trig functions will affect the parametric curves for this type of parametric equations?. Finding Parametric Equations from a Rectangular Equation (Note that I showed examples of how to do this via vectors in 3D space here in the Introduction to Vector Section). Solve trigonometric equations in quadratic form 12. now expanding (x+2)² either by using the identity of a perfect square which x^2 + 2bx + b^2 or simply expanding (x+2)(x+2) you have = x^2 + 4x + 4 and then multiplying by -1 as such because you have -(x+2)² from your original parametric equation -(t²) = -x^2 - 4x - 4. Students model linear situations with parametric equations, including modeling linear motion and vector situations. the domain of the rectangular equation so that its graph matches the graph of the parametric equations. We can express equation (i) in terms of t, therefore we see that t = x + 3,. Complete pp. Download [74. Parametric Equations and Vectors : Questions like write each pair of parametric equations in rectangular form. Example 1: 3, 4 1, for -4 t 2xt y t dd 2 2 Example 2: 2, , ( , )xt y t fortin ff t x y 5/7/19 9. Place the parametric equations in rectangular form. The applet below illustrates parametric coordinate functions for various polygonal trig functions. Definition of Trig Functions Trig Model for Data Graphing Sine and Cosine Functions (3) Relations and geometric reasoning. In the diagram such a circle is tangent to the hyperbola xy = 1 at (1,1). Rectangular - Polar - Parametric "Cheat Sheet" 15 October 2017 Rectangular Polar Parametric Point ( T)= U ( T, U) ( , ) • ( N ,𝜃) or N ∠ 𝜃 Point (a,b) in Rectangular: T( P)= U( P)= < , > P=3𝑟 𝑖 , Q O P𝑖 , with 1 | {
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U) ( , ) • ( N ,𝜃) or N ∠ 𝜃 Point (a,b) in Rectangular: T( P)= U( P)= < , > P=3𝑟 𝑖 , Q O P𝑖 , with 1 degree of freedom (df) Polar Rect. The coordinates are measured in meters. Then find a second set of polar coordinates for the. 7: Complex Numbers, Polar Coordinates, Parametric equationsFall 2014 2 / 17 Complex Numbers - trig form Example (Write the complex number in standard form). Parametric equations are equations that are used to introduce polar coordinates and their relation with rectangular coordinates. Many of the advantages of parametric equations become obvious when applied to solving real-world problems. We will graph polar equations in the polar coordinate system and finally discuss parametric equations and their graphs. Things to try. x = cos 2t, y = sin t, for t in 1-p, p2. Find more Mathematics widgets in Wolfram|Alpha. sssssss ssss sssss Homework Assignment Page(s) Exercises. You will also see how to transform the graph of y = sin(x) to obtain the graph of y = sin[B(x + C)] + D. Linear 64) 2. Trigonometry made completely easy! Our Trigonometry tutors got you covered with our complete trig help for all topics that you would expect in any typical Trigonometry classes, whether it's Trigonometry Regents exam (EngageNY), ACT Trigonometry, or College Trigonometry. Algebra Review: Completing the Square. Get the free "parametric to cartesian" widget for your website, blog, Wordpress, Blogger, or iGoogle. Example $$\PageIndex{7}$$: Eliminating the Parameter from a Pair of Trigonometric Parametric Equations. Comments There are no comments. In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x x and y. However, they used meters instead of feet for gravitational constant. To this point (in both Calculus I and Calculus II) we’ve looked almost exclusively at functions in the form $$y = f\left( x \right)$$ or $$x = h\left( y \right)$$ and almost all of the formulas | {
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in the form $$y = f\left( x \right)$$ or $$x = h\left( y \right)$$ and almost all of the formulas that we’ve developed require that functions be in one of these two forms. An object travels at a steady rate along a straight path $$(−5, 3)$$ to $$(3, −1)$$ in the same plane in four seconds. Write each pair of parametric equations in rectangular form. Find a set of parametric equations to represent the graph of y = (x – 1)2 given the. Search this site. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. asked by Sam on December 9, 2013; geometry. Example 6) Finding parametric equations for a given function is easier. Arc length of a parametric curves. A PLANE CURVE is whereas and are continuous functions on t on an interval and the set of ordered pairs. Therefore, we need to use a change of variables so we can integrate using either cylindrical (polar) or spherical coordinates, or even parametric form. Find more Mathematics widgets in Wolfram|Alpha. How do you convert #r=2sin(3theta)# to rectangular form? Trigonometry The Polar System Converting Between Systems. 2 Plane Curves and Parametric Equations 711 Eliminating the Parameter Finding a rectangular equation that represents the graph of a set of parametric equations is called eliminating the parameter. Pre - Calc. Introduction to Parametric Equations; Parametric Equations in the Graphing Calculator; Converting Parametric Equations to Rectangular: Eliminating the Parameter; Finding Parametric Equations from a Rectangular Equation; Simultaneous Solutions; Applications of Parametric Equations; Projectile Motion Applications; Parametric Form of the Equation of a Line in Space. Graphing a Hyperbola with center at (0 ,0 ). (1) $$f(x,y)=0$$ These are sometimes referred to as rectangular equations or Cartesian equations. Find a rectangular equation for each plane curve with the given parametric equations. The parametric equations are plotted in blue; the. I have the parametric coordinates x=Sec(t) and y=Tan 2 (t) where 0<=t P=3𝑟 N𝑖 , Q O P𝑖 , | {
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blue; the. I have the parametric coordinates x=Sec(t) and y=Tan 2 (t) where 0<=t P=3𝑟 N𝑖 , Q O P𝑖 , with 1 degree of freedom (df) Polar Rect. I'm trying to find the cartesian equation of the curve which is defined parametrically by: $$x = 2\sin\theta, y = \cos^2\theta$$ Both approaches I take result in the same answer:$$y = 1 - \s. Step-by-Step Examples. what is the volume of that rectangular pyramid? asked by riza on August 6, 2012. Angle t is in the range [0 , 2Pi) or [0 , 360 degrees). We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. The rectangular coordinates for P (5,20°) are P (4. x = 7 sin and y = 2 cos 62/87,21 Solve the equations for sin and cos. Parametric and Polar Equations Review Name q) 1. Parametric equation of an ellipse and a hyperbola rectangular hyperbola cartesian and parametric forms examsolutions maths tutorials parametric equations hyperbola hw 1 conic sections in polar parametric forms lesson Parametric Equation Of An Ellipse And A Hyperbola Rectangular Hyperbola Cartesian And Parametric Forms Examsolutions Maths Tutorials Parametric Equations Hyperbola Hw 1 Conic. In , the data from the parametric equations and the rectangular equation are plotted together. Historic applications of parametric equations are discussed so the use of them is realized. SOLUTION The graph of the parametric equations is given in Figure 9. To eliminate the parameter in equations involving trigonometric functions, try using the identities. Textbook Authors: Lial, Margaret L. This is also a great Review for AP Calculus BC. When you first learned parametrics, you probably used t as your parametric variable. Introduction to Parametric Equations; Parametric Equations in the Graphing Calculator; Converting Parametric Equations to Rectangular: Eliminating the Parameter; Finding Parametric Equations from a Rectangular Equation; Simultaneous Solutions; Applications of Parametric Equations; Projectile Motion Applications; | {
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Simultaneous Solutions; Applications of Parametric Equations; Projectile Motion Applications; Parametric Form of the Equation of a Line in Space. A summary of Graphing in Polar Coordinates in 's Parametric Equations and Polar Coordinates. Simply let t x and then replace your y with t. It can handle horizontal and vertical tangent lines as well. One caution when eliminating the parameter, the domain of the resulting rectangular equation may need to be adjusted to agree with the domain of the parameter as given in the parametric equations. 81 KB] Parametric Differentiation : The parametric definition of a curve, differentiation of a function defined parametrically, exercises, …. 3+3𝑖 3√2(cos45°+𝑖sin45°) Write the complex number in the form + 𝒊. Textbook Authors: Lial, Margaret L. This problem is about converting parametric equations to rectangular form. Sketch and identify graphs using parametric equations. This is called a parameter and is usually given the letter t or θ. Find a rectangular equation for each plane curve with the given parametric equations. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 9. The hyperbolic trigonometric functions extend the notion of the parametric equations for a unit circle (x = cos t (x = \cos t (x = cos t and y = sin t) y = \sin t) y = sin t) to the parametric equations for a hyperbola, which yield the following two fundamental hyperbolic equations:. In the entry line, type cos(t) All of the trigonometric functions and the inverse trigonometric functions can be found in the trigonometry section of the catalog. 4 De Moivre's Theorem; Powers and Roots of Complex Numbers 8. For the problems above, let x = t + 2 and find the resulting parametric equations. ACE TRIG Final EXAM REVIEW. Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in [link]. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. Algebra Review: Completing the Square. For instance, you can | {
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[link]. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. Algebra Review: Completing the Square. For instance, you can eliminate the parameter from the set of parametric equations in Example 1 as follows. Plot the points. Sketching the Graph of Trigonometric Parametric Equations. We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. A parametric equation is where the x and y coordinates are both written in terms of another letter. Plot the following points on the polar grid at the right and label each point. Find all solutions to a trigonometric equation 10. Find new parametric equations that shift this graph to the right 3 places and down 2. The coordinates are measured in meters. Polar and Parametric Equations 3. The volume of. 2;5p 3 Give two alternate sets of coordinates for each point. 6 Plane Curves, Parametric Equations. Many of the advantages of parametric equations become obvious when applied to solving real-world problems. But by recognizing the trig identity, we were able to simplify it to an ellipse, draw the ellipse. PreCalculus Class Notes VP5 Converting Parametric and Rectangular Equations Review To convert from parametric equations to rectangular equation: solve the x equation for t, substitute into the y equation Example Rewrite 2 3 2 4 x t y t = − = as a function of x. the process of solving one parametric equation for t so you may substitute that equation into the other parametric equation for t to create a rectangular equation where y=f(x) Example: if x=t-4 and y=¼t solve the first equation for t so t=x-4 then by substitution y=¼(x-4) or y=¼x-1. 95 per month. Thus, keep only the other equation. x: 4 sin (2t) Y : 2 cos (2t) X : 4+2 COS t. Find more Mathematics widgets in Wolfram|Alpha. Change the parametric equations to rectangular form eliminating the variable t and sketch 11. Graph polar equations. x = -2 cos t, y = 2 sin t, 0 lessthanorequalto t lessthanorequalto 2 pi. x = 3t – 1, y = 2t + 1. 4 Trigonometric | {
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y = 2 sin t, 0 lessthanorequalto t lessthanorequalto 2 pi. x = 3t – 1, y = 2t + 1. 4 Trigonometric Functions of Any Angle 497 1-3 4. Find a Cartesian equation for the curve traced out by this function. Trig functions, complex numbers, identities, vectors, and real life applications like circuits are here to energize your classroom. Sketch and identify graphs in polar coordinates. First, we find a vector {c,d} of distance 1 having angle -θ, which is {Cos[-θ], Sin[-θ]}. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 7. Definition of Trig Functions Trig Model for Data Graphing Sine and Cosine Functions (3) Relations and geometric reasoning. Tools We Need x = r * cos θ y = r * sin θ (some trigonometric identities are required) y = r * sin t y = 2 * cos t * sin(cos⁻¹(r/2)). Write the equations of the circle in parametric form. Parametric Equations and Vectors : Questions like write each pair of parametric equations in rectangular form. Parametric and Polar Equations Review Name 1. Challenge Sets. A coordinate system is a scheme that allows us to identify any point in the plane or in three-dimensional space by a set of numbers. Change the parametric equations to rectangular form eliminating the variable t and sketch 11. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate. Graphing parametric equations: The key is to plug in useful points within the specified range of t, not just any points. Show all algebraic support. The Second Fundamental Theorem of Calculus Integration Involving Powers of Trigonometric Functions. This is the a value, this is the b value. A parametric curve in the plane is a pair of functions x = f (t) y = g (t) It is possible to derive the Cartesian equation from the parametric equations. x = cos 2t, y = sin t, for t in 1-p, p2. In the entry line, type cos(t) All of the trigonometric functions and the inverse trigonometric functions can be found | {
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type cos(t) All of the trigonometric functions and the inverse trigonometric functions can be found in the trigonometry section of the catalog. The volume of. The parametric equations are plotted in blue; the graph for the rectangular equation is drawn on top of the parametric in a dashed style colored red. Historic applications of parametric equations are discussed so the use of them is realized. Thank you for purchasing this product! This activity is suitable for PreCalculus - Trigonometry and can be used as a The introduction to concept of Parametric Equations can be difficult for students and yet this topic is so important, especially later in Calculus. Convert the polar coordinates (5 , 2. Parametric Equations: Eliminating Angle Parameters In a parametric equation the parameter can represent anything including an angle. Plot the resulting pairs ( x,y ). Based on your work on the above problems, name another benefit of parametric equations versus rectangular functions. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 9. Example $$\PageIndex{7}$$: Eliminating the Parameter from a Pair of Trigonometric Parametric Equations. Write the equations of the circle in parametric form. Use a table of values to sketch a Parametric Curve and indicated direction of motion. In parametric equations x and y are both defined in terms of a third variable. 1 Answer A. ; Hornsby, John; Schneider, David I. Polar coordinates. 244 Chapter 10 Polar Coordinates, Parametric Equations conclude that the tangent line is vertical. 7-8: 2-8 even, 9-12 all, 14-32 even. In the past, we have seen curves in two dimensions described as a statement of equality involving x and y. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. Things to try. Xmin = -20 Ymin = -12 Xmax = 20 Ymax = 12 Xscl = 5 Yscl = 5. What type of path does the rocket follow? Solution: The path of the rocket is defined by the parametric equations x = (64 cos 30°)t and y = (64 sin 30°)t − 16t2 + 3. Convert to | {
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defined by the parametric equations x = (64 cos 30°)t and y = (64 sin 30°)t − 16t2 + 3. Convert to Polar Coordinates. We will graph polar equations in the polar coordinate system and finally discuss parametric equations and their graphs. Both types depend on an argument, either circular angle or hyperbolic angle. Calculus of a Single Variable. Converting from rectangular to parametric can be complicated, and requires some creativity. Tools We Need x = r * cos θ y = r * sin θ (some trigonometric identities are required) y = r * sin t y = 2 * cos t * sin(cos⁻¹(r/2)). Find a rectangular equation for each plane curve with the given parametric equations. Usually will stand for time. Eliminating the parameter is a method that may make graphing some curves easier. A common parameter used is time (t) or an angle (trig) (x, y) is the place, "t" is the time it is there (at that place) (at that place) Graphing Parametric Equations 2 OPTIONS: (l) Use a chart to find rectangular points. Recall the trig identity d1 Substitute x/r and y/r into the identity: Remove the parentheses: Multiply through by r 2. The curves are colored based on the quadrants. Trig Ratios and Quadrants (9:52) Inverses of Trig Functions and Inverse Trig Functions (20:18) Graphing Sine and Cosine--Amplitude and Period (11:15) Graphing Sine and Cosine--Vertical and Horizontal Shift (6:33) Graphing Sine and Cosine by Hand (29:10) Graphing Tangent by Hand (24:39) Applications (15:50) Unit 4: Triangle Applications of Trig. The parametric equations are plotted in blue; the. A curve is given by the parametric equations: #x=cos(t) , y=sin(2t)#, how do you find the cartesian equation? Calculus Parametric Functions Introduction to Parametric Equations. Homework Statement Reduce these parametric functions to a single cartesian equation:$\displaylines{ x = at^2 \cr y = 2at \cr} \$. Drag points A and B to change the size and orientation of the polygon. now the rest is a matter of simplification:. x-2+t y-2-t, for t in | {
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and orientation of the polygon. now the rest is a matter of simplification:. x-2+t y-2-t, for t in [-2,3] Equation in Rectangular form: (8 points each) b. Graphing a Parabola with vertex at (h ,k ). And then by plotting a couple of points, we were able to figure out the direction at which, if this was describing a particle in motion, the direction in which that particle was actually moving. In parametric equations x and y are both defined in terms of a third variable. It can handle horizontal and vertical tangent lines as well. We show how we can transform between these representations of the same plane. 81 KB] Parametric Differentiation : The parametric definition of a curve, differentiation of a function defined parametrically, exercises, …. Plot the resulting pairs ( x,y ). (θ is normally used when the parameter is an angle, and is measured from the positive x-axis. But how do we write and solve the equation for the position of the moon when the distance from the planet,. In the past, we have been working with rectangular equations, that is equations involving only x and y so that they could be graphed on the Cartesian (rectangular) coordinate system. This calculator converts between polar and rectangular coordinates. Use power reducing formulas 9. now the rest is a matter of simplification:. In the entry line, type cos(t) All of the trigonometric functions and the inverse trigonometric functions can be found in the trigonometry section of the catalog. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. Find exact values of composite functions with inverse trig functions 8. We're converting from rectangular form to trigonometric form and we're starting with the complex number z equals negative root 2 plus i times root 2. Trigonometric substitution. Search this site. I'm not looking for the answer here. 1 Answer Cesareo R. 1 Complex Numbers 8. Parametric curves have a direction of. x =3cosθ and y =4sinθ, 02≤θ≤ π by eliminating the parameter finding the | {
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curves have a direction of. x =3cosθ and y =4sinθ, 02≤θ≤ π by eliminating the parameter finding the corresponding rectangular equation. Write the equations of the circle in parametric form. Thank you for purchasing this product! This activity is suitable for PreCalculus - Trigonometry and can be used as a The introduction to concept of Parametric Equations can be difficult for students and yet this topic is so important, especially later in Calculus. Use a table of values to sketch a Parametric Curve and indicated direction of motion. Graphing a Parabola with vertex at (h ,k ). Convert to Polar Coordinates. Let's look at each solution carefully: The first solution is correct since: r^2 = x^2 + y^2. Subtract 7 7 from both sides of the equation. position time parametric equations path rectangular equation eliminating the parameter square root function direction of motion. now the rest is a matter of simplification:. orgChapter 1. We will then introduce the polar coordinate system, which is often a preferred coordinate system over the rectangular system. Graphing parametric equations: The key is to plug in useful points within the specified range of t, not just any points. Applications. Sketch and identify graphs using parametric equations. Applications of Parametric Equations. This is the a value, this is the b value. interesting variations on the parametric equations. Linear 64) 2. Two versions. Which equation should be solved for the parametric variable depends on the problem -- whichever equation can be most easily solved for that parametric variable is typically the best choice. 5 Parametric Equations part 2. Find the rectangular equation that models its path. Algebra Pre-Calculus Geometry Trigonometry Calculus Advanced Algebra Discrete Math Differential Geometry Differential Equations Number Theory Statistics & Probability Business Math Challenge Problems Math Software. Finding Parametric Equations In Exercises 35 and 36, find two different sets of parametric | {
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Software. Finding Parametric Equations In Exercises 35 and 36, find two different sets of parametric equations for the rectangular equation. Graphs of trigonometric functions in polar coordinates are very distinctive. Lots of vocab in this one -- specific solutions, general solutions, 2npi, 360n -- and lots of factoring to do too. Graph Type⎮ Parametric. Identify the type of graph. Making statements based on opinion; back them up with references or personal experience. A summary of Graphing in Polar Coordinates in 's Parametric Equations and Polar Coordinates. 2 Exercises - Page 365 36 including work step by step written by community members like you. Because the x- and y-values are defined separately in parametric equations, it is very easy to produce the inverse of a function written in parametric mode. Finding cartesian equation from parametric trigonometric equations. This calculator converts between polar and rectangular coordinates. For instance had the problem been y = t -3, and x = t^2 + 5, I hope you see that solving for t in terms of y would make more sense, for exactly the same. Application: Toy Rocket The parametric equations determined by the toy rocket are Substitute from Equation 1 into equation 2: A Parabolic Path 8. Ask Question Asked 5 years, 7 months ago. Sketch and identify graphs in polar coordinates. Find a rectangular equation for each plane curve with the given parametric equations. Find more Mathematics widgets in Wolfram|Alpha. Algebra Pre-Calculus Geometry Trigonometry Calculus Advanced Algebra Discrete Math Differential Geometry Differential Equations Number Theory Statistics & Probability Business Math Challenge Problems Math Software. An alternative approach is two describe x and y separately in terms of a. 1 shows points corresponding to θ equal to 0, ±π/3, 2π/3 and 4π/3 on the graph of the function. Lines in polar coordinates: Let and then the polar equations of the lines x=a and y=b are and for all values of Parametric form of | {
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and then the polar equations of the lines x=a and y=b are and for all values of Parametric form of derivatives:. Then, we do this substitution into the function: x → x c - y d y → x d + y c. The hyperbolic trigonometric functions extend the notion of the parametric equations for a unit circle (x = cos t (x = \cos t (x = cos t and y = sin t) y = \sin t) y = sin t) to the parametric equations for a hyperbola, which yield the following two fundamental hyperbolic equations:. Find parametric equations for 2x 4 a) y = x 2 —2x+3 Note that there are many ways of finding parametric equations for a given function. 2: Trigonometric Functions In this lesson you will use parametric equations to illustrate the connection between the graphs of y = sin(x) and the unit circle. Example 3: Transform the equation x 2 + y 2 + 5x = 0 to polar coordinate form. Just as with non-angle parameters, when the parameter is an angle θ the plane curve can be graphed by selecting values for the angle and calculating the x- and y-values. Covers 55 algebra and trigonometry topics including synthetic division, conics, statistics, quadratics and more. This is also a great Review for AP Calculus BC. Jimmy wants to rewrite the set of parametric equations x = 1/2 T + 3 and y = 2T - 1 in rectangular form by eliminating T. Rewrite the equation as t+7 = x t + 7 = x. now expanding (x+2)² either by using the identity of a perfect square which x^2 + 2bx + b^2 or simply expanding (x+2)(x+2) you have = x^2 + 4x + 4 and then multiplying by -1 as such because you have -(x+2)² from your original parametric equation -(t²) = -x^2 - 4x - 4. Finding all arguments t in 0 <= t <= 4Pi where the parametric graph intersects. You will also see how to transform the graph of y = sin(x) to obtain the graph of y = sin[B(x + C)] + D. Pre - Calc. Find the rectangular coordinates of. The Polar coordinates are in the form (r,q). Then x=f(t) and y=g(t) are called parametric equations for the curve represented by (x,y). Equations | {
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Then x=f(t) and y=g(t) are called parametric equations for the curve represented by (x,y). Equations for and are plotted on the perpendicular and planes as varies from to. Graph polar equations. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. Easy to get confused between these and inverse trig functions! Trig Proofs & Identities. Because r is a directed distance the coordinates (r, θ) and (-r, θ + π). Finding cartesian equation from parametric trigonometric equations. However, they used meters instead of feet for gravitational constant. I don't know what to do from here or if I'm going in the right direction or not. The parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. Homework Statement I have this equation and i need to find the cartesian equation, so i apreciate your help Homework Equations X=cost ' y=2sin2t The Attempt at a Solution I am usign this [/B] Sin2t=2costsint So x+y/2=cost+2costsint But i dont know what to do after, I also try to solve that. Sal gives an example of a situation where parametric equations are very useful: driving off a cliff! Sal gives an example of a situation where parametric equations are very useful: driving off a cliff! If you're seeing this message, it means we're having trouble loading external resources on our website. Calculus of a Single Variable. Rotating a Curve defined by a Equation. The students will analyze and graph Sine, Cosine and Tangent and their inverses. 2sec t and y = -0. It is a parabola with a axis of symmetry along the line $$y=x$$; the vertex is at $$(0,0)$$. We're converting from rectangular form to trigonometric form and we're starting with the complex number z equals negative root 2 plus i times root 2. This is also a great Review for AP Calculus BC. Trig equations are problems where you're solving for X or Theta but they're hidden behind a trig function, like "2sinX-1=0" or "tan 2 X-1=0". Thanks for contributing an answer to Mathematics | {
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a trig function, like "2sinX-1=0" or "tan 2 X-1=0". Thanks for contributing an answer to Mathematics Stack Exchange! Finding cartesian equation for trigonometric parametric forms. Definition of Trig Functions Trig Model for Data Graphing Sine and Cosine Functions (3) Relations and geometric reasoning. Applications of Parametric Equations. 2: Trigonometric Functions In this lesson you will use parametric equations to illustrate the connection between the graphs of y = sin(x) and the unit circle. Thus, keep only the other equation. Algebra Review: Completing the Square. Trigonometric, Parametric, and Polar Graphs OBJECTIVES When you have completed this chapter you should be able to Graph the sine wave, by calculator or manually. What type of path does the rocket follow? Solution: The path of the rocket is defined by the parametric equations x = (64 cos 30°)t and y = (64 sin 30°)t − 16t2 + 3. Two versions. hyperbolic trig identities are similar to the regular trig onesbut beware hyperbolic trig functions have seemingly little in common with regular trig functions so there's no. Finding cartesian equation from parametric trigonometric equations. Making statements based on opinion; back them up with references or personal experience. This is the parameter or a number that affects the behavior of the equation. For each graph you create, identify the specific parametric equations used and the domain for your graph. express the equation in terms of x and/or y. We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. How do I convert from parametric equations to rectangular form? Write the rectangular form for the parametric equation x=cosθ ; y=4sinθ Trig, and Differential. Because the x- and y-values are defined separately in parametric equations, it is very easy to produce the inverse of a function written in parametric mode. Homework Statement I have this equation and i need to find the cartesian equation, so i apreciate your | {
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Statement I have this equation and i need to find the cartesian equation, so i apreciate your help Homework Equations X=cost ' y=2sin2t The Attempt at a Solution I am usign this [/B] Sin2t=2costsint So x+y/2=cost+2costsint But i dont know what to do after, I also try to solve that. Applications. The Organic Chemistry Tutor 247,826 views 33:29. x-2+t y-2-t, for t in [-2,3] Equation in Rectangular form: (8 points each) b. Solve trigonometric equations in quadratic form 12. Basic graphing with direction to them. Videos, examples, solutions, activities and worksheets for studying, practice and review of precalculus, Lines and Planes, Functions and Transformation of Graphs, Polynomials, Rational Functions, Limits of a Function, Complex Numbers, Exponential Functions, Logarithmic Functions, Conic Sections, Matrices, Sequences and Series, Probability and Combinatorics, Advanced Trigonometry, Vectors and. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. Comprehensive End of Unit Review for the Polar, Parametric, and Vectors Sections of PreCalculus or Trigonometry plus Graphic Organizer. Plot the points. 1 Answer A. Trigonometry (10th Edition) answers to Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8. Covers 55 algebra and trigonometry topics including synthetic division, conics, statistics, quadratics and more. Get the free "parametric to cartesian" widget for your website, blog, Wordpress, Blogger, or iGoogle. Fill in the table and sketch the parametric equation for t [-2/6] Problems 2 - 11: Eliminate the parameter to write the parametric equations as a rectangular equation. 933 13 (b) C-/cos(£)) 10. The previous section discussed a special class of parametric functions called polar functions. Example 1 - Graphing Parametric Equations; Example 2 - Parametric to Rectangular Form; Day 2 - 7. Algebra Review: Completing the Square. At any moment, the moon is located at a particular spot relative to the planet. Then, convert each polar coordinate to | {
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moon is located at a particular spot relative to the planet. Then, convert each polar coordinate to rectangular ntoS form (RF). For instance had the problem been y = t -3, and x = t^2 + 5, I hope you see that solving for t in terms of y would make more sense, for exactly the same. uTo graph parametric functions You can graph parametric functions that can be expressed in the following format. 2 Exercises - Page 365 36 including work step by step written by community members like you. 4 De Moivre's Theorem; Powers and Roots of Complex Numbers 8. Trigonometry (10th Edition) answers to Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8. Textbook Authors: Lial, Margaret L. 8/14/2018 12:12 AM §10. Pre-calculus Contents C Parametric Equations Parametric and rectangular forms of equations conversions The parametric equations of a quadratic polynomial, parabola The Trigonometric functions of arcs from 0 to. Parametric Equations: Eliminating Angle Parameters In a parametric equation the parameter can represent anything including an angle. Usually will stand for time. y for values of. 2;5p 3 Give two alternate sets of coordinates for each point. ) Drawing the graphTo draw a parametric graph it is easiest to make a table and then plot the points:Example 1 Plot the graph of the. Lines in polar coordinates: Let and then the polar equations of the lines x=a and y=b are and for all values of Parametric form of derivatives:. Drag P and C to make a new circle at a new center location. 6 Plane Curves, Parametric Equations. parameter t = x – 1. Math Algebra 2 Calculus Precalculus Trigonometry Math Help Complex Numbers Pre Calculus Polar Forms Ellipse High School: Math Derivatives Polar Coordinates Sine Convert Calculus 3 Rectangular Calculus 2 Calculus 1 Polar Equation. Find parametric equations whose graph is an ellipse with center (h,k), horizontal axis length 2a, and. (a) Eliminate the parameter for the curve given by the parametric equations x= 2 t, | {
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length 2a, and. (a) Eliminate the parameter for the curve given by the parametric equations x= 2 t, y= 1 t2 for 1 P=3𝑟 N𝑖 , Q O P𝑖 , with 1 degree of freedom (df) Polar Rect. y = x 2 -2. Graphing parametric equations: The key is to plug in useful points within the specified range of t, not just any points. Trig Functions sine I First, solve for 3. Plot the resulting pairs ( x,y ). It can handle horizontal and vertical tangent lines as well. The Second Fundamental Theorem of Calculus. This problem is about converting parametric equations to rectangular form. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 9. Let A be the point where the segment OB intersects the circle, where point B lies on the line x = 2 a. Rockwall ISD Pre-Calculus Parent Guide 3 Unit 7 Trigonometric Functions In this unit students will continue to apply trigonometric functions including rotation angles, the Unit Circle and periodic functions. A PLANE CURVE is whereas and are continuous functions on t on an interval and the set of ordered pairs. Other Parent Functions C. For the point and So, the rectangular coordinates are See Figure 10. Comments There are no comments. ACE TRIG Final EXAM REVIEW. Calculus Examples. (a) Find a polar equation of the cissoid. Write each pair of parametric equations in rectangular form. There's also a graph which shows you the meaning of what you've found. And Vector Calculus, which tends to be the last chapter for Calculus 3, deals with work on, in, and around a surface which will predominately involve polar coordinates as well. Parametric Equations and Motion Precalculus Vectors and Parametric Equations. I was trying to solve for x for some reason. Use MathJax to format equations. y = x 2 -2. t over the interval for which the functions are defined. Given a point in polar coordinates, rectangular coordinates are given by Given a point in rectangular coordinates,. Trigonometry (MindTap Course List). In many cases, we may have a | {
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point in rectangular coordinates,. Trigonometry (MindTap Course List). In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x x and y. Exchanging x and y 4. The hyperbolic functions represent an expansion of trigonometry beyond the circular functions. For instance, you can eliminate the parameter from the set of parametric equations in Example 1 as follows. To eliminate the parameter in equations involving trigonometric functions, try using the identities. Both types depend on an argument, either circular angle or hyperbolic angle. To this point (in both Calculus I and Calculus II) we’ve looked almost exclusively at functions in the form $$y = f\left( x \right)$$ or $$x = h\left( y \right)$$ and almost all of the formulas that we’ve developed require that functions be in one of these two forms. Finding Parametric Equations for a Graph (Page 799) Describe how to find a set of parametric equations for a given graph. 95 per month. Write each pair of parametric equations in rectangular form. Trigonometric substitution. Example for parametric equations are {eq}\displaystyle x=\sqrt{t},\:y=t-5 {/eq}, what we need to do first is to find x. I'm not looking for the answer here. (a) y = x -3 This is the equation of a line. Finding Parametric Equations from a Rectangular Equation (Note that I showed examples of how to do this via vectors in 3D space here in the Introduction to Vector Section). Converting between polar and rectangular form Converting equations between polar and rectangular form Homework: Finish Day 1 Packet (Optional) Pg. So our parametric equation is x = t and y = 4t. | {
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# Proof of infinitely many prime numbers
Here's the proof from the book I'm reading that proves there are infinitely many primes:
We want to show that it is not the case that there only finitely many primes. Suppose there are finitely many primes. We shall show that this assumption leads to a contradiction. Let $p_1, p_2,...,p_n$ be all the primes there are. Let $x = p_1...p_n$ be their product and let $y=x+1$. Then $y \in \mathbb{N}$ and $y \neq 1$, so there is a prime $q$ such that $q \mid y$. Now $q$ must be one of $p_1,...,p_n$ since these are all primes that there are. Hence $q \mid x$. Since $q \mid y$ and $q \mid x$, $q \mid (y-x)$. But $y-x=1$. Thus $q \mid 1$. But since $q$ is prime, $q \geq 2$. Hence $q$ does not divide 1. Thus we have reached a contradiction. Hence our assumption that there are only finitely many primes must be wrong. Therefore there must be infinitely many primes.
I have a couple of questions/comments regarding this proof. I will use a simple example to help illustrate my questions:
Suppose only 6 primes exist: $2, 3, 5, 7, 11, 13$
Let $x = p_1p_2p_3p_4p_5p_6=30,030$
Let $y = x + 1 = 30,030+1 = 30,031$
1. The proof states there is a prime $q$ such that $q \mid y$ and that $q$ must be either $p_1, p_2, p_3, p_4, p_5,$ or $p_6$. However, none of the 6 primes listed, $(2,3,5,7,11,13)$, divides $30,031$. In fact, the only divisors for $30,031$ are $1, 59, 509$ and $30,031$. Doesn't the proof then break down here since there is no prime $q$ that divides $y$?
2. The prime factorization of $30,031$ is $59 \times 509$. These two numbers are factors of $30,031$ and are in fact primes themselves since they are only divisible by themselves or by 1. Have I shown that there exists $\gt 6$ primes? If so, what can I conclude now that I have shown this? | {
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3. I don't understand why the contradiction $q$ divides $1$ and $q$ does not divide $1$ lead us to the assumption that the finitely many primes must be wrong. I understand how we reached the contradiction. I don't understand why contradiction leads us to the conclusion shows that are assumption that there is only finitely many primes is wrong.
My apologies for the long post. Thanks for any and all help.
• unique factorization requires a unique prime decomposition. If no $p$ exists, there must be a new prime... – Eleven-Eleven Oct 8 '16 at 23:50
• The point of the argument is that there must be some prime not in the list that divides the number you produce. Your example confirms this! You found two such primes $59,509$ neither of which is on the list. Thus, your list can't have been complete. – lulu Oct 8 '16 at 23:51
• The proof makes an assumption that there are finitely many primes, But it then goes on to show, given the conditions, this actually can't be the case. Therefore, the flaw is in your assumption, since all arguments in the proof are mathematically valid. – Eleven-Eleven Oct 8 '16 at 23:53
• As stated above, all of your mathematical arguments are correct when you started from your original assumption. Yet you are led to a point in the proof that is impossible. If the mathematical arguments are sound and correct, what is left for there to be wrong...? The original assumption must therefore be incorrect. – Eleven-Eleven Oct 8 '16 at 23:55
• 2) In this case that there are at least 7 primes. But if you generalize this, that p1p2...pn+1 is never divisible by p1,...,pn then p1,.....pn can not be a list of all primes. No such list can exist. So there can not be a finite number of primes. – fleablood Oct 9 '16 at 0:33 | {
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This is an excellent example of a proof that is traditionally phrased as a proof by contradiction but is much better understood constructively. From a constructive viewpoint, the proof shows that given any list of primes $p_1, \ldots, p_n$ there is a prime $q$ (any prime divisor of $p_1p_2\ldots p_n + 1$) that is distinct from each $p_i$. So given any finite set of primes we can find a prime that is not in that set.
• +1 for emphasizing the constructive aspect of this proof. – P Vanchinathan Oct 9 '16 at 0:44
• Except that this process doesn't actually construct a new prime (the example $y = 30031$ being composite). – Daniel R. Collins Oct 9 '16 at 2:20
• @Rob: I think Daniel’s point is that the proof constructs an integer $y>1$ that is not divisible by $p_1$, $p_2$, …, $p_n$, and then it asserts that there must be a prime $q$ such that $q \mid y$, but it doesn’t construct $q$. – Scott Oct 9 '16 at 4:10
• @Scott But there's a simple algorithm to generate a prime divisor of any given integer. – Jack M Oct 9 '16 at 8:05
• @djechlin This is a constructive proof and you're utterly wrong. The set of divisors of a positive integer is finite. Now just take the minimal element greater than $1$ in the set of divisors; this is obviously prime. Would you say that finding the minimal element in a finite set of integers is not constructive? By the way, the proof by contradiction seems to imply that $p_1p_2\dots p_n+1$ is prime, which is the main source for confusion in newbies. – egreg Oct 9 '16 at 11:20
1. The proof does break down in a sense. You have reached a contradiction, which means the hypothesis that there are finitely many primes can not possibly be true. | {
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2. So you thought you had 6 primes, you found 2 extra. Can you just add these two to your list and get all of the primes? If you repeat the process on these 8 primes, you will find that you will have even MORE primes by considering the product + 1. You can keep going and going and you will never run out of primes. This is the idea of the proof. It uses contradiction because you can do it all in one step and avoid the potential issue of doing the process infinitely many times.
Suppose only 6 primes exist: 2,3,5,7,11,13
However, none of the 6 primes listed, (2,3,5,7,11,13), divides 30,031.
Then we already have a contradiction. Since there are only 6 primes (we supposed that at the beginning) and none of them divide 30,031, then 30,031 must be prime. However, 30,031 is not one of the only 6 primes that exist. So 30,031 cannot be prime, yet it must be prime.
So the proof works. In fact, it works precisely the same regardless of what set of numbers we suppose are the only primes that exist. Thus no finite set of numbers can include all the primes that exist. Thus there are an infinite number of primes.
• I think you mean "... no finite set of numbers ..." – origimbo Oct 10 '16 at 10:48
You ask in #1 if the proof breaks down. No. What breaks down is the assumption that there are no more primes. You assumed you had a complete list of primes. Then you constructed a number that is not divisible by any of the primes in your list. Only two possibilities remain: Either the number you constructed is prime, or it is divisible by a number that is not in your list. Either way your list is not complete, which contradicts your initial assumption.
Simpy put, you assumed you had a complete list and by using that assumption you proved that the list was not complete. The assumption must therefore be wrong.
The answer to #3 is basically the same. By finding a contradiction when you made an assumption, you proved the assumption to be incorrect. | {
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The answer to #2 is no, you did not prove you anything by noting that 59 and 509 are both prime. You are trying to prove that there is a finite list of primes. If you choose a particular set of primes as you did {2, 3, 5, 7, 11, 13} and show that that particular set doesn't hold all the primes, a skeptic would just say that you need to add more primes (like 59 and 509) because you didn't make your set big enough. The proof has to be more general which is why it doesn't say how many primes are in the finite set. The proof is written so that it works no matter how big the finite set is.
Point 1: It's a theorem that any natural number $n>1$ has a prime factor. The proof is easy: for any number $n>1$, the smallest natural number $a>1$ which divides $n$ is prime (if it were not prime, it would not be the smallest).
Point 2: Yes, you have proved there are more than six primes. So what? The proof by contradiction doesn't suppose there are only six, but that there are a finite number of them.
Point 3: Actually, it's not really a proof by contradiction stricto sensu. It is proved that any finite list of primes in incomplete.
Here is a non-mathematical approach to the logic behind the proof by contradiction...
Suppose your assumption is that a suspect in a brutal murder is innocent because he said he couldn't have been at the location the person was murdered. You feel as though he is possibly telling the truth, but what you do know is that there are 5 other suspects. You know with 100% certainty due to the detail of the case there can't be anymore than 5 other suspects via proven information. But, the police can verify using obvious physical evidence such as video cameras that the 5 suspects could not possibly be guilty. Then there is either another suspect or the one assumed innocent is guilty. But we said that it was impossible for any other suspects to be considered. Therefore our original suspect is guilty. | {
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This is exactly what is happening in a proof by contradiction.
• This is good. But I think a better analogy is that you have 6 suspects and you assume that these are the only six possible suspects in the world. Then cameras show each and every one is innocent. Thus our assumption that we have all the suspects is wrong, and therefore there is a seventh possible person who could have (and did) do it. – fleablood Oct 9 '16 at 0:58
• Damn.... you are right. It made logical sense when i wrote it... but yours is better. – Eleven-Eleven Oct 9 '16 at 1:02
• Yours makes sense. And is a proof by contradiction. I just thought for this particular proof this is a closer analogy. I wouldn't say yours was wrong or anything. – fleablood Oct 9 '16 at 1:05
• That was my original goal; understand why contradition works. But yours does that AND is analogous to the problem at hand. Thanks – Eleven-Eleven Oct 9 '16 at 1:07
• I will concede. I get it. – Eleven-Eleven Oct 9 '16 at 23:01
The proof is essentially overexplaining. The point of contradiction would have been much clearer if the author had:
1. Reached it earlier.
2. Introduced less notation.
Since $p_1\cdots p_n+1$ cannot have any of $p_1,...,p_n$ as a prime factor we already have a contradiction.
The rest is just explaining to death why $p_1,...,p_n$ cannot be prime factors of $y=p_1\cdots p_n+1$ which should be clear since $y$ is $1$ off from being a multiple of either of those primes. | {
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• I think the gif dosn't really fit this site, though I can't say if there is a specific rule against such things – Yuriy S Oct 10 '16 at 11:27
• @YuriyS: Hmm ... I am sorry to learn that people may find it inappropriate! That was by no means my intention! Do you think it would work better by removing the gif an leave the link and the figure of speach, or are all of those opfuscating the points I tried to make? – String Oct 10 '16 at 12:21
• @String: your link to a cartoon movie of a dead animal being bludgeoned is disproportionate and offensive. – Rob Arthan Mar 12 '17 at 1:27
• @RobArthan: Sorry, in some parts of the world such a cartoon would be considered merely a funny way to illustrate the saying about beating a dead horse. No offense intended, only a light tone. I cannot help that people do take offense, so I have removed it. Still it puzzles me how a cartoon matching the content of a saying would offend. I am from Denmark, after all. – String Mar 12 '17 at 8:11
• @String: on MSE it's easier just to use neutral language. As I am English (after all), may I point out that the cliched phrase is actually "flogging a dead horse" and it doesn't have the connotations you think it does (it's not "explaining to death", it's using a tired old argument that has lost all interest or relevance). Your cartoon doesn't help with that. – Rob Arthan Mar 13 '17 at 1:55
1. Well... yes... it does break down. You assumed that there are only 6 primes and reached a contradiction. You've successfully proved that there aren't only 6 primes.
2. Under the assumption that there are only 6 primes 30,031 isn't factorizable. | {
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2. Under the assumption that there are only 6 primes 30,031 isn't factorizable.
3. In proof by contradiction you prove proposition A by assuming A is not true, and through a series of logical steps reach an impossibility, thus proving that A must be true. Concurrently, in this proof, we assumed there is a finite number of primes. After a series of logical inferences we reached a contradiction. Since the only assumption we made in the process is that there are a finite number of primes, that assumption must be wrong.
1. Either $6$ primes you considered are not all the primes, or $30031$ is a new prime. Either way assumption is false. Proof by contradiction.
2. No, you haven't shown that two more primes exist. In this example yes, but not in general
3. Again, based on your assumption you reached something impossible that $q$ divides $1$. There is no such $q$ except $1$. | {
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Start with $2,3,5$ being the only prime numbers. Then you would have $31$ as the new number. $2,3,5$ do not divide $31$. But unlike your case, $31$ is a prime number. So the assumption is wrong. | {
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# Spring, weight and potential energy
1. Jan 30, 2014
### Bromio
1. The problem statement, all variables and given/known data
A particle is connected to a spring at rest. Because of weight, the mass moves a distance $\Delta x$. Calculate the value of the elastic constant $k$
2. Relevant equations
$U_{PE} = \frac{1}{2}k\Delta x^2$
$U = mgh$
$F = ma$ (in general)
$F = kx$ (Hooke's Law)
3. The attempt at a solution
I've tried to solve this problem from two points of view, but the result has been different in each one.
First, I've thought that, when the mass is at rest after being connected to the spring and got down to the new equilibrium position:
$mg = k\Delta x \Longrightarrow \boxed{k = \frac{mg}{\Delta x}}$
However, from the point of view of energies:
$mgh_0 = mgh + \frac{1}{2}k\Delta x^2$ (I use $\Delta x$ because the initial position is 0)
As $h_0 - h = \Delta x$, I've got:
$mg\Delta x = \frac{1}{2}k\Delta x^2 \Longrightarrow \boxed{k = \frac{2mg}{\Delta x}}$
Obviously there is a wrong factor of 2.
What's the problem here? Where is the mistake?
Thanks!
2. Jan 30, 2014
### BvU
Energy case is for the situation you let go at h=0. When it is at delta x, it's still moving. Until it is a 2 delta x, when it hangs still again. But now the spring is pulling harder and up she goes!
In other words, before there is rest, some energy has to be dissipated! You now know how much.
3. Jan 30, 2014
### Bromio
Intuitively I understand what you say, but I don't see how to write it analytically.
I mean, as we are in a conservative field, we only have to consider initial and final states. How should I have written the main equation ($mgh_0 = mgh + \frac{1}{2}k\Delta x^2$) to include what you say?
Thanks!
4. Jan 30, 2014
### BvU | {
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Thanks!
4. Jan 30, 2014
### BvU
That would be a little difficult, precisely because it describes an isolated case. It says there is energy left over, that has to go somewhere to get the "in rest" situation. E.g. in overcoming the air resistance which heats up the air a little.
5. Jan 30, 2014
### Bromio
Okay. So the dissipated energy has a value of $\frac{1}{2}k\Delta x^2$, hasn't it?
What I don't understand is why, if we are in a conservative field, we can't just consider initial and final states. Could you explain me the reason?
Thank you.
6. Jan 30, 2014
### BvU
You can, actually. It's just that it isn't a steady state. mg (h-h0) is converted into spring energy plus kinetic energy. If there is no friction, the oscillation goes on forever.
Is there a risk that we are misinterpreting the OP ? My idea was Δx is measured after the thing has come to rest.
7. Jan 30, 2014
### Bromio
No misinterpretation. Now I see all you're trying to explain. For some reason I thought there wasn't friction. It's obvious that, if there isn't it, I can't apply energy equations "at rest" situation, simply because there isn't that situation. Am I right?
So, when I study energies at the beginning ($h = 0$) and at the end ($h = \Delta x$, when entire system is at rest) I should include that dissipated energy.
If $TE_0$ and $PE_0$ are the total mechanical energy and the potential energy at the beginning respectively, $TE$ and $PE$ are the same energies but at the end, $SP$ is the spring energy at the end, and $Q$ is the dissipated energy (say heat), then:
$TE_0 = PE_0$
$TE = SP + PE$
$TE_0 = TE + Q$
Do I agree?
Thank you!
8. Jan 30, 2014
### BvU
Well, there is of course an "at rest" situation, it's just that it doesn't have the same energy as the situation where you attach the weight to the spring. | {
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Yes. in fact, you can probably do that yourself: Suppose you exercise a force to keep the weight in position h0, then gently lower the weight to position h. If you do it slow enough, the force is mg - kx where x runs from 0 to h and h = mg/k. The weight does work, namely ∫ F(x) dx where the integral goes from 0 to h. F(x) = mg - kx. The integral (you can do it) is $\frac{1}{2}k h^2$. Looks familiar ?
9. Jan 30, 2014
### Bromio
When I say that there isn't an "at rest" situation I'm talking about the frictionless case. If there isn't friction, the spring will experiment an non-stop simple harmonic motion, won't it?
Thank you! I've solved that integral and I've got that result. Amazing! | {
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# A rectangular sheet of paper, when halved by folding it at the mid
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A rectangular sheet of paper, when halved by folding it at the mid [#permalink]
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25 Oct 2018, 04:50
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A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in
a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter
sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the
smaller rectangle?
a. $$4 \sqrt{2}$$
b. $$2 \sqrt{2}$$
c. $$\sqrt{2}$$
d. 1
e. 3
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A rectangular sheet of paper, when halved by folding it at the mid [#permalink]
### Show Tags
25 Oct 2018, 06:02
2
RhythmGMAT wrote:
A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in
a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter
sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the
smaller rectangle?
a. $$4 \sqrt{2}$$
b. $$2 \sqrt{2}$$
c. $$\sqrt{2}$$
d. 1
e. 3
$$\frac{l}{b} = \frac{b}{(l/2)}$$ where $$b = 2$$ given
i.e. $$b^2 = l^2 / 2$$
i.e. $$l = 2√2$$
Area of smaller rectangle $$= (l/2)*b = (2√2/2)*2 = 2√2$$
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Re: A rectangular sheet of paper, when halved by folding it at the mid [#permalink]
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### Show Tags
25 Oct 2018, 07:05
2
RhythmGMAT wrote:
A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in
a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter
sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the
smaller rectangle?
a. $$4 \sqrt{2}$$
b. $$2 \sqrt{2}$$
c. $$\sqrt{2}$$
d. 1
e. 3
OA: B
Initial Rectangle Dimensions
longer side $$= x$$
Shorter side$$= 2$$
Dimensions after folding
Longer side $$= 2$$
Shorter side $$= \frac{x}{2}$$
As per question
$$\frac{x}{2}=\frac{2}{\frac{x}{2}}$$
$$x^2=8 \quad;\quad x=2\sqrt{2}$$
Dimensions of Shorter Rectangle
Longer side $$= 2$$
Shorter side $$=\frac{2\sqrt{2}}{2}=\sqrt{2}$$
Area of Shorter Rectangle $$= 2\sqrt{2}$$
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Re: A rectangular sheet of paper, when halved by folding it at the mid [#permalink]
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25 Oct 2018, 14:13
2
RhythmGMAT wrote:
A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in
a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter
sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the
smaller rectangle?
a. $$4 \sqrt{2}$$
b. $$2 \sqrt{2}$$
c. $$\sqrt{2}$$
d. 1
e. 3
$$?\,\, = 2x$$
The rectangles shown above are similar, hence:
$${{2x} \over 2} = {2 \over x}\,\,\,\,\, \Rightarrow \,\,\,{x^2} = 2\,\,\,\,\,\mathop \Rightarrow \limits^{x\,\, > \,\,0} \,\,\,\,x = \sqrt 2 \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 2\sqrt 2$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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Re: A rectangular sheet of paper, when halved by folding it at the mid &nbs [#permalink] 25 Oct 2018, 14:13
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# For positive integers between 999 and 100 inclusive, how many contain the digit 5?
The question comes in two parts:
1. For positive integers between 999 and 100 inclusive, how many contain the digit 5 at least once?
2. For positive integers between 999 and 100 inclusive, how many contain the digit 5 exactly once?
Question 1
Numbers between 999 and 100 inclusive:
999 - 99 = 900
Numbers between 999 and 100 inclusive that do not contain digit 5:
8 * 9 * 9 = 648 (The 8 is because the first digit can't be 5 or 0)
Numbers that contain digit 5 at least once:
900 - 648 = 252
Question 2
Total numbers that contain digit 5 exactly once:
(1*9*9) + (8*1*9) + (8*9*1) = 225
I think I got it wrong but not sure which part. | {
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(1*9*9) + (8*1*9) + (8*9*1) = 225
I think I got it wrong but not sure which part.
• Well, a good way to check is to count the number with exactly two $5's$. $55X$ gives us $9$, $5X5$ gives us $9$, $X55$ gives us $8$. Thus there are $9+9+8=26$ such numbers. Of course $555$ is the unique number in the range with three $5's$. We check that $252=225+26+1$ which seems like pretty good evidence that you are right.
– lulu
Nov 1 '18 at 16:34
• @DavidG.Stork The OP is counting numbers with no digits equal to $5$ so that they can be subtracted from all three-digit positive integers. Nov 1 '18 at 16:38
• @DavidG.Stork OP is using a standard technique of counting how many numbers there are where we ignore the condition, giving $9\times 10\times 10 =900$ numbers in the range, and subtracting the number of numbers which violate the condition, violating the condition in this case meaning "does not have $5$ as a digit", there being $8\times 9\times 9$ such numbers, which will give the number of numbers that satisfy the condition as a result. Nov 1 '18 at 16:38
• @DavidG.Stork The purpose of that step is to get the number which does not contain digit 5 at all, which in this case is 648. My reasoning is that if I subtract 900 with 648, then all the remaining numbers would contain at least one digit 5. Nov 1 '18 at 16:40
• "I think I got it wrong but not sure which part" Looks good to me. Do you have specific reason to doubt the answer? Be more confident, you seem to be doing well from what I can see. Nov 1 '18 at 16:40
There is one number with all three digits equal to 5.
There are 1x10x10 = 100 numbers with the first digit equal to 5.
There are 9x1x10 = 90 numbers with the second digit equal to 5.
There are 9x10x1 = 90 numbers with the third digit equal to 5.
There are 10 numbers with the first and second digit equal to 5.
There are 10 numbers with the first and third digit equal to 5.
There are 9 numbers with the second and third digit equal to 5. | {
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There are 8x9x9 = 648 numbers with no digit equal to 5.
This is enough information to fill in the diagram above.
I got the same answers as you with the (javascript) code
var countFives = function(x){
var count = 0;
for (let c of x.toString(10)){
if (c==="5") count++;
}
return count;
};
var s = 0;
for (let i=100; i<=999; i++) if (countFives(i)>=1) s++;
console.log(s);
252
var s = 0;
for (let i=100; i<=999; i++) if (countFives(i)===1) s++;
console.log(s);
225 | {
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If we define a pure real number as a complex number whose imaginary component is 0i, then 0 is a pure real number. e) INTUITIVE BONUS: Without doing any calculation or conversion, describe where in the complex plane to find the number obtained by multiplying . The real number a is called the real part and the real number b is called the imaginary part. For , we note that . We can picture the complex number as the point with coordinates in the complex … A complex number is a number of the form . Let be a complex number. Here both x x and y y are real numbers. is called the real part of , and is called the imaginary part of . A complex number is the sum of a real number and an imaginary number. Different types of real … As it suggests, ‘Real Numbers’ mean the numbers which are ‘Real’. Keep visiting BYJU’S to get more such maths lessons in a simple, concise and easy to understand way. If a is not equal to 0 and b = 0, the complex number a + 0i = a and a is a real number. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation i2 = −1. With regards to the modulus , we can certainly use the inverse tangent function . A complex number is the sum of a real number and an imaginary number. We start with the real numbers, and we throw in something that’s missing: the square root of . We define the imaginary unit or complex unit to be: Definition 21.2. Yes, because a complex number is the combination of a real and imaginary number. This statement would not make out a lot of logic as when we calculate the square of a positive number, we get a positive result. Move 6 units to the right on the real axis to reach the point ( 6 , 0 ) . Real Numbers and Complex Numbers are two terminologies often used in Number Theory. Python complex number can be created either using direct assignment statement or by using complex () function. This j operator used for simplifying the | {
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assignment statement or by using complex () function. This j operator used for simplifying the imaginary numbers. x x is called the real part which is denoted by Re(z) R e ( z). Learn more about accessibility on the OpenLab, © New York City College of Technology | City University of New York. Example 21.3. Likewise, imaginary numbers are a subset of the complex numbers. The major difference is that we work with the real and imaginary parts separately. basically the combination of a real number and an imaginary number Complex numbers actually combine real and imaginary number (a+ib), where a and b denotes real numbers, whereas i denotes an imaginary number. A complex number is expressed in standard form when written $$a+bi$$ (with $$a, b$$ real numbers) where $$a$$ is the real part and $$bi$$ is the imaginary part. Convert the number from polar form into the standard form a) b), VIDEO: Converting complex numbers from polar form into standard form – Example 21.8. Multiplying Complex Numbers. A complex number is expressed in standard form when written a + bi where a is the real part and bi is the imaginary part.For example, $5+2i$ is a complex number. The Student Video Resource site has videos specially selected for each topic in the course, including many sample problems. From the long history of evolving numbers, one must say these two play a huge role. A complex number is created from real numbers. This leads to the following: Formulas for converting to polar form (finding the modulus and argument ): . is called the real part of , and is called the imaginary part of . But either part can be 0, so all Real Numbers and Imaginary Numbers are also Complex Numbers. Complex numbers are written in the form (a+bi), where i is the square root of -1.A real number does not have any reference to i in it.A non real complex number is going to be a complex number with a non-zero value for b, so any number that requires you to write the number i is going to be an answer to your | {
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value for b, so any number that requires you to write the number i is going to be an answer to your question.2+2i for example. For information about how to use the WeBWorK system, please see the WeBWorK Guide for Students. Perform the operation.a) b) c), VIDEO: Review of Complex Numbers – Example 21.3. Complex numbers are a bit unusual. Imaginary number consists of imaginary unit or j operator which is the symbol for √-1. However, unit imaginary number is considered to be the square root of -1. Imaginary Numbers when squared give a negative result. This question is for testing whether or not you are a human visitor and to prevent automated spam submissions. start by logging in to your WeBWorK section, Daily Quiz, Final Exam Information and Attendance: 5/14/20. If z = 3 – 4i, then Re(z) = 3 and Im(z) = – 4. I – is a formal symbol, corresponding to the following equability i2 = -1. It is important to understand the concept of number line to learn about real numbers. The WeBWorK Q&A site is a place to ask and answer questions about your homework problems. Complex Numbers Complex Numbers 7 + 3 Real Imaginary A Complex Number A Complex Number is a combination of a Real Number and an Imaginary Number Real Numbers are numbers like: 1 12.38 −0.8625 3/4 √2 1998 Nearly any number you can think of is a Real Number! For example, $$5+2i$$ is a complex number. Imaginary numbers are square roots of negative real numbers. In the meantime, ‘Complex Numbers’ as the name refers a heterogeneous mix. Thus, 3 i, 2 + 5.4 i, and –π i are all complex numbers. Therefore, they consist of whole (0,1,3,9,26), rational (6/9, 78.98) and irrational numbers (square root of 3, pi). Similarly, 3/7 is a rational number but not an integer. Read through the material below, watch the videos, and send me your questions. If the formula provides a negative in the square root, complex numbers can be used to simplify the zero.Complex numbers are used in electronics and electromagnetism. We call this the | {
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to simplify the zero.Complex numbers are used in electronics and electromagnetism. We call this the polar form of a complex number. (In fact, the real numbers are a subset of the complex numbers-any real number r can be written as r + 0 i, which is a complex representation.) For example, both and are complex numbers. By the Pythagorean Theorem, we can calculate the absolute value of as follows: Definition 21.6. Points that fall in the right side of origin are considered positive numbers, whereas numbers lying in the left side of origin are considered to be negative. The quadratic formula solves ax2 + bx + c = 0 for the values of x. By definition, imaginary numbers are those numbers which when squared give a negative result. A Complex number is a pair of real numbers (x;y). So, if the complex number is a set then the real and imaginary number are the subsets of it. And actually, the real numbers are a subset of the complex numbers. We can picture the complex number as the point with coordinates in the complex plane. A complex number is represented as z=a+ib, where a … The coordinates in the plane can be expressed in terms of the absolute value, or modulus, and the angle, or argument, formed with the positive real axis (the -axis) as shown in the diagram: As shown in the diagram, the coordinates and are given by: Substituting and factoring out , we can use these to express in polar form: How do we find the modulus and the argument ? To plot a complex number, we use two number lines, crossed to form the complex plane. Let’s learn how to convert a complex number into polar form, and back again. Hence, we need complex numbers, a further extension of the number system beyond the real numbers. Your email address will not be published. How do we get the complex numbers? Give the WeBWorK a try, and let me know if you have any questions. Your email address will not be published. i.e., a complex number is of the form x +iy x + i y and is usually represented by z z. Any number | {
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i.e., a complex number is of the form x +iy x + i y and is usually represented by z z. Any number in Mathematics can be known as a real number. A real number refers to any number that can be found on this number line. They have been designed in order to solve the problems, that cannot be solved using real numbers. Point P is uniquely determined by the ordered pair of a real number(r,θ), called the polar coordinatesof point P. x = r cosθ, y = rsinθ therefore, z=r(cosθ + isinθ) where r =√a2 + b2 and θ =tan-1 =b/a The latter is said to be polar form of complex number. Learn More! HINT: To ask a question, start by logging in to your WeBWorK section, then click “Ask a Question” after any problem. Start at the origin. Complex Numbers are considered to be an extension of the real number system. Its algebraic form is , where is an imaginary number. The set of complex numbers is a field. Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts. Thus, the complex numbers of t… I can't speak for other countries or school systems but we are taught that all real numbers are complex numbers. Therefore, imaginary name is given to such numbers. This .pdf file contains most of the work from the videos in this lesson. A complex number is expressed in standard form when written a + bi where a is the real part and b is the imaginary part. If is in the correct quadrant then . Example 2: Plot the number 6 on the complex plane. The set of complex numbersis, therefore; This construction allows to consider the real numbers as a subset of the complex numbers, being realthat complex number whiose imaginary part is null. Complex numbers are numbers in the form. Every real number is a complex number, but not every complex number is a real number. Many amazing properties of complex numbers are revealed by looking at them in polar form! Similarly, when a negative number is squared it also provides a positive number. A complex number is a number of | {
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